Theme

The thread: The constants decide

A divider's ratio does not predict what happens when something is connected to it — its magnitude does. A filter's family does not say how much delay distortion it costs — a measurement does. The quantity that gets discarded to make a rule memorable is repeatedly the one that decides the outcome.
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.

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 on this site 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 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.

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.

What each family costs, at order 5. Measured on the solved networks. The Chebyshev is 36 dB further down at three times the corner than the Bessel, and pays for it in delay: its group delay varies 49.0% across the passband against the Bessel's 0.06%. Filters, measured not tabulated

What a steep skirt costs

A filter's order buys attenuation at a known rate — twenty decibels per decade per pole, and no arrangement of components changes it. What varies between families is how quickly the slope is reached, and the currency it is paid for in is delay: the steepest of the three distorts delay eight hundred times more than the gentlest.

What coupling buys: the upper edge only. computed by solving, not by drawing. Six couplings from 0.8 to 0.999, each transformer solved and both its edges bisected. The lower edge moves by 1.083× across the whole range — it is set by the magnetising inductance against the source and the reflected load, and the coupling barely enters it. The upper edge moves by 168×, from 4.84 kHz to 814 kHz, because it is set by the leakage — which is what the coupling is. Winding a better transformer widens the band at the top and does nothing at the bottom, where the answer is more inductance or a smaller load. Two windings, and the band between them

What coupling buys, and where it does not

Winding a transformer better is winding it more tightly coupled, and the coupling coefficient is the number a maker works on. Measured across six designs from k = 0.8 to k = 0.999, it moves the upper band edge by 168 times and the lower one by 1.083 — so every hour spent on the winding buys bandwidth at one end of the band and, to within eight per cent, nothing at all at the other.

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.

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 on this site 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.

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.

Group delay across the passband, at order 5. The Bessel filter's delay varies 0.1% below 0.8 of the corner; the Chebyshev's peaks near the band edge and is several times its low-frequency value. Every family here has the same half-power frequency, so this is a difference in behaviour rather than in scaling. Filters, measured not tabulated

Flat magnitude, unflat delay

A filter that passes every frequency in its band at the right amplitude and the wrong time has not passed the signal. Group delay is the measurement that says so, it is absent from the classical comparison, and it varies by fifty per cent across the passband of the two families everybody uses.

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.

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 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.

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.

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.

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.

The millivolts in the wire, which are nobody's signal. 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. Measurement, which is a circuit on a circuit

The millivolts in the wire

Ten millimetres of one-ounce copper is five milliohms and ten nanohenries, and if a hundred-milliamp load and a ten-millivolt sensor both return through it, half a millivolt of somebody else's current is added to the reading — five per cent of it, before anything has been amplified. Above 79.6 kilohertz the error rises a decade per decade with no ceiling, and shortening the shared run moves the whole curve down and the corner not at all.

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.

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.

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.

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.

One bit, oversampled — and where the quantisation noise went. computed by solving, not by drawing. A first-order modulator is marched forward one sample at a time with a one-bit quantiser inside the loop, and the noise inside the band is read out of the transform of the error. It falls by 8.99 dB for every doubling of the oversampling ratio — measured 9.33, 10.12, 6.96, 9.54 — against 3.01 dB for plain oversampling, which is drawn beside it from the same starting point. The loop does not make less noise; it moves the noise out of the band, and at a ratio of 128 one bit is worth 9.18. Where a signal becomes a number

One bit, and where the noise went

Sampling faster spreads a fixed quantity of quantisation noise over a wider band, so the part inside the band of interest falls by 3.01 dB for every doubling — half a bit. Putting the quantiser inside a loop with an integrator does something different in kind: measured on a modulator marched forward one sample at a time, with its test tone inside the band the ratio is quoted over, the in-band noise falls by 8.99 dB per doubling. At an oversampling ratio of 128, one bit is worth 9.18 — and the octaves scatter by a decibel each, which is the loop telling the truth about what its error is made of.

