Concept

Model range — where it appears

The span of frequency, amplitude or size over which a model describes its object to a stated accuracy, and outside which it describes something else. Stating it is the whole of this collection's habit: a model drawn without one is a claim with no way to be wrong.

Named by 219 essays across 14 fields — each of them below, with the objects they name alongside it.

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

Every model has an edge

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

limits · Model edges
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.

The frequency a sample rate invents

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

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

The band a turns ratio holds over

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

magnetics · Transformer
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.

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.

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

Exact outside and wrong within

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

networks · Equivalent circuit
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.

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.

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

The instrument's own rise time

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

instruments · Probe loading
Three balanced rectifier loads conducting 60°, and their neutral. computed by solving, not by drawing. The three phase currents are drawn faint and the neutral heavy. Balanced loads, identical in every respect, and the neutral carries 0.968 A against a line current of 0.559 A — a ratio of 1.7321, where √3 is 1.7321. The pulse trains are disjoint, so the neutral is their union and its mean square is three times one phase's. Rebuilding the same current from the multiples of three in one phase's spectrum gives 0.966 A, 0.13% away, by a route sharing only the waveform.

The neutral that carries more than a line

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

power · Three-phase
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.

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.

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

A band rather than an edge

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

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

The half that never arrives

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

transients · Switching energy
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.

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.

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

The Q the components allow

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

frequency · Resonance
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.

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.

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

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

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

transients · Open circuit time constants
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.

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.

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

The edge that is a region

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

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

The edges that are lengths

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

limits · Model edges
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.

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.

lines · Crosstalk
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 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.

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

The node that is at ground for a while

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

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

The same part written two ways

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

frequency · Series-parallel
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.

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.

networks · Superposition
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.

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.

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

The constant that is a window

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

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

The copy, and its two errors

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

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

The mismatch that the cable hides

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

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

The resistor that is only a resistor

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

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

What actually fills a null

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

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

A boundary in volt-seconds

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

magnetics · Saturation
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.

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.

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

The cancellation that leaves a tail

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

transients · Doublet
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.

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.

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

The tolerance that is not on any part

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

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

A boundary is a model and a tolerance

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

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

The diode that conducts backwards

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

transients · Reverse recovery
The first conduction carries 26× the repetitive peak, and the factor of 7 is the user's. computed by solving, not by drawing. The largest diode current in the first conduction after switch-on, against the phase of the mains at the instant of switching, marched from an empty capacitor through the same netlist the ripple and the conduction angle are read from. The settled circuit's largest current is 1.230 A; the first one is 32.39 A at the worst instant and 4.64 A at the best, so which of them a design has to survive is decided by nothing in the design. The estimate that treats the capacitor as a short circuit gives 34.00 A, 4.7% high, the difference being the charge the capacitor takes during the pulse itself. Holding the first peak to ten times the repetitive one would need 1.90 Ω of winding resistance.

The first cycle, which no steady state contains

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

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

The flux that walks

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

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

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

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

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

The resistor that buys the margin back

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

feedback · Capacitive load
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.

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.

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

The straight lines, and where they are not the curve

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

frequency · Asymptotic approximation
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 LC ladder filter 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.

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 the essays on ladder filters have made from the start and never tested directly — that a doubly terminated ladder filter'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.

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

The ammeter that is not in the circuit

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

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

The band that closes with the order

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

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

The capacitance that is not one number

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

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

The delay that is not one number

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

lines · Dispersion
The inductance divides the current by 7 and leaves the capacitor 29% above the mains peak. computed by solving, not by drawing. The first conduction of a rectifier whose transformer has a leakage inductance as well as a winding resistance, marched from an empty capacitor at the worst instant of the mains. The peak falls from 36.4 A at 20 µH to 5.3 A at 5000, and — unlike the winding resistance measured on the first cycle alone, which limits the current and leaves ∫i²dt exactly where it was — the inductance takes the energy down with it, from 0.370 to 0.073 A²s. What it costs is the second curve: the inductor's current cannot stop at the instant the two voltages are equal, so the capacitor overshoots to 21.86 V at 1000 µH — 28.6% above the 17 V peak of its own supply — and the diodes will not let the charge back out. The overshoot has an interior maximum, because past it the mains reverses before the ring has finished.

The inductance that limits, and lifts

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

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

The input that pushes back

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

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

The sample that is subtracted

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

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

Where the mechanisms are one mechanism

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

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

A resistor made of a clock

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

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

The coefficient that is about one reading

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

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

The inductor that is a capacitor

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

magnetics · Real inductor
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.

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.

instruments · Common-mode rejection
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.

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.

limits · Diode model
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.

Two loops, and one heatsink

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

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

The digits the arithmetic did not have

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

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

The band that does not close

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

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

The corner the instrument has no part in

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

instruments · Common-mode rejection
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.

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.

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

The loop that is worse at full scale

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

digital · Noise-shaping
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.

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.

transients · Reverse recovery
A bridge's reading turns over at 1.885 V, where the test level stops mattering. computed by solving, not by drawing. What a bridge reads is the fundamental of the charge waveform over the fundamental of the voltage, so it is a function of the bias AND of the amplitude, and a data sheet names one point of it: zero bias, one volt. The four curves are four test levels on one part. At zero bias they run from 9.999 µF down to 8.312 µF — the harder the drive the lower the reading, because the capacitance is at its maximum there and a sinusoid spends most of its time off the peak. At the rated 5 volts they run the other way, 2.000 µF up to 2.426 µF, because the curve is convex. Between them is one bias where the two effects cancel: at 1.8848 V a tenfold change of test level moves the reading by nothing at all, and that voltage is 0.6645 of the polarisation's own characteristic voltage.

The reading a data sheet does not take

A class II ceramic's temperature envelope leaves its working capacitance 29 points wide at one end and 79 across, because one printed number cannot pin a two-parameter model. One further bridge reading recovers almost all of it — and where the reading is taken decides everything. Turning the test level down to a fiftieth separates five parts a data sheet cannot tell apart by 17.76 per cent; moving the bias to half the rated voltage separates them by 176.02, and pins the working capacitance to ±0.512 per cent from a reading known to one.

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

The resistance a slow curve cannot see

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

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

What the load sees looking back

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

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

The capacitor that remembers

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

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

The ceiling is not at the output

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

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

The current the instrument draws

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

instruments · Input bias current
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.

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.

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

The point the device is never at

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

semiconductors · Distortion
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.

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.

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

Two parasitics, and the resonance neither of them has

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

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

Where the trouble is at the input

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

feedback · Transimpedance
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.

Terminated at both ends

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

lines · Termination
The temperature through a 20 mm core that makes its own heat. computed by solving, not by drawing. Conduction with volumetric generation, solved on forty-one cells with a surface film at each face and the loss density evaluated at each cell's own temperature — so the middle of the core makes less heat than its faces do, which a closed form for uniform generation cannot express. The peak is 115.05 °C and the surface 112.45: a gradient of 2.59 kelvin, which is 2.9 per cent of the 90.0 kelvin rise. That share is Bi/(Bi + 2) — 3.0 per cent at a Biot number of 0.063 — so it is decided by how well the surface is cooled and not by how much heat is made.

The degrees a thermocouple cannot see

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

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

The inductor that is an amplifier

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

filters · Gyrator
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.

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.

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

The width no load can change

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

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

What the fourth order says about the third

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

semiconductors · Emitter degeneration
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.

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.

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

One inductor, and ten components

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

filters · Gyrator
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 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.

noise · Device noise
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.

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.

feedback · Capacitive load
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.

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.

