Circuit Response
every model has an edge, and the figure carries it
Theme

The thread: Every model has an edge

The ideal amplifier, the small-signal transistor, the lumped element, the capacitor: each is excellent inside a range and wrong outside it, and the range is a number rather than a warning. No figure here is drawn without the frequency, amplitude or size at which the model in it stops being true.
A diode fed from 5 V through 1.0 kΩ. computed by solving, not by drawing. The operating point is where the exponential meets the load line: 0.692544 V and 4.3075 mA, reached in 13 damped Newton steps from a cold start. The one-line Newton on Vs = v + R·i(v), which touches no matrix, gives 0.692544 V. The "drop" is not a constant: it moves 59.53 mV per decade of current, measured between two solved operating points. Devices, and the amplitude they stop being linear at

A bias point is a solution, not a choice

The phrase "the diode drops 0.7 volts" is a constant standing in for the root of a transcendental equation. Solved properly, from a five-volt supply through a kilohm, it drops 0.692544 V — and from forty-eight volts through the same kilohm it drops 0.754459 V, because the drop moves about sixty millivolts for every decade of current through it.

 Where four of these models stop being true. In order: the ideal operational amplifier at 1.42 kHz, a 10 V output at full amplitude at 7.96 kHz, Kirchhoff's laws on 10.0 cm at 3.97 MHz, the ideal 100 nF capacitor at 4.69 MHz. The fifth boundary is an amplitude rather than a frequency and cannot share this axis: a small-signal model is 1% wrong above 7.3 mV, at every frequency there is. Where the models stop

Every model has an edge

Four assumptions the circuits in this field run on, with the frequency at which each stops being true, on one axis. The ordering is not the one most readers would guess — an ordinary amplifier circuit runs out of model at 1.42 kHz, three thousand times sooner than a ten-centimetre circuit board does.

 A single-pole low-pass with its corner at 995 Hz. Solved at 209 frequencies. The straight-line sketch, drawn faintly, is 3.01 dB wrong at the corner and within a tenth of a decibel only below 152 Hz. The phase is already −5.7° a decade before the corner and −84° a decade after it. Frequency, which is the same solve

One solve, read four ways

Reactance, phase, the corner frequency and the roll-off are not four ideas. They are four readings of one complex number, obtained from the same matrix that answers direct-current questions — and the straight-line sketch every engineer draws of them is itself a model, three decibels wrong exactly where it is read.

 One step response, computed twice: from the poles, and by walking the network forward. A damping ratio of 0.22, so the overshoot is 49.2%. The two curves are drawn on top of each other; the panel below is the difference between them, which is the trapezoidal rule's error at 500 steps and reaches 1.70e-3 V. Before the steady state

One step, computed twice

A step response from the poles is exact. The same step walked forward in time is not, and the difference between them is the trapezoidal rule's own error rather than anything about the circuit. It falls by a factor of four every time the step is halved, which is a claim about a method and can be watched.

 A 20 Ω, 50 mH load on 230 V at 50 Hz. computed by solving, not by drawing. The load draws 1636 W and 1285 var, an apparent power of 2080 VA at a power factor of 0.786. The reactive side is confirmed by a route that touches no impedance: 2ω times the energy stored in the inductor gives 1285 var. The cable carries 9.04 A and only 7.11 A of it does anything. Power, and the part that does no work

The current that does no work

A solved network has been reporting its own real power on every page of this collection, as the second of the two checks each answer passes before it is drawn. What that check discards is the imaginary half — the power that flows out to a reactance and back again, does nothing, and is still carried by the cable, still heats the transformer, and is still on the bill.

 The noise of a 1.6 kΩ resistor through a 10.0 kHz filter. computed by solving, not by drawing. A seeded white sequence of 5.06 nV/√Hz marched through the network gives 619.3 nV across six seeds, spread 1.79%. Integrating the same density against the solved |H(f)|² gives 620.6 nV — -0.21% apart, well inside the spread. The noise bandwidth is 15.03 kHz against a −3 dB point of 10.00 kHz. The floor, which bounds from below

The floor a resistor sets

A kilohm at room temperature produces 4.00 nanovolts per root hertz, and it does so because it is warm rather than because of anything about how it was made. That is the first boundary in this collection that bounds a model from below — gain does not help, because gain amplifies it too — and it is the only field here whose figures are samples.

 A 9.0 kHz input sampled at 10 kHz arrives as 1.0 kHz. computed by solving, not by drawing. The dots are the samples. The input at 9.00 kHz is above half the 10 kHz rate, and every dot also lies on the 1.00 kHz curve drawn beside it — the two sample sequences differ by 9.3e-15, which is the arithmetic and not a small effect. Nothing is attenuated and nothing is distorted: the samples are the samples of a different signal, at full amplitude, and there is no measurement of them that could say which one was there. Where a signal becomes a number

The frequency a sample rate invents

Every other boundary in this field is a model getting gradually worse. This one has no gradient at all: below half the sample rate a set of samples has one sinusoid through it, above half the sample rate it has another, and the two sets of numbers are identical to three parts in ten thousand billion. Nothing is attenuated, nothing is distorted, and there is no measurement of the samples that could say which signal was there.

 The closed-loop poles at a gain of 3.05. The locus of the two poles as the amplifier's gain runs from 2.7 to 3.3. It crosses the imaginary axis at a gain of 3.000000 — bisected on the netlist, not quoted — and at 3.05 the real part is 2.500e+2 radians a second, which is an envelope multiplying by 1.17015 every cycle. The crosses are the closed form ω₀(k−3)/2 and they sit on the measured circles. Circuits that do a job, and the range they do it over

The gain that is exactly one

An oscillator is designed by making the loop gain one at the frequency where the phase is zero. The gain at which this circuit's poles reach the imaginary axis is 3.000000000000, bisected on the netlist — an equality, not a range. A gain three per cent high multiplies the envelope by 1.0987 every cycle and reaches the rails in 61 milliseconds; three per cent low divides it by the same factor. A one per cent resistor cannot hold the condition, and neither can any other component.

 Two probes on a 2.0 kΩ source. computed by solving, not by drawing twice per frequency: the node alone, and the node with the probe's elements across it. The one-to-one probe's 115.0 pF makes the reading one per cent wrong at 6.79 kHz. The ten-to-one probe puts 12.8 pF in series with the cable, so its tip sees 11.5 pF and the same error arrives at 69.2 kHz — 10 times further up, bought with a factor of ten in signal — the two edges stand in the ratio of the tip capacitances, 10.00. At direct current neither probe is capacitive at all and the ten-to-one still reads 0.02% low, because 10 MΩ across 2.0 kΩ is a divider. Measurement, which is a circuit on a circuit

The probe is part of the circuit

A one-to-one oscilloscope probe on a two-kilohm source gives a reading that is one per cent wrong at 6.8 kHz. Not because the instrument is inaccurate — it is reading correctly — but because the hundred and fifteen picofarads at the end of the cable are across the node, and above that frequency the trace on the screen is a picture of a circuit that only exists while the probe is attached.

 A 1 V step onto 1.00 m of 50 Ω line into an open circuit. computed by solving, not by drawing as a sum of 81 arrivals. The source drives 0.8333 V into the line immediately — set by 10 Ω against the line's 50 Ω, and not by the load, which it cannot yet know about. One delay of 4.83 ns later the far end reaches 1.6666 V. The staircase settles at 0.999990 V, which is what the resistive divider gives. Lines, where a wire has a length

The staircase in time

A source driving a metre of cable does not know what is on the far end of it for 4.83 nanoseconds. What it drives into during that time is decided by the cable's characteristic impedance and nothing else — and when the far end finally answers, the answer comes back as a staircase whose limit is the resistive divider the circuit was going to be all along.

 Three filter families at order 5, all with the same half-power point. At three times the corner the Chebyshev is -64.0 dB down, the Butterworth -47.7 dB and the Bessel -28.3 dB. The inset is the passband at forty times the vertical magnification, which is the only place the Chebyshev's half-decibel of ripple is visible at all. Filters, measured not tabulated

Three families, one corner

Butterworth is flat, Chebyshev is steep, Bessel has good delay. None of those is a number, so the table they appear in cannot answer the question anybody has. Here each family's poles are computed from its definition, built as an actual network, and then measured — starting with the step every comparison skips.

 Loop gain of a three-pole amplifier closed for a gain of 100. Unity loop gain at 5.73 kHz, where 34.9° of phase remains before −180°. The phase reaches −180° at 89.6 kHz, where the loop gain is 46.1 dB below unity. Feedback, and the margin

What is left at crossover

A feedback loop is stable or not according to one number read at one frequency — how much phase remains before −180° at the point where the loop gain passes unity. The loop gain here is obtained the way it is obtained on a bench: cut the loop, drive one side of the cut, and measure what comes back to the other.

 A 1:1 transformer at k = 0.99, and the band it is a turns ratio over. computed by solving, not by drawing. Two 10 mH windings coupled at 0.99, driven from 50 Ω into 50 Ω, with 0.5 Ω of winding resistance and 100 pF across the secondary. The response is flat at 0.4901 — which is 98.02% of the 0.5000 an ideal transformer of this ratio would give, and that shortfall is the coupling itself: the flat part is k times the turns ratio, times what the two winding resistances leave of the loop, to four figures at every k on the slider — between 400 Hz and 81.3 kHz, which is 2.31 decades. Both edges are bisected on the solved network. Below the first, the magnetising inductance is a short across the source; above the second, the leakage inductance is in series with the load. The slider moves the coupling, and it moves the upper edge only. Two windings, and the band between them

The band a turns ratio holds over

Most circuit models are right below a number or above one. A transformer is wrong at both ends and right in the middle, and the flat part is not the turns ratio either — measured on the solve it is the turns ratio times the coupling, times what the two winding resistances leave of the whole loop — and the first version of that last factor was a coincidence that held for every coupling and broke at a different load.

 A 10:1 divider with 12.8 pF across its top resistor. computed by solving, not by drawing. The divider's resistors set a ratio of 0.10000 and its capacitors 0.10000; the step starts at the second and relaxes to the first over 115 µs. The balance R₁C₁/R₂C₂ is 1.0000, and the edge lands 0.00% away from where it settles. At 12.8 pF the two ratios are the same number and the response is flat. Measurement, which is a circuit on a circuit

A divider with two ratios

Put capacitance in a resistive divider and it divides by resistance at direct current and by capacitance at high frequency, and those are two different numbers unless one equation holds. The adjustable trimmer on every oscilloscope probe exists for that single equation, and the square wave on the instrument's front panel is a display of which of the two ratios is currently winning.

 20 inductor-capacitor sections, against the line they are meant to be. computed by solving, not by drawing by the trapezoidal rule over 2,600 steps. The LC ladder reaches two per cent of full scale at 0.86 delays, before the wave picture says anything can have arrived, and its plateaus are wrong by up to 0.079 V. Neither is a small correction to the wave answer; they are what a network of 20 poles does when asked to be a delay. Lines, where a wire has a length

A ladder is not a line

A transmission line is usually introduced as the limit of a chain of inductors and capacitors as the number of sections goes to infinity. That is true, and it gives entirely the wrong impression of how close a finite chain gets. Forty sections still ring through every plateau by five per cent, and extrapolating the fitted convergence, reaching one per cent would need about nine hundred and sixty.

 What the reconstruction returns, either side of 5.0 kHz. computed by solving, not by drawing. The Whittaker–Shannon sum is evaluated on the samples and compared with two things: the signal that was sampled, and the frequency the samples report. Below 5.00 kHz these are the same curve and the error is 4.88e-3 — the truncation of the sum at sixty-four samples either side, and nothing else. Above it they part: at 9.00 kHz the reconstruction is 1.59e-3 from the alias and 2.000 from the input. The small number is the interesting one. A reconstruction cannot be improved into the right answer, because it is already an exact answer to a different question. Where a signal becomes a number

An exact answer to a different question

The reconstruction that turns samples back into a signal is normally introduced as the thing that recovers what was there. Measured on both sides of half the sample rate it does something more interesting than failing: above the boundary it returns the alias to 1.6 parts in a thousand, which is the same accuracy it returns the input with below the boundary, and it is wrong about the input by twice the amplitude. Its error is not a degradation. It is exactness about something else.

 10.0 cm of track, solved as a lumped circuit and as a line. The two agree to 0.030% at 3.97 MHz, where the track is one degree long, and to 30.1% at 143 MHz, where it is a tenth of a wavelength. Above that the lumped model is not approximately right; it is describing a different object. Where the models stop

Kirchhoff's own frequency

The current law says the current entering a node equals the current leaving it at the same instant, which assumes the signal crosses the circuit in no time. It crosses at about two-thirds the speed of light, so the law has a frequency of its own — set by nothing but the physical size of the board.

 Started from a millivolt at a gain of 3.20. The output grows by 1.8804 a cycle — the factor the poles give — and then stops, at 858 mV of amplitude and 1.58 kHz. The lower panel is the envelope on a logarithmic axis, where the linear model's prediction is the straight line that keeps going. What ends it is the diode pair across the feedback resistor, and no direct-current analysis of this circuit returns that number. Circuits that do a job, and the range they do it over

The amplitude nothing linear predicts

A linear model's poles say the envelope grows by 1.88 a cycle and never say when it stops. Two diodes across the feedback resistor stop it, at 858 millivolts — a number no pole, no transfer function and no bias point contains. Getting it needs a netlist with a nonlinearity in it, marched forward in time, and that pairing is what the whole field is built on.

 A single pole, and the brick wall that passes the same noise. computed by solving, not by drawing and integrated over 267 frequencies. The equivalent noise bandwidth is 1.5706 times the −3 dB point, and π/2 is 1.5708. A noise voltage computed with the corner frequency instead is 20.2% low. The area under the curve and the area of the rectangle are the same number. The floor, which bounds from below

The bandwidth noise sees

A single pole passes π/2 times as much noise power as a brick wall at its own corner frequency, so a noise voltage computed with the −3 dB point is twenty-one per cent low. Measured by integrating the solved response rather than taken from the table it usually comes from, the ratio is 1.5706 and π/2 is 1.5708. A five-pole Chebyshev's is 0.964 — less than one.

 One capacitor of 77.3 µF, against every load it was not sized for. computed by solving, not by drawing. Sized from the 20 Ω load, the capacitor takes the power factor to 1.000000 there and leaves 0.0e+0 var of 1636 VA. At 178 Ω the same installation sits at 0.23 leading. The correction is exact at one point on this axis and nowhere else on it. Power, and the part that does no work

The capacitor that was right once

Cancelling a load's reactive power needs one division and no iteration, and the answer is exact. It is exact for the load it was computed from, at the frequency it was computed at, and the figure here is about what it does everywhere else — which includes making the installation worse than it was before anything was fitted.

 An exponential driven 10.0 mV either side of its bias. computed by solving, not by drawing. A sinusoid in, and out comes a waveform whose peaks are taller than its troughs are deep. The second harmonic is 9.61% of the fundamental, measured by transforming 512 samples and predicted independently as I₂(0.387)/I₁(0.387) = 9.61%. The two routes agree to 5e-10 over the 5 harmonics that stand above the arithmetic's own floor, and share nothing but the amplitude. Devices, and the amplitude they stop being linear at

The distortion a linear model cannot have

A small-signal model's output is a scaled copy of its input by construction, so it has no second harmonic and asking it for one is not a hard question but a meaningless one. Measured on the curve itself, an exponential produces one per cent of harmonic distortion at 1.03 mV of drive — seven times sooner than the 7.30 mV at which its gain is one per cent wrong.

 A 10 kΩ + 10 kΩ divider, solved with its load. The unloaded answer is 6.00 V. It is 1% low at a load of 495 kΩ and 4.00 V at a load equal to the divider's own resistance. The ratio does not predict any of this; the magnitude does. Networks, and how a solve is checked

The divider, and the thing it does not know about

A two-resistor divider's output is set by the ratio of its resistances — with nothing connected. Connect anything at all and what decides the answer is the quantity the ratio was built to discard: the magnitude. Two dividers of identical ratio give six volts and one volt into the same load.

 Three element voltages closing on one source, at 1.59 kHz. Solved at 1.59 kHz, which is 1.00× the frequency at which the two reactances cancel. The three phasors add head to tail to the 1 V source exactly; their magnitudes sum to 5.26 V, which is not the same statement. Frequency, which is the same solve

Three voltages that close on one, and the steady state they assume

Kirchhoff's voltage law drawn as a polygon in the complex plane. The three element voltages of a series circuit add head to tail to the source exactly — while their magnitudes add to five times it. And the whole picture is a statement about a settled circuit, which takes a computable number of cycles to arrive.

 One loop, two measurements of the same margin. The loop gain crosses unity at 5.73 kHz with 34.9° of phase left. The closed-loop step overshoots by 35.2%, which the second-order relation says corresponds to 34.8°. They differ by 0.1°, and the difference is the third pole. Feedback, and the margin

Two measurements of one margin

A phase margin is computed from the loop gain in the frequency domain, without ever looking at a step. An overshoot is measured from the closed-loop step response in the time domain, without ever looking at a Bode plot. Inverting the standard relation on the second returns 34.9° against the first's 34.9°, and the residue is the third pole.

 Two poles at ζ = 0.3, recovered from the matrix. The poles are at -477.5 ± j1518 hertz. Their distance from the origin is the natural frequency to six digits; the cosine of their angle from the negative real axis is the damping ratio. The step response beside them follows. Before the steady state

Where the behaviour is written down

Two numbers in the complex plane contain everything a second-order circuit will ever do. Their distance from the origin is the natural frequency, the cosine of their angle is the damping — and the fastest-settling circuit is not the critically damped one, which is the case the textbooks name.

 Matching 50 Ω to 200 Ω with 51.7 mm of 100.0 Ω line. computed by solving, not by drawing at 261 frequencies. The reflection at the design frequency is 4.6e-17 — nothing, to the arithmetic — against 0.600 for the bare junction, which throws 36% of the power back. It stays under 0.1 from 0.914 to 1.086 of that frequency, a band of 17.1%. Lines, where a wire has a length

A quarter wave, and the path the current takes back

A line a quarter of a wavelength long, whose impedance is the geometric mean of the two it joins, matches them exactly — reflecting 5×10⁻¹⁷ of what arrives, which is the arithmetic's floor. At one frequency. Seventeen per cent either side of it the reflection is back to a tenth, and that band is the whole of what the technique is worth.

 Linearising an exponential at 27 °C, and what it costs. The linear model understates the gain by 1% at 7.30 mV and by 10% at 22.8 mV. The thermal voltage at this temperature is 25.9 mV, so "small compared with Vₜ" is not the criterion — 28% of Vₜ is already 1% wrong. Where the models stop

How small is small signal

Linearising an exponential replaces a curve by its tangent, which is exact at a point and progressively wrong away from it. The amplitude at which it is one per cent wrong is 7.3 millivolts at room temperature — 28 per cent of the thermal voltage, not a small fraction of it, and a good deal smaller than "small signal" suggests.

 A resonant circuit of Q = 8, and its measured bandwidth. The half-power points are 1.50 kHz and 1.69 kHz, a bandwidth of 198.9 Hz. The components predict f₀/Q = 198.9 Hz. They differ by 0.000%. Frequency, which is the same solve

Resonance, and the bandwidth it sets exactly

The half-power bandwidth of a resonant circuit is f₀/Q — not approximately, but to every digit the arithmetic has, which is rare enough to be worth checking. What is not exact, and is drawn as though it were, is the idea that the band sits centred on the resonance. At a quality factor of one its middle is twelve per cent above.

 A gain of 100 asked of an amplifier with 1.00 MHz of gain–bandwidth. The ideal amplifier — a nullor, so the two golden rules exactly — holds 100 at every frequency. The real one is 0.10% low at direct current, 1% low by 1.35 kHz, and 3 dB down at 10.0 kHz. Above 10.0 kHz there is no loop gain left and the ideal answer is not an approximation to anything. Feedback, and the margin

The ideal amplifier, and where it stops being one

An ideal operational amplifier's closed-loop gain is set by two resistors and nothing else — a horizontal line at every frequency. The real one is already a tenth of a per cent low at direct current, one per cent low by 1.35 kHz, and above 10 kHz has no loop gain left, at which point the ideal answer is not an approximation to anything.

 A 9 V source with 500 mΩ inside it. The ideal source is the flat line. The solved terminal voltage leaves it at a rate set entirely by the internal resistance: 1% low at 180 mA, half gone at 9.0 A. Networks, and how a solve is checked

The source that is not a source

An ideal voltage source holds its voltage at any current, which makes it the flattest line in the subject and the most commonly assumed model in it. Its edge is a current, set by one resistance nobody draws — and past that current the model is not approximately right, it is describing a different object.

 Five steps, each divided by its own size, from an amplifier limited to 0.50 V/µs. A linear circuit would put these five curves exactly on top of each other. The 20.0 mV step is linear; everything above 79.6 mV is not, and the largest step takes 16.0 µs to travel a distance the linear model says takes 0.159 µs. Before the steady state

The step that is too big

A linear circuit scales — double the input and the output doubles, exactly. A real amplifier does not, because its output can only move at a fixed rate, and the amplitude at which the two stop agreeing is about eighty millivolts for an ordinary part. No transfer function contains that number, because no transfer function can.

 The neutral of a three-phase supply with one phase 30% off. computed by solving, not by drawing. Balanced, the three line currents sum to 4.6e-16 of one of them and the neutral carries nothing. With one phase 30% heavier the neutral carries 2.65 A against a line current of 11.50 A. The neutral reaches a tenth of a line current at 11.1% imbalance. Power, and the part that does no work

Three phases, and the wire that carries nothing

Three sources a third of a cycle apart, feeding three equal loads, return a current of 5×10⁻¹⁵ amperes down the wire between the star points. That is zero, and the whole of three-phase distribution rests on it. What is worth measuring is how fast it stops being zero, and the answer is that an eleven per cent imbalance in one phase puts a tenth of a line current down a conductor often sized on the assumption that it carries none.

 Measuring with 50 mΩ of lead in each wire. computed by solving, not by drawing at 61 resistances, twice each. The two-wire arrangement measures the leads too, so its error is 2×50 mΩ over whatever is being measured: one per cent at 10 Ω, and 10000% at 1 mΩ. The four-wire arrangement senses on a separate pair that carries almost no current, and its error stays under 1.0e-2% across the whole range. Measurement, which is a circuit on a circuit

Two terminals measure the leads as well

Fifty milliohms in each lead makes a two-wire measurement one per cent high at ten ohms, ten per cent high at one ohm, and a hundred per cent high at a tenth. Not approximately — the reading is the resistance plus the leads, and below about ten ohms most of what is being reported is the wire between the instrument and the thing.

 One exponential and one pair, both driven 20.0 mV. computed by solving, not by drawing. The pair's characteristic is odd, so its even harmonics vanish: the second comes out at 1.5e-16 of the fundamental against 18.88% for the single stage. It is not a small residue but the floor of the arithmetic. The price is the third harmonic, 1.202% against 2.404%, and total distortion of 1.202% against 19.03%. Devices, and the amplitude they stop being linear at

What a pair cancels, and what it only halves

A differential pair's transfer characteristic is an odd function, and an odd function driven symmetrically produces no even harmonics at all. Measured, the second harmonic comes out at 10⁻¹⁶ of the fundamental — the arithmetic's own floor, not a small physical residue. The third harmonic is a different story, and it comes out at exactly half the single stage's, which is a reduction and not a cancellation.

 The sample rate each anti-alias filter demands for 80 dB. computed by solving, not by drawing. A 20 kHz passband, and each filter must be 80 dB down by the frequency that folds back into it. The required rate follows, and it is a property of the filter rather than of the converter: Bessel 3.53× Nyquist, Butterworth 2.08×, Chebyshev 1.53×. The elliptic design at a selectivity of 0.8 is refused: its equiripple stopband has a floor at -74.1 dB, which is above the requirement at every frequency, so no sample rate satisfies it. At a selectivity of 0.5 the same order needs 1.34×. The floor is the selectivity's, not the order's. Where a signal becomes a number

What the filter in front costs

The filter that keeps a converter honest is normally chosen for its skirt. Measured against one requirement — eighty decibels down by the frequency that folds back into a 20 kHz band — the choice is not a decibel or two of skirt but a factor in the clock: Bessel demands 3.53 times Nyquist, Butterworth 2.08, Chebyshev 1.53. And one design is refused outright, because an elliptic stopband is a floor rather than a slope and no sample rate reaches past a floor.

 The limit cycle's spectrum at a gain of 3.20. Ten lines of the settled waveform. The third is 5.692% of the fundamental and the fifth 0.982%; the second and fourth are 6.8e-7, which is the arithmetic's floor and not a small residue. Two diodes facing opposite ways make a symmetric characteristic and a symmetric characteristic produces no even harmonic at all. Circuits that do a job, and the range they do it over

What the limiter charges for

The diodes that set the amplitude are the only nonlinear thing in the loop, so every harmonic in the output is theirs. Across the gain slider the amplitude rises by a factor of 1.51 and the distortion by 15.4 — an amplitude that goes as the 0.11 power of the excess gain and a distortion that goes as the 0.74 power. Two diodes facing opposite ways produce no even harmonic at all, at seven parts in ten million, which is the arithmetic's floor rather than a small residue.

 Which mechanism sets the upper edge, against the load. computed by solving, not by drawing. Two candidate upper edges drawn against the measurement. The one every textbook names is a resonance between the leakage inductance and the winding capacitance; the one that actually binds at ordinary loads is the leakage in series with the load, a first-order corner at R/2πL. At 50 Ω they are 80.4 kHz and 1.13 MHz — a factor of 14 apart — and the measurement follows the first, to 19.5% at worst across nine loads. They swap at about 1500 Ω, above which the resonance is the binding one and the usual picture is right — which is why a transformer feeding a high impedance behaves as the textbooks say and one feeding fifty ohms does not. Two windings, and the band between them

Which picture sets the upper edge

Every account of a transformer's high-frequency limit names the same mechanism: the leakage inductance resonating with the winding capacitance. At fifty ohms that resonance is at 1.13 MHz and the measured edge is at 81.3 kHz — a factor of fourteen away — because what actually binds is the leakage in series with the load, a first-order corner with no resonance in it at all. The two swap at about 1500 Ω, and both accounts are current because both are sometimes right.

 The usable range of one stage, in a 10 kHz measurement. computed by solving, not by drawing. The floor is the Johnson noise of a 1 kΩ source in the measurement's own noise bandwidth — 501.6 nV, using 15.7 kHz rather than the 10 kHz corner. The ceiling is the drive at which an exponential's distortion reaches one per cent, 1.03 mV. Between them is 66.3 dB, and nothing a designer does moves either number without changing the circuit. The floor, which bounds from below

A floor and a ceiling

Every other boundary these essays measure is a ceiling. This one puts a floor underneath and measures the distance between them — 4.00 nanovolts per root hertz at the bottom, one per cent of harmonic distortion at 1.03 millivolts at the top, and 66.3 decibels of range in a ten-kilohertz measurement. Both ends are computed, neither is on a datasheet, and the arrangement of the stages decides which one moves.

 The equivalent of six elements, and the heat it does not account for. computed by solving, not by drawing at 73 loads. The network reduces to 8.56032 V behind 599.768 Ω, obtained twice and agreeing to twelve figures. Into every load on the axis the two put the same voltage to 2.2e-16 of it and deliver the same power to 7.8e-16 — there is no load that can tell them apart. Inside, at a load of 5.62 kΩ, the real network is dissipating 49.168 mW and the equivalent claims 1.1349 mW, a factor of 43.3. With the port open the equivalent says nothing at all is being burned and the network is burning 48.03 mW. The slider moves the resistor bridging the two sources, which joins two ideal sources and so cannot change anything the port can reach: the equivalent stays 8.56032 V behind 599.768 Ω to the last bit at every setting, while the heat inside moves by two and a half times. Networks, and how a solve is checked

Exact outside and wrong within

Six elements reduce to one source and one resistor that no load can distinguish from them: the same voltage into every load across six decades, to the last bit of a double. The reduction is wrong about the heat by a factor of forty-three, and with nothing connected it says the network is dissipating nothing while it burns 48 milliwatts.

 3 quarter-wave sections between 50 Ω and 200 Ω. computed by solving, not by drawing. |Γ| computed by cascading exact line impedances from the load back to the source, so every multiple reflection is in it. The binomial design is exact at the centre and holds |Γ| below 0.10 over 67.9% of the centre frequency, against 17.1% for a single section. The equal-ripple design, found by minimax search rather than from a table, covers 98.1% at the same worst reflection — 45% more — and its ripples come out level to 0.0e+0%, which is the check that the search converged. Lines, where a wire has a length

Several sections, and the band they buy

A quarter-wave transformer is exact at one frequency, and a 4:1 transformation holds |Γ| under 0.1 over 17.1% of it. Several sections whose reflections cancel over a band take that to 47.7, 67.9, 82.1 and 92.6% — which is filter design with the reflection as the shaped quantity. An equal-ripple design found by minimax search buys a further 35 to 46% at the same worst reflection, and its ripples come out level to four decimals.

 A loop whose phase comes back, at a gain of 1.0e+7. computed by solving, not by drawing. Three identical poles take the phase to −270°, two lead sections bring it back to −158°, and their own poles take it down again — so ∠L = −180° at 189 Hz, 3.46 kHz and 23.5 kHz. The gain moves the magnitude curve and not those three frequencies, so it decides only which side of each the unity-gain point falls. At 1.00e+7 the crossover is 10.6 kHz, in the recovered band, and the loop is stable. Feedback, and the margin

Stable, and unstable with less gain

A three-pole loop with two lead sections is stable for gains between 1.81 × 10⁶ and 3.36 × 10⁷ and unstable on both sides of that window. Turning its gain *down* is what breaks it. Neither margin can see this, because a phase margin describes the loop at one frequency and a gain margin at one other, and this loop's phase crosses −180° at three: 189 Hz, 3.46 kHz and 23.5 kHz.

 A 100 nF capacitor, and what it is above 14.5 MHz. The dashed line is 1/(ωC), which is what the symbol means. The solid line is the same part with 30 mΩ of series resistance and 1.2 nH of series inductance, solved. They part company at 4.69 MHz and by a decade above resonance the part's impedance is 99.0× what its capacitance predicts. Frequency, which is the same solve

The capacitor that is an inductor

A hundred-nanofarad capacitor follows 1/(2πfC) for four decades and then turns round and climbs. Above 14.5 MHz it is an inductor, and a decade past that its impedance is ninety-nine times what its capacitance predicts — all of it caused by about a nanohenry of lead and via that nobody chose and nobody drew.

 How long a second-order step takes to arrive inside ±2%. computed by solving, not by drawing from the residue expansion at 260 damping ratios. The fastest is ζ = 0.780 at 3.60/ω₀; critical damping takes 5.83/ω₀, which is 62% longer. Between ζ = 0.775 and 0.780 the time falls by 33% in one step of the sweep, because which excursion is the last one outside the band changes there — the overshoot at the fastest damping is 1.99%, which is the band itself, and one step to the left it is larger. The faint curves are the other bands, each with its own step in a different place. Before the steady state

The cliff before the fastest settling

Settling time against damping is not a smooth curve with a minimum. It falls by a third in one step of a sweep of five thousandths, and the fastest damping sits on the edge of that step — so a design a hundredth of a damping ratio to the left of the optimum settles forty-eight per cent slower, with a waveform that looks no different.

 The floor an amplifier adds, against the source it is given. computed by solving, not by drawing. A part with 4.00 nV/√Hz of voltage noise and 0.60 pA/√Hz of current noise is quietest into 6.67 kΩ, where its noise figure is 1.138 dB. That resistance is the ratio of the two generators and the floor there depends only on their product. Matching the same part for maximum power into its own 1 MΩ input instead — a resistance 150 times larger — costs 12.57 dB. The floor, which bounds from below

The floor a circuit has

A resistor's noise is 4kTR and there is nothing to choose about it. An amplifier adds two generators that belong to the device — 4 nV/√Hz in series with its input and 0.6 pA/√Hz across it — and because one matters most into a small source and the other into a large one, there is a source resistance at which their sum is least. It is 6.67 kΩ, it is the ratio of the two, and it is not the resistance that transfers maximum power.

 Where a converter stops measuring the signal and starts measuring the resistor. computed by solving, not by drawing. The quantisation floor is q/√12 and halves with every bit; the Johnson floor of a 1 kΩ source in 100 kHz is 1.266 µV and does not move. They cross at 18.80 bits. Below that the converter is the limit; above it the resistor is, and a further bit buys a more precise measurement of thermal noise. A resolution quoted without a source impedance and a bandwidth is not a resolution — which is the same sentence the instruments field makes about a probe. Where a signal becomes a number

The floor a converter sets

A converter's resolution is quoted as a number of bits, which is a property of the converter. What it can actually resolve is a property of the circuit in front of it, and the two cross: measured against the Johnson noise of a 1 kΩ source in 100 kHz of bandwidth, the quantiser is the limit up to 18.80 bits and the resistor is the limit above it. Past that crossing every further bit buys a more precise measurement of thermal noise.

 A common-emitter stage with 2.0 pF from collector to base. computed by solving, not by drawing. The stage's midband gain is 144.7 and its −3 dB point is at 504 kHz. The Miller approximation lumps 311 pF at the input and predicts 643 kHz — 21.6% high. The network's second pole is at 336 MHz and its right-half-plane zero at 3.08 GHz, both of which the approximation has no room for. Devices, and the amplitude they stop being linear at

The frequency a device sets for itself

A common-emitter stage's bandwidth is decided by two picofarads between its collector and its base. The Miller approximation says how — lump it at the input, multiplied by one plus the gain — and predicts 643 kHz where the solved network gives 504 kHz. Twenty-two per cent optimistic, and it has no room at all for the second pole or for the zero in the right half-plane that the network also has.

 Two reasons the frequency is not 1/2πRC. The measured oscillation frequency sits below the network's own zero-phase frequency, and by two separate amounts. The amplifier's share falls as 1/ρ — the product of shift and ratio is constant to 6.7% over two decades — and the limiter's share is flat at 0.7713%. They are equal at ρ = 578, and above that a faster amplifier moves the frequency by nothing that matters. Circuits that do a job, and the range they do it over

The frequency that is not the formula

The Wien network's zero-phase frequency is one over two pi RC to every digit the arithmetic has. The circuit does not run there. With a perfect amplifier it runs 0.771 per cent low, because the limiter's harmonics are part of the waveform whose period is being measured; with a real one it runs lower still, by an amount inversely proportional to the gain-bandwidth product. The two are equal at a ratio of 578, and above that a faster amplifier buys nothing.

 How much faster an instrument must be for 10% of inflation. computed by solving, not by drawing. The quadrature rule answers 2.182× and gives the same answer for every instrument, because it contains no instrument. Measured on the solved network, a one-pole front end needs 2.79×, two poles need 3.97×, three need 4.87× and four need 5.62× — between 28% and 215% more than the rule asks for. The rule errs optimistic at every pole count, which is the wrong direction. Measurement, which is a circuit on a circuit

The instrument's own rise time

Rise times add in quadrature, so ten per cent of inflation needs an instrument 2.18 times faster than the edge. That constant contains no instrument. Measured on the solved network it is 2.79 for a one-pole front end, 3.97 for two, 4.87 for three and 5.62 for four — the rule is optimistic at every pole count, which is the wrong direction for a rule of thumb to err in.

 Three balanced rectifier loads conducting 60°, and their neutral. computed by solving, not by drawing. The three phase currents are drawn faint and the neutral heavy. Balanced loads, identical in every respect, and the neutral carries 0.968 A against a line current of 0.559 A — a ratio of 1.7321, where √3 is 1.7321. The pulse trains are disjoint, so the neutral is their union and its mean square is three times one phase's. Rebuilding the same current from the multiples of three in one phase's spectrum gives 0.966 A, 0.13% away, by a route sharing only the waveform. Power, and the part that does no work

The neutral that carries more than a line

Three balanced loads draw currents summing to 5.3 × 10⁻¹⁵ amperes in the neutral. That is a theorem about sinusoids, and it uses nothing except that each current is a single frequency. A harmonic of order three is shifted by 360° between phases, which is no shift at all — so the third harmonics add, and for any conduction angle narrow enough that the pulse trains stay disjoint the neutral carries exactly √3 times a line current.

 What a 0.7 V constant costs, in the quantity it is used to predict. computed by solving, not by drawing by Newton's method on the exponential at 94 supplies through four resistors. The model is exact at 5.748 mA — the current at which the true drop is 0.7 V — and every curve crosses zero there, at four different supplies. Below it the model is low and above it high, and how much depends on the headroom rather than on the diode. Through the 87 Ω curve the drop is 49 mV out at 0.725 V and 147 mV out at 150.7 V — a factor of 3.0 — while the error in the current falls from -66% to 0.10%, a factor of 674, because the headroom underneath it has grown by 2017. On that curve the model is inside one per cent only above 1.12 V. Where the models stop

The one current a constant is right at

Seven-tenths of a volt is the true forward drop at 5.748 milliamperes and at no other current, and every circuit built on it crosses zero error there — four different resistors at four different supplies, all exact at the same current. What decides whether the model is any good is not the diode at all; it is how much of the supply the diode is taking.

 Four families at order 5, and the 2 zeros in the stopband. computed by solving, not by drawing. The elliptic design is drawn at a selectivity of 0.8, so its stopband is asked to begin at 1.250 times the corner, and the degree equation returns 38.68 dB for it at 0.5 dB of ripple. From the stopband edge outward the elliptic response never rises above -38.7 dB; the steepest all-pole family of the same order is only -19.0 dB down there and does not reach -38.7 dB until 1.77 times the corner. That is what 2 transmission zeros buy. The price is in the same picture: the elliptic stopband has a floor and an all-pole one does not, so beyond 2.0×fc the all-pole response is the lower of the two and keeps going. Filters, measured not tabulated

The zeros that buy an order

Butterworth, Chebyshev and Bessel all fall because their denominator grows, so the only way to make one fall faster is another pole. An elliptic filter puts zeros in the stopband instead, and reaches forty decibels at twice the corner with two poles where the steepest all-pole family needs five. What it charges is a stopband that stops falling — measured here at −38.7 dB, and overtaken by an ordinary Chebyshev two corners out.

 The load at which a transformer becomes a resonant circuit, at k = 0.99. computed by solving, not by drawing. The peak output of the same 1:1 transformer against its load, as a multiple of what the turns ratio would give. Below about a kilohm the load damps the leakage resonance, the peak is the plateau, and the ratio is one: there is a band, and it is what the rest of this field measures. Above it the damping goes and the response peaks — 1.18× at 832 kHz into 1500 Ω, rising to 8.80× at the light end. A transformer with voltage gain is not a transformer behaving badly; it is a resonant circuit, and asking for "the band" of one returns the skirts of a resonance. So the measurement reports that the response is peaked, rather than returning two edge frequencies in the wrong order. Two windings, and the band between them

Where the band goes entirely

The three essays before this one measure a transformer's band, and all three assume there is one. Past a few hundred ohms of load there is not: the leakage that sets the upper edge is also what damps the resonance behind it, and a lightly loaded, well-coupled transformer peaks at 3.03 times its own turns ratio. A passive component with voltage gain is not a transformer behaving badly. It is a resonant circuit, and asking for its band returns the skirts of a resonance.

 Where a switch is a switch: a band, and the 6.43 MHz at which it closes. computed by solving, not by drawing. A switch of 0.5 Ω closed, 100 MΩ open and 5 pF across it is within 1.0% of being ideal only for loads between 49.5 Ω and 1.01 MΩ — 4.31 decades, and both edges are the same part. The upper edge is a frequency as well as a resistance, because the off-capacitance shunts the open switch: it falls a decade per decade above 318 Hz and meets the lower edge at 6.43 MHz, where the band closes and no load at all will do. Checked by scanning every load at 1.3 times that frequency and finding the best possible error to be 1.17%. Where the models stop

A band rather than an edge

Every other boundary in this collection is one-sided: a model is true below a frequency, or below an amplitude. A switch is a switch only for loads between 49.5 ohms and 1.01 megohms — bounded at both ends by the same part — and the upper end is a frequency as well as a resistance, so the band narrows as the frequency rises and shuts completely at 6.43 megahertz, above which no load resistance at all will do.

 One per cent on one component, in two realisations of the same order-5 filter. computed by solving, not by drawing. The two realisations agree to 3e-14 dB before anything is moved. Moving each element in turn by 1%, the worst deviation at the 2 ripple peaks is 0.1621 dB for the cascade and 0.0030 dB for the LC ladder — 54 times smaller. Across the whole passband the two are within 3% of each other, because near the band edge both are dominated by the response's own steepness rather than by the realisation. Filters, measured not tabulated

A ladder is not a cascade

The same fifth-order Chebyshev, built two ways. A one per cent component moves the buffered cascade's passband by 0.162 dB at the ripple peaks and the doubly terminated LC ladder's by 0.003 dB — fifty-four times less. The number is not the claim: the claim is the exponent. Fitted over two decades of tolerance the cascade's error grows as the 0.99 power and the LC ladder's as the 2.00 power, because at maximum power transfer the response is stationary in every element it contains.

 How much of the amplifier reaches the answer, at 35° of phase margin. computed by solving, not by drawing by multiplying the forward path's gain by 1 + 10⁻⁵ and solving again. At low frequency 0.0999% of a fractional change in the device reaches the closed-loop gain, which is one over the loop gain of 995.04. It changes sign at 258 Hz — above there a better amplifier gives a smaller gain — and reaches 1.61 at 8.08 kHz, which is worse than having no feedback, and settles at exactly 1 above crossover (5.73 kHz), where there is no loop gain left to spend. The second route, the real part of 1/(1 + T) from the cut loop, agrees to 9.6e-6. Feedback, and the margin

How much of the amplifier gets through

The reason to build an amplifier from a bad amplifier and two resistors is that a change in the device barely reaches the answer — a tenth of a per cent of it at low frequency, measured by making the device better and solving again. Near crossover the same fraction rises above one, so feedback makes the gain more device-dependent than no feedback at all, and below fifty degrees of margin it changes sign on the way.

 The coupling coefficient, from two measurements that do not know it. computed by solving, not by drawing. Two windings in series, connected one way and then the other, each solved as a netlist and its inductance read out of the impedance. The two differ by four times the mutual inductance, so k comes out of the difference and the geometry never enters. The recovered value matches the one stamped into the coupling to 2.4e-15 at eight couplings from 0.1 to 0.99 — which is the second route the new element needed, since neither current law nor the energy balance can see a mutual inductance at all. Their sum stays at L₁ + L₂ throughout, which is the check that the two measurements are of one pair. Two windings, and the band between them

One number from two measurements

Neither of the two standing checks on a solved circuit can see a mutual inductance. A coupling adds no current anywhere, so Kirchhoff's law is unmoved by it; a coupled pair dissipates nothing, so the energy balance is unmoved too. A coupling stamped into the wrong row would produce a well-formed solution to a different circuit and both checks would pass — so the field needed a third route, and the one it uses is the bench method: connect the windings in series one way, then the other, and the difference is four times the mutual inductance.

 6.02 bits + 1.76 dB, measured — and the overload that is blamed on it. computed by solving, not by drawing. A coherent sinusoid is quantised and the ratio is read off the error sequence, never from the formula. One step inside full scale the two agree to 0.15 dB at every bit count from 8 to 16. Driven to exactly full scale the same converter measures 0.83 dB worse at 8 bits when driven to exactly full scale, and the reason is in the count beside it: 327 of 8192 samples hit the top code. A mid-tread converter's largest code is one step below full scale, so a sinusoid that reaches full scale is already overloading — half a step of overload, at the peak, where the error correlates with the signal. The loss halves with every bit for the same reason the clipped count does — 0.83, 0.47, 0.21, 0.06, -0.13 dB — and by sixteen bits it is inside the measurement's own scatter. The formula was never the approximation. Where a signal becomes a number

Six decibels a bit, and the half step blamed on it

Every converter data sheet quotes 6.02N + 1.76 dB and every bench measurement comes up short of it, which is usually explained by calling the formula an approximation. Measured one step inside full scale it agrees to 0.15 dB at every bit count from eight to sixteen. Driven to exactly full scale the same converter loses 0.83 dB at eight bits — because 327 of 8192 samples hit the top code, and a mid-tread converter's largest code is one step below full scale.

 A network the solver will answer, and should not be asked, into 0.01 Ω. computed by solving, not by drawing at 61 spreads. A hundredth of an ohm either side of a wire whose resistance is made smaller and smaller, into 0.01 Ω — an element written where the right answer is no element at all. The solution stays exact: 0.499999998422607 at the last spread before the refusal, against 0.500000000000000. What grows is the current-law residual, as the 0.96 power of the spread. The solve is refused at a spread of 2.5e+8, where the smallest pivot falls under 1e-12; the residual tolerance of 1e-7 would have been reached at about 3.3e+9. Two guards written for unrelated reasons, arriving within a decade of one another. Networks, and how a solve is checked

The answer that is perfect and absurd

A network with a wire written into it as a small resistance returns the right node voltage to fifteen figures, passes both verifications with a residual of two parts in ten to the sixteenth, and reports two hundred thousand amperes. The solver refuses it one decade further on, and by then it has been answering for eight decades.

 Averaging a white sequence, and averaging a pink one. computed by solving, not by drawing. Both sequences are the same seeded white stream, one of them put through the 1/f network. Averaged in non-overlapping blocks, the white one's spread falls as n to the -0.510 ± 0.006 across five seeds — the √N law — and the pink one's as n to the -0.087 ± 0.013, which is very nearly not at all. A thousand-sample average buys a factor of 36.9 on the first and 2.1 on the second. The floor, which bounds from below

The corner where averaging stops working

Average N samples and the noise falls by √N. That is a statement about independent samples, and flicker noise's samples are not independent — its correlation extends over every time scale, which is what a spectrum with no bottom means. Measured on the same seeded stream filtered and not: the white sequence falls as the −0.510 power of the block length and the pink one as the −0.087 power, so a thousand-sample average buys a factor of 36.9 on one and 2.1 on the other.

 1000 µF across a 100 Ω load, rectified from 17 V peak. The output sits at 15.69 V with 1.331 V of ripple, against the 1.569 V the expression I/2fC gives — 15.1% high, because the capacitor is being recharged for part of the cycle rather than discharging throughout it. The lower panel is why: the diode conducts for 28.8° of each half cycle and carries 2.10 A at the peak, which is 13.4 times the 157 mA the load draws. Circuits that do a job, and the range they do it over

The direct voltage that is a sawtooth

A rectifier and a reservoir capacitor make what everybody calls a direct voltage. Marched with the diodes in the netlist, a thousand microfarads across a hundred ohms gives 15.69 volts with 1.33 volts of ripple on it, against the 1.57 the textbook expression predicts. The expression is high by the fraction of the cycle the diode conducts for — measured at 0.94 to 0.96 of it across two sweeps — and it has no opinion at all about the quantity that actually sizes the transformer, which is a peak diode current 13.4 times the current the load draws.

 What a charge through 1 kΩ costs, against how long it is given. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. A step loses 1.00000 of ½CV² — 12.5 µJ here — and that number is the same through 10 Ω and through 100 kΩ, which is what the faint curves show. What the resistance decides is where the fall starts: the loss goes as 2τ/T once the ramp is long against τ = 1 ms, with a fitted exponent of -0.994 and no floor beneath it. The circles are marched; the line through them is a closed form that never sees a netlist. Before the steady state

The half that never arrives

Charging a capacitor from a step loses exactly as much energy as it stores, and the resistance it is lost in does not appear in the answer — the same 12.5 microjoules through ten ohms and through a hundred kilohms, to nine figures. Drive the same network with a ramp instead and the loss falls as two time constants over the ramp, with no floor beneath it at all.

 The load that takes the most power, and the load that wastes the least. computed by solving, not by drawing at 71 load values. The power into the load peaks at a ratio of 1.000000, which is the magnitude of the source impedance to six figures, and the efficiency there is 0.500000 — the source dissipates as much as the load receives. Ninety per cent efficiency needs a ratio of 9.7 and delivers 34% of the available power. Power, and the part that does no work

The load that takes the most

A load equal to the source resistance takes more power than any other, and it does so at exactly fifty per cent efficiency — the source burns as much as the load receives. Ninety per cent efficiency needs a load nine times the source and delivers 36% of what was available, and a load half the source resistance delivers exactly as much as one twice it.

 A resonator's Q against its inductor's, with a capacitor of Q 1581. computed by solving, not by drawing. The dashed line is what the resonator's Q would be if the inductor were its only loss; the solid one is what it is with a capacitor of Q 1581 beside it. They part company where the inductor stops being the worst component. At the marked point the inductor's Q is 79.06, the capacitor's is 1581.1, the reciprocals predict 75.2923 and the solved network measures 75.2923 — 2.2e-7% apart, by two routes that share only the element values. The resonance stays at 1/2π√(LC) to a part in a million throughout. Frequency, which is the same solve

The Q the components allow

Resonance and its bandwidth measured a half-power width of exactly f₀/Q at every Q tried, for a circuit whose only resistance was the one deliberately put there. Real components arrive with resistance of their own, and the consequence is a ceiling rather than a penalty. The reciprocals of the component quality factors add, so the total sits below the smallest of them: an inductor of 79 beside a capacitor of 1 581 gives a resonator of 75. The worst component decides and the best one cannot help.

 A diode's drop from 250 to 400 K, at 1.00 mA. computed by solving, not by drawing. Thirty-one operating points, each Newton's method on the exponential at its own temperature. The drop falls at 1.828 mV/K measured against 1.830 mV/K from the closed form — falls, although the thermal voltage in the exponent rises, because the saturation current rises faster. Over the same range the slope per decade of current goes the other way, from 49.6 mV to 79.4 mV, because that one is Vₜ ln 10 and nothing else. Devices, and the amplitude they stop being linear at

Two millivolts a kelvin, and the wrong sign

Every number in the semiconductor field was computed at 300 K, and the model had no temperature in it at all. Putting it in moves a diode's drop by 1.828 mV/K — downwards, although the thermal voltage in the exponent is rising, because the saturation current rises by nine orders of magnitude across the same range. Two temperature dependences of one device, of opposite sign, from one solve.

 The return under 10.0 cm of track, 200 µm above the plane. computed by solving, not by drawing, as an estimate from the geometry: a low-frequency path assumed to be three track-widths wide, 83.3 mΩ, and the parallel-plate inductance µ₀h/w, 125.7 nH, which is the limit for a track much wider than its height. The estimated resistance and reactance are equal at 106 kHz. Solved across the plane instead of assumed, the return does not change path at one frequency: it gathers beneath the track across a band about three decades wide, and this corner falls inside that band. Above the band the loop is the track's length times its height, 20.0 mm², and a milliamp round it at 100 MHz radiates -1.1 dBµV/m at three metres. Lines, where a wire has a length

Where the current comes back

The return current under a track spreads out at low frequency and runs directly beneath it at high. Estimated from two paths chosen in advance, the change is a corner at 106 kilohertz for any track two hundred micrometres above a half-milliohm plane, with neither the length of the track nor its width in it. Solved, it is a band three decades wide rather than a corner, and the width enters it after all — but the length is still absent and the stack-up still sets the top, which is the part of the argument a designer needs.

 A five-volt regulator's output impedance, with 1.00 Ω of series resistance. 0.430 mΩ at direct current, 1.95 Ω at 10.0 kHz — a factor of 4.52e+3 — and it has already doubled by 4.81 Hz. The upper curve is the same circuit with its loop opened, and the ratio between them is the loop gain. A regulator is a voltage source below a frequency and the datasheet's milliohms are the value at the bottom of it. Circuits that do a job, and the range they do it over

A source below a frequency

A five-volt regulator's output impedance is 0.43 milliohms, which is the number a datasheet quotes. It has doubled by 4.8 hertz, is ten times worse by 27, and reaches 1.95 ohms at ten kilohertz — four and a half thousand times its own specification, and higher there than the same circuit with its feedback loop cut. Two routes to that curve, sharing only the netlist, agree to a part in ten to the thirteenth.

 A sum that is exact, and the bandwidth estimate that is not. computed by solving, not by drawing, at 28 spreads of the three capacitor values in a resistor chain. The sum of the open-circuit time constants — each capacitor's own value times the resistance seen at its terminals with the other two removed — is 600.00 µs here, and it equals the ratio of the first two coefficients of the denominator to 2.0e-9 and the sum of the negated reciprocal poles to 2.0e-9. That much is a theorem. What is an estimate is the bandwidth: one over 2πΣτ gives 265.3 Hz against a measured 309.2 Hz, low by 14.2%. It is low at every spread on the axis — the estimate is never optimistic — and comes within ten per cent only once one of the three time constants is 7.48 times the others. Before the steady state

A sum that is exact, and the estimate that is not

Add each capacitor's value times the resistance seen at its own terminals with the others removed, and the total is the ratio of the first two coefficients of the denominator polynomial — a theorem, holding to a part in a billion at every spread tested. Divide one by two pi times it and you have a bandwidth estimate that is 14 per cent low with three equal capacitors and never once optimistic. Two settings of the slider have the same three time constants and bandwidths two per cent apart, which is why the sum can never be more than an estimate.

 A fifth-order chebyshev delay, and 1 all-pass section in front of it. computed by solving, not by drawing at 60 frequencies across the passband. The delay varies by 891 µs from end to end; 1 all-pass section, with pole pair found by search rather than taken from a table, brings that to 498 µs — a factor of 1.79. The price is that everything is later: 722 µs more at direct current, which is more than the 392 µs of variation removed. Filters, measured not tabulated

Flat delay, bought with more delay

An all-pass section has a magnitude of one at every frequency — measured here on a solved network as 1 to within 9 × 10⁻¹⁶ over six decades — which makes it the only thing that can change a filter's delay without touching its magnitude response. It flattens by adding. A fifth-order Chebyshev's 891 microseconds of delay variation comes down to 498, and everything leaves 722 microseconds later than it did.

 A 100 ms pulse through a 0.159 Hz corner, 9.52% shorter by the end of it. computed by solving, not by drawing. Marched. The dashed line is the pulse that was sent. The solid line is what a 1 MΩ input with 1 µF in front of it receives: the top decays as exp(−t/RC) for the whole 100 ms, ending 9.515% down, and the trailing edge undershoots by exactly the same amount. The closed form for the same two components gives 9.516%. The input's specification is a corner at 0.159 Hz; a top flat to one per cent needs a pulse shorter than 10.1 ms, which is a rate 625.2 times the corner — a constant with no component in it, and the reason a low-frequency specification says nothing useful about an edge. Measurement, which is a circuit on a circuit

The corner that says nothing about an edge

An AC-coupled input is sold on a low-frequency corner, and a corner is a statement about steady sinusoids. What an instrument is usually shown is a pulse, and for a pulse the number is a sag: a hundred-millisecond pulse through a 0.159 hertz corner comes out 9.5 per cent shorter than it went in. A one per cent flat top needs a pulse rate 625 times the corner, which is a constant with no component in it — and above fifty per cent duty an AC-coupled pulse never reaches half its own height at all.

 The ideal amplifier is good to 1% over a region, and its corner is 21% inside the specifications. computed by solving, not by drawing. The 1 per cent contour of the ideal-amplifier model for a non-inverting stage of gain 2 built from a 10 MHz part, drawn over frequency and output amplitude at once. Each point is bisected on a marched circuit: the error is the root-mean-square difference between the marched output and 2 times the input, which counts the gain that is low, the phase that is late and the peak that is flat. Three mechanisms bound the region — finite gain–bandwidth on the left, the input pair's slew rate on the diagonal, and the rails at 12.19 V along the top. The two dashed lines are the numbers a data sheet gives: a small-signal edge at 48.8 kHz with no amplitude in it, and a full-power bandwidth of slew rate over 2πV̂ with no gain–bandwidth in it. They cross at 10.60 V and 48.8 kHz; the measured contour passes 38.5 kHz at that amplitude, which is 0.790 of it. Where the models stop

The edge that is a region

Every boundary drawn so far in this subject is a number on one axis, and the figure that gathers four of them admits in its own caption that the fifth is an amplitude and cannot go there. Drawn on both axes at once, the ideal amplifier's one per cent boundary is a region with three sides and a corner — and the corner sits at 38.5 kilohertz where the two numbers a data sheet quotes cross at 48.8, because the two mechanisms are lags on the same waveform and add as magnitudes rather than in quadrature.

 Where four of these models stop being true. In order: the ideal operational amplifier at 1.42 kHz, a 10 V output at full amplitude at 7.96 kHz, Kirchhoff's laws on 10.0 cm at 3.97 MHz, the ideal 100 nF capacitor at 4.69 MHz. The fifth boundary is an amplitude rather than a frequency and cannot share this axis: a small-signal model is 1% wrong above 7.3 mV, at every frequency there is. Where the models stop

The edges that are lengths

Almost every boundary in this collection is a frequency or an amplitude, and both of those are things a circuit designer chooses. A handful are lengths — the 0.60 millimetres a gap's field reaches into a window, the 200 microns between a track and its plane, the 10 centimetres at which Kirchhoff's laws are a degree out — and they behave differently in one way that matters: nobody chooses them at the schematic, they are set by whoever builds the thing, and they appear in no netlist at all.

 Two tracks, and a far end that cancels exactly when the field is all in one material. computed by solving, not by drawing, on 12 coupled sections of a 100 mm pair terminated in 50 Ω at all four ends. A mutual capacitance injects a current proportional to dV/dt and splits it towards both ends of the quiet track; a mutual inductance injects a voltage proportional to dI/dt and drives the two ends in opposite directions. So the near end goes as Cm/Ct + Lm/Lt and the far end as their difference, with the same constant in front of both — measured here as 1.048e-2 either way, over a slider that moves the ratio by five times. The consequence is that the far end is not a smaller effect but a cancellation: at a ratio of one it is 7.52e-19 of the drive, which is zero to the last bits of a double, while the near end is 1.048e-3. That is why a stripline has no far-end crosstalk and a microstrip has some — what shows up there measures the field that is in air, not the spacing. The model is lumped and stops where it says: a section is one degree long at 50.0 MHz. Lines, where a wire has a length

The far end that cancels

Two mechanisms couple two parallel tracks: a mutual capacitance injecting a current and a mutual inductance injecting a voltage. They add at the near end of the quiet track and subtract at the far end, with the same constant in front of both — measured here as 1.048 times ten to the minus two either way, across a slider that moves their ratio by five times. So the far end is not a smaller effect: when the two couplings are equal it is 3.5 times ten to the minus nineteen of the drive, which is zero to the last bits of a double.

 Two floors on one axis, and the 50.0 mV between them. computed by solving, not by drawing. The flat line is the Johnson current noise of 1 kΩ, √(4kT/R), which has no current in it. The rising line is shot noise, √(2qI), which has no resistance in it. They cross at 50 µA — and the direct voltage across the resistance there is 49.981 mV, which is 2kT/q and contains neither quantity. The slider moves the resistance over six decades; the crossing moves with it and the voltage at the crossing does not move at all, to the last bit of a double. The floor, which bounds from below

The floor a current sets

A resistor's noise contains no current and a current's noise contains no resistance, and the two are equal when the direct voltage across the thing carrying the current is 2kT/q — 50.0 millivolts at 290 kelvin, whatever the resistance and whatever the current. It is the only boundary in this collection whose axis is a direct voltage across an element. And a forward-biased junction, which has the same dynamic resistance as some resistor, produces exactly half its noise power at every current.

 What the summing junction of an inverting amplifier actually is, at 1.00 MHz of gain–bandwidth. computed by solving, not by drawing by driving a current into the node and reading the voltage. It is 100 mΩ at direct current, rises 1.000 decades per decade of frequency, and settles at 909.5 Ω — which is the 1 kΩ and 10 kΩ in parallel, with the amplifier contributing nothing. It passes one per cent of the input resistor at 995 Hz, a factor of 1,005 below the gain–bandwidth. The second route — the open-loop impedance over one plus the return ratio from the cut loop — agrees to 0.045%. Feedback, and the margin

The node that is at ground for a while

An inverting amplifier's summing junction is held at ground by the loop, so it is at ground exactly as well as the loop is strong. Driven with a current and measured, it is a tenth of an ohm at direct current, ten ohms at a kilohertz, and 909 ohms above a megahertz — which is the two feedback resistors in parallel, with the amplifier contributing nothing.

 A 100 nF capacitor with 100 mΩ in series, written the other way round. computed by solving, not by drawing. At 100 kHz the series pair and the parallel pair are the same impedance to 8.7e-19 of itself — the arithmetic's floor, not a tolerance — with Rp = 2.533 kΩ against Rs = 0.100 Ω and Cp = 99.996 nF against Cs = 100 nF. Away from it they part company at a rate set by Q = 159.2: the substitution costs one per cent below 48.2 kHz and above 207 kHz, a band of 4.3 to one. Frequency, which is the same solve

The same part written two ways

A capacitor's loss is quoted either as a resistance in series with it or as one across it, and the pair of expressions that converts between them is exact at one frequency and at no other. How wide the band is around that frequency is set entirely by the quality factor: an octave and a half at Q of eight, a thousand to one at Q of three thousand.

 Superposition holds for the solution and not for its square. computed by solving, not by drawing, at 81 settings of the second source. Two ten-volt sources reach one hundred-ohm load through a hundred ohms each. Every node voltage and every branch current is the sum of the two single-source solves to 3.8e-16 of itself — superposition, exactly. The power is not: the difference is 2·Re(I₁·conj(I₂))·R to 1.0e-15, and at 0° between the sources and equal size it is 444.4 mW against the 222.2 mW that adding gives. Adding the two powers is within one per cent of the truth only when one source is below 0.00505 of the other. The slider is the angle between them: at 90° the two curves coincide to the last bit, and at 180° the true power falls to zero while the sum does not. Networks, and how a solve is checked

Two solves that add, and the one that does not

Every node voltage and every branch current in a linear network is the sum of the per-source solves, here to the last bit of a double at eighty-one settings. The power is not, and the gap is not a correction: two equal sources in antiphase put nothing at all into a load while adding their powers gives 222 milliwatts, and the sum is within one per cent of the truth only when one source is two hundred times the other.

 Two meters, one current, and neither of them measuring the heat. computed by solving, not by drawing, at 35 conduction angles. The first curve is an average-responding meter: it rectifies, averages and multiplies by 1.1107, which is exactly right for a sinusoid — -7.8e-5% here — and exactly 11.07% high on a square wave, because the error is the ratio of two form factors and contains neither the amplitude nor the frequency. On a rectifier drawing its 100 W in sixty degrees of conduction it is -35.90% low. The second curve is a true-RMS meter that reaches 9 harmonics, which has no shape assumption in it and a bandwidth instead: it returns the root-sum-square of the lines it can see, and is one per cent low below every angle here of conduction. The crest factor at sixty degrees is 1.733, which is inside every instrument's rating — neither meter is failing because the peak is large. One is failing because the shape is not a sinusoid and the other because the spectrum is wider than it is. Power, and the part that does no work

What a meter multiplies by

An average-responding meter rectifies, averages and multiplies by 1.1107, which makes it exactly right for a sinusoid and wrong for everything else by the ratio of two form factors — 11.07 per cent high on a square wave and 35.9 per cent low on a rectifier drawing its current in sixty degrees. It is also exactly right at one other waveform, a 145.90 degree conduction angle, which is nobody's sinusoid. Beside it a true-RMS meter that reaches nine harmonics is two per cent low on a square wave and never within one per cent of anything narrower.

 What a 5% mismatched pair leaves behind, across 150 K. computed by solving, not by drawing. A saturation-current mismatch of 5.0% appears as an input offset of Vₜ·ln(m) — 1.2613 mV at 300 K — which is proportional to absolute temperature and therefore drifts at 4.2044 µV/K, exactly the offset divided by the temperature. That is 3333 ppm per kelvin at every mismatch on the slider, because the ratio is 1/T and contains nothing about the device. One junction on its own drifts 1.828 mV/K, 435 times harder. Devices, and the amplitude they stop being linear at

What matching does about temperature

A pair cancels the 1.8 mV/K that broke the previous essay, and what it leaves behind is exact: a saturation-current mismatch of m shows up as an offset of the thermal voltage times ln(m), which is proportional to absolute temperature and therefore drifts in proportion to itself. 3 333 parts per million per kelvin, at every mismatch on the slider, because the ratio is 1/T and contains nothing about the device at all.

