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The thread: The parasitic is the component

A capacitor is a capacitance, a resistance and an inductance, and above a computable frequency the third one is the whole part. A source is an electromotive force and a resistance. The thing nobody draws is usually the thing that decides the answer, and it is always a number.

Where the models stop

Every model has an edge

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

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

Two windings, and the band between them

The band a turns ratio holds over

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

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

Two windings, and the band between them

What coupling buys, and where it does not

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

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

Networks, and how a solve is checked

The source that is not a source

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

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

Frequency, which is the same solve

The capacitor that is an inductor

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

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

Where the models stop

A band rather than an edge

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

Measurement, which is a circuit on a circuit

The millivolts in the wire

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

Frequency, which is the same solve

The Q the components allow

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

Where the models stop

The edges that are lengths

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

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

Frequency, which is the same solve

The same part written two ways

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

Two windings, and the band between them

The energy is in the gap

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

An inverting unity gain, and the 100 pF that only the loop can see. computed by solving, not by drawing. Two ten-kilohm resistors around a 10 MHz amplifier make a gain of 1.00, and a loop that closes against 2.01 — one plus the ratio, not the ratio. Adding 100 pF at the summing junction leaves the closed-loop gain at a kilohertz unchanged — 0.99998051 against 0.99997988, three parts in a million at the far end of the slider — and takes the phase margin from 90.0° to 14.4°, because the noise gain now rises a decade per decade and the loop closes at forty decibels per decade instead of twenty. Forty-five degrees is reached at 9.00 pF, bisected on the netlist. The capacitance is not part of the signal path and does not appear in any expression for the gain.

Feedback, and the margin

The gain the loop closes against

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

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

Lines, where a wire has a length

The mismatch that the cable hides

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

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

Frequency, which is the same solve

The resistor that is only a resistor

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

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

Filters, measured not tabulated

What actually fills a null

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

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

Devices, and the amplitude they stop being linear at

The buffer that is not a buffer

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

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

Before the steady state

The cancellation that leaves a tail

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

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

Feedback, and the margin

The load that gets inside the loop

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

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

Before the steady state

The diode that conducts backwards

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

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

Feedback, and the margin

The resistor that buys the margin back

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

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

Filters, measured not tabulated

The same filter a thousand times larger

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

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

Filters, measured not tabulated

The band that closes with the order

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

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

Frequency, which is the same solve

The capacitance that is not one number

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

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

Lines, where a wire has a length

The delay that is not one number

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

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

Power, and the part that does no work

The inductance that limits, and lifts

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

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

Devices, and the amplitude they stop being linear at

The input that pushes back

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

Two windings, and the band between them

The resistance that grows with frequency

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

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

Two windings, and the band between them

The inductor that is a capacitor

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

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

Lines, where a wire has a length

The millimetre that becomes common mode

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

A bulk capacitor and a ceramic, and the peak between them at 6.52 MHz. computed by solving, not by drawing. Each capacitor is three elements — its capacitance, its series resistance and its series inductance — and a one-amp source drives the node, so the node voltage is the impedance. Alone, each dips to its own series resistance at its own self-resonance and rises on either side. Together they do not: between the two resonances the bulk part is an inductor and the ceramic is still a capacitor, and an inductance across a capacitance is a parallel resonance. The pair reaches 1.187 Ω at 6.52 MHz, where the bulk alone would give 0.2023 Ω and the ceramic alone 0.2055 — 5.87 times worse than either. The dashed curves are the two parts on their own; the solid one is what the load actually sees.

Power, and the part that does no work

The pair that is worse than either

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

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

Filters, measured not tabulated

The band that does not close

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

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

Measurement, which is a circuit on a circuit

The corner the instrument has no part in

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

One part, two corners: 3.98 kHz to the common mode and 7.86 MHz to the signal. computed by solving, not by drawing. Two windings on one core with a coupling of 0.999, driven twice from the same netlist — once with the two conductors in opposition, which is the signal, and once with them in parallel, which is everything the cable picked up. The mode that goes the same way round both windings meets (1+k)L and is down three decibels by 3.98 kHz; the mode that goes opposite ways meets the leakage, (1−k)L, and is untouched until 7.86 MHz. The ratio is 1975, which is 2/(1−k) and contains no inductance at all. Neither number is computed here: both modes are driven and the answer is read.

Lines, where a wire has a length

The inductor one mode cannot see

Two windings on one core present a millihenry to a current that goes the same way round both and a microhenry to one that goes opposite ways, so the same component has a corner at 3.98 kHz and another at 7.86 MHz — a ratio of two thousand, which is 2/(1−k) and contains no inductance at all. It is bought to remove the conversion the previous essay measured, and five picofarads across each winding turn it over at 1.59 MHz and leave it worth 0.02 decibels by ten gigahertz.

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

Where the models stop

The resistance a slow curve cannot see

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

Three nanohenries of copper move the peak to 5.63 MHz and raise it to 1.29 Ω. computed by solving, not by drawing. The same two capacitors, with and without the inductance of the way to them: one nanohenry of mounting loop per part and two nanohenries of plane between the bank and the load. The dashed curve is the bank with no inductance in the way, peaking at 1.187 Ω at 6.52 MHz; the solid one is what the load sees, peaking at 1.293 Ω at 5.63 MHz. The peak moves down because the branch that is inductive at that frequency got more inductive, and it rises for the same reason. Above about twenty megahertz the two part company entirely: the bank is still falling toward its parts' own resistances and the load is rising on two nanohenries that no capacitor is across.

Power, and the part that does no work

The capacitor that is not where the load is

A decoupling bank is usually drawn as two capacitors joined to a load through nothing, which is a stated simplification. Put three nanohenries of ordinary copper in — one of mounting loop per part and two of plane between the bank and the load — and the anti-resonance moves down to 5.63 megahertz and up to 1.29 ohms, a probe touching the ceramic reads a twelfth of what the load sees at 16.7 megahertz and three and a half times too much at 8.35, and the twentieth capacitor is worse than the second.