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

The gap a derivative needs

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

transients · Poles
What one temperature costs the loop gain of a part that has a gradient. computed by solving, not by drawing. The thermal loop gain of a 30 mm core, solved as a body with its own internal temperature profile and again as a single lump at that profile's mean, against the Biot number. Both are negative, so the core is a stabilising feedback either way — but the body's loop is the more negative of the two at every point, by 3.0 per cent at a Biot number of 0.108 and 38 per cent at 10.8. A lumped calculation therefore understates how stable a wound part is, and the amount it understates by is not a property of the material but of how well the surface is cooled relative to how well the inside conducts. Below a Biot number of about a tenth it is worth under two per cent and the lump is the right model; at the cooled end the part has 7 kelvin inside it and half the feedback is invisible to a single temperature.

The loop gain one temperature understates

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

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

The margin the straight lines report

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

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

The order that stops helping

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

semiconductors · Emitter degeneration
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.

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.

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

Where an open switch leaks to

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

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

The resistor that is right in size and wrong in angle

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

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

The cable that hides two things

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

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

The capacitance a third switch moves

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

limits · Ideal switch
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.

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.

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

The duty cycle that costs nothing

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

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

The filter that samples

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

filters · Switched-capacitor
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.

The half a switch keeps

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

transients · Switching energy
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 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.

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

The rail the load moves

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

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

The voltmeter four wires do not remove

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

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

Ten seconds, and fifteen minutes

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

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

The corner that is three decades wide

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

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

The errors that arrive before the gain

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

instruments · Common-mode rejection
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.

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.

limits · Ideal switch
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.

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.

semiconductors · Cascode
A true-RMS reading of a sine: ripple a second filter removes, and a bias it cannot. An explicit converter — square, average through a one-pole of τ = 100 ms, take the root — in steady state on a sine of unit root-mean-square value, integrated exactly over a period at 91 frequencies and by a fourth-order march of its own equation at 6, which agree to 1.9e-8. The upper curve is half the ripple on the reading and the lower one the amount by which its mean is low. The reading is low at every frequency, because the square root is concave; it is 1% low below 1.86 Hz, while the ripple is inside ±1% only above 39.8 Hz. The dashed curve is the small-ripple form, an eighth of the averaged square's ripple power, which the bias approaches as the ripple shrinks.

The average a square root pulls low

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

power · Rms and average
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.

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%.

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

The cure that becomes a different circuit

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

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

The current inside the iron

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

magnetics · Magnetic loss
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.

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.

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

The load a curve recommends

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

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

The Q the amplifier decides

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

filters · Q enhancement
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 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.

noise · Shot noise
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.

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.

digital · Quantisation
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.

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.

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

The loop that never crosses

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

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

The resistor that is not made of the resistors

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

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

The sign of what the guard gives back

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

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

The straight line between two models

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

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

Two ladders the terminals cannot tell apart

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

transients · Reverse recovery
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.

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.

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

Only the real part is warm

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

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

Shorted instead of opened, and the error changes sign

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

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

The dip whose area is fixed

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

lines · Reflections
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.

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.

feedback · Capacitive load
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%.

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.

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

The optimum that hands back a bandwidth

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

instruments · Current sensing
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.

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.

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

The floor that is only a floor while nothing flows

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

noise · Johnson noise
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.

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.

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

The rail that is an input error

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

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

The resistance that lowers the ripple

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

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

The stage that is wrong is the far one

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

networks · Divider
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%.

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.

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

The assumption that is a geometry

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

magnetics · Winding field
The current divider, and the resistance that is not in the branch. computed by solving, not by drawing. A current source into two parallel branches, the metered one 10.0 kΩ and the other 1.00 kΩ. The metered branch takes 0.090909 of the current, which is the OTHER branch's resistance over the sum; writing the subscripts the way a voltage divider writes them gives 0.90909, a different number at every ratio but one. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with the roles exchanged: one per cent of error at 111.1 Ω, which is that resistance over ninety-nine to 1.7e-12%. And the headline of the loaded divider holds in the dual too — two current dividers of identical ratio read 0.04762 and 0.09090 into one hundred-ohm meter — while a perfect ammeter reads them identically.

The branch the other resistance decides

A voltage divider's output is set by the resistance the output is taken across; a current divider's is set by the resistance the current does not go through. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with one word changed: one per cent at a meter resistance of R/99 where R is what the meter looks back into — and removing an ideal current source means OPENING it, so that R is the two branches in series, 11 kΩ here, not the 909 Ω of their parallel combination — a factor of twelve in the same construction on the same network.

networks · Divider
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.

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.

frequency · Asymptotic approximation
With its own capacitance at B|Z| = 0.2, a line's nose for a unity-power-factor load moves from 0.618 to 0.658 of a matched line's power, at 0.635 of the source. computed by solving, not by drawing: a unity-power-factor load swept in size from open circuit to short circuit on a line of 50 Ω + j100 Ω carrying its own shunt capacitance as a π, half at each end, with the total susceptance stated as B|Z|. At B|Z| = 0 the nose is 0.6180 of a matched resistive line's power with 0.5878 of the source voltage left, and the unloaded far end reads 1.0000. At B|Z| = 0.1 the nose is 0.6376 of a matched resistive line's power with 0.6108 of the source voltage left, and the unloaded far end reads 1.0465. At B|Z| = 0.2 the nose is 0.6582 of a matched resistive line's power with 0.6353 of the source voltage left, and the unloaded far end reads 1.0969. At B|Z| = 0.4 the nose is 0.7025 of a matched resistive line's power with 0.6895 of the source voltage left, and the unloaded far end reads 1.2107. Each nose is found by golden-section search on the solved network and agrees with the nose of the Thevenin equivalent V/(1 + jBZ/2) behind Z/(1 + jBZ/2).

The headroom that is the line's own charge

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

power · Voltage regulation
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.

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.

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

The resistance that depends on the reading

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

instruments · Current sensing
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.

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.

transients · Step response
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.

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.

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

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.

filters · Gyrator
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.

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.

frequency · Asymptotic approximation
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.

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.

transients · Step response
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.

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.

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

The resistance that bends the signal

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

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

The turns nearest the gap

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

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

The node that does not care how many

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

feedback · Virtual earth
A switching load's comb in a 10 mV reading: every tooth the same height. computed by solving, not by drawing. The harmonics of a 50-per-cent trapezoid of 100 mA at 100 kHz with a 10 ns edge, each multiplied by the shared conductor's impedance at its own frequency. Below 79.6 kHz the conductor is 5.00 mΩ of resistance and the teeth fall as 1/n. Above it the conductor rises as n while the harmonics still fall as 1/n, so the two cancel and every tooth from 239 kHz to 10.6 MHz is 0.828 mV — flat to 5.4 per cent across 52 of them. The comb stops at 31.8 MHz, which is the edge's own corner, so the total of 12.6 mV is decided by how many teeth there are: fitted over five edge rates it grows as the edge time to the power -0.497, against the −½ that a quadrature sum of equal teeth gives. A faster edge raises no harmonic and adds teeth. What a slow measurement keeps is a different term entirely: the zeroth harmonic, which is the 100 mA average through 5.00 mΩ of resistance — 500.0 µV, with no inductance in it and no edge, against the comb's 12.6 mV.

Every tooth the same height

A shared return conductor's impedance rises a decade per decade above 79.6 kilohertz and a switching load's harmonics fall a decade per decade, so the two cancel exactly: every harmonic of a hundred-milliamp square wave puts 0.828 millivolts into a ten-millivolt reading, from the conductor's own corner up to the edge's. The comb is flat rather than falling, so the total is decided by how many teeth there are — and since they add in quadrature it grows as the square root of the edge rate, 39.7 millivolts at a nanosecond against 4.03 at a hundred.