 The amplitude at which a converter's floor becomes distortion. computed by solving, not by drawing. The quantisation error is transformed and the power in harmonics of the input is measured directly, at the bins those harmonics occupy. Undithered, the share rises from 0.29% at 230 levels to 76.5% at 1.3: the error has stopped being spread and has become a deterministic staircase locked to the signal. The sweep is drawn against the levels the input uses rather than against volts because it is then the same sweep for every converter — a twelve-bit part and a sixteen-bit part agree here to 0.0e+0, so what decides the character of the error is how many levels the signal crosses and not how many the converter owns. With 1 least significant bit of dither the share stays under 0.34% at every amplitude, for 3.03 dB of signal to noise. Adding noise to a converter's input improves what comes out of it, which is true and sounds like it should not be — and the amount is one whole step, not "some": a quarter of a step leaves 52.7% and a half leaves 25.6%, because a dither smaller than a step cannot make the quantiser cross one. Where a signal becomes a number

When a floor stops being a floor

Quantisation error is treated as noise and behaves like noise while the input crosses many levels. As the amplitude falls it stops: measured at the bins its harmonics occupy, the share of the error's power sitting in harmonics of the input rises from 0.29% at 230 levels to 76% at 1.3. The floor has not moved. It has become a distortion product locked to the signal, and the fix is to add noise on purpose — one whole least significant bit, and no more.

 One diode curve, eight one-decade fits, and eight different ideality factors. computed by solving, not by drawing. A junction with two conduction mechanisms — recombination near n = 2 at low current, diffusion near n = 1 above it — and a series resistance, which is what a real diode is. Fitting ln(i) against v over each decade in turn returns an ideality factor for each, and they run from 1.227 to 1.984 without being monotonic: the factor rises through the recombination region, falls through the diffusion region, and rises again where the series resistance takes over. Two of the windows are straight to a few parts in a thousand, so the residual gives no warning. The bars are what each fit predicts for the forward voltage at 1 mA: the worst is out by -186 millivolts, which is a current 0.01 times the truth. Where the models stop

The constant that is a window

A diode's ideality factor is quoted as a number and defined as a derivative, which means it has a value at every current and no value anywhere. Eight one-decade fits to one curve return factors from 1.23 to 1.98, two of them straight to a few parts in a thousand — so the residual gives no warning at all. Asked for the forward voltage at a milliamp, the window containing it is right to a third of a millivolt and the worst is out by 186, which is a current a hundredth of the truth.

 A copy out by 1.3% for the reason everybody names, and 11% for the one nobody does. computed by solving, not by drawing at 60 output voltages, with the Early conductance iterated to self-consistency against the current that sets it. Two base currents are stolen from the reference, so the copy is β/(β+2) of it — 1.32% low at β = 150 — and that is exact at exactly one output voltage, 0.7043 V, which is 9.39 mV under the reference's own base-emitter voltage of 0.7137 V — a displacement that goes as 1/(β+2), so that the product of the two is 1.427 V at every β the slider offers. Everywhere else the Early effect is larger: the current rises at 1.21% per volt, so moving the output from one volt to ten changes it by 11.2%. One per cent holds over 0.810 V, which is the Early voltage over a hundred and contains neither the current nor any resistor. The slider is β: it moves the first error by fifty times and the second by nothing at all. Devices, and the amplitude they stop being linear at

The copy, and its two errors

Every account of a current mirror leads with the base currents: two are stolen from the reference, so the copy is beta over beta plus two, which is 1.32 per cent at beta of 150 and is what a third transistor is spent on. The Early effect is a footnote and is nine times larger over any useful swing — 11.2 per cent between one volt and ten. Moving beta from 20 to 1000 changes the first by a factor of forty-five and the second by nothing at all.

 50 Ω + j100 Ω of line, and the load angle past which the far end rises. computed by solving, not by drawing at 71 load resistances and four load angles. A source of 50 Ω + j100 Ω feeding loads of the same resistance and different power factor: at unity power factor the voltage across the load climbs towards the source's and stops there, reaching 0.9675 of it at the largest load drawn. A lagging load leaves less. A leading one leaves more, and past a computable angle it leaves more than the source has: the condition is 2Rₗ(Rₛ + Xₛ·tanφ) + |Zₛ|² < 0, which for the largest load here is 28.35° of lead — bisected on the solve at 28.35° — tending to atan(Rₛ/Xₛ) = 26.57° as the load grows. So the edge is a property of the line and the load angle together, and "voltage regulation" quoted as a percentage carries neither. Power, and the part that does no work

The far end that rises

Voltage regulation is quoted as a percentage: how far the voltage at the end of a line falls when the load is applied. The percentage carries neither of the two things that decide it. Past a computable angle of leading load the voltage at the far end goes above the source's — 28.35 degrees for a line of fifty ohms and a hundred of reactance — and with no reactance in the line there is no such angle at all, because the rise is a partial resonance and needs both halves.

 A 4.0:1 load reads 1.13:1 through twenty metres of cable. computed by solving, not by drawing. A 200 Ω load on a 50 Ω line — a standing-wave ratio of 4.00 at the load itself — measured from the other end of a length of cable that attenuates 0.500 dB per metre at 1 GHz. The reading falls as the length grows, exactly as |Γ| times ten to the minus twice the one-way loss over twenty, and reaches 1.5 at 9.5 m and 1.128 at twenty metres, which is 24.4 dB of return loss and would pass any acceptance test. The load is unchanged; the instrument is looking at it through 10.0 dB of attenuator. The length that hides it falls 3.16× per decade of frequency, which is the root of ten and is the attenuation's own law. Lines, where a wire has a length

The mismatch that the cable hides

A lossless line carries a reflection back unchanged, so the standing-wave ratio at the instrument is the standing-wave ratio at the load. A real line does not, and the departure is exact: ten decibels of one-way loss improves any mismatch by twenty. A four-to-one load at the end of twenty metres of ordinary coaxial cable measures 1.13 at the near end, a return loss of 24 decibels, and passes an acceptance test the load could never pass. The boundary is a loss rather than a length, which makes it a frequency: 30.2 metres at 100 megahertz, 9.5 at a gigahertz, 3.0 at ten.

 A passive network, read from each end — and the two readings are one number. computed by solving, not by drawing. A five-element RLC ladder with a current injected at one port and the voltage read at the other, then the two exchanged. With no controlled source the two readings agree to 4.9e-14 of themselves over four decades, which is the arithmetic's noise rather than a physical difference — the network cannot tell which way round it is being used. A mutual inductance keeps that: a 1 mH and a 4 mH winding at k = 0.7 give 0.0e+0, because the coupling puts the same entry in both halves of the matrix. A transconductance does not, and the departure is proportional to it with a fitted exponent of 1.000 over four decades — so there is no small amount of gain that is harmless. It passes the arithmetic's own floor at 0.781 femtosiemens, and the smallest transistor in this collection is nine orders above that. Networks, and how a solve is checked

The reading that does not care which way round it is

Inject a current at one end of a network and read the voltage at the other; then swap the two. A passive network gives back the same number — not a similar one, but the same to five parts in ten to the fourteenth over four decades of frequency, and for a coupled pair of windings the same double. Put a transconductance of eight tenths of a femtosiemens anywhere in it and the two readings part company.

 The same millivolts, subtracted — and what four resistors leave behind. computed by solving, not by drawing at 145 frequencies. A 10 mm run of one-ounce copper carries the return of a 100 mA load and the reference of a 10 mV sensor. Its 5.00 mΩ and 10.0 nH put 500.0 µV in series with the sensor at low frequency — 5.00% of the reading — rising a decade per decade above 79.6 kHz until at 1.59 MHz the error is the whole signal. The full solve and the interfering current times the shared impedance agree to 4.4e-7. The slider is the length of the shared run: both the resistance and the inductance are proportional to it, so every point on the curve moves down together and the corner stays at 79.6 kHz. The second curve is the same measurement made differentially: a unity-gain difference amplifier across the sensor, its four resistors carrying the worst-case skew 0.1% allows. It divides the error by 500.5 — 53.99 dB against a closed-form (1 + G)/(4·tolerance) of 53.98 dB — at every frequency, because four resistors have no frequency in them. What is left is 998.9 nV, which is not zero: the amplifier subtracts the interference and the resistors decide how much of it survives. Measurement, which is a circuit on a circuit

The rejection four resistors decide

Measuring a ten-millivolt sensor against a ground that somebody else's hundred milliamps is also using puts five hundred microvolts of their current into the reading. Subtracting the two ends of that conductor with a difference amplifier removes it — by 53.99 decibels with one-tenth-per-cent resistors, against a closed form of 53.98, which is a factor of five hundred and not a removal. The amplifier has nothing to do with it: the number is one plus the gain over four times the resistor tolerance.

 A 10 kΩ resistor, and the 19.3 MHz it is one below. computed by solving, not by drawing. The dashed line is R, which is what the symbol means. The solid line is the same part with 8.0 nH of lead inductance in series and 0.40 pF across the body, solved as a three-element network and checked against the closed form for the same three elements to 3.3e-16. It is ten per cent below its own value by 19.3 MHz, and which of the two parasitics does that depends on the resistance: the shunt capacitance wins above 91.02 Ω and the lead inductance below it. The slider is the resistance, and the departure frequency it moves is not monotonic — it rises a decade per decade of resistance, peaks near 91.02 Ω at 2.00 GHz, and falls a decade per decade after that. Frequency, which is the same solve

The resistor that is only a resistor

A capacitor becomes an inductor above a frequency its leads decide, and an inductor becomes a capacitor. The third member of that family is the one nobody draws, and it is the only one whose edge is not monotonic in its own value: a ten-megohm resistor stops being one at 19 kilohertz, a ten-ohm resistor at 92 megahertz, and between them sits a resistance whose impedance is flat to fourth order — 91.02 ohms here, which is the square root of L over C divided by the root of one plus root two.

 Five resistances, five corner frequencies, and one total on the capacitor. computed by solving, not by drawing. The noise density at a 1 pF capacitor charged through 100 Ω, 1 kΩ, 10 kΩ, 100 kΩ and 1 MΩ, at 290 K. The densities are 100 times apart and the noise bandwidths 1.0e+4 times apart, in opposite directions, so the area under every curve is the same: 63.2762 µV against 63.2762 µV, and √(kT/C) is 63.2762 µV. The resistance has cancelled out of the answer, and the reason is that ½C⟨v²⟩ is the ½kT a degree of freedom in contact with a bath holds — which no arrangement of resistors can change. The claim is about the whole frequency axis and nothing less: inside a 15.9 MHz band the same five networks give 5.05 µV to 63.07 µV, a factor of 12.5. The floor, which bounds from below

The total that has no resistor in it

A larger resistor is noisier and makes a narrower filter, and the two dependences are exactly reciprocal: the density goes as the square root of the resistance and the noise bandwidth as its inverse. Five decades of resistance charging one picofarad therefore give five decades of corner frequency, two and a half decades of density, and one total — 63.2762 microvolts at every one of them, which is the square root of kT over C and contains no resistance at all.

 A step on a series RLC at ζ = 0.079, and the two numbers read off H(s). computed by solving, not by drawing. A 50.3 kHz series RLC driven by a one-volt step, with the capacitor voltage and the inductor voltage drawn together. Two limits of the transfer function are two points of the waveform and neither needs the waveform: H(0) = 1.000000 is where the capacitor ends up, and H(∞) across the inductor is 1.0000, which is what it does at the first instant — the expansion gives 1.000000 for it at t = 0. Here the damping ratio is 0.0791, the response is inside ±2% after 7.6 cycles, and 100.0% of the last twenty-four cycles sit there. The poles are at a real part of -7.91e-2 of ω₀, which is the condition the final-value theorem actually has — not a property of H but of where sY(s) has its poles. Before the steady state

Two numbers without solving for the waveform

Where a step response starts and where it ends are two limits of the transfer function, and neither needs the waveform. Both are exact here — 1.000000000 volts at the end and the whole step at the first instant — and one of them is a lie waiting to happen: take the damping to zero and the final-value theorem still returns 1.000000 for a response that swings between 0 and 2 for ever. Its condition is not on the transfer function but on where the poles are, and the practical condition is narrower still: at five ohms the poles are safely in the left half-plane and sixty cycles is not enough time.

 The window a 10 µF capacitor leaves. Above 939 mΩ the loop holds 45° of margin; below it the regulator rings and then oscillates. The droop after a 100 mA step is smallest at 817 mΩ — 129 mV — and by 19.9 Ω it is 1.468 V, because at the first instant of a step the capacitor cannot move and the whole step falls across its series resistance. Two requirements, opposite directions, and the useful values are between them. Circuits that do a job, and the range they do it over

Two requirements pulling one capacitor

The output capacitor's series resistance is a stability requirement and a transient requirement at once, and they pull it in opposite directions. Below 939 milliohms this loop has less than 45 degrees of margin; above about 850 the droop after a load step starts to grow, because at the first instant of a step the capacitor cannot move and the whole step falls across that resistance. The droop is smallest 10 per cent inside the unstable region, which means the best transient this design can have is one it must not be built with.

 What a mistuned arm leaves at 1300 Hz. computed by solving, not by drawing. Neither curve is the null's depth — the null is still bottomless, it has simply moved — but the depth at the frequency the notch was designed for, which is the number a filter is bought for. One per cent components leave -32.9 dB if both errors go the same way and -78.8 dB if they oppose, a factor of 197 from the same tolerance on the same two parts. The slopes are 20.0 and 40.0 decibels per decade: first order in the error on the product LC, second order in the error on the impedance level. Filters, measured not tabulated

What actually fills a null

A ten per cent error in the two components of a notch's arm leaves the null three hundred decibels deep — it moves it rather than filling it. What fills it is loss, at twenty decibels per decade of arm resistance exactly. And the depth at the frequency the notch was designed for splits into two orders depending on which way the two errors go: one per cent parts leave 79 decibels one way and 33 the other.

 The voltage a winding may carry, which is a volt-second limit read at a frequency. computed by solving, not by drawing. The dots are bisections on a marched flux — the voltage integrated sample by sample until the peak excursion reaches 0.35 T — and the line is N·Ae·Bsat·2πf. They agree to 0.001% over three decades, and the fitted slope is 1.000000: exactly proportional, because flux is the integral of voltage and nothing else. The quantity that belongs to the core is the 3.500 mWb-turn, which has no frequency in it. A transformer "rated for 50 Hz" is a transformer whose volt-second product was divided by 2π × 50 once. Two windings, and the band between them

A boundary in volt-seconds

A core saturates on the integral of the voltage applied to it, not on the current through it and not at a frequency. The quantity that belongs to the core is N·Ae·Bsat — 3.500 mWb-turn here — and it has no frequency in it at all. Everything a data sheet says about a transformer's frequency rating is that one number divided by 2πf once: measured on a marched flux, the voltage a winding may carry is proportional to frequency to a fitted exponent of 1.000000.

 The best shunt drops 7.75 mV, whatever the current is. computed by solving, not by drawing. Two errors on one axis, both from solved networks: the shunt's own drop, which lowers the current that was to be measured, and the amplifier's 5.0 µV of offset divided by the voltage the shunt develops. The first rises with the burden voltage and the second falls, so the worst case has an interior minimum at 7.7460 mV — the geometric mean of the offset and the 12 V supply — where the error is 0.1291%, being twice the root of the offset over the supply. Neither the shunt's resistance nor the current appears in either number: at 1 A the answer is 7.75 mΩ, and at a hundred times the current it is the same burden voltage across a hundredth of the resistance. What does depend on the current is the 7.7 mW the shunt then dissipates, and 40 K of self-heating at 50 ppm/K is 0.2000% on its own. The third curve is the shunt's own Johnson noise in a kilohertz of measurement bandwidth, as a fraction of the current: it is 4.5e-6% at the best burden and is the only line here that moves with the current at all, falling as one over its square root — so above about an ampere it leaves the bottom of these axes entirely and is drawn nowhere rather than flattened onto the floor. Measurement, which is a circuit on a circuit

The ammeter that is a resistor

Every direct measurement of a current is a measurement of a voltage across something the current was made to flow through, so the instrument has two errors pointing opposite ways: a larger shunt changes the current, a smaller one leaves less for the amplifier's offset to be compared with. Written in the burden voltage they are the burden over the supply and the amplifier's offset over the burden, and the best of them is at the geometric mean — 7.75 millivolts on a twelve-volt rail, with a worst-case error of 0.129 per cent. Neither number contains a resistance, and neither contains the current: ten milliamps and a hundred amps want the same 7.75 millivolts.

 A follower's output impedance from 1 kΩ of source, bare and with 100 pF on it. computed by solving, not by drawing, on a small-signal follower at 2.0 mA with β = 150 and fT = 560 MHz. At 100 Hz the emitter presents 19.08 Ω against a textbook 1/gₘ + Rₛ/(β+1) of 19.55 Ω — the expression is an upper bound here and at every source resistance on the slider, 2.4% high at this one. What it cannot describe is the frequency axis: the β that divided the source resistance down is itself falling, so the impedance rises, and the reactance at 3 MHz is 4.3 Ω — an inductance of 0.229 µH against Rₛ/ωT = 0.284 µH. With 100 pF hung on the output that impedance peaks at 67.0 Ω at 29.3 MHz, 3.51 times its own low-frequency value: an inductive source and a capacitive load are a resonant circuit, and this one is inside a part whose output impedance is quoted as a single number. Devices, and the amplitude they stop being linear at

The buffer that is not a buffer

An emitter follower is reached for when something has to be driven without being loaded: unity gain in, high impedance seen, low impedance presented. The last of those is a number with a range, and the range is narrow. At a kilohm of source the emitter presents 19.08 ohms at low frequency and 67 ohms at 29 megahertz, because the current gain that made it small is falling — and the peak is worst in the middle of the slider, so it cannot be avoided by making the source stiffer or softer.

 A 1.0% doublet: 0.078 dB in the magnitude, 36× the settling time. computed by solving, not by drawing. Above, the magnitude of a fast circuit followed by a pole and a zero that were meant to cancel and miss by 1.00%, against the same circuit with the cancellation exact: the worst disagreement anywhere up to the fast corner is 0.0777 dB. Below, the error left in the step response, in units of the tail's own amplitude of 0.909%. Settling to 0.10% takes 245.2 fast time constants against 6.9 with the cancellation exact, and the closed form τ·ln(A/B) gives 245.2 — a time that contains nothing of the fast circuit at all. Before the steady state

The cancellation that leaves a tail

A pole and a zero placed on top of each other disappear from the response. Miss by one per cent and the magnitude changes by 0.078 decibels, which no measurement would report as a fault, while the time to settle to a thousandth goes from 6.9 time constants to 245 — thirty-six times longer. The settling time has a closed form containing neither the fast circuit nor the doublet's separation as such, and its consequence is blunt: settling to a part in ten thousand needs a cancellation good to a part in ten thousand, however fast the amplifier in front of it is.

 Linearising an exponential at 125 °C, and what it costs. The linear model understates the gain by 1% at 9.69 mV and by 10% at 30.2 mV. The thermal voltage at this temperature is 34.3 mV, so "small compared with Vₜ" is not the criterion — 28% of Vₜ is already 1% wrong. Where the models stop

The edges that move with the room

Every boundary in this collection is quoted at one temperature and most of them are functions of it. The small-signal edge is proportional to the thermal voltage, so it runs from 5.67 millivolts at −40 degrees to 9.69 at +125 — a factor of 1.71, the ratio of the absolute temperatures exactly. A realised Q is 1.54 per cent high at one end of that range and 2.58 at the other. The numbers are right; the condition attached to them was left off, and it is the same condition every time.

 An inverting unity gain driving 2.2 nF, and the pole that is inside the loop. computed by solving, not by drawing. Two ten-kilohm resistors around a 10 MHz amplifier make a gain of 1.00, and a loop that closes against 2.01 — one plus the ratio, not the ratio. Hanging 2.2 nF on the output leaves the closed-loop gain at a kilohertz unchanged — 0.99998002 against 0.99997988 — and takes the phase margin from 90.0° to 30.1°. The mechanism is at the other end of the amplifier from the summing-junction case and the arithmetic is the same: the load works against the amplifier's own fifty ohms of output resistance, which puts a second pole in the forward path — inside the loop, where the feedback has to live with it — while the gain the loop closes against does not move at all. Forty-five degrees is reached at 905 pF, bisected on the netlist. The capacitance is not part of the signal path and does not appear in any expression for the gain. Feedback, and the margin

The load that gets inside the loop

Hanging a capacitor on an amplifier's output changes nothing a reader can find in any expression for its gain, and takes the phase margin of a unity-gain inverter from ninety degrees to thirty. The mechanism is fifty ohms of output resistance that no data sheet page puts next to the stability page: the load works against it, the pole that results is in the forward path, and forty-five degrees arrives at 905 picofarads — which is a metre of coaxial cable.

 kT/C, on a circuit whose resistance is a clock and moves by 1e+4. computed by solving, not by drawing. The noise a switched-capacitor low-pass leaves on its holding capacitor, against the ratio of the two capacitors, measured by marching a seeded sequence through the recursion the circuit obeys. It is 6.328 µV at every ratio drawn — kT/Ch, with the holding capacitor and nothing else in it — while the resistance the arrangement behaves as moves from 1.00 kΩ to 10.0 MΩ across the same axis. The two lines below it are the two sampling events that make it: what the input switch leaves on the switched capacitor, and what the sharing switch leaves behind when it opens. Neither is the answer and their sum is, exactly, because the pole's own bandwidth factor is the same expression. The floor, which bounds from below

The noise a clock does not make

A hundred-megohm resistor has a noise density of 1.27 microvolts per root hertz. A switched capacitor that behaves as a hundred megohms has none of it: the noise on the capacitor it charges is kT/C, with the holding capacitor in it and nothing else — not the clock, not the switched capacitor, not the on-resistance. And it is exact rather than asymptotic, at every capacitor ratio from a thousandth to ten.

 One magnitude curve, two phase curves, and the one of them the magnitude decides. computed by solving, not by drawing. A passive lead network — a resistor with a capacitor across it, over a second resistor, its zero at 1.00 kHz — and the same network followed by a first-order all-pass. The two magnitudes agree to the last bits of a double at every one of the 1601 frequencies sampled, and the phases differ by as much as 180°. Bode's gain–phase integral, fed the magnitudes alone with their phases discarded, returns 39.29° at the corner against a solved 39.29°, and tracks the minimum-phase curve to 0.02° across the band — while being wrong about the second network by the all-pass's own phase, which is what excess phase means. The edge here is the span of the sweep rather than a frequency of the circuit: ±3 decades of magnitude carries 99.92% of the integral's weight, and what is left out is the tail of a logarithm. Frequency, which is the same solve

The phase the magnitude already knows

For one class of network the phase is not an independent measurement: it is fixed everywhere by the magnitude, through an integral Bode wrote down in 1945. Fed nothing but the magnitudes of a solved lead network, that integral returns 39.289 degrees at the corner against a solved 39.289, and tracks the whole curve to two hundredths of a degree. Cascade an all-pass and the magnitudes agree to the last bits of a double while the phases part by 180 degrees — so the recovery is exact for one of the two and cannot see the other at all.

 25% compensation: 4.26% regulation, and a resonance at 25.0 Hz. computed by solving, not by drawing. The current a 230 V feeder draws, swept from two hertz to a hundred and fifty, with 25% of the line's reactance cancelled by a series capacitor. The regulation falls from 5.17% to 4.26%, which is what the capacitor was fitted for. What comes with it is a series resonance at 9.59 Hz with the load connected and 24.99 Hz with the far end shorted — the latter being exactly f₀√k = 25.00 Hz, a frequency the compensation fraction chooses on its own. The line's inductance does not appear in it and neither does the voltage. It is always below the fundamental, which is where the machines are. Power, and the part that does no work

The reactance cancelled, and the resonance it buys

Putting a capacitor in series with a feeder cancels part of its reactance and the far end falls less: five per cent of regulation becomes four and a quarter at a quarter compensation and 2.7 per cent at seventy per cent. What comes with it is a series resonance that was not there before, at the line frequency times the root of the fraction cancelled — so a quarter compensation resonates at exactly half the line frequency and a ninth at exactly a third. The line's own inductance is not in that answer and neither is the voltage.

 A series-terminated net holds half a swing for 1.0 delays at 50% along it. computed by solving, not by drawing. What a receiver 50% of the way along one net sees under three terminations, from a lattice evaluated at that point rather than at the ends. The shaded strip is the interval in which a logic input has no defined answer — between 30% and 70% of the swing. The series-terminated net sits in it for 4.83 ns, which is 2(1 − x) delays exactly and is zero only at the far end; the unterminated net sits in it for 4.83 ns and then overshoots by 82%; the parallel-terminated net never does, and draws 60 mA down the line for as long as the level is held. Lines, where a wire has a length

The resistor at the wrong end

A lattice diagram is read at the two ends of a line, and that is where the two respectable terminations look identical: a clean step, one delay late, at the receiver. Anyone standing halfway along a series-terminated net sees half the swing held for a full round trip, which for a logic input is not a level at all. The interval is 2(1−x) delays exactly, it is zero only at the far end, and the scheme that never has it draws sixty milliamperes for as long as the level is held.

 The droop a zero-order hold imposes at 48 kHz. computed by solving, not by drawing. Holding each sample for a clock period convolves the output with a rectangle, so the spectrum is multiplied by a sinc: -0.143 dB down at a tenth of the sample rate, -0.912 at a quarter and -3.922 at half — which is exactly 20 log(2/π) and contains no design decision at all. The dots are the amplitude of the fundamental read out of the transform of the staircase itself, agreeing with the closed form to 0.008%. There is also half a sample of delay, 10.417 µs here, which is the reason a held reconstruction is not a droopy copy of the signal but a droopy copy that has moved. Where a signal becomes a number

The staircase on the way out

A converter does not emit impulses. It holds each sample for a whole clock period, which is a convolution with a rectangle and therefore a multiplication by a sinc — 0.14 dB down at a tenth of the sample rate, 0.91 at a quarter, and 3.92 at half, which is exactly 20 log(2/π). Nobody chose that droop and it is nearly eight times the half-decibel ripple of a Chebyshev passband. Measured on the transform of the staircase itself, it agrees with the closed form to 0.008%.

 A divider of 4 equal 1.0% resistors, solved 3000 times. computed by solving, not by drawing. Every resistor drawn from its tolerance band and the divider solved, 3000 times. The worst case is ±1.000% — the part tolerance itself, and it does not improve when the divider is built from more parts — while the measured spread is 0.2944% and the worst of 3000 draws reached 82% of the bound. The root-sum-square, offered as though it were a standard deviation, is 1.70 of one here: for uniformly distributed parts it is √3 σ, a coverage of about 92%. Networks, and how a solve is checked

The tolerance that is not on any part

Four one per cent resistors in a divider give an answer whose worst case is one per cent, whose measured spread is 0.29 per cent, and whose root-sum-square bound — offered everywhere as though it were a standard deviation — is 1.70 of one. Adding parts does not move the worst case at all and shrinks the spread as one over their root, so the gap between the promise and the fact widens with every resistor. And the same arithmetic draws a boundary in tolerance rather than in frequency: an R–2R ladder is a twelve-bit converter only while its resistors are inside 0.14 per cent.

 An order-5 ladder driven from 2× the 1 kΩ it was designed between. computed by solving, not by drawing. A doubly-terminated Butterworth ladder is a two-port designed between two stated resistances, and the resistances are part of the design rather than the environment it happens to be used in. At match it loses 6.021 dB — exactly half the voltage — and its passband has no peak anywhere in it. Driving the same five reactances from 2× that resistance moves the shape by 2.345 dB and the insertion loss to 9.542 dB. The band inside which the shape is right to 0.5 dB runs 0.8857× to 1.1371× — a window of 25 per cent on a quantity usually written down as a round number. And the two sides are not alike: at twenty times the design resistance the departure has settled at 5.33 dB, while at a twentieth of it the passband is 15.51 dB out with 7.21 dB of peaking on it, so driving a ladder from too low an impedance is worse than driving it from too high a one. Filters, measured not tabulated

The two resistors a ladder was designed between

A passive ladder filter is not a transfer function with some resistors attached; it is a two-port designed between two stated resistances, and the resistances are as much part of the design as the inductors. Driving an order-five Butterworth from anything outside 0.886 to 1.137 times its design resistance puts more than half a decibel of error on the passband — a tolerance tighter than the resistor is usually specified to — and the two sides of that window are not alike.

 Rejection is not a property of the loop. The same regulator with the same loop gain and the same 46.5° of phase margin, drawn twice. With the amplifier's output referred to ground the rail reaches the output at -67.3 dB at the bottom and 5.79 dB at 10.0 kHz — where it is amplified. Referred to the rail instead, every point is 60.00 dB lower, which is 20 log(gm·ro) for the pass device and not a design choice. Circuits that do a job, and the range they do it over

What gets through from the rail

A regulator's job is to hold its output still while its input moves, and the measurement says it does that well at ten hertz, badly at a kilohertz, and not at all at ten kilohertz — where this one puts out 1.9 times what arrives. Then the same netlist with one node moved, the same loop gain and the same 46.5 degrees of margin, rejects 60.009 decibels better at every frequency in six decades. The sixty decibels is the pass device's own intrinsic gain and it is not a design choice.

 Five boundaries, one tolerance, and three exponents. computed by solving, not by drawing. Each of five model boundaries re-solved at forty-one tolerances from 0.1% to 30%, divided by its own value at 0.1% so that an amplitude in millivolts and four frequencies can share one axis — an exponent has no units. Fitted over the two decades to 10%: Kirchhoff's laws 1.000, the ideal amplifier 0.513, the small-signal model 0.497, the ideal capacitor 0.500, and the full-power bandwidth 0.000. A boundary set by a first-order departure moves in proportion to the tolerance, one set by a second-order departure moves as its square root, and a refusal does not move at all — so relaxing the tolerance from 0.1% to 10% buys a factor of 100 on the board and 10.0 on the capacitor. Where the models stop

A boundary is a model and a tolerance

Every edge in this collection is computed from a fraction of error nobody states, and the four on its opening axis use three different ones. Swept over two decades, each boundary moves as a power of that fraction — Kirchhoff's laws exactly as the first power, the amplifier and the capacitor and the small-signal model as its square root to within three per cent, and a full-power bandwidth not at all. The exponent identifies the mechanism, and it re-orders the axis twice: the board fails before the capacitor below 0.424 per cent, and the output before the amplifier above 21.7.

 The switches set the floor below 60.2 MHz and the amplifier sets it above. computed by solving, not by drawing. The two contributions to a switched-capacitor stage's noise floor, against the clock frequency. √(kT/C) is 63.3 µV on a 1 pF hold capacitor and does not move with the clock at all. The amplifier's own 4 nV/√Hz is white and is sampled, so all of it folds into the band: settling to 12 bits in half a clock period needs 8.32 time constants, the closed loop's noise bandwidth is then 8.32 clocks over two, and the number of folds is 8.32 — the same number, exactly. The marched route is a seeded white sequence put through the settling exponential and sampled once per clock, and it agrees with the closed form to 0.05% of variance. The two floors are equal at 60.2 MHz, above which a larger capacitor buys nothing. The floor, which bounds from below

The amplifier inside the sample

kT/C is exactly independent of the clock, of the capacitor ratio and of the switch resistance — two essays measured that and found it identically true rather than nearly so. The amplifier in the same loop behaves in the opposite way in every respect: its noise is white, it is sampled, and the number of times it folds into the band is exactly the number of time constants the settling needs. So the switches set the floor below sixty megahertz and the amplifier sets it above, and asking for two more bits of settling costs fifteen per cent more noise before anything else has changed.

 A 4 kHz Butterworth, mapped to 48 kHz two ways. computed by solving, not by drawing. The upper panel is the digital response on the unit circle; the lower one is where each analogue frequency lands. The bilinear transform compresses the axis as it approaches half the sample rate — -3.11% at a tenth of the rate and -15.2% at a quarter — so a design mapped straight through is -3.432 dB at its own corner instead of -3.010. Pre-warping puts that one frequency back exactly and no other: above it the pre-warped curve is the further of the two from the analogue prototype. It is a choice of where to be right, not a correction. Where a signal becomes a number

The corner that moved

The bilinear transform has to fit an infinite frequency axis onto a circle, so something must be compressed, and what is compressed is everything near half the sample rate. A 4 kHz Butterworth mapped to a 48 kHz clock is 3.432 dB down at its own corner instead of 3.010, and at a quarter of the sample rate the axis is 15.2% out. Pre-warping puts one frequency back exactly and no other — it is a choice of where to be right, not a correction.

 The reverse peak reaches the forward current at 12.6 A/µs. computed by solving, not by drawing. The peak current a diode conducts backwards while its stored charge is removed, against the rate the external circuit drives its current down. The charge equation is integrated in closed form and marched trapezoidally, and the two agree to a part in a thousand. Drawn over it are the two expressions that describe the limits: √(2·IF·a·τ), which is what a reference gives and which is 16.7% high at this slope, and aτ, which contains no forward current at all and is what the peak approaches when the ramp is slow. At 100 A/µs the junction goes on conducting for 48.3 ns and reaches 3.83 A backwards — 3.83 times the current it was carrying forwards — and the two are equal at 12.6 A/µs. Before the steady state

The diode that conducts backwards

Every diode in this collection is an instantaneous function of its own voltage, which is exact for an operating point and has no time in it at all. A conducting junction holds a charge, and until that charge is gone it cannot block: drive its current down at a hundred amperes a microsecond and it conducts 3.83 amperes backwards for 48 nanoseconds, against the one ampere it was carrying forwards. The expression every reference gives for that peak is 17 per cent high there, and is right to a per cent only above sixteen thousand amperes a microsecond.

 The first conduction carries 26× the repetitive peak, and the factor of 7 is the user's. computed by solving, not by drawing. The largest diode current in the first conduction after switch-on, against the phase of the mains at the instant of switching, marched from an empty capacitor through the same netlist the ripple and the conduction angle are read from. The settled circuit's largest current is 1.230 A; the first one is 32.39 A at the worst instant and 4.64 A at the best, so which of them a design has to survive is decided by nothing in the design. The estimate that treats the capacitor as a short circuit gives 34.00 A, 4.7% high, the difference being the charge the capacitor takes during the pulse itself. Holding the first peak to ten times the repetitive one would need 1.90 Ω of winding resistance. Power, and the part that does no work

The first cycle, which no steady state contains

Every number this field computes about a rectifier — the ripple, the crest factor, the conduction angle, the power factor — is read from the settled state, and the march that produces them starts from an empty capacitor and throws the first cycle away. That first conduction carries 32.4 amperes against a repetitive peak of 1.23, it is 26 times larger than anything the circuit ever does again, and how large it is depends on when somebody's hand closed the switch.

 A 1% imbalance, and the 64 cycles it survives. computed by solving, not by drawing. The upper panel is the peak flux density, marched cycle by cycle, under a square drive whose positive half is 1% larger in area than its negative half. It does not settle. It walks, by the same area every cycle, and reaches 0.35 T after 64 cycles — 1280 ms at 50 Hz — against a closed form of 63.7. The lower panel is the count against the imbalance, and it rises without bound and never becomes infinite. Halving the drive gives 128 cycles, which is exactly twice: reducing the amplitude buys time and not safety, and there is no amplitude at which this design is inside a limit. Two windings, and the band between them

The flux that walks

The previous essay's saturation limit is an amplitude, and an amplitude can be respected. This one cannot. A drive whose two half-cycles differ in volt-seconds by one per cent adds the same small area to the flux every cycle, so it reaches saturation after 64 cycles — and halving the drive gives 128, and a tenth of it gives 637. Reducing the amplitude buys time in exact proportion and removes nothing. There is no amplitude at which the design is inside a limit.

 A first stage of 100 buys 40.0 dB of rejection, and gives it back above 10.0 kHz. computed by solving, not by drawing. The rejection of a three-amplifier instrumentation amplifier against frequency, beside the one-amplifier difference stage it is built around. At low frequency the two differ by 39.99 dB against 20 log 100 = 40.00 dB, and the reason is that the input stage passes a common-mode voltage at exactly unity: the common-mode gain of the whole instrument is 1.998 mV/V, which is the difference stage's own. So the four resistors around the last amplifier decide the rejection and the two that set the gain do not — ten per cent between them moves it by less than a hundredth of a decibel. What ends it is bandwidth: above 10.0 kHz, which is the amplifier's gain–bandwidth divided by the gain that bought the rejection, the differential gain falls and the rejection falls with it at twenty decibels a decade. Measurement, which is a circuit on a circuit

The four resistors that decide, and the two that do not

A difference amplifier's rejection is decided by four resistors and one-tenth-per-cent parts give 54 decibels. Putting a two-amplifier stage in front adds exactly twenty times the log of its gain — 94 decibels at a gain of a hundred — and the reason is not that the input stage rejects anything. It passes common mode at exactly unity, so the common-mode gain of the whole instrument is 1.998 millivolts per volt at every gain tried, and the improvement is entirely the differential signal arriving larger. The two resistors that set that gain may be ten per cent apart without moving the answer a hundredth of a decibel.

 Fourteen decades of imbalance, fourteen digits gone, and a matrix in perfect health. computed by solving, not by drawing. A Wheatstone bridge walked towards balance, with the relative error of the solved output against a closed form that cannot lose digits. The condition number of the nodal matrix is 505.0 at every imbalance and the smallest pivot is 2.0e-3 of the matrix norm — orders above the 1e-12 at which this solver refuses to answer at all. Neither number moves, and the answer still loses one digit per decade of imbalance, reaching 33% at δ = 1e-15. The bound drawn over it is the round-off divided by the imbalance, which the measurement stays under at every point. The third curve is the same closed form written as ½ − 1/(2+δ) — algebraically identical, and it loses its digits at the same rate, which is where the loss lives: in the subtraction of two nearly equal numbers, not in the matrix. Networks, and how a solve is checked

The matrix that is ill, and the answer that is not

Every other essay here treats the solve as exact, and it is not. A bridge walked towards balance loses one digit per decade of imbalance and has none left at a part in 10¹⁵ — on a matrix whose condition number never moves and whose smallest pivot stays four orders above the threshold this solver refuses at. A feedback amplifier does the opposite: the solver declines to answer at a gain of 10⁹, one decade after it returned an answer that was exact to the last bit.

 A stub holds the far end at two thirds for twice its own delay. computed by solving, not by drawing. A series-terminated net with a branch on it, marched as waves on a delay grid. Three lines of equal impedance meet at the junction, so each presents the others with Z₀/2 and a wave arriving is reflected by exactly −1/3 with two thirds going on. The far end therefore receives 66.7% of the swing at one line delay instead of all of it, and is held there for 0.400 line delays — twice the stub's own delay of 0.41 ns, being the round trip to its open end and back. The same net without the branch is drawn beside it and settles in one round trip, which is what a series termination is for. Each further round trip of the stub divides what is left of the error by three and turns it over, because the returning wave doubles at the open far end — so the receiver approaches its level alternately from below and from above. Lines, where a wire has a length

The receiver that is a branch

A lattice diagram has two ends, and an interior receiver is not a point on a net — it is a short piece of track leading off it to a pin, open at the far end. Three lines of equal impedance meeting at a junction present each other with half the impedance, so a wave arriving is reflected by exactly minus a third and two thirds goes on: the far end receives two thirds of the swing and sits there for twice the stub's own delay, whatever the net is terminated with and wherever on it the branch is.

 9.9 Ω restores 45°, 23 Ω restores 60°, and the load pays for it in ohms. computed by solving, not by drawing. Phase margin against the resistor placed between a unity-gain inverter's output and 2.2 nF of load capacitance, with the feedback taken from the amplifier's own side of it. With no resistor the margin is 30.11°; 9.90 Ω restores 45° and 23.20 Ω restores 60°. The lower curve is the same resistor with the feedback taken from the load instead, where it makes every value worse — the pole is then inside the loop rather than outside it, and at 220 Ω the margin is 13.4°. The rising curve is what it costs: the loop no longer regulates the load's node, so 1 kΩ of resistive load pulls the output down by 0.990% at 10 Ω, uncorrected, at direct current. Feedback, and the margin

The resistor that buys the margin back

Two point two nanofarads takes a unity-gain inverter's phase margin from ninety degrees to thirty. Ten ohms between the amplifier and the load restores forty-five, twenty-three restores sixty, and it works for a reason that reads as a cheat: the feedback is taken from the wrong side of the resistor, so its pole is outside the loop. Take the feedback from the load instead — which is what anyone controlling the load would do — and the same resistor makes every value worse. What it costs is that the loop no longer regulates the load's node at all: ten ohms is one per cent of error into a kilohm, at direct current, uncorrected.

 One Sallen-Key design at 10 kΩ, and the band of impedance levels it survives. computed by solving, not by drawing. A 10.0 kHz unity-gain Sallen-Key section realised at seven impedance levels three decades apart, with every resistance multiplied and every capacitance divided by the same factor. The design is exactly invariant: the solved magnitudes agree to 1.0e-15 of each other across all seven and every frequency in the passband, which is the last bits of a double rather than a good approximation. The realisation is not, because two of its quantities do not scale — 50 Ω of amplifier output resistance and 2 pF of stray capacitance to everywhere — and they bind at opposite ends. Inside 0.1 dB the band runs 31.6 Ω to 31.6 kΩ, with the least departure of 0.0133 dB at 1000 Ω; at this setting it is 0.036 dB at 20.0 kHz. The consideration that does not appear on either axis is noise: the resistors' density goes as √R, so across the six decades drawn the noise moves by 1000 times and points at the low end of the band. Filters, measured not tabulated

The same filter a thousand times larger

Multiply every resistance by a thousand and divide every capacitance by a thousand and the response does not change — not approximately, but to a part in ten to the fifteenth, which is the last bits of a double. So a designer has a free parameter that the design says nothing about, and what decides it is the two quantities that refuse to scale: fifty ohms of amplifier output resistance at one end and two picofarads of stray at the other. Between them the realisation survives over three decades of impedance level and nowhere else.

 A pair at Q = 0.7071: the sketch is -3.01 dB out at the corner. computed by solving, not by drawing. The solved magnitude against the two straight lines that stand in for it. At a conjugate pair the error at the corner is 20 log Q = -3.010 dB, which is unbounded in both directions, and the worst error anywhere is 3.010 dB at 1.00× the corner. No damping brings it inside 0.770 dB: that is the minimax, at Q = 0.9152, where the corner error and an interior maximum are equal. Frequency, which is the same solve

The straight lines, and where they are not the curve

Two straight lines through a corner is the most-used approximation in this subject and almost the only one with no number attached. It has one, and it is exact: at a single real pole the sketch is 3.0103 decibels high at the corner and nowhere worse, and its error is the same a factor above the corner as the same factor below — a symmetry the construction does not suggest. A pole pair has no such bound at all, and the best any two-slope sketch can do on one is 0.770 decibels, at a quality factor of 0.9152.

 A threshold crossed once, in 20k samples of noise. With no hysteresis the comparator changes its mind 22.7 times on average and as many as 29, on a signal that crosses the threshold once. The vertical bars are the range over twelve seeds. The expected number of extra transitions falls below a tenth at 3.57 standard deviations — so the hysteresis a threshold needs is set by the noise under it and not by the signal over it. Circuits that do a job, and the range they do it over

Two thresholds because there is a floor

A comparator with one threshold, watching a slow signal cross it once, changes its mind 22.7 times on average and as many as 29 — because there is noise under the signal and no threshold is ever crossed once. Hysteresis fixes it, and how much is needed is a multiple of the noise's own standard deviation rather than a voltage: 3.57 of them here. The multiple grows with how long the threshold is watched, and slowly — a hundredfold longer record needs three times the hysteresis, not a hundred times.

 An exponential with 6× of degeneration: both edges move, and not together. computed by solving, not by drawing at 61 amplitudes. Total harmonic distortion reaches one per cent at 36.6 mV and the gain falls one per cent short of its small-signal value at 85.0 mV. An emitter resistor dividing the gain by 6 moves the distortion edge by 35.4 times — the square of the factor, because the resistor both divides the drive reaching the junction and linearises what the junction does with it — while the gain edge moves by only 11.6 times. So the two edges close up: 2.32 times apart here against 7.06 bare, and a well-degenerated stage stops being limited by its linearity and starts being limited by how accurately its gain is known. Devices, and the amplitude they stop being linear at

What a resistor in the emitter buys

Degeneration is described as a trade: give up gain, get linearity. Measured on the transfer curve, the two sides of that trade are not the same size. Dividing the gain by six moves the amplitude at which distortion reaches one per cent by thirty-five times — the square of the factor — and the amplitude at which the gain is one per cent out by only twelve. So the two edges close up, and a well-degenerated stage stops being limited by its linearity and starts being limited by how accurately its gain is known.

 A part in ten thousand of ratio, and half a degree that costs 18% of a power reading. computed by solving, not by drawing. The two errors of a 1000:1 current transformer into a 10 Ω burden, read off one solve of a netlist driven by a current source. The magnetising inductance puts a high-pass under the measurement with a corner at 0.456 Hz, so the ratio error falls as the frequency rises — and stops falling at 100 parts per million, which is 1 − k and is a coupling rather than a frequency. The phase error falls too and is worth far more: 0.522° at 50 Hz is 0.0142% of the current and 18.2% of the power at a power factor of 0.05. What bounds it from above is not on this plot: the burden voltage is integrated by the core, so the largest primary current is 916 A at 50 Hz and proportionally more at 400. Measurement, which is a circuit on a circuit

The ammeter that is not in the circuit

A shunt measures a current by putting a resistance in the circuit, and every objection to it follows from that. A current transformer puts nothing in the circuit at all — a thousand-turn secondary reflects twelve microhms into the primary — and charges for it in a different currency: no response at direct current, a ratio error that stops falling at one minus the coupling, and a phase error of half a degree at fifty hertz that costs eighteen per cent of a power reading at a power factor of 0.05.

 An order-6 cascade at 10 kΩ: a band 10× wide. computed by solving, not by drawing. A 10.0 kHz unity-gain Butterworth of order 6, 3 Sallen-Key sections in cascade, realised at seven impedance levels three decades apart, with every resistance multiplied and every capacitance divided by the same factor. The design is exactly invariant: the solved magnitudes agree to 4.0e-15 of each other across all seven and every frequency in the passband, which is the last bits of a double rather than a good approximation. The realisation is not, because two of its quantities do not scale — 50 Ω of amplifier output resistance and 2 pF of stray capacitance to everywhere — and they bind at opposite ends. Inside 0.1 dB the band runs 316 Ω to 3.16 kΩ, with the least departure of 0.0763 dB at 1000 Ω; at this setting it is 0.182 dB at 12.9 kHz. The consideration that does not appear on either axis is noise: the resistors' density goes as √R, so across the six decades drawn the noise moves by 1000 times and points at the low end of the band. Filters, measured not tabulated

The band that closes with the order

One Sallen-Key section is inside a tenth of a decibel of its own design over three decades of impedance level, bounded below by fifty ohms of amplifier output resistance and above by two picofarads of stray. Give it three more sections and the band is one decade; give it four and there is no impedance level at all that meets a tenth of a decibel. Every section brings three more nodes each carrying their own stray and one more amplifier carrying its own output resistance, so the floor rises with the order until it crosses the tolerance — a boundary in the order rather than in the impedance.

 Three capacitances, all correct: 2.000 µF, 5.814 µF and 2.105 µF at 5 V. computed by solving, not by drawing. The charge is C∞·v + Q_s·tanh(v/V_k) — a linear backbone and a polarisation that saturates — with both parameters pinned by the capacitance at zero volts and at the rated voltage, so there is no third degree of freedom to tune the answer with. The three curves are three questions. The small-signal value is the slope at the bias, which is what a ripple sees. The charge-average is the total charge moved from zero divided by the voltage, which is what a reservoir or a hold capacitor obeys. What a bridge reads is neither: it is the fundamental of the charge waveform under a one-volt test, which is a measurement condition. At 5 V they are 2.000 µF, 5.814 µF and 2.105 µF — a factor of 2.91 between the extremes, and every one of them is the capacitance. Frequency, which is the same solve

The capacitance that is not one number

A ten-microfarad ceramic at its rated five volts is 2.000 µF as a slope, 5.814 µF as a charge average and 2.105 µF as a bridge reads it — three answers to three different questions, all correct, all called the capacitance. At zero bias the standard one-volt test alone reads 5.2 per cent low. Marched in a circuit the same part distorts as the square of the drive with no bias and in proportion to it with a bias, because the bias is what puts a second harmonic there.

 Below 910 kHz a trace is a diffusion, not a line — and its velocity goes as √f. computed by solving, not by drawing. The phase velocity of an ordinary FR-4 trace against frequency, computed from γ = √((R + jωL)(G + jωC)) with a series resistance that rises as √f above its skin-effect corner and a shunt conductance proportional to frequency. Above 910 kHz the velocity is 0.4767c and does not move, which is the number every other essay in this field uses. Below it the series resistance dominates the reactance, the line is a diffusion, and the velocity falls as the square root of frequency — measured at the 0.467 power. The characteristic impedance is not a constant down there either: 1508 Ω at a kilohertz against 50.0 Ω at ten gigahertz. Lines, where a wire has a length

The delay that is not one number

Nine essays in this field quote a delay: a length divided by a velocity, the same for every frequency, and the edge that comes out is the edge that went in. A real trace has a series resistance, and below the frequency where the reactance overtakes it — 910 kilohertz for ordinary copper — the line is a diffusion rather than a wave, with a velocity proportional to √f. What survives is that the arrival is still exactly linear in the length. What does not is the rise time, which grows as the square of it.

 Nine tenths of the heat is in the switch, and above 1.28 MHz there is no temperature at all. computed by solving, not by drawing. The junction temperature a recovering diode settles at, against switching frequency, with the carrier lifetime rising as the 1.8 power of absolute temperature so that the recovery gets worse as the junction gets hotter. Each event costs 15.43 µJ when cold, of which 90% is dissipated in whatever is pulling the current down rather than in the diode: while the junction is still conducting it holds almost no voltage, and the recovered charge is delivered through the switch at the full 100 V. That energy equals the diode's own conduction loss at 29.2 kHz. The fixed point T = Ta + Rth·P(T) is iterated from the ambient upward; the junction passes its rated 150 °C at 904 kHz, and above 1.28 MHz there is no temperature that satisfies it at all — the loop has gain and the solver reports the refusal rather than the last iterate. Before the steady state

The heat a recovery leaves behind

The essay below this one measured how much current a diode conducts backwards and for how long, and stopped there. Both numbers are multiplied by a voltage somewhere, and the surprise is where: while the junction is still conducting it holds almost nothing, so nine tenths of the energy is dissipated in the transistor pulling the current down and not in the diode. Repeat it a hundred thousand times a second and it is 1.5 watts, the lifetime rises with temperature, and above 1.28 megahertz the diode's own loop has no fixed point at all.

 The inductance divides the current by 7 and leaves the capacitor 29% above the mains peak. computed by solving, not by drawing. The first conduction of a rectifier whose transformer has a leakage inductance as well as a winding resistance, marched from an empty capacitor at the worst instant of the mains. The peak falls from 36.4 A at 20 µH to 5.3 A at 5000, and — unlike the winding resistance measured on the first cycle alone, which limits the current and leaves ∫i²dt exactly where it was — the inductance takes the energy down with it, from 0.370 to 0.073 A²s. What it costs is the second curve: the inductor's current cannot stop at the instant the two voltages are equal, so the capacitor overshoots to 21.86 V at 1000 µH — 28.6% above the 17 V peak of its own supply — and the diodes will not let the charge back out. The overshoot has an interior maximum, because past it the mains reverses before the ring has finished. Power, and the part that does no work

The inductance that limits, and lifts

Adding winding resistance to a rectifier limits the first peak and does not reduce the energy at all — the essay below measured ∫i²dt as two per cent apart over a factor of four in the resistance. Adding leakage inductance does both: it divides the peak by seven and the energy by five, and dissipates nothing to do it. What it buys instead is a rectifier whose output sits 29 per cent above the peak of its own supply, permanently, which every steady-state expression in this field says cannot happen.

 A follower with 1000 pF on it looks like -1182 Ω of negative resistance. computed by solving, not by drawing. The impedance looking into the base of an emitter follower carrying 5.0 mA, with 1000 pF on its emitter. The real part is negative from 1.25 MHz upward and reaches -1182 Ω at 3.40 MHz: the load's reactance multiplied by a complex current gain, with nothing added to the model. A negative resistance is not an oscillator until a reactance cancels, and the base lead supplies it — the total loop reactance passes through zero at a frequency the inductance chooses, and the loop resistance there goes negative above 74.9 nH with 10 Ω of source, which is a few centimetres of wire. A hundred ohms of source raises that to 913 nH: the repair is a resistor in the base, and it works by making the source worse. Devices, and the amplitude they stop being linear at

The input that pushes back

An emitter follower with a capacitor on its emitter has a negative resistance looking into its base — 1182 ohms of it at 3.4 megahertz for a nanofarad, with nothing added to the model. A negative resistance is not an oscillator until a reactance cancels, and the base lead supplies it: with ten ohms of source the loop goes unstable above 74.9 nanohenries, which is seven centimetres of wire. The repair is the opposite of the instinct — a hundred ohms of source raises the threshold to 913 nanohenries, so the fix for a follower that oscillates is to make the thing driving it worse.

 The second path costs nothing at 12 pF and an order at 100 pF. computed by solving, not by drawing. Settling time to 0.01% of final value, marched on the closed loop, against the value of the second feedback path's capacitor — with the phase margin of the same circuit divided by ten drawn on the same axis so the two can be compared. The direct-current error the previous rung recorded as the isolation resistor's cost, 0.99% into 1 kΩ, falls to 1.20e-4% with the second path in. What the second path costs instead is a range: at 12 pF the circuit settles in 0.745 µs against the isolation resistor's 1.419 µs — faster than the thing it repairs — and at 100 pF it takes 9.18 µs, 12 times longer, at a phase margin of 47.9° that reports nothing whatever about it. What it is settling by there is one exponential of time constant 0.99 µs, which is the feedback network's own RC and contains no amplifier. Feedback, and the margin

The path that buys the error back

An isolation resistor restores a capacitively loaded amplifier's phase margin and costs it the thing feedback was for: the loop stops regulating the node the load is on, and a kilohm of load pulls the output down by a per cent. The standard repair is a second feedback path, and its cost is not an error or a margin — it is a range. At twelve picofarads it settles to a hundredth of a per cent in 0.745 microseconds, faster than the circuit it repairs; at a hundred it takes 9.18, and the phase margin there is better.

 A comparator's delay, at a divider ratio of 0.50. Each nanosecond of comparator delay adds 2.650 nanoseconds to the period — measured as a slope between two delays, both exact multiples of the marching step — against the 2.667 that 4/(1+β) gives and the 2 that counting the delay twice gives. A 50 ns comparator therefore holds the frequency to one per cent only below 75.0 kHz. Circuits that do a job, and the range they do it over

The period a delay lengthens

Feed a comparator's output back through a resistor to the capacitor on its own input and the two thresholds stop defending a decision and start setting a period. The closed form is two RC times the log of one plus beta over one minus beta, and a marched circuit recovers it as the step shortens. A comparator that responds fifty nanoseconds late does not add fifty nanoseconds to each half cycle: it adds 4/(1+beta) times the delay to the period, 2.65 here against the 2 that counting it twice gives, because during the delay the capacitor keeps going the way it was going.

 A 0.5 mm conductor's resistance against frequency, exact and asymptotic. computed by solving, not by drawing. The exact ratio is computed from the Kelvin functions by their series; the dashed curve is the asymptote everybody quotes, which treats the current as flowing in one skin depth of the rim and is drawn only where that annulus is inside the wire. At 17.4 kHz, where the skin depth equals the radius and the rule of thumb says the effect "starts", the asymptote says 1.0000 — no effect at all — and the exact answer is already 1.0208. The rule of thumb names a frequency the effect has passed, which is the same shape as the tenth-of-a-wavelength criterion marking a point at which the lumped model is already 30% wrong. Two decades above, the two agree to 0.00%, which is what makes it an asymptote rather than a formula. Two windings, and the band between them

The resistance that grows with frequency

The rule of thumb names the frequency at which the skin depth equals the conductor's radius as the point where the effect begins. Computed exactly from the Kelvin functions, the resistance is already 2.05% up there — and the rule's own asymptote says 1.0000, no effect at all. Two decades higher the two agree to 0.01%, which is what makes it an asymptote rather than a formula, and what makes the frequency it names the wrong one to design at.

 A 8th-order Butterworth at 12 bits, as a cascade and as one polynomial. computed by solving, not by drawing. The open circles are the poles as designed, on the z-plane with the unit circle drawn. Filled marks are where they go once the coefficients are stored in 12 bits. A cascade of biquads keeps two coefficients per pole pair, so a rounding error moves that pair and nothing else: 8.43e-3, largest radius 0.97478. A direct form keeps one denominator whose coefficients are symmetric functions of every pole, so one rounding error moves all of them: 6.21e-1, largest radius 1.54393 — outside the unit circle, which is not an inaccurate filter but an unstable one. The two are the same filter until they are written down. Where a signal becomes a number

The same filter, rounded twice

The filters field measured two realisations of one analogue response and found the passband error growing as the 0.99 power of a component tolerance in a cascade and the 2.00 power in a ladder. The digital version of that argument comes out harder. At eighth order and sixteen bits, a cascade of biquads moves its poles by 6.6 × 10⁻⁴ and stays at a radius of 0.9748; the same filter written as one polynomial moves its poles to a radius of 1.4536, which is not an inaccurate filter but an unstable one.

 Subtracting removes kT/C entirely and doubles the amplifier — worth 31× at a megahertz and a loss above 60 MHz. computed by solving, not by drawing. The noise on one sample of a switched-capacitor stage, and on the difference of two samples taken a settled interval apart, against clock frequency. The reset level is the same number in both samples and cancels exactly; the amplifier's own noise is two independent samples and its variance doubles, measured at 2.000 against the 2 the correlation predicts. At a megahertz that is 63.8 µV down to 11.53 — 31 times in power. The two curves cross at 60.2 MHz, which is where the amplifier's own noise equals kT/C, and above it the subtraction costs more than it removes. The floor, which bounds from below

The sample that is subtracted

Three earlier measurements found floors that no gain moves and no filter reaches, because both arrive as numbers already sampled. One of them can be subtracted: the reset level a capacitor holds is the same number in two consecutive samples and cancels exactly. What it costs is that the amplifier's own noise is not — two samples of it are independent, so its variance doubles. That is thirty times better at a megahertz, a loss above sixty, and the crossing is the one an earlier essay computed for a different question.

 Two of the three are one mechanism at 19.0°; the third arrives at ninety. computed by solving, not by drawing. The cosine between each pair of error waveforms at 30.0 kHz, against how much output the ideal model is asked for. Bandwidth and slewing sit at 0.9454 — 19.0 degrees — and close only slowly, reaching 0.8813 at 16 V. Clipping does not exist below 12.00 V, where its error is 4.0e-9 per cent of the signal and its direction is the direction of rounding; above it the mechanism is real — 18.1 per cent at the top of the sweep — and its cosine against both of the others stays under 0.0025. The bandwidth error is 0.615 per cent at every amplitude here, unchanged to 5.3e-15, because a linear stage's fractional error has no amplitude in it. Where the models stop

Where the mechanisms are one mechanism

An amplifier is said to run out of three separate things — bandwidth, slew rate and rails — and errors from separate mechanisms add in quadrature while errors from one mechanism add as magnitudes. Measured as waveforms rather than as numbers, two of the three sit 18.4349 degrees apart, which is exactly the angle between a sinusoid and its own cube, and the third sits at ninety: its cosine against both of the others stays under 0.0025 wherever it exists. So the arithmetic is neither of the two anybody reaches for, and a budget built the right way is within 2.9 per cent where quadrature is 17 per cent low and a straight sum 37 per cent high.

 What 10 ps of aperture jitter is worth, in bits. computed by solving, not by drawing. Samples are taken at instants displaced by a seeded Gaussian of 10 ps and the error is measured against the same sinusoid sampled exactly. The line is −20 log(2π f × jitter), which the measurement matches to 0.12 dB across three decades. The penalty is exactly twenty decibels a decade of input frequency, because the error is the signal's slope times the timing error and nothing else — so a converter holds 16 bits only up to 199 kHz and 12 bits up to 3.18 MHz. An aperture figure quoted without an input frequency states no resolution at all. Where a signal becomes a number

A picosecond, read as bits

Every other boundary in this collection has a frequency, an amplitude or a size on its axis. This one has a duration. A converter that samples at t + δ instead of t gets a value wrong by the slope times δ, so the damage is proportional to input frequency and to nothing else about the part: ten picoseconds holds sixteen bits up to 199 kHz and twelve bits up to 3.18 MHz, falling at exactly twenty decibels a decade. An aperture figure quoted without an input frequency states no resolution at all.

 A switched capacitor is a resistor below a ratio, not below a frequency. computed by solving, not by drawing. The exact response of a capacitor shuttled between the input and a holding capacitor at 1.00 MHz — a difference equation with one pole, evaluated on the unit circle — against the continuous R–C its equivalent resistance is supposed to make. The corner is 3.15 kHz against the model's 3.18 kHz, 0.99% out, and the discrepancy is set by the capacitor ratio alone: one per cent needs a ratio under 0.0201, which is a clock 315 times the corner. The second difference has no counterpart at all — the sampled response repeats at the clock, so the image rising on the right of this plot is signal at 997 kHz arriving as though it were at the corner. The third is settling: 1 pF charged through 1 kΩ gets 500.0 time constants a half period at this clock, and being short of full charge raises the equivalent resistance by 0.000%, which puts a ceiling on the clock at 108 MHz. Filters, measured not tabulated

A resistor made of a clock

A capacitor shuttled between two nodes at a megahertz behaves as a megohm, and a tenth of a picofarad shuttled at ten kilohertz behaves as a gigohm — which is how a filter with a one-hertz corner fits on a chip. What the equivalence costs is three conditions, and they bind on three different quantities: a capacitor ratio under 0.0201, a signal below half the clock, and a clock below the frequency at which the charge stops arriving.

 A 350 Ω bridge, with ideal leads. The straight line is the expression every textbook gives, Vδ/4; the curve is the solve. They part company at half the fractional change — 0.498% at 1.0% — so the tangent is worth one per cent only up to 2.020%. Driving the bridge from a current source instead halves the departure at every point and moves that edge to 4.040%. Circuits that do a job, and the range they do it over

The bridge that is linear near one point

The expression for a Wheatstone bridge's output is V delta over four, and the solve says it is V delta over two times two plus delta — low by half the fractional change, exactly, at every change tested. So the tangent is worth one per cent only up to a two per cent change and a tenth of a per cent only up to two tenths. Driving the bridge from a current source instead halves the departure at every point and doubles both edges, which is a change of one component and no change at all to the four resistors being measured.

 A ±15% envelope on what a bridge reads permits 29 points of working capacitance. computed by solving, not by drawing. The charge-average capacitance between zero and the rated voltage — the number a reservoir or a hold capacitor obeys — against how the data sheet's stated temperature change is divided between the model's two parameters. Every point honours the envelope exactly: the measured value at zero bias with a one-volt test signal is 15 per cent from nominal at each point on each curve, by construction. The working capacitance is not. At the cold end it is anywhere from -42.9 to -13.9 per cent — 29.0 points of ambiguity at a temperature where the measured value is pinned exactly — and at the hot end from 13.9 to 36.2. Across the whole envelope that is 79.0 points against the 30 the specification bounds, a factor of 2.63. The left-hand end of the upper curve is missing because it is impossible: with the amplitude fixed, no characteristic voltage makes the measured value exceed the zero-bias capacitance. Frequency, which is the same solve

The coefficient that is about one reading

A class II ceramic's temperature coefficient is a third printed number, and it is a coefficient of the one capacitance a data sheet reports: what a bridge sees at zero bias with a one-volt test. The model behind the part has two parameters, one number does not determine two, and every way of dividing a ±15 per cent envelope between them honours the envelope exactly while putting the working capacitance anywhere across twenty-nine points — and two parts a bridge cannot tell apart differ by 1.80 at the voltage they are used at.

 A 10 mH inductor with 8 pF across it, and where ωL stops being its impedance. computed by solving, not by drawing. The dashed line is ωL, which is what an inductor is supposed to be; the solid one is the impedance of the same inductor with 8 pF of winding capacitance across it. They part company at 170 kHz, which is ten per cent, and the impedance peaks at 563 kHz and falls thereafter — above which the component is a capacitor. The ratio between the two is 3.317, and the slider shows it is the same ratio at every capacitance: the shape of the departure belongs to the resonance rather than to either part. This is the capacitor essay with the components exchanged, and it comes out with the same structure and a different number. Two windings, and the band between them

The inductor that is a capacitor

The frequency field's second essay measures where a capacitor stops being one, because its own leads are an inductance. This is the same measurement with the components exchanged, and it comes out with the same structure and a different number: the ten-per-cent departure from ωL sits at f₀/3.317, and it sits at f₀/3.317 at every winding capacitance and every inductance tried. The shape of the departure belongs to the resonance rather than to either part.