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

Before the steady state

The capacitor that remembers

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

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

Measurement, which is a circuit on a circuit

The current the instrument draws

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

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

Frequency, which is the same solve

Two parasitics, and the resonance neither of them has

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

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

Feedback, and the margin

Where the trouble is at the input

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

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

Filters, measured not tabulated

The inductor that is an amplifier

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

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

Where the models stop

The width no load can change

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

The bank reaches 1.500 mΩ and the load sees 141.5 mΩ at that same frequency. computed by solving, not by drawing. One bulk part and 20 ceramics, with a nanohenry of mounting loop each and two nanohenries of plane between the bank and the load, solved once per frequency and read at both nodes. The dashed curve is the bank's own node — what a probe on the parts measures. The solid one is the load. The bank's least impedance is 1.500 mΩ at 11.3 MHz, and at that frequency the load sees 141.5 mΩ, which is 94.3 times more, against 141.5 mΩ of plane reactance at that frequency. Whatever the parts do, the load's reading cannot fall below the reactance of the copper in front of them, and the parts reach their best by moving up the frequency axis into it.

Power, and the part that does no work

The floor and the ceiling move apart

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

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

Where the models stop

Where an open switch leaks to

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

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

Frequency, which is the same solve

The resistor that is right in size and wrong in angle

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

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

The capacitance a third switch moves

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

Correcting a 20 Ω, 50 mH load to unity, and what it stores. computed by solving, not by drawing. A capacitor across a 230 V, 50 Hz supply is swept from nothing to 300 µF against a load drawing 1636 W and 1285 var. The reactive power falls through zero at 77.31 µF and keeps going; the energy stored in the installation rises from 2044.9 mJ to 4089.7 mJ at that point — exactly twice, because unity power factor means the two stores are equal — and goes on rising afterwards. Only the cable current has a least value, 7.113 A against 9.044 A.

Power, and the part that does no work

The energy a unity power factor doubles

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

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

The floor below any load

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

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

Measurement, which is a circuit on a circuit

The cure that becomes a different circuit

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

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

Lines, where a wire has a length

The depth a resonance does not have

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

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

Devices, and the amplitude they stop being linear at

The resistance that is below zero

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

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

Measurement, which is a circuit on a circuit

The sign of what the guard gives back

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

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

The floor, which bounds from below

Only the real part is warm

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

Measurement, which is a circuit on a circuit

The optimum that hands back a bandwidth

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

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

The floor, which bounds from below

The floor that is only a floor while nothing flows

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

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

Circuits that do a job, and the range they do it over

The resistance that lowers the ripple

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

Power, and the part that does no work

The headroom that is the line's own charge

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

The same 100 mΩ in the winding instead of the capacitor: 1.317 V of ripple against 1.335 V, at the same crest factor of 11.15. computed by solving, not by drawing, marched with the diodes in the netlist: a 1000 µF reservoir behind a centre-tapped rectifier, with a series resistance from 1 mΩ to 1 Ω placed either in the capacitor or in each half-winding. The crest factor is the same in both places to two parts in a thousand at every resistance — 13.34, 13.26, 13.03, 12.49, 11.15, 9.069, 6.548 — because both limit the charging current alike. The ripple is not: in the capacitor it has a minimum and rises to 1.711 V at an ohm; in the winding it falls throughout, to 1.172 V, against 1.331 V with no resistance. At an ohm the winding costs 534 mV of mean output and the capacitor 380 mV. The diode's root-mean-square current at an ohm is 0.3449 A with the resistance in the winding and 0.3484 A with it in the capacitor.

Circuits that do a job, and the range they do it over

The resistance that belongs in the winding

A reservoir capacitor's series resistance lowers the ripple to a minimum of 1.3303 V at 17 mΩ and raises it past that. The same resistance moved into the transformer's winding limits the peak current by the same amount — the crest factor agrees to two parts in a thousand at every value from a milliohm to an ohm — and the minimum is gone: the ripple falls throughout, to 1.172 V at an ohm against 1.711 V in the capacitor. The two resistances each carry a current the other does not, and that one asymmetry decides where a deliberate one should go.

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

Where the models stop

The shunt switch the source sizes

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

A via's area changes sign at 44.7 Ω, and two impedances give its 0.5 pF and 1 nH back. computed by solving, not by drawing, as a cascade of two-ports: a via of 0.25 pF, 1 nH and 0.25 pF, met by an edge of 59 ps from reference lines of 20 to 150 Ω. The area under the reflection is −Z₀C/2 + L/2Z₀ at every impedance: a bump below 44.7 Ω, where the inductance's term is the larger, nothing at it, and a dip above. From 50 Ω the area is 2.5 ps of dip, which a single-capacitance reading calls 0.100 pF. From 50 and 75 Ω together the two areas give 0.5000 pF and 1.0000 nH. An error of 50 fs on each area moves them by up to 0.8% and 1.5%.

Lines, where a wire has a length

The via two lines can weigh

The area under a reflection is a property of the discontinuity rather than of the edge, and for a via it is one number made of two: −Z₀C/2 from its pads and +L/2Z₀ from its barrel, with opposite signs. A via of half a picofarad and a nanohenry, seen from fifty ohms, leaves 2.5 picoseconds of dip — which a reading that assumes a capacitor calls 0.100 pF, a fifth of what is there. From fifty and seventy-five ohms together the two areas give 0.5000 pF and 1.0000 nH back, and fifty femtoseconds of error on each costs 0.8 per cent of the capacitance and 1.5 of the inductance. A second line at fifty-five ohms costs five times as much.