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

The peak that only has a bound

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

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

The floor every filter has

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

digital · Anti-alias
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.

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.

limits · Ideal 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.

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.

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

The product that is not the third

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

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

The wire that is not a foil

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

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

The zero that lifts the lines

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

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

Two returns in one plane

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

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

The window the square-root law has

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

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

The node that is an inductance

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

feedback · Virtual earth
Two sensors on one return: 511 µV in each reading, 100 nV in the difference. computed by solving, not by drawing. Two sensors referenced to the same local node, each reaching it through 2 mm of its own copper and each drawing 1 mA through that copper, with a 100 mA load returning through the 10 mm they all share. The interfering voltage is one voltage on one node, so it is added to both readings identically and the difference between them has none of it: 511.0 µV in each reading against 100.0 nV in the difference at a kilohertz, which is 74.2 dB. No resistor tolerance enters it — with the two stubs equal the rejection is exact to the solver, 166 decibels. What is left is the 10 per cent MISMATCH between the stubs times the sensors' own current, which the closed form puts at 100.0 nV, and it is proportional to the mismatch: 10.0 nV at 1%, 30.0 nV at 3%, 100 nV at 10%, 300 nV at 30%.

One voltage added to two readings

A difference amplifier across a shared return rejects somebody else's current by 54 decibels, and the number is four resistors rather than the amplifier. Two sensors referenced to the same node do better by a different mechanism entirely: the interfering voltage is one voltage on one node, so it is added to both readings identically and the difference between them has none of it — 511 microvolts in each reading against two and a half picovolts in the difference, which is 166 decibels and is exact rather than good. What is left is not a tolerance but a mismatch between the two sensors' own return paths, ten nanovolts per per cent of it.

instruments · Common-impedance
Zeroing the dependent source too: an answer that does not depend on the gain. computed by solving, not by drawing. Two independent sources reach a node through resistors; a voltage-controlled source of gain −10 is driven from that node and fed back to it through 10 kΩ. Solved whole, the output is -25.0000 V. Solved by superposition with the dependent source LEFT IN each of the two solves it is -25.0000 V — the same number to 1.1e-16, so superposition is exact here as it is everywhere else in a linear network. Zeroing the dependent source along with each independent one gives nought volts at the output and 4.06250 V at the node, against the node's true 2.50000 V — 62.5 per cent out — because a controlled voltage source set to zero is a short circuit and the three solves are then of three different circuits. What makes it unmistakable is that the wrong answer is 4.06250 V at EVERY gain from one to three hundred: the route has deleted the only element that knows what the gain is, while the true answer moves by more than a factor across the same range.

The source that must not be zeroed

Superposition says to zero all the sources but one, and the care usually taken over that instruction is that zeroing is not deleting — a voltage source with no voltage is a short circuit and it stays in the netlist. A dependent source needs the opposite care: it must not be zeroed at all, because it is an element whose law involves another node rather than a source the theorem is about. Zeroed, it is a short circuit, and the answer that comes back is 4.0625 volts at every gain from one to three hundred — while the true answer moves from 3.82 volts to 0.206.

networks · Superposition
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%.

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.

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

The gap that is bigger than it is

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

magnetics · Magnetic path
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.

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.

semiconductors · Current mirror
An integrator's summing junction is flat at 15.7 Ω, and worst at direct current. computed by solving, not by drawing. A current is driven into the node and the voltage read. The dashed curve is the same amplifier with a 10 kΩ resistor as its feedback element: a tenth of an ohm at direct current, rising a decade per decade, 909.5 Ω at the top. With a 10 nF capacitor instead, the loop gain has no frequency in it — the amplifier's gain falls as 1/f while the feedback factor rises as f — so it is flat at 62.83 and the node is flat with it, at Rin/(1 + 2π·GBW·C·Rin) = 15.666 Ω, measured 15.668 Ω and holding to 0.50 per cent from 100 Hz to 5.31 kHz — between the amplifier's own pole and the capacitor's corner, which is where a loop gain with no frequency in it lives. Both ends are the other way round from the resistive case: at direct current the capacitor is an open circuit, there is no loop at all, and the node is 843.0 Ω — a factor of 8388 worse than the resistor's 100 mΩ. Above the amplifier's crossover the capacitor is a short and the node is 47.6 Ω against 909.5. The two cross at 1.56 kHz, which is the frequency above which an integrator is the better virtual earth of the two.

The shelf a capacitor makes

An inverting amplifier's summing node rises a decade per decade to the two resistors in parallel. Replace the feedback resistor with a capacitor and it does neither: an integrator's loop gain has no frequency in it, so the node is flat at Rin/(1 + 2π·GBW·C·Rin) — 15.67 ohms over four decades, to half a per cent. Both ends invert. At direct current the capacitor is an open circuit, there is no loop at all, and the node is 843 ohms against the resistor's tenth of one; above the amplifier's crossover the capacitor is a short and the node is nineteen times better.

feedback · Virtual earth
A star ground is a 0.50 mm daisy chain, not a point. computed by solving, not by drawing. 8 circuits of 100 mA each meeting at one pad, which reaches the plane through a via of 0.25 mΩ and 0.35 nH. The via is shared by every one of them, so the star's residual is 7 other currents through it — 175.0 µV at 1.00 kHz against the 1.800e+4 µV a daisy chain of 10 mm hops would give at its far end, a factor of 103. Measured as a length of the same track, the via's resistance is worth 0.500 mm and its inductance 0.350 mm — two different lengths, so a via has a corner of its own at 114 kHz where a track's is at 79.6 kHz wherever it is cut. The faint curves are other circuit counts. The two grow differently: fitted over counts from 2 to 32 the star's error goes as N^1.22 and the chain's as N^1.87, because a chain's hop carries every circuit beyond it — so the star's advantage grows with the count rather than shrinking. And a star has no position in it: a chain's far end reads 4.5 times its near end, which is a diagnostic no real fault should have.

A star is half a millimetre long

The repair for a shared return is a star ground, and it is described as making the shared length zero. It does not: every return meets at a pad, the pad reaches the plane through a via, and a via of a quarter-milliohm and 0.35 nanohenries is worth half a millimetre of one-ounce track by its resistance and 0.35 by its inductance — two different lengths, so a via has a corner of its own at 114 kilohertz where a track's is at 79.6. What that buys is a change of law rather than a factor: a daisy chain's error grows as the square of the circuit count and a star's as the first power, so the advantage runs from sixty times at two circuits to three hundred and forty at thirty-two.

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

The clock that is too fast

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

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

The boundary that is a rise time

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

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

The slowest one decides it

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

transients · Doublet
A twin-T's null: everything fills it, and at one order. computed by solving, not by drawing. A twin-T of 10 kΩ and 10 nF nulls at 1591.5 Hz, where a low-pass T of two resistors and a high-pass T of two capacitors deliver equal and opposite contributions. With exact components the null is -Infinity dB, which is the arithmetic's floor. A tolerance fills it at 20.0 decibels per decade — first order, where an LC notch splits into twenty and forty depending on which way two errors go, because there a product can be left alone and here there is no product. Loss fills it at 20.0 decibels a decade too, and barely moves it — 1591.55 to 1595.52 Hz across four decades of series resistance — because a resistance of r in a capacitor branch is a fractional error of r/R and nothing else: one ohm gives -88.5 dB against a hundredth of a per cent of tolerance's -92.0. So a notch's two mechanisms are not exchanged here — they are one mechanism, because a twin-T's null is stationary in nothing and an LC notch's frequency is stationary in the product of its two components.