 The same three stages, in two orders. computed by solving, not by drawing by Friis's cascade. With the low-noise amplifier first the chain's noise figure is 3.31 dB; with the mixer first it is 12.01 dB. The gain is identical either way — 48.0 dB — so the 8.70 dB is bought with nothing but an ordering. Everything after the first stage contributes 26.0% of the total. The 2.0 dB of cable in front has gain below one, so it multiplies every later stage's contribution rather than dividing it. The floor, which bounds from below

The loss in front, counted twice

A decibel of cable before an amplifier costs a decibel of signal, which everybody expects, and a decibel of noise figure, which is a separate decibel arriving from a separate place. Measured across the slider it is exact: 1.31 dB of chain noise figure becomes 2.31 with one decibel of cable and 9.31 with eight, every time, while the 8.70 dB that stage ordering is worth does not move at all. A lossless reactive network in the same position costs nothing, because only the real part of an impedance is warm.

 What 1.5 mm of length mismatch does to a differential pair. computed by solving, not by drawing. Two lines of the same impedance and different lengths, driven differentially. The solid rising curve is what arrives as common mode; the dashed one beside it is sin(ωΔτ/2), which is what a lossless pair gives and is the same curve until the null. The flat curve at the top is the differential signal, and it is the point: at 3.04 GHz a tenth of the launched amplitude is common mode and the differential has lost 5011 parts per million of itself. The conversion is first order in the skew and the loss is second order, so the error is not missing from the signal — which is why a pair can pass its own eye and fail an emissions test. At 95.3 GHz the closed form has a null and the real pair does not: the longer conductor is also the lossier one, and an amplitude imbalance has no null in it. Lines, where a wire has a length

The millimetre that becomes common mode

A pair carries two modes rather than two signals, and a length mismatch between its halves converts one into the other. A millimetre and a half of skew is ten picoseconds, a tenth of the signal is common mode by three gigahertz, and the differential signal has lost five thousand parts per million of itself getting there — so the error is not missing from the signal, which is why a pair can pass its own eye and fail an emissions test. The product in the answer is ωΔτ, which is what the instruments field's rejection corner is one over.

 A bulk capacitor and a ceramic, and the peak between them at 6.52 MHz. computed by solving, not by drawing. Each capacitor is three elements — its capacitance, its series resistance and its series inductance — and a one-amp source drives the node, so the node voltage is the impedance. Alone, each dips to its own series resistance at its own self-resonance and rises on either side. Together they do not: between the two resonances the bulk part is an inductor and the ceramic is still a capacitor, and an inductance across a capacitance is a parallel resonance. The pair reaches 1.187 Ω at 6.52 MHz, where the bulk alone would give 0.2023 Ω and the ceramic alone 0.2055 — 5.87 times worse than either. The dashed curves are the two parts on their own; the solid one is what the load actually sees. Power, and the part that does no work

The pair that is worse than either

A bulk capacitor and a ceramic are fitted together because each is good where the other is not, and between them is a frequency at which the pair presents six times the impedance either one does alone. The peak is a parallel resonance between one part's inductance and the other's capacitance, its height is one over the series resistance every data sheet asks to be minimised, and at it the two capacitors exchange 5.87 amps for every amp the load draws.

 Matched parts cost nothing; a 2 dB difference between them sets a 112 dB ceiling. computed by solving, not by drawing. The common-mode rejection of a three-amplifier instrumentation amplifier against the gain of its input stage, with each amplifier's own rejection in the netlist as an input-referred error of the common-mode voltage over the rejection. The architecture's own figure rises decibel for decibel with the gain, because the difference stage sees a larger differential signal beside the same common-mode one. The parts' contribution does not rise with anything, and the part of it that matters is not their rejection but the difference between their rejections: two amplifiers of 98 dB that are identical cost 0.000 dB, while 100 dB against 98 dB leaves a ceiling of 111.7 dB with no gain in it. The two mechanisms cross: below a gain of 1903 the four resistors decide everything, and above it more gain buys no more rejection at all — 111.9 dB at a gain of 100000, where the arrangement alone would have been worth 148. The one place the instrument beats its own floor is a gain of 1000, where the two errors cancel; that is a coincidence of signs and not something a design can hold. Measurement, which is a circuit on a circuit

The rejection the parts have

Two essays measured the architecture: four resistors decide an instrumentation amplifier's rejection, two do not, and the answer is the one-amplifier figure plus twenty log of the first stage's gain — exactly, with amplifiers of infinite rejection. Give each amplifier its own and something unobvious happens: two matched but individually mediocre parts cost nothing at all, because their error is a common-mode signal at the difference stage and is rejected there. What costs is the difference between them, and it sets a ceiling with no gain in it.

 Two currents called saturation: one doubles every 4.49 K, the other every 8.98 K. computed by solving, not by drawing. The two current scales of one model junction against temperature, on a logarithmic axis. The saturation current of the exponential law goes as the square of the intrinsic carrier density — a cube of the temperature and the whole band gap in a Boltzmann factor — and doubles every 4.489 K at 300 K. The generation current a reverse-biased junction actually conducts goes as the density itself, with half the band gap, and doubles every 8.978 K: exactly twice as long, at every temperature. The dashed line is "doubles every ten kelvin" drawn through the generation scale, which is the current the rule belongs to. At 300 K this junction's two scales are 10 fA and 2 nA, which are its own parameters and not a property of silicon, and they become equal only at 616.8 K, or 343.6 °C. Where the models stop

Two currents with one name

A junction's saturation current is two currents with one name. The one in the forward law doubles every 4.49 kelvin; the one a reverse-biased junction actually conducts is generated in its depletion region, doubles every 8.98, and on this model junction is 3.06 × 10⁵ times larger at a volt of reverse bias. Doubles every ten kelvin is the second current's rule, and applied to the first it turns the forward drop's −1.81 millivolts per kelvin into +0.39.

 The quietest capacitor is 18× the fastest one, and the margin prefers neither. computed by solving, not by drawing. The total noise at the load of a capacitively loaded stage against its compensation capacitor, with the settling time on the same axis at ten microseconds to the microvolt. Three independent sources are put in the netlist and solved separately — the amplifier's own 4 nV/√Hz at its input, and √(4kTR) in series with each of the two feedback resistors — and added in power. The noise falls monotonically with the capacitor, from 50.7 µV at 1 pF to 12.1 µV at 220 pF. The peak in the noise gain falls with every larger capacitor and is gone entirely from 12 pF upward, where the uncompensated stage's peaks at 3.85 times its own low-frequency value. What the capacitor costs is settling: the fastest is 12 pF at 0.74 µs — the same capacitor that flattens the noise gain, because one handover decides both — and the quietest takes 20.3 µs, at a margin above 40° everywhere in that range. Feedback, and the margin

What the second path costs at the floor

The arrangement that repaired a capacitively loaded amplifier was suspected of paying for itself in noise, because that is how compensations usually pay. It does not: it has no peak in its noise gain at all, and the total at the load falls from 54.9 microvolts to 28.9 as the capacitor is added. What it costs is settling, and the capacitor that is quietest is eighteen times the capacitor that settles fastest — a trade the phase margin says nothing about, because the margin is comfortable at both.

 An exponential with 6× of degeneration: both edges move, and not together. computed by solving, not by drawing at 61 amplitudes. Total harmonic distortion reaches one per cent at 36.6 mV and the gain falls one per cent short of its small-signal value at 85.0 mV. An emitter resistor dividing the gain by 6 moves the distortion edge by 35.4 times — the square of the factor, because the resistor both divides the drive reaching the junction and linearises what the junction does with it — while the gain edge moves by only 11.6 times. So the two edges close up: 2.32 times apart here against 7.06 bare, and a well-degenerated stage stops being limited by its linearity and starts being limited by how accurately its gain is known. Devices, and the amplitude they stop being linear at

Where the two exponents come from

What a resistor in the emitter buys measured two exponents and could explain neither: the distortion edge moves as the square of the degeneration factor and the gain edge as its 0.950 power, and at a factor of exactly three halves the gain error vanished. The degenerated transfer curve has no closed form forwards and an exact one backwards, and reverting that series gives all three. The distortion exponent is exactly two; the gain edge is proportional to D squared over the root of the absolute value of three minus twice D, which is infinite at three halves and tends to a three-halves power; and the two edges are 7.07 apart on the bare device and 1.62 at a factor of eleven, against 7.06 and 1.61 measured.

 Two loops on one heatsink give out at 135 kHz, and it is the switch that goes. computed by solving, not by drawing. The junction temperatures of the diode and the switch against switching frequency, with each device's own thermal resistance to a case they share. Each has a positive temperature loop and they are different loops — the diode's runs through its carrier lifetime and its recovery, the switch's through its on-resistance and its conduction — and the electrical coupling goes one way, since the charge the switch has to take at full supply is the diode's. The pair has no settled temperature above 135 kHz and the component that gives out is the switch, which has no exponential in it and is taking 84 per cent of the heat. The same two devices with the same total thermal resistance and no case in common survive to 485 kHz; the diode on its own to 1.28 MHz. Before the steady state

Two loops, and one heatsink

Nine tenths of a reverse recovery's energy is dissipated in the transistor and not in the diode, and the obvious calculation of the diode's junction temperature puts all of that energy in it anyway. Repaired, the diode alone survives to 1.28 megahertz instead of 128 kilohertz — a factor of exactly the ninety per cent. What replaces the number is the arrangement that exists: two devices with two different positive temperature loops on one piece of aluminium, giving out at 135 kilohertz, and it is the switch that goes.

 Sixteen more digits move the boundary by sixteen decades and leave it exactly where it was. computed by solving, not by drawing. The earlier essay's bridge, walked towards balance and solved twice: once in double precision and once with a pair of doubles carrying about 31 decimal digits, against a closed form that cannot lose any. The 33 per cent error at an imbalance of 10⁻¹⁵ becomes 7.0e-18 — so that loss was the arithmetic's and not the network's, which is what the earlier essay could not say. Each arithmetic's error is its own round-off divided by the imbalance, drawn as the two straight lines, so the second boundary is the first one moved by exactly the extra digits. The condition number is 505 in both cases and at every point, which is the diagnostic being blind twice over. Networks, and how a solve is checked

The digits the arithmetic did not have

Bounding the arithmetic of double-precision circuit solving finds two boundaries that cannot be attributed to it: a bridge with no correct figures left at an imbalance of 10⁻¹⁵, and a filter synthesis that stalls at order 14. An ill-conditioned problem stays ill-conditioned however many digits are used, and a well-conditioned one computed badly gets better — so adding digits is the experiment that tells them apart. The bridge's loss is entirely the arithmetic's. The synthesis's is mostly the data's, and doubling the digits makes it worse.

 At the order a cascade runs out, a ladder still has 4.2 decades of level. computed by solving, not by drawing. How many decades of impedance level each structure can be built at while staying inside 0.1 dB of its own design, against order. Both carry two picofarads of stray at every node. The cascade also carries fifty ohms of amplifier output resistance, a fixed resistance, which binds it from below; its floor rises 8.5× over the four orders drawn, to 0.114 dB at order eight. The ladder's inductors carry a fixed resistance per henry instead — a fixed quality factor, which scales with the design and bounds nothing — so what limits it from below is a few milliohms of track, and at order nine it still has 4.20 decades with a floor of 0.0337 dB, 3.3× its own floor at order three against the cascade's 8.5×. Filters, measured not tabulated

The band that does not close

A cascade of active sections can be built at three decades of impedance level at second order, one at sixth, and none at all at eighth — the band shuts by half a decade per order because two fixed quantities bind it from opposite ends. A doubly terminated ladder has only one of those quantities, because an inductor's loss is a fixed quality factor rather than a fixed resistance and therefore scales with the design. At order nine it still has four decades.

 Alternating-current resistance against foil thickness, 4 layers. computed by solving, not by drawing. The falling dashed curve is the direct-current resistance, which is what more copper buys. The solid curve is the alternating-current resistance at 100 kHz for a portion of 4 layers, and it turns over: past ξ = 0.663 skin depths, thicker foil has MORE resistance, not less. The minimum sits at 1.3368 times the direct-current resistance of the same foil, which is four thirds and is the same number for every layer count above one. The resistance per turn there is 2.016 against √m = 2.000, which is the law the layer count obeys. Two windings, and the band between them

The copper that makes it worse

The rung below measured one conductor pushing its own current to its rim, and there is nothing to optimise in it: thicker wire is always less resistance. Stack the conductors and the quantity changes character. Each layer sits in the field of the ones below it, the loss that field drives has no upper bound in the thickness, and the product turns over — so a portion of four layers has a best foil thickness, and above it more copper is more resistance. The best thickness is the fourth root of three over the square root of the layer count, in skin depths, and the penalty at it is four thirds for every layer count above one.

 95 dB of instrument, 290 Hz corner — and the corner belongs to the source. computed by solving, not by drawing. The common-mode rejection of the same three-amplifier instrument measured earlier measured, with 1 kΩ of imbalance between the two source resistances and 10 pF at each input. The instrument's own curve is drawn beside it. Below 290 Hz the two agree; above it the measurement falls at twenty decibels a decade while the instrument does not, reaching 84.0 dB at a kilohertz against the instrument's 95.0. What converts common mode into differential is the difference of the two input time constants — 10.0 ns here — and once it is differential no rejection repairs it. Measurement, which is a circuit on a circuit

The corner the instrument has no part in

Three earlier essays measured a three-amplifier instrumentation amplifier's rejection at direct current and found 95 dB, of which the resistors' matching decides one part and the amplifiers' own mismatch another. Connect it to a source with a kilohm of imbalance and ten picofarads at each input and the rejection has a corner at 290 Hz and falls twenty decibels a decade after it — reaching 84 dB at a kilohertz on an instrument that is still doing 95. What converts common mode to differential is the difference of two time constants, and the cure is a capacitor on the quiet input.

 The square law is within 1% over a factor of 1.06 in overdrive. computed by solving, not by drawing. One field-effect device drawn against the two models it is between: the square law, which is zero below threshold and rises as the square of the overdrive, and the weak-inversion exponential, which rises at 77.4 mV per decade. The device is neither and approaches both. The two errors point opposite ways — the subthreshold current lifts it above the square law below, and velocity saturation holds it below above — so the square law is exact at 156.5 mV and the shaded band is where it is inside 1%: 152.0 mV to 161.7 mV, a factor of 1.06. The slider is the velocity-saturation voltage, which is the channel length times a critical field, so what it moves is the band's width and not its position. Devices, and the amplitude they stop being linear at

The exponent that is a square

Every device in this collection so far has been an exponential, and the whole of its arithmetic — 59.5 millivolts a decade, a distortion edge at 1.03 millivolts, 3,333 parts per million a kelvin — comes out of that one law. A field-effect device obeys a different one, and the interesting part is that it obeys both: an exponential below threshold and a square above it. The square law is within one per cent over a factor of 1.06 in overdrive at half a micron, and the two errors that bound it point opposite ways.

 A second integrator is 5.8 more decibels an octave, and a third 5.4 more. computed by solving, not by drawing. Modulators of order 1, 2, 3 marched one sample at a time with a one-bit quantiser, their in-band noise read from the transform of the error with the test tone a quarter of the way up the band. The measured slopes are 8.17, 13.93, 19.33 decibels per doubling of the oversampling ratio, against the 9, 15, 21 the white-noise argument predicts and the 3.01 plain oversampling gives. Each order is short of its own prediction by about a decibel, in the same direction, which is the error in the band not being white. The straight lines are each order's prediction drawn through its own first point, so what is compared is a slope against a slope. Where a signal becomes a number

The loop that is worse at full scale

A second integrator in a one-bit loop takes the shaping law from nine decibels an octave to fourteen, and a third to nineteen. What it charges is not in decibels at all: past six tenths of full scale the ratio starts falling, and a fourth integrator's states run away at seven tenths. The best input to a second-order modulator is 0.6 of the reference it is measured against — an amplitude boundary of exactly the kind this collection is built for, on the one object in the field that has no continuous output to draw.

 The frequency at which a pulse train becomes an average. computed by solving, not by drawing. The same 5 watts of average dissipation at every frequency, delivered 2 per cent at a time. The flat line is the steady-state answer, which does not know about the frequency. The falling curve is the marched peak junction temperature, which does. They meet at 308 Hz, and that frequency is not a property of the converter: it is a fraction of one junction time constant per period — f·τ = 0.738 at this duty, with τ = 2.40 ms, and between 0.78 and 0.56 across the duties on the slider. A hundred-kilohertz converter fits 240 periods inside that time constant, and at the top of the sweep — 10.0 kHz — the steady state is already exact to 0.46 per cent, so the averaged-power fixed point is right and this is the measurement that says why. The march puts 48 steps inside each pulse, which is what the answer is sensitive to: at six it put the boundary 19 per cent too high. Before the steady state

The pulse the heatsink does not feel

A thermal resistance iterated to a fixed point with a diode or a switch is a statement about a power — so it assumes that a hundred and fifty watts for two per cent of the time is three watts. The die's own heat capacity decides whether that is true, and it decides it at a frequency: above 308 hertz the junction integrates, by a hundred kilohertz the fixed point is exact to five parts in ten thousand, and at one hertz the same average power on the same heatsink puts the junction three hundred kelvin hotter.

 A fit to the held curve reads the series resistance falling to nothing at 532 K/W. computed by solving, not by drawing. Each point is a three-parameter fit — a constant, an ideality factor and a series resistance — to the held forward curve between 10 and 100 mA, for a junction built with 0.6 Ω and no temperature coefficient on it, mounted at the thermal resistance on the axis. With no thermal resistance the fit returns 0.580 Ω with a residual of 43.9 µV. At 350 K/W it returns 0.184 Ω, an ideality of 1.086 and a residual of 34.0 µV. The resistance it reports reaches zero at 531.7 K/W and is negative beyond. Where the models stop

The resistance a slow curve cannot see

A diode's series resistance is read off the top of its forward curve, and a bench curve is a slow one: each point is held until the junction has warmed to it. Through 350 kelvin per watt the held curve sits 39.7 millivolts below the pulsed one at 100 milliamps, and the three-parameter fit that reads 0.580 ohms from the pulsed curve reads 0.184 from the held one — with a smaller residual. The fitted resistance reaches zero at 531.7 kelvin per watt, and the resistance it hid is what keeps the junction from folding back: with 0.05 ohms instead of 0.6 the held curve turns over at 87.6 milliamps.

 The load sees 10.0 Ω, 1.2e-3 Ω or 1.2e-3 Ω at direct current, and the peak is lowest for the arrangement with both paths. computed by solving, not by drawing. The impedance at the load node of all three arrangements, measured by grounding the input and driving a unit current into the load. Feedback from the amplifier leaves the load looking at the isolation resistor — 10.0 Ω, with no loop gain in it at all. Feedback from the load gives 1.2e-3 Ω, and the two-path arrangement has the same, which is what its direct-current path is for. All three resonate with the load capacitance near 3.2 MHz, and the two-path arrangement's peak is the lowest — 27.6 Ω against 37.4 and 59.2. What it gives up is between: above the 159 kHz handover it has let go of the load node. Feedback, and the margin

What the load sees looking back

Four rungs of this argument have measured what the amplifier does to the signal — the margin, the settling, the error, the noise. None has asked the question from the other end. A load that draws its own current sees an impedance looking back, and with the feedback taken from the amplifier that impedance is the isolation resistor, with no loop gain in it whatever: ten ohms, and a load step leaves an error that never goes away. The two-path arrangement recovers to a thousandth of it and charges for that in a quantity none of the four rungs below measured.

 The charge that comes back: a 0.2% dielectric, 10 s shorted, read at 900 s. computed by solving, not by drawing. The capacitor is charged to 10 V until every relaxation is complete, shorted for 10 seconds, then opened and watched. It climbs back to 20.00 millivolts — 0.2000 per cent of where it was — and the shape is the finding: it is a straight line on a logarithmic time axis, gaining 0.097 per cent of the charging voltage per decade. There is no time constant after which it is over, because there is no single time constant: one branch of the model comes to equilibrium per decade, for as many decades as the dielectric has. A decade before the reading it was at 0.1033 per cent. Before the steady state

The capacitor that remembers

Charge a capacitor, short it for ten seconds, open it, and it climbs back to a fifth of a per cent of where it was. Nothing leaked and nothing was gained: some of the dielectric had not finished discharging. The same defect measured as an admittance says the part is 0.593 per cent more capacitance at a tenth of a millihertz than at a kilohertz, and measured in a sample-and-hold it says a millisecond of hold costs a hundred parts per million — thirteen bits, on a part specified at nothing.

 Every node of a Chebyshev 5, and the one that clips first. computed by solving, not by drawing. The largest signal each node of the realised network ever carries, over the whole frequency sweep, relative to the input. The output reaches 1.000 times the input and F0b reaches 4.537, so on a ±15 V supply the input can be driven to 3.31 V rather than 15.00 before something clips — and the thing that clips is not the output. With the floor at 249.7 nV that is 142.44 dB of range against the 155.57 dB an instrument on the output would report, a difference of 13.14 dB that no measurement at the output can see. The floor, which bounds from below

The ceiling is not at the output

A fifth-order Chebyshev's noisiest node is its output and its largest signal is not. F0b carries 4.537 times the input, so on ±15 V the range is 142.44 decibels and not the 155.57 an instrument on the output reports. Across fifteen realised filters the output reading spans 3.97 dB and the range they actually have spans 22.34 — and one family of the three has no internal peaking at all, at any order.

 Where an amplifier's reading comes from, against the source it is reading. computed by solving, not by drawing. Three errors with three different dependences on the source, each measured by a solve with the other two set to zero. The offset voltage is flat — 50 microvolts wherever the source is. The bias current times the imbalance is linear in the source and is what balancing removes. The offset current times the source is linear too and is what balancing leaves. Unbalanced, the current overtakes the voltage at 1.77 kΩ; balanced, at 10.0 kΩ, which is the offset voltage divided by the OFFSET current and is the ratio of the two currents further along. Below about a kilohm, balancing makes the reading worse — the feedback network is already the larger resistance, and equalising means adding to the source. Measurement, which is a circuit on a circuit

The current the instrument draws

Every amplifier in this collection has had inputs that take no current, and that is not an idealisation of a small quantity — it is an idealisation of one whose size is decided by something outside the part. Fifty nanoamps is nothing until it flows in a megohm, and then it is fifty millivolts. The classical cure balances the two resistances and removes the bias current, leaving the offset current: worth a factor of ten, not a thousand, and it costs forty per cent of the noise density to get.

 The floor rises as the 1.06 power of the order, and the arithmetic gives out first. computed by solving, not by drawing. The least departure a doubly terminated ladder can achieve at any impedance level, against order — the floor, which is what is left when the tolerance the band is drawn at is taken away. It rises from 0.0087 dB at order three to 0.0389 dB at order 13, as the 1.06 power of the order, so the order at which it reaches the 0.1 dB the band is drawn at is near 32. Drawn over it are the same networks with one imperfection removed at a time, and they do not add: the winding loss alone leaves a larger departure than the complete realisation does, because the track resistance is in series with the load and lifts the passband exactly where the winding loss droops it. The stray capacitance contributes nothing at all, since the level that minimises the departure is a few ohms. What ends the sweep is neither: the continued-fraction synthesis loses its leading coefficients to cancellation and stalls at order 9 for a Butterworth and 14 for a Chebyshev. Filters, measured not tabulated

The floor that outlives the arithmetic

A band of impedance levels is what a tolerance allows; a floor is what a structure has. Measured at six orders, a doubly terminated ladder's floor rises as the 1.06 power of the order and would not reach a tenth of a decibel until order thirty-two — an order nobody builds. What stops the sweep is neither the structure nor the parasitics: the continued fraction that turns a reflection polynomial into element values loses its leading coefficients to cancellation and stalls at order nine for a Butterworth and fourteen for a Chebyshev.

 One part, three saturation currents: 1.90 A, 2.40 A, 2.69 A. computed by solving, not by drawing. B(H) is μ₀H plus a saturating magnetisation, written as a flux linkage, and the inductance drawn here is dλ/di — the slope of that flux, which is what a small signal on a direct current actually meets. It is 18.92 µH at no current, 16.73 µH at two amps and 10.43 µH at three, and it never reaches zero: the vacuum is still there, so the part falls to its air-core 0.05 µH and stays. The three marks are the ten, twenty and thirty per cent drops different manufacturers print as the saturation current — 1.896, 2.395, 2.694 amps, a spread of 42 per cent on one part. Two windings, and the band between them

The inductance the current decides

The rung below bounded the flux and the boundary is exact: the volt-seconds decide the flux swing whatever the material does, and a cycle whose current ripple runs from 519 mA to 75 A has the same flux excursion to better than two per cent. What saturation breaks is the relationship between that flux and the current — so a ripple the design expression puts at 514 mA is 1,022 mA at 2.56 A of load, the peak reaches 3.57 A where the part is at a fifth of its nameplate inductance, and the same part has three saturation currents depending on which per cent it was quoted at.

 At 300 K one junction holds a logarithm to ±1% over 2.4 decades as a diode and 8.5 at its collector. computed by solving, not by drawing. The voltage of one model junction against the logarithm of the current it carries, as a percentage error of that current from a straight line fitted over the widest range that stays within ±1%. Taken as a diode — both mechanisms and 0.6 Ω of series resistance — the range is 2.40 decades, from 50.1 nA to 12.6 µA, and its slope is an ideality of 1.982. Taken at the collector, where the recombination current is supplied from the base and 1.604 Ω remains, it is 8.50 decades, from the axis's own end at 1 pA to 316 µA, at an ideality of 1.0001. Nothing arrives beside the collector current, so its lower end on this axis is the axis. Where the models stop

The logarithm is in the collector

A diode is the textbook logarithm, and a real junction holds one to within one per cent over only 2.40 decades — from 50 nanoamps to 12.6 microamps, at an ideality of 1.98 — because two mechanisms and a series resistance share its terminals. The same junction read at a transistor's collector, with its recombination current supplied from the base, holds 8.50 decades at an ideality of 1.0001. How far the logarithm reaches is decided by which terminal the current is taken from, and at the bottom of the range by a leakage current a millivolt is enough to switch on.

 A pair's third-order intercept is 7.8 dB above anything it can produce. computed by solving, not by drawing. Two equal tones through a differential pair, transformed coherently so every product lands in a bin of its own. The fundamental rises with slope 1.000 and the third-order product with slope 3.000, both fitted over the decade marked, and the dashed extensions are the extrapolation a specification quotes. They meet at a drive of 4.00 thermal voltages and an output of 2.00 — against a largest output of 0.8108, which is 8/π² and is what two equal tones give through a limiter. The intercept is 7.84 dB above it, which is π²/4 exactly. Devices, and the amplitude they stop being linear at

The point the device is never at

A device's linearity is specified by one number, and that number is a place on no curve. The third-order intercept is where two straight lines would cross if both went on being straight, and neither does. For a differential pair the crossing sits π²/4 — 7.84 decibels — above the largest output the device can produce at any drive whatever, and the arithmetic that says so contains no tail current and no temperature.

 An inrush limiter's steady state, and how little of it is still a limiter. computed by solving, not by drawing. A negative-temperature-coefficient thermistor in series with a supply, at 1 ampere of load current. The falling curve is what it dissipates at a temperature — I²R with R following the two-point β fit a catalogue prints — and the rising line is what its mounting removes. They cross once, at 83.1 degrees, and the loop gain there is -1.374: negative, so the part is stable at every current rather than below a boundary. What is left of its cold 10 ohms at that temperature is 1.937 — 19.4 per cent. The slider moves the load current, and more current leaves less resistance. Power, and the part that does no work

The protection that is gone by the second time

An inrush thermistor is ten ohms cold and holds the first cycle down; then the load current warms it and it settles at 83 degrees and 1.94 ohms, which is 19 per cent of what was bought. That is the design working. It is also a part that takes 198 seconds to recover half its cold resistance, against a reservoir capacitor that empties in tens of milliseconds — so a mains dip in that window hands the rectifier an unlimited inrush into an empty capacitor, which is the exact event the part is on the bill of materials for.

 The growth per cycle, and the form that is a fifth low at the top of the range. computed by solving, not by drawing. The factor the envelope is multiplied by each cycle, against the gain. The solid curve is exp(π(k−3)/√(1 − ((k−3)/2)²)), which is what the characteristic equation gives and what the netlist's own poles return to twelve digits; the dashed one is exp(π(k−3)), which drops the denominator. The circles are the marched envelope, fitted over the cycles that are still small — 80 of them at k = 3.01 and 4 at k = 3.2, and none at all above that. The two expressions differ by 3.9e-7 at k = 3.01 and by 20.462% at k = 3.8, so the approximation fails exactly where nothing is left to check it against. Circuits that do a job, and the range they do it over

Two exponentials, and where they meet

An oscillator's envelope grows by exp(π(k−3)/√(1 − ((k−3)/2)²)) a cycle, and the form usually quoted drops the denominator — exact to four parts in ten million at a hundredth above three, and 20.462 per cent low at 3.8, which is precisely where too few small cycles are left to measure it. Where the growth stops is the diode's own exponential: 108.5 millivolts of amplitude for every decade of saturation current, proportional to the ideality to four parts in a thousand. Above 60.121 nanoamperes the limiter is already conducting at zero signal and there is no oscillation at all.

 What a cascaded modulator is worth, against how well its two paths match. computed by solving, not by drawing. Two first-order loops marched sample by sample at an oversampling ratio of 64, with the first stage's quantisation error taken as the difference between what its comparator said and what was presented to it — nothing here reads a state a real converter could not. Perfectly matched, the cascade gives 72.5 decibels against the first stage's own 47.5: second-order shaping out of two first-order loops, neither of which can be unstable. The gain with which the first error reaches the second stage is then given an error, and the flat left-hand half of the curve is the arrangement working. It costs three decibels at 5.29 per cent, which is a capacitor ratio — achievable, and not free, and not something the digital side can measure or correct. Where a signal becomes a number

Two loops, and the mismatch between them

Single loops of second, third and fourth order, marched alone, each stop working above an amplitude that falls with the order — which is why nobody builds a fourth-order single loop. The standard answer is two first-order loops with the first one's error fed to the second and differentiated back out, giving second-order shaping out of parts that cannot be unstable. The cancellation is between an analogue path and a digital one, and it is worth 25 decibels until the two differ by five per cent.

 A coil of 500 nF and a capacitor of 100 µH give the loop three features, not one. computed by solving, not by drawing. The current round the loop for a volt across it, with the coil carrying 500 nanofarads of its own capacitance and the capacitor 100 microhenries of its own inductance. Three features rather than one: the resonance the two nameplate values set, at 4.02 kHz; the coil's own self-resonance at 7.12 kHz, which is a parallel tank and so a null in a series loop, 5.67e+3 times below the peak beside it; and a second series resonance at 28.2 kHz that belongs to neither component. Above the null the coil is a capacitance, that capacitance is in series with the tuning capacitor, and the capacitor's own inductance resonates with the pair — which is why removing either parasitic removes this peak and neither alone can produce it. Frequency, which is the same solve

Two parasitics, and the resonance neither of them has

A resonator's quality factor is supposed to sit below the worst of its components, because reciprocals add. Give the capacitor half a millihenry of its own inductance and the loop measures 92.21 against a coil that allows 64.55 and a reciprocal sum that predicts 62.47 — the ceiling passed exactly where ESL/L crosses ESR/DCR, at 5.00 per cent. Add the coil's own capacitance beside it and the loop grows a second resonance at 28.2 kHz that neither part has alone, taller than the first by 40.4, and the half-power level is then crossed four times.

 A photodiode's own capacitance sets the bandwidth, as its -0.50 power. computed by solving, not by drawing. The bandwidth of a 1.0 MΩ transimpedance stage against the capacitance of the diode driving it, with the feedback capacitor at each point bisected to give exactly forty-five degrees of phase margin on the solved loop. The classical expression √(GBW/2π·rf·cd) is drawn over it: the right shape, and conservative by about a fifth at every capacitance. The bandwidth falls as the -0.497 power of the capacitance — a square-root law, so a diode of four times the area costs half the bandwidth rather than three quarters of it. At 30 pF the compensation is 0.598 pF against the expression's 0.725, and the bandwidth 295 kHz against 220. Feedback, and the margin

Where the trouble is at the input

Every other arrangement in this field has its difficulty at the output — a capacitive load, an isolation resistor, a load that draws current. A photodiode amplifier has it at the input, and the capacitance causing it is not a parasitic: it is the diode's junction, which is the price of its area, and area is what a photodiode is bought for. The feedback resistor's own noise is 127 nV/√Hz against the amplifier's 4, and the amplifier is still ninety-five per cent of the noise at a large diode.

 Same 2 ps of jitter: a floor at -121 dB, or a line at -91. computed by solving, not by drawing. A 3.337 MHz sinusoid sampled at 10 MHz with two clocks of identical root-mean-square jitter. One clock's timing error is independent gaussians and the other's is a sinusoid at 130 kHz. The total error power is the same — -87.43 and -87.55 decibels below the signal, against the closed form's -87.55 — and the pictures are not. The random clock spreads it over 2048 bins, -120.5 dB each; the modulated one puts it into two lines at 3.337 ± 0.130 MHz, -91.5 dB down. A specification in picoseconds does not choose between them. Where a signal becomes a number

A floor, or a line

Two clocks with identical two-picosecond jitter sample the same sinusoid, and the total error power is the same to a tenth of a decibel — the closed form an earlier essay computed, right for both. One of them puts that error across two thousand bins at −120 dB each; the other puts it into two lines at −91.5. The gap is the processing gain, it grows with the record length because a density falls and a line does not, and nothing in a specification quoted in picoseconds root-mean-square distinguishes the two cases.

 Terminated at both ends: no interval anywhere, at 1.65 V of 3.3. computed by solving, not by drawing. The same net as the three-way comparison, with a fourth trace: a series resistor at the driver AND a parallel one at the receiver. Both reflection coefficients are zero, so the wave that arrives at a receiver 50% of the way along is already the final value and there is no second arrival to wait for — the departure after the first edge is 0.0e+0 per cent, which is the arithmetic's floor. What it costs is the level: 1.650 V of 3.3, exactly half, because two equal resistances divide the supply and nothing reflects to double it back. The series scheme in the same place is undefined for 1.00 delays and the unterminated one overshoots by 82 per cent. The hold current is 33.0 mA against the parallel scheme's 60, because the path to ground now has two resistances in it. Lines, where a wire has a length

Terminated at both ends

A series resistor at the driver and a parallel one at the receiver cost exactly half the swing — 1.650 volts of 3.300 — and no reflection ever gives it back, because there is no reflection. What the half buys is measured rather than assumed: a driver thirty per cent off its assumed impedance rings a series-terminated net by 16.3 per cent and a doubly terminated one by nothing at all, and an interior receiver on a series-terminated net sits in the undefined band for every far-end resistance above 125 ohms, which is Z₀/(1−2b) and contains no length, no driver and no frequency.

 A major hysteresis loop at 9.0 A/m of coercivity, and the anhysteretic curve it closes onto. computed by solving, not by drawing. The B–H loop of a core driven sinusoidally to ±400 A/m, marched through a superposition of twenty-four play operators and drawn over the single-valued magnetisation curve with no hysteresis in it. The loop encloses 12.481 joules per cubic metre per cycle, which is the core loss and which no single-valued model can produce, because a curve has no area. The coercivity is 8.96 amperes per metre and the remanence 22.3 millitesla; both are read off the marched descending branch rather than handed in. The slider takes the threshold spread down to zero, where the two branches become one, the area falls to 9.8e-15 J/m³, and the object is exactly the core the field already had. Two windings, and the band between them

The area a curve cannot have

Every magnetic model in this collection is a single-valued B(H), and a single-valued B(H) cannot dissipate: ∮H dB around a curve is zero, so the transformers and inductors here have all run cold. Giving the material a second branch costs one object — a play operator, which lags the field by a threshold and is otherwise nothing — and a superposition of twenty-four of them produces a loop with 12.481 joules per cubic metre in it, a coercivity of 8.958 amperes per metre and a remanence of 22.3 millitesla, none of which was handed in. At zero threshold the whole thing collapses onto the curve the field already had, to fifteen figures.

 A clean board is 14.0 mV of error, and a guard makes it 1.0 nV. computed by solving, not by drawing. Two boards differing by one wire, solved with the leakage present and again with it removed, so the number plotted is the leakage's own contribution and nothing else. Unguarded, a teraohm across the laminate from a 15 V rail through a 1 GΩ source is 13.99 millivolts — and a humid morning takes that resistance down two decades, which is the left-hand end of this axis. Guarded, the ring is held at the input's own potential by the amplifier, so what is across the leakage is the amplifier's own error and the result is 1.00 nanovolts. The guarded line is flat in the rail and proportional to the signal: the offset has become a gain error of 1.00 parts per billion, and the reading is low by it rather than high: the ring sits a little below the input, so the last of the leakage pulls the input down. Measurement, which is a circuit on a circuit

The current that does not reach the input

A teraohm across a board from a fifteen-volt rail is fourteen millivolts of error through a gigohm source, and a humid morning takes that resistance down two decades. A ring held at the input's own potential leaves a nanovolt — proportional to the signal rather than to the rail, so an offset has become a gain error of one part in a billion — and the same wire multiplies the input resistance by the loop gain, which makes it 10¹⁸ Ω at direct current and 10¹² Ω at a megahertz.

 The temperature through a 20 mm core that makes its own heat. computed by solving, not by drawing. Conduction with volumetric generation, solved on forty-one cells with a surface film at each face and the loss density evaluated at each cell's own temperature — so the middle of the core makes less heat than its faces do, which a closed form for uniform generation cannot express. The peak is 115.05 °C and the surface 112.45: a gradient of 2.59 kelvin, which is 2.9 per cent of the 90.0 kelvin rise. That share is Bi/(Bi + 2) — 3.0 per cent at a Biot number of 0.063 — so it is decided by how well the surface is cooled and not by how much heat is made. Power, and the part that does no work

The degrees a thermocouple cannot see

Every thermal answer in this collection has been one temperature, and a core makes its heat in its volume and loses it from a surface, so it has two. Solved as a conduction problem, a twenty-millimetre core in still air is 2.59 kelvin hotter in the middle than on the outside — 2.9 per cent of a ninety-kelvin rise, which is why the lumped answer has been good enough. Cool the same core on a plate and the gradient does not shrink; it grows to 3.37 kelvin and becomes 78 per cent of what is left.

 A one-henry inductor with nothing magnetic in it, good for 3.6 decades. computed by solving, not by drawing. The impedance at the input of an Antoniou impedance converter, read as an inductance: a current source drives the node and the voltage is solved for, and the imaginary part divided by ω is what is plotted. Five components — four resistors of 10 kΩ and a 10 nF capacitor — behave as 1000 mH, which as a wound coil would be several henries of wire. It is that inductance to within one per cent from 1.00 Hz to 3.65 kHz, 3.56 decades, and the upper edge belongs to the amplifiers rather than to the arrangement: with ideal ones in the same netlist the inductance is exact everywhere drawn. Nothing in it stores energy in a magnetic field — the current lags because an amplifier is holding a capacitor's voltage somewhere else in the loop. Filters, measured not tabulated

The inductor that is an amplifier

Four resistors and a capacitor, arranged around two amplifiers, present one henry at a node — an inductance with nothing magnetic in it, which as a wound coil would be several henries of wire. It is that inductance to within one per cent over three and a half decades, and its series resistance goes negative at 63 hertz, which is well inside the band where it is still an excellent inductor. A resonator built around it there does not have a high quality factor; it has a negative loss, and starts on its own noise.

 A tenth of a per cent of lead is 999 µε of strain that is not there. computed by solving, not by drawing. A 350 Ω quarter bridge at zero load, against the resistance in each lead. With both leads in the changing arm the output is 4.99500 mV at 350 mΩ — an apparent fractional change of 0.1998%, which at a gauge factor of 2 is 999 microstrain. With one lead in that arm and one in the arm beside it the output is zero to the last bit, at every lead resistance drawn. The lower panel is what the second arrangement costs: the sensitivity falls as 1/(1 + Rlead/R), which is 0.100% at the same lead and is a calibration constant rather than a drift. Circuits that do a job, and the range they do it over

The leads that are in the bridge

A strain gauge on the end of two long wires cannot be told from a strain gauge under load: both leads in the changing arm is bit for bit the same netlist as a quarter bridge whose fractional change is larger by twice the lead over the gauge, which at 350 milliohms on a 350 ohm gauge is 999 microstrain that is not there. Moving one of those leads into the arm beside it leaves the output at exactly zero for every lead resistance drawn, and turns a twenty-kelvin drift of 76.8 microstrain into 0.077. That factor is two over the strain being read — 999 at a thousand microstrain and 9990 at two hundred — and it costs 0.100 per cent of sensitivity.

 40.0 dB of gain change and 180.0 degree-decades of phase, peaking at 109.8°. computed by solving, not by drawing. The gain and the phase of two leads of 10:1, together, each section buffered from the next. The gain changes by 40.000 dB from one end of the sweep to the other. The area under the phase, integrated on the solve over seven decades beyond the outermost corner, is 180.000 degree-decades, against 180.000 for ninety per twenty decibels of gain change. The peak is 109.806° at 3.16 kHz, below the 131.93° no arrangement of real leads sharing that rise can pass. Frequency, which is the same solve

The phase a decibel buys

The area under a minimum-phase network's phase, counted in degrees across decades of frequency, is fixed by how far its gain moves from one end of the spectrum to the other: ninety degree-decades for every twenty decibels, however the network is arranged. One 100:1 lead peaks at 78.579 degrees and two 10:1 leads stacked together at 109.806, and both enclose 180.000. The peak is a design decision and the area is a bill — and a network that gives its gain back gives its phase back with it.

 The noise bandwidth of four families at 8 orders, and the one that has none. computed by solving, not by drawing. Every one of the 32 entries is an integral of the realised network's own squared magnitude, divided by that network's own measured −3 dB point. At order one the four families are the same filter and return 1.5706, which is π/2 — the calibration the rest of the table is quoted against. Only Butterworth then does what the ratio is usually said to do: it falls at every order, to 1.0065 at 8. Bessel is least at order 5 (1.0385) and rises to 1.0441; Chebyshev alternates with parity, 0.9637 at five against 1.0686 at six; and an even-order elliptic has no noise bandwidth at all, because its stopband comes back up to a constant — its magnitude at the top of the range moves by 0.00 decades per decade of frequency, so the integral grows with whatever limit it is stopped at. The floor, which bounds from below

The ratio that does not walk to one

A single pole passes π/2 times as much noise as a brick wall at its corner, and every account of it says the ratio falls towards one as the skirt steepens. Over thirty-two realised filters only Butterworth does that. Bessel is least at order five, 1.0385, and rises again; Chebyshev alternates with parity and the two branches separate only above 0.1968 decibels of ripple; and an even-order elliptic has no noise bandwidth at all, its integral returning 380 or 38,005 depending on where it was stopped.

 The step at which the output impedance stops being a number. computed by solving, not by drawing. The excursion divided by the step, against the step. The flat line is the linear model, and it is flat to 0.0 parts per million across four decades — which is what an impedance is. The rising curve is the same netlist with the differential pair's tanh in the transconductor, and it leaves at 10.6 mA: the input error there is 3.63 thermal voltages, so the boundary is an amplitude in the pair's own units rather than a current with the amplifier's name on it. At 300 mA the ratio is 54.5 Ω against the linear 23.6 — 131 per cent, and it is no longer a property of the circuit at all. The slew rate that decides it is 3.25 V/µs, which is twice the thermal voltage times the gain-bandwidth in radians, and contains no design choice. Feedback, and the margin

The step too large to have an impedance

The rung below drove the load node with a current step and reported an impedance: a voltage divided by a current, which is a number only if the ratio does not depend on the current. Give the amplifier the differential pair's own tanh in place of a linear transconductor and it is a number up to 10.6 milliamps and not above — where the input error is 3.63 thermal voltages, and where the slew rate that decides it is twice the thermal voltage times the gain-bandwidth in radians, containing no design choice at all.

 A hold capacitor's band closes at 6.43 MHz, where a resistor's does. computed by solving, not by drawing. The switch of 0.5 Ω closed, 100 MΩ open and 5 pF across it, driving a capacitor. The lower edge is the smallest capacitance onto which the open switch feeds through no more than 1%: 495 pF far above 318 Hz, rising as the reciprocal of frequency below it because the leakage charges the capacitor. The upper edge is the largest capacitance the closed switch tracks to 1%, counted as a vector. The two meet at 6.43 MHz; a resistor on the same switch closes at 6.43 MHz, and counted as a magnitude the capacitor's band closes at 91.6 MHz. At the closure the closed switch's phase is 0.57 degrees. Where the models stop

The width no load can change

A switch's band was drawn against a load resistor and closed at 6.43 megahertz. Put a hold capacitor where the resistor was and the band changes axis and shape — a diagonal below 318 hertz, a floor of 495 picofarads above it — and stays exactly as wide: 1.8083 decades at 100 kilohertz for both loads, closing at 6.43 megahertz for both. Count the capacitor's error as a magnitude, as the resistor's always was, and the band appears to stay open to 91.6 megahertz. That extra room is 8.11 degrees of lag the count cannot see.

 The fourth-order term explains 95% of what the third leaves out. computed by solving, not by drawing. Two closed forms for the amplitude at which a degenerated stage reaches 1% of second harmonic, each against the same measurement — a bisection on the harmonic content of a Newton-solved curve, which shares no arithmetic with either. The leading expression is 4Vₜ·t·D², derived by expanding the transfer curve and exact in the limit of small drive; its error grows as the 1.96 power of the drive it is evaluated at. At a degeneration of eleven that drive is 4.53 thermal voltages and the expression is 6.89% optimistic. Carrying the reversion one order further takes it to 0.362%. Devices, and the amplitude they stop being linear at

What the fourth order says about the third

A closed form derived by neglecting the fourth-order term is a claim with an error, and the error is a quantity the fourth-order term can be asked about. Carried one order further, the degenerated exponential's reversion predicts 7.33 per cent where the leading expression was measured to be 6.89 per cent optimistic — and takes the residue to 0.362 per cent. The correction has its own edge, in the same quantity, and it is measured too.

 Three cliffs, not one, and the fastest damping is on the last of them. computed by solving, not by drawing. Settling time against damping for a third-order response — a complex pair at unit natural frequency and a real pole at 3 — with the second-order case behind it. Both are staircases: the settling time is set by the last excursion outside the band, so there is one step for each excursion that stops happening, and there are 3 of them between 0.3 and 0.98. They are at 0.378, 0.522, 0.773, with jumps of 1.24, 1.30, 1.42. The earlier essay found the last and largest of them and did not look below it. The fastest damping is 0.775, sitting on the edge of the last step, and a design a hundredth to the left of it settles 42 per cent slower. Before the steady state

Three cliffs, and where they are

Sweep a second-order step's damping and the settling time falls by a third in one step of a five-thousandth sweep, which is usually called the cliff. There are three of them between 0.3 and 0.98, one for each excursion that stops leaving the band, and adding a third pole moves all three left and makes all three shallower — so the classic 0.78 for fastest two per cent settling is a second-order number, and at a third pole one and a half times the natural frequency the answer is 0.745 and 0.78 is on the wrong side of the step.

 The null is a V and not a bowl: one per cent of ratio error is 5.0e-3 of the near end. computed by solving, not by drawing. The same twelve coupled sections read at 10.0 MHz, with the ratio of the two couplings swept across the null rather than sat at one setting. The far end divided by the near end is |1 − r|/(1 + r) at every point — a straight-sided V through zero, first order in the departure with a coefficient of one half, and not a rounded minimum with a flat bottom. So there is no tolerance band: a ratio one per cent off gives 4.98e-3 of the near end and ten per cent off gives 4.76e-2, and the exchange rate between them is fixed. The upper trace is the near end over the same sweep, which moves by 11 per cent while the lower one moves through 14 decades — the two ends are the same coupling read as a sum and as a difference, which is why one of them has a zero in it and the other cannot. Lines, where a wire has a length

How wide a null is

A far end at 3.5×10⁻¹⁹ of the drive is a statement about arithmetic until somebody asks how far the two couplings may differ before it comes back. The answer has no flat bottom in it: the far end divided by the near end is |1−r|/(1+r) exactly, so the null is a V and a ratio one per cent off returns 4.98×10⁻³ of the near end. On the axis a board is built to that is a difference of 0.0081 between the two modes' effective permittivities, out of 3.99 — two parts in a thousand, and 0.675 picoseconds of mode skew over a hundred millimetres.

 One component of a gyrator moves the response by 0.50; the passive ladder's inductor moves it by 3e-8. computed by solving, not by drawing. The magnitude sensitivity at the passband maximum of a gyrator ladder, against the resistance scale the gyrators are built at, for a single component and for the combination of both halves that actually changes the synthesised inductance. The passive ladder realising the identical response is at 3.2e-8. A single component sits at about a half whatever the resistance scale is, and the combination falls as one over it — the 0.97 power over 3 decades. So the stationarity has not been destroyed, it has become a statement about a combination of components rather than about a component, and independent parts do not come in combinations. Filters, measured not tabulated

One inductor, and ten components

The rung below built an inductor out of an amplifier and ended with a boundary that is not a frequency: one end of it is soldered to ground. A ladder's series inductors are floating, so making one takes four amplifiers rather than two — and the four-amplifier version is exactly floating, to five figures, and reproduces the ladder's response to a hundredth of a decibel. It does not reproduce its stationarity. Each of a gyrator's ten components moves the response by exactly a half, where the inductor it replaced moved it by ten to the minus eight.

 Two curves that only rise, and the gap between them that has a minimum. computed by solving, not by drawing. The source's own Johnson density, √(4kTR), and the amplifier's total input-referred density, √(4kTR + eₙ² + (iₙR)²), for a part with 4 nV/√Hz and 0.6 pA/√Hz. Neither curve has a minimum: the total is 4 nV/√Hz at a source of nothing, is 4.196 at 100 Ω, and rises without limit. What has a minimum is the ratio, at 6.67 kΩ, where the noise figure is 1.138 dB and the total density is 11.780 nV/√Hz — 2.81 times noisier in volts than at 100 Ω, where the noise figure reads 10.41 dB. The two statements are about different questions and the figure is what stops them being confused. The floor, which bounds from below

The bowl, and the bottom of it

An amplifier's noise figure has a minimum against source resistance and its input-referred noise has none: the 4 nV/√Hz part reads 1.138 dB into 6.67 kΩ and 10.41 dB into 100 Ω, and is 2.81 times noisier in volts at the first. The bowl is one shape scaled by its own depth, so the quieter the part the flatter it is — ±30.1 times for a decibel on the best of four, ±2.16 on the worst — and three parts of equal eₙiₙ share a floor of 0.3138 dB at optima 16 times apart.

 Which limit binds is a property of the load, and they change places near 22 nF. computed by solving, not by drawing. Each limit measured on its own, as the departure of its march from the linear one, at a load step of half the output stage's rating. The input pair's departure falls with load capacitance — a bigger reservoir holds the node while the loop responds, which is the sixth rung's own result — and the output stage's does not fall nearly as fast, because what it has to supply is the charge the capacitor wants. Below about 22 nanofarads the thermal voltage decides the answer and above it the output stage does, and nothing about the amplifier changed. Feedback, and the margin

The current above which there is no impedance

The sixth rung found the impedance leaving at 10.6 mA, where the input pair's own tanh takes over and the slew rate is twice the thermal voltage times the gain-bandwidth in radians, with no design choice in it. A real output stage has a second limit that is nothing but design choice, and the two do not bind at the same load: at 0.47 nF the input pair's departure is 19.4 per cent against the output stage's 4.2, at 22 nF it is 0.9 against 2.3, and above the output stage's rating the excursion does not come back at all — 2,254 Ω for a quantity that was 37.

 The crest factor is 13.4 and the winding is sized by 3.01. computed by solving, not by drawing. Three ratios of the same settled march, against the reservoir. The crest factor — the peak diode current over the load's direct current — runs 6.43 to 24.25. The form factor, which is the root-mean-square current over the same direct current and is what a winding heats by, runs 2.114 to 4.077. Its square is the copper loss against a winding carrying the direct current alone, and that runs 4.47 to 16.62. The first ratio is 3.04 times the second at 220 µF and 5.95 times at 4700, so quoting one of them tells a reader nothing about the other. Circuits that do a job, and the range they do it over

The current that sizes the transformer

A reservoir's crest factor is 13.374 at a thousand microfarads and the winding is not sized by it. The root-mean-square of the same marched current is 3.0069 times the load's direct current, so the copper dissipates 9.0417 times what it would carrying the direct current alone — and the two ratios diverge, from 3.04 apart at 220 microfarads to 5.95 apart at 4700. The expression for the mean output is wrong in three places whose signs differ, and at 313.9 microfarads they cancel to six microvolts while the ripple expression inside it is still 32.3 per cent high.

 The Steinmetz exponent is a local slope, and how far it moves is a property of the material. computed by solving, not by drawing. Loss per cycle against peak flux density over three decades, marched on a play-operator core, with the local exponent d ln W / d ln B drawn across the top of the same frame. It is not a constant anywhere: 2.797 at 5.5 millitesla, heading for the three that Rayleigh's law gives, and 1.462 near saturation where the material has run out of magnetisation to give — a range of 1.420. How wide that range is is itself a property of the material: over the same amplitudes a soft core's exponent moves by 1.73 and a hard one's by 0.21. A single power law fitted across the whole range returns β = 2.518 and misses by 72.2 per cent; the same law fitted over the quarter of it from 9.7 to 24 millitesla returns 2.743 and misses by 0.97. Below 0.58 millitesla this discretisation has no loss at all, which is the finite operator count showing and not the material; the sweep starts above it. Two windings, and the band between them

The exponent nobody put in

A catalogue prints core loss as a coefficient times the frequency raised to one power and the flux to another, and the two exponents look like material constants. Neither is. On a loop built from play operators the frequency exponent is exactly one — a theorem, not a fit, because a rate-independent locus has the same area however fast it is traced — and the flux exponent is a local slope that runs from 2.94 at half a millitesla to 1.46 near saturation, so five windows on one measured curve give β from 1.58 to 2.84 and predictions three times apart at a hundred and fifty millitesla.

 The bank reaches 1.500 mΩ and the load sees 141.5 mΩ at that same frequency. computed by solving, not by drawing. One bulk part and 20 ceramics, with a nanohenry of mounting loop each and two nanohenries of plane between the bank and the load, solved once per frequency and read at both nodes. The dashed curve is the bank's own node — what a probe on the parts measures. The solid one is the load. The bank's least impedance is 1.500 mΩ at 11.3 MHz, and at that frequency the load sees 141.5 mΩ, which is 94.3 times more, against 141.5 mΩ of plane reactance at that frequency. Whatever the parts do, the load's reading cannot fall below the reactance of the copper in front of them, and the parts reach their best by moving up the frequency axis into it. Power, and the part that does no work

The floor and the ceiling move apart

A decoupling bank is judged by two numbers — the lowest impedance it reaches and the highest frequency at which it still meets its target — and with no copper between the parts and the load both improve together as capacitors are added, 5.000 milliohms down to 1.500 and 19.8 megahertz up to 162. Three nanohenries of ordinary board separate them. The bank's own floor still falls 3.33 times while the load's falls 1.49, and the ceiling read at the parts climbs to 82.4 megahertz while the load's peaks at 8.06 and falls to 5.05. At twenty parts the two nodes disagree by a factor of 94 about the same solve.

 The sensitivity of a pole against the room it has. computed by solving, not by drawing. A series R–L–C whose damping is walked from 0.3 to 0.999999, which slides its two poles together along a straight line and changes nothing else. The exact derivative of a pole with respect to the capacitor climbs from 0.5241 to 353.6 as the gap between them falls from 19078 to 28.28 radians a second. The fitted exponent over the closest four is -1.0000, and the product of the two is the natural frequency itself — 9999.6894 against 9999.6894, at every damping drawn and not merely in the limit, which a closed form gives and this computation never sees. The resistor's curve runs at 2ζ times the capacitor's — below it at 0.3 and at twice it by the time the poles have met — and the inductor's lies exactly under the capacitor's throughout. Before the steady state

The gap a derivative needs

The derivative of a pole is exact and has no step size in it, and beside the formula sits a sentence nobody had measured: it divides by a quantity that vanishes when two poles meet. Driven together, the sensitivity climbs as the reciprocal of the gap — fitted exponent −1.0000, the product a constant 1.00000 times the natural frequency — while the largest change it still describes falls as the gap *squared*. A one per cent capacitor is outside first order once the poles are 3194 radians a second apart, which is an ordinary critically damped design.

 What one temperature costs the loop gain of a part that has a gradient. computed by solving, not by drawing. The thermal loop gain of a 30 mm core, solved as a body with its own internal temperature profile and again as a single lump at that profile's mean, against the Biot number. Both are negative, so the core is a stabilising feedback either way — but the body's loop is the more negative of the two at every point, by 3.0 per cent at a Biot number of 0.108 and 38 per cent at 10.8. A lumped calculation therefore understates how stable a wound part is, and the amount it understates by is not a property of the material but of how well the surface is cooled relative to how well the inside conducts. Below a Biot number of about a tenth it is worth under two per cent and the lump is the right model; at the cooled end the part has 7 kelvin inside it and half the feedback is invisible to a single temperature. Power, and the part that does no work

The loop gain one temperature understates

Every thermal loop gain this collection has computed was computed at a single temperature, because a lumped fixed point has only one — and the essay that measured the gradient inside a core recorded, without measuring it, that this makes each of those numbers a lower bound. It is a lower bound by three per cent where a ferrite usually sits and by thirty-eight per cent at the well-cooled end, always in the direction that makes the part safer than the calculation said. The obvious candidate for what decides it is refused: three geometries at one Biot number are 3.3 times apart.

 The straight lines report 45.00° of margin, and the solved loop has 51.83°. computed by solving, not by drawing. The gain and phase of a loop made of an integrator and one pole, the corner at 1.00 kHz, with the integrator set so that the straight-line asymptotes cross unity at 1.00 kHz. The solved loop crosses at 786 Hz instead. Read at the lines' crossover, with the phase taken from the solve, the margin is 45.00°; at the loop's own crossover it is 51.83°. Closed, the loop is stable: the largest real part among its closed-loop poles is -5.00e-1 of the corner's angular frequency. The solved gain never rises above its asymptotes, so the lines cannot report more margin than the loop has. Frequency, which is the same solve

The margin the straight lines report

A phase margin read off a sketch is read where the straight lines cross unity, and that is not where the loop does. For an integrator and one real pole the lines report 45.00 degrees on a loop that has 51.83 — short, and always short, because a real pole's response never rises above its asymptotes. A pair that peaks reverses the sign and removes the bound: at a quality factor of two the lines report 71.57 degrees on a loop that has no margin at all, and at five they report 82.41 on a loop that is unstable.

 Every order buys less range than the one before, and above 9 Vₜ the sixth is worse than the fourth. computed by solving, not by drawing. The error of the same expression truncated at three orders, against the drive it is evaluated at, with the measurement it is chasing being a Newton-solved transfer curve that knows about no series at all. Each truncation's error grows as its own order in the drive — fitted at 2.00, 4.00, 5.88 against 2, 4 and 6 — so each buys a further range at a stated accuracy: inside 1% the leading expression is good to 1.12 thermal voltages, the fourth order to 3.88 and the sixth to 7.03, factors of 3.46 and 1.81. Beyond all of them the series stops helping: at 8.9 thermal voltages, where the second harmonic is 11.4%, the sixth-order expression is exactly as wrong as the fourth and is worse above it. What a designer does there is bisect the curve. Devices, and the amplitude they stop being linear at

The order that stops helping

Three essays in this field have derived expressions for the amplitude at which a degenerated stage's distortion reaches a target, each one order longer than the last, and each one nearer the measurement. This is where that stops. The error of an expression truncated at order m grows as the m-th power of the drive — 2.00, 4.00 and 5.88 measured — so every added order buys a range that ends sooner than the last one bought, and above 8.9 thermal voltages the six-term expression is further from the device than the four-term one.

 What each factor of attenuation buys on a 2.0 kΩ source. computed by solving, not by drawing at 12 probe ratios: the one-per-cent frequency bisected on the node with and without the probe, against the frequency a tip capacitance alone would predict. A one-to-one probe reaches 6.79 kHz and a hundred-to-one 692 kHz. The first step, from 1× to 2×, multiplies the bandwidth by 2.03 for a factor of two in signal; the two routes differ by at most 1.8% across the sweep, and they differ at all only because the probe's 1.0 MΩ is already 0.20% of the reading before any frequency is applied. Measurement, which is a circuit on a circuit

The probe that takes a tenth

A ten-to-one probe buys an order of bandwidth for a tenth of the signal, and on a two-kilohm source the bandwidth is exact: 6.79 kHz becomes 69.2 kHz. The tenth of the signal is not a tenth of the signal-to-noise ratio. Solved resistor by resistor, the noise referred to the tip goes from 1.782 µV to 55.78 µV — a factor of 31.3 — because the divider that does the attenuating is nine megohms and a megohm, and √(n(n−1)kT/C) on the cable's own capacitance has no source resistance in it at all.

 Where 8 channels leak to: 91.0 MHz from buffered sources, 901 kHz from 50 Ω. computed by solving, not by drawing. The frequency at which the open channels of a multiplexer built from the 0.5 Ω, 100 MΩ, 5 pF switch leak 1% of the signal onto the shared output, against the impedance of the source driving the selected channel, into 1 MΩ. 2 channels: 637 MHz buffered, 6.30 MHz from 50 Ω; 8 channels: 91.0 MHz buffered, 901 kHz from 50 Ω; 16 channels: 42.4 MHz buffered, 420 kHz from 50 Ω. Each falls as the reciprocal of the source impedance plus the on-resistance, and the single switch's own band closes at 6.43 MHz. Where the models stop

Where an open switch leaks to

A lone switch has a band whose width no load can change, because its open state leaks into the load. In a multiplexer the seven open channels leak into a node the selected channel holds, so the load leaves the answer and the source takes its place: one per cent of leak at 91.0 megahertz from buffered sources and 901 kilohertz from fifty ohms, moving as the first power of the tolerance rather than the second. Adding the channels' capacitance into one forty-picofarad switch puts it at 804 kilohertz, near the fifty-ohm figure by coincidence and a hundred and thirteen times low for a buffered one.

 Fed nothing, an eighth-order cascade sits 88 least significant bits from zero. computed by solving, not by drawing. Where each section of a 12-bit rounded cascade settles with zero input and a seeded state. Nothing decays to zero: a constant state survives whenever rounding returns it to itself, which needs only |y·(1 + a₁ + a₂)| ≤ q/2, and that denominator is small precisely because the corner is far below the sample rate. Each section's own band is 30, 31, 32, 33 least significant bits and the settled offsets are -27, -50, -70, -88 — accumulating, because a section's dead-band output is the next section's input and the next section passes direct current. In volts the offset halves with every bit added; in least significant bits it does not move at all. Where a signal becomes a number

Zero in, and not zero out

Rounding a filter's coefficients and marching it in double precision leaves out the other half: the products are rounded too. Put that in and a twelve-bit eighth-order cascade fed nothing at all settles two per cent of full scale away from zero and stays there — and the offset does not shrink with the word length, it grows as the square of how far the corner sits below the sample rate, reaching eighteen per cent at a hundred and twenty-eight times.