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

Where the models stop

The resistance that bends the signal

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

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

Two windings, and the band between them

The turns nearest the gap

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

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

Lines, where a wire has a length

The via that is a piece of line

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

Two 1000 µF parts of 100 mΩ each against one 2000 µF part of 100 mΩ: 709 mV against 728 mV of ripple, and a crest factor of 14.41 against 12.61. computed by solving, not by drawing, marched with each capacitor behind its own series resistance. Two identical 500 µF, 60 mΩ parts give 1.330303 V of ripple and one 1000 µF, 30 mΩ part 1.330303 V — the same network. Two 1000 µF parts of a stated resistance each, against one 2000 µF part of the same resistance: at 10 mΩ, 704 mV against 704 mV of ripple and a crest factor of 17.40 against 16.93; at 30 mΩ, 704 mV against 705 mV of ripple and a crest factor of 16.51 against 15.48; at 100 mΩ, 709 mV against 728 mV of ripple and a crest factor of 14.41 against 12.61; at 300 mΩ, 758 mV against 867 mV of ripple and a crest factor of 11.45 against 9.490; at 1000 mΩ, 1.013 V against 1.334 V of ripple and a crest factor of 8.170 against 6.620. The pair's ripple is lower only where one part's resistance is past the ripple's minimum, and its peak current is higher at every resistance.

Circuits that do a job, and the range they do it over

Two capacitors that are one

Two identical reservoir capacitors in parallel are not a new circuit to be marched: 500 µF at 60 mΩ twice gives 1.330303 V of ripple and so does 1000 µF at 30 mΩ once, to the sixth decimal. So a pair against one part of the same capacitance is one curve read at two resistances, and the pair always sits at the lower one — which lowers the ripple only past the curve's minimum, 709 mV against 728 at 100 mΩ a part, and raises the peak current at every value, a crest factor of 14.41 against 12.61. Mismatch the pair and the current still divides by capacitance, so a part with twice the resistance carries 2% less current and 1.93 times the heat.

Measurement, which is a circuit on a circuit

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.

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

The floor every filter has

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

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

Where the models stop

The leak no switch can hold

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

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

Two returns in one plane

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

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

Feedback, and the margin

The node that is an inductance

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

Measurement, which is a circuit on a circuit

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.

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

Circuits that do a job, and the range they do it over

The carrier a cable allows

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

A 10 ps stub under a balanced via, and the pad it has to come out of. computed by solving, not by drawing, as a cascade. The reflection from a via of 0.5 pF of pad and 1 nH of barrel on its balancing line of 44.72 Ω, met by an edge of 59 ps, four ways: with no stub, whose largest excursion is 24.4 mV; with a 10 ps open stub below its bottom pad, 90.5 mV and an area of 5 ps; with the stub's 0.224 pF taken out of both pads, 56.1 mV; and taken out of the bottom pad alone, 24.6 mV. Both trims balance the area again. Only the second restores the doublet.

Lines, where a wire has a length

The pad the stub comes out of

A via used to a middle layer leaves barrel hanging below the connection: an open stub. Under a via balanced on its √(L/C) line, a 10 ps stub is 0.224 pF of extra capacitance, 5 ps of area, and a reflection of 90.5 mV where the balanced via gave 24.4. Taking that capacitance out of the pads balances the area again whichever pad it comes from. Only one restores the doublet: out of both pads the via still reflects 56.1 mV, out of the bottom pad the stub hangs from, 24.6. The trim works for every stub whose capacitance fits inside that pad, delays up to Z·C/2 = 11.2 ps, whose quarter wave is above 22.4 GHz. A longer stub has nothing left to come out of. At 40 ps it is a notch in the transmission at 6.25 GHz.

A lossy cancelling capacitor beats a resistor alone only above Q = 3.236 if it cancels everything, and only above 1.618 at all. computed by solving, not by drawing. A source of 50 Ω + j100 Ω, its reactance cancelled by a capacitor whose quality factor is Q, so that the capacitor carries a series resistance of its reactance over Q. Cancelling fully, with the load resistor matched to the source and capacitor resistances together, the load takes 1/(1 + x/Q) of the available power, at exactly half efficiency; a resistor alone takes 0.6180, 2/(1 + √(1 + x²)). They are equal at Q = 3.2361, twice the golden ratio. Cancelling only part of the reactance does better than either: it helps at all above Q = (√(1 + x²) + 1)/x = 1.6180, and at Q = 3 takes 0.6650 against 0.5971 cancelled fully.

Power, and the part that does no work

The reactance that is not free

A conjugate match cancels the source's reactance with an opposite one and takes all the available power, where a resistor alone takes 2/(1 + √(1 + x²)), 0.618 of it when the reactance is twice the resistance. The cancelling capacitor is not free. With quality factor Q it carries a resistance of its reactance over Q, and cancelling fully then delivers 1/(1 + x/Q). That beats the resistor alone only above Q = 2(√(1 + x²) + 1)/x, 3.236 here, twice the golden ratio. But full cancellation is not the best use of a lossy capacitor. Cancelling part of the reactance beats both, and helps at all above Q = (√(1 + x²) + 1)/x, the golden ratio itself. Below that, the right amount to cancel is none.

Measurement, which is a circuit on a circuit

The resistor that moves the lag

A guarded input driven by a one-megahertz follower presents a negative resistance of −1/(ωₜC), −1.59 kΩ behind 100 pF of cable. The standard remedy is a resistor between the follower and the ring. It weakens the negative resistance, to −6.5 kΩ at 1 kΩ and −403 kΩ at 100 kΩ, but it does not remove it: the resistor and the ring's own capacitance to ground make a lag of their own. With a large enough resistor the worst value is −R(C + Cₛ)²/(C·Cₛ) for a drive resistor R — minus four times the resistor when the ring is as capacitive to ground as to the input — and the same at every length of one cable. The length moves somewhere else instead. The frequency where the bootstrap is given up, 1/(2πR(C + Cₛ)) for a drive resistor R, falls with every metre.