The null that is stationary in nothing

An LC notch has two mechanisms and they do different things: a tolerance moves its null rather than filling it, because the resonance depends on a product two errors can leave alone, and loss fills it at twenty decibels per decade of arm resistance. A twin-T has no resonance and no arm. Measured, everything fills it at twenty decibels a decade and nothing moves it — one ohm in series with its capacitors gives 88 decibels where a hundredth of a per cent of tolerance gives 92, because a series resistance in a capacitor branch is a fractional error of r over R and nothing else.

filters · Null depth
4 branches read with 10 Ω ammeters: one at a time every reading is low, all at once they add up and the small branches are high. computed by solving, not by drawing. A 1 A source into 4 parallel branches of 100 Ω, 1 kΩ, 10 kΩ, 100 kΩ, each read with an ammeter of 10 Ω. With one ammeter moved from branch to branch, each reading is low — by −0.989%, −0.902%, −0.099%, −0.010% — exactly r over the resistance that meter looks back into plus r, and the four readings add to 990.275 mA rather than 1000.000. With an ammeter in every branch at once the readings add to 1000.000 mA and are wrong by −0.892%, +7.939%, +8.910%, +9.008%: the meter in the largest branch pushes current out of it, and the small branches take it.

The readings that add up and are wrong

Four branches of 100 Ω, 1 kΩ, 10 kΩ and 100 kΩ share a 1 A source. Read them one at a time with a single 10 Ω ammeter moved from branch to branch, and every reading is within one per cent of the truth, but the four add to 990.275 mA. Put a 10 Ω ammeter in every branch at once and the four readings add to 1000.000 mA exactly, satisfying Kirchhoff's current law to the last digit. Yet three of them are 7.9 to 9.0 per cent high. The check a reader would run on the readings is passed by the wrong set and failed by the right one.

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

How many straight lines a diode is

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

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

The count under the density

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

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

The boundary that is a starting point

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

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

The optimum that does not move

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

magnetics · Winding
With 100% of its capacitance to its mounting, a 141.4 Ω helix is 1.37° out at 100 MHz, not 2.7 × 10⁻³°. computed by solving, not by drawing. The angle of the impedance of a 141.4 Ω resistor — √(L/C) for 8 nH of lead and 0.4 pF — against frequency: the three-element part with all 0.4 pF across its film (dashed), and a helix of 32 turns with the same totals (solid), 0% of the capacitance between adjacent turns and 100% from the turns to the part's mounting, with one end grounded. At √(L/C) the three-element part has no first-order angle and rises as the cube of frequency. The helix is not flat there: its capacitance to the mounting counts only a third, so its own flat resistance is √(3L/C) = 244.9 Ω, and at 141.4 Ω its angle is first order in frequency — 1.37° at 100 MHz against 2.7 × 10⁻³°.

The resistor that is a helix

A spiral-cut film resistor is a helix, and its capacitance is distributed along the cut rather than lumped across the ends. Where it goes decides whether the three-element model survives. Capacitance between adjacent turns bridges each turn's own piece of film, and a helix of any number of turns built that way is the three-element part exactly — to 8×10⁻¹³ at 64 turns. Capacitance from the turns to what the part is mounted over is different: its current returns through only part of the film, it weighs (4N² − 1)/(12N²) of its value, a third in the limit, and the resistance with the flattest angle moves from √(L/C) = 141.4 Ω to √(3L/C) = 245 Ω. At 141.4 Ω such a part is 1.37° out at 100 MHz where the lumped model says 0.0027°. The third is a first-order rule; above about a gigahertz the helix is a line.

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

The load that rings with the line

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

limits · Lumped element
The winding window solved electrostatically, in two portions. computed by solving, not by drawing. The same cross-section the loss solve reads, read with ∇·(ε∇φ) = 0 instead. Two things are the opposite way round from the magnetic problem and both are the whole difference. The iron is now a Dirichlet boundary rather than a Neumann one — an earthed core is an equipotential, so the field meets it at right angles instead of running along it — and a conductor carries a prescribed potential rather than a prescribed current. The thin curves are equipotentials, which are contours of φ, so equal spacing is equal potential step and crowded curves are a strong field. The copper is shaded by the potential each foil sits at, which rises along the winding rather than being one number. Winding to winding this window is 926.9 picofarads a metre, and 89 per cent of the energy is inside insulation that occupies a fraction of the window.

The other half of the same window

The two-dimensional solve that settled what a winding's alternating-current resistance really is computed one of the window's two parameters and never mentioned the other. Read with Laplace instead of the vector potential, the same cross-section returns 926.9 picofarads a metre — and 89 per cent of that energy sits inside films that occupy 14.1 per cent of the window. The instrument agrees with a layered slab to three parts in ten thousand billion and converges on a real winding at order 1.34, and the reason for the shortfall is not the arithmetic but the corner of a conductor.

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

Interleaving is a choice, not an improvement

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

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

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.

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

The tail with no slowest relaxation

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

transients · Doublet
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 of the anhysteretic model.

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.

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

The pole that crosses with the resistance

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

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

The conductor made causal too

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

lines · Dispersion
Two junctions at 8 to 1 in area, at four currents from 10 µA to 10 mA. computed by solving, not by drawing: each drop is the diode law solved at its own temperature, from 200 to 400 K, with a saturation current that rises 15.0 orders of magnitude over the range. The two drops themselves sit 179 mV apart across the four currents; their difference does not move at all. It is Vₜ·ln 8 to 0.40 parts in a million at 10 mA and to 0.040% at the worst point, 10 µA at 400 K, where the larger junction's saturation current is no longer negligible against the current — 53.758 mV at 300 K, rising 179.19 µV/K — and the straight line through it reaches 0.001 µV at absolute zero (dashed).

The voltage at which the slope is zero

Two junctions carrying one current, one of them eight times the other's area, differ by exactly the thermal voltage times ln 8 — 53.758 mV at 300 K, rising 179.19 µV/K along a line that reaches zero at absolute zero to a thousandth of a microvolt — while each drop alone moves 1.9 mV/K. Weighted by 10.72 and added to one drop, the difference cancels it, and the sum is then 1.171704 V. That voltage is not a coincidence of this design: every weighting of the sum leaves the band gap at absolute zero, so a reference's slope at 300 K is its distance from 1.1717 V divided by 300 K, and two hundred references built wrong three different ways sit on that one line to 0.16 nV/K. A reference is trimmed by its voltage, at one temperature, and the voltage it is trimmed to is a statement about how its current is made.

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

The time constant one trimmer cannot reach

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

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

The rate each reading needs

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

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

The bow a trim leaves

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

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

The instant a transformer is worst at

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

power · Inrush
The quantisation error's power as a function of where the input sits, for four dithers. computed by solving, not by drawing; the error's mean square given the input, integrated over each dither's density through the converter's own quantiser, and checked against seeded draws to 6 × 10⁻³. Undithered it is the square of a sawtooth, x² up to half a step and (1 − x)² beyond. One uniform step makes it x(1 − x): zero at a code, a quarter of a step² between codes — the error's power is carried by the signal. Two steps (triangular) hold it at ¼ everywhere, and one step subtracted after the converter at 1/12 everywhere. Averaged over the input the prices are 3.01, 4.77 and 0.00 dB against the undithered 1/12.