A gapped core: where the inductance goes, and where the energy is. computed by solving, not by drawing. Reluctance in series — the gap's lg/µ₀Ae and the core's le/µ₀µᵣAe — with the inductance N²/ℛ and the share of the stored energy in each proportional to its share of the reluctance. At two hundred microns on a µᵣ = 2000 core the inductance has fallen from 41.89 mH to 5.366, and 87.2% of the energy is in the gap — which is air. The share is lg/(lg + le/µᵣ), so it contains neither the turns nor the area and what decides it is µᵣ·lg against the path length; the slider shows the same gap holding 40% at µᵣ = 200 and 98% at 15,000. That is the reason a gap is a design parameter: it is the part of the magnetic circuit whose properties do not drift, do not saturate and do not depend on temperature. Two windings, and the band between them

The energy is in the gap

A ferrite core is chosen for its permeability and then deliberately cut, and the cut is not a compromise. Reluctance adds in series, so a two-hundred-micron gap in a µᵣ = 2000 core holds 87.0% of the stored energy while the ferrite holds thirteen — and the share is lg/(lg + le/µᵣ), a ratio of two lengths, containing neither the turns nor the area. The material chosen for its permeability holds almost none of what the component stores.

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.

An inverting unity gain, and the 100 pF that only the loop can see. 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. Adding 100 pF at the summing junction leaves the closed-loop gain at a kilohertz unchanged — 0.99998051 against 0.99997988, three parts in a million at the far end of the slider — and takes the phase margin from 90.0° to 14.4°, because the noise gain now rises a decade per decade and the loop closes at forty decibels per decade instead of twenty. Forty-five degrees is reached at 9.00 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 gain the loop closes against

An inverting amplifier with two equal resistors has a gain of one and a loop that closes against two, so it has half the bandwidth of a follower built from the same part — 4.99 megahertz against ten. Nine picofarads at the summing junction, less than a scope probe, takes the phase margin from ninety degrees to forty-five, and a hundred picofarads puts twelve decibels of peaking on a response whose designed gain is nought decibels and whose measured gain at a kilohertz has not moved by three parts in a million.

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.

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.

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.

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.

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.

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.

An order-5 ladder driven from 2× the resistance 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.

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 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.

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 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.

A difference quotient is best at a step of 1e-5, and is 1e+6 times worse at 10⁻¹¹. computed by solving, not by drawing. The worst disagreement between the adjoint network's derivatives and a central difference quotient of the same quantities, against the fractional step the quotient is taken with, on a 7-element ladder at 1000 Hz. The curve has a minimum because two errors pull opposite ways: the curvature the quotient neglects falls as the square of the step, and the digits its subtraction destroys rise as one over the step. The best it reaches is 1.4e-9, against the 2.3e-11 that the two-thirds power of the machine epsilon predicts. The exact route costs 2 solves against 15, and has neither error term. Networks, and how a solve is checked

Every derivative, and the one that is zero

How much does this response move if that capacitor is one per cent out? A difference quotient answers it one component at a time, in two solves each, and its best possible accuracy is four parts in a hundred million. Transposing the matrix and solving once more answers it for every component at once, exactly. Pointed at a claim this collection has made since its ladder essay and never tested directly — that a doubly terminated ladder's response is stationary in every element at its passband maxima — it returns two parts in ten billion, where the cascade realising the identical response returns 0.72.

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.

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.

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.

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.

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

The rung below this one found that nine tenths of a reverse recovery's energy is dissipated in the transistor and not in the diode, and then computed the diode's junction temperature with all of that energy in it. 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.

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.