Measurement, which is a circuit on a circuit

The source that rings against its guard

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

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

Devices, and the amplitude they stop being linear at

The noise the cascode device does add

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

1 nF at the node lifts the inverting stage's virtual earth 1.28-fold at its resonance, the integrator's 1.09-fold. computed by solving, not by drawing. The impedance looking into the summing junction of an inverting stage (1 kΩ in, 10 kΩ feedback) and an integrator (1 kΩ in, 10 nF feedback) on the same 1 MHz amplifier, bare (dashed) and with 1 nF of sensor capacitance from the node to ground (solid). The inverting stage's node rises as an inductance, and a capacitance across an inductance is a resonance: its node reaches 909 Ω at 126 kHz, 1.28 times the bare node there. The integrator's node is a resistance, the shelf of 15.9 Ω, and a capacitance across a resistance is a corner: the loaded node is at most 1.086 times the bare one, and only where the amplifier's own output resistance is part of the loop.

Feedback, and the margin

The sensor an integrator does not see

A sensor's capacitance at an inverting stage's summing junction sits across a node that rises like an inductance, and makes a resonance: 10 nF lifts the node 2.50-fold at 39.8 kHz, the lift is the resonance's Q, and the loop's margin falls from 90° to 24.8°, and to 8.0° at 100 nF. The same capacitance at an integrator's node sits across a resistance — the shelf, 1/(2π·GBW·Cf) — and the shelf does not move, because the capacitance enters the open node and the loop gain together and cancels. With an ideal output the integrator's margin is 90.91° at every sensor from 1 pF to 100 nF. What it does see is its amplifier's output resistance: 50 Ω driving the feedback and sensor capacitors in series is a pole, worst when the two are equal, 58.3° at 10 nF, and it recovers on both sides.

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

Before the steady state

The parasitic that moves the start

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

Lines, where a wire has a length

The slot that does not separate

Cutting a slot in the plane between two tracks is the common response to coupling between their returns, and in a cross-section it looks as if it separates them. It does not, if the plane is one ground: the halves stay joined where the run ends, so at direct current both returns still spread across the whole plane and share all of its resistance — 1.0000 with a 0.4 mm slot — and above the band the slot makes the sharing worse, 4.98% against 4.03%, rising to 15.7% as the slot widens to 1.6 mm, because it takes away the copper between the tracks and pushes each return towards the other. Only a split into two grounds removes the shared resistance at DC, at the price of each loop's resistance multiplied by 2.10, and it leaves the in-band coupling 29% lower, not gone: most of two tracks' coupling at 100 MHz is their own magnetic field, which the plane's copper does not carry.

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°.

Frequency, which is the same solve

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.

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

Devices, and the amplitude they stop being linear at

The resistor the upper base is held through

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

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

Circuits that do a job, and the range they do it over

The capacitance that is already a difference

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

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

Two windings, and the band between them

The gap that is three gaps

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

A mirror, cascoded or not, delivers two transistors' shot noise at low frequency — and cascoding it does not add a third. computed by solving, not by drawing. The output noise current of a 1 mA NPN mirror into a short, in units of one device's collector shot noise 2qI, against frequency, for a plain two-transistor mirror, the stacked cascode mirror and the wide-swing cascode with its upper base held by a separate diode at 100 µA. At 1 kHz they deliver 1.986, 1.960 and 1.965: the reference transistor's shot noise, copied, and the output transistor's own. At 100 MHz the mirror's own pole and the capacitance at the output transistor's collector have begun to take noise away, and the three read 1.681, 1.196 and 1.512.

Devices, and the amplitude they stop being linear at

The base current the mirror counts twice

A current mirror delivers two transistors' shot noise — the reference device's, copied, and the output device's own — and cascoding it does not add a third: 1.960 and 1.965 of one device's noise for the stacked and wide-swing cascodes against 1.986 for the plain mirror. What the cascode adds is its upper output transistor's base noise, and there the two arrangements differ by a factor of four. In the wide-swing mirror the base current goes to a separate bias and counts once, one part in β. In the stacked mirror it goes into the reference branch, which mirrors it back into the node it left, and it counts 3.82 times at β = 150 — the same base current that took 43 per cent of that mirror's output resistance. The reference resistor's noise, meanwhile, adds almost nothing even at 1.8 V, because it takes its place in the diode's node rather than adding to it.

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

Measurement, which is a circuit on a circuit

The rejection two capacitors lose at their corner

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

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

Where the models stop

The load that rings with the line

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

An unregulated supply's output against its load: 3.38 Ω of slope with perfect parts, 3.64 with 0.3 Ω in the winding, 3.32 with it in the capacitor. computed by solving, not by drawing, marched with the diodes in the netlist: a 17 V peak centre-tapped rectifier into 1000 µF, loaded from 200 Ω down to 25 Ω. The mean output with perfect parts falls from 15.99 V at 80 mA to 14.33 V at 573 mA, an average slope of 3.380 Ω against the textbook's 1/(4fC) = 5 Ω, drawn dashed without its diode drop. With 0.3 Ω in each half-winding the slope is 3.642 Ω; with the same 0.3 Ω in the capacitor it is 3.317 Ω. The slider is the resistance.

Circuits that do a job, and the range they do it over

The resistance a supply's regulation counts

An unregulated supply's output falls as its load rises, and the slope is its output resistance: 4.17 Ω at 80 mA and 2.80 at 575 for a thousand microfarads, below the textbook's 1/(4fC) = 5 Ω and falling, because the diodes conduct for longer as the load grows. A winding resistance adds to that slope at about its own value, 0.94 to 1.18 of it; a capacitor's adds −0.14 to −0.25. The guess that each should count by the share of the cycle it carries current in — a sixth and five sixths — is wrong in both places, and the right weighting is the mean current each carries: all of the load's through the winding, none through the capacitor.

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

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.