One step of dither per moment

A whole step of uniform dither makes a quantiser's error average to zero wherever the input sits, and leaves its power at x(1 − x) steps²: zero at a code and a quarter between codes, so the noise rises and falls with the signal. Two steps — a triangular dither — hold the power at a quarter everywhere and leave the third moment swinging by 0.096 steps³; three steps fix that too. Each added step fixes one more moment of the error and costs a twelfth of a step² of power: 3.01 dB for one, 4.77 for two. Subtracted after the converter, one step fixes the error's whole distribution for 0.00 dB.

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

The edge of a chain

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

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

The window that closes from the top

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

noise · Shot noise
The second harmonic in each cycle of a transformer's inrush, as a share of the fundamental. computed by solving, not by drawing, the harmonics taken per cycle of the marched current. The first cycle carries 79% second harmonic and the second 93%. The share falls below the 15% a differential relay is commonly set to restrain at only in cycle 30, when the peak is 205 mA against a first peak of 88 A — the harmonic outlasts the inrush it identifies.

The harmonic that names an inrush

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

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

The envelope that leaves early

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

filters · Group delay
What balancing the source resistances is worth, against how correlated the two input currents are, for a bias-cancelled input. computed by solving, not by drawing; 400 seeded parts at each correlation, each solved unbalanced and balanced at a 1 MΩ source, input currents of zero mean and 1 nA spread. The ratio of rms errors (dots) follows 1/√(2(1 − ρ)) (line): 0.72 for independent currents — balancing makes it 38% worse — one at ρ = ½, and 6.5 at 0.99. The cure removes what the two currents share, and is a cure only if they share more than half.

The cure that needs one current

Balancing the resistances an amplifier's inputs look into removes whatever its two input currents have in common and leaves their difference. On a plain bipolar input the two are one current at two pins, correlated at 0.991, and the cure is worth ten. On an input whose bias current has been cancelled they are two residues of either sign, correlated at 0.10, and over 400 seeded parts the cure makes the error 38 per cent worse. It is worth 1/√(2(1 − ρ)) for a zero-mean current, nothing at ρ = ½, and in general exactly the input current over the offset current.

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

One margin, and the gain under it

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

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

The node the bandwidth belongs to

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

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

The gain an amplifier asks for, squared

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

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

The table that has to be refitted

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

limits · Diode model
A chain of identical one-pole stages: the time its step takes to settle within 1%, against the number of stages. computed by solving, not by drawing: the chain's step response is the Erlang distribution's, and its 1% settling bisected on it, held against a fourth-order march of the chain at one, four and sixteen stages. One stage settles in 4.61τ, the familiar ln 100; four in 10.0τ, sixteen in 26.7τ, a thousand and twenty-four in 1099.9τ. The frequency edge's rule, √N stretched by the bandwidth that holds 1% (faint), says sixteen stages need 4.02 times one stage's time; they need 5.81. For a long chain the settling time approaches the delay, N time constants (dashed): 1.074 of it at 1024 stages. For one stage (τ = 1.59 µs at 100 kHz) the 1% time is 7.33 µs.

The delay a chain adds

A chain of one-pole stages leaves its flat band as the square root of the number of stages, and it leaves its settling time faster: four stages take 10.0 time constants to settle within 1 per cent against one stage's 4.61, sixteen take 26.7 — 5.81 times one stage, where the frequency rule allows 4.02. The difference is a delay. The chain's step is the sum of N exponential lags, so it is an edge of mean N time constants and spread √N; the magnitude response sees only the spread. At a thousand stages the 1 per cent time is 1.074 of the delay. A chain budgeted to stay flat to 1 per cent settles to 1 per cent 45 per cent late.

limits · Model edges
A gain of ten round an amplifier with a second pole: its half-power bandwidth and its noise bandwidth, each over what the gain–bandwidth product predicts. computed by solving, not by drawing: the stage as a netlist round a two-pole amplifier of 1 MHz, the second pole placed at r times GBW/G. The half-power bandwidth agrees with the second-order closed form √r·g(√r/2) to two parts in a thousand at every placement: 1.000 at r = 0.5, 1.272 at r = 1, 1.414 at r = 2, 1.231 at r = 5, 1.110 at r = 10, 1.010 at r = 100 times GBW/G. It is 1.414 times the product's prediction at r = 2 and within ten per cent only beyond r = 10. The noise bandwidth is (π/2)·GBW/G to half a per cent at every r: the second pole moves the one and leaves the other where it was.

The bandwidth the product still gets right

Gain times bandwidth is a constant for an amplifier with one pole, and real amplifiers have a second. Put it at r times the closed-loop bandwidth the product predicts and the half-power bandwidth becomes √r·g(√r/2) times GBW/G — 41.4 per cent wide at r = 2, within ten per cent only beyond r = 10 — so a follower round a 1 MHz part with its second pole at 2.5 MHz runs 40 per cent wider than the rule says. The noise bandwidth does not move at all. A second-order loop's noise bandwidth is set by one coefficient the second pole cannot touch, and it stays (π/2)·GBW/G at every r and every gain, to half a per cent on the netlist. A third pole is what breaks it.

feedback · Gain-bandwidth
A power ferrite driven one way to 0.3 T: ungapped, every pulse starts at 90.8 mT; with a 0.1 mm gap, at 22.0 mT. computed by solving, not by drawing, marched on the hysteretic core. A ferrite with 90.6 mT of remanence and a coercive field of 38 A/m, 100 turns on 100 mm² and 60 mm of path, its flux raised to 0.3 T by a pulse and then released by letting the winding current fall to zero, cycle after cycle until the starting flux stops moving. Ungapped (solid) the release ends at H = 0 on the descending branch, 90.8 mT: each pulse can add only 209.2 mT. With a 0.1 mm gap (dashed) the current is zero where le·H + lg·B/μ₀ = 0 — the shear line (dotted) — at H = −29.0 A/m and 22.0 mT, under the bound μ₀·Hc·le/lg = 30.2 mT; each pulse can add 278.0 mT.

The flux the reset leaves behind

A transformer driven one way only — a forward converter, a single-ended pulse transformer — is reset each cycle by letting its magnetising current fall to zero, and zero current is not zero flux. Marched on the hysteretic core, a power ferrite with 90.6 mT of remanence driven to 0.3 T starts every pulse but the first at 90.8 mT, and keeps 34.9% of the swing a symmetric drive would get, where half is the most a one-way drive could; a square-loop core starts at 212 mT and keeps 14.7%. A gap shears the loop: the current is zero where H = −lg·B/(μ₀·le), so the start is bounded by μ₀·Hc·le/lg and falls onto the bound as the gap grows — 22.0 mT at 0.1 mm, recovering 69 mT of swing for 239 mA more magnetising current. The first pulse from a demagnetised core is the only one that starts at zero, and it is the one a bench test sees.

magnetics · Saturation
A 63.7 Ω resistor in series with a gyrator's capacitor moves its series resistance's sign change from 63 Hz to 754 Hz, and 70.0 Ω removes it below the amplifiers' edge. computed by solving, not by drawing. The one-henry synthetic inductor of four 10 kΩ resistors, a 10 nF capacitor and two 1 MHz amplifiers with a gain of 10⁵, its series resistance against frequency with a resistor r₄ in series with the capacitor of 0 (grey), 0.9, 1 and 1.1 times 4/(ωₜ·C) = 63.66 Ω. Without it the resistance is −0.153 Ω at 100 Hz and −25.2 Ω at 1 kHz, negative from 63.0 Hz. r₄ adds ω²·L·C·r₄, the same frequency law, so at exactly 4/(ωₜ·C) the resistance is 0.099 Ω at 100 Hz and −0.075 Ω at 1 kHz and changes sign only at 754 Hz; at 1.1 times it stays positive throughout: 2.44 Ω at 1 kHz.