Twenty nanoseconds of skew is 0.28% of duty, not 0.40. computed by solving, not by drawing. The duty cycle of a relaxation oscillator whose comparator takes longer to go one way than the other, marched, against the expression that lengthens each half cycle by its own delay. The two part company immediately and by a constant factor of about 1.42: during a delay the capacitor keeps charging past the threshold it already crossed, so the next half cycle starts further out and takes longer, and the two halves partly cancel. At 200 ns of skew on a 2585 ns period the duty is 52.805 per cent where the expression says 54.004. The open circles are the same quantity in closed form — the overshoot is V(1 − (1 − β)e^(−d/τ)) and the next half starts from it — which the march reproduces to parts in ten thousand. Circuits that do a job, and the range they do it over

The delay that is two delays

Modelling one comparator delay applied to both transitions makes the two half cycles equal by construction, and the fix is a change of one line. Made, the duty cycle moves by 0.28 per cent for twenty nanoseconds of skew rather than the 0.40 the obvious expression gives, more hysteresis improves the duty cycle and worsens the volt-seconds at the same time, and ten nanoseconds of skew saturates a hundred-turn core in 231 cycles.

Where one pole goes when each component is 5% high. computed by solving, not by drawing. A series R–L–C, its poles recovered by rooting the determinant, and the derivative of the upper one with respect to each element taken exactly from the two null vectors at the pole. The dashed lines are the first-order prediction for a 5 per cent change; the filled circles are where the root actually goes when the element is changed and the determinant re-rooted. The three directions are the argument: the resistance moves the pole along a circle of constant radius, because the natural frequency does not contain it — its normalised sensitivity has a real part of 3.1e-16. The inductance and the capacitance each carry exactly −½ of the radius, and imaginary parts that are exact negatives. At 5 per cent the prediction is out by 0.122 per cent of the pole's own magnitude. Networks, and how a solve is checked

The derivative of a root

The rung below turns one transposed solve into the derivative of a response with respect to every element, and found a doubly terminated ladder stationary at its ripple peaks to a part in ten to the eighth. A pole is a different object — a value of s at which the matrix loses rank — and its derivative comes from two null vectors and a division. Pointed at the same two realisations, the ladder's advantage is a factor of 2.17, not eight orders of magnitude: what is stationary is the magnitude at one frequency, and it says nothing about where the poles are.

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.

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.

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.

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.

At a stationary point a tolerance has a mean, not a spread. computed by solving, not by drawing. Six hundred ladders with every reactance drawn independently from ±1 per cent, measured at the ripple peak and at a frequency between the peaks. Away from the peak the distribution is centred on nominal and 325 of 600 are above it. At the peak none is: the whole distribution lies below, with a mean of -0.0030 per cent and a worst case of -0.0140. A yield calculation that assumes a symmetric spread at a frequency that has a stationary point is wrong in both directions at once — it allows parts above a limit that cannot exist, and it misses that the whole batch has moved. Networks, and how a solve is checked

The tolerance that can only take away

At a doubly terminated ladder's ripple peak the first derivative of the magnitude with respect to every reactance is zero to ten digits, which the rung below measured and which says nothing about how much the response moves. This says how: every second derivative is negative, so of six hundred ladders built from one per cent components not one is above nominal, the mean has shifted rather than the spread having grown, and doubling the tolerance quadruples the damage instead of doubling it.

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

The rung below marched single loops of second, third and fourth order and found the amplitude at which each stops working, falling 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.

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 asserted: 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.

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.

Five tolerances, and the response moves in two directions. computed by solving, not by drawing. The eigenvalues of the relative second-derivative matrix of a fifth-order 0.5 dB Chebyshev ladder's magnitude at its lower ripple peak, over its five reactances. Two are of order one — -0.9473 and -0.8051, both negative — and the other three are 3.9e-9, which is zero at the precision the arithmetic has. So the quadratic form is negative semi-definite of rank two, and there is a three-dimensional subspace of component errors that the peak cannot see. The open circles are the same matrix computed by four re-solves per pair, sharing no adjoint arithmetic with the filled ones: they agree to parts in ten thousand on the two that are there and place the three zeros about two decades higher, which is the price of differencing a difference. Networks, and how a solve is checked

The three tolerances that do nothing

The rung below computed every second derivative of a ladder's magnitude at a ripple peak, found them all negative, and built six hundred ladders to argue that no combination of tolerances could raise the response. The whole matrix says so outright — and says something six hundred samples could not have found, because a sample of a five-dimensional box never lands on a three-dimensional subspace: two of the five eigenvalues are of order one and the other three are nine decades down.