Routed along the glass weave, a length-matched pair with one conductor over a bundle and one over a gap builds 63.6 ps of skew in 20 cm — the skew nobody routed. computed by solving, not by drawing by integrating each conductor's delay along its route: a length-matched pair of 0.25 mm pitch over a glass weave of 0.5 mm pitch whose effective permittivity swings between 4.2 and 4.6 across it, positioned so that one conductor runs over a bundle and the other over the gap. Routed along the weave (dashed) the skew grows in proportion to the length, 0.318 ps a millimetre, to 63.62 ps in 200 mm. The slider rotates the route.

Lines, where a wire has a length

The skew in the glass

Match a differential pair's two lengths exactly and it can still have tens of picoseconds of skew, because a circuit board is woven glass in resin and the two conductors need not sit over the same one. Over a weave of half a millimetre whose effective permittivity swings from 4.2 to 4.6, a pair of quarter-millimetre pitch with one conductor over a bundle and the other over a gap builds 0.318 ps of skew every millimetre — 63.6 ps in 20 cm, the equivalent of 13.6 mm of routed mismatch by 30 cm, and 68 per cent common mode at 2.5 GHz. Where the pair falls across the weave is a lottery from nothing to that. Rotate the route by ten degrees and the skew never exceeds 0.146 ps, however long the pair: each conductor crosses bundle and gap in turn, and the difference averages out.

Before the steady state

The tail the source decides

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

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

Circuits that do a job, and the range they do it over

What a calibration at the carrier removes

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

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

Two windings, and the band between them

Interleaving is a choice, not an improvement

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

An amplifier gain of 1000 caps a switched-capacitor integration at 8000 cycles' worth of averaging. computed by solving, not by drawing. The noise, referred to the input, of the estimate a switched-capacitor integrator makes by integrating a steady input for N cycles, with 1 pF sampling and 4 pF feedback capacitors, against N, for amplifier gains of 100, 1,000, 10,000 and infinity. Each cycle adds 2kT/Cs — 89.5 µV rms — and an ideal integrator averages it as 1/√N without limit. A finite gain leaks, and the noise stops falling at √(kT/(A·Cf)) = 1.000 µV for A = 1000, reached after about 8000 cycles. The closed form is checked against a seeded march.

The floor, which bounds from below

The average an integrator can hold

A switched-capacitor integrator is an averager whose samples are exactly independent, because every cycle resets the sampling capacitor — and it pays for that with two switch phases, 2kT/Cs a cycle, 89.5 µV rms with a picofarad. An ideal integrator averages that as 1/√N without end. A real one leaks through its amplifier's finite gain, and the leak caps the average at 2A·Cf/Cs cycles: 8,000 for a gain of a thousand and a 4 pF feedback capacitor, where the noise stops at √(kT/(A·Cf)) = 1.00 µV. That floor contains the feedback capacitor and the gain and not the sampling capacitor at all, which decides only how long the integration takes to reach it.

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

Circuits that do a job, and the range they do it over

The sensor that is the capacitance

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

Lines, where a wire has a length

The guard that is part of the return

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

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

Measurement, which is a circuit on a circuit

The second pole the ringing guard sees

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

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

Devices, and the amplitude they stop being linear at

The bias point that heats itself

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

With a leak of a hundred periods the flux settles to a spread of 1.22% of the swing at β = 0.5 — 0.759 of what independent duty errors would give. Seeded: three runs of sixty thousand periods of the event map, 5 mV of threshold noise, β = 0.5, driving a magnetising branch whose flux leaks with τ = L/R. The stationary spread of the period-end flux in volt-nanoseconds against τ in periods, quoted here as a percentage of one half cycle's swing V·T/2 = 5493 V·ns: 1 → 0.146% (closed form 0.145%), 3 → 0.233% (closed form 0.232%), 10 → 0.401% (closed form 0.397%), 30 → 0.685% (closed form 0.673%), 100 → 1.25% (closed form 1.22%), 300 → 2.16% (closed form 2.10%). The closed form linearises one period, keeping the leak inside each half and the covariance −AB between consecutive periods' imbalances; dropping that covariance gives the upper curve, 1.61% at a hundred periods. For a slow leak the spread is V·σ·|B − A|·√(τ/T), the drawn asymptote.

Circuits that do a job, and the range they do it over

The spread a winding resistance holds

A relaxation oscillator's threshold noise walks an ideal core's flux without limit. Put the winding's resistance back and the walk stops at a spread: 1.22% of a half cycle's swing when L/R is a hundred periods, growing as the square root of L/R, while a comparator skew's offset grows as L/R itself. The anticorrelation between consecutive duty errors is worth 24% of that spread at β = 0.5 and 62% at β = 0.2 — but only in a core slow enough to remember several periods. And the resistance that bounds the walk is a boundary in its own right: with a millihenry and a fifth of the swing to spare, an ohm tolerates 12.9 mV of noise, and a hundred picoseconds of skew needs at least 0.14 Ω.

$With inductors of Q 100 the filter absorbs up to 10.5% of the available power, and the loss rate times its delay predicts the curve's shape — 11.2% at the same frequency. computed by solving, not by drawing. The fifth-order 0.5 dB Chebyshev ladder filter of the reflection essay, each inductor given a series resistance ωL/Q at the corner, so that each loses energy at the rate σ = 2πfc/Q. The power absorbed in the inductors, as a fraction of the available power, against frequency; beside it the lossless ladder filter's delay, D = |t|²τ₂₁ + |ρ|²τ₁₁, times σ — the stored energy over the available power, times the loss rate. At Q 100 the absorption peaks at 10.53% at 0.957 of the corner, where σ·D is 11.24%; between 0.3 and 0.95 of the corner the two agree to 7%, the rest being the split between magnetic and electric energy and the terms of second order in the loss. The slider is the inductors' Q.$ Filters, measured not tabulated

The heat a delay predicts

A lossy LC ladder filter absorbs power where it stores energy, and a lossless ladder filter stores energy in exact proportion to its delay — the transmitted delay weighted by what it transmits plus the reflected delay weighted by what it reflects. So the droop of a lossy filter is its loss rate times that delay: with equal losses in inductors and capacitors, to 0.09% at Q 10,000. With inductors of Q 100 in a fifth-order Chebyshev, 10.5% of the available power is absorbed at the band edge against 11.2% predicted. Inductor loss alone misses only by the split between magnetic and electric energy — near one in the passband, 4.6 to one at 1.6 times the corner — and the dual ladder filter mirrors the split. The law is first order, and it overstates the band-edge peak by 17.5/Q.