 Flat in angle at √(L/C), 141.42 Ω, and flat in size at 91.018 Ω. computed by solving, not by drawing, at 127 resistances on the closed form the network was checked against. The upper curve is the frequency at which the part's size is 1% away from R, the lower one the frequency at which its angle reaches 1°. Both are V-shaped and their points are in different places: the angle's first-order term vanishes at √(L/C) = 141.42 Ω, bisected on the measured slope to ten figures, and the size's second-order term at 91.018 Ω. The widest 1° band is 1.13 GHz, at 150.69 Ω; the widest 1% band is 1.66 GHz, at 95.806 Ω. At 91.018 Ω, flattest in size, the angle reaches 1° by 53.9 MHz. The flat line is 139 MHz, where a 6 mm body is one degree long: it binds the angle's edge from 118.72 Ω to 168.88 Ω and the size's from 43.947 Ω to 359.53 Ω. Frequency, which is the same solve

The resistor that is right in size and wrong in angle

A resistor's impedance departs from its value in size as the square of frequency and in angle as the first power, so the angle always leaves first: at ten milliohms and at a megohm alike, where the angle has reached a degree the size is still only 152 parts per million out. One time constant, L/R − RC, sets that degree — 4.00 nanoseconds and 695 kilohertz at ten kilohms, 800 nanoseconds and 3.47 kilohertz for a ten-milliohm shunt. The resistance flattest in angle is exactly √(L/C), 141.42 ohms, the value the size question rejected; at the 91.02 ohms flattest in size the angle reaches a degree by 53.9 megahertz, and no resistance is flat in both.

 Twenty metres buys 20 dB of apparent match and costs 10 dB of noise figure. computed by solving, not by drawing. The same cable and the same 200 Ω load as the reading, with the amplifier that is actually behind the instrument. The rising trace is the return loss the instrument reads, which is the load's own 4.44 dB plus twice the one-way loss. The lower pair is the chain's noise figure: a 2 dB amplifier with the cable in front of it, counted as a matched attenuator whose noise factor is its loss, and counted honestly from the available gain of a lossy line driven by a source that reflects 0.60. The first says the exchange rate is exactly two decibels of match per decibel of noise figure, at every length here. The second is higher everywhere — by (1−Γ²u²)/(1−Γ²), which is 0.028 dB at five centimetres and 1.938 dB, the load's own mismatch loss, once the cable is long enough to have absorbed the reflection. At twenty metres the instrument reads 24.4 dB and the chain costs 13.92 dB against the amplifier's own 2. Lines, where a wire has a length

The cable that hides two things

A length of cable improves a return-loss reading by twice its loss and raises a noise figure by once it, so the rule of thumb is two decibels of apparent match per decibel of floor. Both halves are owned here and neither essay had the other. Put together they say what an acceptance limit costs: making a 4.0:1 load read 1.50:1 spends 6.53 decibels of noise figure, against a mismatch that was itself costing 1.938 — and the exchange rate is not two but 2(1−Γ²)/(1+Γ²), which is 0.94 where a pad is actually short.

 From 50 Ω into 50 Ω: a T isolates to 643 MHz, a changeover to 6.34 MHz. computed by solving, not by drawing. The fraction of the drive that arrives with the path open, against frequency, from a 50 Ω source into 50 Ω, for the 0.5 Ω, 100 MΩ, 5 pF switch used three ways. A T reaches 1% at 643 MHz; a changeover reaches 1% at 6.34 MHz; one switch reaches 1% at 3.18 MHz. Where the models stop

The capacitance a third switch moves

A changeover's open channel leaks into the source of the channel that is closed, so its isolation into fifty ohms falls from 643 megahertz with a buffered source to 6.34 megahertz with a fifty-ohm one. Put a third switch to ground between two series switches and the leak lands on half an ohm of closed switch instead: 814, 643 and 2,240 megahertz from sources of nothing, fifty ohms and a kilohm, rising forty decibels a decade where a changeover's rises twenty. The price is the shunt switch's own capacitance, moved under the closed path, which makes the T one per cent wrong as a waveform at 6.59 megahertz beside the changeover's 6.61.

 A cascode multiplies rₒ by β, not by gₘrₒ — and the two are 21× apart. computed by solving, not by drawing. The output resistance of a cascode stage, measured by driving the output node with a current source and reading the voltage, against the current gain of the upper device. The plain stage's is 80 kΩ — rₒ and nothing else. The cascode's is 11.5 MΩ at β = 150, which is βrₒ to within a tenth and is 21 times below the gₘrₒ² every reference gives. The reason is in the netlist rather than in the algebra: the upper device's base draws current, so its rπ sits from the lower device's collector to signal ground and shunts the node the feedback works through. What the arrangement buys therefore scales with β and stops when β does, and the curve is the two expressions drawn against the measurement. Devices, and the amplitude they stop being linear at

The device that never sees the swing

A second transistor standing between the first and the load does two things that every reference gives one expression each for, and one of the two expressions has no ceiling in it. The output resistance is not gₘrₒ² — that is 248 megohms here, and the measurement is 11.5 — it is βrₒ, because the upper device's base draws current and shunts the very node the feedback works through. The bandwidth really is fourteen times better, and what it costs is two volts of a five-volt supply.

 A 4.0:1 flux slope ratio and a sinusoid enclose the same loop to 0.00%. computed by solving, not by drawing. A core driven in flux rather than in field — the way a winding drives it, by integrating a rectangular voltage — around a triangle of ±100 millitesla at a duty cycle of 0.2, whose two slopes differ by 4.00 to one. The loop it traces encloses 2.1006 joules per cubic metre, against 2.1007 for a symmetric triangle and 2.1006 for a sinusoid of the same peak: the same number to 0.001 per cent. That is not an approximation, it is a theorem about the model — a rate-independent locus depends on where the flux went and not on how fast — and it is the prediction that real cores disagree with by tens of per cent. The disagreement is the measurement of what the model has left out. Two windings, and the band between them

The duty cycle that costs nothing

A converter drives its core with a rectangular voltage, so the flux is a triangle whose two slopes differ by nineteen to one at a five per cent duty. The play-operator core charges exactly the same for all of them — 2.1006 joules per cubic metre at every duty and for a sinusoid of the same peak, to three parts in ten thousand — because a rate-independent locus depends on where the flux went and not on how fast. Real cores charge tens of per cent more, and the standard correction hides its entire waveform dependence in α − 1, which is the one term a rate-independent model has none of.

 Correcting a 20 Ω, 50 mH load to unity, and what it stores. computed by solving, not by drawing. A capacitor across a 230 V, 50 Hz supply is swept from nothing to 300 µF against a load drawing 1636 W and 1285 var. The reactive power falls through zero at 77.31 µF and keeps going; the energy stored in the installation rises from 2044.9 mJ to 4089.7 mJ at that point — exactly twice, because unity power factor means the two stores are equal — and goes on rising afterwards. Only the cable current has a least value, 7.113 A against 9.044 A. Power, and the part that does no work

The energy a unity power factor doubles

Reactive power can be computed three ways, and their agreement is easily called a verification. Two of the three are one theorem written twice and cannot disagree about anything; only the third is independent, and what it computes is a difference. Correcting a 20 Ω, 50 mH load to a power factor of 1.000000 takes its reactive power from 1,285 var to nothing and takes the energy stored in the installation from 2,044.9 mJ to 4,089.7 mJ — exactly twice, at every load and every frequency.

 A switched-capacitor low-pass driven past half its own clock. computed by solving, not by drawing. The clock is 1.00 MHz and the corner the earlier essay fitted is 1.59 kHz. The falling dashed curve is that continuous model, which knows nothing about a clock and goes on falling. The circles are the marched circuit, read at the frequency the output actually appears at. They part company past half the clock and by 992 kHz the model is 42.1 decibels wrong — an input just below the clock arrives just above direct current, in the middle of the passband, with the passband's own gain. The third curve is the exact discrete transfer function evaluated at the folded frequency, and it agrees with the march to 0.26 decibels, which is what says the march is measuring the folding rather than an artefact of itself. Filters, measured not tabulated

The filter that samples

The rung below built a resistor out of a clock and measured two ways it is not one: a settling time, and a corner that is a capacitor ratio rather than an R–C product. Both are errors in a value and both get smaller as the design gets better. This is an error of a different kind — the arrangement is not a continuous system at all, and nothing below half the clock shows it. An input at 992 kilohertz arrives at 7.8 kilohertz with the passband's own gain, where the continuous model the rung below fitted says it is 56 decibels down.

 A staircase costs one Nth, computed rather than quoted. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. The charge is broken into N equal risers, each held for 16 time constants so that it completes. The measured losses are 1.00000, 0.500001, 0.250000, 0.125000, 0.0625000, 0.0312500 of ½CV² — which is 1.000004, 1.000002, 1.000001, 1.000001, 1.000000, 1.000000 times 1/N, so the law is exact to four parts in a million at the worst rather than approximately true. The fitted exponent is -1.00000 and the energy account closes to 2.17e-6 at the worst. Before the steady state

The half a switch keeps

Charging a capacitor from a step loses half the delivered energy whatever the resistance, and a ramp takes the loss down as 2τ/T with no floor. A staircase of N settled risers costs one Nth of the step, exact to four parts in a million, and the law ends at a dwell of 5.272 time constants. A switch is the other half of the same product and buys nothing at all: with the supply held at five volts and the channel conductance ramped over a thousand time constants, the loss is 1.00000000 of ½CV².

 A junction's noise against its own resistance's: exactly one at zero volts, and a half only far from it. computed by solving, not by drawing. A junction carries two currents at once, Is·e^(V/nVt) forwards and Is backwards, and each has its own shot noise. Their noise over the Johnson noise of the junction's own conductance is n(1 + e^−u)/2. At n = 1 it is 1.000000 at zero volts, 0.5676 at 50 mV, within one per cent of 0.50 only above 115.1 mV — where the forward current is ninety-nine saturation currents — and 2.978 at −40.00 mV of reverse bias. The half the forward-biased junction is known for is the limit of this curve, not its value. The floor, which bounds from below

The junction that is a resistor at zero volts

A forward-biased junction makes half the noise power of a resistor of its own dynamic resistance, and that half is a limit rather than a value. Kept with the saturation current that flows backwards across it, the ratio is one exactly at zero volts, 0.5676 at 50 millivolts, and within one per cent of the half only above 115.1 — at ninety-nine saturation currents, which is a picoampere on a small silicon diode and a microampere on a leaky one. A photodiode held at zero volts has the Johnson noise of its shunt resistance and nothing else, and it becomes shot-noise-limited at 49.981 millivolts of photocurrent drop.

 One channel's load step reaches another through the supply, and the compensation decides by 4497×. computed by solving, not by drawing. Two identical amplifiers on one rail — sharing no signal node — with an ampere of load step pulled from the first and the second's output read. With the wiring left out the coupling is exactly zero, which is what the seven rungs below this one computed. With 30 nanohenries and fifty milliohms of rail and 10 microfarads of decoupling it is not: 3.84 microvolts per ampere at 271 kHz if the compensation capacitor returns to ground, and 17.29 millivolts per ampere at 2.33 MHz if it returns to the rail. That is a factor of 4497 decided by a modelling choice, which is why both are drawn. The channel that caused the step is unaffected: its own loop corrects the disturbance along with everything else, and the crosstalk is entirely a problem for the channel that did not. Feedback, and the margin

The rail the load moves

Seven rungs of this ladder end by saying the same thing: the supply is an ideal voltage source, so a load step is drawn from a node that cannot be disturbed. Giving the rail an impedance turns out to change the disturbing channel's own output impedance by three parts in ten million — its loop corrects the supply along with everything else — and to open a path to a second amplifier that shares nothing with it but a wire. How large that path is is a modelling choice: 3.84 microvolts per ampere with the compensation capacitor returned to ground, 17.3 millivolts with it returned to the rail, a factor of four and a half thousand.

 The ripple 1000 µF leaves, through the regulator: 60.62 mV rather than 21.11 mV. One settled cycle of the reservoir's output — 1.331 V peak to peak across 1000 µF — split into 1000 Fourier lines, each passed through the solved regulator's rail-to-output response, and summed. The output is 60.62 mV peak to peak. The ripple times the rejection at 100 Hz, -36.00 dB, gives 21.11 mV, which is 2.872 times too little: the rejection worsens at twenty decibels a decade, so each harmonic of the sawtooth arrives at nearly the size of the first. The 100 Hz line carries 23.4% of the output's mean square and the lines above 1 kHz 10.4%; the first three arrive at 7.731 mV at 100 Hz, 7.200 mV at 200 Hz, 6.388 mV at 300 Hz. The dashed curve is the 100 Hz line alone. Circuits that do a job, and the range they do it over

The ripple that arrives as a comb

The rejection essay multiplied two numbers: 1.331 volts of reservoir ripple and the regulator's 36.0 decibels of rejection at a hundred hertz, for 21 millivolts at the output. The ripple is a sawtooth, a comb of lines at every multiple of a hundred hertz, and the rejection worsens at twenty decibels a decade — so the second line arrives at nearly the size of the first, and the third too. Summed with their phases, the output is 60.62 millivolts, 2.87 times the estimate, and the hundred-hertz line carries only 23.4 per cent of it. A reservoir twenty-one times larger cuts the rail's ripple 14.8 times and the regulated ripple 5.95.

 Four wires against a 10 MΩ voltmeter. computed by solving, not by drawing at 81 resistances, twice each, with a voltmeter of 10 MΩ and 50 mΩ in every lead. The four-wire error is not zero: it is the voltmeter's own divider, −(R + 2R_lead)/(R + 2R_lead + R_m), which grows with the resistance being measured rather than shrinking. The two-wire error is that same quantity plus the leads, so it passes through zero at 1000 Ω — where the reading is right to 1.8e-12 while the four-wire reading is 0.0100% low — and above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance. Measurement, which is a circuit on a circuit

The voltmeter four wires do not remove

A four-terminal measurement is described everywhere as removing the leads from the answer. It moves them. What is left is the voltmeter's own input resistance, and it grows with the resistance being measured rather than shrinking: with a ten-megohm voltmeter and fifty milliohms of lead, the four-wire reading is 0.0100 per cent low at a kilohm, where the two-wire reading is exactly right — 1.8 × 10⁻¹² — because its lead error and its loading error cancel. Above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance.

 Inside the passband the matrix is worst at the edge, and at no ripple peak. computed by solving, not by drawing. The condition number of a 5th-order Chebyshev filter's nodal matrix at every frequency from a hundredth of its cutoff to ten times it. It is 4.004e+3 at direct current, rises to 1.5868e+7 at 964.0 Hz — located by golden section rather than read off the sweep — falls to 4.970e+6 at 1520 Hz and rises again through the stopband. The maximum is above every ripple peak (the highest is at 897.9 Hz) and within 3.6 per cent of the half-power frequency: the worst-conditioned place in the passband is where the filter stops passing and starts blocking, which is a place the response curve has no feature at. It is a local maximum: above the band κ climbs again, reaching 2.702e+7 at ten times the cutoff, because the susceptances in the matrix grow with the frequency and κ counts them. Nothing about any of these numbers says whether a digit is actually lost anywhere. Networks, and how a solve is checked

Where the matrix is worst, and where the answer is not

When a solve stops being exact has been answered with two networks at direct current. Swept along the frequency axis, a resistive chain's nodal matrix has a condition number of 4.00×10³ at direct current and 1.59×10⁷ at 964 Hz — a maximum at the band edge, at no ripple peak and at no feature the response has. It predicts nothing. The solution vector is right to three units of round-off everywhere, the response taken out of it loses four decades into the stopband, and the same filter written at 400 kΩ instead of 10 Ω has a condition number 1.4×10⁹ times larger and returns the same twelve digits.

 What the accuracy costs: dynamic range against the resistance scale. computed by solving, not by drawing. The earlier essay found the active realisation's response converging on the passive one's as the resistance scale rises. This is the price. The floor rises as the square root of the scale — fitted at 0.500 — because the resistors are the noise. The largest internal swing rises as the scale itself — fitted at 1.012 — because each gyrator forces the inductor's own current through its own resistors, so an amplifier inside it carries that current times R. Dynamic range on a ±15 V supply therefore falls as the three-halves power: 78 dB at 100 kΩ and 18 dB at 10 MΩ. The passive ladder realising the same response has 141 dB, and its worst internal node carries 1.10 times the input. Filters, measured not tabulated

Eight amplifiers, and what they add

The rung below realised a Chebyshev ladder out of floating gyrators and found the response converging on the passive one's as the resistance scale rises — twenty-two decibels out at ten kilohms, a twentieth of a decibel at ten megohms. It closed by naming two quantities it had not measured. They are the same quantity: the resistors that buy the accuracy are the noise, and the amplifiers inside the gyrators carry the inductor's own current through them, so the floor rises as the square root of the scale and the ceiling falls as the scale.

 The reading is a count of decades: 1.0288 parts per thousand of them. computed by solving, not by drawing. 17 marched tests, three families of absolute time — a tenth of a second, one second and ten seconds of short — plotted against the number of decades between the short and the reading. The families lie on one another, which is the finding: the answer is not a property of the part alone and not a property of either duration, it is a count of the decades of relaxation time the test leaves in. The line is a least-squares fit through the origin at 1.0288e-3 per decade; the model's own capacitance per decade of relaxation time is α = 1.0343e-3, which nothing in the fit was told — the fit sits 0.53 per cent under it, because the charge that comes back is shared with the slow branches it came off. The worst residual is 4.80 per cent, at the narrowest ratio drawn, and 1.23 per cent over the 8 tests that are two decades wide and read before the slowest relaxation the model has; the 3 read after it fall away to 4.49 per cent, which is where the law ends. Families a hundred times apart in absolute time differ by at most 0.84 per cent, which is the whole of the collapse. Before the steady state

Ten seconds, and fifteen minutes

A data sheet's dielectric absorption is quoted as a property of the part. It is not: the same modelled capacitor reads 0.4050 per cent with a tenth-of-a-second short and 0.0305 per cent with a thousand-second one, and 0.0047 against 0.2948 depending on when the reading is taken. Seventeen marched tests collapse onto one line — the recovery is 1.0288 parts per thousand for every decade between the two durations — and four dielectrics the specified test declares identical read a factor of 3.31 apart one decade away from it.

 Switched on at 0.95× resonance, a Q 50 capacitor reaches 1.550 times its settled voltage. computed by solving, not by drawing. The capacitor voltage of the series circuit, switched on from rest at the crest of the drive, drawn as the tip of its arrow in the frame that turns with the drive, so that the settled state is a fixed point — the arrow from the centre, 10.067 V long, with the circle of that radius around the centre. The path is the settled arrow plus a second one turning at the circuit's own frequency and shrinking, drawn until the second is a hundredth of its first length. The voltage reaches 15.603 V in cycle 9, 1.5499 times the settled amplitude, and the tip's farthest point is 1.5499 times it. The approximation 1 + exp(−π/2Q|δ|) gives 1.5335. Marched in time by the trapezoidal rule, the network agrees with the exact solution to 1.2e-3 of the settled amplitude over its first 11 cycles. Frequency, which is the same solve

The arrow that goes past where it settles

A phasor is where a driven resonator ends up, and counting the cycles it takes to get there says nothing about the path. Switched on from rest a little away from resonance, the capacitor's arrow circles its settled tip instead of approaching it: a Q of 50 driven at 0.8 of resonance reaches 1.854 times its settled voltage, and a Q of 200 at 1.25 switched on through zero reaches 2.182. At resonance exactly, where the settled voltage is largest, the arrow never passes its mark. So a resonance read by the largest voltage after switch-on is 1.20 times wider than its phasor says.

 The capacitor across the upper resistor: 90.9° of margin at 836.5 pF, and less ripple past it. The regulator's phase margin against a capacitor across the upper divider resistor, with the output ripple the 1000 µF reservoir leaves beside it. With no capacitor the margin is 46.47° at a crossover of 9.73 kHz, the output impedance at 10 kHz is 1.95 Ω, the worst rail rejection is 5.79 dB and the ripple 60.62 mV. The margin is greatest, 90.929°, at 836.5 pF — a zero at 6.34 kHz and a pole at 25.4 kHz around a crossover moved to 16.9 kHz — where the output impedance at 10 kHz is 828 mΩ, the worst rail rejection -1.50 dB and the ripple 52.33 mV. At 10 nF the margin has fallen back to 66.34° and the ripple is 35.40 mV. Circuits that do a job, and the range they do it over

The capacitor across the upper resistor

The rejection essay said a regulator reproduces its reference times its divider's four, and that a capacitor across the lower divider resistor brings that down to one at high frequency. Measured, the loop peaks the reference's gain to 5.68 near its crossover before any capacitor is added; a nanofarad across the lower resistor raises the peak to 11.2; and the capacitor that brings it down belongs across the upper resistor, where a nanofarad keeps the gain from ever exceeding four. The same capacitor is a lead pair in the loop: 836.5 picofarads takes the phase margin from 46.5 to 90.9 degrees, and ten nanofarads, past that optimum, still holds 66 while cutting the output ripple from 60.6 millivolts to 35.4.

 The return under a track gathers from 26.3 kHz to 1.42 MHz, not at one frequency. computed by solving, not by drawing, on a cross-section of a 50 mm plane cut into 120 strips, each with its resistance and its partial inductance to every other strip and to the track. The solid curve is the share of the return current inside one track-height of the point beneath the track; the second is the share inside ten heights. At direct current the return spreads evenly — 0.8 per cent within one height — and far above the band it is the image-current distribution, 48.7 per cent, which the closed form gives as 48.7. Between them it gathers across three decades: a tenth of the way by 26.3 kHz, half by 283 kHz, nine tenths by 1.42 MHz, shaded. The resistance of the path equals its reactance at 1.59 kHz, where the return has not yet moved. The single corner estimated from a path three track-widths wide and a parallel-plate inductance is 106 kHz, 28 per cent of the way through the band. Lines, where a wire has a length

The corner that is three decades wide

Where the current comes back put the change in a return current's path at 106 kilohertz, from a low-frequency path assumed three track-widths wide and an inductance taken from a parallel-plate formula. Solved across a plane cut into a hundred and twenty strips, the loop's resistance equals its reactance at 1.59 kilohertz, where the current has not moved at all, and the return then gathers beneath the track over three decades — half of the way by 283 kilohertz, nine tenths by 1.42 megahertz. The single corner is a point about a quarter of the way through a band.

 Two of the four errors are divided by the gain; the best the instrument gets is 110.1 dB, at ×776. computed by solving, not by drawing. The four mechanisms that limit a three-amplifier instrumentation amplifier's common-mode rejection, each measured alone against the gain of its input stage and then all four together, at 50.0 Hz with 1 kΩ of imbalance between the source resistances and 10 pF at each input. The difference stage's four resistors and the difference amplifier's own rejection are injected after the gain, so their common-mode gain is a constant — 1.998 mV/V and 0.0100 mV/V — and the rejection they allow rises decibel for decibel with the gain. The input pair's mismatch and the source's time-constant gap are injected before it, so they are amplified by exactly the gain the signal is and the rejection they allow is flat. The four add as complex numbers: the two largest are real and of opposite sign, they cancel at a gain of 776, and what is left there is the source's 3.142 µV/V, which is purely imaginary because it is ωΔτ. The instrument's best is 110.07 dB against the source's own 110.06, and above that gain more of it buys nothing. Measurement, which is a circuit on a circuit

The errors that arrive before the gain

Four earlier measurements each found a different owner of one instrument's common-mode rejection and each measured it alone. Solved together, the four are complex numbers that add — to 1.25 parts in ten thousand — and two of them carry a factor of the first stage's gain while two do not. The two that do cancel the two that do not at a gain of 776, and what is left there is 110.066 decibels, which is exactly the number the cable sets. Better-matched amplifiers move that gain from 93 to 3392 and do not move the ceiling by a hundredth of a decibel.

 One switch is never better than 70.71 ppm; a T of three reaches 10 ppb only into 200 MΩ. computed by solving, not by drawing, at direct current. The worse of a switch's two errors — closed, the fraction the load fails to receive; open, the fraction it receives anyway — against the load, for one 0.5 Ω, 100 MΩ switch and for a T of three, from a buffered source. The lone switch is best at 7.07 kΩ, the geometric mean of its two resistances, where both errors are 70.71 ppm, 13.79 bits: no load does better. The T has no best load. Its worse error falls with the load towards Rₒₙ/(Rₒₙ + Rₒff) = 5 ppb, the square of the lone switch's resistance ratio rather than its root; it is within twice that from 200 MΩ, it passes the lone switch's floor only above 14.1 kΩ, and into 7.07 kΩ it is 141.4 ppm, worse than one switch. Solved on the network up to 1000 MΩ and continued, dashed, from the closed form it matches. Where the models stop

The floor below any load

A switch of half an ohm closed and a hundred megohms open is within one per cent of ideal for loads between two edges, and the edges close on each other as the tolerance tightens. At direct current they meet at 70.71 parts per million, into 7.07 kilohms: no load makes that switch better, which is 13.79 bits and a boundary with no frequency in it. A T of three such switches has no best load at all. Its error falls with the load towards five parts per billion — the square of the lone switch's resistance ratio rather than its root — and reaches ten only into two hundred megohms. Into the 7.07 kilohms that suited one switch, the T is worse than one switch.

 The two sequences a neutral current says nothing about. computed by solving, not by drawing at 61 imbalances. Three 20 Ω loads on a 230 V, 50 Hz star supply, one of them raised by a fraction of itself, with the neutral in place. The zero-sequence current is the one the neutral carries three times and is the only one a neutral ammeter reads; the negative sequence is a balanced set of three phasors rotating the other way. At 30.0 per cent imbalance it is 0.8846 A against 10.6154 A of positive sequence, 8.333 per cent, against 8.333 per cent from x/(3 + 2x). Two per cent arrives at 6.250 per cent imbalance, bisected on the network. Power, and the part that does no work

The half the neutral does not carry

A star load unbalanced in one phase produces two things, not one. The neutral carries three times the zero-sequence current, which is the half a neutral ammeter reads; the other half is a negative-sequence set of exactly the same size, rotating backwards, that the neutral never sees. With 0.5 Ω of line in front of 20 Ω loads, losing a phase entirely puts 50.00 per cent negative sequence in the current and 0.8265 per cent in the voltage a switchboard meter reads.

 The ratio of the two readings, drawn where it lives. computed by solving, not by drawing. The same five-element RLC ladder at 10 kHz, read from each end, with the quotient of the two readings plotted in the complex plane. A reciprocal network sits at the point 1. One transconductance moves it along a straight ray, 1 − gm·Z, whose direction is the phase of the single branch between the source's control node and its output node and whose length is gm|Z| — 62.83 Ω for the 1 mH inductor, 120 Ω for a resistor, 72.34 Ω for 220 nF, all at 10 kHz. A mirrored pair of transconductances stays at 1 to 2.3e-15; reversing one of them runs the ratio around the unit circle to 8.9e-16, where the two readings are the same size and differ only in phase. The straight-ray law needs the two nodes joined by exactly one branch, and a second path between them takes it away by a factor rather than by a percentage. Networks, and how a solve is checked

The reading that does care which way round

An earlier essay measured a network's departure from reciprocity and left its size as a constant — 62.8 per siemens, for that network at that frequency. It is not a constant and it is not the network's: it is 2πfL for the single inductor between the controlled source's control node and its output node, 62.832 ohms at ten kilohertz, and the ratio of the two readings is 1 − gm·Z to 8.8 parts in 10¹⁴ over ninety-nine readings. So 4.5455 millisiemens across a 220 ohm branch makes a ladder that transmits a hard zero forwards at every frequency at once, and 206.13 ohms back at ten kilohertz.

 A junction and its resistor in one loop: equal shares at 12.50 mV, and quietest against both at 49.98 mV. computed by solving, not by drawing. A junction carrying 25 µA in series with a resistor, the loop closed into a short and solved as a netlist with each noise current injected across its own element. Against the drop across the resistor: the bare junction's 2qI, the bare resistor's 4kT/R, each one's share of what reaches the outside, and the total. The shares are equal at 12.50 mV (500 Ω), not at the 49.98 mV where the bare floors cross; there the resistor supplies 80.0 per cent and the total is 0.5556 of either floor. The total is below both floors at every drop. The floor, which bounds from below

The resistor in the same loop

A resistor's noise and a junction's are equal as bare densities at 49.98 millivolts of drop, and a junction in series with the resistor that carries its current is the arrangement every current source is built from. In one loop each noise current has to cross the other element, so the two supply equal shares at 12.50 millivolts, a quarter of the crossing; at the crossing itself the resistor supplies 80 per cent and the loop is 2.553 decibels below both floors, which is further than it gets anywhere else. The same resistor multiplies the stage's input-referred noise by five.

 A cascoded mirror is 90× the output resistance, and 43% of it goes back into the reference. computed by solving, not by drawing. The output resistance of a two-transistor mirror and of the same mirror with a cascode on each branch, measured by moving the output a little either side of its operating point and reading the current, against the current gain of every device. The plain mirror sits at rₒ = 89 kΩ and does not move. The cascoded one reaches 7.39 MΩ at β = 150 and rises with β until β stops being the smaller of the two quantities, where it saturates on gₘrₒ² = 268 MΩ. The third curve replaces the diode-connected upper device with a held voltage at the same potential and recovers 1.76 times the resistance, which is the upper device's base current being charged a second time — to the reference branch, where it moves the mirror's own bias. Devices, and the amplitude they stop being linear at

The source that holds to the supply

Putting a second transistor on each branch of a current mirror is always described as buying output resistance and costing headroom, and both halves of that are measured here rather than repeated. The resistance goes from 82 kΩ to 7.39 MΩ, the floor rises by 0.71 volts — and the range over which the current is actually what it was set to goes from 1.70 volts to 9.09, because a plain mirror's current never stops climbing. Forty-three per cent of the resistance that should be there is missing, and it is in the reference branch.

 The resistors own the floor between 2.91k Ω and 85.9k Ω, and the part owns it outside. computed by solving, not by drawing. The noise at the load of the two-path compensation against the impedance of its own feedback network, with the resistors scaled together and the compensation capacitor taken down in proportion so that Rf·Cf — the handover between the two feedback paths — does not move. Four contributions are integrated over 10 Hz to 100 MHz: the amplifier's 4 nV/√Hz, fitted as the 0.005 power of the impedance and so flat; the two resistors' √(4kTR), the 0.501 power; and the amplifier's 0.60 pA/√Hz flowing in the feedback resistor, the 0.997 power. Two different powers of one quantity cross twice. The resistors carry more than half the power only between 2.91k Ω and 85.9k Ω; outside that window, in both directions, the part does. The part's share is least at 15.8k Ω, which is not eₙ/iₙ — it is that ratio multiplied by the noise gain of 2.000 and again by 1.187, the square root of the ratio of the bandwidths the two generators actually see; there it carries 26.26 per cent. The model stops where the amplifier's output current does: at 100 Ω the feedback resistor alone draws 10 mA a volt. Feedback, and the margin

The window the resistors own

Eight essays have priced one compensated stage, and the fourth of them left two of the amplifier's own generators named and uncounted. With the current generator put in the netlist the floor at ten kilohms goes from 28.88 microvolts to 30.13, and the resistors carry more than half the noise power only between 2.91 kΩ and 85.9 kΩ — outside that window, in both directions, the part does. The flicker corner turns out to be worth 1.00009 in this stage's own band, and 3.474 one band away.

 The same core at the same current has two inductances, 1.80 times apart. computed by solving, not by drawing. The small-signal inductance of a sixty-turn winding on a core walked down from 400 amperes per metre, measured by pushing the excitation up and by pushing it down at each bias. They are never the same: 14.899 millihenries against 8.297 at zero bias, a factor of 1.796, and up to 1.796 across the sweep. The mechanism is that an operator held inside its own backlash contributes nothing to dB/dH, so a reversal is measured by whichever operators are still moving, and that is a different set in each direction. The single-valued curve the field already had is drawn above both, which is what it is: an optimistic reading of an object with two answers. Two windings, and the band between them

Two inductances at one current

A data sheet prints one L(i) curve and there are two. An operator sitting inside its own backlash contributes nothing to dB/dH, so a small excitation sees only the operators still moving — every one at the tip of a loop, none just after a reversal — and the same core at zero bias measures 14.90 millihenries pushed downward and 8.30 pushed upward, a factor of 1.80. The same split decides what a converter's ripple costs: held at the flux swing volt-seconds actually fix, a twenty-millitesla ripple costs twelve times more at a hundred and seventy-five amperes per metre of bias than at none, and above a hundred and eighty-three the question has no answer at all.

 The direct form works at 28 bits and fails again at 29. computed by solving, not by drawing. The largest pole radius of an order-8 Butterworth at a 1 kHz corner and a 48 kHz clock, at every coefficient word length from 6 to 40 bits, for both realisations. Above the unit circle the filter is not inaccurate but unstable. The direct form is outside at 23 of the 35 word lengths drawn; the cascade at none of them, its worst radius being 1.000000. The failures are not an interval: 28 bits works and 29 bits does not, so the smallest word length that works is 1 below the largest that does not and a search for the crossing has no crossing to find. The direct form's curve is clipped at 1.24: a radius of 4.5 and a radius of 1.01 are the same verdict. Where a signal becomes a number

The word length that is not a threshold

The essay below this one measured an eighth-order direct form at five word lengths, found its poles outside the unit circle at every one, and left behind a phrase — the coefficient resolution required. Sixty word lengths later there is no such resolution. That filter is stable at 28 bits, unstable at 29 and stable again at 30, and the smallest word length a search would return is not one anybody can use. What survives is a law with a slope: 4.53 bits of coefficient for every pole added, rising to 47 bits at twelfth order, and none at all for a cascade.

 Against its own shaping the error is 17.4× tonal, not 51×. computed by solving, not by drawing. The share of an order-1 loop's in-band error sitting in its five largest lines, from 0.02 to 0.9 of full scale, with both nulls drawn. The lower level is 1.953 per cent — five lines of a FLAT error over 256 — and it is what this measurement has always been quoted against. The upper level is 5.74 per cent, which is what five lines of the loop's OWN shaping hold with no tone anywhere in them. Measured against the first the error is 51 times tonal at 0.02 of full scale and 26 at 0.9; against the second, 17.4 and 8.9. The direction survives the correction and the size does not. Where a signal becomes a number

A floor, or five tones

The field's sharpest statement about a one-bit loop is that three quarters of its in-band error sits in five lines, against the 1.953 per cent a white error would put in any five — a factor of thirty-eight. The comparison is to a white error, and a shaping loop exists to make its error anything but white. Measured against the loop's own transfer function, which the loop reproduces line by line to a part in five hundred when it is dithered, the same error is 17.4 times tonal at a fiftieth of full scale and 8.9 times at nine tenths. And 99.5 per cent of it at the quiet end is two harmonics of the input.

 A Q 10 resonance read by a stepped sweep is within 1% of its width after 1.65Q cycles a step. computed by solving, not by drawing. A stepped sweep switches the series circuit on from rest at each of 201 frequencies, waits a stated number of cycles of resonance, and reads the largest capacitor voltage in the last cycle of the wait; the resonance's half-power width is then bisected on those readings and compared with the settled width, 0.1003 of the resonant frequency. After Q cycles the reading is 19.6% too wide and its peak 4.66% low. The width stays within 1% from 1.65Q cycles on and the peak from 1.50Q, against the ln(100)/π = 1.466Q cycles the second arrow takes to fall to a per cent. A reading that holds its largest value instead is 20.2% too wide at any dwell, since the overshoot it holds has already happened. Frequency, which is the same solve

How long a sweep waits at each step

A resonance measured by a stepped sweep that starts each frequency from rest and reads the last cycle of its wait comes out 19.6 per cent too wide after Q cycles a step, and within one per cent of its width only from 1.65Q cycles at a Q of ten and 1.60Q at fifty. That is longer than the 1.47Q the transient takes to fall to a per cent, and the reason is the centre rather than the skirts: the width's error is the peak's shortfall read twice, while at the half-power frequencies the transient swings through its settled value and partly cancels itself. A reading that holds its peak instead is twenty per cent wide however long it waits.

 A true-RMS reading of a sine: ripple a second filter removes, and a bias it cannot. An explicit converter — square, average through a one-pole of τ = 100 ms, take the root — in steady state on a sine of unit root-mean-square value, integrated exactly over a period at 91 frequencies and by a fourth-order march of its own equation at 6, which agree to 1.9e-8. The upper curve is half the ripple on the reading and the lower one the amount by which its mean is low. The reading is low at every frequency, because the square root is concave; it is 1% low below 1.86 Hz, while the ripple is inside ±1% only above 39.8 Hz. The dashed curve is the small-ripple form, an eighth of the averaged square's ripple power, which the bias approaches as the ripple shrinks. Power, and the part that does no work

The average a square root pulls low

A true-RMS converter squares, averages and takes the root, and the root of a quantity that ripples averages below the root of its mean. With a hundred-millisecond averager a sine is read one per cent low below 1.86 hertz, where the ripple is still ±20 per cent — and a second filter that steadies the display takes the ripple away and leaves the reading exactly as low as it was. A square wave is read exactly at any averaging time; a rectifier current conducting for twenty degrees needs 2.41 times the averaging a sine does. The implicit converter is the explicit one at half the time constant, and a reading falls 1.38 times slower than it rises.

 The fastest damping is a surface, and the band is worth 5 times the third pole. computed by solving, not by drawing. Each point is the last settling cliff, bisected — the damping at which the first overshoot's peak lands exactly on the band's edge, which is where the fastest settling is. Across the five bands the optimum moves by 0.231 of damping ratio; across a third pole from 1.5 times the natural frequency out to a second-order response it moves by 0.047. The two axes are worth 5.0 to one, and the expensive one is the specification rather than the parasitic. The classic 0.78 for fastest two per cent settling is the second-order curve's value at ±2%, 0.7797; at ±1% the same response wants 0.8261. Before the steady state

The best damping is not the one to build

The fastest settling damping is the right-hand limit at a discontinuity, so two thousandths below it costs 41 per cent and two thousandths above it costs 0.34 — a ratio of 120 in the penalty for the same error. With ±2 per cent on the damping ratio the nominal that minimises the worst case is 0.7927 rather than the optimum's 0.7734, and it guarantees 4.243/ωₙ against 5.943. The band moves the optimum by 0.231 of damping ratio and the third pole by 0.047, and 0.78 is exact at ±2% and 55 per cent slow at ±1%.

 The cure changes shape at 909 Ω, which is a property of the feedback network and of nothing else. computed by solving, not by drawing. What balancing actually does to the circuit, against the source resistance it is done for, at a gain of 11 with a 1.0 kΩ bottom resistor. The inverting input looks back into 909 Ω — the bottom resistor times (G−1)/G — and that number is the whole of the knee. Below it the cure is a resistor in series with the source and the feedback network is untouched. Above it there is no resistor to add, and the network is scaled up to meet the source instead: 1100× at 1.0 MΩ, which puts 11 MΩ in the feedback path. The scaled feedback resistor is the source resistance times the gain exactly, so the network's own size has left the answer — it decided where the knee was and nothing after it. Measurement, which is a circuit on a circuit

The cure that becomes a different circuit

The classical cure for an amplifier's input current is to make the two resistances its inputs look back into equal, and it reads as one instruction. Solved, it is two circuits meeting at 909 ohms — the feedback network's bottom resistor times (G−1)/G — and above that knee there is no resistor to add: the network is scaled to the source, which at a megohm means 11 megohms of feedback and at a gain of 1001 means 1001. Above the knee three different networks become one instrument to twelve figures, the noise penalty settles at 1.41420 against a √2 of 1.41421, and the benefit at 10.49 against two currents whose ratio is ten.

 The eddy term is f² below the skin-depth frequency and f^1.5 above it, both exactly. computed by solving, not by drawing. Eddy-current loss in a 0.35 millimetre lamination held at a mean flux of 1 tesla, against frequency, with the classical uniform-flux expression drawn beside it and the local exponent across the top. Below the frequency at which the sheet is two skin depths thick — 465 Hz here — the two agree and the exponent is 2.0000. Above it the flux is confined to a layer whose thickness falls as one over the square root of the frequency, and the exponent is 1.5000: three halves, exactly. It does not arrive there monotonically — it undershoots to 1.4847 at ξ = 3.28 and comes back up, which is the bounded cosine term the asymptotic statement drops. At 1.00 kHz the classical term is already 1.74 times the truth. The solve and the closed form agree to 7.5e-3 per cent across six decades. Two windings, and the band between them

The current inside the iron

Every core-loss law has an eddy term of the form d²f²B²/6ρ, and it is derived by assuming the flux is uniform across the lamination — an assumption that is a frequency and that the expression does not carry. Solved instead as a diffusion, the exponent is exactly 2 below the frequency at which the sheet is two skin depths thick, exactly 1.5 above it, and it undershoots to 1.485 on the way. For a 0.35 mm sheet the crossing is 465 hertz, so at a kilohertz the classical term is 1.74 times the truth and at ten kilohertz it is forty-seven times.

 A core loss of 100 kΩ gives the resonance a floor at −66 dB, and by 10.0 GHz the choke is worth 0.02 dB. computed by solving, not by drawing. The common mode at the load, in decibels against the same circuit with no choke in it, with 5 pF across each winding and a core loss of 100 kΩ across each, drawn over the lossless curve. At the winding's resonance, 1.59 MHz, the inductance and the capacitance cancel and what is left is the loss, so the deepest point is 66.02 dB — 20·log(1 + (Rp/2)/(Zs + Zl)), with no inductance and no capacitance in it. Above the resonance the part is a capacitor across the path it was fitted to block, and by 10.0 GHz it attenuates 0.02 dB. Lines, where a wire has a length

The depth a resonance does not have

The inductor one mode cannot see printed its choke's best attenuation as 102 decibels at 1.59 megahertz. Swept with 1,999, 2,000 and 2,001 samples the same lossless part reports 110.0, 96.3 and 103.6 decibels, because a resonance with nothing to dissipate has no bottom and a sweep reports how near its nearest sample fell. Give each winding a core loss and the depth is 20·log(1 + (Rp/2)/(Zs + Zl)) — 66.02 decibels at a hundred kilohms, twenty more per decade of loss, with no inductance and no capacitance in it, and half of it belonging to the circuit the choke sits in.

 300 mirrors built to one design, with 2% device mismatch. computed by solving, not by drawing. Every pair in the population is a full Newton solve of the same netlist with two saturation currents drawn from a normal distribution, the Early conductances iterated to self-consistency for each. The mean is 3.836 per cent, which is the systematic error the rung below computed with identical devices (3.937 per cent) — the mismatch does not move it. The spread about it is 1.955 per cent, which is the device mismatch arriving with nothing dividing it, and the worst pair of the 300 is 8.43 per cent out. A design whose specification is the mean has specified the one mirror nobody has. Devices, and the amplitude they stop being linear at

The error that is a distribution

The rung below solved a mirror and separated two errors — one that falls with beta and one that does not. Neither is what limits a real mirror. Two transistors on the same die differ, a fractional difference in saturation current is a fractional difference in collector current with nothing dividing it, and the honest object is a spread rather than a number: mean 3.84 per cent, standard deviation 1.96, worst of three hundred 8.43. Degeneration divides it by one plus gm·R and stops at the resistors' own tolerance, and where it stops is a voltage — a hundred millivolts, containing nothing but the ratio of two tolerances.

 One capacitor moves three quantities, and the expression's own answer peaks by 1.18 dB. computed by solving, not by drawing. The bandwidth, the peaking and the total output noise of a 1.0 MΩ transimpedance stage against its feedback capacitor, swept from 0.30 to 4.20 times what the classical expression asks for. Over that factor of fourteen the bandwidth falls from 334 to 55.4 kHz, the total noise from 230 to 65 µV, and the peaking from 10.0 dB to nothing. The expression's own answer sits at 1.18 dB of peaking, 0.82× is where the loop reaches forty-five degrees, and √2× is where the response is flat — so the choice usually quoted as maximally flat is neither of the two conditions it is quoted for. The faint families are the same three quantities at 3 pF and 300 pF of diode: normalised this way they are one curve, so the diode sets the scale and the multiple sets the shape. Feedback, and the margin

The factor the expression leaves out

The classical compensation for a photodiode amplifier is quoted both as the forty-five degree choice and as the maximally flat one, and it is neither: it leaves 1.18 dB of peaking and 52.4° of margin. Flat is at exactly √2 times it — fitted at 1.4186 against 1.4142, at every detector from 3 pF to 1 nF. And the third quantity the capacitor is supposed to trade, the noise, does not move at all inside the signal band: three compensations spanning a factor of fourteen give 8.37 against 8.36 µV in a 4.36 kHz measurement and 230 against 65 µV over the whole plane.

 The load that takes the most power is 3.67 Ω, and the open-circuit voltage over the short-circuit current is 3.97 Ω. computed by solving, not by drawing. The terminal characteristic of a photocurrent with a junction across it, with the power along it drawn on the same axes and scaled to fill them. The most power, 80.99 W, is delivered at 17.244 V and 4.696 A, which is a load of 3.67 Ω. The incremental resistance of the source there — the negative of the characteristic's own slope — is 3.67 Ω, the same number to four figures. A straight line drawn between the two end points has a resistance of 3.97 Ω and would promise 24.81 W. The ratio of the true maximum to that promise is 3.264, and its reciprocal is the fill factor, 0.8160. Networks, and how a solve is checked

The load a curve recommends

The maximum power theorem says to match the load to the source's internal resistance, and for a straight-line source the internal resistance, the slope and the open-circuit voltage over the short-circuit current are one number. On a photovoltaic panel they are three: the slope at the maximum is 3.6717 ohms, the load there is 3.6717, and the open-circuit voltage over the short-circuit current is 3.9709 — and a quarter of the open-circuit voltage times the short-circuit current, which is what a straight line would deliver, is 24.81 watts against the 80.99 that is there. The theorem survives with the incremental resistance in place of the internal one, and it survives only where the characteristic has a slope: on a supply whose current limit folds back, the most power is delivered at a corner.

 An amplifier a hundred times the corner moves the unity-gain section's Q by 2.0%. computed by solving, not by drawing. A unity-gain Sallen–Key section designed for Q = 2 at 1.00 kHz, built with a one-pole amplifier, and its poles recovered by rooting the determinant. The section as drawn has two poles; as built it has three, and the pair is not where it was put. At a gain-bandwidth of a hundred times the corner — which is the rule of thumb — the quality factor is 1.97 per cent high and the pole frequency is 1.97 per cent low. The two are the same number with opposite signs over the window where the amplifier is well clear of the section and its own pole is still resolvable, and the number is the designed quality factor, 2, divided by the ratio: 2.00 per cent at 100 times the corner. Filters, measured not tabulated

The Q the amplifier decides

A second-order section's quality factor is set by a capacitance ratio and its pole frequency by a product of four passive values, and neither expression contains the amplifier. Build it with one that has a gain-bandwidth a hundred times the corner — the usual rule — and the Q comes out 1.97 per cent high while the pole comes out 1.97 per cent low, the same number in both directions, and the number is the designed Q divided by the ratio. A fifth-order half-decibel Chebyshev built that way has 2.1 decibels of ripple.

 A transistor's two noise generators are one current: their product is 0.8008 nV·pA/Hz at every bias. computed by solving, not by drawing. The input voltage noise of a bipolar stage, √(2kT·rₑ), is its collector current's shot noise referred through gₘ, and falls as the current rises; its input current noise, √(2qI_C/β), is its base current's, and rises. At β = 100 their product is 2kT/√β = 0.8008 nV·pA/Hz at every collector current from a microampere to ten milliamps, so the best noise figure, 0.4139 dB, does not depend on the bias. What does is where it is: the optimum source resistance times the current is √β·Vt = 249.9 mV, and the two lines cross where that resistance is a kilohm, at 250 µA. The floor, which bounds from below

The two generators that are one current

An amplifier's noise is two generators, a voltage in series with its input and a current across it, and the essays on its noise figure treat them as independent numbers. In a bipolar input stage they are one current's shot noise divided two ways, by the collector and the base, and their product is 2kT/√β at every collector current: 0.8008 nV·pA per hertz at a current gain of a hundred, from a microampere to ten milliamps. The best noise figure, 0.4139 dB, does not depend on the bias. The bias decides only where the best source is, and 50 ohms of base resistance decides what the best actually is below 500.

 The winding current a hysteretic core asks for, and the current the same circuit's single-valued core asks for. computed by solving, not by drawing. One netlist — a 22.6 volt peak sinusoid at 5.00 kHz through 4 ohms into a sixty-turn winding — marched twice, once with the core as a superposition of play operators and once with the single-valued saturating curve every earlier figure in this field used. The two currents differ in shape and not only in size: the hysteretic one leads the flux by an angle that is not ninety degrees, which is the whole of the core loss, and its peak is 59.45 milliamperes against 50.12. The energy the source delivers over a complete cycle is 17.19 microjoules for the loop and 2.5e-5 for the curve, which is zero to the resolution of the march. The slider takes the flux the drive demands from a twentieth of the material's saturation to most of it. Two windings, and the band between them

The core the solver has to remember

A saturating inductor's state is one number, because the current is a function of the flux linkage. A hysteretic core's is not: half an amp is one flux on the way up and a different flux on the way down, and there is no function of the current that returns the flux. So the march's state vector gains twenty-four more numbers, one per play operator, and the Newton loop is forbidden to touch them. The circuit and the bench then agree on the loss to three parts in ten million — one integral taken at two wires, the other inside the material.

 Two networks of one magnitude deliver the same energy, and the minimum-phase one delivers half of it 5.61 times sooner. computed by solving, not by drawing. A low-pass with poles at 1.00 kHz and 10.0 kHz and a zero at 3.00 kHz, and the same network with an all-pass behind it that moves the zero into the right half-plane. Their magnitudes agree at every frequency sampled to 4.4e-16. The energy of each impulse response, from the residues in closed form, is 6029.319 for both, and the integral of |H|² over frequency gives 6029.305. What differs is when it arrives: the minimum-phase network has delivered half its energy by 11.27 µs and its mirror by 63.23 µs; by 20 µs the fractions are 0.658 and 0.352, by 100 µs 0.918 and 0.676; and at no instant has the mirror delivered more. Frequency, which is the same solve

The energy that arrives first

Two networks with the same magnitude at every frequency have the same impulse-response energy, and Parseval's theorem says so before either is solved. They do not deliver it on the same schedule. A low-pass with poles at one and ten kilohertz and a zero at three delivers half its energy by 11.27 microseconds; the same network with its zero mirrored into the right half-plane, which changes no magnitude anywhere, takes 63.23, and at no instant has it delivered more. Its step response starts the wrong way, to −0.170 of the final value, before it turns round. Minimum phase is minimum delay, and the delay is in the energy rather than in any one number a frequency plot shows.

 A capacitor across the upper divider resistor removes the output capacitor's resistance floor. computed by solving, not by drawing. Two series resistances against the capacitance across the upper divider resistor: the smallest the loop tolerates at 45° of margin (lower curve), and the one that gives the smallest droop after a 100 mA load step (upper). With no capacitor they are 939 mΩ and 885 mΩ — the second BELOW the first, which is the conflict this design has: the best transient is one the loop refuses. The floor falls as the capacitor grows and between 500 and 836.5 pF it leaves the sweep altogether, so every series resistance down to a milliohm is stable. Past about 5000 pF the floor climbs back and overtakes the optimum again. The shaded band is where the design a transient wants is one the loop allows. Circuits that do a job, and the range they do it over

The floor a second capacitor removes

Two requirements pulling one capacitor found a regulator whose best transient is one it must not be built with: below 939 milliohms of output-capacitor series resistance the loop has under 45 degrees of margin, and the droop is smallest at 885. The capacitor across the upper divider resistor, added for the reference's sake, dissolves that conflict. At 836.5 picofarads the 45-degree floor leaves the sweep entirely — every series resistance down to a milliohm is stable — and the droop falls 40 per cent at the same time. The band of capacitances that do it runs from 100 picofarads to 5 nanofarads, and above it the conflict returns.

 50 Ω + j100 Ω of line: every power below the nose at two voltages, and a leading load's nose at 1.055 of the source. computed by solving, not by drawing: a load of fixed angle swept in size from open circuit to short circuit on a line of 50 Ω + j100 Ω, with the power it takes against the voltage it is left with, for four load angles. Each curve rises to a most power and turns back while the voltage keeps falling, so every smaller power is delivered at two voltages and every larger one at none. A 30° lagging load reaches 0.4222 of a matched resistive line's power with 0.5221 of the source voltage left; a unity power factor load reaches 0.6180 of a matched resistive line's power with 0.5878 of the source voltage left; a 30° leading load reaches 0.8240 of a matched resistive line's power with 0.7293 of the source voltage left; a 60° leading load reaches 0.9960 of a matched resistive line's power with 1.0553 of the source voltage left. Each nose is found by golden-section search on the solved network, agrees with V²cos φ/(2|Z|(1 + cos(θ − φ))), and falls where the load impedance's magnitude equals the line's. Power, and the part that does no work

The load that has two voltages or none

A load that takes a fixed power takes more current as its voltage falls, and on a line with impedance in it every power below a limit is delivered at two voltages and every power above it at none. The limit sits at the load the maximum-power theorem describes, half the source voltage on a resistive line. On fifty ohms and a hundred of reactance a load leading by sixty degrees reaches that limit with its far end at 1.055 of the source, and at nine tenths of it reads 1.172 — so a far end that reads high is not a far end with margin. On a direct-current bus the lower of the two voltages is not a state at all: one per cent below it the bus runs down to nothing in 3.48 milliseconds.

 Loop gain of a two-pole amplifier closed for a gain of 100. Unity loop gain at 5.73 kHz, where 35.0° of phase remains before −180°. The phase never reaches −180° at any frequency, so there is no gain margin to quote: 2 poles contribute at most 180° and the last of it arrives only at infinity. Feedback, and the margin

The loop that never crosses

Every loop this field draws carries two margins, and one of them is not always a number. Take the third pole out of the standard loop and its crossover moves by 2.45 parts per million and its phase margin by 0.164 degrees — the third pole's own arctangent there, to five decimal places — while its gain margin goes from 46.06 decibels to no number at all. The phase reaches −180° only where the magnitude has already reached −62 decibels, and the two instruments that are supposed to notice report 3.19 × 10⁻⁶ either way.

 The "offset" a switch leaves moves by 1.863 mV across a volt of signal. computed by solving, not by drawing. The charge a switch leaves on a 1 pF hold capacitor, as the input it was sampling changes. It is W·L·C_ox·(V_gs − V_th) with the clock's high level for the gate, so it falls as the input rises for two reasons at once: less gate drive, and a threshold that rises with the source potential through the body effect. The dashed line is the part that does not know what the signal is — the clock coupling through the gate overlap capacitance, -1.649 mV — and it is the only part of this that is honestly an offset. The curve stops where the switch does: an n-channel device passes nothing within a threshold of its own gate, and the model refuses rather than returning a zero. Filters, measured not tabulated

The offset that knows the signal

The charge a switch leaves on the capacitor it was sampling is the gate oxide capacitance times the channel area times the overdrive, and both terms in that overdrive depend on the input — one directly and one through the body effect. So the 5.67 mV a data sheet would call an offset moves by 1.86 mV across a volt of signal, which is a gain error of 0.186 per cent and only 3.4 µV of anything else. Opening the summing-node switch first divides the gain error by the amplifier's own gain, exactly, and what is left goes from twenty-nine times the sampled-noise floor to nineteen times below it.

 A follower fed through 100 nH has an output resistance of -21.9 Ω. computed by solving, not by drawing. The real part of the impedance looking into the emitter, driven by a current source and read, at every frequency. At direct current it is 5.50 ohms, which is the same rₛ/(β+1) + 1/gₘ the follower's first essay computed. Between 110 MHz and 301 MHz it is negative: rπ and Cπ delay the current the transistor sources into the emitter, and past a quarter of a cycle of delay pushing the emitter up makes the device push it up as well. The dashed curve is the same follower with no inductance between the source and the base, and it never goes below zero — the sign belongs to the wire and the transistor together, and to neither alone. Devices, and the amplitude they stop being linear at

The resistance that is below zero

An emitter follower's output resistance is 5.5 Ω at direct current and −21.9 Ω at 257 MHz, and the sign is not the transistor's: with an ideal source at the base there is no negative band at all, and a hundred nanohenries of wire between the source and the base produces one from 110 to 301 MHz. A capacitance resonating inside that band is a resonator with loss of the wrong sign, so 4.7 to 100 pF on the emitter oscillates while 1 pF and 470 pF do not — a band of load capacitance with quiet ground on both sides of it.

 One gain takes the equivalent resistance from 5 kΩ through infinity to negative. computed by solving, not by drawing. A 5 V source drives a node through 10 kΩ; the node also reaches 10 kΩ whose far end is held at A times the node's own voltage. The Thévenin resistance looking into that node is r1 in parallel with r2/(1 − A), which the solve returns to a part in a billion without being told: 5 kΩ at no gain, 10 kΩ at unity where r2 takes no current at all, and unbounded at A = 2.00 where the two conductances cancel. Above that it is negative. The open-circuit voltage follows it, because the short-circuit current is 500.0 µA at every gain — a short across the controlling node leaves the dependent source nothing to be controlled by — so the open-circuit voltage is simply the short-circuit current times whatever the resistance is, and reaches 150.0 volts from a five-volt source inside the range drawn. Networks, and how a solve is checked

The resistor that is not made of the resistors

Exact outside and wrong within reduced six elements to one source and one resistor and found the resistor two ways that agreed to the last bit. Put a dependent source in the network and one of those routes stops working, because setting the sources dead kills the independent ones and leaves the dependent one where it is. On a bootstrap of two ten-kilohm resistors the Thévenin resistance runs from five kilohms through infinity to minus ten, the open-circuit voltage of a five-volt source reaches 225, and above one gain the equivalent's resistor is negative — which the netlist refuses to stamp, correctly, because a negative resistance is a controlled source and not a resistor.

 The guard leaves a negative resistance, and it reaches −1.59 kΩ. computed by solving, not by drawing. The magnitude of the conductance a source sees looking into the input, guarded and not, with 100 pF of cable and a 1.00 MHz amplifier. The unguarded input's conductance is positive everywhere — a capacitance to ground and a leakage to a rail are both losses. The guarded one is negative above 0.0404 Hz, and its magnitude rises as the square of frequency: −15.9 MΩ at 10 kHz, −161 kΩ at 100 kHz, −3.18 kΩ at a megahertz. Above the amplifier's gain-bandwidth product it flattens at ωₜ·C, which is −1.59 kΩ. That is the same input the guard raises to 10¹⁸ Ω at direct current, and nothing about the leakage the guard was installed for appears in it: the negative resistance is a product of the amplifier's bandwidth and the cable it is driving. Measurement, which is a circuit on a circuit

The sign of what the guard gives back

A guard ring is sold on two numbers and they are both about magnitudes: a teraohm of leakage multiplied to 10¹⁸ ohms, and a hundred picofarads of cable bootstrapped out of the way. The guard is also driving that capacitance with a copy of the input that lags it, and a capacitance driven by a lagging copy of its own voltage takes current out of phase with the voltage across it. What the guarded input presents is a negative conductance rising as the square of frequency — −15.9 megohms at ten kilohertz, −3.18 kilohms at a megahertz, flattening at the gain-bandwidth product times the capacitance — and a faster amplifier makes it worse.

 Over one decade a constant is out by ±29.76 mV and a drop plus a resistance by ±8.002 mV. computed by solving, not by drawing. The forward drop of a pure exponential junction over the decade from 1e-3 to 1e-2 amperes, with the best constant drop and the best drop-plus-resistance drawn across it. Both are fitted minimax — the model whose WORST error over the window is smallest, which is what a design has to tolerate — rather than by least squares. The constant is 684.6 mV and is out by ±29.76 mV; the line is 656.2 mV plus 6.61 Ω and is out by ±8.002 mV, which is 3.72 times better. The best constant needs no search: it is the midpoint of the window's highest and lowest voltage, and its error is half their difference. Where the models stop

The straight line between two models

Between a constant seven-tenths of a volt and an exponential sits the model a designer actually reaches for: a drop plus a resistance. Fitted so that its worst error over a decade of current is as small as it can be, it is out by ±8.00 millivolts where the best constant is out by ±29.76 — and both numbers are the same over every decade, because a decade of a logarithm is the same shape wherever it is taken. On a real two-mechanism junction the line does best in the top decade, ±8.71 against a constant's ±58.53, because a series resistance is exactly the term the line has and the constant has none. And the resistance the fit returns is the part's plus 701 milliohms that is not there.

 The thermal network's step response, and the sum of its own stages — 0.95 per cent apart at worst. computed by solving, not by drawing. A step of power into a three-stage thermal ladder, and the junction's rise divided by it. The solid curve is exact: the impedance is a continued fraction in s, its denominator has 3 real negative roots, and the partial-fraction expansion of Z(s)/s is a sum of that many ordinary exponentials — no march, no step size. The dashed curve is the sum every account of a thermal path writes, each stage's own resistance times 1 − exp(−t/RC) with its own local time constant, and it is an approximation because the stages load each other. What that costs is 0.950 per cent, once, at 12.9 ms — between the fastest stage's 2.4 ms and the next one's 200 ms, which is the only place two stages are moving together. It is one-sided: the sum never reads low. Before the steady state

Two ladders the terminals cannot tell apart

A thermal path drawn as a ladder and the same path drawn as a sum of exponentials are called different models of one object, and the difference between them has never been priced because pricing it needs an exact answer. Solved in closed form, the sum is 0.950 per cent high at worst and never low; the marched netlist is right to a part in 21,169; and the largest disagreement in the picture was 2.919 per cent that has nothing to do with heat at all, which reading the curve one sample differently removes.

 A track needs about three heights of copper beside it, and it is the resistance that says so. computed by solving, not by drawing at 100 MHz, each point a strip solve of its own on a 50 mm plane of the same area, moved sideways. The horizontal axis is where the track's centre sits relative to the plane's edge, in units of the track's height above it; negative is a track hanging past the edge with no copper beneath it. With the centre directly over the edge the loop's inductance is 1.161 times its centred value and its resistance 2.90 times, because the return has to crowd into the last few hundred micrometres of copper. Three heights in, the inductance is 1.006 times and the resistance 1.09; ten heights in, both are within 0.7 per cent. Three heights past the edge the inductance is 1.82 times. At direct current every point on this axis is exactly one, because the copper has been moved and not removed. Lines, where a wire has a length

Where the plane runs out

The corner that is three decades wide solved a return current over a plane that extends well past the track on both sides. Where it does not, the two costs arrive at opposite ends of the band: at direct current a plane that ends under the track costs 27 per cent of inductance and not one part in a million of resistance, and above the band it costs 16 per cent of inductance and 199 per cent of resistance. Three track-heights of copper beside the track removes almost all of both, and the number three has no millimetres in it — sixteen times the whole cross-section gives the same ratios to a part in a billion.

 What is warm in a capacitor, by its two loss models. computed by solving, not by drawing. One 100 nF capacitor of loss tangent 0.02, written as a 3.183 Ω resistance in series with it and as a 7.958 kΩ resistance across it — the pair that converts exactly at 10.0 kHz and nowhere else. The noise at the terminals is 4kT times the real part of the impedance, so the two models give the same density at 10.0 kHz and are 33.0 dB apart at 100 Hz and 40.0 dB apart at a megahertz. The dots are the same quantity computed the other way — the resistor split out of the netlist, a source put in its place and the network re-solved — agreeing to 3.3e-16. The reactance itself contributes nothing at either end: a lossless capacitor has no real part and is not warm. The floor, which bounds from below

Only the real part is warm

Johnson's 4kTR is the special case of a statement about impedances: the noise across any passive two-terminal in equilibrium is 4kT·Re{Z}, so a reactance contributes nothing however large it is. That turns a modelling convenience into a noise figure. A 100 nF capacitor of loss tangent 0.02 written as 3.183 Ω in series and as 7.958 kΩ across it — the pair that converts exactly at 10 kHz — gives 0.226 and 10.10 nV/√Hz at 100 Hz, 33 dB apart, and 40 dB apart the other way at a megahertz.

 The lower corner, estimated from short-circuit time constants. computed by solving, not by drawing, on three coupling capacitors and three shunt resistors at 28 spreads of the capacitor values. Each capacitor's short-circuit time constant is its own value times the resistance between its terminals with the other two shorted; the sum of the RECIPROCALS is 60000 s⁻¹ here, and it equals the ratio of the denominator's two highest coefficients to 1.4e-12 and the negated sum of the poles to 1.4e-12. That much is the same theorem as the other end. What is an estimate is the corner: 9549 Hz against a measured 8192 Hz, high by 16.6%. It is high at every spread drawn — the error reverses direction with the construction, so both ends of a band are estimated inwards. Before the steady state

Shorted instead of opened, and the error changes sign

The same construction with the other capacitors shorted rather than removed sums the reciprocals of the products, and that sum is the ratio of the denominator's two HIGHEST coefficients — the negated sum of the poles, exact to a part in 10¹². Divided by 2π it estimates the lower corner of a band, and it is 16.6 per cent HIGH with three coupling capacitors and never once low. Two settings of the slider give the same three time constants in a different order, the same sum, and corners two per cent apart.