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 rung below 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

The rung below sweeps a second-order step's damping, finds the settling time falling by a third in one step of a five-thousandth sweep, and calls it 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.

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 most sensitive direction at one ripple peak is not the one at the next. computed by solving, not by drawing. The eigenvector of the largest eigenvalue of the relative curvature matrix, at each of the 2 ripple peaks of a 5th-order 0.5 dB Chebyshev ladder, drawn as its components over the reactances in order along the ladder. Every one is symmetric under the ladder's own reversal and no two of them are the same direction: the largest overlap between any pair is 0.0603, which is 86.5 degrees apart. The eigenvalue that belongs to them grows from -0.9473 at 554.9 Hz to -12.005 at 897.9 Hz, a factor of 12.7, so the peak nearest the band edge is both the most sensitive place in the passband and sensitive to a different combination of parts. Each direction is confirmed by a second computation of the whole matrix — four re-solves per pair, sharing no adjoint arithmetic — which returns the same direction to 7.6e-7 radians. Networks, and how a solve is checked

The direction a response is most sensitive to

The rung below diagonalised a ladder's curvature at one ripple peak and read only the eigenvalues: rank two of five, three directions of nothing. The eigenvectors say what the two directions are, and doing it at every peak rather than one changes the conclusion. The most sensitive combination at 554.9 Hz and the one at 897.9 Hz are 86.5 degrees apart, the curvature that belongs to them grows from −0.556 to −75.4 across a ninth-order passband, and exactly one combination survives the whole band at every order — which turns out to be the ripple depth.

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.

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.

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 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

The rung below found that charging a capacitor from a step loses half the delivered energy whatever the resistance, and that 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.

Twenty-four orderings, and 11 of them are choices. computed by solving, not by drawing. Every ordering of the four sections of an eighth-order 0.5 dB Chebyshev, at 16 bits, drawn against the two things an ordering decides. The horizontal axis is the round-off floor the arrangement adds — 24.2 least significant bits at best and 99.8 at worst. The vertical axis is the largest value any section's output reaches, which is what decides whether a word overflows: 0.088 of full scale at best and 0.699 at worst, a range of 18.0 decibels. 13 of the twenty-four are beaten on both counts by another ordering and are simply mistakes; the 11 on the lower-left frontier are the actual choices, and no one of them is best. Where a signal becomes a number

Which section goes first

A cascade of four biquads can be assembled in twenty-four orders, all of which realise exactly the same transfer function. They do not cost the same: the round-off floor runs from 24 to 100 least significant bits and the largest value any section reaches runs over eighteen decibels — and the two go opposite ways, so eleven of the twenty-four are genuine choices and thirteen are beaten on both counts by another arrangement.

What the accuracy costs: dynamic range against the resistance scale. computed by solving, not by drawing. The rung below 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.

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.

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.

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 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 period's spread is 1.20 times what counting two crossings gives. computed by solving, not by drawing. 19999 periods of a relaxation oscillator with 5 mV rms of noise on its thresholds, computed from the exact flip instants rather than marched, at β = 0.5. The measured standard deviation is 3.4 ns and the closed form — three partial derivatives of the period with respect to the three draws it depends on — gives 3.4 ns. The estimate that counts two threshold crossings and divides the noise by the slope at each gives 2.83 ns, which is 17 per cent low. The curve is the closed form's Gaussian, drawn on the measured histogram rather than fitted to it. Circuits that do a job, and the range they do it over

The decision taken where the ramp is slowest

A relaxation oscillator decides at its thresholds, and a threshold is the one place on a charging exponential where the slope is smallest. Noise there costs 3.399 units of period against the 2.828 that counting two crossings gives, because a draw moves the crossing it is armed for and the level the next ramp starts from. Consecutive periods share that draw, so they are positively correlated and the jitter accumulates at 3.771 per root period rather than at 3.399. And at a fixed frequency there is a best hysteresis: β = 0.648, where β·ln((1+β)/(1−β)) = 1.