With the weft counted as well as the warp, the skew a pair can reach is least at 45.0° to the warp — 128 fs — and turning further off the warp makes it worse again. computed by solving, not by drawing, to first order in the permittivity's swing, the summed bound held against an integration over 64 registrations of the pair to the cloth at six angles. A 0.25 mm pair over a cloth whose warp has a 0.5 mm pitch and whose weft has a 0.5 mm pitch, each swinging the permittivity by ±0.2 about 4.4. The warp's contribution falls as the route turns off it, as the earlier figures found; the weft's rises, because the pair's projection across the weft grows as d·sin θ while its yarns are crossed less often. The worst skew is least at 45.00°, 128.3 fs. At 10°, where the warp alone allows 291.4 fs over every registration — twice the earlier figure's 146 fs, which was for a route starting over bundle and gap — the weft adds 13.8 fs. Integrated: 5° 587.6 fs, 15° 215.1 fs, 30° 139.3 fs, 40° 126.3 fs, 60° 139.2 fs, 80° 304.9 fs. The slider is the weft's pitch.

Lines, where a wire has a length

The angle two weaves allow

Turning a differential pair a few degrees off a board's glass weave stops its skew growing with length — but a cloth has two weaves, and turning off the warp turns onto the weft. Counting both, the worst skew over every registration of the artwork to the cloth is a sum of two bounds, one falling as the route leaves the warp and one rising as it meets the weft, and it is least at an angle the cloth decides: 45° and 128 fs for two identical half-millimetre weaves and a quarter-millimetre pair, against 305 at 10°; 52° for a finer weft; 67° for a flatter one. Doing the count properly also finds the earlier bound a factor of two low, and finds the obvious answer a trap: on a square weave's exact diagonal the two weaves' product stops averaging and the skew grows again, 2.2 fs a millimetre, while a degree either side it stays at 128.

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

Measurement, which is a circuit on a circuit

The ring a blocking capacitor buys

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

A 50 Ω source over 20.0 Hz–20.0 kHz: the best turns ratio is 10.7, giving 1.30 dB; the white optimum's 11.5 gives 1.31, and no transformer at all 13.2. computed by solving, not by drawing, the band integrals checked by quadrature. The drifting part of the essay before — 4 nV/√Hz and 0.6 pA/√Hz white, flicker corners at 10 Hz and 1 kHz — fed from a 50 Ω source through an ideal step-up transformer of ratio n, so it sees n²·50 Ω. The noise figure over 20.0 Hz to 20.0 kHz against n: least at n = √(√(E/I)/Rₛ) = 10.73, where the amplifier sees 5.76 kΩ, at 1.298 dB. The ratio that matches the white optimum, 11.55, gives 1.310 dB; the source connected directly, 13.23 dB. The slider is the band.

The floor, which bounds from below

One turns ratio for a band

A 50 Ω source and an amplifier whose optimum is 6.67 kΩ want a step-up transformer, and the ratio that matches the white optimum, 11.5, is the one a design note computes. Over the audio band it is nearly right: 1.31 dB against 1.30 at the band's own best ratio, 10.7. Over 1 Hz to 100 Hz it is badly wrong — 9.2 dB where a ratio of 4.8 gives 5.4 — because the band's noise lives in its bottom decade, where the flicker has moved the optimum down. One ratio for a two-decade band costs at most 0.24 dB against a ratio that could follow the optimum. The transformer's winding resistance multiplies the whole noise factor by one plus the winding's resistance over the source's, not just its own share: a tenth of the source costs 0.41 dB where its own noise alone would cost 0.31. And 115 amplifiers in parallel are an ideal 10.7:1 transformer, noise for noise.

In the stopband the series-first ladder filter is an inductance — 5.24 R₀ at three times the corner — and its dual a capacitance, 0.191 R₀: the same filter, opposite loads. computed by solving, not by drawing. The input impedance of the order-5 0.5 dB Chebyshev ladder filter realised with a series inductor first and with a shunt capacitor first, both terminated in the design resistance R₀, against frequency from a tenth of the corner to ten times it. In the passband both sit near R₀ and are nearly resistive. Above the corner they part: at 1.2 times the corner 1.33 R₀ at 89.4° against 0.749 R₀ at −89.4°; 2 times the corner 3.19 R₀ at 90.0° against 0.314 R₀ at −90.0°; 3 times the corner 5.24 R₀ at 90.0° against 0.191 R₀ at −90.0°; 5 times the corner 8.90 R₀ at 90.0° against 0.112 R₀ at −90.0°. The product of the two is R₀² at every frequency to 2e-15, which is what duality means for an input impedance. The slider is the order.

Filters, measured not tabulated

The load a filter is in its stopband

A fifth-order Chebyshev ladder filter can begin with a series inductor or, as its dual, with a shunt capacitor, and between its design resistances the two transmit identically. In the stopband they are opposite loads: at three times the corner the first is 5.24 R₀ of inductance and the second 0.191 R₀ of capacitance, their product R₀² at every frequency. That stops mattering only while the source is the design resistance. From a tenth of R₀ the shunt-first ladder filter rejects 55.9 dB at three times the corner instead of 64.0, draws 14 times the passband current at five times the corner against the series-first ladder filter's 0.24, and with a lead inductance of a tenth of R₀ over the corner's angular frequency grows a hole in its stopband — −33.8 dB where a matched source gives −54.0. A soft source does the mirror image to the series-first ladder filter. The first element has to be the one the source cannot short or open.