 1 pF across a 50 Ω line: a dip of 0.320 V and an area of 25 ps. computed by solving, not by drawing as a cascade of two-ports, with a raised-cosine edge of 59 ps sent into it. The incident edge is the faint curve; what comes back is the shaded dip and what goes on is the third. The dip reaches -0.3202 V and its area is 25 ps, which is Z₀C/2 to a part in ten thousand. Driven by an edge fifty times faster the same cascade returns the single exponential the closed form gives, to 9.3e-6 of a volt. The transmitted edge leaves at 81.1 ps, against 59 ps arriving. Lines, where a wire has a length

The dip whose area is fixed

A picofarad across a 50 Ω line makes a dip in what comes back. Its depth is 0.833 volts to a six-picosecond edge and 0.0196 volts to a nanosecond one, forty-two times less; its area is 25 picoseconds to both, to six parts in a hundred thousand, because the area is Z₀C/2 and contains nothing about the edge. Two half-picofarad discontinuities too close to tell apart read as exactly one picofarad, and so do two far enough apart to be separate — the area is additive where the depth is not. And what a reflectometer calls the capacitance of an impedance step is the step's real excess capacitance times 1 + Z/Z₀.

 Two large-signal limits, each alone and then both, at 20 mA and half of it. computed by solving, not by drawing. Four marches of one netlist at each load: neither limit, the input pair's tanh alone, the output stage's 20 mA alone, and both, driven by a 10 mA step. The three curves are each limit's departure from the linear march and the departure with both present; the faint line is the two singles added. Both lies on the sum and a little above it — 1.112 times it at 0.47 nF and 1.022 at 47 nF — so the limits are present together rather than taking turns. Where the two singles cross, near 10 nanofarads, the pair costs 1.88 times what the worse of them costs alone. Feedback, and the margin

The load that neither limit owns

Nine rungs of this argument asked which of an amplifier's two large-signal limits binds, and drew the load capacitance where the answer changes hands. Both are present at every load: the excursion with both in the netlist is the two departures added and between 2 and 12 per cent more, never the larger of them. So the crossing is not a handover but a maximum — at 12 nanofarads the pair costs 2.084 times what the worse of them costs alone, against 1.35 at 2.2 nanofarads and 1.07 at 47 — and the same peak sits on the resistance axis at 20 ohms and the gain-bandwidth axis at 50 megahertz.

 The heat a core makes against the heat its path removes, and the two temperatures where they are equal. computed by solving, not by drawing. The rising straight line is what the thermal path can carry away at a temperature — (T − 25)/45 watts, a line because a thermal resistance is a resistance. The curve is what the wound part actually dissipates at that temperature, marched from a hysteresis loop at a material whose saturation flux and permeability both move with temperature. They cross twice. The lower crossing at 88.8 degrees is the operating point and its loop gain is -0.192 — negative, so the core is a stabilising feedback and not a destabilising one. The upper crossing at 191.1 degrees is an ignition temperature: above it the part cannot get rid of what it makes. The slider moves the thermal resistance. Two windings, and the band between them

The loss that depends on what it causes

Every thermal figure in this collection has had the power handed in. A ferrite's has no business being: its saturation flux falls with temperature, its permeability rises, and both move the loss. Closing that loop makes the temperature a fixed point rather than a product — and the fixed point has a stable root at 89 degrees whose loop gain is negative, an ignition root at 191 whose loop gain is 120, and a thermal resistance of 183 kelvin per watt at which the two touch and neither exists.

 An L-section from 50 Ω to 1 kΩ: Q 4.359, fixed by the two resistances, and a band of 4.73%. computed by solving, not by drawing. A series inductor and a shunt capacitor matching 50 Ω to 1 kΩ at 1.00 MHz, their values from the series–parallel conversion: the load with the capacitor across it is 50 Ω in series with a reactance of 217.9 Ω at the design frequency, and the inductor cancels the reactance. The reflection there is 3.6e-16. The section's Q is √(20 − 1) = 4.3589 and no choice of parts changes it. |Γ| stays under a tenth from 976 kHz to 1.02 MHz, 4.73% of the design frequency, against 0.2/Q = 4.59%; and under half the power from 727 kHz to 1.21 MHz, 48.53%, against 2/Q = 45.88%. Frequency, which is the same solve

The match with no knob

An L-section — a series inductor and a shunt capacitor — is the series–parallel conversion used on purpose: a load with a capacitor across it is, at one frequency, the source's resistance in series with a reactance an inductor cancels. Matching fifty ohms to a kilohm that way reflects 3.6 × 10⁻¹⁶ at its design frequency and has a Q of √19 = 4.359 that no choice of parts can change, so it holds its reflection under a tenth over 4.73 per cent of band whatever it is built from. The band is 0.2/Q to within three per cent, it depends on nothing but the ratio, and only splitting the match widens it: 14.98 per cent in two sections, 30.34 in three — and 30.83 in four.

 The images a zero-order hold leaves, at 0.222 of the sample rate. computed by solving, not by drawing. A 10.67 kHz tone held at 48 kHz, with every line read out of a transform of the staircase itself. The sampled spectrum repeats at every multiple of the clock and the hold multiplies all of it by one sinc, so each image survives scaled by the sinc at its own frequency: the fundamental at -0.72 dB, the largest image (1fs−f, 37.3 kHz) at -11.60 dB, which is 10.88 dB of rejection. The sinc's nulls are exactly at the multiples of the clock and the two first-order images straddle the first of them without touching it — so the hold's rejection is 25.6 dB for a tone at 0.05 fs and 1.74 dB for one at 0.45, falling to nothing at half the clock. Measured and closed form agree to 0.026%. Where a signal becomes a number

The nulls are where nothing is

A zero-order hold multiplies the whole repeated spectrum by one sinc, so it attenuates every image at the image's own frequency and its nulls land exactly on the multiples of the clock. Nothing is ever at a null: the two first-order images straddle it, and the closer the signal comes to half the clock the closer they come to each other. The hold gives 25.6 dB of image rejection to a tone at a twentieth of the clock, 1.74 dB at 0.45 of it, and nothing at all at half — which is the frequency the band most needs it at.

 The corner a 2 nH shunt has, against the current it is sized for. computed by solving, not by drawing. A shunt held at the best burden voltage of 7.75 mV has R = u⁄I, so its own 2 nH of series inductance puts a corner at u⁄(2πLI) — 616 kHz at an ampere and 6.16 kHz at a hundred, for the same piece of metal. The optimum that contains no current at all therefore hands the bandwidth a current dependence: the corner falls in exact proportion. At 1 A the shunt is 7.75 mΩ with a time constant of 258.2 ns, so a 10 ns edge is read 2.58e+3% high and a 1 ns edge 259.20 times too large. A resistor and a 258.2 pF capacitor across it — the value found by search on the solved response, agreeing with L/(R·Rc) to 7.9e-5% — flatten the reading to 7.8e-5% across six decades, and a fifth too much makes it ten times worse. The dots are the corner bisected on the solved impedance rather than taken from R/2πL. Measurement, which is a circuit on a circuit

The optimum that hands back a bandwidth

The best burden voltage across a shunt is 7.75 mV and contains neither the current nor the resistance, which is what made it worth having. A shunt has two nanohenries whatever it is made of, so holding the burden fixed fixes the resistance at u*/I — and the corner R/2πL then falls in exact proportion to the current: 6.16 MHz at a tenth of an ampere, 616 kHz at one, 6.16 kHz at a hundred. A resistor and a 258.2 pF capacitor across it, found by search on the solved response, flatten the reading to 8×10⁻⁵ per cent across six decades.

 One output in saturation takes 6.5% off another that is at five volts. computed by solving, not by drawing. A three-transistor mirror: a reference, an output taken down into saturation, and a third output held at five volts throughout. The upper curve is the saturating output's own loss and the lower one is the sibling's. At 50 mV the saturating output is 27.3 per cent down and the sibling, which is nowhere near saturation, is 6.54 per cent down. The reference current moves by -0.0188 per cent, which is nothing: a base current is a hundred and fiftieth of a collector current and the reference is set by a resistor from the supply. What does move is the base-emitter voltage every output shares — by -1.75 millivolts — and every output is an exponential of it. Devices, and the amplitude they stop being linear at

The refusal, and what it was protecting

A current mirror's model has declined to answer below two hundred millivolts of collector-emitter voltage since it was written, because it has no base-collector junction and would return a forward-active current for a saturated transistor. Put the junction in and the refusal turns out to have been placed where the model it protects is still right to seven parts in ten thousand — and the mirror's real failure is somewhere else entirely: a saturated output takes six and a half per cent off an output that is sitting at five volts.

 A half-decibel filter that meets one decibel at 25 °C and does not at 89. computed by solving, not by drawing. The passband ripple of a fifth-order 0.5 dB Chebyshev built from three Sallen–Key sections, against temperature, with a part whose gain-bandwidth at 300 K is 300 kHz against a 1 kHz corner. Not one passive component has a temperature coefficient in this model. The ripple is 0.824 dB at −40 °C and 1.049 at +125, and it crosses a one-decibel specification at 88.6 °C — a boundary in temperature, which every other edge in this collection is not. The dashed line is the same filter with a tail current proportional to absolute temperature: 0.916 dB at both ends, and no crossing anywhere. Filters, measured not tabulated

The ripple that is a temperature

An amplifier a hundred times the corner leaves a section's quality factor two per cent high. That two per cent has a temperature in it: with a tail current a resistor sets, the transconductance falls as one over absolute temperature, and a fifth-order half-decibel design whose passives have no temperature coefficient at all goes from 0.82 decibels of ripple at minus forty to 1.05 at a hundred and twenty-five — crossing a one-decibel specification at 89 °C. With the other bias it does not move at all.

 Three amplitudes, all of them "one per cent wrong". computed by solving, not by drawing. An exponential driven by a sinusoid has I₀(a) as its mean, 2I₁(a) as its fundamental and 2Iₙ(a) as its harmonics, all checked here against a numerical transform of the waveform itself, agreeing to 9.0e-11. Each gives a different one-per-cent boundary at 27 °C: 1.03 mV for the second harmonic, 5.17 mV for the shift in the operating point the model was linearised about, and 7.30 mV for the gain — which is 1 : 5 : 5√2 at this criterion, and the largest of them is the one usually quoted. The spacing is not a property of the device: the second harmonic is first order in the amplitude and the other two are second, so tightening the criterion to a part in ten thousand spreads the same three to 1 : 50.0 : 70.71. At a drive of one thermal voltage the bias current is 26.6% above quiescent, which is the boundary nobody counts because it moves the thing the model was built at rather than what the model predicts. Where the models stop

Three amplitudes, all of them one per cent

The amplitude at which linearising an exponential is one per cent wrong is 7.30 mV, and that is a statement about the gain. Two other quantities are also one per cent wrong somewhere: the second harmonic reaches one per cent at 1.03 mV and the shift in the operating point the model was linearised about reaches it at 5.17 — which is √2 below the gain boundary exactly, because the mean goes as a²/4 and the fundamental as a²/8. And the spacing is not a property of the device: tighten the criterion to a part in ten thousand and the same three spread to 1 : 50 : 70.7.

 What flatness costs, in the two places it can be bought. computed by solving, not by drawing. The hold's sinc across a band ending at 0.40 of the sample rate, and the same sinc with a one-zero one-pole shelf fitted to its reciprocal over that band. The droop to be removed is 2.420 dB. Corrected digitally the band comes flat exactly and the flat level sits 2.420 dB below what the uncorrected converter gave at direct current, because nothing may exceed full scale — the price is the disease. Corrected in analogue the band comes flat to 0.2308 dB and the shelf is still rising where the images are, so the worst image at 0.60 fs comes up by 3.799 dB, which is 1.57 times the droop it removed — and that boost is between 3.4 and 4.7 dB at every band on the slider, while the droop it cures runs from 0.58 to 3.75. Neither correction changes the signal-to-noise ratio, because the droop never cost any. Where a signal becomes a number

Flatness, and the two currencies it is bought in

The hold's droop takes the signal and everything arriving with it down together, so it costs no signal-to-noise ratio at all — a fact that is never stated and settles what correcting it can possibly be worth. Corrected digitally the price is headroom and is exactly the disease: 2.42 dB of flatness for 2.42 dB of output level. Corrected by an analogue shelf the price is image rejection and is nearly a constant: between 3.4 and 4.7 decibels whatever the band, so it is eight times the droop at a fifth of the clock and nine tenths of it at forty-nine hundredths.

 Four threshold densities, all matched to the same measured coercivity. computed by solving, not by drawing. Every hysteresis figure below this one used a uniform density of play-operator thresholds, and the reason given was that a uniform density makes the small-signal loss law come out cubic — Rayleigh's law, which is what soft ferrites do. That is a real constraint and it is a weak one. These four are all bisected to a measured coercivity of 9 amperes per metre on a deeply driven loop, and they need threshold spreads from 30.1 to 76.0 amperes per metre to get there. The slider moves the coercivity they are all matched to. Two windings, and the band between them

The distribution the bench cannot see

Six rungs of this ladder assumed a uniform density of operator thresholds, and the reason given was that a uniform density makes the small-signal law come out cubic. That reason is weaker than it looks: every density finite at the origin does the same. Matched on a measured coercivity, four distributions whose spreads differ by 2.5 times give the same remanence to 0.6 per cent, the same loop area to 1.0, and the same branch-slope ratio to 0.54 — and small-signal exponents from 2.81 to 3.81. The major loop cannot see the shape and the small signal cannot see anything else.

 What an L-section costs when its parts have a quality factor of 100. computed by solving, not by drawing. The same L-section as the ideal one, with each component given a series resistance of its own reactance over 100, and the efficiency read off a solve rather than from an expression. The section circulates Qₛ times the load's current through its own parts, and Qₛ is √(ratio − 1) with nothing left to choose, so the loss is TWICE Qₛ/100 — once in the inductor and once in the capacitor — to 4.40% wherever it is small. The consequence is that a match starts to cost something at a ratio nobody would call demanding: one per cent at a ratio of 1.253 — which is fifty ohms to 62.7, and is 1 + (Q/200)² to 0.25%. Splitting the match buys efficiency only above a ratio of 10.0: at a ratio of three one section loses 2.76% against 3.35% in two, and at a hundred 16.6% against 11.2%. The best number of sections for efficiency is 2 at a ratio of twenty and 4 at a ratio of a thousand — which is not the answer the band gives, where the band keeps improving with every section. Frequency, which is the same solve

The efficiency a fixed Q costs

An L-section's quality factor is √(ratio − 1) with nothing left to choose, and the same fixed Q that decides its band decides what it dissipates. The circulating current is Q times the load's and it goes through both components, so the loss is twice Qₛ over the components' own Q — measured to 0.03 per cent. With parts of Q 100 that is one per cent at a resistance ratio of 1.253, which is fifty ohms to sixty-three. And splitting the match buys efficiency only above a ratio of 10.02: below it a second section adds two more lossy parts for less than it takes off anybody's Q.

 The direct-current error a bigger feedback resistor does not fix. computed by solving, not by drawing, at direct current with the amplifier in the netlist. A 1.0 nA photocurrent into 100 MΩ gives -0.100 V. The bias current flows in the same resistor, so its contribution is the ratio of the two currents — 0.1000% at 25 °C, and the same percentage at 1 MΩ and at 1 GΩ, which is checked by changing the resistor rather than by reading an expression. The amplifier's 100 µV of offset behaves the other way: it is multiplied by one plus the resistor over the diode's own leakage, so it grows with the resistor and with temperature. Both terms double every ten kelvin, for two different reasons, and at 85 °C they are 6.40% and 0.740% against 0.100% and 0.110% at 25. Feedback, and the margin

The error a bigger resistor cannot help

Every other quantity in a photodiode amplifier improves with a larger feedback resistor: the signal grows as R and the resistor's own noise as √R, so the ratio goes as √R. The bias current does not behave like either — it flows in the same resistor the signal does, so its contribution is the ratio of the two currents with the resistance cancelled, 0.1 per cent at 25 °C and the same 0.1 per cent at a megohm and at a gigohm. The offset behaves the other way, growing as one plus the resistor over the diode's own leakage, and at 85 °C the two are 6.40 and 0.74 per cent against 0.10 and 0.11 at room temperature.

 The floor a biased resistor is not standing on. computed by solving, not by drawing. A 100 kΩ resistor with 10 V across it. Its Johnson noise is 40.02 nV/√Hz and does not depend on the voltage; its excess noise is a 1/f density proportional to that voltage, 0.1 µV per volt per decade here, and the two are equal at 271 Hz. Below that the resistor is dominated by a generator that is a property of how it was made rather than of its resistance, and the crossover moves with the SQUARE of the voltage. Splitting the same total resistance into eight parts in series leaves the Johnson noise exactly where it was and divides the excess by √8 = 2.828, taking the total at 1 Hz from 660.2 to 236.4 nV/√Hz. With no voltage across it the second generator is absent rather than small. The floor, which bounds from below

The floor that is only a floor while nothing flows

Johnson noise depends on a resistance and a temperature and on nothing else, which is what makes it a floor. A real resistor has a second generator that depends on how it was made and on the voltage across it: 0.1 µV per volt per decade for a metal film, rising as 1/√f. On 100 kΩ with 10 V across it the two are equal at 271 Hz, and the crossover moves with the SQUARE of the voltage — 0.678 Hz at half a volt, 2.44 kHz at thirty. Splitting the same total resistance into eight parts in series divides the excess by exactly √8 and leaves the Johnson noise where it was.

 A permittivity quoted as one number falls 0.645 across five decades. computed by solving, not by drawing. The real part of the relative permittivity against frequency for FR-4 (ε′ = 4.4 at 1 GHz, tanδ = 0.02), from a continuum of relaxations spread uniformly in log-frequency — the arrangement that makes the loss tangent flat. The dashed line is the single number a datasheet quotes. FR-4: 4.787 at 1 MHz and 4.142 at 100 GHz, a fall of 14.7 per cent. Nothing here is fitted: the slope is what a flat loss tangent forces. Lines, where a wire has a length

The permittivity a loss forbids

A datasheet quotes a relative permittivity and a loss tangent as two independent numbers, and they are not two numbers. A material that dissipates has a permittivity that falls logarithmically with frequency at a rate its own loss fixes — 0.129 of permittivity a decade for FR-4, so the 4.4 quoted at a gigahertz is 4.79 at a megahertz and 4.14 at a hundred. Three hundred millimetres of track loses 62.9 picoseconds of delay between 100 MHz and 10 GHz, which a constant permittivity puts at 1.3; and the constant-permittivity model smears an edge backwards, taking 180 picoseconds to reach half height and 133 more to reach nine tenths.

 What a high-side shunt's optimum is made of, at 100 dB of rejection. computed by solving, not by drawing. The same two errors as a low-side shunt, with the amplifier now standing at the rail rather than at the return. 100 dB of rejection turns 12 V of common mode into 120.0 µV of equivalent input error, which is 24 times the amplifier's own 5.0 µV of offset. The optimum keeps its form — the geometric mean of an input error and the supply, golden-sectioned on the solved worst case rather than substituted — and changes its value: 38.73 mV of burden and 0.6440% of error, against 7.75 mV and 0.129% low-side. A shunt sized by the low-side answer reads 1.678% wrong. With the common-mode term removed the optimum returns to the low-side value exactly, which is what says the term is the whole of the difference. Measurement, which is a circuit on a circuit

The rail that is an input error

The best burden voltage across a shunt is the geometric mean of the amplifier's offset and the supply, and moving the shunt to the high side does not change that form — it changes what the offset is. A hundred-decibel amplifier on a twelve-volt rail turns the rail into 120 µV of equivalent input error, twenty-four times its own five, so the optimum moves from 7.75 mV and 0.129 per cent to 38.73 mV and 0.644. The two are equal only at 128 dB, and with the common-mode term removed the optimum returns to the low-side value exactly.

 The series resistance that makes the ripple smaller. computed by solving, not by drawing, marched with the diodes in the netlist. A 1000 µF reservoir with 30 mΩ of its own series resistance. The resistance adds a step of ESR times the diode's peak current to the output and at the same time limits that peak current, and the two nearly cancel: the ripple has an interior minimum of 1.330 V at 17.0 mΩ, BELOW the 1.331 V a perfect capacitor gives, and rises to 1.711 V at an ohm. What the resistance buys monotonically is the peak current: the crest factor falls from 13.37 to 6.550 at an ohm, which more than halves the current that sizes the transformer, for 380 mV of mean output and a root-mean-square diode current that falls from 0.4717 A to 0.3484. The textbook ripple expression says 1.568 V here and moves by 2.4% across the whole axis, because it has no term for a series resistance at all. Circuits that do a job, and the range they do it over

The resistance that lowers the ripple

Two earlier essays here marched a reservoir with a perfect capacitor. A real one has tens of milliohms of its own, and the obvious expectation — that the resistive step it adds makes the ripple worse — is wrong in an interesting direction: the resistance also limits the charging current, and the ripple has an interior minimum of 1.3303 V at 17.0 mΩ, below the 1.3312 V a perfect capacitor gives. What the resistance buys monotonically is the peak current, which falls from 13.37 times the load's to 6.55 at an ohm, for 380 mV of mean output.

 A diode thermometer measuring its own sense current, at 600 K/W. computed by solving, not by drawing. The same fixed point as the core and the thermistor, on a junction: the dissipation is I·V and V falls with temperature, so the loop gain is negative and the equation has one root at every current. What it costs is two errors. The junction sits above ambient by 0.29 millikelvin at a microamp and 4.26 kelvin at ten milliamperes, which a calibration removes; and a kelvin of ambient produces less than a kelvin of junction, by 1/(1 − R_th·dP/dT), which it does not. After a calibration at 25 degrees the reading at 85 is out by -583 millikelvin at ten milliamperes and -0.09 at a microamp. The coefficient itself moves too — -2.403 against -1.613 millivolts a kelvin — so a quoted tempco carries a sense current as well as a junction. Devices, and the amplitude they stop being linear at

The sensor inside its own answer

A junction driven from a current source cannot run away, because its forward voltage falls with temperature and its loop gain is therefore negative. What that costs is a thermometer that is warmer than what it is measuring by 4.26 kelvin at ten milliamperes, and — the part a calibration cannot remove — under-reports every change in ambient by 9,584 parts per million, because the sense current's own dissipation falls as the reading rises. Calibrated at 25 degrees, it is out by 583 millikelvin at 85.

 Three dividers in a row, and where the error actually is. computed by solving, not by drawing. Three two-resistor dividers cascaded with nothing between them. The product of their ratios is 0.1250 and the solved output is 0.076923, 38.5% low at a staircase of ×1. Decomposed stage by stage with the rest of the chain in place — and the product of those three is the answer exactly — they are 0.3846, 0.4000, 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is at the front, which is the opposite of where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion, fitted exponent -0.993, so a decade a stage is within 4.88% and two decades within 0.50% — which is why a chain that has to be right is built as a staircase and not out of one value repeated. Networks, and how a solve is checked

The stage that is wrong is the far one

Three identical ten-kilohm dividers in a row give 0.076923 rather than the product of their ratios, 0.125 — 38.5 per cent low. Decomposed stage by stage with the rest of the chain in place, and the product of those three is the answer exactly, they are 0.3846, 0.4000 and 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is entirely at the front, which is the opposite end from where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion — fitted exponent −0.993.

 What a pair does to the three boundaries: removes two, moves one. computed by solving, not by drawing, with every harmonic taken from a transform of the waveform rather than from a series. A differential pair's transfer is an odd function, so the mean and every even harmonic are zero — -3.2e-17 and 9.1e-17 at a drive of two thermal voltages, which is absent rather than small. Two of the three one-per-cent boundaries therefore do not exist for a pair at all. The third, the loss of gain, moves by √2 in the limit of a tight criterion and by 1.42610 at the one per cent quoted here — 10.42 mV against 7.304 mV at 27 °C, against √2 = 1.41421 and 1.41433 at a criterion a hundred times tighter. So a pair reached for as headroom has bought forty per cent of it and a pair reached for as linearity has bought something else entirely. Twenty millivolts of imbalance brings the even orders straight back, which is what says the cancellation belongs to the symmetry rather than to the topology. Where the models stop

Two boundaries removed, and one moved

A differential pair's transfer is odd, so its mean and every even harmonic are zero — −3.2×10⁻¹⁷ and 9.1×10⁻¹⁷ at a drive of two thermal voltages, absent rather than small. Two of the three one-per-cent boundaries therefore do not exist for a pair at all. The third, the loss of gain, moves by √2 in the limit and by 1.42610 at the one per cent usually quoted: 10.42 mV against 7.304. So a pair reached for as headroom has bought forty per cent of it, and twenty millivolts of imbalance brings the even orders straight back.

 Where the bandwidth estimate stops being conservative. computed by solving, not by drawing. A Sallen–Key low-pass at unity gain, its quality factor swept by the ratio of its two capacitors. The sum of its open-circuit time constants is 2RC₂ and nothing else — the feedback capacitor sees zero resistance — so the estimate is 7957.7 Hz at every setting while the measured corner walks down past it. Below a quality factor of √2 the estimate is low, as it is on every network with real poles; above it the estimate is HIGH, by 6.45 times at a Q of ten. The crossing, bisected on the solved response, is at 1.414213032 against √2 = 1.414213562, and the estimate is at its worst at the Butterworth value 1/√2 where it is low by exactly 1 − 1/√2 = 29.29%. Before the steady state

Where the estimate stops being a bound

The sum of open-circuit time constants is never optimistic on a network with real poles, and the claim is about the network rather than about the theorem. On a second-order section the ratio of the estimate to the truth is Q/√(k + √(k²+1)) with k = 1 − 1/2Q², which is exactly 1/√2 at the Butterworth quality factor — its worst point, 29.29 per cent low — and exactly 1 at a quality factor of √2. Above that the estimate is high, by 6.45 times at a Q of ten, and the crossing bisected on the solved response is 1.414213 against 1.414214.

 What an oversampling ratio buys, and at two different rates. computed by solving, not by drawing. A 20 kHz band on a 48 kHz base clock, interpolated by ratios from 1 to 64. The hold's droop at the band edge falls with the SQUARE of the ratio — the fitted exponent over six doublings is -2.0113 — from 2.640 dB at the Nyquist rate to 0.0097 at sixteen times it. The nearest image moves out with the FIRST power, exponent 1.0222, from 1.40 times the band edge to 37.4. So one decision buys two things at rates differing by a factor of two in the exponent, and the third quantity — the poles a reconstruction filter needs for sixty decibels — collapses from 20.5 to 3.21 by a ratio of four alone. Where a signal becomes a number

One knob, and the two exponents it turns

Oversampling is quoted as buying one thing and buys two that improve at different rates. The hold's droop at the band edge falls with the SQUARE of the ratio — fitted exponent −2.011 over six doublings, from 2.640 dB at the Nyquist rate to 0.0097 at sixteen times it — while the nearest image moves out with the first power, exponent 1.022. The third quantity, the poles a reconstruction filter needs for sixty decibels, collapses from 20.5 to 3.21 by a ratio of four alone, because it is a logarithm of the second.

 The winding window solved in two dimensions, copper filling 100% of it. computed by solving, not by drawing. The grey frame is iron of infinite permeability, which in this formulation is a Neumann boundary — flux enters it at right angles and pays nothing. The thin curves are flux lines, which are contours of the vector potential, so equal spacing is equal flux. The copper is shaded by its own share of the loss. At 100 per cent fill the solved ratio is 16.280 against Dowell's 16.382, and the difference is entirely the flux that curls round the ends of the foils — which the one-dimensional model has no way to hold. Two windings, and the band between them

The assumption that is a geometry

Every alternating-resistance number this collection has computed for a winding rests on one sentence — the field is parallel to the layers everywhere — and the sentence has never been tested, because testing it needs a field. Solved as one, a portion of foils that fills its window returns Dowell's expression to 0.155 per cent; the same copper filling a quarter of it returns 9.00 against the expression's 16.38, and dissipates 0.528 watts a metre against 0.232. The ratio falls by 45 per cent and the loss more than doubles.

 The straight lines report 6.02 dB of gain margin on a loop that has none. computed by solving, not by drawing. The gain and phase of a loop made of an integrator and a pair at Q = 2, the corner at 1.00 kHz, the integrator set so that the straight-line asymptotes cross unity at 500 Hz. The loop's phase passes −180° at 1.00 kHz. There the lines put the loop gain at −6.02 dB and the solve at 0.00 dB, so the gain margin they report is 6.02 dB against 0.00 dB. The difference is 6.0206 dB, which is 20 log Q exactly with Q = 2, at any integrator gain. Closed, the loop is on the edge: the largest real part among its poles is -5.87e-17 of the corner's angular frequency. Frequency, which is the same solve

The gain margin the straight lines get exactly wrong

A phase margin read off the straight lines is wrong by an amount that depends on where the loop crosses unity. A gain margin read off them is not: for an integrator and a pole pair the phase passes −180° at the pair's own frequency whatever the gain, and the lines are out there by exactly 20 log Q — 6.0206 dB for two real poles, 13.98 dB at a quality factor of five, at every integrator gain drawn. The stability condition for the loop turns out to be the same inequality: it is stable exactly when the gain margin the lines report exceeds that error.

 With its own capacitance at B|Z| = 0.2, a line's nose for a unity-power-factor load moves from 0.618 to 0.658 of a matched line's power, at 0.635 of the source. computed by solving, not by drawing: a unity-power-factor load swept in size from open circuit to short circuit on a line of 50 Ω + j100 Ω carrying its own shunt capacitance as a π, half at each end, with the total susceptance stated as B|Z|. At B|Z| = 0 the nose is 0.6180 of a matched resistive line's power with 0.5878 of the source voltage left, and the unloaded far end reads 1.0000. At B|Z| = 0.1 the nose is 0.6376 of a matched resistive line's power with 0.6108 of the source voltage left, and the unloaded far end reads 1.0465. At B|Z| = 0.2 the nose is 0.6582 of a matched resistive line's power with 0.6353 of the source voltage left, and the unloaded far end reads 1.0969. At B|Z| = 0.4 the nose is 0.7025 of a matched resistive line's power with 0.6895 of the source voltage left, and the unloaded far end reads 1.2107. Each nose is found by golden-section search on the solved network and agrees with the nose of the Thevenin equivalent V/(1 + jBZ/2) behind Z/(1 + jBZ/2). Power, and the part that does no work

The headroom that is the line's own charge

A line of 50 + j100 ohms delivers at most 0.6180 of a matched resistive line's power to a unity-power-factor load, with 0.5878 of the source voltage left. Give the line its own shunt capacitance — a π, half at each end, B|Z| = 0.4 in all — and the nose moves to 0.7025 at 0.6895: 13.7 per cent more power and 17.3 per cent more voltage. It is headroom, but not the headroom the far end advertises, which with no load rises 21.1 per cent. The capacitance turns the source and line into a Thevenin equivalent with more voltage behind more impedance, and it moves a voltage threshold's meaning in opposite directions depending on whether it is referred to the source or to the unloaded far end.

 The band closes over a stage's own bias at 7.93 microns. computed by solving, not by drawing. The overdrive over which the square law is within 1.0 per cent, against the channel length that sets it — the velocity-saturation voltage is Ec·L, so the axis is a size and not a bias. The shaded region is the band; the curve through it is the overdrive at which the square law is exact, which exists at every length because the subthreshold and velocity-saturation errors have opposite signs. Both edges move: the lower one from 79.6 mV to 252.9 mV and the upper from 81.2 mV to 1200.0 mV, so the band is a factor of 1.021 at 0.050 µm and 4.7 at 30 µm. The fourth curve is the overdrive a common-source stage with a fixed gate voltage and a fixed source resistor solves to, which barely moves at all; it leaves the band at 7.929 microns and is outside it for every shorter device. What the square law would have said about that stage is the last two rows: 220 per cent too much current on the 0.050 µm device and 53 per cent too much efficiency, against 0.15 and 0.66 per cent at 30 µm. Devices, and the amplitude they stop being linear at

The length that is a voltage

The square law's band is closed from above by a parameter that is not a bias, a current or a temperature: it is the channel length, wearing a voltage's units. Swept, the band goes from a factor of 4.74 on a thirty-micron device to 1.021 at fifty nanometres — and the overdrive a stage actually biases itself to barely moves at all, so the two cross at 7.93 microns and every shorter device is biased outside the band. The band was also measured in the wrong quantity: the square law is exact in the current somewhere at every length, and its error in the transconductance is never below 42.5 per cent at fifty nanometres and reaches one per cent only above 6.502 microns.

 The shunt's resistance as a function of what it is measuring. computed by solving, not by drawing, as a fixed point: the shunt dissipates I²R, its temperature rises by 20 K per watt, and at 50 ppm/K its resistance rises with its temperature — so the resistance the reading is divided by depends on the reading. Iterated to convergence it agrees with the closed form R₀/(1 − αθI²R₀) to 2.2e-16. Along the burden-voltage optimum, where R = u⁄I, the dissipation is I·u rather than I²R, so the temperature rise is 155 mK per ampere and the error is the FIRST power of the current — fitted exponent 1.0007 over five decades. That is the only one of the shunt's errors with the current in it, and it puts a term in I² into the reading, which is a curvature no single-current calibration removes. The upper curve is a shunt of fixed resistance, where the error is quadratic. The fixed point stops existing at 129 kA and never at a current a shunt will see. Measurement, which is a circuit on a circuit

The resistance that depends on the reading

Three of a shunt's errors are free of the current being measured, which is the whole content of the burden-voltage optimum. The fourth is not: the shunt dissipates, warms, and its resistance rises — so the divisor the reading uses is a function of the reading. Solved as a fixed point it agrees with R₀/(1 − αθI²R₀) to 2×10⁻¹⁶, and along the optimum, where the dissipation is I·u* rather than I²R, the error is the FIRST power of the current: 7.75 ppm at an ampere, 775 at a hundred, fitted exponent 1.0007.

 Stepped at 20 of its time constant, a 1 µs pole rings between 1.818 and 0.331 V, and needs 23 steps to settle. Marched with the trapezoidal rule at a step of 20.0 µs. A 1 µs pole (1 kΩ, 1 nF) drives, through a unity buffer, a 1 ms pole (1 kΩ, 1 µF). The fast node's exact response reaches its final volt within a few microseconds; the march's first values are 1.8182, 0.3306, 1.5477, 0.5519, 1.3666 V. Its distance from its final volt is multiplied by (1 − h/2τ)/(1 + h/2τ) = −0.8182 every step, measured and checked against that form, so it changes sign every step and takes 23 steps to fall below 1% — 460 µs. The slow node it drives is 1.23e-5 V from exact at 1 ms, because a 1 ms pole averages an alternation at half the stepping rate to nothing. Before the steady state

The ringing that belongs to the rule

The trapezoidal rule is stable for every stable circuit and every step size, and it is not damping. March a one-microsecond pole with twenty-microsecond steps and its node reads 1.818, 0.331, 1.548, 0.552 volts — an oscillation at half the stepping rate, its distance from the final volt multiplied by exactly −0.8182 every step, taking twenty-three steps to fall below one per cent. The slow node that pole drives is right to 1.2 × 10⁻⁵ V at a millisecond. One backward-Euler step at the discontinuity cuts the first swing from 0.818 V to 0.048 and two to 0.0023, because backward Euler multiplies the same error by 1/(1 + h/τ) and the trapezoidal rule by (1 − h/2τ)/(1 + h/2τ), which approaches −1.

 The best shunt switch for a T is 3.8×, 0.33×, 0.082× a series switch from sources of 0 Ω, 50 Ω, 1 kΩ. computed by solving, not by drawing. The frequency at which the band of a T closes — where no load leaves it within 1% of ideal in both states, the closed state counted as a shortfall in amplitude — against the size of its shunt switch, as a multiple of the 0.5 Ω, 100 MΩ, 5 pF series switches, with every conductance and the capacitance scaled together. From a 0 Ω source the band closes latest with a shunt 3.775 times the series switch, at 2.40 GHz, against 702 MHz with three identical switches — 3.42 times later. From a 50 Ω source the band closes latest with a shunt 0.325 times the series switch, at 243 MHz, against 89.8 MHz with three identical switches — 2.71 times later. From a 1 kΩ source the band closes latest with a shunt 0.082 times the series switch, at 59.4 MHz, against 4.53 MHz with three identical switches — 13.12 times later. A larger shunt holds the open node harder and hangs more capacitance on the closed path, and the source decides where the two meet. Where the models stop

The shunt switch the source sizes

A T is two series switches and a third to ground, and it is always drawn with three of the same part. The shunt switch pulls its own size two ways: larger, it holds the open node harder; larger, it hangs more capacitance on the closed path. The size at which the band closes latest is a balance of the two — the fourth root of 2/ε times √(Rₒₙ/(Rₛ + Rₒₙ)) — when the closed state is counted as an amplitude — 3.8 times a series switch from a buffered source, a third of one from fifty ohms, a twelfth from a kilohm — and it buys a band 3.4, 2.7 and 13 times wider. Counted as a waveform the root of ε goes, and from a buffered source the best shunt is exactly the series switch.

 What a feedback tee buys, and the single thing it charges. computed by solving, not by drawing. Two 50 kΩ resistors with a 6.250 kΩ tap give 500.0 kΩ of transimpedance, which is R₁(1 + R₂/R₃) + R₂ solved on the netlist. That expression has no term for the noise gain, and the tap sets it to 1 + R₁/R₃ = ×9.00 — against exactly one for a single resistor of any value, because at direct current the source is a capacitor. One quantity then does all three things at once. The signal and R₁'s own noise are multiplied together, so the tee buys no signal-to-noise ratio at all: 1846 against 5582 in a hertz at a nanoamp for one resistor of the same transimpedance, a factor of 3.02 which is the square root of the noise gain. The amplifier's own voltage noise and offset are multiplied where they were not before. And the phase margin RISES, from 1.8° to 15.9°, because starting the noise gain high shortens its climb to the peak — so the tee is a compensation that charges for itself in noise, and a capacitor across the feedback is the same compensation for nothing. With no tap the three effects are absent rather than small. Feedback, and the margin

The tee that charges for its own compensation

A feedback tee makes a large transimpedance out of small resistors, and R₁(1 + R₂/R₃) + R₂ is the whole of what is usually said about it. The expression has no term for the noise gain, and the tap sets that to 1 + R₁/R₃ — nine, where a single feedback resistor of any value gives exactly one, because at direct current the source is a capacitor. One quantity then does everything: the signal and R₁'s own noise are multiplied together so the tee buys no signal-to-noise ratio at all, and the phase margin RISES from 1.8° to 15.9°.

 The sign change follows the 0.51 power of the amplifier, not the inductor. computed by solving, not by drawing. Both of the arrangement's frequency boundaries against the gain–bandwidth of the two amplifiers in it, over three decades. The lower curve is the frequency at which the series resistance changes sign, bisected on the sign of the real part; the upper one is where the inductance leaves one per cent. The crossing grows as the 0.513 power of the gain–bandwidth, and the dashed prediction over it is ½√(fc·fₚ) — half the geometric mean of the arrangement's own corner r/2πL = 1.59 kHz and the amplifier's open-loop pole f_t/A₀ — which is inside one per cent while that pole is at least fifteen times below the corner and 8.7 per cent out at the top of the sweep, where it is not. The two boundaries stay between 1.61 and 2.30 per cent of one another throughout, so a faster amplifier moves the active region rather than removing it. Filters, measured not tabulated

The boundary that improves when the part gets worse

A synthetic inductor's series resistance changes sign at 63.0 Hz with one-megahertz amplifiers, and that frequency is not a property of the inductor. It is half the geometric mean of the arrangement's own corner and the amplifier's open-loop pole — half the square root of their product, which the bisection confirms to a part in a thousand — and it therefore falls as the amplifier's direct-current gain rises, from 686 Hz at a gain of a thousand to 19.9 Hz at a million. The quantity that decides whether a resonator starts does not move at all: it is the transition frequency over four times the Q, 2.50 kHz for a tank of a hundred.

 An order-8 Butterworth: the whole sketch is 3.0103 dB out at the corner, and its 4 sections −5.85 to +8.17 dB. computed by solving, not by drawing. The error of the straight-line sketch against the solved response — for each buffered section of an order-8 Butterworth lowpass at 1.00 kHz, and for the whole cascade. At the corner the sections are out by −5.852 dB (Q = 0.5098), −4.418 dB (Q = 0.6013), −0.915 dB (Q = 0.9000), +8.175 dB (Q = 2.5629), which add to −3.0103 dB: the whole filter's error, the same 10 log 2 as a single pole. The whole sketch is never further out than that anywhere; the section with the highest quality factor is +8.343 dB out at 1.04 kHz. The quality factors multiply to 1/√2. Frequency, which is the same solve

The corner error a filter hides in its sections

The straight lines of an eighth-order Butterworth filter are 3.0103 dB out at the corner and nowhere worse — the same as one pole, at every order. The lines of the four sections it is built from are out by −5.85, −4.42, −0.92 and +8.17 dB there, which must add to the whole because the sections multiply. The whole sketch never gets worse with order and the worst section's error grows as 20 log(n/π). And the section the sketch misrepresents most is the one whose frequency error moves the filter most: 0.312 dB for one per cent, against the 0.173 dB any section's stopband shift gives.

 At 20 steps a cycle, ten cycles of an undamped LC: the trapezoidal rule keeps the amplitude and falls 29.2° behind; backward Euler keeps 0.0082% of it. Marched, both rules, against 1 − cos ωt for a 1 kHz inductor–capacitor pair stepped with no resistance at all. At 20 steps a cycle the trapezoidal march's amplitude stays at 1.00000 a cycle and its frequency is slow: it loses 2.918° a cycle, measured from the march's own recurrence, against 2π − 2N·atan(π/N) = 2.918°, so after ten cycles it is 29.2° behind. Backward Euler keeps 0.3901 of its amplitude a cycle, against (1 + (2π/N)²)^(−N/2) = 0.3901, so 0.0082% is left after ten, and it loses 11.19° a cycle. No resistance is in the circuit; every loss is the rule's. Before the steady state

The phase the rule loses

An inductor and a capacitor with no resistance ring for ever, and two ways of marching them disagree about how. The trapezoidal rule keeps the amplitude exactly — its factor per step has a magnitude of one — and loses phase instead: 2π − 2N·atan(π/N) a cycle, 2.918° at twenty steps a cycle, so ten cycles later it is 29.2° behind the circuit. Backward Euler keeps 0.3901 of the amplitude a cycle at the same step, and after ten cycles 0.0082 per cent of the ringing is left, in a circuit that has no loss. The two errors fall at different rates: the trapezoidal rule's phase as the square of the steps a cycle, backward Euler's amplitude as the first power. A hundred cycles to within one per cent needs 182 steps a cycle of one and 196,404 of the other.

 A regulator holding 10 V on a load that asks for 101% of the nose power collapses it in 256 s; at 110%, in 79 s. Marched with a fourth-order rule. A 10 V source behind 1 Ω feeds a load resistance through an ideal ratio n, and a regulator raises n at 0.05 per volt-second of error to hold the load at 10 V. The load's resistance is chosen so that at 10 V it takes the stated fraction of the most the line can deliver, 25 W. At 90% the regulator settles at n = 1.5195, below the nose ratio √(Rₗ/R) = 2.1082. At 99% the regulator settles at n = 1.8182, below the nose ratio √(Rₗ/R) = 2.0101. At 101% the voltage climbs to 9.950 V — the most the line allows that load, V₀√(Rₗ/R)/2 — at 101.7 s, and falls below half the setpoint at 256.3 s while the regulator keeps raising the ratio. At 110% the voltage climbs to 9.535 V — the most the line allows that load, V₀√(Rₗ/R)/2 — at 23.5 s, and falls below half the setpoint at 78.6 s while the regulator keeps raising the ratio. The regulator's gain, the slope of the load's voltage against the ratio, is positive below √(Rₗ/R) and negative above it. Power, and the part that does no work

The regulator that pushes past the nose

A regulator that raises a ratio whenever its load's voltage is low is a stabiliser only while raising the ratio raises the voltage, and on a line that stops being true at exactly the nose: the load's voltage, n·V₀ times the load resistance over n²R plus that resistance, peaks at a turns ratio of the square root of the load over the line and falls beyond it. Ask the load for 90 per cent of the nose power and the regulator settles at n = 1.519 — unless it starts above n = 2.925, where the same setpoint is met on the wrong side of the peak, and then it collapses the voltage. Ask for 101 per cent and the voltage climbs to 9.950 volts, the most the line allows, and is below half its setpoint 256 seconds later. Near the nose the collapse takes a time that grows as the inverse square root of the excess: 2,521 seconds at a hundredth of a per cent.

 An on-resistance 10% highest at mid-range: 74.16 ppm uncalibrated, 13.89 ppm once the straight line is removed. computed by solving, not by drawing, at direct current, at 41 levels across the range. A 0.5 Ω, 100 MΩ switch whose on-resistance moves by 10%, highest at mid-range, against the load. Uncalibrated, the worse of its closed error at the worst level and its open leak is least at 7.42 kΩ, 74.16 ppm — 13.72 bits, the lone switch's floor at its largest on-resistance, against 70.71 ppm for a constant one. With the gain and offset calibrated away, what is left of the closed error is the curvature, and against the leak it is least at 1.39 kΩ: 13.89 ppm, 16.14 bits. Where the models stop

The resistance that bends the signal

A switch of half an ohm and a hundred megohms has a floor of 70.71 parts per million, 13.79 bits, because its on-resistance and its off-resistance cannot both be small beside one load. Most of that floor is a gain error, and a gain error calibrates away. Give the on-resistance a realistic ten per cent of movement across the signal range and the uncalibrated floor slips to 74.16 parts per million, while the part no calibration can touch — the curvature — balances the leak at 13.89 parts per million, 16.14 bits, into 1.39 kilohms. The floor was never set by the on-resistance. It is set by how much the on-resistance moves, as its square root.

 The fringing field, and the turns standing in it. computed by solving, not by drawing. The slot on the left is the gap, cut through the centre leg to the core's own symmetry plane where the potential is zero. Flux crossing it does not stay in the slot: it bulges into the window and crosses the copper at right angles to the layers, which is the one direction Dowell's expression and every one-dimensional winding model assumes has no field in it. The turns are shaded by their own loss. The worst is turn 4, level with the gap, at 32.5 times its direct-current dissipation; the best is 1.39 times. Same wire, same current, same winding, and a spread of 23.4 between them. Two windings, and the band between them

The turns nearest the gap

A gapped inductor's flux does not turn a corner into the iron on its way out of the gap; it bulges into the window and crosses the copper at right angles to the layers. Four tenths of a millimetre from a one-millimetre gap, the worst turn of an eight-turn winding dissipates 37.5 times its direct-current loss and the winding as a whole 14.1 times. Move the same winding three millimetres further out and those become 2.9 and 2.4 — and the distance that governs it is 0.60 millimetres, which is not the gap length and does not scale with it.

 At 44.7 Ω the via's area is zero; its reflection is a doublet falling as the 1.99 power of the edge, not the first. computed by solving, not by drawing, as a cascade. A via of 0.25 pF, 1 nH and 0.25 pF on a line of √(L/C) = 44.72 Ω, met by an edge of 59 ps, against the same capacitance alone on the same line. The via's reflection is a doublet — a dip and a bump of equal area — whose largest excursion is 24.4 mV against 166 mV for the capacitance alone. Over edges from 5.9 ps to 295 ps the capacitance's reflection falls as the −0.97 power of the edge and the via's as the −1.99 power: 24.4 mV at 59 ps, 994 µV at 295 ps. The via delays the edge going past it by 22.4 ps, against √(LC) = 22.4 ps. Lines, where a wire has a length

The via that is a piece of line

A via of half a picofarad of pad and a nanohenry of barrel puts no area under its reflection from a line of √(L/C) = 44.72 ohms, from any edge. It has not vanished. It delays the edge going past it by 22.4 picoseconds, which is √(LC) exactly, and it still reflects: a doublet whose largest excursion falls as the −1.99 power of the edge where a lone capacitance's falls as the −0.97 — 24.4 millivolts for a 59-picosecond edge against 166 for the pads alone, and 994 microvolts for a 295-picosecond one against 35. The balanced via is a short piece of line, and the reflection it leaves is the reflection of its length rather than of its size.

 The bandwidth a windowed density is divided by, against the one bin an analyser assumes. Computed from the weights, from the coefficients in closed form, and out of the bins by Parseval — three routes agreeing to a part in 10¹². A record of N samples weighted by a window and transformed gives bins whose noise bandwidth is N·Σw²/(Σw)² in units of the bin spacing: 1 exactly for an unwindowed record, 1.5000 for Hann — three halves, exactly, because (a₀² + ½a₁²)/a₀² is (0.25 + 0.125)/0.25 with no record length in it — 1.3628 for Hamming, 1.7268 for Blackman and 2.0044 for Blackman–Harris. A density divided by one bin instead is wrong by the figure beside each bar. The bandwidth is not ordered with the sidelobe: Hamming's first sidelobe is 11.2 dB below Hann's and it costs less bandwidth as well, so the received sequence of windows fails at its second entry. What Hann buys is the far skirt instead, and that is a different measurement. The floor, which bounds from below

The bandwidth a bin is not

A density read off a transform is divided by a bandwidth, and the bandwidth belongs to the window rather than to the bin spacing — exactly three halves of a bin for a Hann window, so a density divided by one bin is 1.761 decibels high in power. Measured three ways that share only the weights. And the received sequence of windows fails at its second entry: Hamming's first sidelobe is 11.2 decibels below Hann's and it costs less bandwidth as well, with Hann collecting its debt at the fourth sidelobe and every one after it.

 A summing junction with 8 inputs on it, and the ceiling that fell. computed by solving, not by drawing by driving a current into the node and reading the voltage, with the input source zeroed, at each of 6 channel counts. With one input the node's ceiling is 909.5 Ω — the input and feedback resistors in parallel — and with 32 it is 31.15 Ω, a factor of 29.19 stiffer for 32 times the inputs — less than the count, because the feedback resistor is in the parallel combination too. Only the ceiling moves: the direct-current value and the whole rise are the same to 5.4 per cent at every count, because the open-loop impedance and the loop gain both scale as 1/N and cancel. So what the channel count does is bring the ceiling down to meet the rise earlier, and the per-pair leakage saturates rather than reaching unity — -0.8 dB at one input against -30.1 dB at 32. The total into every other channel together does not saturate: it is (N − 1)·Z/Rin, and with Z ≈ Rin/N that tends to unity — 0.966 at 32 inputs, 0.476 at two. A summing junction above its loop hands over essentially all of one input's signal to the others collectively, and the count decides only how it is divided. Feedback, and the margin

The node that does not care how many

Put thirty-two inputs on a summing junction instead of one and its ceiling falls from 909 ohms to 31.2 — the input resistors in parallel with the feedback resistor, twenty-nine times stiffer. Every other part of the curve is unchanged to five per cent, because the open-loop impedance and the loop gain both scale as one over the count and cancel. So the stiffer node buys nothing: the per-channel leakage saturates instead of reaching unity, and the leakage into every other channel together tends to exactly one, whatever the count.

 One rms value, and a peak that moves 41 per cent with a phase. A 10 V fundamental and a 3.33 V 3th harmonic, summed, with the harmonic's relative phase swept. The two are orthogonal over a period, so the root-mean-square value is the quadrature sum 7.45356 V at every phase — to 6.2e-15, which is the one case superposition allows and is a theorem rather than an approximation. The peak is not a sum of anything: it runs from 9.428 V to 13.33 V, a factor of 1.414, against a bound of 13.33 V that it reaches exactly at 180° — where this harmonic's own crest lands on the fundamental's, which for n ≡ 3 (mod 4) is that end of the sweep and not the other. At the far end it flattens the crest to 9.428 V, and there the peak has no closed form at all. So one signal has one heating and a headroom requirement that varies by 41 per cent, decided by a phase that no magnitude spectrum carries. Networks, and how a solve is checked

The peak that only has a bound

Power does not superpose and the gap has a closed form — two times the real part of one current times the conjugate of the other, exact at every setting. A peak has nothing of the kind. A ten-volt fundamental with a third harmonic a third its size has one root-mean-square value, 7.454 volts at every relative phase because the two are orthogonal, and a peak running from 9.428 to 13.333 volts across the same sweep — reaching the sum of the two peaks exactly at 180 degrees, where the harmonic's own crest lands on the fundamental's, and having no closed form at all at the other end.

 An all-pole filter's stopband floor: -120 dB, from 10 fF nobody drew. computed by solving, not by drawing. A butterworth 8 realised at a 20 kHz passband, drawn twice: from its poles, which fall at 48 decibels an octave for ever, and from the same netlist with 10 femtofarads of stray capacitance from its input to its last shunt node — a fraction of a picofarad between two tracks. The second stops falling at -120.0 dB at 136 kHz, because the stray bypasses every pole in front of it and what is left is its ratio to the shunt capacitance it divides against. The poles alone predict -293 dB a decade further out, which is 173 decibels lower. So an all-pole design has a stopband floor exactly as an elliptic one does — -140 dB at 1 fF, -130 dB at 3 fF, -120 dB at 10 fF, -110 dB at 30 fF, -100 dB at 100 fF, which is 20.0 decibels per decade of stray — and above that frequency a faster sample rate buys nothing at all, which is the refusal recorded for the elliptic row and not for this one. Where a signal becomes a number

The floor every filter has

The elliptic design was refused because its equiripple stopband is a floor and no sample rate reaches past a floor, while an all-pole design falls at six decibels an octave per pole for ever — so a faster clock is always an answer for one and never for the other. On paper. Ten femtofarads of stray capacitance from a filter's input to its output, which is a fraction of a picofarad between two tracks, puts the eighth-order Butterworth's stopband at −120 decibels from 136 kilohertz onwards where its poles predict −293. Every all-pole design has a floor, and it is a layout rather than a design.

 The resistance that just stabilises it is 17 times smaller than the one that damps it. computed by solving, not by drawing. Two consequences of one base resistor, against how much of it there is, for a follower fed through 100 nH of wire with 47 pF on its emitter. The falling curve is the Q of the worst pole pair, rooted from the determinant so that no frequency grid is involved; the rising one is the output impedance the stage presents at low frequency. Bisecting on the SIGN of the pole's real part returns 69.7 Ω — at which the pair is stable with a Q of 1.7e+15, which is to say no damping and a peak whose height belongs to the arithmetic. A Q of one needs 1.15 kΩ, 16.6 times more, and that resistor takes the output impedance from 5.66 to 13.10 Ω — exactly R/(β+1) added, which is the bare follower's own expression with the base resistance in the place of the source resistance. Devices, and the amplitude they stop being linear at

What the cure at the base costs

Bisecting for the smallest base resistor that stops an emitter follower oscillating gives 8 to 79 ohms. That bisection stops at the sign change, so at the value it returns the pole pair sits on the imaginary axis with a real part of 10⁻⁷ per second and a quality factor of 1.7 × 10¹⁵ — stable, and undamped. A quality factor of one needs 1.15 kΩ at 47 pF, sixteen times more, and that resistor takes the output impedance from 5.66 to 13.10 ohms. The other cure the model has always accepted and nothing has ever used is a resistor at the emitter: it reaches the same damping with 5.68 ohms and costs half the signal.

 With 100 pA of junction leakage at 25 °C, a T keeps 2.8 bits over one switch and loses them all by 91 °C. computed by solving, not by drawing, at direct current, at every five kelvin from 0 to 150 °C. The floor — the least worse-of-two error any load gives — of a 0.5 Ω, 100 MΩ switch alone and as a T of three, from a buffered source, with a junction leakage of 100 pA at 25 °C on every terminal, doubling every 10 K, the worse sign taken. At 25 °C the lone switch's floor is 71.06 ppm (13.78 bits) and the T's 9.998 ppm (16.61 bits). The T is worse than one switch above 91.4 °C, where the junction current equals the off-resistance's conductance at one volt. The lone switch drops below 13 bits at 101.3 °C and below 12 at 125.9 °C; the T below 16 at 37.2 °C. Where the models stop

The leak no switch can hold

A T of three switches reaches five parts per billion because its shunt switch holds the node a leak has to cross. A junction leakage does not cross anything: it flows out of the outer switch's terminal straight into the load. With 100 picoamperes of it at 25 °C, doubling every ten kelvin, the T's floor is 9.998 parts per million rather than five parts per billion — 16.61 bits, not 27.6 — and it has a best load again, at 100 kilohms. A lone switch loses nothing at room temperature. Above 91.4 °C, where the junction current reaches the off-resistance's conductance at one volt, the T is worse than one switch.

 Degeneration removes the second-order product 9 times less well than the third. computed by solving, not by drawing. Both intermodulation products of a degenerated stage at 5 mV a tone, against the degeneration factor. The second-order product falls as D⁻² — the straight reference is exactly that law, anchored at D = 1 — and the fitted exponent is -2.000. The third-order product falls faster, as D⁴/|3 − 2D|, which is one more power of D at large factors, and it collapses altogether at D = 1.5 where its coefficient changes sign. So the ratio between the two goes from 20.7 at D = 1 to 183 at D = 16: the more linear the stage is made, the more completely its distortion is the product a third-order account never measures. Devices, and the amplitude they stop being linear at

The product that is not the third

Every distortion result in this field is odd-order, and every two-tone calculation behind those results computed the second-order product and threw it away. On a bare exponential it is the drive over twice the thermal voltage — 1.934 × 10⁻² of the fundamental at a millivolt, against 1.870 × 10⁻⁴ for the third-order product, a ratio of 4Vₜ/a and a hundred and three to one. A differential pair puts it at 6.2 × 10⁻¹⁶. And degeneration removes it as D⁻² where it removes the third order as D⁴/|3 − 2D|, so a stage linearised until its third-order product is negligible is a stage whose distortion is almost entirely the one nobody measured.

 A step through r sections starts as (t/τ)^r: it reaches 1% at 10.1 µs, 105 µs, 243 µs, 380 µs, 508 µs for r = 1 to 5. Solved, and expanded two ways. The step response of buffered RC sections of time constants τ, τ/2, … τ/r, with τ = 1 ms, on logarithmic axes. The relative degree of the recovered transfer function is r, so the first r − 1 derivatives of the step are zero at the start and the r-th is lim s^r·H(s) = r!/τ^r, read off the network solved far above its poles and off the expansion of H about infinity; the step therefore starts as (t/τ)^r, a straight line of slope r. The expansion about infinity and the residue expansion agree to a part in a million where both are well conditioned. The output reaches 1% at 10.1 µs (r = 1), 105 µs (r = 2), 243 µs (r = 3), 380 µs (r = 4), 508 µs (r = 5), and half its final value at 693 µs, 1.23 ms, 1.58 ms, 1.84 ms, 2.04 ms. For these time constants the whole step is (1 − e^(−t/τ))^r, checked against both expansions, so the time to a fraction ε is −τ·ln(1 − ε^(1/r)). Before the steady state

The start a step takes from infinity

The initial-value theorem reads where a step starts off H at infinite frequency. Apply it again to s·H, s²·H and on, and it reads how the step starts: the first r − 1 derivatives are zero for a network r degrees more poles than zeros, and the r-th is the ratio of the leading coefficients. So a step through r sections begins as a power of time — for sections of τ, τ/2, … τ/r, exactly (t/τ) to the r — and reaches one per cent at 10.1 µs through one section, 105 µs through two and 508 µs through five. Put a zero anywhere, even a thousand times above every pole, and the step starts linearly instead, with a slope of twice the zero's time constant over τ² that is the larger term for the first two of them.

 A porosity of 0.50, with the field the substitution smooths away. computed by solving, not by drawing. The flux lines between the conductors are the whole difference. The porosity substitution replaces this layer with a foil of the same direct-current resistance spread over the full breadth, in which the field is parallel to the layers by construction; here it is not, and it crowds between the turns. The solved ratio is 5.816 against the substitution's 6.212, 6.4 per cent apart. The copper is shaded by its own loss, which is what says the turns inside a layer are not alike either. Two windings, and the band between them

The wire that is not a foil

Almost no winding is made of foil, and the closed form for a winding's alternating-current resistance is about foils. The bridge between them is a substitution — squeeze the layer's conductors together, spread the result back across the breadth, divide the conductivity by the porosity — and it replaces a two-dimensional geometry with a one-dimensional one. Solved as a field it is exact where it must be, at a porosity of one, and 7.2 per cent high at a porosity of 0.40. It errs on the safe side, which is the half of the answer nobody could have assumed.

 Lead zero at 500 Hz: the lines over-report by at most 3.12° with the loop's phase, and short by up to 20.6°. computed by solving, not by drawing. The error in the phase margin read off the straight lines of an integrator, two poles at 1.00 kHz and a lead section with its zero at 500 Hz and its pole 10 times higher, against where the lines cross unity, measured in decades from the zero. Gaps are placements where the lines are flat at unity or the loop crosses more than once. Read with the loop's own phase the reading is over by at most +3.12°, with the lines crossing −0.01 decades from the zero on a loop with 73.4° of margin, and never over on any loop with 60° or less; it is short by as much as 20.65°. Read with the phase off its straight lines too, on loops with 60° or less, it is never over. Frequency, which is the same solve

The zero that lifts the lines

A phase margin read off the straight lines of an all-pole loop can only be short, and the proof takes three steps. A lead section breaks two of them: a zero's response lies above its lines, and between a zero and its pole the phase rises. The two breaks pull opposite ways, and measured across every placement of the crossing, in every sweep drawn, they leave an over-report of at most 6.58°, on a loop with 87.5°; on loops with sixty degrees or less it never exceeds 1.52°. The phase sketch is another matter: with the lines crossing at a lead zero above the plant's corner, it reports 17.91° on a loop with 1.17°.

 Two tracks 1 mm apart share all of the plane's resistance at direct current and 14.5% of it above the band. computed by solving, not by drawing, across a 50 mm plane cut into 239 strips, with two tracks 200 µm above it and 1 mm apart — 5.0 heights. One track carries the current and the voltage along the other's loop is measured. The shared resistance, as a fraction of the driven loop's own, is 1 at direct current, where both returns spread across the whole plane and share its 10 mΩ/m; it falls through a half at 183 kHz and settles at 0.1454 above 10.0 MHz, where each return has gathered under its own track. The shared inductance is 0.398 of the loop's own at direct current and 0.0341 above the band. Lines, where a wire has a length

Two returns in one plane

Two tracks over one plane share the whole of its resistance at direct current, however far apart they are routed: 10 milliohms a metre on a fifty-millimetre plane, from tracks a millimetre apart or ten. The sharing ends across the same band a single return gathers over, and it ends sooner the farther apart the tracks are — through a half at 525 kilohertz for tracks three heights apart and at 9.88 kilohertz for fifty. Above the band what is left is the overlap of two image distributions, 4h²/(4h² + d²): 14.5 per cent of the resistance at a millimetre, 0.68 at five. And in the middle of the band the two loops' mutual inductance changes sign.

 Where "the noise is the density over twice the window" is true for a mean of 600 samples. The exact sampled autocorrelation of a one-pole front end, summed in closed form and checked against seeded samples through the same filter. The solid curve is the reading's noise power divided by what the rule gives. It crosses one at 0.1355 of half the sample rate and nowhere else, so the rule is right at one front-end bandwidth rather than over a range: it is within one per cent from 0.0712 to 0.205 of half the sample rate, a window a factor of 2.88 wide. Above it the front end admits noise from past half the sample rate and the sampler folds it in, the ratio tending to (x/2)·coth(x/2) with x = 1/(τ·fₛ), which has no N in it — so the sample rate must be 9.7 times the front end's noise bandwidth for one per cent, not twice it. Below it the front end has already removed noise the rule charges for, and the faint curve — the √N law's own error, which is a claim about the samples rather than about a density — shows why: consecutive samples are correlated by ρ = e^(−1/(τ·fₛ)) and a mean of 600 of them is a mean of fewer things. The floor, which bounds from below

The window the square-root law has

Averaging N readings divides the noise power by N only if the readings are independent, and samples of a filtered noise are independent only over a band. Measured on the exact sampled autocorrelation, the rule that a reading's noise is the density over twice the window is right at one front-end bandwidth and wrong on both sides — a factor of three wide at six hundred samples, asking the sample rate to exceed the front end's noise bandwidth by about ten rather than by two. Above it the folded noise arrives at full amplitude; below it the front end has removed noise the rule is still charging for.