The resistor that pays the amplifiers back

A synthetic inductor's negative resistance grows as the square of frequency, and so does the real part a resistor in series with the gyrator's capacitor adds: Zᵢₙ = sL/(1 + sCr₄) ≈ sL + ω²LCr₄. Fitted on solves across six designs, the resistor that cancels the amplifiers' term is 4/(ωₜ·C) to 1.8% — 63.7 Ω for 10 nF and 1 MHz amplifiers, whatever the resistance level or the open-loop gain. Put in, it moves the sign change from 63 Hz to 754 Hz; a tenth more removes it below the amplifiers' edge and gives a quality factor of 2,579 at 1 kHz where the bare inductor had −250. The catch is that the cancellation is as good as the gain–bandwidth is known: amplifiers 10% slower than the resistor was chosen for leave −2.9 Ω at 1 kHz. Chosen a tenth over the slowest amplifier a ±30% spread allows, the inductor's Q at 1 kHz runs from 312 to 1,847 and is never negative.

filters · Gyrator
A 1 MHz sine of 1 V into an emitter follower with 1 nF on its emitter and a 2 mA sink: the input falls at up to 6.28 V/µs, the output can fall at only 2.0, and the transistor switches off. Marched, the base driven by a stiff source and the emitter current Is·exp((vᵢₙ − vₒᵤₜ)/Vₜ). The input (dashed) and the output (solid) over the last of four cycles. The input's steepest fall is 2πfA = 6.28 V/µs; the load can be discharged no faster than the sink allows, I/C = 2.00 V/µs. On every falling stretch the base–emitter junction reverse-biases, the emitter current drops below a hundredth of the sink's (dots), and the output falls as a straight line of slope −2.000 V/µs until the input comes back up to meet it. On the rising half the transistor pushes whatever current the capacitor needs and the output follows a Vbe below the input.

The follower that lets go on the way down

An emitter follower can push current into its load without limit and pull it out only through whatever sinks the emitter. Marched as a large-signal circuit — a 2 mA sink, 1 nF on the emitter — a 1 V, 1 MHz sine asks the output to fall at up to 6.28 V/µs, the sink allows 2.00, and on every falling stretch the junction reverse-biases, the emitter current drops below a hundredth of its quiescent value for 57% of the cycle, and the output falls as a straight line of exactly −I/C. The boundary is a slope, 2πfA ≤ I/C: 2.87 V at 100 kHz, 0.309 V at 1 MHz, creeping above the line only as the period approaches the junction's own 12.9 ns. Past it the follower rectifies — its output's average rises 415 mV at 1 V of drive and 1.30 V at 2 V — and with a resistor in place of the sink the fall slows as the output descends, from 2.41 V/µs to 1.99. None of it is visible to the small-signal model the earlier essays used.

semiconductors · Emitter follower
One step of refinement in the same double precision returns the stopband's lost digits: 1.1e-12 at 10 kHz before, under 2.4 units of round-off after. computed by solving, not by drawing. The 5th-order Chebyshev ladder solved at every frequency in double precision (solid), and again after one step of iterative refinement — the residual b − A·x formed in double, the correction solved with the same elimination and added (dashed) — each against the same network eliminated in a pair of doubles. Plain, the response's error tracks the round-off divided by the response (dotted), from 1.1e-16 at 10 Hz to 1.1e-12 at 10 kHz, where the output is 6.78e-7 of the input. One step brings every frequency to within 2.4 units of round-off of the truth. No more digits were used: the loss belonged to the elimination's errors in the small component, and the residual, computed against the original matrix, sees them.

The step that returns the digits

A fifth-order ladder filter solved in double precision loses four decades in its stopband — 1.1 × 10⁻¹² of relative error at ten times the cutoff — on a solution vector that is right to round-off. One step of iterative refinement, in the same double precision, returns all of it: form the residual b − A·x, solve for a correction with the same elimination, add it, and every frequency is within 2.4 units of round-off of a 31-digit solve. The reason is that the plain solve is stable by the norm and not by the row — its componentwise backward error is 5.3 × 10⁻¹³ in the stopband rows — and one step makes it componentwise stable. The bridge walked to balance cannot be helped at all. Its node voltages are already right to the last bit; its digits are lost storing two numbers whose difference is the answer, and again storing its conductances, which moves the network itself by round-off over δ before anything is solved.

networks · Conditioning
A stray of a tenth of the sampling capacitor moves a one-plate integrator's coefficient by 9.85%; switched at both plates, by 0.14%, nearly all of it the amplifier's own gain error. computed by solving, not by drawing. A switched-capacitor integrator with a 1 pF sampling capacitor, 4 pF of feedback and 0.5 pF at the amplifier's input, and a stray capacitance to ground at the sampling capacitor's switched plate; the error of the per-cycle coefficient against Cs/Cf, against the stray as a fraction of Cs. With one plate switched (solid) the stray is charged to the input and dumped with the capacitor, so it is part of the capacitor: the error is the stray's fraction less the amplifier's small gain error, 9.85% at a tenth. With both plates switched (dashed, A = 1000; dotted, A = 10⁵) the plate that meets the summing node moves between ground and −Vo/A, so the stray's charge carries a factor 1/A: its sensitivity falls from 1/Cf to Cs/(A·Cf²), 4000 times smaller, and what is left, 0.14%, is the amplifier's gain error (1 + (Cx + Cs)/Cf)/A, which the one-plate arrangement has as well.

The capacitance at the switched plate

A switched-capacitor filter's corner is a ratio of two capacitors, and every plate on a chip has a capacitance to the substrate that is on no schematic. Switch one plate of the sampling capacitor and that stray is charged to the input and dumped with it: a stray a tenth of the capacitor is a tenth of error, and a bottom plate's fifteen per cent is fifteen per cent at any size. Switch both plates and integrate into a virtual ground and the stray at the plate that meets the summing node moves only between ground and −Vₒᵤₜ/A — its sensitivity falls from 1/Cf to Cs/(A·Cf²), four thousand times less at a gain of a thousand, and what is left is the amplifier's own gain error, 0.14%. The stray has not vanished: it has moved out of the coefficient that sets the corner and into the integrator's leak, where the direct-current gain becomes A·Cs/(Cs + Cp).

filters · Switched-capacitor
Asked for 105% of the nose power, a tap changer that blocks on a wrong-signed step stops at 9.759 V and holds it; one that does not keeps raising, and the voltage keeps falling. Marched. The 10 V source behind 1 Ω of the regulator essays, its load drawing 105% of the line's 25 W nose power at the 10 V setpoint, regulated by a tap changer that raises the ratio by 1.25% every 5 s while the load is more than 20 mV low. With no blocking rule (dashed) it climbs past the nose ratio √(Rₗ/R) = 1.9518 and keeps going, to n = 3.250 and 8.61 V by 900 s, still falling; the integrating regulator of the earlier essay (dotted) collapses the same way, faster. With the rule (solid) the tap at n = 1.9625 is the first to lower the voltage, the regulator stops at 385 s, and the load holds 9.759 V — within a tap of the most this line can give it, 9.759 V. The setpoint is out of reach either way; the rule converts a collapse into a stall.