An L-section whose coil has a Q of 50 and whose capacitor has a Q of 1000. computed by solving, not by drawing at 89 ratios, the loss read as one minus power into the load over power into the section. It follows Qₛ(1/Qcoil + 1/Qcap) (dashed) to 3% wherever it is under 2%: at a ratio of 20 it is 8.39% against 9.15%. Two parts of Q 95.2 each — the harmonic mean of 50 and 1000 — lose 8.40% there (faint), and within 0.4% of the unequal pair at every ratio: the section hardly cares which part is lossy, only what the two reciprocals sum to.

Frequency, which is the same solve

The part that decides the match

An L-section's loss is not twice its Q over the components' Q unless the two components are equally good. Given a coil of Q 50 and a capacitor of Q 1000, a 50 Ω to 1 kΩ match loses 8.39%, which is what two parts of Q 95.2 — the harmonic mean — would lose: the reciprocals of the two Qs add, each weighted by the loaded Q, and 95% of the heat is in the coil. Both parts carry the same circulating energy, so they share the loss in the inverse ratio of their Qs, which makes the coil the one to spend on: Q 50 to 100 halves the loss, and past the capacitor's own Q a better coil buys almost nothing. The high-pass section, with each part in the other arm, loses the same to 0.01%; per hundred watts delivered the coil runs at 4.36 W; and the ratio above which splitting the match pays, 9 to 10, does not move with the parts at all.

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

Where the models stop

The load spread along the track

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

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

Lines, where a wire has a length

The conductor made causal too

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

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

Two windings, and the band between them

The walk that stops

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

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

Measurement, which is a circuit on a circuit

The time constant one trimmer cannot reach

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

A pi network matching 50 Ω to 1 kΩ at a loaded Q of 8, coil Q 100, capacitor Q 1000. computed by solving, not by drawing at 10 design Qs from √19 = 4.36 — the least a pi network can have, at which it is the L-section and loses 4.58% — upwards. At a Q of 8 it loses 9.47% (solid) and its band for |Γ| under 0.1 is 2.3% of the frequency (dashed, scaled to the left end), against 4.7% for the L-section. At 30 it loses 28.73% for a band of 0.6%. The chosen Q is paid for in both currencies at once.

Frequency, which is the same solve

The Q a pi network makes a designer choose

An L-section has no choice of Q; a pi network adds a third part and can be built at any loaded Q above √(ratio − 1). Matching 50 Ω to 1 kΩ with coils of Q 100 and capacitors of Q 1000, it is the L-section at its least Q of 4.36 — 4.58% lost, a band of 4.7% — and at a Q of 8 it loses 9.47% for a band of 2.3%; at 30, 28.7% for 0.6%. What the extra loss buys is rejection: the second harmonic 16.0 dB down at the least Q, 29.8 at Q 10, 39.1 at 30. A tee network, with two coils where the pi has one, loses within 0.8% of the pi at every Q, because the loss follows the energy the Q makes the network circulate rather than the number of coils that carry it. And the freedom runs one way only: no pi or tee is less selective than the L-section; a lower Q needs more sections.

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

Lines, where a wire has a length

The skew that moves with frequency

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

Two windings, and the band between them

The leakage a series reading cannot see

The series-aiding and series-opposing readings recover a coupling coefficient to fifteen figures on a netlist, and on a bench with a 0.1% meter they recover it to about 0.001 at every coupling — which is useless exactly where transformers live. At k = 0.999 the leakage, L₁(1 − k²), is two thousandths of the winding, and the series route leaves it 133% uncertain: 88 of 400 seeded benches report a negative leakage. Shorting the other winding reads the leakage directly, to the meter's own 0.1%, and recovers k 943 times more precisely at 0.999. The short is a short only where the shorted winding's reactance dwarfs its resistance by 10k/√(1 − k²), and the winding to short is the one with the longer time constant.

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

Circuits that do a job, and the range they do it over

The inductance that flattens the supply

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

The Q a 10 mH coil with 1.2 Ω of winding and 8 pF across it shows, against frequency. computed by solving, not by drawing, the coil's impedance at each frequency and its reactance over its resistance. The straight ωL/R line (pale) rises without limit; the coil's Q follows it at first, peaks at 11340 at 325 kHz — 0.5773 of the 563 kHz self-resonance, which is 1/√3 — and falls to zero at the resonance. The peak is 0.3849 of what the line reaches at resonance, 2/(3√3).

Two windings, and the band between them

The Q that peaks before resonance

A coil's quality factor is quoted as ωL/R and drawn rising with frequency, and its own winding capacitance turns it over: Q peaks at the self-resonance over √3 and falls to zero at the resonance. Eighteen coils from 1 mH with 300 pF to 100 mH with 1 pF all peak at 0.5774 of their resonance, at 2/(3√3) of the line's value there. Skin effect moves the peak down toward the resonance over √5. And the falling Q is a property of the coil read alone: in a tank tuned at 0.8 of its self-resonance, the same coil gives the full ωL/R, 23,570 against the 8,485 a meter reads off the bare part.

A 3.3 V edge from a 10 Ω driver down a 50 Ω line of 1 ns, received through 50 Ω in series with four capacitors, beside a plain 50 Ω and an open end. computed by solving, not by drawing: the line marched by its characteristics. The open end rings up to 5.50 V; the plain resistor holds 2.750 V, the launched level, and never reaches the swing. With the capacitor the far end arrives at the same 2.75 V and then climbs as the capacitor charges and stops terminating: with a time constant of half a round trip (20.0 pF) it overshoots to 4.48 V; with one, 3.83 V; with three (120 pF) it rises without overshoot to 3.30 V by 24 ns; with ten, 3.13 V by 24 ns, still on its way.