 A summing junction with a second pole above it: 3.1 dB of peak, and inductive. computed by solving, not by drawing on a two-pole amplifier — a gain block, two sections and a buffer, so the feedback network loads the output exactly as it does with one pole. The faint curve is the one-pole case, which rises to 909.1 Ω and stops. With a second pole at 100 kHz the junction reaches 1304 Ω at 112 kHz — 3.13 dB above the two resistors in parallel — because the division is by |1 + T| and near crossover that is smaller than one. A third of a decade below the peak the node's impedance leads its current by 84.4°, which is an inductance: the node is no longer a resistance that a capacitance loads, it is a reactance that a capacitance resonates with. The slider puts capacitance on the node and moves the peak. Feedback, and the margin

The node that is an inductance

A summing junction with a one-pole amplifier round it rises to the two feedback resistors in parallel and stops there, and that ceiling is the whole of what a virtual earth can be. Give the amplifier a second pole and the node overshoots it — 1074 ohms against 909 with the second pole at 300 kilohertz, 1868 with it at thirty — and below the peak the impedance leads its current by more than twenty degrees, which is an inductance. So a stray capacitance on the node is no longer a load on a resistance; it is a resonance, and the peak grows with it rather than being damped by it.

 Summed over whole periods 1.3% too long, a sine is read to ±0.65% for up to 38 periods, whatever their number. Integrated exactly over each window, the worst over every starting phase. The error in a root-mean-square summed over N assumed periods 1.3% too long, against N, for a sine and a 60° rectifier current, beside an explicit converter averaging over a comparable time, τ of N/2 periods. For the sine the worst error is 0.643% at one period and stays near δ/2 until N approaches 1/(2δ) = 38; it vanishes where Nδ is a whole number of half-periods of the square, and beyond it is bounded by 1/(4πN). The rectifier current's is 1.274% at one period, near δ(CF² − 1)/2 with a crest factor of 1.732. The converter at τ = N/2 periods is low by 15.8 ppm on the sine at N = 10, with a ripple of ±0.796%; on the rectifier current, low by 46.3 ppm with ±1.665%. Power, and the part that does no work

The cycle a converter has to know

Summing a waveform's square over a whole number of periods reads its root-mean-square exactly: no averager, no ripple, no bias. It needs the period, and a period known one per cent long puts a hundredth of a period too much into the window. Wherever that extra piece falls, the reading moves — on a sine by up to 0.50 per cent over one period, and by 0.46 per cent over ten, because the extra piece grows with the window as fast as the window does. The worst error is δ(CF² − 1)/2, set by the crest factor and the period error and not by how many periods are summed, until the excess reaches half a period. A square wave is read exactly from any window, and a 60° rectifier current twice as badly as a sine.

 How much larger a gap is than its own length says. computed by solving, not by drawing. A magnetic circuit prices a gap as g/(µ₀A) and everybody knows that is low, because the flux bulges out of the sides. The usual repair is to add one gap length to each dimension of the gap's area, which is the dashed line. The measurement is the solid one: at a 0.3 mm gap in a 6 mm leg the true correction is 1.100 and the rule offers 1.050, so the rule supplies 50 per cent of a correction worth 10 per cent of the inductance; at 1.7 mm it supplies 85 per cent. The rule is not wrong so much as it is a rule whose accuracy depends on the thing it is correcting. Two windings, and the band between them

The gap that is bigger than it is

A magnetic circuit prices a gap as g/µ₀A and everybody knows that is low, because the flux bulges out of the sides. The usual repair — add one gap length to each dimension of the gap's area — supplies half the correction at a 0.3 mm gap and 85 per cent at 1.7 mm, on a correction worth 10 per cent of the inductance at the first and 33 at the second. It is not a rule that is right or wrong; it is a rule whose accuracy is a function of the very thing it is correcting.

 20 sections imitating a 1 m line: its delay is 1% long at 318 MHz and its group delay at 185 MHz, and it passes nothing above 1.32 GHz. computed by solving, not by drawing, as a chain of 20 series inductors and shunt capacitors carrying the inductance and capacitance of a metre of 50 Ω line of delay 4.83 ns, terminated in 50 Ω at both ends, beside the Bloch phase of an endless chain, 2·arcsin of ω over the cutoff, a section. The chain's cutoff is the cutoff 2/√(LₛCₛ), 1.32 GHz. Its phase delay is too long by arcsin(x)/x − 1 with x = f over that cutoff, 1% at 318 MHz (x = 0.2417); its group delay by 1/√(1 − x²) − 1, 1% at 185 MHz (x = 0.1404). Well below cutoff the solved chain's delay follows the closed form; nearer it the fifty-ohm terminations, which are not the LC ladder's own impedance there, add a ripple. Ten sections per wavelength is 414 MHz for this chain. Lines, where a wire has a length

The sections a wavelength needs

A ladder of inductors and capacitors is a line only below its own cutoff, 2/√(LC) of one section, and in the frequency domain how far below can be written down exactly: its delay is too long by arcsin(x)/x − 1 and its group delay by 1/√(1 − x²) − 1, where x is π over the number of sections per wavelength. One per cent of delay needs 13.0 sections per wavelength; one per cent of group delay, 22.4. The rule of ten per wavelength is 1.72 per cent slow in phase and 5.33 in group delay. Twenty sections imitating a metre of cable are a line to a per cent of group delay up to 185 megahertz and pass nothing at all above 1.32 gigahertz.

 Sampling 100–120 MHz: 6 allowed windows and 5 forbidden gaps. A band 20 MHz wide centred at 110 MHz, and the sample rates at which it and all its images land without overlapping. Twice the top of the band is 240 MHz and every rate above that works, which is the familiar answer. Below it there are 5 more windows, the slowest at 40.00 MHz — 1.00 times twice the band's WIDTH, which is the quantity that bounds a sample rate when the signal is not at baseband. Between the windows are 5 forbidden gaps, shaded: at 219.1 MHz the band's images overlap it, while 200.00 MHz works and 240.0 works. So a clock raised out of a working window has been made worse by being made faster, and "a faster clock is always an available answer" is a statement about baseband and not about sampling. Where a signal becomes a number

The clock that is too fast

Every sample rate in the requirement is a lower bound: the filter must be down by the frequency that folds back, and a faster clock is always at least as good. That holds for a signal at baseband. A band from 100 to 120 megahertz can be sampled at 240 and above, or at 120 to 200, or 80 to 100, or 60 to 66.7, or 48 to 50, or 40 — six disjoint windows with five forbidden gaps between them, so 110 megahertz fails while 100 works and 120 works. The bound is twice the band's width, 40 megahertz, and it is six times below twice its top.

 A cable's carrier limit is its conductors' matching, not its capacitance. computed by solving, not by drawing. A 350 Ω quarter bridge at 1000 µε, excited at a carrier frequency, with 10 metres of cable at 100 pF a metre on each output terminal. Each terminal sees the two arms meeting there in parallel against that capacitance, so each lags by arctan(2πf·(r/2)·C) with a corner at 909 kHz — and with the two conductors matched the lag CANCELS in the difference: -0.117° where each terminal has -5.71°. A mismatch does not cancel, and it arrives multiplied by the ratio of the excitation to the signal, which at 1000 µε is 2000: at 1 per cent the phase is 40.3° at a hundred kilohertz. So the excitation a 10-metre cable allows for 200 parts per million falls from 205 kHz matched to 1.82 kHz at 1 per cent. And it depends on the strain being measured — 183 Hz at 100 µε, 547 Hz at 300 µε, 1.82 kHz at 1000 µε, 5.45 kHz at 3000 µε — because the amplification is the reciprocal of the signal. Circuits that do a job, and the range they do it over

The carrier a cable allows

A carrier-frequency amplifier removes an instrument's offset drift by exciting the bridge with alternating current, and the cable that was a resistance at direct current becomes a capacitance. Both output terminals see the same arms and the same capacitance, so the lag cancels in the difference: ten metres matched leaves 0.15 degrees at a hundred kilohertz where each terminal has 6.3. What does not cancel is the mismatch between the two conductors, arriving multiplied by the ratio of the excitation to the signal — two thousand at a thousand microstrain — so one per cent of it takes the allowed carrier from 205 kilohertz to 1.8.

 10.0 cm of track leaves 40 per cent on a 1 ns edge. The far end of 10.0 cm of 50 Ω track, driven from 10 Ω into 200 Ω, marched as the exact staircase of reflections it is and driven by a ramp rather than by a step. The staircase converges on the resistive divider the circuit was going to be all along, to a part in 10⁹, which is the number the lumped model returns immediately and with no staircase. What is left on the edge is an overshoot: 40.0 per cent at 0.1 ns, which is the first wave's own reflection coefficient and contains no rise time at all, falling to 0.399 per cent at 100 ns where the staircase happens inside the edge. It reaches ten per cent at an edge of 4.24 ns and one per cent at 40.0 — which is 28.6 times the 1.40 ns round trip. Nothing in the measurement is a repetition rate: the boundary is a transit time, a mismatch and a rise time, and the clock the signal repeats at does not enter. Where the models stop

The boundary that is a rise time

Kirchhoff's own frequency is 3.97 megahertz for ten centimetres of track, and a one-kilohertz clock is four thousand times below it — which is the reading that concludes wrongly. Marched as the exact staircase of reflections it is, the same track leaves forty per cent of overshoot on a one-nanosecond edge and 0.40 per cent on a hundred-nanosecond one, reaching one per cent at 40.0 nanoseconds. The boundary is a rise time, it is twenty-nine times the conductor's round trip, and the repetition rate is not in it anywhere.

 8 doublets over 2 decades settle 31.6 times slower than one. A tail of 1.0 per cent split into 8 doublets whose time constants are spread over 2 decades from 0.100 ms, each carrying 0.125 per cent. Settling to 0.10 per cent takes 7.278 ms against 0.2303 for the same total tail at one time constant — a factor of 31.6 — and it is between τₘₐₓ·ln((A/N)/B) = 2.231 and τₘₐₓ·ln(A/B) = 23.03 ms, which says the SLOWEST doublet decides it whatever its share. The spread is what costs and the number is not: four decades of spread multiply the settling by 2.04e+3 while sixteen times the count changes it by 0.386. And the trimming result, halving the tail for τ·ln 2, now buys 3.86 ms — the slowest time constant's ln 2 rather than the fastest's. Before the steady state

The slowest one decides it

One doublet's settling is its own time constant times the log of the tail over the band, so halving the mismatch buys ln 2 of it and no more. Split the same one-per-cent tail into eight doublets spread over four decades and the settling goes from 0.23 milliseconds to 470 — a factor of two thousand — while sixteen times as many doublets over the same spread changes it by less than two. It is the spread that costs and not the number, because the slowest doublet decides the settling however small its own share, and its share enters only through a logarithm.

 Where a transformer's leakage inductance actually is. computed by solving, not by drawing. Both windings carry the same ampere-turns in opposite directions, which is the short-circuit condition a leakage measurement is made under, so the flux drawn here is the flux that fails to link the two — the leakage field, and nothing else. It is largest in the insulation between the portions, where the magnetomotive force is at its full value and there is no copper to be in. The energy in this window is 2.058 microjoules per metre, which is 4.116 microhenries per metre referred to the primary against a closed form of 5.213. Two windings, and the band between them

The inductance that is a shape

Leakage inductance is the one transformer parameter that belongs to the geometry rather than to the material: twice the magnetic energy in the window under equal and opposite ampere-turns, divided by the square of the current. Solved as a field it is 3.086 microhenries a metre against a closed form's 3.128 when the copper fills the window, and 4.608 against 6.255 when it fills half of it. Interleaving is worth 3.11 times and not the four it is quoted as, and the missing 0.89 is the insulation nobody puts in the formula.

 The gain recovered from the phase alone, right in shape, with its level supplied by the gain at direct current. computed by solving, not by drawing. The lead, 1.00 kHz to 10.0 kHz, solved at 1601 frequencies over sixteen decades, and its gain rebuilt by Bode's companion integral from the phase alone over ±3 decades of 1.00 kHz. The integral returns a difference of log-magnitudes and nothing else, so the curve it gives has no level: drawn as if the direct-current gain were one, it is the dashed curve. Supplying the direct-current gain, −20.000 dB, places it. The placed curve lies within 0.055 dB of the solved gain across the drawn band, worst at 794 kHz. Frequency, which is the same solve

The level the phase cannot know

Bode's gain–phase relation runs both ways. Given only the phase of a lead network, its companion integral returns the gain's shape across six decades to 0.055 dB. It cannot return the gain's level, because multiplying a network by a constant changes no phase anywhere. The level is one number, −20.000 dB here, and it has to come from outside the phase: one gain reading at any frequency supplies it. The same integral fed an all-pass reads a flat gain as a double pole, 6.021 dB down at the all-pass's frequency and 40.086 dB down a decade above it.

 BDF2 loses amplitude as N⁻³ and phase as N⁻², four times the trapezoidal rule's: a hundred cycles to 1% needs 363 steps a cycle, against 182 and 196,404. Each rule's error a cycle on an undamped inductor–capacitor pair, against the steps a cycle, measured from BDF2 marches and drawn with the trapezoidal rule's and backward Euler's closed forms. BDF2 loses amplitude a cycle as the −2.91 power of N and phase as the −1.95 power, its phase loss 3.988 times the trapezoidal rule's at 128 steps a cycle. To keep 99% of the amplitude over a hundred cycles it needs 157 steps a cycle, and to stay within 1% of a cycle in phase 363; the trapezoidal rule needs 182 for the phase and loses no amplitude, and backward Euler needs 196,404 for the amplitude. Before the steady state

The rule between the two

The trapezoidal rule keeps an undamped ringing's amplitude and loses its phase, and it preserves a stiff pole's error with its sign alternating. Backward Euler damps the stiff error at once and damps the ringing too, keeping 0.0082 per cent after ten cycles at twenty steps a cycle. The second-order backward difference sits between them in a precise sense. It damps the stiff error by 1/√(3 + 2h/τ), 0.152 a step at twenty time constants, and settles in three steps where the trapezoidal rule needs twenty-three. It loses a ringing's amplitude only as the cube of the steps a cycle, and its phase as the square with a constant four times the trapezoidal rule's. A hundred cycles to one per cent costs 363 steps a cycle, against 182 and 196,404.

 Losing capacitance to bias raises this loop's margin; the corner that binds is the ceramic at full value. computed by solving, not by drawing. Phase margin of the five-volt regulator with a 5.00 mΩ ceramic output capacitor, against the capacitance that capacitor keeps at the working voltage, from the full 10 µF down to 1.5 µF, for five capacitors across the upper divider resistor. With none, the margin climbs from 18.0° at full value to 43.6° at 15% kept, so the lost capacitance is stabilising rather than the reverse. 630 pF holds at least 49.7° across the whole range. 836.5 pF, the value that removed the resistance floor at the nominal capacitance, gives 48.8° at full value and 44.3° at 15%: the capacitor that has lost the most is the one it fails on. Circuits that do a job, and the range they do it over

The capacitance the bias takes away

A ceramic output capacitor has a few milliohms of series resistance, which a capacitor across the upper divider resistor made legal, and a capacitance that falls with the voltage across it — a ten-microfarad part at five volts is two microfarads to a small signal. Swept from the full ten down to that fifth, the regulator's margin does not collapse. It rises: with no divider capacitor from 18.0° to 38.5°, because the output pole climbs away from the amplifier's. The corner that binds is the capacitor at full value. One divider capacitor, 423 to 808 pF, holds 45° at every capacitance between. The 836.5 pF that removed the resistance floor sits just above that band, and the lost capacitance is paid for in droop: 103 mV at full value, 202 mV at a fifth.

 6 straight lines over six decades of a diode: ±8.002 mV everywhere, and they meet at every join. computed by solving, not by drawing. A pure exponential junction from a nanoamp to a milliamp, modelled by 6 drop-plus-resistance segments each spanning a current ratio of 10.00, each fitted minimax on its own window. The model's error swings between +8.002 mV and −8.002 mV in every segment, which is the closed form for that ratio, because a span of a logarithm is the same shape wherever it is taken. The dots are the joins: at each one the two neighbouring segments are off by the same amount in the same direction, so they meet, and the chain is continuous although no segment was asked to be. Where the models stop

How many straight lines a diode is

One drop-plus-resistance line fitted minimax over a decade of a diode's current is out by ±8.00 millivolts, and it is the same over every decade. Chain such segments over six decades, a nanoamp to a milliamp, and the error of each is a closed form in its current ratio alone. Minimax segments meet at their joins without being asked to, so the chain is a continuous piecewise-linear model for free. The error falls as the inverse square of the segment count: one segment is out by 132 mV, six by 8.00, eighteen by 0.94. A millivolt takes 18 segments and a tenth of one takes 56. A real junction needs more, 23 and 72, because recombination gives it an ideality near two over most of the range and each segment narrows by the square root of the local ideality.

 Inside a loop of gain 1000, a twin-T error of +0.10% raises the peak's Q from 250 to 333, and the same error the other way lowers it to 200. computed by solving, not by drawing. An amplifier of gain 1000 with a 10 kΩ, 10 nF twin-T as its feedback network, whose closed-loop gain A/(1 + A·T) peaks at the null, 1591.5 Hz. With exact components the peak is 1000 and its quality factor 250.2, a quarter of one more than the gain. With the twin-T's shunt elements −0.10% out the other way the peak falls to 800 and the Q to 200.2; +0.10% raises them to 1333 and 333.5, and +0.30% to 3973 and 994. At +0.40% the loop oscillates. Filters, measured not tabulated

The sign the null leaves behind

Put a twin-T in the feedback path of an amplifier of gain A and its null becomes a peak with a quality factor of (A + 1)/4. A filled null is not a small phasor of any phase. It is a real number, minus a quarter of the error in the shunt elements, and once the loop multiplies it by A its sign decides everything. At a gain of a thousand a 0.1 per cent error lowers the Q from 250 to 200 one way and raises it to 334 the other, and at 0.4 per cent the second direction oscillates. Every gain falls on one curve, the exact-null Q divided by 1 − A·t/4. So a Q is set to ten per cent only by a null deeper than 20 log(44·Q): one per cent parts allow a Q of 9.6.

 1 pA in 10.0 kHz: 312 carriers a window, and a Gaussian of the same variance that looks right and is not in its tails. computed by solving, not by drawing. A direct current of 1 pA counted in windows of 1/(2B) for a noise bandwidth of 10.0 kHz: the mean count is N = I/(2qB) = 312.1, and the count is Poisson, drawn as bars, against a Gaussian of the same mean and variance, drawn as a curve. The variance is N exactly, so 2qI is right; the signal-to-noise ratio is √N = 17.67 and the skewness 1/√N = 0.0566. The count exceeds its mean by five standard deviations with probability 7.84 × 10⁻⁷, against the Gaussian's 2.87 × 10⁻⁷. A seeded arrival train counted in 20,000 windows gives a mean of 312.200 and a variance of 314.461. The floor, which bounds from below

The count under the density

Shot noise's density, 2qI, is exact at every current, because it is the variance of a count of independent carriers, and that count is Poisson at any size. Reading that variance as a Gaussian floor is not exact. A picoampere in a 10 kHz noise bandwidth is 312 carriers a window, and it passes five standard deviations above its mean 2.74 times as often as a Gaussian of the same variance says. At 10 femtoamperes it is 359 times as often, and 4.4 per cent of the windows are empty. A single pole reads a third more skewness than a count does, (4/3)/√N against 1/√N. The density is right and the shape is wrong, and the shape is what a threshold uses.

 Held one drop and 300 mV up, the cascode's floor falls from 0.832 V to 0.403 V. computed by solving, not by drawing. The output current of two cascode mirrors against their output voltage near the bottom of the supply, each divided by its own value at 5 V, on a model whose transistors saturate. The stacked mirror holds its upper device's base two drops up and stays within one per cent down to 0.832 V; the wide-swing mirror holds it one drop and 300 mV up, which leaves the lower device 0.302 V of collector-emitter voltage, and stays within one per cent down to 0.403 V. Its output resistance at 5 V is 13.2 MΩ against 7.39 MΩ, and its copy ratio 0.9758 against 0.9747. Devices, and the amplitude they stop being linear at

The drop the lower device did not need

A cascode mirror's floor is a whole base-emitter drop plus the upper device's own minimum, because the stack that biases the upper base holds it two drops above ground. The lower device needs only a few hundred millivolts. Hold the upper base one drop and 300 mV up instead, and on a model whose transistors saturate the one-per-cent floor falls from 0.832 V to 0.403 V. The output resistance does not fall with it: it rises from 7.39 to 13.2 MΩ, because the upper device's base current no longer comes out of the reference. The cost has moved into the margin itself. Below about 150 mV the lower device saturates, and at 50 mV the resistance is 0.51 MΩ and the copy ratio 0.7665.

 The temperature a part cannot come back from, and how long it takes to leave. computed by solving, not by drawing. The same fixed-point equation as the rung below, marched in time with a thermal capacitance rather than solved for its steady states: C dT/dt = P(T) − (T − T_a)/R_th, stepped adaptively on the temperature change because dT/dt goes through zero at each fixed point. Every trajectory starting below 191.1 °C returns to 88.8, however far above the operating point it began; every one starting above it leaves the material's range entirely, the closest in 0.2 minutes. The two nearest starts are 3.0 kelvin apart. The ignition temperature is a boundary in the STARTING CONDITION, and no steady-state analysis contains one. Two windings, and the band between them

The boundary that is a starting point

A wound part with a stable operating point at 88.8 degrees and an ignition temperature at 191.1 will never reach the second, because nothing takes it there. Marched in time rather than solved for its steady states, the same equation says what does: a trajectory starting at 189.6 degrees settles back and one starting at 192.6 leaves the material's range in twelve seconds — two starts three kelvin apart. And an overload of four times the normal loss is survivable for ever, while seven times is survivable for seventeen minutes.

 Four networks of one magnitude: the first and last are fixed at every instant, and the two between swap at 12.64 µs. computed by solving, not by drawing. A low-pass with poles at 1, 10 and 100 kHz and zeros at 3.00 kHz and 30.0 kHz, and the three networks that share its magnitude with one or both zeros moved into the right half-plane by an all-pass. The magnitudes agree to 6.7e-16 and each impulse response carries 9261.008 of energy. The fraction delivered by each instant: the all-left network is never behind any other and the all-right network never ahead of any other. The two between are not ordered. Half the energy arrives at 2.13, 2.80, 10.1, 11.2 µs in the order listed, so mirroring the higher zero is the later; nine tenths arrives at 50.2, 156, 60.9, 167 µs, so mirroring the lower zero is the later. The curves cross at 12.64 µs. Frequency, which is the same solve

The two mirrors that change places

A low-pass with two zeros has four networks that share its magnitude, one for each choice of side for each zero. The one with both zeros on the left delivers its energy first at every instant, and the one with both on the right delivers last. The two in between are not ranked at all. With zeros at 3 and 30 kHz, mirroring the higher zero delays half the energy more and mirroring the lower zero delays nine tenths more. They swap at 12.64 µs, and that instant depends only on the product of the two zeros: it is where the all-pole response is rising at a rate of √(ω₁ω₂). And the network that delivers last does not start the wrong way at all.

 A guarded input driven through an inductance oscillates once the source's L/R passes 1/ωₜ. computed by solving, not by drawing. The real part of the rightmost natural frequency of a guarded input — a 100 pF cable bootstrapped by a 1.00 MHz follower, 10 pF of input capacitance — driven from sources of 10 Ω, 100 Ω, 1 kΩ with an inductance in series, against the inductance. Above zero the input rings with growing amplitude. Each source crosses at 1.75 µH, 17.6 µH, 187 µH, close to R/ωₜ — 1.59, 15.9, 159 µH — and oscillates at 11.4 MHz, 3.59 MHz, 1.14 MHz. The same source without the guard is stable. Measurement, which is a circuit on a circuit

The source that rings against its guard

A bootstrapped cable is a capacitance in series with a negative resistance that falls as the square of frequency, −1/(ω²·C/ωₜ). Drive it from a source with resistance R and inductance L and the loop's reactances cancel at 1/√(LC). Its resistances cancel where R = L·ωₜ, so the input oscillates once the source's own time constant L/R passes the follower's 1/ωₜ. The poles of the whole netlist say 17.6 µH for a 100 Ω source behind a 1 MHz follower, against 15.9 from the expression. The cable's length moves that by at most a factor of two and sets only the frequency it rings at. The same source without the guard is stable. A resistor in the guard drive raises the limit, to 164 µH through 1 kΩ, and pays for it in bootstrap bandwidth.

 At the same rise time a lossy edge draws 2.9–4.0 times more from a balanced via than a clean one, and the square law survives. computed by solving, not by drawing. The largest excursion of the reflection from a balanced via and from a 0.5 pF capacitive via against the 10–90 rise time of the edge that reaches them: behind 5 to 50 cm of lossy line, with a 20 ps edge sent in, and from clean raised-cosine edges of the same rise times on a lossless line. Behind the loss the balanced via's reflection falls as the −1.97 power of the rise time and the capacitive one's as the −0.82. Against clean edges of the same rise time the lossy edge draws 2.93 to 3.97 times as much from the balanced via and 1.07 to 1.54 times as much from the capacitive one. Lines, where a wire has a length

The edge the loss delivers

A balanced via's reflection falls as the square of the edge's rise time, so the loss in front of it should hide it. A 20 ps edge sent down 20 cm of lossy line arrives with a 10–90 rise time of 194 ps, and the via's reflection falls to 9.1 mV. But a clean edge with that same 194 ps rise time draws only 2.29 mV from the same via. The lossy edge excites it 3.97 times as much, because loss makes an edge steep in the middle and slow at both ends, 3.07 times steeper at its centre than a raised cosine of the same 10–90 time. Across 10 to 50 cm the square law survives, an exponent of −1.97. It survives shifted: the 10–90 time overstates, by a factor of three or four, how slow the edge is to anything that reads its curvature.

 Matched at 1 kHz, the conjugate load wins only from 730 Hz to 1.84 kHz; the resistor-only load is a low-pass, not a broadband answer. computed by solving, not by drawing. A source of 50 Ω and an inductance of 100 Ω at 1 kHz, with the conjugate load for 1 kHz — 50 Ω and a capacitor of −j100 Ω — and the resistor-only load for 1 kHz, 111.8 Ω, both held fixed while the frequency moves. The conjugate match takes all the available power at 1 kHz and stays within a decibel of it from 777 Hz to 1.29 kHz, a band of 50.9% of the centre, 1.0176/x. The resistor-only load takes 0.6180 at 1 kHz, 0.8509 at 100 Hz, and less at every frequency above: a low-pass whose peak is at direct current. The conjugate load is the better of the two only between 730 Hz and 1.84 kHz. Power, and the part that does no work

The load for a source that will not hold still

A conjugate match is exact at one frequency. Held fixed while the frequency moves, it stays within a decibel of the available power over 1.0176/x of the centre: 51 per cent of it when the source's reactance is twice its resistance, 12.7 per cent at eight times. It beats the resistor-only load only between 0.73 and 1.84 of the centre frequency. The resistor-only load is not a broadband answer, as it is usually called. It is a low-pass whose best frequency is direct current. And for a source whose reactance could be anywhere within ±Δx of a centre, the fixed load that guarantees the most power cancels the centre exactly and takes a resistance of √(1 + Δx²) times the source's. That guarantees 2/(1 + √(1 + Δx²)): the resistor-only rule, applied to the uncertainty instead of to the reactance.

 The cascode device's collector noise is cancelled below 7.22 MHz and its base noise never is. computed by solving, not by drawing. The output noise current of a cascode stage at 1 mA and β = 150, driven from 1 kΩ, into a short, source by source, with 20 pF at the node between the two devices. At low frequency the upper device's base shot noise contributes 0.667% of the lower device's collector noise power, 1/β, and its collector shot noise 4.88 × 10⁻⁵, about 1/β². The collector's contribution overtakes the base's at 7.22 MHz, where the admittance reaches gₘ/√β — 7.21 MHz from the lower half's solved admittance, against 25.1 MHz from the node capacitance alone, because the lower device's Cμ, seen from its collector, is multiplied by its own gain; it passes the lower device's own collector noise near 316 MHz. Devices, and the amplitude they stop being linear at

The noise the cascode device does add

A cascode's upper device is said to add no noise, because it is a common-base stage fed from a high impedance and its own noise current has nowhere to go but round itself. That is true of its collector noise, which reaches the output at 4.88 × 10⁻⁵ of the lower device's, about one part in β². It is false of its base noise. The partition of the emitter current into collector and base cannot cancel, so it arrives whole: one part in β, 0.667 per cent at a current gain of 150. The cancellation also needs the node between the devices to stay high-impedance. The lower device's own base-collector capacitance, seen from its collector, is multiplied by its gain, so the upper device's collector noise comes through above 7.2 MHz with 20 pF at that node, and at 100 MHz the cascode stage is up to 0.55 dB noisier than the plain one.

 A field-effect input beats a bipolar one biased for each source from 2.83 kΩ to 5 GΩ. computed by solving, not by drawing. The noise figure a field-effect input shows a source — drain current 1 mA, voltage noise 2.31 nV/√Hz from its channel, current noise 0.566 fA/√Hz from 1 pA of gate leakage — against a bipolar input of β = 100 and 50 Ω of base resistance with its collector current set for each source, whose best is 1 + rb/Rₛ + 1/√β. The field-effect input is quieter between 2.83 kΩ and 5 GΩ; the closed forms √β·(Rₙ − rb) and 2Vₜ/(√β·Iₗ) give 2.83 kΩ and 5 GΩ. Its own best, 0.0007 dB, is at 4.08 MΩ; the bipolar floor never goes below 0.414 dB. The floor, which bounds from below

The window a field-effect input wins in

A bipolar input's two noise generators are one current's shot noise split two ways, so their product, and with it the best noise figure, is fixed at 1 + 1/√β, 0.414 dB at a gain of a hundred, whatever the bias. A field-effect input's are two unrelated things. Its voltage noise is its channel's and its current noise is the shot noise of its gate leakage, so their product is a design variable: 0.0007 dB at a milliampere of drain and a picoampere of leakage. Against a bipolar input biased for each source, the field-effect input wins in a window of source resistance with two ends in closed form, from 2.83 kΩ, set by the two inputs' noise resistances, up to 5 GΩ, set by the gate leakage. A nanoampere of leakage brings the top down to 5 MΩ, and 438 nA closes the window.

 On a 300 kHz amplifier the stage lags 27.9° at the null: the Q stays at 25.4 and the peak moves 1.02% below it. computed by solving, not by drawing. The 10 kΩ, 10 nF twin-T, exact, as the feedback network of a stage of gain 100 whose own bandwidth is the amplifier's gain–bandwidth over 100, so that at the null, 1591.5 Hz, it lags 27.95° (solid); dashed is an ideal stage. The quality factor is 25.440 against 25.247 — hardly moved — and so is the peak's height, 99.7 against 100.0. What moves is where the peak is: 1.024% below the null, where 2·f₀/GBW is 1.061%. At the null itself the gain is 88.3, 0.886 of the peak. With the shunt elements 1% out, the Q is 33.72 against 33.50. Filters, measured not tabulated

The peak the amplifier moves

A twin-T around a stage of gain A makes a peak of quality factor (A + 1)/4, and a real stage has a bandwidth of its own, so its gain at the null is complex. The expectation is that the lag turns the loop's gain and the Q rises by 1/cos φ, doubling at 60°. It does not: a single pole that turns the gain by φ also shrinks it by cos φ, and the two cancel in the width of the peak — the Q holds to about one part in the gain up to 60°, and the loop still oscillates at the same shunt error, 4.1% at a gain of 100, to three per cent. What the stage's bandwidth moves is the peak itself: 2·f₀/GBW below the null, with the gain nowhere in it. At 45° of lag the frequency the circuit was built to select sits on its own −3 dB edge.

 A line cut into 8 LC sections passes 10% at 0.863 and 50% at 1.043 of its time of flight. computed by solving, not by drawing, marched at 8000 steps. A lossless line of 1 µH and 100 pF — a 100 Ω line with a 10 ns time of flight — built from 8 equal LC sections and driven through 100 Ω into 100 Ω, so the ideal line would deliver a clean half-volt step at exactly one time of flight (dashed). The ladder network passes 10% of its final value at 0.8630, 50% at 1.0428 and 90% at 1.1592 of the time of flight: a delay quoted at 10% is 18.0% of a flight time shorter than one quoted at 50%. It starts as the 16th power of time, for 16 poles and no zeros, and rings after the edge because a lumped ladder network cannot pass what lies above its sections' cut-off. Before the steady state

The threshold a delay is quoted at

A step through a network of r poles starts as the r-th power of time, so a delay measured at 10% moves with the network's order more than one measured at 50%. Cut a line into more sections and the order grows without limit, and what happens to the two delays depends on the line. A lossless LC line converges on its time of flight at every threshold: the spread between its 10% and 50% delays falls from 0.656 of a flight time with one section to 0.070 with thirty-two, as the −0.68 power of the section count, the −2/3 of an Airy front. An RC line converges too, but its 10% and 50% delays converge to 0.134 and 0.391 of R·C, a third of each other for good. On a propagating line a delay belongs to the path; on a diffusing one, two-thirds of it belongs to the threshold.

 At full capacitance the margin falls with the load, and below a tenth of an ampere no divider capacitor gets it past 36.9°. computed by solving, not by drawing. Phase margin of the five-volt regulator with a 5 mΩ ceramic keeping 10 µF, against the load current from 5 mA to 1 A, for four capacitors across the upper divider resistor. 630 pF, which held 45° at every capacitance at an ampere, gives 49.7° at 1 A and 36.0° at 5 mA. With no capacitor the margin at 5 mA is 0.1°. The dotted level is 36.87°, the most phase a zero and a pole four times apart can ever lead by, and every curve ends under it. The crossover stays between 12.9 kHz and 13.2 kHz over the whole range. Circuits that do a job, and the range they do it over

The lead a divider can give

At an ampere one capacitor across the upper divider resistor holds this ceramic regulator above 45° at every capacitance the part keeps. At 5 mA nothing does. The crossover stays at 13 kHz while the output pole falls from 3.20 kHz to 31.8 Hz, so the loop arrives at unity with every lag it has and the only phase left is the lead the divider capacitor makes — and that lead has a ceiling, arcsin((G − 1)/(G + 1)), set by the divider's ratio alone. For a five-volt output from a 1.25 V reference it is 36.87°, and the best capacitor reaches 36.83°. Over the rectangle from 5 mA to an ampere and ten microfarads to two, the best single capacitor gives 32.4°. Putting 416 mΩ back in series with the ceramic restores 45° over all of it, and lowers the droop rather than raising it.

 At gₘRₑ = 1 a pair holds its gain to one per cent up to 29.2 mV, a single transistor to 29.9 mV. computed by solving, not by drawing, every point a waveform solved sample by sample and its fundamental read from its transform. The fractional departure of the fundamental from the small-signal gain against the peak drive, for a differential pair (solid) and a single transistor (dashed), each device with an emitter resistor of 1 times its own 1/gₘ at 27 °C. The pair reaches one per cent at 29.2 mV; the cubic law 0.4·Vₜ·(1 + gₘRₑ)^(3/2) says 29.3 mV, which holds while the drive leaves most of the tail unsteered and overstates the boundary beyond; the single transistor at 29.9 mV, a ratio of 0.974. Without degeneration the ratio was √2 in the limit; with heavy degeneration it approaches two; and near gₘRₑ = ½, where the single transistor's cubic vanishes, the single transistor briefly holds its gain longer. Where the models stop

Symmetry and a resistor together

A pair linearises by symmetry and an emitter resistor by feedback, and the two are normally used together. Solved together, the pair's one-per-cent gain boundary is 0.4·Vₜ times (1 + gₘRₑ) to the three-halves power up to gₘRₑ = 3: 10.4 mV bare, 29.2 mV at gₘRₑ = 1, 80.9 at 3. The √2 by which it outlasted a single transistor does not survive: at gₘRₑ = 1 the two are within three per cent, near ½ the single transistor outlasts the pair fourfold, because its cubic vanishes there and a pair's never can, and with heavy degeneration the ratio climbs to 1.95 at gₘRₑ = 50 — two, for a reason that has nothing to do with junctions: a single transistor cuts off when the drive reaches Ic·Re, and a pair has steered its whole tail at 2·Ic·Re, its two emitter resistors in series. What symmetry alone buys is the even orders, at every degeneration.

 The one number the window does not move. computed by solving, not by drawing. Loss against foil thickness, at three window fills, with each curve's minimum located by a parabola through its three lowest points rather than read off the grid. The optimum sits at 0.654 skin depths at full fill, 0.708 at forty per cent, against the closed form's 0.663 — a drift of 8.3 per cent while the ratio the same winding carries moves by eighty. The alternating-current resistance at the optimum is 1.340, 1.351, 1.406, against four thirds. What did move is the loss it costs: 0.0555 watts a metre at full fill and 0.1383 at forty per cent, for the same current in the same number of layers. Two windings, and the band between them

The optimum that does not move

A foil winding has a best thickness — past it, more copper is more resistance — and at that thickness the alternating-current resistance is four thirds of the direct-current resistance, whatever the layer count. Both of those are one-dimensional results, and this ladder has spent three rungs finding that the one-dimensional picture is 82 per cent wrong about the resistance ratio. Solved as a field, the optimum drifts by 8.3 per cent between a full window and a quarter-full one, and four thirds becomes 1.34, 1.35, 1.41. The trade barely moves while everything it is made of moves a great deal.

 With the crossover at 10% of the clock the straight line overshoots 13.7%, the hold with its pole 32.8%. computed by solving, not by drawing, marched at 160 steps a clock. A unit step through a loop whose plant is an integrator crossing over at 10% of the sample rate, reconstructed three ways: held (dotted), joined by straight lines (solid), and held then passed through the analogue pole that attenuates the nearest image, at 90% of the clock, exactly as much as the straight line's second sinc does (dashed) — a pole at 9.90% of the clock, at the crossover itself. Overshoot: 0.0%, 13.7% and 32.8%. The straight line buys the same image rejection for half a clock of delay; the pole buys it for forty-five degrees. Where a signal becomes a number

Half a clock against a pole

Inside a digitally controlled loop a converter's delay is phase at the crossover, and a straight-line output pays one more half clock than a hold: 180 degrees times the crossover over the clock. The analogue pole that rejects the nearest image as much as the straight line's second sinc does turns out to sit at the crossover itself — 0.990 of it when the loop crosses over at a tenth of its clock, sinc(r) of it in general — so it costs forty-five degrees however slow the loop is. By the phase estimate the half clock is the cheaper filter below 26% of the clock. Marched as a sampled loop the answer is 17.6%, and at a tenth of the clock a step overshoots 13.7% through the straight line against 32.8% through the hold and its pole, for the same image rejection measured on both waveforms.

 A parasitic of 0.01 ms changes the start only inside itself: t³ until 30.0 µs with a pole, t until 20.0 µs with a zero. Expanded about infinity, and by residues after. The start of the step through two buffered RC sections (τ = 1 ms and 0.5 ms) on logarithmic axes: bare, a straight line of slope two (dashed); with one more pole of 0.01 ms (solid), slope three until the asymptotes cross at (r + 1)τp = 30.0 µs; and with a zero of the same time constant (dotted), slope one until r·τz = 20.0 µs. Well inside the parasitic's time constant the pole multiplies the bare step by t/((r + 1)τp) and the zero by r·τz/t, both checked on the expansions; outside it both curves rejoin the bare one, displaced. A parasitic a hundred times above the network's bandwidth decides the start's power law for its own few time constants and no longer. Before the steady state

The parasitic that moves the start

A pole far above a network's bandwidth raises its relative degree by one, so the initial-value theorem says its step now starts as tʳ⁺¹ instead of tʳ; a zero lowers it. Both are true and both expire. Added to a two-section step, a 10 µs parasitic pole holds the start at the cube of time until the asymptotes cross at (r + 1)τp = 30 µs and a 10 µs zero holds it at the first power until r·τz = 20 µs; thirty time constants later the local power is back within 0.07 of the bare step's. After that each does one thing only: it moves every threshold by its own time constant, the pole later and the zero earlier. Since the time is the same at every threshold, the fraction it costs is not — 9.1% of a 1% delay, 0.82% of a 50% one — and a pole and a zero together keep the start's power, scale it by τz/τp, and move every later threshold by τp − τz.

 A 1 ms pulse of 100 A warms the element 98.7 mK: 4.94 ppm at its end, 2.86 ppm averaged, against 775 ppm held. computed by solving, not by drawing, from the exact step response of the shunt's thermal network — element, terminations and board, 0.2, 4 and 15.8 K/W with 5 mJ/K, 0.1 J/K and 2 J/K — expanded into its Foster modes. A 100 A pulse of 1 ms through the 77.5 µΩ shunt the burden optimum picks, dissipating I·u* = 0.775 W: the element rises 98.7 mK by the pulse's end and cools after it. At 50 ppm/K that is 4.94 ppm of error at the end of the pulse and 2.86 ppm averaged over it, where the same current held long enough for the board to settle would read 775 ppm high. Measurement, which is a circuit on a circuit

The pulse that ends before the heat

A shunt's self-heating error is a fixed point in the steady state — 775 ppm at 100 A for the shunt the burden optimum picks, through 20 K/W. A pulse never reaches it. Heat leaves the element through a ladder network — the element's own 5 mJ/K, its terminations, the board — and a pulse short against the element's millisecond heats it adiabatically, by P·t over its heat capacity alone: 0.737 ppm for 100 µs at 100 A, a thousandth of the held figure, 10.7 ppm for 10 ms. Along the burden optimum the error is linear in the current, so the current a pulse may carry for 100 ppm is exact: 12.9 A held, 936 A for 10 ms, 13.6 kA for 100 µs. A reading averaged over the pulse carries half the error of one taken at its end while the heating is adiabatic, and the section-by-section thermal model a data sheet quotes is 10% wrong at 7 ms.

 A 1 kΩ bias resistance at the upper base is attenuated to 7.97 × 10⁻⁶ at low frequency and passes the lower device's noise at 15.2 MHz. computed by solving, not by drawing. The output noise current of the cascode stage at 1 mA and β = 150, driven from 1 kΩ, source by source, with the upper device's base fed through 1 kΩ instead of held at signal ground. At low frequency the resistance's noise reaches the output as 7.97 × 10⁻⁶ of the lower collector's, about 2R/(gₘrₒ²): its voltage is followed to the node between the devices and becomes a current only through the lower device's rₒ. The upper base's own shot noise, flowing in the same resistance, is 0.683%. The resistance's noise passes the lower device's at 15.2 MHz, and at 100 MHz the stage's noise figure is 6.07 dB worse than with the base held. Devices, and the amplitude they stop being linear at

The resistor the upper base is held through

A cascode's upper base is drawn at signal ground, and on a board it is held by a bias network with a resistance of its own. At low frequency that resistance's noise is harmless for the reason usually given — the upper device follows its base and turns the voltage into a current only through the lower device's output resistance — so 1 kΩ reaches the output at 8 × 10⁻⁶ of the lower device's noise. But the attenuation belongs to rₒ, and rₒ is shunted by the node's capacitance from 23.5 kHz, three hundred times below the frequency at which the upper collector's own noise stops cancelling. Unbypassed, 1 kΩ costs 6.07 dB at 100 MHz, more than 100 Ω or 100 kΩ does, and 100 pF across it makes any resistance free.

 A tenth of a per cent of screen mismatch on the gauge's side cancels the pair's own phase: 18.2 kHz becomes 143 kHz. computed by solving, not by drawing. A 350 Ω quarter bridge at 1000 µε read through ten metres of screened pair: 100 pF a metre from each conductor to the screen, Cs, and 50 pF a metre between the two, Cp. The loss of the synchronously demodulated reading, 1 − cos φ, against the excitation frequency: for the capacitance between the conductors alone, for the screen capacitances alone with a mismatch of 0.1% on the gauge's side, for both together, and for a matched screen with nothing between. For a 200 ppm budget they allow 18.2 kHz, 18.3 kHz, 143 kHz and 205 kHz. The pair's capacitance lags the reading; extra screen capacitance on the gauge's own terminal leads it, by an amount divided by the signal, so at a mismatch of δ·Cp/Cs the two cancel to first order. Circuits that do a job, and the range they do it over

The capacitance that is already a difference

A screened pair has capacitance from each conductor to the screen, which cancels in a bridge's difference when the two match, and capacitance between the conductors, which is a difference from the start and cannot cancel. Ten metres of it between a 350 Ω quarter bridge and its carrier amplifier allows 18.2 kHz for a 200 ppm loss, against 205 kHz for the screen alone. The two phases are ωR·ΔC/δ and ωR·C, the same slope with one divided by the signal, so a screen mismatch of δ·Cp/Cs on the gauge's own terminal cancels the pair's lag exactly: 0.10% at 1000 µε takes the limit to 143 kHz. That is what a carrier amplifier's capacitance balance finds when it nulls the quadrature — and because the mismatch's phase changes sign with the strain and the pair's does not, the balance holds in tension above half the strain it was set at and never in compression.

 With 1 mV of offset the pair's second harmonic returns in proportion to the drive, and the third harmonic reaches one per cent first. computed by solving, not by drawing, every harmonic read from a solved waveform. A differential pair at 27 °C with 1 mV of input offset: its second harmonic (solid) and third (dashed) over its fundamental, against the peak drive, beside a single transistor's second harmonic (dotted). The third is the balanced pair's own, u²/12, and reaches one per cent at 17.9 mV whatever the offset. The second is (u/2)·tanh(Vos/2Vₜ) at small drive, the single transistor's second harmonic times tanh(Vos/2Vₜ) = 1.93e-2, and never reaches one per cent: it peaks at 0.74% as the pair begins to saturate and falls again. Where the models stop

The offset that brings the even orders back

A balanced pair has no second harmonic. An offset gives it one, and the closed form is exact: the second harmonic over the fundamental is (u/2)·tanh(Vos/2Vₜ), with u the drive over 2Vₜ — a single transistor's second harmonic times the fraction of the tail the offset steers aside, 1.93% of it at a millivolt, not the twentieth usually estimated. It is first order in the offset up to 9.0 mV. It does not grow with the drive for ever: it peaks as the pair saturates and falls again, so below 1.36 mV of offset it never reaches one per cent at any drive, and above 3.14 mV it reaches one per cent before the pair's own third harmonic does, at 17.9 mV. Read the other way, a matching specification is a distortion specification: −60 dB of second harmonic at 10 mV of drive allows 0.535 mV of offset.

 3 gaps of 0.40 mm, where one of 1.20 would do. computed by solving, not by drawing. The same total gap, the same turns, the same core, and very nearly the same inductance — cut into 3 instead of one. Each gap now drops one part in 3 of the magnetomotive force, so the field it throws into the window is that much weaker; and because the field's own energy goes as its square, the loss it causes falls faster than the field does. The winding dissipates 3.00 times its direct-current loss here against 6.52 with a single gap, and its worst turn 3.7 against 13.3. Two windings, and the band between them

The gap that is three gaps

Reluctances in series add, so one gap of 1.2 millimetres and three of 0.4 are the same magnetic circuit: the same inductance, the same saturation current, the same energy in the air. They are not the same field. Solved in two dimensions, the winding beside the single gap dissipates 6.52 times its direct-current loss and its worst turn 13.3; beside three gaps those are 3.00 and 3.7. And there is a best number of gaps rather than a monotone gain — past three, spreading them along the leg brings each one close to a different part of the winding.

 The gain's sensitivity to the device and to the divider add to one at every frequency, and at crossover each is exactly one half (34.9° of margin). computed by solving, not by drawing, three times at each of 241 frequencies: nominal, with the amplifier's gain raised by one part in 10⁵, and with the feedback fraction raised by one part in 10⁵. The fraction of the device's change reaching the closed-loop magnitude is 9.97 × 10⁻⁴ at 10.0 Hz and 1.60 at its peak near 7.94 kHz; the fraction of the divider's change is 0.99899 at 10.0 Hz and falls to −0.60 at 7.94 kHz. They add to one within 1.16 × 10⁻⁵ everywhere, and at the loop's crossover, 5.73 kHz, they are 0.5000 and 0.5000: the real parts of 1/(1 + T) and T/(1 + T), which are one half at |T| = 1 for any phase. The cut loop, a second route, agrees to 9.55 × 10⁻⁶. Feedback, and the margin

Half to the amplifier, half to the resistors

A feedback amplifier is supposed to hand its accuracy to the resistors, and at low frequency it does — 99.9 per cent of a change in the divider reaches the gain and 0.1 per cent of a change in the device. The two fractions are measured here by nudging each and solving again, and they add to one at every frequency, to five parts in a hundred thousand. At the loop's crossover each is exactly one half, whatever the margin: the gain's magnitude answers the amplifier and the resistors equally. As complex numbers both are 1/(2 sin(φ/2)) there, above one for any margin under sixty degrees, and around the crossover lies a band where the gain is more sensitive to both than it would be with no feedback at all.

 Two coupling capacitors 1% apart reject common mode by only 46.0 dB at their corner, 0.159 Hz — and perfectly at direct current and far above it. computed by solving, not by drawing at 281 frequencies: one common-mode source driving two coupling networks, each 1 µF into 1 MΩ, with the capacitors 1% apart. The difference between the two outputs, per volt of common mode, agrees with the difference of the two first-order high-passes to 6.9e-14. It is zero at direct current, where both block, and zero far above the corner, where both pass; between, it peaks at the corner, 0.159 Hz, at 5.000e-3 — half the mismatch — which is 46.0 dB of rejection. Behind a 90 dB amplifier (grey), the coupling network decides the rejection from 0.000503 Hz to 50.3 Hz. Measurement, which is a circuit on a circuit

The rejection two capacitors lose at their corner

AC-couple both inputs of a differential instrument and the two coupling networks become part of its common-mode path. Two 1 µF capacitors one per cent apart turn common mode into difference at exactly half their mismatch at their shared corner — 46.0 dB of rejection at 0.159 Hz, from a pair that is perfect at direct current and far above. Behind a 90 dB amplifier they decide the instrument's rejection over five decades around that corner. A 10 V common-mode step arrives as a 36.8 mV differential pulse one time constant later. Four hundred pairs of 20 per cent electrolytics reject a median 25 dB. Coupling the difference instead of each side removes the dip and hands the common mode to the amplifier.

 Cancelling 50% of the line's reactance: a 0.3 lagging load's regulation goes from 7.78% to 4.18%, a unity one's from 2.17% to 1.96%, a 0.7 leading one's from −3.78% to −1.26%. computed by solving, not by drawing, at 50 Hz: a 230 V feeder of 0.05 + j0.2 Ω carrying 20 kVA, its regulation against the load's power factor from 0.7 leading to 0.3 lagging, with no compensation (dashed) and with 50% of the line's reactance cancelled by a series capacitor (solid). The solve agrees with the voltage divider in closed form to 9e-10 of a per cent. At 0.3 lagging the regulation falls from 7.781% to 4.176%; at 0.9 lagging, from 5.168% to 3.381%; at unity, only from 2.171% to 1.960%; and at 0.7 leading the far end, above the source at −3.778%, comes down to −1.255%. The capacitor cancels what the load's reactive current does in the line's reactance, in either direction, and pulls every load towards the regulation the line's resistance alone would give. Power, and the part that does no work

The drop a series capacitor can reach

A series capacitor cancels part of a line's reactance, so it can only cancel the part of the drop that the reactance makes — and that part is set by the load, not the line. On a 230 V feeder of 0.05 + j0.2 Ω carrying 20 kVA, half compensation takes a 0.3 lagging load from 7.78% regulation to 4.18% and a unity load only from 2.17% to 1.96%, and it brings a 0.71 leading load's raised far end back down. Cancel all of it and a leading load and its lagging twin sit at exactly the same voltage. Cancel more and the far end rises with its load; at exactly 100% the resonance is the fundamental and a far-end fault draws 4,600 A. Where the capacitor goes, the load cannot tell.

 10 pF on the far end of 10 cm of 50 Ω track: a 1 ns edge overshoots by 47.5% and rings near 174 MHz, against 40.0% with no capacitance. computed by solving, not by drawing, by the method of characteristics: 10 cm of 50 Ω track on FR-4 (a 0.70 ns transit), driven from 10 Ω by a 1 V edge rising in 1 ns into 200 Ω with 10 pF across it, and the far end's voltage against time. With no capacitance (dashed) the far end is the staircase of the rise-time essay, overshooting by 40.0% and settling on the divider, 0.952 V. With 10 pF the reflection is no longer a number: the capacitor first holds the far end down and then lets it swing, the overshoot is 47.5%, and the ringing is near one over 2π√(LC) = 174 MHz, with L the track's own 35.0 nH and C its own 14.0 pF plus the load's. Both settle on the divider. Where the models stop

The load that rings with the line

Put 10 pF on the far end of ten centimetres of 50 Ω track and the reflection stops being a number. A one-nanosecond edge from 10 Ω into 200 Ω overshoots by 47.5 per cent instead of 40.0, and rings near 174 MHz — the track's own 35 nH against its 14 pF and the load's 10 together. The one-per-cent boundary, a rise time of 39.9 ns on a bare track, becomes 50.4 ns with 10 pF and 83.4 with 50: a rise time and a capacitance together. As the load outweighs the track the lumped model comes back, missing the line's waveform by 55 per cent of the settled value with no load capacitance and by 2.2 with 200 pF. And a source matched to the line turns every capacitance's ring into a delay of about 2.2·Z₀·C.

 Three imbalances, three slopes: a 2% amplitude imbalance is 1% common mode at every frequency, 1 mm of skew overtakes it above 455 MHz, and a 5 ps rise-time mismatch climbs at forty decibels a decade. computed by solving, not by drawing, from the transforms of the two conductors' edges and checked against the sampled edges transformed by FFT to two per cent: the common mode at the far end of a pair, as a fraction of the differential, for three separate imbalances between its halves. An amplitude imbalance of 2% gives 1.00% at every frequency. A skew of 1 mm on FR-4, 7.00 ps, gives sin(ωΔτ/2), rising at twenty decibels a decade: 2.20% at 1 GHz. Edges of 50 and 55 ps, perfectly timed and equal in size, give a common mode rising at forty decibels a decade: 0.043% at 1 GHz and 5.38% at 10 GHz. The skew passes the amplitude imbalance near 455 MHz, which is a/(2πΔτ). Lines, where a wire has a length

The common mode that has three slopes

A differential pair turns into common mode wherever its two halves differ, and they can differ in three ways — in size, in timing and in edge rate — each with its own slope. A 2 per cent amplitude imbalance is 1 per cent common mode at every frequency. A millimetre of skew rises at twenty decibels a decade and overtakes it above 455 MHz, a/(2πΔτ). Edges of 50 and 55 picoseconds, perfectly timed, rise at forty: 0.043 per cent at a gigahertz and 5.4 at ten. In time they are a step that stays, a pulse with an area, and a doublet with none — each slope is a moment that vanishes. And an amplitude imbalance and an edge-rate mismatch of the right signs cancel at one frequency, down 74 dB, which half a picosecond of skew fills back to −25.

 Settling to 0.01% takes 1.47 µs for any step up to 79.6 mV, and twice that only from 1.02 V — the slew limit arrives in the settling time gently. computed by solving, not by drawing, marched and in closed form: an amplifier of 1 MHz gain–bandwidth and a 0.5 V/µs slew rate, at unity gain, settling to 0.01% of its step, against the step's size. Up to SR·τ = 79.6 mV it never slews and settles in τ·ln(1/ε) = 1.466 µs whatever the size. Above it, it slews until its error is SR·τ and settles exponentially from there, in (V − SR·τ)/SR + τ·ln(SR·τ/εV): 1.56 µs at 200 mV, 2.01 µs at 500 mV, 2.90 µs at 1 V, 4.79 µs at 2 V, 10.65 µs at 5 V, 20.54 µs at 10 V. The march and the closed form agree to a per cent. The curve leaves the flat line with zero slope and joins the pure slewing line V/SR (dashed) from below; the settling time has doubled only at 1.02 V, 12.8 times the linear boundary. Before the steady state

How long a large step takes to settle

An amplifier of 1 MHz gain–bandwidth and 0.5 V/µs slew rate stops being linear above an 80 mV step. Its settling time does not notice for a long while: to 0.01 per cent it is 1.47 µs for any step up to 79.6 mV, 1.56 at 200 mV, 2.90 at a volt, and twice the linear figure only from 1.02 V, thirteen times the boundary. The reason is that a large step slews until its error is SR·τ and then settles exactly as a small one does, so the extra time is τ(x − 1 − ln x) — second order at the boundary and the whole slewing time far beyond it. A data sheet's 20.5 µs for a 10 V step therefore says almost nothing about a 1 V step, which takes 2.9, and nothing about 100 mV.

 At 100 Hz the bowl's bottom is at 2.11 kΩ and 3.10 dB, against 6.67 kΩ and 1.14 dB in the white region. computed by solving, not by drawing: the spot noise figure against source resistance of an amplifier with 4 nV/√Hz and 0.6 pA/√Hz of white noise, whose voltage noise has a flicker corner at 10 Hz and whose current noise has one at 1 kHz. At 100 Hz the spot densities are 4.20 nV/√Hz and 1.99 pA/√Hz, so the optimum is 2.11 kΩ — found by search on the drawn curve at 2.11 kΩ — with a floor of 3.102 dB. Dashed: the white-region bowl, 6.67 kΩ and 1.138 dB. A source left at the white optimum reads 4.492 dB at this frequency. The slider is the frequency. The floor, which bounds from below

The source resistance the flicker moves

An amplifier's optimum source resistance is the ratio of its voltage noise to its current noise, and both densities rise below their flicker corners — which are rarely the same frequency. With the voltage noise's corner at 10 Hz and the current noise's at 1 kHz, a part whose optimum is 6.67 kΩ at high frequency wants 2.11 kΩ at 100 Hz and 699 Ω at 1 Hz, tending to the white value times the square root of the ratio of the corners. Over the audio band the drift costs 0.012 dB; over a band from 1 to 100 Hz, choosing the white optimum costs 3.77 dB, and the spot optimum at the band's centre recovers all but a tenth of a decibel.

 A second coupling stage with the same time constant: the top sags to 0.303 instead of 0.607, and the tail crosses zero at 2.27 s and overshoots to +0.025. computed by solving, not by drawing, marched and checked against the closed form to 3.0e-4: a half-second pulse through a 1 µF, 1 MΩ coupling stage (dashed), and through that stage followed by a buffered second one with a time constant of 1 s (solid). One stage sags to 0.6065 by the pulse's end and its undershoot decays monotonically to zero. Two stages sag to 0.3033, undershoot to −0.6966, cross zero 2.271 s after the pulse began and overshoot positive — the second capacitor removes the first moment of the waveform as well as its area, so the tail has to change sign. Measurement, which is a circuit on a circuit

The second capacitor, and the tail that changes sign

One coupling stage turns a pulse into a sagging top and an undershoot that returns to zero from below. Add a second stage with the same one-second time constant and a half-second pulse sags to 0.303 of itself instead of 0.607, undershoots to −0.697, crosses zero 2.27 s after it began and overshoots positive before settling. The step response becomes (1 − t/τ)·exp(−t/τ), through zero at exactly one time constant and down to −e⁻² at two. The reason is arithmetic: one capacitor removes the waveform's area, two remove its area and its first moment, and a pulse with a one-signed tail cannot have both. Built to the same stated corner, two stages still sag 1.29 times as fast, and a 1% flat top needs a pulse rate 806 times the corner rather than 625.

 Two 230 V sources joined by 0.05 + j0.2 Ω: the most that can be received is 194 kW, at 76.0°; with 50% of the reactance cancelled it is 262 kW, at 63.4°. computed by solving, not by drawing, at 50 Hz: two 230 V sources, the sending one leading the receiving one by an angle δ, joined by a line of 0.05 Ω and j0.2 Ω with a series capacitor cancelling a fraction of the reactance; the real power absorbed at the receiving end against δ. Uncompensated (dashed) the maximum is 194.37 kW at 75.96°, the line's impedance angle. With 50% cancelled (solid) it is 261.55 kW at 63.43°. Both maxima match V² cos β (1 − cos β)/R, found by a search on the solve to a part in a million. Past its maximum the received power falls as the angle grows: a larger angle drives more current, and past the impedance angle the extra current's loss outgrows what it carries. Power, and the part that does no work

Sixty degrees, and the limit compensation stops raising

Series capacitors are put in transmission lines to raise how much power can pass between two sources before the angle between them runs out. The lossless textbook limit, V²/X(1 − k), rises without bound as the reactance is cancelled. With the line's resistance kept, the limit is V² cos β (1 − cos β)/R for an impedance angle β, and it peaks where β is exactly sixty degrees: 56.7 per cent cancellation and 264.5 kW on a 230 V tie of 0.05 + j0.2 Ω, against 194.4 uncompensated — then falls, to nothing at full cancellation. A line of R/X 0.1 can nearly triple its limit; one above 1/√3 cannot raise it at all. And at the best compensation the line delivers exactly half of what it is sent.

 A 1 ns edge from 10 Ω into 200 Ω on 50 Ω track stays inside one per cent of overshoot only up to 2.5 mm of track. computed by solving, not by drawing, the staircase of reflections marched against a ramp: the far end's overshoot against the track's length for a 1 ns edge, 50 Ω track on FR-4 driven from 10 Ω into 200 Ω. 2 mm: 0.79%; 5 mm: 1.98%; 10 mm: 3.98%; 20 mm: 8.07%; 50 mm: 22.93%; 100 mm: 40.00%; 200 mm: 40.00%; 500 mm: 40.00%. The longest track inside one per cent, bisected, is 2.51 mm — a round trip of 35 ps, the edge being 28.5 times it — and the rise-time boundary of that length is 0.994 ns, this edge given back. Where the models stop

How long a track may be

The lumped-element boundary read the way a layout is designed: given the edge, how long may the track be? For 50 Ω track driven from 10 Ω into 200 Ω, a 1 ns edge stays inside one per cent of overshoot for 2.51 mm, and every nanosecond of edge buys 2.52 mm more — proportional to within half a per cent from 0.1 ns to 30. The mismatch decides the rest. From 45 Ω the same edge may cross 23.5 mm; from 50 Ω, or anything above it into this load, no length overshoots at all; and into an open end, a CMOS input, 1.80 mm. The common rule — a track is short while its round trip is under the edge — allows 49 per cent of overshoot from a 5 Ω driver, and a tenth of its length still allows 3.3.

 The same 0.1 ps of period jitter: 107.0 dB at any record length if each edge is independent, 80.3 dB over 4096 samples if the edges walk. computed by solving, not by drawing, averaged over 8 seeded clocks at each of 5 record lengths. A 10.01 MHz tone sampled at 160 MHz by a clock whose every period has a spread of 0.1 ps, in two ways: each edge independently displaced (white), and each edge displaced from where the last one landed (a free-running oscillator's walk). The sampled tone is fitted by least squares for amplitude, delay and a small frequency offset, and what is left is the error. White jitter gives 107.2, 107.0, 107.0, 107.1, 107.0 dB — the closed form 107.0 dB, at every length. The walk gives 92.6, 86.4, 80.3, 74.5, 68.2 dB for 256, 1024, 4096, 16384, 65536 samples, falling 10.1 dB a decade, as the period spread times √(n/15): a jitter with no value until the length of the look is named. Where a signal becomes a number

A jitter that grows with the look

Two clocks with the same 0.1 ps spread in every period sample the same 10 MHz tone. If each edge is displaced independently, the error is 107.0 dB down at any record length. If each edge is displaced from where the last one landed — a free-running oscillator — the edges walk, and the error after fitting the best delay and frequency is 92.6 dB down over 256 samples, 80.3 over 4,096 and 68.2 over 65,536: ten decibels a decade of record, as the period spread times √(n/15). The one oscillator honestly has jitter figures from 0.10 to 4.7 ps depending on the length named. A locking loop is what gives the jitter a value: with a time constant of a hundred periods it stops growing at 87.0 dB.

 Driven through 1kΩ, the absorbing capacitor's tail is 2.6e-5 of the step, not 1%. computed by solving, not by drawing. The error at a capacitor's terminal after a step applied through 1k ohms, where the capacitor is 1 nF with eight absorption branches spread over two decades from 0.1 ms, together holding one per cent of its capacitance. The poles are found by bisection and the residues in closed form, and a trapezoidal march of the same netlist agrees. The tail the branches leave is 2.58e-5 of the step — the source resistance over the branch resistances in parallel — and settling to 1e-5 takes 173µs against 11.5µs with no branches; the model that weights each branch by its share of the capacitance says 48.4ms. Before the steady state

The tail the source decides

A capacitor with one per cent of dielectric absorption does not leave a one-per-cent tail in the circuit it is in. Driven through a kilohm, the eight slow branches that model the absorption leave 2.58×10⁻⁵ of the step at its terminals — the source resistance over the branch resistances in parallel — and settling to 10⁻⁴ takes 9.49 µs against 9.21 with no absorption at all. The share of the capacitance is not the amplitude; it is a ceiling the circuit reaches only when the source's own time constant falls inside the spread of the branches, and there the cost is 2.2 times the settling at 10⁻⁴ and 16 times at 10⁻⁵.