The rule that reads the sign of the gain

A regulator that raises a ratio whenever its load is low collapses the voltage once the load asks for more than the line's nose power, because past the nose raising the ratio lowers the voltage. The standard protection is a rule: after each tap, compare the voltage with the one before, and if it fell, stop. On a 10 V line asked for 105% of its 25 W, a 1.25% tap changer with the rule stops at n = 1.9625, the first tap past the nose, and holds 9.759 V — the most the line can give — where without it the ratio climbs to 3.25 and the voltage is 8.61 V and still falling. The rule is exact without noise and fragile with it, because it reads a difference that vanishes at the nose: with 3 mV of measurement noise it wrongly stops a raise that could have reached the setpoint 0.3% of the time at 98% of the nose and 15% at 99.5%. Trusting it to 1% costs margin — 97.9% of the nose at 3 mV, 79.9% at 10 mV — or coarser taps.

power · Voltage regulation
Three waveforms at 16 samples a period, sampled at the worst instant: the largest sample of a sine is 1.92% short, of a triangle 12.5%, of a clipped sine 0. computed by solving, not by drawing: three waveforms with the same period and the same peak, sampled 16 times a period at the phase that puts the peak midway between two samples (dots). The sine leaves its peak as the square of the time, the triangle in proportion to it, and the sine clipped at nine-tenths of its height does not leave it at all for as long as it stays flat. The largest sample therefore falls short by 1.92% for the sine, 12.5% for the triangle, 0 for the clipped sine, at one rate, for three waveforms with the same period.

The peak that is a corner

The largest of a set of samples falls short of a sine wave's peak as the square of the sample interval, and that law belongs to the sine. Measured over every sampling phase, a triangle's corner falls short as the first power, 12.5 per cent at sixteen samples a period against the sine's 1.92, a peak whose curvature vanishes as the fourth power, and a sine clipped at nine-tenths of its height reads exactly above seven samples a period. The order in which a waveform leaves its peak is the whole of the law, and it also fixes how much better the average reading is than the worst. A real corner is round at some scale, so the law bends where the samples resolve its top harmonic, and a parabola through three samples helps a round peak, barely touches a corner and makes a flat one worse.

digital · Sample rate
A stage with its second pole at 3 times the bandwidth the product predicts settles within 0.1% in 5.06τ, against 6.91τ for the one-pole stage the product describes. computed by solving, not by drawing: the step of a stage whose amplifier has a second pole at r = 3 times GBW/G (solid), against the one-pole stage of the same product (dashed). Time is in units of τ = G/(2π·GBW). Settled within 0.1% at 5.055τ, against ln(1000) = 6.908τ, with the netlist of a gain-of-ten stage agreeing at 5.057τ. The stage with the second pole is usually faster, whatever it does on the way.

The gain that stops making a stage faster

The gain–bandwidth product says a stage of gain G settles in time proportional to G, and a stage with a second pole does something else. Below r = 4 the closed loop's poles are a complex pair whose real part is half the sum of the open loop's, π times the second pole's frequency, with neither the gain nor the product in it: a 1 MHz amplifier with a 0.5 MHz second pole takes 4.10 µs to settle within 0.1 per cent at unity gain, 4.04 at two and 4.22 at five, where the product says 1.10 to 5.50. Above the knee at G = 4·GBW/f₂ the product returns. In between lies the fastest stage, r = 3.32 for a 0.1 per cent band, which settles in 3.44 time constants against the one-pole stage's 6.91: half the time, and not at the r that gives the widest band.

feedback · Gain-bandwidth
A megohm part with one picofarad to its mounting, heavier at one end: 0.433 pF with that end driven and 0.233 pF turned round, 1.56° against 0.84° at 10 kHz. computed by solving, not by drawing: the angle of the impedance of a 1 MΩ resistor as 200 sections, with 1 pF in total from its film to its mounting tapered as c(x) ∝ 1 + a(1 − 2x), a = 0.6, read at the terminal with the far end grounded. Driving the end nearer the mounting (solid) it presents 0.433 pF and 1.56° at 10 kHz; turned round (dashed), 0.233 pF and 0.84°. The first-order weight is ∫c(x)(1 − x)²dx, which is C(1/3 + a/6) and C(1/3 − a/6). The part's total capacitance has not changed, nor has its resistance, and neither of the readings is that total.

The resistor turned round

A resistor's stray capacitance is usually drawn as one capacitor across its ends, and a stray to the part's mounting is not one number. Read at the terminal with the far end grounded it is weighted by the square of the distance from ground, so a uniform stray counts as a third of itself, and turned round a part with a tapered stray reads differently: 0.433 pF one way and 0.233 pF the other from one picofarad in all. In the current through the part the weight is x(1 − x), a sixth, and the current reaching the far end is 3 dB down at ωRC = 9.18 rather than at one. Two readings of one part separate the total from the taper, and a distributed stray's phase never reaches a capacitor's: it stops at 45 degrees.

frequency · Real resistor
Eight stages of damping 0.8 against eight one-pole stages of the same bandwidth: settled within 1% at 18.75τ against 16.00τ, with 2.02% of overshoot. computed by solving, not by drawing: the step of eight identical second-order stages of damping ζ = 0.8, each with its half-power bandwidth at ω = 1, marched (solid), against eight one-pole stages with the same bandwidth from the closed Erlang step (dashed). Within 1% at 18.75τ against 16.00τ, the overshoot being 2.02% and each stage's delay 1.393 against one. At ζ = 1 the march agrees with the Erlang step of order sixteen.

The stage that overshoots a little

A stage with a little overshoot settles sooner than a one-pole stage of the same bandwidth, 3.53 time constants against 4.61 within one per cent, and a chain of them should do better still, since a stage at the right damping adds almost no spread. It does not. The saving is 17 per cent at two stages, 11 at four and 2 at eight, and at sixteen the chain is 5 per cent slower than the one-pole chain, because a stage's extra delay adds as the number of stages and the spread it removes adds only as the root. The best damping stays near 0.83 at every length, just short of letting the overshoot out of the band. In frequency the same stages are five times flatter at their best, at a different damping, and so the two edges of a chain ask for two different stages.

limits · Model edges
Scaling a Chebyshev 5 with real 4 nV/√Hz amplifiers everywhere: 118.18 dB at equal maxima and 118.32 dB with no gain stages at all, so the gains cost 0.14 dB, where silent ones promised 2.16 dB. computed by solving, not by drawing: the dynamic range on ±15 V of a fifth-order Chebyshev with gains d^α in front of each section and after the last, against α. Every amplifier has a voltage noise of 4 nV/√Hz, the sections' own and the stages', and each stage carries a kilohm in its divider or lower feedback arm (solid). With every amplifier and gain silent (dashed), equal maxima recovers 2.16 dB over no scaling, 142.44 dB to 144.60 dB. With real ones: 118.32 dB with no stages, and with stages 115.75 dB at α = 0.2, 116.27 dB at α = 0.4, 116.86 dB at α = 0.6, 117.50 dB at α = 0.8, 118.18 dB at α = 1. Merely having the stages is a step down at the smallest α, since each is a real amplifier in the path, and the curve climbs from there to equal maxima. The response is unchanged to a part in a billion at every α.

The floor the gains bring

Equal-maxima scaling recovers 2.16 decibels of a fifth-order Chebyshev's range when its gains make no noise and its amplifiers make none either. Give every amplifier in the circuit a voltage noise of 4 nV/√Hz, the sections' own and the stages', and the unscaled filter reads 118.32 dB over a 1 MHz band and the scaled one 118.18: the last gain has nothing after it to filter its noise, so it multiplies 4 µV of wide-band noise by 2.83, and the lower ceiling is spent paying for it. At 1 nV/√Hz scaling costs 6.61 dB, and at 16 nV/√Hz it gains 0.99, so a noisier amplifier makes it look better. A narrower measuring band does not rescue it. On a Chebyshev 7, with a node 13.26 times the input, the same arrangement gains 6.15 dB.

noise · Dynamic range
A Gaussian edge read at 3 samples per rise, at the worst sampling phase: straight chords between the samples say the rise is 8.78% from its true value. computed by solving, not by drawing: a Gaussian edge (solid), its samples at 3 per 10–90% rise at the sampling phase that reads worst (dots), and the straight lines an instrument's linear interpolation draws through them (dashed). The chord crosses 10 and 90 per cent earlier and later than the edge does, because a Gaussian edge curves in opposite senses at the two thresholds. The reading is 1.0878 rises against 1.