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.

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.

A whole number of steps: 1, 2, 3 and 4 agree to 7% and a step and a half is 1.9× worse. computed by solving, not by drawing. The upper panel is the share of the quantisation error sitting in harmonics of the input against the amount of dither added — the upper curve the largest share anywhere on the amplitude sweep, with the spread across the five tones each point averages drawn as a bar, and the lower curve that sweep's mean. Undithered the worst is 76.5 per cent. It falls steeply up to one whole step (4.70 per cent at three quarters, 0.34 at one) and then stops improving — but only AT whole steps. The sweep means at 1, 2, 3, 4 steps are 0.283, 0.297, 0.297, 0.301 per cent, flat to 7 per cent; at 1.5, 2.5, 3.5 they are 0.526, 0.345, 0.311, each above both whole steps beside it. The lower panel is what each costs in signal-to-noise ratio, with 10·log₁₀(1 + L²) drawn through it — the measurement is that curve to 0.118 dB everywhere, so the price is known in advance and only the benefit has to be measured. The choice is a corner and a comb: nothing here is minimised, something stops improving, and between the places where it has stopped it is worse again. Where a signal becomes a number

The dither that is a decision

One whole least significant bit is quoted everywhere as the dither, which makes a decision look like a constant. Swept, the axis is a corner and a comb. An eighth of a step leaves 64.5 per cent of the error locked to the signal and one whole step leaves 0.34; above that the sweep mean is 0.283, 0.297, 0.297 and 0.301 per cent at one, two, three and four steps and 0.526 at a step and a half, which fails at exactly the small amplitudes dither exists for. The price is 10·log₁₀(1 + L²) to 0.118 of a decibel, and four steps cost 12.41 for nothing.

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.

Three averagers passing the same noise, and three different half-power points. Integrated by eight-point quadrature on every lobe, with the tail past the last lobe in closed form. A mean over 20 ms has the response sin(πfT)/(πfT), and the area under its square is 25.000000 Hz against the 25 Hz of 1/(2T) — the brick wall drawn shaded. A mean over the window passes half its power at 22.147 Hz, so the noise bandwidth is 1.12880 times that frequency; A one-pole averager passes half its power at 15.915 Hz, so the noise bandwidth is 1.57080 times that frequency; Two means in cascade pass half their power at 23.919 Hz, so the noise bandwidth is 1.04521 times that frequency. Every curve drawn encloses the same area; they differ in where they spend it. The floor, which bounds from below

The filter an average is

A mean taken over a window is a filter, and the area under its squared response is exactly one over twice the window — 25 hertz of noise bandwidth for twenty milliseconds, passing half its power at 22.15. Built to the same noise, a one-pole averager passes half its power at 15.92 hertz and takes 2.33 times as long to settle to one per cent, and two means in cascade pass half at 23.92 and take 1.25 times as long. Between its nulls a mean rejects the mains no better than the one-pole does, and one per cent off a null it rejects it by forty decibels however many cycles the window holds.

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.

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.

The load that may be complex, and what a resistor alone gives up. computed by solving, not by drawing, with the best load searched over BOTH of its parts on the solved network rather than substituted. Against a source of 50 Ω + j100 Ω the search returns 50.00 − j100.0 Ω — the conjugate — which delivers the available 500.00 mW at exactly 50.00% efficiency, and does so at every source reactance. A load that may only be a resistance takes 2/(1 + √(1 + x²)) of that, matching the solved answer to 4.4e-16 at ten reactances — and at x = 2 that fraction is exactly the golden ratio less one, 0.618034. The resistor-only load is the MORE efficient of the two at every non-zero reactance, rising towards one while the conjugate match sits at a half for ever, so the familiar "maximum power at fifty per cent" belongs to the conjugate and not to the load. Power, and the part that does no work

The load that may be complex

Freed of the constraint that it be a resistance, the best load is the source's conjugate — found here by a two-dimensional search on the solved network rather than assumed — and it takes the available power at exactly fifty per cent efficiency whatever the source's reactance. A load that may only be a resistance takes 2/(1 + √(1+x²)) of that, and at a source reactance of twice its resistance that is exactly the golden ratio less one, 0.618034. The resistor-only load is also the MORE efficient of the two, rising towards one while the conjugate sits at a half for ever.