 With 1 RC of tracking, sixteen samples average like 7.8 independent ones. computed by solving, not by drawing. The rms noise on the average of N consecutive held samples of a 1 pF capacitor, against N, for track times of 0.25, 1, 3 and 10 RC between samples, with the 1/√N law for independent samples dotted. A sample shares exp(−Tₜ/RC) of its value with the one before, so at 1 RC the correlation is 0.368 and sixteen samples are worth 7.81 independent ones; 1,024 are worth 473.6. The closed form is checked against a seeded march of the held sequence. The floor, which bounds from below

The average that is not N times quieter

Averaging sixteen samples of a 1 pF capacitor is supposed to take its 63.3 µV of kT/C noise down by four. With one RC of tracking between samples it takes it down by 2.8: the switch has not been closed long enough for the capacitor to forget the last sample, each shares 37 per cent of its value with the one before, and sixteen samples are worth 7.8 independent ones. At a quarter of an RC they are worth 2.6. Sampling faster never helps past a ceiling: however quickly the samples come, an average gathers at most one independent sample per 2RC of time — the continuous average's own limit. And a sampler that settles its signal to 8 bits has already waited long enough for its noise.

 Calibrated at the carrier, a matched cable allows 121 kHz where its phase allowed 16.5 kHz. computed by solving, not by drawing. A 350 Ω quarter bridge read through ten metres of screened pair, excited over leads of 0.35 Ω and 3 µH each with 500 pF between them. Solid: the worst error of a synchronous reading over strains from −2000 to +2000 µε after a calibration at +1000 µε at the operating carrier, with the reference taken from the source at the instrument; dashed, from the bridge's own terminals on sense wires; dotted, the phase measure 1 − cos φ the earlier readings of this cable used. For a 200 ppm budget: 16.5 kHz by phase, 119 kHz calibrated against the source and 121 kHz calibrated against the bridge. Circuits that do a job, and the range they do it over

What a calibration at the carrier removes

The phase a cable puts on a carrier-excited bridge is a loss of 1 − cos φ, and a loss that is the same at every strain is a gain error — which a shunt calibration at the operating carrier exists to remove. Calibrated there, a synchronous reading of a quarter bridge through ten metres of matched screened pair allows 121 kHz for 200 ppm over ±2000 µε, where the phase measure allowed 16.5; the capacitance between the conductors, which no matching could improve, stops mattering. Where the reference is taken — at the instrument or from sense wires at the bridge — moves that by 1.2 per cent. What calibration cannot touch is a screen mismatch: its leak is a fixed voltage, not a fixed gain, and a hundredth of a per cent of it takes 121 kHz back to 11.5.

 The trade, in the plane where both halves of it live. computed by solving, not by drawing. Leakage inductance across, interwinding capacitance up, both solved on the same cross-section with the same cells. The faint diagonals are lines of constant leakage-times-capacitance, so a design action that runs along one of them has bought nothing and only moved where the energy is kept. Interleaving runs at slope -0.78, which is nearly along them: six sections cut the leakage 21.3 fold and multiply the capacitance 11.0 fold, and the product moves by 1.94. What it does change is the winding's characteristic impedance, 95 ohms down to 6.2 — a factor of 15. Thickening the interlayer instead runs at -4.8, steeply across the diagonals, and moves the product 5.2 fold over the same sweep. It is the cheaper action by that measure and it is not free either: the millimetre it spends is a millimetre of window that is not copper. Two windings, and the band between them

Interleaving is a choice, not an improvement

Splitting a transformer's windings into six sections divides its leakage inductance by 21.4 and multiplies its winding-to-winding capacitance by 11.0. The product of the two — which is what sets the frequency the part stops being a transformer at — moves by 1.94, and the resonance it decides goes from 1.804 to 2.515 megahertz for all that work. What interleaving really changes is the winding's characteristic impedance, 95.2 ohms down to 6.2, and nobody quotes it.

 3 relaxations over two decades come back within 0.22% from readings good to 1e-5. computed by solving, not by drawing. A one-per-cent tail divided equally among 3 branches spread over two decades from 0.1 ms, read at 80 times with 1e-5 of seeded noise on each reading, and the amplitudes recovered by least squares with the time constants given to the fit. Dots: one recovery; bars: the rms spread over 200. The condition number of the fit is 8.29e+0, and the worst branch comes back with an rms error of 0.224 per cent. Before the steady state

How many relaxations a settling curve holds

A settling curve read at eighty times, each good to ten parts in a million of the step, returns the amplitudes of three relaxations spread over two decades to 0.22 per cent and of eight to 46 — even when the fit is handed every time constant exactly. The condition number of the inversion grows by a factor of 7.7 for each extra relaxation packed into a decade, so each tenfold improvement in the reading buys a little more than one more relaxation a decade and no more. A settling measurement can count the dielectric's slowest relaxations; it cannot resolve a crowded spectrum, and how crowded is readable before any amplitude is.

 Three bias points below a coupling of 56.6 kΩ, one above — and the middle one is where Newton from nought lands. computed by solving, not by drawing. Two transistors, each with 1 kΩ to a 5 V supply and each biased from the other's collector through a coupling resistor into a 10 kΩ base resistor, solved by Newton on the netlist from six starting guesses and de-duplicated. Below a coupling of 56.6 kΩ there are three operating points: two latched, with one collector at 0.061 V and the other at 4.615 V at 10 kΩ, and a symmetric one at 1.702 V whose Jacobian determinant is negative, so the circuit cannot rest there. Above it the three merge into one symmetric point, and the circuit is an amplifier rather than a memory. Devices, and the amplitude they stop being linear at

The circuit with three bias points

Two transistors, each biased from the other's collector, have three operating points: one collector at 0.061 V and the other at 4.615, the same the other way round, and both at 1.702. All three solve the circuit to a part in 10⁹, and Newton started from nought — where a simulator starts — finds the symmetric one, which is the one the circuit can never sit at: its Jacobian determinant is negative, so any positive capacitance at the nodes makes it run. Weaken the coupling past 56.6 kΩ and the three merge into one, the loop gain around the symmetric point having fallen through one. Which answer a solver returns depends on where it begins, and from 1,681 starting guesses at 10 kΩ, 437 return the one that is not a state.

 Returning the shunt resistor to k times the output divides the residue by up to 30 and never takes it through zero. computed by solving, not by drawing. The residue of an RC bridged-T with R₁/R₂ = 100 at its null frequency, with its shunt resistor returned to k times its own output rather than to ground, against 1 − k: solved (dots) and 2/(2 + n(1 − k)) (line). An inverting return raises the effective ratio to n(1 − k) — at k = −15 it is 1600 — so the residue falls to 0.00125. But the null's slope falls with it, and the loop's ceiling, the slope over twice the residue, is 4.994 against 4.902 earthed: it approaches √n/2 = 5.000 of the resistors' own ratio and never passes it. At no k below one does the residue reach zero, because the trim is a ratio and the residue is positive for every ratio. Filters, measured not tabulated

The residue that falls with its slope

A bridged-T's residue at its null is 2/(2 + n), and a loop around it stops at a Q near √n/2. Returning its shunt resistor to −15 times the output instead of to ground divides the residue by sixteen, to 0.00125 — and the loop's ceiling stays at 5.00, because the slope through the null falls with the residue. It has to: at a transmission zero the output is zero, so a trim that acts through the output leaves the zeros where they were, and the ceiling of a loop with unlimited gain is the zeros' own Q. What the trim does change is the gain the ceiling needs — nine-tenths of it at A = 9/ρ, 449 earthed and 14,400 at k = −31. No return gain below one takes the residue through zero, and a non-inverting one lowers the ceiling.

 A 1% mismatch in the cable reads as 3.0% of full-scale displacement, at every frequency. computed by solving, not by drawing. A capacitive displacement sensor — two 10 pF plates driven at ±5 V in antiphase facing a common electrode — read by a charge amplifier through two metres of three-conductor cable with 30 pF a metre between each excitation conductor and the sense conductor. With the two cross-capacitances mismatched by 1 per cent and the sensor centred, the reading's in-phase part (solid) is 3.00 per cent of full scale at every carrier from 100 Hz to 1 MHz: the cable is a sensor capacitance, and its mismatch a displacement. The quadrature part (dashed), which the cable's resistance makes, rises with frequency and a synchronous detector rejects it. With each conductor in its own screen the cross-capacitance is gone and so is the error. Circuits that do a job, and the range they do it over

The sensor that is the capacitance

Read a capacitive displacement sensor through the same kind of cable that limited a strain-gauge bridge's carrier, and the roles exchange. On the bridge, a 1% mismatch in the cable's capacitance was a quadrature leak that grew as the square of the carrier and could be escaped by a lower one. On a sensor of two 10 pF plates it is 3.0% of full-scale displacement, in phase, at every carrier from 100 Hz to 1 MHz — because a capacitance between an excitation conductor and the sense conductor is in parallel with the measurand and is a sensor capacitance. No detector phase separates it and no carrier escapes it; only geometry does. Screen each conductor separately and there is nothing left to mismatch.

 A tee of factor 10 turns 1 µV/K of offset drift into 65.1 pA of error at −40 °C, 10 times the single resistor's. computed by solving, not by drawing. The direct-current error of a 10 MΩ photodiode amplifier calibrated at 25 °C, as the input current that would give the same output, against temperature: the part from the amplifier's offset drift of 1 µV/K (solid) and the part from its input bias current, 1 pA at 25 °C and doubling every 10 K (dashed), for a tee of factor 10 (heavy) and a single resistor (light). The offset reaches the output multiplied by the direct-current noise gain — one for the single resistor, since the source is a capacitor, and 10 for the tee — while the bias current sees the same transimpedance through both. The bias term overtakes the offset term at 85 °C with the tee and 40 °C with one resistor; below that the tee's error is the offset's, 10 times as large. Feedback, and the margin

The drift a tee multiplies

A feedback tee reaches a large transimpedance with small resistors, and on a photodiode its price has been counted in noise: at direct current its noise gain is its own factor, where a single resistor's is one. The same noise gain multiplies the amplifier's offset. Calibrated at 25 °C and read at −40, a 10 MΩ amplifier with 1 µV/K of offset drift is 6.56 pA wrong through one resistor and 65.1 pA through a tee of factor ten — ten times, exactly, while the input bias current's error is identical through both. So the tee's drift penalty is hidden wherever the bias current dominates, which for a field-effect input doubling every 10 K is the hot end, and fully exposed at the cold end. The worst error over −40 to 85 °C comes within a factor of two of the single resistor's only above 10 MΩ for a tee of ten, and above 100 MΩ for a tee of a hundred.

 A 0.6 mm guard between the tracks leaves all the resistance shared at DC and moves the in-band share from 4.03% to 3.50%. computed by solving, not by drawing, across the plane and the guard in strips. Two tracks 200 µm above a 50 mm plane and 2 mm apart, one driven and the other listened to, with a grounded trace of 0.2, 0.6, 1.2 mm centred between them at the tracks' height and joined to the plane at the ends of the run: the share of the driven loop's resistance that the quiet loop's return also carries, against frequency. The guard is part of the return net, so at direct current it changes nothing — the whole plane's resistance is still shared. Above the band a 0.6 mm guard moves the share from 4.03% to 3.50%. Lines, where a wire has a length

The guard that is part of the return

Where a slot between two tracks makes their returns share more of the plane, adding copper instead — a grounded trace between them — makes them share less. But not much, and not at every width. At direct current a guard joined to the plane at the ends of the run is part of the same return and changes nothing: the whole plane's resistance is still shared. At 100 MHz a guard 1.4 mm wide in the 2 mm gap cuts the shared resistance from 4.03 to 2.55 per cent and the mutual inductance by half; a guard 0.1 mm wide raises the sharing to 4.44 per cent, since a thin trace beside both tracks carries both returns through its own small cross-section. The guard works by taking return current the quiet track's plane would have carried — 7 per cent of the neighbour's for a 0.6 mm guard — and it is bounded by exactly that.

 A second pole at 3.00 MHz makes a 9.49 Ω source 4.1 times safer and a 949 Ω one 20% less safe. computed by solving, not by drawing. The source inductance above which a guarded input oscillates — a 100 pF cable bootstrapped by a 1.00 MHz follower with 10 pF of input capacitance — bisected on the sign of the rightmost pole of the whole netlist, against the source resistance, for a follower with no second pole and with one at 30 MHz, 3 MHz and 1.5 MHz; the thin line is R/ωₜ. With a second pole at 3.00 MHz the critical inductance is 6.85 µH at 9.49 Ω against 1.66 with one pole, and 141 µH at 949 Ω against 177: a low-resistance source rings high above the follower's crossover, where the second pole takes gain away, and a high-resistance one rings near it, where the pole adds phase. Measurement, which is a circuit on a circuit

The second pole the ringing guard sees

A guarded input driven through an inductive source oscillates once L/R passes 1/ωₜ — for a follower with one pole. Give the follower the second pole every real amplifier has and the rule stops being a bound in either direction. A 9.49 Ω source that rang above 1.66 µH now needs 6.85 µH with a second pole at 3 MHz, four times safer; a 949 Ω source that rang above 177 µH now rings above 141, a fifth worse. The dividing line is the frequency the input rings at: above the geometric mean of the follower's crossover and its second pole, the pole takes gain from the bootstrap and helps; below it the pole adds phase and hurts. On three poles a decade apart the change of sign falls within 2.4 per cent of √(fₜ·f₂).

 Degenerated to gₘRₑ = 1, a pair with 1 mV of offset peaks at 0.41% of second harmonic, against 0.74% bare. computed by solving, not by drawing, every harmonic read from a solved waveform. The second (solid) and third (dotted) harmonics, as fractions of the fundamental, of a differential pair with 1 mV of input offset and emitter degeneration gₘRₑ = 1, against the peak drive; the same pair undegenerated in grey (dashed: second; dot-dash: third). Degeneration lowers both and moves both to larger drives. At small drive the second harmonic follows (u/4)·g″/g′² at the offset's operating point, which for a small offset is (u/2)·(Vos/2Vₜ)/(1 + gₘRₑ)³. Its largest value over all drives is 0.406 per cent at 177 mV, against 0.736 per cent at 76 mV undegenerated. Where the models stop

Two decibels of matching for one of gain

An offset gives a differential pair back the second harmonic its symmetry removed. Put resistors in the emitters and that second harmonic falls as the cube of one plus gₘRₑ at a fixed input and as its square at a fixed output — measured slopes of −3.00 and −2.00 — so each decibel of gain given up buys two decibels of distortion, or two decibels more offset at the same distortion. A matching specification that allowed 0.535 mV for −60 dB at 10 mV of output allows 2.14 mV at gₘRₑ = 1. The worst case over all drives relaxes more slowly: the offset below which the second harmonic can never reach one per cent goes from 1.36 mV to 4.39 mV as gₘRₑ goes to three, because the peak it limits sits where the pair saturates, and degeneration only moves that further out.

 Over ±25% of the centre, three elements deliver 97.3% at worst; Bode and Fano allow 99.8%. computed by solving, not by drawing. The fraction of the available power a source of 2 times its resistance in inductive reactance delivers to a resistive load through the best matching network of 0, 1, 2 and 3 reactive elements for a band of ±25 per cent about the centre, each network the one whose worst transfer in the band is largest; the vertical lines mark the band. The worst in-band transfer is 54.2%, 82.8%, 89.5%, 97.3%; the Bode–Fano limit for any lossless network, reached only with infinitely many elements, is 99.81%. Power, and the part that does no work

What each element buys against the limit

A source whose resistance sits behind twice as much inductive reactance can deliver, over ±25 per cent of its centre frequency, 54.2 per cent of its available power to the best plain resistor and at worst 82.8 per cent through the best single capacitor. That single-element answer is exactly the fixed load for an uncertain reactance: the capacitor makes the reactance swing ±1 across the band and the load is √2 times the source. Two elements lift the worst case to 89.5, three to 97.3, and a fourth, in a chain of single elements, to 97.4. Bode and Fano bound every lossless network at 99.81 per cent over that band — a limit three elements come within a tenth of a decibel of and no finite network reaches.

 At 60 dB of rejection an eighth-order filter keeps 3 of the brick wall's 6 windows; a fourth-order one keeps 2. computed by solving, not by drawing. The sample rates at which a band from 100 to 120 MHz can be sampled without anything the filter in front passes at more than −60 dB folding onto the band, for a band with hard edges and for Butterworth bandpass filters of order 12, 8, 6 and 4 on the band's own edges; each row is labelled with the span, in megahertz, where the filter is still above −60 dB. A real filter moves each window's edges inward by half its transition band. The window from 120 to 200 MHz narrows to 124.4–193.2, 127.9–188.4, 132.8–182.4, 149.7–166.9 MHz as the order falls, and the lower windows close altogether. Every window is checked by folding every frequency the filter passes. Where a signal becomes a number

The windows a real filter leaves

A band from 100 to 120 MHz can be sampled at any clock in six disjoint windows — for a band with hard edges. Put a real bandpass filter in front and ask that whatever folds onto the band be 60 dB down, and every window's edges move inward by half the filter's transition band: an eighth-order Butterworth keeps three of the six, a fourth-order one two, and the window from 120 to 200 MHz shrinks from ±25 per cent of clock tolerance to ±19.1 per cent at order eight and ±5.4 per cent at order four. The windows depend on the rejection only through the rejection per order: 40 dB at order four, 60 at order six and 80 at order eight all leave the same ±15.7 per cent. So a filter's order is bought in clock tolerance, at one order per ten decibels.

 Swept at η = 1, a resonance reads 93.6% of its height, 0.45 bandwidths late and 1.26 bandwidths wide. computed by solving, not by drawing. The envelope of a series resonator's capacitor voltage, Q 50 at 1591.5 Hz, while its drive sweeps linearly upwards through resonance, plotted against the frequency the drive has reached, for sweep rates η = r/B² of 0.1, 0.3, 1 and 3; dashed, the steady-state response the sweep is chasing. At η = 1 — 1013 Hz a second, Q cycles for every bandwidth crossed when η is one — the reading peaks at 93.6 per cent of the true height, 0.445 bandwidths above resonance, and is 1.263 bandwidths wide. Swept at η = 0.03 it is the steady curve to 0.2 per cent. Frequency, which is the same solve

A sweep that never stops

Sweep a drive continuously through a resonance and read the envelope against the frequency the drive has reached, and the reading depends on one number: the rate in hertz a second divided by the square of the bandwidth. At that rate equal to one — Q cycles for every bandwidth crossed, the dwell a stepped sweep needed — a Q-50 resonance reads 93.6 per cent of its height, 0.445 bandwidths late and 1.263 bandwidths wide, and at Q 10 or 200 the same within a per cent. A tenth of that rate gets within 0.3 per cent of the height and costs ten times the time. The peak always lands beyond resonance in the direction of travel, so an upward and a downward sweep together find it to 0.0075 bandwidths where either alone is off by nearly half.

 The best foil thickness for 4 layers, for three currents with the same fundamental. computed by solving, not by drawing. The loss of a portion of 4 layers against foil thickness, with the loss weighted by the current in each harmonic rather than computed for one frequency. A sinusoid wants 0.6631 skin depths and lands at 1.3368 times the direct-current resistance — four thirds, the single-frequency constant, reproduced. A triangular ripple wants 0.6432, which is the same answer to within 3.0 per cent, so a winding carrying one needs none of this. A square current of the same fundamental wants 0.3838 — thinner by a factor of 1.728 — and lands at 1.8313, which is not four thirds and is not any constant the geometry knows. Building to the sinusoid's answer costs 16.8 per cent more loss. Two windings, and the band between them

The optimum a spectrum moves

The best foil thickness for a winding is derived for one sinusoid and quoted as a property of the geometry: a minimum at four thirds of the direct-current resistance, whatever the layer count. Weight the loss by the current in each harmonic instead and a square current of the same fundamental wants foil 1.728 times thinner and lands at 1.83, and a narrow pulse wants it 3.68 times thinner. Four thirds is a property of the current. The constant that replaces it for an ideal square edge is exactly two, and a real winding sits between them at a place its edge rate decides.

 Weighted towards the fast end as τ^(−1), the tail falls as a power of time and each decade of band costs 9.9× the time. computed by solving, not by drawing. A one-per-cent tail spread densely over five decades of time constant from 0.1 ms to 10 s, with each branch's amplitude weighted as τ to the power −m, for m = −0.5, 0, 0.5 and 1. The drawn weighting is m = 1: it settles to 10⁻³, 10⁻⁴, 10⁻⁵ and 10⁻⁶ in 972µs, 9.71ms, 96.3ms, 891ms. With m above zero the fast end carries more and the error between the ends is a power of time; at or below zero the slow end decides. Before the steady state

The tail with no slowest relaxation

Spread a one-per-cent tail densely over five decades of time constant and weight each relaxation as τ to the power −m. With m = 1 the tail falls as one over time and settles to 10⁻⁴ in 9.71 ms and to 10⁻⁶ in 891 ms — each tenfold tighter band costing ten times the time, not the fixed τ·ln 10 one relaxation charges. With even weighting it takes 14.4 s to reach 10⁻⁴; weighted towards the slow end, 26.5 s. The exponent decides whether any relaxation is the slowest one that matters: above zero none is, the settling is a power of the band with slope −1/m, and the spread's far end is never reached.

 At θ = 4 K/W the transistor holds at 311.7 K; the second crossing at 333 K runs away. computed by solving, not by drawing. The power a transistor dissipates against its junction temperature (solid), each point an electrical operating point solved at that temperature with the saturation current and thermal voltage it has there, for a transistor biased from a stiff 1 Ω Thévenin source to carry 100 mA from 15 V at 300 K with no emitter resistor; and the heat a sink of 4 K/W removes (dashed). Operating points are the crossings: 311.7 K with a thermal loop gain of 0.64 and 332.5 K with a thermal loop gain of 1.43. Above θ* = 4.43 K/W the two meet and vanish. Devices, and the amplitude they stop being linear at

The bias point that heats itself

A transistor driven from a stiff base voltage to carry 100 mA from 15 V dissipates 1.5 W, and every kelvin that heats it raises the current it carries. On a 4 K/W heat sink the heat balance has two solutions — 311.7 K, which holds, and 332.5 K, which runs away — and above 4.43 K/W it has none. A tenth of an ohm in the emitter moves that edge to 7.69 K/W, and from a quarter of an ohm there is no edge below 520 K at all. The designer's estimate from the loop gain at room temperature puts the edge at 11.04 K/W, two and a half times too high, because the loop gain grows as the part warms.

 A 10 µs delay in the loop: the peak rises to 13.53 dB and the sensitivity above crossover becomes a comb, and the two areas are still equal — 7.14 kHz·Np each side. computed by solving, not by drawing, from the loop's closed form, held against the netlist's cut loop to a part in a billion. The three-pole amplifier of the area essay (second pole 4 kHz) with a pure delay of 10 µs in its loop. Its margin falls from 34.9° to 14.3° at an unchanged crossover of 5.73 kHz, and the peak of |1/(1 + T)| rises from +5.70 to +13.53 dB. Above crossover the delay keeps turning the loop's phase, so the sensitivity rises and falls about zero with a period of 100 kHz. The area below zero is 7.144 kHz·Np, the area above 7.144, net 9.00 × 10⁻⁵ Hz·Np — integrated to 50 MHz on a grid that takes sixty steps a tooth. Feedback, and the margin

The phase a delay never stops adding

Put a pure delay in a feedback loop and Bode's sensitivity integral does not notice: with 10 µs in the three-pole amplifier's loop the logarithm of the sensitivity still has equal areas above and below zero, 7.14 kHz·Np each side. What changes is where they are. The delay keeps turning the loop's phase above crossover, so the sensitivity becomes a comb with a tooth every 100 kHz; the desensitisation shrinks from 6.74 to 6.48 kHz·Np, the main peak rises from 5.70 dB to 13.53 and overpays it by 7%, and the comb hands the excess back below zero. The margin reads the peak from below by about 1.4 dB at every delay. Past 16.9 µs the loop is unstable and the integral stops balancing — by exactly the growth rate of the pole pair that has crossed.

 From a 10 Ω driver with a 1 ns edge, a CMOS input of 1.5 pF allows only 0.86 mm of track — less than the bare open end's 1.80 — and one of 100 pF allows 10.9 mm. computed by solving, not by drawing, bisected on the march by characteristics: the longest 50 Ω track a 1 ns edge from a 10 Ω driver can cross into an open end with a capacitance across it and overshoot by less than one per cent, against the capacitance. With none it is 1.80 mm. 0.3 pF: 1.28 mm, 0.5 pF: 1.04 mm, 1 pF: 0.88 mm, 1.5 pF: 0.86 mm, 2 pF: 0.88 mm, 3 pF: 0.96 mm, 5 pF: 1.15 mm, 10 pF: 1.68 mm, 20 pF: 2.71 mm, 50 pF: 5.70 mm, 100 pF: 10.94 mm. The limit falls to its least, 0.86 mm, near 1.5 pF, and from there rises nearly in proportion to the capacitance (slope 0.87 between 20 and 100 pF), because the driver's resistance damps the ring the track's inductance makes with a heavy load. The slider is the driver's resistance. Where the models stop

The heavier input that allows a longer track

Put a capacitance across a CMOS input and the longest track a 1 ns edge from a 10 Ω driver may cross inside one per cent of overshoot does not shrink steadily. It falls from the bare open end's 1.80 mm to 0.86 mm at 1.5 pF, and then it rises — 2.71 mm at 20 pF, 10.9 at 100 — nearly in proportion to the capacitance. A heavy load turns the track into the inductor of a series RLC whose damping, (Rₛ/2)·√(C/L), the driver's resistance supplies, and at every heavy load the one-per-cent length sits at the same damping, about 0.8. So the limit scales as Rₛ²·C: a slower edge buys a 20 pF input only the 0.4 power of its length, and a stronger driver — 2 Ω instead of 10 — cuts the track it may drive from 2.71 mm to 0.20. Into a heavy input, a weak driver is the clean one.

 With 0.05 Ω of machine reactance at each end the sixty-degree peak needs 107% of the line cancelled — out of reach — so the limit rises all the way to full cancellation, 261.6 kW against 145.3. computed by solving, not by drawing, at 50 Hz, searched over the angle between the two EMFs on the solve (dots) and from V²·cos β·(1 − cos β)/R with the totals in it (line): the 230 V tie of 0.05 + j0.2 Ω with each source a machine whose EMF is held behind 0.05 Ω of its own reactance, against the fraction of the LINE's reactance cancelled. The capacitor cannot reach the machines' reactance, so the total impedance angle reaches sixty degrees only at k* = (X + 2Xₘ − √3·R)/X = 106.7%, beyond full cancellation; the limit rises from 145.34 to 261.55 kW and never turns. Dashed: the bare tie, peaking at 56.7%. The slider is the machines' reactance. Power, and the part that does no work

The reactance the capacitor cannot reach

A series capacitor raises a tie's transfer limit until the loop's impedance angle is sixty degrees, and on a bare 0.05 + j0.2 Ω line that is 56.7% cancellation. Put a machine at each end, its EMF held behind a reactance of its own, and the capacitor still cancels only the line: the sixty degrees now needs the line's reactance plus twice a machine's, less √3 times the resistance, as a fraction of the line's reactance, which passes 100% as soon as each machine's reactance exceeds √3/2 of the line's resistance — 0.043 Ω here. Beyond that the limit rises at every step to full cancellation, and the lossless estimate of the gain, wildly high on a bare line, becomes nearly right: 1.43 against 1.50 at 0.2 Ω. The higher limit is reached with the receiving terminal lower — 180 V falling to 148 at 0.1 Ω — and the capacitor's own resistance is negligible for any power capacitor: a Q of 222 keeps the peak within 1%.

 With a 300 ps skew and 5 mV of noise a core with 100 ns of headroom saturates after 502 periods on average — the 500 the skew alone gives — and one drive in a hundred by 184. Seeded: 2,000 drives of an ideal core by the event map at β = 0.5, 5 mV of threshold noise and a comparator skew of 300 ps, each run until the summed high-minus-low time leaves ±100 ns. A histogram of the counts on logarithmic bins, against the closed-form first-passage distribution, between two barriers, of a walk with drift 0.200 ns a period and spread 1.89 ns per root period. Mean 501.9 against 499.9; quantiles 1% 184, 10% 275, 50% 463; the empirical and closed-form cumulative distributions differ by at most 0.014. Circuits that do a job, and the range they do it over

The count a core survives

A relaxation oscillator with a 300 ps comparator skew and 5 mV of threshold noise saturates an ideal core with 100 ns of volt-second headroom after 502 periods on average — the 500 the skew alone gives — and one drive in a hundred by 184. The mean does not see the noise at all until the noise is large enough to saturate the core the other way first; the early tail sees nothing else. Every skew and noise level falls on one curve in one ratio, the skew's count over the count at which its ramp overtakes the walk. Read as a burst, a thousand periods with one failure in a thousand allows 150 ps of skew with no noise, 93.5 ps at a millivolt, and none at all above 2.41 mV.

 With a 100 kHz flicker corner the 0.1 ps oscillator's ratio stops falling ten decibels a decade of capture and starts falling twenty, past 2,277 samples — 1.42 periods of the corner. computed by solving, not by drawing, from the exact expected residual of a summed process, with seeded clocks as dots (24 at each length to 16,384). A free-running oscillator with a 0.1 ps period spread and a flicker of its frequency whose timing spectrum meets the walk's at 100 kHz, sampling a 10.01 MHz tone at 160 MHz. Signal-to-timing-error after removing each capture's best delay and frequency: the walk alone falls ten decibels a decade, 85.7 dB at 1024, 67.6 dB at 65536; with the flicker, 84.1 and 52.9 dB. The flicker's share equals the walk's at 2,277 samples, 14.2 µs, which is 1.42 periods of the corner frequency; beyond it the ratio falls 19.5 dB a decade. Removing a quadratic as well buys 3.80 dB at 65,536 samples and changes no slope. Where a signal becomes a number

A frequency that wanders

Below its flicker corner an oscillator's phase noise falls thirty decibels a decade instead of twenty, because the oscillator's frequency itself wanders. A clock with 0.1 ps of period spread and a 100 kHz corner samples a 10 MHz tone at 84.1 dB over 1,024 samples and 52.9 over 65,536 — 14.7 dB worse than the walk alone — because past 2,277 samples its error grows as the capture rather than its square root. That crossing is 1.42 periods of the corner frequency for every corner. Fitting a parabola as well as a line buys 3.88 dB at every length and changes no slope; each further term buys a fixed fraction. And a first-order locking loop cannot stop it: the locked clock's variance grows with the logarithm of the record, 1.92 dB lost between 4,096 and 262,144 samples, until a loop with an integrator holds it at 87.5.

 Sampled at 100 kHz, the loop repays its whole debt below 50.0 kHz: 6.60 kHz·Np each side, with a peak of 8.72 dB — the continuous loop had repaid 89% by then. computed by solving, not by drawing, from the loop's exact zero-order-hold equivalent. The three-pole amplifier of the area essay (second pole 4 kHz), closed through a sampler and hold at 100 kHz. The device sensitivity, in decibels, against linear frequency from zero to half the sample rate, where the discrete response ends. Margin 24.7° at a crossover of 5.71 kHz, against 34.9° for the continuous loop; peak +8.72 dB at 6.45 kHz. The area below zero is 6.598 kHz·Np and the area above, all of it inside the band, 6.598; net 1.84 × 10⁻² Hz·Np. The continuous loop, by 50.0 kHz, had repaid 88.6% of its own. Feedback, and the margin

The band a sampled loop repays in

Close the three-pole amplifier's loop through a sampler and hold at 100 kHz and Bode's sensitivity integral still balances — but only over the band below 50 kHz, because that is all the band a sampled loop has. The whole debt, 6.60 kHz·Np, is repaid there; the continuous loop had repaid 88.6% of its own by the same frequency and spends decades above on the rest. Below a tenth of the sample rate the sampled loop is the continuous loop with half a sample of delay, 24.7° of margin against 24.6°. And the finite band turns the equality into a floor: hold the sensitivity 20 dB down over a third of the band and no loop can peak below 10 dB. This loop's floor is low and its margin sets its peak; the sample rate at which the peak is 3 dB above the continuous loop's is 18 times the crossover.

 With the receiving end at 0.9 of the sending voltage the limit peaks at 62.6% cancellation, where the impedance angle is 56.3° — and the peak is still Vₛ²/4R = 264.5 kW. computed by solving, not by drawing, at 50 Hz, searched over the angle between the sources on the solve (dots) and from (Vᵣ·cos β/R)·(Vₛ − Vᵣ·cos β) (line): the 0.05 + j0.2 Ω tie with the sending source at 230 V and the receiving one at 207 V, against the fraction of the reactance cancelled; dashed, equal ends. The limit is a parabola in cos β with its top at cos β = Vₛ/(2Vᵣ) = 0.5556, an impedance angle of 56.25°, reached at k = 62.58%; its height there is Vₛ²/4R = 264.50 kW, the same as for equal ends. Uncompensated, this tie carries 180.53 kW at most. The slider is the receiving voltage as a fraction of the sending. Power, and the part that does no work

The peak a sagging receiver keeps

The sixty degrees at which series compensation stops raising a tie's transfer limit is a statement about two ends at equal voltage. With the receiving end lower, the limit is a parabola in the cosine of the impedance angle whose top is at cos β = Vₛ/2Vᵣ: 56.3° and 62.6% cancellation for a receiver at 0.9 of the sender, 33.6° and 83.4% at 0.6, full cancellation at one half. But the height of that top is Vₛ²/4R whatever the receiving voltage — 264.5 kW for the 0.05 + j0.2 Ω tie from a receiver at 1.1 of the sender down to one at 0.5 — and at it the receiving end takes exactly half of what is sent, to four decimal places, for every receiving voltage tried. The best compensated tie is the sending source driving a matched load, and the receiving source's voltage only decides how much capacitor that takes.

 With a 0.30 µF coupling capacitor and a stiff driver the transformer's damping is 0.289: the pulse's tail crosses zero 8 times in sixty units of L/Rₗ. computed by solving, not by drawing, marched and checked against the closed form to 2.3e-4: a pulse two L/Rₗ long through a series coupling capacitor into a 1 mH magnetising inductance with 100 Ω reflected across it, from a stiff driver. The capacitor is 0.30 µF, x = CRₗ²/L = 3. The poles are those of 3.000e-10·s² + 1.000e-5·s + 1, a damping of 0.2887 and a natural frequency of 9.19 kHz. The top ends at 0.128 of the high-frequency gain, and after the pulse the tail crosses zero 8 times, with lobes above a thousandth of the pulse, in the sixty units drawn. The slider is the capacitor. Measurement, which is a circuit on a circuit

The ring a blocking capacitor buys

A series capacitor in front of a transformer stops its flux from walking: a 55%-duty drive that walks a bare core by 0.40 units of volt-time a period settles within a few dozen periods once the capacitor takes the drive's average. The price is a second-order high-pass with the magnetising inductance, and an inductance lets its poles go complex where two resistive coupling stages never could. Between equal resistances the damping cannot fall below 1/√2, and the poles are complex only for a capacitor between 0.17 and 5.8 times L/R². From a stiff driver the damping is √(L/C)/2R and has no floor: a capacitor large enough to make the droop small — the droop is 1/RC, the capacitor's and not the transformer's — makes the tail ring, from one zero crossing at a tenth of L/R² to thirty-five at a hundred.

 At 20 steps a cycle backward Euler cannot show a quality factor above 3.27 and BDF2 none above 77.9; the trapezoidal rule shows every Q, 1.6% high. computed by solving, not by drawing: a series resistor–inductor–capacitor resonance at 1 kHz, its quality factor set by the resistor, marched at 20 steps per undamped cycle with each rule; the quality factor the march shows — its decay per radian of its own ringing — against the circuit's, from 1 to 500. Lines are each rule's factor per step in closed form; dots are marched and measured from the record at Q = 3, 30, 300, agreeing to 0.3%. Backward Euler's shown Q flattens at 3.272 and BDF2's at 77.9, whatever the circuit's; the trapezoidal rule's sits 1.64% above the diagonal at every Q. Q 3: trapezoidal 3.05, BDF2 3.06, backward Euler 1.62; Q 30: trapezoidal 30.5, BDF2 22.5, backward Euler 2.97; Q 300: trapezoidal 305, BDF2 62.6, backward Euler 3.24. Before the steady state

The Q a rule can show

A resonance marched forward in time decays at the circuit's rate plus whatever the marching rule adds, and the rule's share has a ceiling built into it. At twenty steps a cycle backward Euler cannot show a quality factor above 3.27 and BDF2 none above 77.9, whatever resistor the circuit has; the ceilings rise as N/2π and N³/4π³. The trapezoidal rule has no ceiling at all — it shows every Q 1.64 per cent high at twenty steps a cycle, a ratio fixed by the step alone. So the steps a cycle needed to read a resonance's Q to one per cent are about 26 for the trapezoidal rule at any Q, 99 for BDF2 at Q = 100 and 62,202 for backward Euler. Marched at Q = 50, the circuit has 0.533 V of ringing left after ten cycles, the trapezoidal rule 0.536 and BDF2 0.346.

 The bootstrap with 1 nF at its node: one natural frequency, against the gain. computed by solving, not by drawing: the natural frequency of the network found as the root of its determinant at each of 31 gains. It is −G/C to a part in a million at every gain, where G is the Thévenin conductance the reduction gives: −200 thousand per second at A = 0, zero at A = 2 where the resistance passes through infinity, and +100 thousand at A = 3 — a growth rather than a decay. The gain at which the pole crosses into the right half-plane is the gain at which the resistance changes sign, and the capacitor's size moves the pole's speed and nothing else. Networks, and how a solve is checked

The pole that crosses with the resistance

A bootstrap's Thévenin resistance passes through infinity and turns negative at a gain of two. Put one nanofarad at its node and the network has one natural frequency, −G/C to a part in a million: −200 thousand per second with no gain, zero at A = 2, +100 thousand at A = 3. The gain at which the pole crosses is the gain at which the resistance changes sign — for this network. Loaded by a resistor it crosses at one plus R₂ times the sum of the conductances to ground, so a negative resistance is stable across a smaller load and not a larger one; a lag in the source slows the growth and does not move the crossing; a tank on the node turns the runaway into a 30.5 kHz oscillation once its coil is below about 1.5 Ω; and with an inverting source the resistance is positive at every gain and the node still oscillates beyond A = −16.

 The longest track a 1 ns edge from a 10 Ω driver may cross inside 1% overshoot, with its load lumped at the end and spread over 32 taps. computed by solving, not by drawing: a network of line segments marched by its characteristics, the load spread over equal taps with the last at the far end, the length bisected at each total capacitance. Lumped at the end (faint, the earlier essay's curve) the limit falls from 1.83 mm to 0.88 mm and rises to 10.9 mm at 100 pF. Spread over 32 taps (solid) the valley bottoms at 1.14 mm, and the limit is 29.5 mm at 100 pF — 2.70 times the lumped length — and 74.2 mm at 300 pF. Where the models stop

The load spread along the track

A hundred picofarads at the end of a track ring with its inductance as a series circuit, and a 1 ns edge from a 10 Ω driver may cross only 10.9 mm inside 1% of overshoot. The same hundred picofarads spread over 32 inputs along the track allow 29.5 mm, and 300 pF spread allow 74.2. Spread capacitance is not a load at the end; it lowers the line's own impedance to Z₀ divided by the square root of one plus the ratio of the added to the line's own capacitance, and slows it, and once the loaded line's impedance falls below the driver's — 8.2 Ω for 150 pF over 30 mm — every reflection at the source returns with the same sign and the far end climbs a staircase that cannot overshoot. The valley the lumped load has at a picofarad or two survives, shallower: 1.14 mm at its lowest, against 0.88. Two taps are no better than one; from eight the track is a loaded line.

 The delay of 50 Ω FR-4 track against frequency, with nothing dispersing, the dielectric causal, and the conductor causal as well. computed by solving, not by drawing at 108 frequencies on one track: 4.4 and a loss tangent of 0.02 at a gigahertz, 2 Ω/m of copper rising as √f above 1 MHz. Held constant (faint), the delay is flat above the resistance's corner. With the permittivity allowed to disperse (dashed), it falls 209.6 fs/mm from 100 MHz to 10 GHz. With the conductor's internal reactance put beside its resistance (solid), it falls 464.4 fs/mm over the same two decades: 7.389 ps/mm at 100 MHz, 7.094 at 1 GHz, 6.925 at 10 GHz. Lines, where a wire has a length

The conductor made causal too

A dielectric that loses energy must disperse, and the essay before this one made FR-4's permittivity do so: the track's delay then falls 209.6 fs/mm from 100 MHz to 10 GHz. The copper has the same obligation. A conductor whose resistance rises as √f has an internal impedance (1 + j) times that rise, and putting the reactance beside the resistance more than doubles the fall, to 464.4 fs/mm — the conductor's share largest at low frequency, 418 of 584 fs/mm between 10 and 100 MHz, and the dielectric's constant at about 100 a decade. On a 30 ps edge after 300 mm the toe sharpens from 74.5 to 55.6 ps and the tail stretches from 221 to 398: the rise time is 453 ps against 295, and the tail-over-toe ratio a scope can read is 8.0 at 400 mm against 2.6.

 The same imbalance at five winding resistances, and the fixed point three of them reach. computed by solving, not by drawing. The peak flux density in each cycle, against the cycle, under a square drive whose positive half carries 1 per cent more volt-seconds than its negative one, for winding resistances of 0, 0.05, 0.15, 0.4, 1.5 ohms. With none the flux walks to 0.35 T in 15 cycles, which is the walk an earlier measurement of this drive found. With 1.5 Ω it settles at an offset of 5.67 millitesla and stays there, because the offset draws a direct magnetising current and that current's drop across the winding opposes the imbalance. The resistance at which the two outcomes change places is 0.1014 ohms, bisected on whether saturation is reached at all. Every curve carries the identical drive; the resistance is the only difference between them. Two windings, and the band between them

The walk that stops

A drive whose two half-cycles differ in volt-seconds walks the flux to saturation in a count of cycles, and the essay that measured it concluded that no amplitude puts the design inside a limit. That model has a stiff source and a winding of no resistance. With resistance in the loop the walk has a fixed point, held by an identity the material is not in — the mean magnetising current is the drive's direct component divided by the resistance, to seven parts in 10¹³ — and the boundary becomes a resistance rather than a time: 0.1014 ohms bisected, at a one per cent imbalance and half the volt-second limit.

 A gain of −1 on a part stable only at gains of ten, with 1.25 kΩ across its inputs. computed by solving, not by drawing. A 100 MHz amplifier whose second pole is at 20 MHz, wired as an inverter with two 10 kΩ resistors. Its signal gain is one at a kilohertz on every curve. Its noise gain is two with nothing across the inputs (dashed): 35.0° of margin and 4.4 dB of peaking. A resistor from the summing junction to ground raises the noise gain without touching the signal: at a noise gain of 10 the margin is 65.6°, the peak is 0.00 dB and the −3 dB point is 14.1 MHz. Feedback, and the margin

The gain added to buy a margin

An amplifier sold as stable only at gains of ten — 100 MHz of gain–bandwidth, its second pole at 20 MHz — has 35.0° of margin and 4.4 dB of peaking as a gain of −1, because that gain closes its loop against two. A 1.25 kΩ resistor from the summing junction to ground raises the noise gain to ten without touching the signal gain, which stays one to a part in ten thousand, and the margin becomes 65.6°: still 14.1 MHz of bandwidth, 2.19 times what a unity-stable part from the same die gives at its natural gain. The price is everything the amplifier contributes, multiplied by ten instead of two — 64.0 nV/√Hz at the output against 20.5 — and a capacitor in series with the resistor buys back the low-frequency half of it and none of the wideband total.

 A 10:1 divider into 100 pF and 15 pF more behind 1.0 kΩ, trimmed for the calibrator, on a 1 ns edge. computed by solving, not by drawing from the residues of its four poles, on a time axis from 0.1 ns to 1 ms. The trimmer is set to 12.8 pF, where a 1 kHz square wave reads flat: the upper arm then matches the lower arm's total of 115 pF. On a 1 ns edge the output overshoots to 1.128 times where it settles, because for the first few nanoseconds the late 15 pF is not yet part of the divider, and falls back over that branch's own time constant, 15 ns, within 1% only by 35.5 ns. With a second branch in the upper arm, 9.0 kΩ and 1.7 pF — a scaled copy of the late branch — the same edge is flat to 7.0e-11 (dashed). Measurement, which is a circuit on a circuit

The time constant one trimmer cannot reach

A ten-to-one divider whose lower arm is 100 pF of instrument input and 15 pF more reached through a kilohm reads flat on a 1 kHz calibrator with its trimmer at 12.8 pF — the same setting as a lower arm of 115 pF lumped — and overshoots 12.8% on a 1 ns edge, settling within 1% only after 35.5 ns. No setting of the one trimmer is flat: the best judged on a fast edge is 6.1%, at 11.9 pF, and a trimmer set by eye so the fast edge lands flat leaves the calibrator's square wave 9.9% low with a 114 µs slope on its top. Flat at every time needs the upper arm to be a scaled copy of the lower one — a second branch of 9 kΩ and 1.7 pF — and then a 1 ns edge is flat to 10⁻¹⁰, with a window of −18 to +24% on that resistor and ±6% on its capacitor.

 3 equal lags in a loop closed by an inverting gain k, from k = 0 to 12.8. computed by solving, not by drawing: the closed-loop poles, rooted from the netlist's determinant at 61 gains. At k = 0 the 3 poles coincide at −1/τ; as k rises they leave along 3 straight lines at angles π(2j + 1)/3, each ∛k/τ from the start, to the imaginary axis at k = sec(π/3) to the power 3 = 8.000. A gain of a hundredth moves them 21.5% of their own frequency; a single lag would move 1%. Before the steady state

The root a ring of equal stages takes

Two coincident poles separate as the square root of whatever splits them; three as the cube root; m as the m-th. In a loop of m equal lags closed by a gain k, the m-fold pole leaves along an m-pointed star, each point the m-th root of k from where it started — fitted exponents 1.0000, 0.5000, 0.3333 and 0.2500 over six decades — so a loop gain of a hundredth moves a triple pole 21.5% of its frequency where a single lag would move 1%. The star's nearest points reach the imaginary axis at k = secᵐ(π/m): 8 for three lags, 4 for four, 1.884 for eight, falling towards one. Staggering three lags by a factor of four raises the gain they tolerate from 8 to 26.6, and at the same gain cuts a step's overshoot from 53.9% to 20.6% — or, at the same gain margin, buys nothing.

 The Chebyshev 5's sections, each node's peak before and after equal-maxima scaling. computed by solving, not by drawing. Unscaled (lighter), section by section the largest signal anywhere inside reaches 4.537×, 3.109×, 1.000× the input, so the ceiling is 4.537× at F0b. With a gain of 0.220 at the input, 1.000 and 1.602 between sections and 2.833 at the output, every section peaks at 1× (darker) and the response is unchanged. The floor rises from 249.7 to 883.8 nV, so the range gains 2.16 dB of the 13.14 dB the internal peak was costing. The floor, which bounds from below

The ceiling a gain gives back

A fifth-order Chebyshev loses 13.14 dB of its range because its first section's capacitor node carries 4.537 times the input. The textbook repair is to scale: divide the signal in front of the loud section, restore it later, so that every node peaks at the same level. Done exactly, with silent gains of 0.220, 1.000, 1.602 and 2.833 and the response unchanged to the last digit, it lowers the ceiling to the input and raises the floor from 249.7 to 883.8 nV — and gives back 2.16 dB of the 13.14. Every section between the division and the restoration has its noise multiplied by what the signal was divided by. On the even-order Chebyshevs scaling loses half a decibel; on the seventh-order it recovers 3.77 of 22.45. Reordering the sections and scaling them together gives 4.75 dB. The rest of the ceiling belongs to the high-Q section and no gain moves it.

 A 1 ns edge from a 10 Ω driver, a trunk, and a fork into two branches each ending in 2 pF. computed by solving, not by drawing: the network of trunk and two branches marched by its characteristics, the branch length bisected for 1% overshoot at either end. Forked at the driver, each branch may be 2.33 mm, against 1.05 mm for one track carrying both inputs at its end (dashed). Every millimetre of trunk costs more than a millimetre of branch: 0.0 mm of trunk, 2.33 mm branches; 0.2 mm of trunk, 1.89 mm branches; 0.4 mm of trunk, 1.43 mm branches; 0.6 mm of trunk, 1.01 mm branches; 0.8 mm of trunk, 0.57 mm branches; 1.0 mm of trunk, 0.11 mm branches. Where the models stop

The track that forks

Two 2 pF inputs on one track from a 10 Ω driver with 1 ns edges may sit only 1.05 mm away inside 1% overshoot if both hang from its end. Fork the track at the driver and each branch may run 2.33 mm — more than twice as far — because two fifty-ohm branches in parallel are a 25 Ω line, half as mismatched to the driver. Every millimetre of trunk before the fork costs more than a millimetre of branch: with a millimetre of shared trunk the branches may run only 0.11 mm. Equal branches are the only good ones: a 2.2 mm branch overshoots 0.91% beside an equal partner and 1.7% beside a short one. And a star reaches further the more inputs it serves — 2.33 mm for two, 11.85 for four, where four branches in parallel make 12.5 Ω — while a daisy chain of the same inputs reaches 2.54.

 A pair with one conductor over glass-rich laminate (4.6, tan δ 0.015) and one over resin-rich (4.2, tan δ 0.025): the skew per millimetre against frequency. computed by solving, not by drawing: each conductor as its own causal line — permittivity dispersing with its own loss tangent, copper with its internal reactance — and the skew as the difference of their delays. With both permittivities held at their 1 GHz values the skew would be 0.318 ps/mm at every frequency (level line). Dispersing, it is 0.272 ps/mm at 100 MHz, 0.319 at 1 GHz and 0.366 at 10 GHz: the resin, lossier, loses permittivity faster with frequency, and the gap between the two halves widens. Lines, where a wire has a length

The skew that moves with frequency

A pair with one half over glass-rich laminate and one over resin-rich builds 0.318 ps of skew a millimetre when each permittivity is held at its 1 GHz value. But the two regions lose differently — tan δ 0.015 against 0.025 — so their permittivities fall at different rates, and the skew is a curve: 0.272 ps/mm at 100 MHz, 0.319 at 1 GHz, 0.366 at 10 GHz. A deskew set at one frequency leaves ±2.35 ps in 50 mm at the other two. The resin half also loses 3.7 dB more by 10 GHz over 200 mm, so its edge arrives 63.7 ps early and rises in 257 ps against 202; the halves differ in timing and in edge rate at once, and leave a common-mode pulse of 151 mV for each volt.

 A sinusoid sampled 8 times a cycle, at the phase that puts its peak between two samples. computed by solving, not by drawing: the samples of a unit sinusoid at 8 a cycle, synchronous, at the sampling phase that hides its peak worst. 8 samples a cycle is 4 times the Nyquist rate and holds the waveform completely; the largest sample is still 0.9239, 7.61% below the peak, which is 1 − cos(π/8). Read straight off the samples the peak needs 22.2 samples a cycle for 1% and 70.2 for 0.1%. Where a signal becomes a number

The rate each reading needs

Twice the frequency is the rate at which a set of samples holds a sinusoid completely, and it is not the rate at which any particular property can be read off them. Eight samples a cycle — four times the Nyquist rate — hold the waveform, and their largest is still 7.61% below the peak; the largest sample needs 22.2 a cycle for 1% and 70.2 for 0.1%, because its error falls only as the square of the rate. A parabola through the three largest gets 1% at 8. A zero crossing found by a straight line between two samples is a hundredth of a cycle right at 5 a cycle, its error falling as the cube, because a sine is straight at its zero. The rms of whole cycles is exact from three a cycle — and at exactly two it reads anything from zero to √2. A slow rate with a long record of a steady signal does what a fast rate does, once the samples have slipped a whole interval past it: 42 cycles at 4.0236 a cycle.

 A reference flattened at 300 K, from −40 to 125 °C, with its core current as T to the 1. computed by solving, not by drawing every 1.25 K. The reference has zero slope at 300 K by construction and its output there is 1.17170 V. It is never flat anywhere else: it falls 1.391 mV by −40 °C and 2.526 mV by 125 °C, a bow of 2.526 mV that no choice of the weight can remove, and the residue matches −(xti − α)·Vₜ·[x ln x − (x − 1)] with x = T/300 K (dashed) to within the diode law's "+1". Devices, and the amplitude they stop being linear at

The bow a trim leaves

A band-gap reference trimmed to zero slope at 300 K is flat at exactly one temperature. Everywhere else it falls: 1.39 mV by −40 °C and 2.53 mV by 125 °C for a core biased proportional to absolute temperature, 3.79 mV for one run at a constant current, and nothing at all for one whose current rises as the cube of temperature. The residue is (xti − α) thermal voltages times x ln x − (x − 1), with x the temperature over the trim temperature, and six references of different bias, exponent and current fall on that one curve. The least box, 1.9 mV, comes from making the slope zero at 39 °C rather than at room temperature; quoted per kelvin it is 9.8 ppm/K over the automotive range and 4.2 over the commercial one. A third junction supplies the bow's own shape, and subtracting it leaves 18 µV — at an output that is then the band gap itself.

 A 230 V transformer's primary switched on at 0° of the mains, secondary open. computed by solving, not by drawing, marched with the core's saturating flux law and checked at the voltage zero against the netlist march. Settled, the primary draws a magnetising current of 111 mA peak. Switched on at 0° it draws 87.6 A in the first three cycles — 791 times the settled peak. At the voltage zero the first peak is 87.6 A, 791 times; at the voltage peak there is no inrush at all. Power, and the part that does no work

The instant a transformer is worst at

A capacitor-input rectifier draws its worst inrush when it is switched on at the peak of the mains. A transformer's own magnetising branch is the other way round: switched on at the voltage zero, a 230 V primary worked at 1.4 T draws 87.6 A in its first half-cycle, 791 times its settled magnetising current, and at the voltage peak it draws nothing extra at all. The core's flux has to start from zero, the voltage does not, and how close the working flux sits to half of saturation decides the size.

 Near-end crosstalk from a 200 ps edge, 100 mm of coupled track against 10 mm. computed by solving, not by drawing, the pair solved mode by mode and returned to time from its harmonics. Over 100 mm the near end rises with the edge, holds a plateau of 12.51 mV per volt for twice the line's delay, 1.33 ns, and falls. Over 10 mm the round trip is shorter than the edge and the pulse never reaches the plateau: 8.34 mV. The plateau is Kb, a quarter of the sum of the two couplings, on the 0.5 V the driven line carries: 12.50 mV. Lines, where a wire has a length

The near end that stops growing

Near-end crosstalk is a length only until the coupled run's round trip equals the edge's rise time; past that it is a constant. Solved mode by mode and returned to time, a 200 ps edge on 100 mm of coupled track gives a plateau of 12.51 mV per volt — Kb, a quarter of the sum of the two couplings, on the driven line's half-volt — held for the 1.33 ns round trip. The knee is at v·tr/2: 7 mm for a 100 ps edge, 75 mm for 1 ns. Past it a longer run adds width and not height, and the pulse's area keeps growing as Kb times the round trip.

 One series resonator of Q = 1, read five ways. computed by solving, not by drawing; every reading from the solved network, none from its formula. The half-power bandwidth, the phase slope at resonance and the energy ratio driven at resonance give 1.0000, 1.0000 and 1.0000 — Q exactly. The ring-down's logarithmic decrement gives 0.8660, and the largest capacitor-voltage magnification 1.1547. Frequency, which is the same solve

Five readings of one Q

A resonator's Q can be read from its bandwidth, from the slope of its phase, from the energy it stores against the energy it loses, from how fast its ringing dies and from how high its response peaks. On one series resonator at Q = 1 the first three give 1.0000 and the last two 0.8660 and 1.1547. Bandwidth, phase slope and energy are Q exactly at every Q, because they are one equation read three ways; the ring-down reads Q√(1 − 1/4Q²) because a damped cycle is longer than a natural one, and the peak reads the reciprocal because the peak is not at resonance. Each is within 1% of Q only above 3.5.

 Where a chain of 4 stages stops obeying its model, against where one stage does. computed by solving, not by drawing. amplifier, 1%: one stage 14.2 kHz, 4 stages 7.06 kHz, a ratio of 0.498; Kirchhoff, 1°: one stage 3.97 MHz, 4 stages 993 kHz, a ratio of 0.250; small-signal, 1%: one stage 7.30 mV, 4 stages 7.26 µV, a ratio of 0.000995. The amplifier's edge moves as N to the −½ because its error grows as the square of frequency; the board's as N to the −1 because a phase grows in proportion; and the small-signal edge collapses, because the last stage sees the gain of all the others. Where the models stop

The edge of a chain

A system built from stages that are each inside their own edge is not inside its own: the stages share one error budget. Four ×10 stages from a 1 MHz amplifier are 1% from ideal at 7.06 kHz, not at the single stage's 14.2 kHz — the edge moves as N to the −0.502. A signal path through four ten-centimetre boards is a degree long at 993 kHz, a quarter of one board's edge, as N to the −1. The chain's exponent is minus the edge's own exponent in its tolerance, measured on the line y = −x. And the small-signal edge of a chain with gain belongs to its last stage: four junction stages of ten are 1% off at 7.26 µV, a thousandth of one stage's 7.30 mV.

 The supply's mean output against its load, with no leakage inductance, 0.3 mH and 1 mH in the winding. computed by solving, not by drawing, marched with the diodes in the netlist: 17 V peak into 1000 µF, 50 mΩ of winding throughout. Without inductance the output falls 1.69 V from 200 Ω to 25 Ω, an average slope of 3.43 Ω. With 0.3 mH it falls 0.62 V, 1.14 Ω, from a higher start; with 1 mH, 0.42 V, 0.77 Ω. An element that carries the load's current and drops no mean voltage makes the regulation better, not worse. Circuits that do a job, and the range they do it over

The inductance that flattens the supply

An unregulated supply's output resistance counts each resistance by the mean current it carries, and an inductance in the winding carries the whole mean current and drops no mean voltage — so by that rule it should count for nothing. Marched, it counts for less than nothing: 0.3 mH of leakage takes the output resistance at 100 Ω from 3.88 Ω to 0.67, and the mean output falls 0.62 V from 200 Ω to 25 Ω instead of 1.69. The effect has an interior optimum — 0.03 mH makes it slightly worse, 0.3 mH best, 3 mH back to 2.86 Ω — because the inductance first rings the reservoir up and then turns the rectifier toward choke input.

 A field-effect input against a bipolar one biased for each source, at 25 °C. computed by solving, not by drawing. The field-effect input (dark) has 1 mA of drain current and a gate leakage that is 1 pA at 25 °C, from the junction's own law; the bipolar input (pale) is at its best current for each source. It wins between 2.83 kΩ and 5.14 GΩ. At 25 °C the ceiling was 5.14 GΩ; it falls in proportion to the leakage, which the slider's temperature multiplies. The floor, which bounds from below

The window that closes from the top

A field-effect input beats a bipolar one biased for each source between 2.83 kΩ and 5.14 GΩ at 25 °C, and the top of that window is twice the thermal voltage over √β times the transistor's gate leakage: it falls as fast as the input junction leaks. Carried through the junction's own law, the ceiling is 39.5 GΩ at 0 °C, 913 MΩ at 50 and 5.79 MΩ at 150 — while the floor does not move from 2.83 kΩ until 200 °C. A 1 GΩ sensor leaves the window at 48.6 °C, a 10 GΩ one at 16.4, about 33 kelvin a decade. The leakage's doubling interval is not ten kelvin but 7.5 at 0 °C and 15.0 at 150, so the rule of thumb is right at one temperature only.

 An integrator of 100 kΩ and 1 µF with its input grounded: what its own input currents do to it. computed by solving, not by drawing, marched for ten seconds from rest. With 50 nA flowing out of each input and the + input grounded, the output ramps at 50.0 mV/s — the bias current over the capacitance, to six figures — and reaches 0.500 V. With 100 kΩ in the + input to balance the resistances, only the 5 nA of offset current is left and the ramp is 5.00 mV/s. Neither stops: an integrator has no gain to quote, because its gain is time. Measurement, which is a circuit on a circuit

The current a capacitor adds up

An amplifier's input current in a resistor is an offset; in a capacitor it is a rate, and time is the gain. An integrator of 100 kΩ and 1 µF with its input grounded ramps at 50.0 mV/s from 50 nA, and balancing the inputs leaves the 5.00 mV/s of the offset current, still unbounded. A leak resistor bounds it at the current times the leak and costs the lowest frequency the circuit integrates. A hold capacitor's droop times its acquisition time is fixed, so 50 nA through a 100 Ω switch needs 655 nF and 0.77 ms for half a 16-bit step. A charge amplifier's offset times its low corner is Ib/2πCf, whatever its bleed.

 The second harmonic in each cycle of a transformer's inrush, as a share of the fundamental. computed by solving, not by drawing, the harmonics taken per cycle of the marched current. The first cycle carries 79% second harmonic and the second 93%. The share falls below the 15% a differential relay is commonly set to restrain at only in cycle 30, when the peak is 205 mA against a first peak of 88 A — the harmonic outlasts the inrush it identifies. Power, and the part that does no work

The harmonic that names an inrush

A transformer's differential relay sees an inrush as current going in and nothing coming out, which is what an internal fault looks like, and it tells them apart by the second harmonic. Marched cycle by cycle, an inrush starts at 79% second harmonic and stays above the usual 15% restraint for thirty cycles — long after its current has fallen from 88 A to 205 mA, because the current unwinds in a few cycles and the flux offset under it takes seconds. A fault's offset is a smooth exponential and peaks at 14.6% only when X/R is as low as three.

 A terminated ladder network's output noise three ways: one solve per resistor, one solve from the output, and 4kT·Re{Z}. computed by solving, not by drawing, at 290 K. The line is 4kT·Re{Zout}. The dots sum each resistor's 4kT/R through its own transfer to the output, one solve per resistor. The rings get the same transfers from a single solve with an ampere injected at the output — reciprocity — and agree to 4.4e-16; the sum agrees with Nyquist's expression to 4.4e-16. 4.60 nV/√Hz at 100 Hz, 3.82 at 1 MHz. Networks, and how a solve is checked

The resistance colder than the room

In a passive network the noise at a port is 4kT times the real part of its impedance, and reciprocity is why: the transfer from each resistor to the port equals the transfer from the port to that resistor, so one solve from the output gives every resistor's share and the shares add to Nyquist's expression, here to 4.4 × 10⁻¹⁶. With gain in the network both fail. A 50-ohm termination made by feedback around an amplifier of gain A presents 50.000 Ω and has the noise of a 50-ohm resistor at 290/(1 + A) kelvin — 2.90 K at a gain of 99 — but only where the amplifier still has its gain.

 Both ends of the quiet line, 100 mm of microstrip-like coupling, edge 200 ps. computed by solving, not by drawing. The near end (pale) is the saturated plateau, 16.28 mV per volt. The far end (dark) is a single pulse arriving with the edge after one delay, 0.676 ns, as wide as the edge and 25.16 mV per volt deep — negative, because the inductive coupling exceeds the capacitive. Its depth is the coupling difference, the delay and the edge's slope multiplied: 24.96 predicted. Lines, where a wire has a length

The far end that is a derivative

Where the two couplings are unequal the far end does not cancel, and in time it is the derivative of the edge: a single pulse as wide as the rise time, arriving after one delay, its height the coupling difference times the delay times the edge's slope. On 100 mm of microstrip-like coupling a 200 ps edge gives −25.16 mV per volt against 24.96 predicted, and the height falls as one over the rise time with a fitted slope of −1.000 while the saturated near end sits still. It grows with length without a knee until the two modes separate by a whole edge — 249 mm for a 50 ps edge — and then stops at half the driven line's voltage.