The samples a rise time needs

A rise time is read from two thresholds, and an instrument that draws straight chords between its samples is wrong at each by an amount that goes as the square of the interval. For a Gaussian edge at the worst sampling phase that is 5.02 per cent at four samples per rise and 0.818 at ten; the rule of thumb of three to five is a five per cent rule, and one per cent takes nine. The reading is not monotonic in the rate: above 20.9 per rise a first-order edge reads exactly at whole numbers and 0.046 per cent at 24.5, because its two errors subtract at one phase, and a Gaussian edge reads worse there because its two add. Reconstructing before reading needs 1.87 per rise for one per cent, and at four times the edge's 3 dB bandwidth the reading is still 11.7 per cent out.

digital · Sample rate
Two loops with exactly 45° of margin and the same crossover: with 100% of the extra lag a delay, the step overshoots 28.7%, against 23.3% for two poles. computed by solving, not by drawing: the closed-loop step of a loop K·exp(−sD)/(s(1 + s/p)) with its crossover at one, its phase margin exactly 45°, and 100% of the 45° beyond the integrator supplied by the delay (solid), against the same margin from a lag alone (dashed). Delay 0.785, lag's pole absent. Overshoot 28.7% against 23.3%; settled within 2% at 7.03 against 7.06; gain margin 6.0 dB against no limit. Marched with a history buffer, and at δ = 1 checked against the exact series of the method of steps.

The margin a delay spends

A phase margin can be made of a lag or of a delay, and the number is the same either way. Forty-five degrees from a lag overshoots 23.3 per cent, and forty-five degrees from a delay overshoots 28.7; at sixty degrees the sign reverses, 8.77 against 5.57, and the two cross near 54 degrees. The overshoot is the smaller difference. A two-pole loop's gain margin has no limit at all, and five per cent of its extra lag as delay brings that to 25.5 dB; with all of it delay the gain margin is 90°/(90° − margin), which is 6.0 dB at 45°, and nothing else is in the expression. Across twenty-one such loops the margin alone misses the overshoot by 13.1 points. The closed-loop peak, which already reads the whole band, misses by 1.2.

feedback · Margin overshoot
A tank with its loss in its coil, Q = 3: the voltage is largest at 0.9972 of 1/√(LC) and in phase with the current at 0.9428, and neither is where the poles sit. computed by solving, not by drawing, by closed form and by a netlist of L, Rs and C. The impedance of a coil of inductance L and series resistance Rs across a capacitor, against frequency, for Q = √(L/C)/Rs = 3. The poles are the series circuit's, at 1/√(LC) with a damping of Rs/2L. The coil's own zero at −Rs/L moves the peak to ωₘ = 0.9972·ω₀ and the zero-phase point to ωᵣ = 0.9428·ω₀, and the ring is at 0.9860·ω₀. Four frequencies, which a series circuit's voltage across its resistor would have as one.

The reading a coil leaves nearly exact

On a series resonator three readings of Q agreed exactly and two did not. A parallel tank whose loss is in its coil has no exact readings at all, and the order is not the one its schematic suggests. Its poles are the series circuit's, but the coil's own zero puts the peak, the zero-phase point and the ring at three different frequencies, and all of them below 1/√(LC). At Q = 3 the phase slope reads 16.2 per cent low, the stored energy 5.72 low and the peak height 5.42 high, while the bandwidth reads only 0.307 low: the loss that a parallel resistance stands in for tilts across the band, and the tilt is nearly antisymmetric about the centre, so the reading that spans it cancels it. The instrument's own source then decides what is being read.

frequency · Resonance
An exponential stage at 20% output, with a loop gain of 1: second 2.5% from 4.97%, third 0.0415% from 0.165%, and fifth 0.68 ppm from 0.814 ppm. computed by solving, not by drawing: the harmonics of an exponential stage with emitter resistance, gm·Re = T, solved point by point on the implicit curve and transformed, at a fixed output fundamental of a fifth of the quiescent current, for T = 1 (dots a) against no feedback (dots b). 2: 2.5%, against 4.97%; 3: 0.0415%, against 0.165%; 4: 5.36 ppm, against 40.9 ppm; 5: 0.68 ppm, against 0.814 ppm; 6: 0.0202 ppm, against 0.0135 ppm; 7: 0.000245 ppm, against 0.000192 ppm; 8: 0.0000478 ppm, against 0.00000239 ppm; 9: 0.00000178 ppm, against 2.62e-8 ppm. The series gives the second and third at small drive: the implicit function's Taylor coefficients are 1/(1 + T), 1/2(1 + T)³ and (1 − 2T)/6(1 + T)⁵, and they agree. The third therefore has a null at T = ½, and past that it has the opposite sign, and feedback that reduces the low orders puts some of them back into the high ones.

The fifth a resistor adds

Local feedback's reputation is that it reduces distortion, and on the second order it does: 4.97 per cent falls to 2.5 at a loop gain of one and 0.46 at ten, with the output held fixed. The third has a null at a loop gain of exactly a half and comes back with its sign changed. And from the fifth upward, a harmonic rises above its open-loop value before it falls: the fifth 1.42 times, the sixth 5.23, the seventh 24.6, the ninth 960, peaking at a loop gain between six and twenty. The second order, carried back through the resistor, mixes with the signal and makes them. A total-harmonic reading shows only improvement throughout, because the second dominates it. A differential pair has no second order to carry back, and its fifth never passes its open-loop value.

semiconductors · Distortion
A 0.2% part sampling 0 V after long at 10 V, holding 1 ms: the first reading is 109 ppm of the step, and the hundredth still 4.5 ppm, every one of them made from the same branches. computed by solving, not by drawing: the reading at each sample, as a fraction of the step, when a sample-and-hold with a 1 µF part of 0.2% absorption samples 0 V after a long time at 10 V, acquiring for 100 µs and holding for 1 ms. The first reading agrees with a march of the netlist. It is 109 ppm, the tenth 46.7 ppm, the hundredth 4.5 ppm and the four-hundredth 0.904 ppm. The earlier samples are still being read, and for far longer than an acquisition suggests, because each acquisition can reach a branch only in proportion to the fraction of its time constant it lasts.

The sample before the sample

A sample-and-hold whose capacitor has 0.2 per cent of dielectric absorption reads a 10 V step 109 parts per million wrong on the first sample after it, 46.7 on the tenth, and 4.5 on the hundredth, because every earlier sample is still in the dielectric. That error has been described as distortion, since it grows with the signal. It is not. The state is linear and one period is a single matrix, so the reading is a fixed weighted sum of earlier samples, a filter with a very long tail: a sine reads 28.1 parts per million low and its largest harmonic is 1.5 × 10⁻¹² of the fundamental. A filter can be undone. Subtracting fifty earlier samples takes a stepped signal from 51.6 parts per million to 1.61, and with the absorption known to 20 per cent it stops at 5.47.

transients · Dielectric absorption

Named alongside it

The objects these essays reach for when they reach for this one.

Design tradeoffVerificationParasiticsMeasurement conditionLoop gainLoadingGain–bandwidth productComponent toleranceMarchingSettling timeThermal voltageOutput impedance

All concepts