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.

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

The rung below found that 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.

Two leads nobody counts cost 1995 parts per million, and two more leads remove all of it. computed by solving, not by drawing. The error in the reported strain against the resistance in each of the two excitation leads, at 1000 µε on a quarter bridge of 350 Ω excited at 10 V. With four wires the instrument takes the excitation to be the supply's voltage, so every reading is scaled by R/(R + 2r): 1995 parts per million at 0.35 Ω, 54029 at 10. Two more leads brought back from the bridge's own terminals, carrying only the instrument's input current, leave 3.5e-4 parts per million. Exciting with a current instead of a voltage does the same thing with no extra leads at all, because the lead resistance is in series with a source that does not care. Circuits that do a job, and the range they do it over

The two leads nobody counts

The leads that are in the bridge moved a lead out of the changing arm and turned a 999-microstrain error into nothing. The two leads carrying the excitation are still there, and they scale every reading: 0.35 ohms each on a 350-ohm bridge is 1,995 parts per million, exactly −2r/(R + 2r), the same on a full bridge as on a quarter one, and drifting 0.153 microstrain over twenty kelvin — twice what the three-wire fix left behind. Two more wires brought back from the bridge's own terminals leave 0.00035 parts per million. So does exciting the bridge with a current, which needs no extra wires 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.

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.

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.

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.

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 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.

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.

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.

An order buys 38.7 dB at 1.25× the corner and 56.6 at 1.67×. computed by solving, not by drawing. The degree equation for an order-5 elliptic filter, swept over the two quantities a designer sets. The horizontal axis is where the stopband is required to begin; the five curves are passband ripples from 0.01 to 3 decibels. Nothing on this page is a choice: pick a ripple and a transition width and the attenuation is decided. At a half decibel of ripple, order 5 gives 38.68 dB with the stopband beginning at 1.25 times the corner and 66.09 dB with it beginning at twice — a factor of two in transition width for 27.4 decibels. The vertical spacing between the curves is the ripple's own term and is the same at every transition width: relaxing from a half decibel to three buys 9.12 dB wherever it is spent. Filters, measured not tabulated

The selectivity that is not free

The filter trade is normally drawn with two quantities in it. It has three, and an order fixes a relation between all of them: at order five and half a decibel of ripple, a stopband asked to begin at twice the corner is worth 66.1 decibels and one asked to begin at 1.25 times is worth 38.7. The exchange is exact addition in decibels — relaxing the ripple from a half to three buys 9.12 dB wherever it is spent — and there is a fourth price nobody writes down: settling to a tenth of a per cent goes from 9.04 milliseconds to 23.23 while the overshoot does not move.

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.

The net noise power between two resistors at two temperatures. computed by solving, not by drawing. A 1 kΩ resistor at 400 K joined to a second one at 290 K, whose resistance runs over six decades. Each delivers 4kTR₁R₂/(R₁+R₂)² per hertz to the other; the net is 1.5187 zW/Hz at the match, which is exactly kΔT and contains no resistance — a gigohm pair at the same two temperatures exchanges the same. Away from the match the exchange falls, to 0.0596 zW/Hz at a ratio of a hundred, so the maximum-power argument applies to noise as it does to a signal. At one temperature the net is zero at every ratio, to 7.5e-37 W/Hz, which is the statement a wrong noise model would break. The floor, which bounds from below

Which way the noise goes

Two warm resistors joined together each drive the other, and the net flow is 4kΔT·R₁R₂/(R₁+R₂)² per hertz. At the match that is kΔT exactly — 1.5187 zeptowatts per hertz between 400 K and 290 K — and a kilohm pair and a gigohm pair at the same two temperatures exchange the same, which is why noise is quoted as a temperature. At one temperature the net is zero at every ratio to a part in 10³⁷, and that zero is the second law rather than a tolerance.