 A 1 kHz tone burst through a band-pass of Q = 10, centred on its resonance. computed by solving, not by drawing, marched twice — sine and cosine carriers — so the output's envelope is the modulus of the two, exactly. The input envelope (pale) peaks at 240 ms; the output's (dark) at 243.171 ms, 3.171 ms late. The group delay at resonance is 2Q/ω₀ = 3.183 ms. The phase there is zero, so the carrier itself is not delayed at all. Filters, measured not tabulated

The envelope that leaves early

A band-pass's group delay at its centre is the delay of an envelope, not of a carrier. Marched twice — on a sine and a cosine carrier, so the envelope is their modulus exactly — a 1 kHz burst through a band-pass of Q = 10 comes out 3.171 ms late against 2Q/ω₀ = 3.183, while the carrier's zero crossings come out within 6.4 µs of where they went in. Through a notch, whose group delay at its centre is −3.810 ms, the envelope peaks 3.746 ms before the input's. Neither is a paradox, and both hold only for bursts long against the inverse bandwidth: 60.6 per cent wrong at 0.6 of it, 0.7 per cent at 20.

 Three loops with exactly 45° of phase margin, and their step responses. computed by solving, not by drawing; three lags spaced by a ratio r, the amplifier's gain bisected in each for a 45.000° margin, the step from the closed loop's residues. Lags a hundredfold apart need a loop gain of 141 and overshoot 23.6%. Spaced by three they need 4.7 and overshoot 34.5%; coincident, 2.83 and 41.0%. The same margin, and overshoots 1.73 times apart. Feedback, and the margin

One margin, and the gain under it

A phase margin of 45° is quoted as an overshoot of 23 per cent, and on a loop with plenty of gain below its crossover it is: three lags a hundredfold apart, closed at 45°, overshoot 23.6%. Bring the same three lags together and the loop reaches 45° at a loop gain of 2.83 instead of 141, and overshoots 41.0% — the same margin, 1.73 times the ringing. The closed-loop peak follows the overshoot where the margin does not: predicted from the peak the worst error over 27 loops is 3.1 points, from the margin 22.2. The gain margin moves with it, from 37.2 dB to 9.0.

 The common-emitter stage from a 100 Ω source: the solved response against the Miller estimate. computed by solving, not by drawing. The stage of 1 mA, 5 kΩ load, Cπ 20 pF and Cμ 2 pF has a gain from base to collector of 182.0. From 100 Ω the network's bandwidth is 3.41 MHz; the Miller estimate — Cμ moved to the input, multiplied by one plus the gain — says 4.23 MHz, 1.24 times it. The estimate has one pole, at the input; the network has two, and from a low source the second is the one that sets the bandwidth. Devices, and the amplitude they stop being linear at

The node the bandwidth belongs to

The Miller estimate puts a common-emitter stage's bandwidth at its input node, and whether that is right depends on the source. From 10 kΩ it is within 1% at every load from a gain of 19 to 1,190; from 100 Ω it is 24% optimistic and from 10 Ω 3.41 times, because the input node has become fast and the output node sets the bandwidth. The stage's two poles split from 92 to 2,993 times apart as the source rises. A load capacitance the estimate cannot see takes the bandwidth from 12.1 MHz to 1.49 MHz from 10 Ω and moves it 6% from 100 kΩ. And the multiplier is the gain from base to collector, 182 — not the 144.7 from the source.

 The gain a Wien oscillator needs above three, against its amplifier's gain–bandwidth product. computed by solving, not by drawing: the closed-loop gain at which the oscillating pair reaches the axis, bisected on the netlist with a one-pole amplifier of 100 dB direct-current gain, for equal 10 kΩ and 10 nF parts. At 10 times the frequency, 3.15339; at 30 times the frequency, 3.02349; at 100 times the frequency, 3.00257; at 1000 times the frequency, 3.00012. Every point agrees with 3(1 + δ) + 3ε²/(1 + δ + 3ε), where ε = kf₀/GBW and δ = k/a0, to two parts in ten million. The bandwidth's share falls as the square of the ratio; below it sits a floor of 9.00e-5 from the finite direct-current gain, and the two are equal near a ratio of √(3a0) = 548. Circuits that do a job, and the range they do it over

The gain an amplifier asks for, squared

A Wien oscillator built round a real amplifier needs more than a gain of three, and the excess is easy to overrate: the amplifier's lag moves the frequency in proportion to the lag and the required gain only in proportion to its square. At a gain–bandwidth product a hundred times the oscillation frequency the frequency is 4.22 per cent low and the gain needed is 3.00257 — fifty-one times smaller in proportion. The square is also what makes the bias spread: an amplifier 30 per cent slow asks for 1.94 times the typical excess. Against 1 per cent parts the amplifier matches the network's share at a ratio of 16; against a network matched to 0.05 per cent, not until 86.

 A second-order loop at a ratio of 64, with a quantiser of one to five bits: the ratio in the band against the input amplitude. computed by solving, not by drawing: the loop whose noise transfer is exactly (1 − z⁻¹)², marched with a sine inside the band, the error measured through a squared Hann window. At half scale, 1 bit 69.4 dB, 2 bits 82.5 dB, 3 bits 90.3 dB, 4 bits 96.8 dB, 5 bits 102.0 dB. The first extra bit adds 13.1 dB because a one-bit quantiser's error is larger than a step's uniform error; after that each adds six or seven. The amplitude each does best at moves up with the bits — 0.6 for 1, 0.6 for 2, 0.8 for 3, 0.85 for 4, 0.95 for 5 — because a fine quantiser is never driven past its last level by the shaped error. Where a signal becomes a number

The levels the loop cannot shape

Give a second-order noise-shaping loop a quantiser of several bits instead of one and every bit beyond the second buys six or seven decibels, the best input amplitude climbs from 0.6 of full scale to 0.95, and loops of third and fourth order that will not run on one bit run on three or four. What it charges is outside the loop's reach. The levels the quantiser feeds back enter where the input does, so their errors are passed rather than shaped: a three-bit loop loses its lead over a one-bit one when each level is out by 0.027 per cent, its error stops falling with oversampling, and the error arrives as harmonics. One bit is immune because two levels always lie on a line.

 A zero on the right at 1.00 kHz: the straight lines report 46.47° of margin, and the loop has 18.19°. computed by solving, not by drawing, the closed form held to the netlist — an inverting differentiator added back to its input makes the zero — at three frequencies and to the closed loop's poles. An integrator with a zero in the right half-plane at 1.00 kHz, the integrator set so that the straight lines cross unity at 950 Hz. The magnitude, and so the lines, are those of a zero on the left; the phase falls by the zero's angle where the left zero's would rise (dashed). The loop crosses at 3.04 kHz. Read at the lines' crossing with the loop's own phase the margin is 46.47°, and 133.53° with the phase the magnitude implies; at the loop's own crossing it is 18.19°. Frequency, which is the same solve

The zero the lines put on the wrong side

A zero in the right half-plane has exactly the magnitude of its mirror on the left, so it has exactly the same straight lines, and nothing drawn from a magnitude can say which side it is on. The phase can: it falls where the mirror's rises. An integrator and such a zero, read off its lines, reports 90° − atan x of margin on a loop that has acos x, where x is where the lines cross as a fraction of the zero — so the lines say 45° of a loop with none. Behind one real pole the over-report reaches 37.6° on a loop of 2.1° when the zero sits below the pole, and vanishes when it sits above it: the reading is unsafe exactly when the zero comes first.

 An eighteen-segment diode table built at 300 K, carried to other temperatures four ways: the worst error over six decades of current. computed by solving, not by drawing: the two-mechanism junction's law moved with temperature, eighteen equal-ratio minimax segments from 100 nA to 100 mA fitted at 300 K (worst 1.886 mV). Unchanged, the table is out by 163.0 mV at 250 K and 265.5 mV at 400 K. One shift of every drop leaves 47.14 mV and 38.32 mV at 250 and 350 K. A shift per segment leaves 3.768 mV and 3.839 mV — and 14.67 mV at 400 K. Refitting every segment on the same breakpoints holds 1.573 mV to 2.379 mV at every temperature: the breakpoints never need to move. Where the models stop

The table that has to be refitted

Eighteen straight segments model a real diode from 100 nA to 100 mA to within 1.886 mV at 300 K. Carried to 350 K unchanged they are 149 mV out; moved by one common shift, 38.3 mV, because the drops at the bottom of the table fall 3.31 mV a kelvin and those at the top 1.54; moved by a shift each, 3.84 mV. That last correction holds within twice the table's own error over fifty kelvin either way and fails at 400 K, where the bottom of the table has dropped below the junction's own recombination current and is no longer an exponential at all. Refitted on the same breakpoints the table holds 1.57 to 2.38 mV at every temperature: the breakpoints never need to move, and the lines always do.

 The droop of the ceramic regulator against the rise time of a 100 mA load, with the capacitor at full value, half and a fifth. computed by solving, not by drawing, marched from the operating point: 10 µF, 5 mΩ of series resistance, 630 pF across the upper divider resistor, the 100 mA arriving as a straight ramp. A step droops 103 mV at 100%, 137 mV at 50%, 203 mV at 20% of the capacitance. A ramp of a millisecond droops 2.98 mV, 2.56 mV, 2.56 mV, and the slowest ramps approach the loop's inductance, 25.2 µH, times the ramp's rate (dashed) — a quantity with no capacitance in it — times 1.18, 1.03, 1.03, the share the loop's ringing adds. The full-value and fifth-value curves cross at 23 µs; slower than that, the larger capacitor droops more. Circuits that do a job, and the range they do it over

The load that rises slower than the loop

Every droop a regulator is sized against is a load step with an infinitely fast edge, and a ceramic regulator droops 103 millivolts on its full 10 µF and 203 on a fifth of it. Give the same 100 mA a rise time and the ordering does not survive: past 23 microseconds the full capacitor droops more than the fifth, and a slow ramp's droop is the loop's own 25.2 µH times the load's rate of change — a number with no capacitance in it, raised 18 per cent on the full capacitor by the impedance peak its slower loop leaves. For a 10 mV budget the load must take 298 µs to arrive on the full capacitor and 253 on the fifth.

 A second-order four-bit loop whose integrators are built round amplifiers of finite gain: the ratio in the band against oversampling. computed by solving, not by drawing: each integrator's pole at 1 − 1/A, half-scale input, half a step of shaped dither, three seeds averaged, squared Hann window. With perfect integrators the ratio rises 15.1 dB an octave, to 138.9 dB at 512. With A = 1000: 138.5 dB at 512, 0.4 below; with A = 100: 122.3 dB at 512, 16.6 below; with A = 30: 102.9 dB at 512, 35.9 below; with A = 10: 85.0 dB at 512, 53.9 below. Each curve leaves the ideal one near a ratio of about πA; for A = 10 the last octave gains 2.7 dB against the ideal's 15.1 — toward the three an octave a floor that is not shaped improves at. Where a signal becomes a number

The integrator that leaks the zero away

A noise-shaping loop puts its error's zero at zero frequency because its integrators never forget. An integrator built round an amplifier of gain A forgets a part in A every sample, and the zero moves off zero frequency to a floor. The cost is a function of one number, the oversampling ratio over π times the gain: for two leaky integrators it is 10 log(1 + (10/3)x² + 5x⁴), which the marched loops follow to within two decibels from gains of 10 to 300, and keeping it under a decibel needs a gain of 1.2 times the oversampling ratio. A first-order loop pays a second way: inputs within exactly 1/(2A − 1) of zero leave its output where it was.

 A resonator of Q 10 driven at 1.2 times its frequency, the drive switched off at the capacitor's crest and at its zero. computed by solving, not by drawing: the series circuit's exact free response from its settled state, held to a fourth-order march of the same circuit. Switched off at the capacitor's crest the ring starts at 1.0013 of the settled amplitude and never passes it; switched off at its zero it starts at 1.2015 — ω/ωd — and its first crest reaches 1.1120. The settled arrow turning at the drive's frequency is gone at the switch; what is left turns at the circuit's own, and its length is set by where in its cycle the settled arrow was. Frequency, which is the same solve

The ring a drive leaves behind

Stopping a resonator's drive removes the arrow that turned at the drive's frequency and leaves the one that turns at the circuit's own, and how long that arrow is depends on the instant the drive stopped. Its length is the energy the circuit held at that instant, and off resonance the energy is not constant through a cycle: it breathes between the capacitor's peak and the inductor's, which differ by (ω/ω₀)². So above resonance a drive stopped as the capacitor's voltage passes through zero leaves a ring 1.2015 times the settled amplitude at 1.2 times resonance, and its first crest passes the settled value; below resonance no instant does.

 A chain of identical one-pole stages: the time its step takes to settle within 1%, against the number of stages. computed by solving, not by drawing: the chain's step response is the Erlang distribution's, and its 1% settling bisected on it, held against a fourth-order march of the chain at one, four and sixteen stages. One stage settles in 4.61τ, the familiar ln 100; four in 10.0τ, sixteen in 26.7τ, a thousand and twenty-four in 1099.9τ. The frequency edge's rule, √N stretched by the bandwidth that holds 1% (faint), says sixteen stages need 4.02 times one stage's time; they need 5.81. For a long chain the settling time approaches the delay, N time constants (dashed): 1.074 of it at 1024 stages. For one stage (τ = 1.59 µs at 100 kHz) the 1% time is 7.33 µs. Where the models stop

The delay a chain adds

A chain of one-pole stages leaves its flat band as the square root of the number of stages, and it leaves its settling time faster: four stages take 10.0 time constants to settle within 1 per cent against one stage's 4.61, sixteen take 26.7 — 5.81 times one stage, where the frequency rule allows 4.02. The difference is a delay. The chain's step is the sum of N exponential lags, so it is an edge of mean N time constants and spread √N; the magnitude response sees only the spread. At a thousand stages the 1 per cent time is 1.074 of the delay. A chain budgeted to stay flat to 1 per cent settles to 1 per cent 45 per cent late.

 A gain of ten round an amplifier with a second pole: its half-power bandwidth and its noise bandwidth, each over what the gain–bandwidth product predicts. computed by solving, not by drawing: the stage as a netlist round a two-pole amplifier of 1 MHz, the second pole placed at r times GBW/G. The half-power bandwidth agrees with the second-order closed form √r·g(√r/2) to two parts in a thousand at every placement: 1.000 at r = 0.5, 1.272 at r = 1, 1.414 at r = 2, 1.231 at r = 5, 1.110 at r = 10, 1.010 at r = 100 times GBW/G. It is 1.414 times the product's prediction at r = 2 and within ten per cent only beyond r = 10. The noise bandwidth is (π/2)·GBW/G to half a per cent at every r: the second pole moves the one and leaves the other where it was. Feedback, and the margin

The bandwidth the product still gets right

Gain times bandwidth is a constant for an amplifier with one pole, and real amplifiers have a second. Put it at r times the closed-loop bandwidth the product predicts and the half-power bandwidth becomes √r·g(√r/2) times GBW/G — 41.4 per cent wide at r = 2, within ten per cent only beyond r = 10 — so a follower round a 1 MHz part with its second pole at 2.5 MHz runs 40 per cent wider than the rule says. The noise bandwidth does not move at all. A second-order loop's noise bandwidth is set by one coefficient the second pole cannot touch, and it stays (π/2)·GBW/G at every r and every gain, to half a per cent on the netlist. A third pole is what breaks it.

 The group delay of fifth-order filters along the line from Bessel to Butterworth, and of a 0.1 dB Chebyshev, all with one half-power point. computed by solving, not by drawing: the delay as a sum over each pole set's poles, held to the slope of each realised network's own phase at four frequencies. At zero frequency the delay is 0.386, 0.413, 0.449, 0.488, 0.515, 0.633 periods of the corner; across the passband it varies by 0.40%, 5.7%, 16.4%, 34.1%, 59.6%, 122.7% of that. The Bessel is flat to the corner and beyond; every step toward Butterworth bends the delay up into a peak near the corner, and the Chebyshev's peak is the largest. Filters, measured not tabulated

The attenuation a flat delay cannot buy

Along the fifth-order line from Bessel to Butterworth the group delay's spread across the passband grows at every step, from 0.40 per cent to 59.6, and on through the Chebyshevs to 191 — monotone, like overshoot and settling, and lost fastest in the first eighth of the way. The third axis turns the frontier into something sharper than a trade. More order flattens a Bessel's delay four to seven times an order and buys it no attenuation at all: 14.17 dB at twice the corner at order six is the most any Bessel reaches, and from there the attenuation falls toward the 12.04 dB of the Gaussian every delay-flat filter tends to. A flat delay and a steep skirt cannot both come from poles.

 A 3.3 V edge from a 10 Ω driver down a 50 Ω line of 1 ns, received through 50 Ω in series with four capacitors, beside a plain 50 Ω and an open end. computed by solving, not by drawing: the line marched by its characteristics. The open end rings up to 5.50 V; the plain resistor holds 2.750 V, the launched level, and never reaches the swing. With the capacitor the far end arrives at the same 2.75 V and then climbs as the capacitor charges and stops terminating: with a time constant of half a round trip (20.0 pF) it overshoots to 4.48 V; with one, 3.83 V; with three (120 pF) it rises without overshoot to 3.30 V by 24 ns; with ten, 3.13 V by 24 ns, still on its way. Lines, where a wire has a length

The termination that runs out

A 50 Ω resistor at the end of a line terminates every edge and draws current for as long as the line is held high: 75.6 mW on an alternating pattern from a 3.3 V, 10 Ω driver. Put a capacitor in series with it and the standing current is gone, and so, eventually, is the termination: the far end reflects nothing as an edge arrives and the whole of it once the capacitor has charged, by one less the exponential of −t/2Z₀C exactly. Under two round trips of time constant the receiver overshoots — 54.8 per cent at a quarter of one — and above it the receiver waits on the capacitor to reach the swing. On a pseudo-random pattern the levels are tightest at three round trips, and on a fast pattern the capacitor holds the line's mean and the resistor still burns half of what a plain one does.

 Three trackers finding a panel's maximum power from open circuit, against the number of readings each takes. computed by solving, not by drawing: the single-diode panel of 81.0 W at 17.244 V, every tracker starting at 0.995 of the 19.85 V open-circuit voltage. Perturb-and-observe with 0.2 V steps walks down and then oscillates about the maximum, within two parts in a thousand of it from reading 39. Newton on I + V·dI/dV, two readings a step for the current and its slope, is within a part in a million from reading 10. The fractional open-circuit rule needs one reading and lands exactly, because its fraction, 0.8686, was taken at this light. Networks, and how a solve is checked

How many readings a maximum takes

A panel's maximum power sits where I + V·dI/dV is zero, and a tracker finds it by reading the panel. From open circuit, perturb-and-observe in 0.2 V steps takes 39 readings to come within two parts in a thousand and then oscillates about the maximum forever, giving up 7|P''|δ²/24 of it; Newton's method on the condition itself takes 10 readings to come within a part in a million, the digits doubling at each step. Newton converges only from a window: above 15.82 V on a panel whose maximum is at 17.24, because below it the condition is nearly the short-circuit current and nearly flat, and one step leaves the characteristic. And the one-reading rule — a fixed fraction of the open-circuit voltage — loses nothing at the light it was set at and 17 per cent at a fiftieth of it.

 A half-micron field-effect device's current against its gate voltage at 250, 300, 350 and 400 K, and the gate voltage at which they agree. computed by solving, not by drawing: mobility falling as T^−1.5, threshold falling 2 mV a kelvin, velocity saturation at 2 V. Below the crossing a warmer device draws more, because its threshold has fallen; above it less, because its mobility has. The curves agree at 1.7000 V, 1.0000 V of overdrive — the closed form's 2κT·vc/(1.5·vc − κT) — where 300 K draws 0.667 mA and 250 and 400 K draw 0.669 and 0.673 mA. Devices, and the amplitude they stop being linear at

The overdrive at which temperature cancels

A field-effect device's current has two temperature coefficients pulling opposite ways: its threshold falls about two millivolts a kelvin, which raises the current, and its mobility falls as the temperature to the −1.5, which lowers it. They cancel at one overdrive, (4/3)κT = 0.800 V for a long channel, and the device's curves at every temperature cross there. Velocity saturation moves the crossing up — to 1.000 V at half a micron and 2.400 V at 0.15 µm — and, if the saturation velocity does not itself change with temperature, removes it altogether below a hundred nanometres. Below the zero a warmer device takes more current, the bipolar runaway in a device said to be free of it: 26 per cent more for 10 K at 0.1 V of overdrive.

 A 10 µs lead-inductance resonance holds the start at t⁴ until √12·τr = 34.6 µs, at a damping of 0.1 or any other. Expanded about infinity, and by residues after. The start of the step through two buffered RC sections (τ = 1 ms and 0.5 ms) on logarithmic axes: bare, a line of slope two (dashed), and with a buffered series resistance and lead inductance into a stray capacitance of natural time constant τr = 10 µs and damping 0.1 (solid). The relative degree rises by two, so the start is the bare step times t²/(12τr²) — the faint line, checked on the expansion to three per cent a hundredth of τr in — until the two meet at √12·τr = 34.6 µs. The damping does not enter that coefficient: it is the inductance and the capacitance, through τr² = LC, and nothing else. After it the curve rejoins the bare one, displaced by a delay that the damping does set. Before the steady state

The delay an inductance does not add

Lead inductance against a stray capacitance is a resonance, and far above a network's bandwidth it raises the relative degree by two: a two-section step starts as t⁴ instead of t² until √12·τr — 34.6 µs for a 10 µs resonance — at any damping, because the coefficient is LC and nothing else. After that the resonance moves a half-way threshold by R·C, its first moment, and the inductance is not in it. It moves every threshold by R·C only at ζ = 1/√2, where its second moment is a pure delay's, and below that its variance is negative: at ζ = 0.1 the 1% point moves by 1.48 R·C while the 10–90 rise time gets shorter. The ring that reaches the output is the network's own gain at the resonance, 2.0 × 10⁻⁴ at ω0τ = 100, and the quality factor decides only how long it lasts. Lightly damped, the start is a staircase that touches the bare step once a period and never overtakes it.

 A cascode is 15.9 times wider than the plain stage from 100 kΩ and 1.30 times from 10 Ω, with a 5 kΩ load. computed by solving, not by drawing, each bandwidth bisected on the solved magnitude. The common-emitter stage of 1 mA, Cπ 20 pF and Cμ 2 pF into 5 kΩ, alone and with a second device stacked on its collector, across four decades of source resistance. The plain stage falls from 12.1 MHz to 110 kHz as its Miller-multiplied input node slows. The cascode runs from 15.8 MHz, where its output node sets the bandwidth — Cμ2 against 5 kΩ, 15.9 MHz — to 1.75 MHz, where its input node does: rπ in parallel with the source, against Cπ and twice Cμ, 1.78 MHz. Neither single-node figure is ever below the stage's own bandwidth. The faint line is the cascode with the upper device's own Cπ left out of the middle node, and it is never below the solid one and at most 0.74% above it. Devices, and the amplitude they stop being linear at

The nodes a cascode hands the bandwidth to

A cascode is said to remove the Miller effect, and into 5 kΩ it does: from a 100 kΩ source it is 15.9 times wider than the plain stage. From 10 Ω it is 1.30 times wider, because there the bandwidth was never the input's — it belongs to the output node, Cμ2 against the load, 10.0 ns from every source. From a high source it belongs to the input, which is now rπ against Cπ and twice Cμ rather than against a multiplied Cμ, 74.7 ns of Cπ beside 15.4 of Cμ. The node the cascode adds holds 0.545 ns and moves the bandwidth less than one per cent. Its three poles cross rather than split, the sum of time constants is worst — 24% low — at the source where two nodes are equally slow, and the removal is not unconditional: the middle node's gain is 1.01 at a 1 kΩ load, 12.3 at 1 MΩ, and above 3.11 MΩ the cascode is narrower than the stage it replaced.

 A complex pair of zeros at 30.0 kHz, damping 0.5, and its mirror: one magnitude, and half the energy 10.6 µs later — 4ζ/ω₀ is 10.6. computed by solving, not by drawing. A low-pass with poles at 1, 10 and 100 kHz and a pair of zeros at 30.0 kHz with damping ζ = 0.5, against the same network with the pair mirrored into the right half-plane; the fraction of the impulse response's energy delivered by each instant, from the residues, with the mirrored network also marched from its polynomials (dots, within 6e-7). Both carry the same energy. The left-half-plane network is ahead at every instant; half its energy arrives at 75.4 µs against 86.0, a delay of 10.62 µs, where the mirror's all-pass delays a low frequency by 4ζ/ω₀ = 10.61 µs. Frequency, which is the same solve

The zeros that have no side

A complex pair of zeros can be mirrored into the right half-plane without changing a magnitude anywhere, and what the mirror adds is a second-order all-pass whose delay is 4ζ/ω₀ at low frequency. Above the band that is the whole story: with the pair at 30 kHz on a 1 kHz low-pass, half the energy arrives 0.21 µs later at ζ = 0.01, 2.12 µs at 0.1 and 21.2 µs at 1, linear in the damping — so the zeros of a notch, which sit on the axis, have no side worth choosing. Inside the band the all-pass's delay peaks at a millisecond exactly where the notch has removed the energy, and the energy's centroid moves by the delay averaged over |H|², which for fifteen pairs from 2 to 100 kHz matches a quadrature over frequency to five parts in a hundred thousand. And unlike one mirrored zero, a mirrored pair never starts a step the wrong way: it is only late.

 One loop, cut at its output wire with 160 nF on it: the voltage injection crosses unity twice, the current injection reads 57 dB too much at low frequency, and together they are the loop gain. computed by solving, not by drawing. A three-pole amplifier closed for a gain of a hundred, with 100 Ω of output resistance and 160 nF of load. The loop gain read at the controlled source's own input, which draws no current (solid); read by a voltage injected in series with the output wire (dashed); and by a current injected into the same wire (dotted). The true loop crosses unity at 5.31 kHz with 8.9° of margin. The voltage reading first crosses at 5.04 kHz with 15.1°, and rises through it again at 10.6 kHz, because above 9.95 kHz the load is a lower impedance than the source. The current reading crosses at 5.82 kHz with 19.8°. Combined through 1/(1 + T) = 1/(1 + Tv) + 1/(1 + Ti) (dots), the two give the controlled source's reading to 3e-9 at every frequency. Feedback, and the margin

Two wrong readings and one loop gain

A loop gain measured on a bench is measured at a wire, and a wire has an impedance on each side. Cut an amplifier's output with 100 Ω behind it and 160 nF ahead, inject a voltage in series, and the reading is the loop gain plus (1 + T)·Zs/Zl — 15.1° of margin where the loop has 8.9°, and a second unity crossing at 10.6 kHz that the loop does not have. Inject a current into the same wire and the error is (1 + T)·Zl/Zs: 19.8°, and 57 dB too much loop gain at low frequency. The two errors are reciprocal, so the readings combine exactly: 1/(1 + T) = 1/(1 + Tv) + 1/(1 + Ti) returns the loop gain at the controlled source's own input to a few parts in a billion, at every frequency and every load.

 A zero at 1ωₙ moves the fastest-settling damping from 0.782 to 0.873 and the fastest settling from 3.62 to 2.38/ωₙ. computed by solving, not by drawing. The 2% settling time of a second-order step against its damping ratio, with no zero (grey), a zero at 2ωₙ (dashed) and at 1ωₙ (solid), each located by residues on the recovered polynomial and refined by bisection; the step by residues equals the plain step plus (1/b) times its derivative to 2e-16. zero none: fastest at ζ = 0.7825, 3.623/ωₙ; zero 2ωₙ: fastest at ζ = 0.7975, 3.028/ωₙ; zero 1ωₙ: fastest at ζ = 0.8725, 2.376/ωₙ. The zero adds the step's own derivative, which adds overshoot, so more damping is needed to keep the last excursion inside the band — and it speeds the rise, so the best settling is faster than with no zero at all. The cliffs in the curve are still there, moved. Before the steady state

The damping a zero asks for

The fastest-settling damping of 0.78 belongs to a step with no zero in it. Put a zero at the natural frequency and the step becomes the plain one plus its own derivative: the best damping moves to 0.873 and the best settling falls from 3.62/ωₙ to 2.38/ωₙ. A zero five times further out moves nothing, and a zero at three times moves the optimum by only 0.005, yet keeping 0.78 there costs 52 per cent, because the optimum sits on a cliff. A zero in the right half-plane moves the best damping by 0.015 and costs 20 per cent in settling that no damping can buy back.

 Forty thousand periods at β = 0.5: each period's error against the same period's duty error, correlated by 0.008. Seeded: forty thousand periods of the event map, β = 0.5, 5 mV of threshold noise; one in twenty is drawn, each scaled by its own standard deviation (3.4 ns for the period, 2.49 ns for the high-minus-low time). A period is three draws with coefficients A, A + B, B and its duty error the same three with A, B − A, −B, so their covariance is σ²[A² + (A + B)(B − A) − B²], which is zero at every β. Measured, the correlation is 0.0080, inside the ±0.0100 that forty thousand samples resolve. With normal draws, uncorrelated is independent: the cloud is round. Circuits that do a job, and the range they do it over

Every draw shared, and nothing in common

A relaxation oscillator's period error and its volt-second error are made of the same threshold draws, and they are exactly uncorrelated. A period weighs its three draws A, A + B and B; the high half less the low weighs the same three A, B − A and −B; the products sum to A² + B² − A² − B², which is zero at every hysteresis. Over forty thousand periods the measured correlation is 0.008 against a resolution of 0.010. What the two do share is one draw with the period on either side, ±0.157 at β = 0.5, and over any whole burst those two lags cancel term for term. Four thousand bursts judged against twenty-five pairs of limits fail both ways exactly as often as the product of the two rates says: the margins specified separately double-count nothing.

 A power ferrite driven one way to 0.3 T: ungapped, every pulse starts at 90.8 mT; with a 0.1 mm gap, at 22.0 mT. computed by solving, not by drawing, marched on the hysteretic core. A ferrite with 90.6 mT of remanence and a coercive field of 38 A/m, 100 turns on 100 mm² and 60 mm of path, its flux raised to 0.3 T by a pulse and then released by letting the winding current fall to zero, cycle after cycle until the starting flux stops moving. Ungapped (solid) the release ends at H = 0 on the descending branch, 90.8 mT: each pulse can add only 209.2 mT. With a 0.1 mm gap (dashed) the current is zero where le·H + lg·B/μ₀ = 0 — the shear line (dotted) — at H = −29.0 A/m and 22.0 mT, under the bound μ₀·Hc·le/lg = 30.2 mT; each pulse can add 278.0 mT. Two windings, and the band between them

The flux the reset leaves behind

A transformer driven one way only — a forward converter, a single-ended pulse transformer — is reset each cycle by letting its magnetising current fall to zero, and zero current is not zero flux. Marched on the hysteretic core, a power ferrite with 90.6 mT of remanence driven to 0.3 T starts every pulse but the first at 90.8 mT, and keeps 34.9% of the swing a symmetric drive would get, where half is the most a one-way drive could; a square-loop core starts at 212 mT and keeps 14.7%. A gap shears the loop: the current is zero where H = −lg·B/(μ₀·le), so the start is bounded by μ₀·Hc·le/lg and falls onto the bound as the gap grows — 22.0 mT at 0.1 mm, recovering 69 mT of swing for 239 mA more magnetising current. The first pulse from a demagnetised core is the only one that starts at zero, and it is the one a bench test sees.

 Three matches of one length, half a wavelength at f₀: two quarter-wave sections are bad again at 2f₀ and 4f₀; an exponential and a Klopfenstein taper are not bad again at all. computed by solving, not by drawing, as cascades of exact line input impedances. 50 Ω into 200 Ω through three matching sections each half a wavelength long at f₀: a two-section binomial transformer (dashed), an exponential taper (solid) and a Klopfenstein taper designed for 5% ripple (dotted), the tapers as four hundred thin slices. The stepped transformer is a band-pass: matched at f₀, and at 2f₀, where each section is half a wavelength and transparent, |Γ| is the bare mismatch, 0.600, again — and at 4f₀, and every even multiple. The tapers are high-pass: the exponential is 0.0194 at f₀ and under 0.131 everywhere above 1.6f₀; the Klopfenstein holds 0.050 from 1.1f₀ up. Lines, where a wire has a length

The match with no upper edge

A stepped quarter-wave transformer is a band-pass in disguise: at twice its design frequency every section is half a wavelength and transparent, and a 50-to-200 Ω match reflects the bare 0.600 again. Let the steps multiply over a fixed length and the staircase of n steps returns the mismatch at n·f₀; in the limit the upper edge has gone, and what is left is a taper — a high-pass match. An exponential taper half a wavelength long is 0.019 at f₀ and never above 0.131 past 1.6f₀. The Klopfenstein profile, the Chebyshev design's continuous limit, holds 0.050 from 1.1f₀ up, needs half the exponential's length for |Γ| ≤ 0.1 and an eighth of it for 0.02, and is not continuous: it starts with a step from 50 to 52.6 Ω. And the small-reflection formula that defines all of them as a Fourier transform overstates a 10:1 taper's reflection by 41% — the line gives tanh of it.

 The bank's impedance peaks at 0.966 Ω at 7.12 MHz; a one-ampere load step moves the supply by 132.4 mV. computed by solving, not by drawing. The bulk-and-ceramic pair of the decoupling essays — 10 µF with 5 mΩ and 5 nH, 100 nF with 30 mΩ and 1 nH — standing on a regulator of 2 mΩ and 20 nH, and the voltage at the bank when the load's current steps by one ampere, from the impedance's own step response by residues (solid). The impedance this bank presents peaks at 0.966 Ω at 7.12 MHz (upper line), the parallel resonance of the bulk part's inductance with the ceramic. The step rings at that frequency and its largest swing is 132.4 mV, 31.3 ns after the step — about a quarter period — which is 14% of the peak and 59% of the resonance's characteristic impedance √(L/C) = 0.224 Ω (lower line). Power, and the part that does no work

The peak a load step does not see

A decoupling bank is specified by its impedance against frequency, and the number that fails a target is the anti-resonance peak — 0.966 Ω at 7.12 MHz for the bulk-and-ceramic pair standing on a regulator. A load that steps by one ampere moves that supply by 132 mV, 14% of what the peak suggests: a step rings the resonance once, and its first swing is set by the resonance's characteristic impedance √(L/C) = 0.224 Ω, not by its Q. The peak is the voltage per ampere of a load that sits at the resonance; switched at 7.12 MHz the same load builds over six cycles to ±611 mV, the peak times the current's fundamental. So damping, the cure for the peak, is a cure for that load only: sixteen times the ESR cuts the peak 8.9-fold and leaves a step's swing between 120 and 155 mV. What decides a step is its edge — 132 mV for anything faster than a few nanoseconds, 28 mV for a microsecond.

 At a fixed output a 0.5 mV junction mismatch's second harmonic falls as (1 + gₘRₑ)⁻² and a 1% resistor mismatch's as (1 + gₘRₑ)⁻¹: past gₘRₑ = 1.93 the resistors set it. computed by solving, not by drawing, every waveform solved arm by arm in currents. A pair on a 1 mA tail, its output held at what the bare, balanced pair gives for 10 mV of peak drive; the second harmonic over the fundamental from a 0.5 mV VBE mismatch alone (dashed), from a 1% mismatch between the emitter resistors alone (dotted), and from both in the same sense (solid), against one plus the degeneration. Between gₘRₑ = 5 and 20 the junction's falls with slope −2.00 and the resistors' with −0.89, still approaching −1, since the local slope of gₘRₑ/(1 + gₘRₑ)² is 1/gₘRₑ − 1. The resistors' mismatch is an input offset of Vₜ·gₘRₑ·δ, which grows with the degeneration that divides it, so the two cross where gₘRₑ = ΔVBE/(Vₜδ) = 1.93; beyond it each decibel of gain given up buys one decibel of distortion, not two. Where the models stop

The resistor mismatch that is only an offset

Two emitter resistors that differ by δ unbalance a degenerated pair, and with an ideal tail they do it in the simplest way possible: the arm currents always sum to the tail, so the mismatch adds the constant (Iₜ/2)·R·δ = Vₜ·gₘRₑ·δ to the input and nothing else. Solved arm by arm, a 1% mismatch at gₘRₑ = 3 is the balanced pair moved by 0.776 mV to 4 × 10⁻¹⁶ of the tail, and a trim of that size leaves no second harmonic at any drive. What separates it from a junction's offset is that it grows with the degeneration meant to divide it: at a fixed output its second harmonic falls as gₘRₑ/(1 + gₘRₑ)², one decibel per decibel of gain rather than two, and it takes over from 0.5 mV of base–emitter mismatch at gₘRₑ = 1.93. For −80 dB at 10 mV of output the resistors must match to 0.823% at gₘRₑ = 1 even with perfect transistors.

 Fed nothing, a 12-bit eighth-order cascade sits 88 least significant bits from zero; with its rounding error fed back, never more than 1. computed by solving, not by drawing. The same cascade with its corner at 1 kHz and its clock 48 times higher, each section's state seeded at eight least significant bits and every product rounded to 12 bits, marched with no input. Plainly rounded, the sections settle at 27, 50, 70, 88 least significant bits — a constant state survives whenever rounding returns it to itself, and each section's offset is the next one's input. With each section's last two rounding errors added back with weights −2 and +1, the error reaches the output through (1 − z⁻¹)²/A(z), a constant state y needs (1 + a₁ + a₂)·y = 0, and the only one is zero: the largest excursion over the last two thousand samples is 1, 1, 1, 1 least significant bit, single isolated steps rather than a settled offset. Where a signal becomes a number

The rounding a section keeps

A fixed-point biquad that remembers its last two rounding errors and adds them back with weights −2 and +1 before it rounds again sends its error to the output through (1 − z⁻¹)²/A(z) instead of 1/A(z). For a low-pass clocked far above its corner that double zero lands exactly on the recursion's largest gain. Fed nothing, an eighth-order 12-bit cascade at 48 times its corner sat 88 least significant bits from zero; with the error kept, a constant state needs (1 + a₁ + a₂)·y = 0, its only solution is zero, and the output never strays more than one bit at any clock rate from 8× to 128×. The round-off floor at 128× falls 89 times, 6.5 bits, for two additions and a shift per section — and the white-noise model, which the plain cascade broke by a factor of ten at eight bits, holds to 2.3% from six bits to twenty.

 A 1 MHz sine of 1 V into an emitter follower with 1 nF on its emitter and a 2 mA sink: the input falls at up to 6.28 V/µs, the output can fall at only 2.0, and the transistor switches off. Marched, the base driven by a stiff source and the emitter current Is·exp((vᵢₙ − vₒᵤₜ)/Vₜ). The input (dashed) and the output (solid) over the last of four cycles. The input's steepest fall is 2πfA = 6.28 V/µs; the load can be discharged no faster than the sink allows, I/C = 2.00 V/µs. On every falling stretch the base–emitter junction reverse-biases, the emitter current drops below a hundredth of the sink's (dots), and the output falls as a straight line of slope −2.000 V/µs until the input comes back up to meet it. On the rising half the transistor pushes whatever current the capacitor needs and the output follows a Vbe below the input. Devices, and the amplitude they stop being linear at

The follower that lets go on the way down

An emitter follower can push current into its load without limit and pull it out only through whatever sinks the emitter. Marched as a large-signal circuit — a 2 mA sink, 1 nF on the emitter — a 1 V, 1 MHz sine asks the output to fall at up to 6.28 V/µs, the sink allows 2.00, and on every falling stretch the junction reverse-biases, the emitter current drops below a hundredth of its quiescent value for 57% of the cycle, and the output falls as a straight line of exactly −I/C. The boundary is a slope, 2πfA ≤ I/C: 2.87 V at 100 kHz, 0.309 V at 1 MHz, creeping above the line only as the period approaches the junction's own 12.9 ns. Past it the follower rectifies — its output's average rises 415 mV at 1 V of drive and 1.30 V at 2 V — and with a resistor in place of the sink the fall slows as the output descends, from 2.41 V/µs to 1.99. None of it is visible to the small-signal model the earlier essays used.

 At 1.00 A forward the silicon-carbide Schottky hands the switch more charge than the silicon PN diode above 245 V. computed by solving, not by drawing. The charge each diode returns through the switch when it is forced to block, against the reverse voltage, at 1.00 A forward and 100 A/µs. The silicon-carbide Schottky's is its depletion capacitance charged to the rail, closed form checked against Simpson's rule: 11.8 nC at 100 V, 25.0 nC at 400 V, 30.9 nC at 600 V, the same at any current. The ultrafast silicon diode's is its stored minority charge, 18.1 nC at this current from the recovery's closed form, plus 1.59 nC of its own capacitance at 400 V. The two are equal at 245.3 V. Before the steady state

The recovery a Schottky does have

A Schottky diode stores no minority charge and is quoted as having no reverse recovery. It has a depletion capacitance instead, and to block it must be charged to the rail through the switch: a 650 V silicon-carbide part holds 25.0 nC at 400 V whatever current it was carrying. An ultrafast silicon diode with a 20 ns lifetime returns 19.7 nC at 1 A, so below 1.42 A forward the diode said to have no recovery hands the switch more charge, and at 125 °C the boundary falls to 632 mA. In energy the capacitive charge costs 64.9 per cent of V·Q rather than all of it, and the boundary is 799 mA.

 One step of refinement in the same double precision returns the stopband's lost digits: 1.1e-12 at 10 kHz before, under 2.4 units of round-off after. computed by solving, not by drawing. The 5th-order Chebyshev ladder solved at every frequency in double precision (solid), and again after one step of iterative refinement — the residual b − A·x formed in double, the correction solved with the same elimination and added (dashed) — each against the same network eliminated in a pair of doubles. Plain, the response's error tracks the round-off divided by the response (dotted), from 1.1e-16 at 10 Hz to 1.1e-12 at 10 kHz, where the output is 6.78e-7 of the input. One step brings every frequency to within 2.4 units of round-off of the truth. No more digits were used: the loss belonged to the elimination's errors in the small component, and the residual, computed against the original matrix, sees them. Networks, and how a solve is checked

The step that returns the digits

A fifth-order ladder filter solved in double precision loses four decades in its stopband — 1.1 × 10⁻¹² of relative error at ten times the cutoff — on a solution vector that is right to round-off. One step of iterative refinement, in the same double precision, returns all of it: form the residual b − A·x, solve for a correction with the same elimination, add it, and every frequency is within 2.4 units of round-off of a 31-digit solve. The reason is that the plain solve is stable by the norm and not by the row — its componentwise backward error is 5.3 × 10⁻¹³ in the stopband rows — and one step makes it componentwise stable. The bridge walked to balance cannot be helped at all. Its node voltages are already right to the last bit; its digits are lost storing two numbers whose difference is the answer, and again storing its conductances, which moves the network itself by round-off over δ before anything is solved.

 A 100 µs pulse through the transformer whose band runs 400 Hz to 81.3 kHz: it rises in 4.03 µs, droops 21.6% along its top, and swings 21.6% below zero after it. computed by solving, not by drawing, twice. A 1 V, 100 µs pulse into the 1:1 transformer of the band essays — 10 mH windings, coupling 0.99, 50 Ω each side — from residues (solid) and from a trapezoidal march of the netlist (dots, within 0.34% of the plateau). The upper edge, 81.3 kHz, is a rise of 4.03 µs from 10 to 90%; the lower edge, 400 Hz, is a droop with time constant 1/(2πfₗ) = 398 µs, and the top ends at 0.7842 of the 0.4901 V plateau against exp(−T/τₗ) = 0.7778. When the pulse ends the output falls by the full plateau and goes negative by what the top lost, then recovers with the same time constant. Two windings, and the band between them

The band counted in rise times

A transformer's band is two frequencies and its pulse response is two times, and they are the same two facts. Through the 1:1, k = 0.99 transformer whose band runs 400 Hz to 81.3 kHz, a 100 µs pulse rises in 4.03 µs — 0.35 over the upper band edge — and droops 21.6% along its top with a time constant of one over 2π times the lower edge, 398 µs; solved by residues and by a march of the netlist, they agree to a third of a per cent. Coupling buys rise time and nothing else: from 30.8 µs at k = 0.9 to 422 ns at 0.999, while the droop's time constant moves from 414 to 396 µs. The longest pulse that droops no more than 10%, over the rise, is 0.048 times the ratio of the upper band edge to the lower — the band ratio counted in rise times, 10.4 at k = 0.99 and 98.8 at 0.999. And the droop is not lost: a transformer has no gain at direct current, so the output swings below zero afterwards by exactly what the top lost, and the two areas match to 0.08%.

 Asked for 105% of the nose power, a tap changer that blocks on a wrong-signed step stops at 9.759 V and holds it; one that does not keeps raising, and the voltage keeps falling. Marched. The 10 V source behind 1 Ω of the regulator essays, its load drawing 105% of the line's 25 W nose power at the 10 V setpoint, regulated by a tap changer that raises the ratio by 1.25% every 5 s while the load is more than 20 mV low. With no blocking rule (dashed) it climbs past the nose ratio √(Rₗ/R) = 1.9518 and keeps going, to n = 3.250 and 8.61 V by 900 s, still falling; the integrating regulator of the earlier essay (dotted) collapses the same way, faster. With the rule (solid) the tap at n = 1.9625 is the first to lower the voltage, the regulator stops at 385 s, and the load holds 9.759 V — within a tap of the most this line can give it, 9.759 V. The setpoint is out of reach either way; the rule converts a collapse into a stall. Power, and the part that does no work

The rule that reads the sign of the gain

A regulator that raises a ratio whenever its load is low collapses the voltage once the load asks for more than the line's nose power, because past the nose raising the ratio lowers the voltage. The standard protection is a rule: after each tap, compare the voltage with the one before, and if it fell, stop. On a 10 V line asked for 105% of its 25 W, a 1.25% tap changer with the rule stops at n = 1.9625, the first tap past the nose, and holds 9.759 V — the most the line can give — where without it the ratio climbs to 3.25 and the voltage is 8.61 V and still falling. The rule is exact without noise and fragile with it, because it reads a difference that vanishes at the nose: with 3 mV of measurement noise it wrongly stops a raise that could have reached the setpoint 0.3% of the time at 98% of the nose and 15% at 99.5%. Trusting it to 1% costs margin — 97.9% of the nose at 3 mV, 79.9% at 10 mV — or coarser taps.

 A Wien oscillator with a lamp for its lower gain arm, τ = 100 ms: the envelope rings 63.6 times in ten time constants at 63.6 Hz and settles at 1.000 V. computed by solving, not by drawing: three state equations marched through every cycle of the 1.59 kHz carrier. The peak at the network's node, over its settled value (solid), from a millivolt with a cold filament, against time in units of the filament's time constant τ = 100 ms. Cold, the 50 Ω filament asks for a gain of 11 and the output sits on the 12 V rail until the filament has warmed. It leaves the rail after 1.99 ms, falls through the operating point, overshoots and rings at 63.6 Hz with a damping ratio of 0.0225. The dashed bound is the linear loop's decay, exp(−(1 + m)t/2τ) with m = 0.8, and it contains neither the carrier's frequency nor the amplitude. The settled amplitude, 1.000 V at the node and 3.00 V at the output, is the one at which the filament is at 250 Ω. Circuits that do a job, and the range they do it over

The lamp that has to be slow

A small lamp in a Wien oscillator's gain network adjusts the gain to whatever each unit needs, so the margin a limiter spends is no longer spent, and the price is an amplitude loop of its own. The envelope is an integrator and the filament is a lag, which makes a second-order loop that rings at the square root of 2ω₀m/τ and decays at (1 + m)/2τ. The filament's ripple at twice the carrier is the only distortion, 0.201/ω₀τ. So the slowness that lowers the distortion lengthens the settling in exact proportion: distortion times settling is 0.163 cycles at any time constant, and 0.01 per cent takes a second at 1.59 kHz and 82 seconds at 20 Hz.

 Three waveforms at 16 samples a period, sampled at the worst instant: the largest sample of a sine is 1.92% short, of a triangle 12.5%, of a clipped sine 0. computed by solving, not by drawing: three waveforms with the same period and the same peak, sampled 16 times a period at the phase that puts the peak midway between two samples (dots). The sine leaves its peak as the square of the time, the triangle in proportion to it, and the sine clipped at nine-tenths of its height does not leave it at all for as long as it stays flat. The largest sample therefore falls short by 1.92% for the sine, 12.5% for the triangle, 0 for the clipped sine, at one rate, for three waveforms with the same period. Where a signal becomes a number

The peak that is a corner

The largest of a set of samples falls short of a sine wave's peak as the square of the sample interval, and that law belongs to the sine. Measured over every sampling phase, a triangle's corner falls short as the first power, 12.5 per cent at sixteen samples a period against the sine's 1.92, a peak whose curvature vanishes as the fourth power, and a sine clipped at nine-tenths of its height reads exactly above seven samples a period. The order in which a waveform leaves its peak is the whole of the law, and it also fixes how much better the average reading is than the worst. A real corner is round at some scale, so the law bends where the samples resolve its top harmonic, and a parabola through three samples helps a round peak, barely touches a corner and makes a flat one worse.

 A stage with its second pole at 3 times the bandwidth the product predicts settles within 0.1% in 5.06τ, against 6.91τ for the one-pole stage the product describes. computed by solving, not by drawing: the step of a stage whose amplifier has a second pole at r = 3 times GBW/G (solid), against the one-pole stage of the same product (dashed). Time is in units of τ = G/(2π·GBW). Settled within 0.1% at 5.055τ, against ln(1000) = 6.908τ, with the netlist of a gain-of-ten stage agreeing at 5.057τ. The stage with the second pole is usually faster, whatever it does on the way. Feedback, and the margin

The gain that stops making a stage faster

The gain–bandwidth product says a stage of gain G settles in time proportional to G, and a stage with a second pole does something else. Below r = 4 the closed loop's poles are a complex pair whose real part is half the sum of the open loop's, π times the second pole's frequency, with neither the gain nor the product in it: a 1 MHz amplifier with a 0.5 MHz second pole takes 4.10 µs to settle within 0.1 per cent at unity gain, 4.04 at two and 4.22 at five, where the product says 1.10 to 5.50. Above the knee at G = 4·GBW/f₂ the product returns. In between lies the fastest stage, r = 3.32 for a 0.1 per cent band, which settles in 3.44 time constants against the one-pole stage's 6.91: half the time, and not at the r that gives the widest band.

 Eight stages of damping 0.8 against eight one-pole stages of the same bandwidth: settled within 1% at 18.75τ against 16.00τ, with 2.02% of overshoot. computed by solving, not by drawing: the step of eight identical second-order stages of damping ζ = 0.8, each with its half-power bandwidth at ω = 1, marched (solid), against eight one-pole stages with the same bandwidth from the closed Erlang step (dashed). Within 1% at 18.75τ against 16.00τ, the overshoot being 2.02% and each stage's delay 1.393 against one. At ζ = 1 the march agrees with the Erlang step of order sixteen. Where the models stop

The stage that overshoots a little

A stage with a little overshoot settles sooner than a one-pole stage of the same bandwidth, 3.53 time constants against 4.61 within one per cent, and a chain of them should do better still, since a stage at the right damping adds almost no spread. It does not. The saving is 17 per cent at two stages, 11 at four and 2 at eight, and at sixteen the chain is 5 per cent slower than the one-pole chain, because a stage's extra delay adds as the number of stages and the spread it removes adds only as the root. The best damping stays near 0.83 at every length, just short of letting the overshoot out of the band. In frequency the same stages are five times flatter at their best, at a different damping, and so the two edges of a chain ask for two different stages.

 Scaling a Chebyshev 5 with real 4 nV/√Hz amplifiers everywhere: 118.18 dB at equal maxima and 118.32 dB with no gain stages at all, so the gains cost 0.14 dB, where silent ones promised 2.16 dB. computed by solving, not by drawing: the dynamic range on ±15 V of a fifth-order Chebyshev with gains d^α in front of each section and after the last, against α. Every amplifier has a voltage noise of 4 nV/√Hz, the sections' own and the stages', and each stage carries a kilohm in its divider or lower feedback arm (solid). With every amplifier and gain silent (dashed), equal maxima recovers 2.16 dB over no scaling, 142.44 dB to 144.60 dB. With real ones: 118.32 dB with no stages, and with stages 115.75 dB at α = 0.2, 116.27 dB at α = 0.4, 116.86 dB at α = 0.6, 117.50 dB at α = 0.8, 118.18 dB at α = 1. Merely having the stages is a step down at the smallest α, since each is a real amplifier in the path, and the curve climbs from there to equal maxima. The response is unchanged to a part in a billion at every α. The floor, which bounds from below

The floor the gains bring

Equal-maxima scaling recovers 2.16 decibels of a fifth-order Chebyshev's range when its gains make no noise and its amplifiers make none either. Give every amplifier in the circuit a voltage noise of 4 nV/√Hz, the sections' own and the stages', and the unscaled filter reads 118.32 dB over a 1 MHz band and the scaled one 118.18: the last gain has nothing after it to filter its noise, so it multiplies 4 µV of wide-band noise by 2.83, and the lower ceiling is spent paying for it. At 1 nV/√Hz scaling costs 6.61 dB, and at 16 nV/√Hz it gains 0.99, so a noisier amplifier makes it look better. A narrower measuring band does not rescue it. On a Chebyshev 7, with a node 13.26 times the input, the same arrangement gains 6.15 dB.

 A Gaussian edge read at 3 samples per rise, at the worst sampling phase: straight chords between the samples say the rise is 8.78% from its true value. computed by solving, not by drawing: a Gaussian edge (solid), its samples at 3 per 10–90% rise at the sampling phase that reads worst (dots), and the straight lines an instrument's linear interpolation draws through them (dashed). The chord crosses 10 and 90 per cent earlier and later than the edge does, because a Gaussian edge curves in opposite senses at the two thresholds. The reading is 1.0878 rises against 1. Where a signal becomes a number

The samples a rise time needs

A rise time is read from two thresholds, and an instrument that draws straight chords between its samples is wrong at each by an amount that goes as the square of the interval. For a Gaussian edge at the worst sampling phase that is 5.02 per cent at four samples per rise and 0.818 at ten; the rule of thumb of three to five is a five per cent rule, and one per cent takes nine. The reading is not monotonic in the rate: above 20.9 per rise a first-order edge reads exactly at whole numbers and 0.046 per cent at 24.5, because its two errors subtract at one phase, and a Gaussian edge reads worse there because its two add. Reconstructing before reading needs 1.87 per rise for one per cent, and at four times the edge's 3 dB bandwidth the reading is still 11.7 per cent out.

 Two loops with exactly 45° of margin and the same crossover: with 100% of the extra lag a delay, the step overshoots 28.7%, against 23.3% for two poles. computed by solving, not by drawing: the closed-loop step of a loop K·exp(−sD)/(s(1 + s/p)) with its crossover at one, its phase margin exactly 45°, and 100% of the 45° beyond the integrator supplied by the delay (solid), against the same margin from a lag alone (dashed). Delay 0.785, lag's pole absent. Overshoot 28.7% against 23.3%; settled within 2% at 7.03 against 7.06; gain margin 6.0 dB against no limit. Marched with a history buffer, and at δ = 1 checked against the exact series of the method of steps. Feedback, and the margin

The margin a delay spends

A phase margin can be made of a lag or of a delay, and the number is the same either way. Forty-five degrees from a lag overshoots 23.3 per cent, and forty-five degrees from a delay overshoots 28.7; at sixty degrees the sign reverses, 8.77 against 5.57, and the two cross near 54 degrees. The overshoot is the smaller difference. A two-pole loop's gain margin has no limit at all, and five per cent of its extra lag as delay brings that to 25.5 dB; with all of it delay the gain margin is 90°/(90° − margin), which is 6.0 dB at 45°, and nothing else is in the expression. Across twenty-one such loops the margin alone misses the overshoot by 13.1 points. The closed-loop peak, which already reads the whole band, misses by 1.2.

 The same 15 pF spread along 2 kΩ of line, on a 1 ns edge: with 1 of its branches copied the output is flat to 1.07%, against 11.2% with one trimmer alone. computed by solving, not by drawing: the response to a 1 ns edge of a ten-to-one divider whose lower arm is 100 pF direct and 15 pF along a 2 kΩ resistive line, with one trimmer (dashed) and with the first 1 branch of the line's exact partial-fraction expansion copied into the upper arm (solid). The expansion's time constants are 12.2 ns, 1.35 ns, 486 ps and so on, as 1/(2k − 1)². Inverted numerically by Talbot's method, and the line itself checked against a netlist of sections. Measurement, which is a circuit on a circuit

The branches a square root needs

A resistive cable's late charge is a continuum and not one branch, and the partial fractions of its tanh are an exact description of it: 81.1 per cent of the capacitance behind a time constant of 12.2 ns, 9.01 per cent behind a ninth of that, 3.24 behind a twenty-fifth, all through the same kilohm. So a divider's upper arm can copy it one piece at a time. On a 1 ns edge one trimmer alone is 11.2 per cent out, one copied branch 1.07, two 0.252 and four 0.0353. The first branch does most of the work and the uncopied capacitance falls only as one over the count; what saves the rest is that a branch faster than the edge is averaged away. Faster edges ask for more branches, and the tolerance on the first ends the series before the arithmetic does.

 A follower section of Q = 5 on an amplifier 100 times the corner: +4.79% as designed from the formula; designed for that amplifier, exact at it, and +1.64% and −0.966% on parts 30% slow and fast. computed by solving, not by drawing: the realised Q of a unity-gain Sallen–Key section against the actual amplifier's gain–bandwidth, relative to the typical value it was chosen for, from the poles of the netlist. Designed from the textbook formula (solid): +6.51% at 0.7, +4.79% at 1, +3.78% at 1.3. Predistorted, with its capacitance ratio and resistors found by iteration on the netlist so that the typical part is exact (dashed): +1.64% at 0.7, exact at 1, −0.966% at 1.3. Predistortion moves the line down without changing its slope much: it removes the error the amplifier has in common and leaves what differs between amplifiers. Filters, measured not tabulated

The Q designed for the amplifier

A follower section's Q comes out 4.79 per cent high at Q = 5 with an amplifier a hundred times the corner, and nothing stops a designer asking for 4.771 so that the real circuit lands on 5.000. That removes the error the amplifier has in common and leaves what differs between amplifiers: parts 30 per cent slow and fast read +1.64 and −0.97 per cent. So the exchange rate is the spread itself, and it is large. One per cent of Q across ±30 per cent asks for 725 times the corner from the formula and 196 designed for the part. The design runs out at a definite ratio, 7.2 times at Q = 5, below which no value of its parts reaches the Q wanted. On a whole Chebyshev 5 the passband error falls from 1.43 dB to nothing on a typical part and 0.55 on a slow one.

 A line ten rises long stood in for by 20 LC sections, matched at both ends: the edge overshoots 17.5% and takes 1.66 rises to climb, where the line delivers it unchanged. computed by solving, not by drawing: an edge of 10–90% rise tr at the far end of a chain of 20 LC sections with a total delay of 10 tr, matched at both ends, marched (solid), against the same edge delayed by exactly 10 tr, which is what a line delivers (dashed). Overshoot 17.5%, rise 1.66 tr. With no reflections at either end, everything here is the chain's own dispersion: a section has a group delay that rises with frequency, so the fast end of the edge arrives late and rings behind it. Lines, where a wire has a length

The sections an edge needs

A chain of LC sections rings on an ideal step however many sections it has, because a step has energy all the way to the chain's cut-off. An edge has a top frequency, and for a matched chain carrying one the section count has a clean answer: for one per cent of overshoot, 6 sections on a line one rise long, 183 on ten rises and 5,760 on a hundred, 5.8 times the length in rises to the three-halves power, within a few per cent from end to end. So sections per rise is not a constant. A section's phase error goes as the cube of frequency and the inverse square of the count, and it accumulates along the line. A 1 ns edge on a metre of cable asks for 62 sections, a 100 ps edge for 2,048, and the delay is right long before the edge is.

 A shorted reading at k = 0.99: within 1% of the leakage from 279 Hz to 112 kHz, too high below through the winding's resistance and above through the leakage's resonance with 100 pF, at 1.13 MHz. computed by solving, not by drawing, in closed form, and by a netlist with a coupling element at three frequencies. The reading Im(Z)/ω at the driven terminals of a 10 mH and 40 mH pair, with the other winding shorted through its own 1 Ω, 0.5 Ω in the driven winding and 100 pF across it, against frequency, as a fractional distance from the leakage L₁(1 − k²) = 199.00 µH. Below the band the short does not cancel the shared flux. Above it the leakage and the capacitance resonate, at 1.13 MHz, and the reading rises to meet them. Band: 279 Hz to 112 kHz, a ratio of 402.1. Two windings, and the band between them

The window that moves

A shorted winding reads its own leakage only above the frequency at which the short really cancels the shared flux, and only below the leakage's resonance with whatever capacitance is across it. A tighter coupling raises the first edge as one over the root of 1 − k², and it was expected to pull the second down and close the window. It raises the second as well, by the same factor. So the band is 82 Hz to 36 kHz at k = 0.9, 279 Hz to 112 kHz at 0.99 and 2.81 kHz to 1.12 MHz at 0.9999, about four hundred times wide throughout. Its width is 0.01·L₂/(R₂k√(L₁C)), with the leakage nowhere in it, so what closes the window is the winding and not the coupling. Reading the real part as well is exact on paper and limited by phase.

 What a simulator's small conductance reports: at 10⁻¹² S a resistive divider reads 5 V, 1 µA into a capacitor reads 1 MV with a residual at machine precision, and the slope says which of those belongs to the circuit. computed by solving, not by drawing on the nodal matrix at direct current, with a conductance g from each node to ground. A divider of two 1 kΩ on 10 V (flat, 5 V at every g); 1 µA into a capacitor, which the solver refuses without g and which reads I/g with it (slope −1, 1 MV at 10⁻¹² S, 100 mV at 10⁻⁵ S); a biased stage behind a coupling capacitor, 100 mV (flat until g is comparable with its 1 MΩ); and two capacitors in series on 10 V, whose node reads 0 mV at every g, since a conductance to ground is the only thing at that node and it has nothing to divide against. Every one of these passes Kirchhoff's law with g included to machine precision. The residual has no way to tell these answers apart. Networks, and how a solve is checked

The conductance added so it would invert

A solver that refuses a node with no direct-current path can be made to answer by adding a tiny conductance from every node to ground, which is what a circuit simulator does. Kirchhoff's law then holds to machine precision, and the numbers are of three kinds. A real divider's node is exact at any conductance. 1 µA into a capacitor reads a megavolt at 10⁻¹² S, ten times that at 10⁻¹³, and is purely the reciprocal of what was added. A floating 9 V source reads ±4.5 V, a centre tap nobody built, and follows the ratio of two conductances that do not exist. Scaling the conductance finds the first kind and misses the third. One test current per node finds them all, because a node held only by what was added has a resistance to ground near one over it.

 An exponential stage at 20% output, with a loop gain of 1: second 2.5% from 4.97%, third 0.0415% from 0.165%, and fifth 0.68 ppm from 0.814 ppm. computed by solving, not by drawing: the harmonics of an exponential stage with emitter resistance, gm·Re = T, solved point by point on the implicit curve and transformed, at a fixed output fundamental of a fifth of the quiescent current, for T = 1 (dots a) against no feedback (dots b). 2: 2.5%, against 4.97%; 3: 0.0415%, against 0.165%; 4: 5.36 ppm, against 40.9 ppm; 5: 0.68 ppm, against 0.814 ppm; 6: 0.0202 ppm, against 0.0135 ppm; 7: 0.000245 ppm, against 0.000192 ppm; 8: 0.0000478 ppm, against 0.00000239 ppm; 9: 0.00000178 ppm, against 2.62e-8 ppm. The series gives the second and third at small drive: the implicit function's Taylor coefficients are 1/(1 + T), 1/2(1 + T)³ and (1 − 2T)/6(1 + T)⁵, and they agree. The third therefore has a null at T = ½, and past that it has the opposite sign, and feedback that reduces the low orders puts some of them back into the high ones. Devices, and the amplitude they stop being linear at

The fifth a resistor adds

Local feedback's reputation is that it reduces distortion, and on the second order it does: 4.97 per cent falls to 2.5 at a loop gain of one and 0.46 at ten, with the output held fixed. The third has a null at a loop gain of exactly a half and comes back with its sign changed. And from the fifth upward, a harmonic rises above its open-loop value before it falls: the fifth 1.42 times, the sixth 5.23, the seventh 24.6, the ninth 960, peaking at a loop gain between six and twenty. The second order, carried back through the resistor, mixes with the signal and makes them. A total-harmonic reading shows only improvement throughout, because the second dominates it. A differential pair has no second order to carry back, and its fifth never passes its open-loop value.

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