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√(f_c·f_p) — 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.

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.

A true-RMS converter reads noise low by 1/(16Bτ): 0.600% at Bτ = 10 and 672 ppm at 100, where a sine at fτ = 100 is 0.0396 ppm. Seeded, and measured. An explicit true-RMS converter — square, one-pole average of time constant τ, root — reading Gaussian noise of unit power, against the product of the noise's bandwidth and τ, from 1 to 100. Each point is 2²¹ samples at eight times the bandwidth, with the reading compared against the record's own root-mean-square and its standard error from batch means. Bτ = 1: 4.550% ± 150 ppm (band from zero), 4.081% (band of the same width about 3B). Bτ = 2: 2.598% ± 114 ppm (band from zero), 2.411% (band of the same width about 3B). Bτ = 5: 1.149% ± 75.4 ppm (band from zero), 1.098% (band of the same width about 3B). Bτ = 10: 0.600% ± 53.1 ppm (band from zero), 0.576% (band of the same width about 3B). Bτ = 20: 0.309% ± 39.1 ppm (band from zero), 0.295% (band of the same width about 3B). Bτ = 50: 0.127% ± 26.3 ppm (band from zero), 0.119% (band of the same width about 3B). Bτ = 100: 672 ppm ± 19 ppm (band from zero), 623 ppm (band of the same width about 3B). The dashed line is 1/(16Bτ), an eighth of the averaged square's variance, which the readings approach above Bτ = 10 and fall short of below it. A sine read by the same converter at the same product of frequency and τ is low by 3.96 ppm at 10 and 0.0396 ppm at 100 — the square of the noise's rate rather than its first power. Power, and the part that does no work

The noise a true-RMS meter reads low

A true-RMS converter reads a sine low by an amount that falls as the square of its frequency, and for anything but a slow sine that amount vanishes: 3.96 parts per million at ten times the averager's corner. Noise is not a sine. Its square fluctuates at every frequency down to zero, and the averager passes a share of that set by its own bandwidth against the noise's, so the reading is low by 1/(16Bτ) — the first power, not the second. Measured on seeded noise at Bτ = 10 it is 0.600 per cent low against a predicted 0.625; at 100, 672 parts per million. One reading scatters by eighteen times that, so no single reading shows the bias and the mean of a few hundred is nothing but bias.

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.

The load alternates with parity between 1 and 1.9841, at every order. computed by solving, not by drawing. The last quotient of the continued fraction that turns a reflection polynomial into element values, which is the load resistance the design demands, drawn against order. A Butterworth returns exactly one at every order. A Chebyshev alternates: one at every odd order and 1.984056 at every even one, the same number each time, and it is the closed form (√(1+ε²)+ε)² to a part in 10¹⁵. The mechanism is in the last two rows of the panel. A lossless ladder is two resistances at direct current, so it must deliver the maximum available power there, which requires that zero be one of the frequencies the design reflects nothing at — and an even-order Chebyshev's reflection zeros are the roots of an even Chebyshev polynomial, none of which is zero. The synthesis stalls at order 9 for a Butterworth, which is why that series stops at eight. Filters, measured not tabulated

The termination an even order cannot have

Every even-order Chebyshev ladder terminates in 1.984056 times its source resistance at half a decibel of ripple, the same number at orders two, four, six and eight, and it is (√(1+ε²)+ε)² to a part in 10¹⁵. Building one between equal terminations instead — which is what a table of g-values and a matched pair gives — turns 0.5000 dB of ripple into 1.8123 at order four, deletes one of the passband maxima outright, and moves the worst tolerance corner from the 2.000 power of the component tolerance to the 1.109 power: a factor of 95 at one per cent parts.

At β = 0.5, the duty error scatters by 2.49 ns a period against the period's 3.39 ns — and consecutive duty errors are anticorrelated, −0.217. Seeded: forty thousand periods of the event map with 5 mV of threshold noise, β = 0.5. The period scatters by 3.39 ns against σ√(A² + (A+B)² + B²) = 3.4 ns; the high half less the low half scatters by 2.49 ns against σ√(A² + (B−A)² + B²) = 2.49 ns, with A = RC/V(1+β) and B = RC/V(1−β) — 0.667 and 2.000 in units of RC/V. The draw both halves share enters the period with A + B and the difference with B − A. Consecutive periods correlate by 0.107 (closed form 0.115); consecutive duty errors by −0.217 (closed form −0.214). Circuits that do a job, and the range they do it over

The walk the core sees

Threshold noise in a relaxation oscillator walks its timing at 3.77 nanoseconds per root period at β = 0.5, because the draw two half cycles share adds. A transformer driven by the same square wave sees the difference of the halves instead, where the shared draw subtracts, and that walks at 1.89 — exactly β times the timing, 18.3 ns against 35.9 after a hundred periods over six hundred seeded runs. The per-period duty error does not vanish with the hysteresis and the walk does, consecutive duty errors are anticorrelated where consecutive periods are not, a comparator skew outruns the walk after 2(σB/d)² periods, and at a fixed frequency the core wants less hysteresis than the clock does.

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.

Three mismatches, and only one of them reaches the output. computed by solving, not by drawing. Each of the three quantities that can differ between the two transistors is given a spread of its own, one at a time, and 200 pairs are solved at each. The saturation currents produce a spread that follows them exactly — exponent 0.998, so 1.910 per cent of copy error for two per cent of mismatch. The current gains produce a line of slope 2.003, which is second order rather than first, and land at 2.20e-4 per cent for the same two. There is no third line because there is no third component: with no emitter resistors there is nothing for a resistor tolerance to be a tolerance of, and matching a mirror is a statement about emitter area and about nothing else. Devices, and the amplitude they stop being linear at

The mismatch that cancels itself

A current mirror's copy error is spread by three things the two transistors can differ in, and the population that measures it has always drawn all three at once. Turned on one at a time, a two per cent spread of saturation currents gives 1.910 per cent of copy error and a two per cent spread of current gains gives 0.00022 — because the gains enter only as a sum of reciprocals, which has no first derivative where they are equal. Then the standard cure un-cancels it, by a factor of 86.

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.

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 rung below's 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.

A core walked from lossless to lossy, and the three straight lines it walks along. computed by solving, not by drawing. Loop area, measured coercivity and measured remanence against the threshold spread the model was handed, over five decades of it, driven sinusoidally to ±400 A/m. None of the three is an input: the area is ∮H dB round the marched loop, the coercivity is interpolated where the descending branch crosses zero, and the remanence is read at zero field. Over the lowest four decades all three are exactly proportional to the spread — 0.625961 joules per cubic metre per ampere-metre of spread, a coercivity 0.4499 of it and a remanence of 1.1304 millitesla per ampere-metre — and at 80 A/m the area is 2.69 per cent below the line and the remanence 16.42 per cent, because the pinned operators have reached the flat of the magnetisation curve. At zero the area is 9.8e-15 J/m³, which is the single-valued core the rungs below this one measured. Two windings, and the band between them

One dissipation, two exponents

The core this ladder built takes two numbers — a threshold spread and the fraction of the magnetisation that follows the field with no threshold — and seven rungs moved the first and left the second at 0.55 without ever saying why. The first decides how much loss there is: area, coercivity and remanence are all exactly proportional to it over four decades, at 0.625961 joules per cubic metre, 0.449775 and 1.130408 millitesla per ampere-metre of spread, the last two of which are closed forms. The second decides nothing about the loss at all — it is single-valued, so it contributes exactly zero to the loop area, to twelve digits — and it moves the Steinmetz exponent from 1.5042 to 2.9860.

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