Lines, where a wire has a length

The termination that runs out

A 50 Ω resistor at the end of a line terminates every edge and draws current for as long as the line is held high: 75.6 mW on an alternating pattern from a 3.3 V, 10 Ω driver. Put a capacitor in series with it and the standing current is gone, and so, eventually, is the termination: the far end reflects nothing as an edge arrives and the whole of it once the capacitor has charged, by one less the exponential of −t/2Z₀C exactly. Under two round trips of time constant the receiver overshoots — 54.8 per cent at a quarter of one — and above it the receiver waits on the capacitor to reach the swing. On a pseudo-random pattern the levels are tightest at three round trips, and on a fast pattern the capacitor holds the line's mean and the resistor still burns half of what a plain one does.

A 10 µs lead-inductance resonance holds the start at t⁴ until √12·τr = 34.6 µs, at a damping of 0.1 or any other. Expanded about infinity, and by residues after. The start of the step through two buffered RC sections (τ = 1 ms and 0.5 ms) on logarithmic axes: bare, a line of slope two (dashed), and with a buffered series resistance and lead inductance into a stray capacitance of natural time constant τr = 10 µs and damping 0.1 (solid). The relative degree rises by two, so the start is the bare step times t²/(12τr²) — the faint line, checked on the expansion to three per cent a hundredth of τr in — until the two meet at √12·τr = 34.6 µs. The damping does not enter that coefficient: it is the inductance and the capacitance, through τr² = LC, and nothing else. After it the curve rejoins the bare one, displaced by a delay that the damping does set.

Before the steady state

The delay an inductance does not add

Lead inductance against a stray capacitance is a resonance, and far above a network's bandwidth it raises the relative degree by two: a two-section step starts as t⁴ instead of t² until √12·τr — 34.6 µs for a 10 µs resonance — at any damping, because the coefficient is LC and nothing else. After that the resonance moves a half-way threshold by R·C, its first moment, and the inductance is not in it. It moves every threshold by R·C only at ζ = 1/√2, where its second moment is a pure delay's, and below that its variance is negative: at ζ = 0.1 the 1% point moves by 1.48 R·C while the 10–90 rise time gets shorter. The ring that reaches the output is the network's own gain at the resonance, 2.0 × 10⁻⁴ at ω0τ = 100, and the quality factor decides only how long it lasts. Lightly damped, the start is a staircase that touches the bare step once a period and never overtakes it.

A milliohm read four-terminal with a drifting thermoelectric EMF: the error against the test current, read three ways, and the shunt's own heating. computed by solving, not by drawing: an EMF of 0.5 µV settled plus 1 µV relaxing with a minute's time constant, read thirty seconds after a disturbance with readings a second apart; the shunt 20 ppm/K and 20 K/W. At 10 mA one polarity is out by 11.1%, a reversal by 501 ppm, the three-reading delta by 4.14 ppm. Every EMF error falls as 1/I; the heating, 0.400 ppm at an ampere, rises as I². The best current is 11 A for one polarity (149 ppm in all), 1.8 A reversed (4.08 ppm) and 0.37 A with the delta (0.167 ppm).

Measurement, which is a circuit on a circuit

The voltage four wires carry in

Four wires remove a resistance measurement's leads by keeping the sense loop free of current, and that loop is exactly where a thermoelectric voltage lives. Ten milliamps through a milliohm is ten microvolts, so every microvolt where a copper lead meets a brass terminal moves a single-polarity reading by a tenth. Reversing the current removes any EMF that holds still and leaves the change over one interval; the + − + delta leaves the second difference; four readings leave the third — 501 ppm, 4.14 and less at 10 mA while a disturbed junction settles. Against the shunt's own heating each scheme has a best current, and the least error rises only as the cube root of the shunt's temperature coefficient: the reading method moves the error by orders of magnitude, the alloy by a cube root.

The Q of a 10 mH coil with 1.2 Ω of winding and 8 pF across it, with cores whose own Q is 1000, 300, 100 and 50. computed by solving, not by drawing: the core's loss a resistance ωL·Qc across the inductance, constant loss tangent, the coil solved at each frequency. Without a core loss the Q peaks at 11340 at 325 kHz, the self-resonance over √3. With the cores: 825.6 at 136 kHz for Qc = 1000, 274.9 at 94.0 kHz for Qc = 300, 95.9 at 66.2 kHz for Qc = 100, 48.7 at 52.8 kHz for Qc = 50. The peak falls to just under the core's own Q for a poor core and further under it for a good one, and moves down from the self-resonance over √3.

Two windings, and the band between them

The Q a core allows

A coil's Q, limited by its winding and its own capacitance, peaks at the self-resonance over √3. Put a core with a loss of its own across the inductance and the peak falls and moves: with a core whose own Q is 100, the 10 mH coil that peaked at 11,340 near 325 kHz peaks at 95.9 near 66 kHz. The peak sits where x³ = a/2 — x the frequency over the self-resonance, a the core's Q over the winding's — and reaches the core's Q less three times the two-thirds power of a/2 of it. At that peak the core carries 97 per cent of the loss, so a better winding hardly moves it. A core whose loss is a fixed resistance instead puts the peak at the geometric mean of two corners, as high as half the square root of the core's resistance over the winding's; a loss tangent that rises with frequency drags the peak down to the material's corner.

A 20 dB line from 300 K down to 4 K, fed from a 300 K source: how much of the noise at its cold end each piece of it makes. computed by solving, not by drawing: the line as forty matched pads, each at the temperature of its own place on a linear gradient, each pad's law held to the resistor-by-resistor sum. The noise temperature at the cold end is 67.7 K — the continuous line's closed form gives 67.63 K — of which 3 K is the source's, attenuated, and the rest the line's own. Each piece emits (1 − 1/L)T at its own temperature and is attenuated by every piece after it, so the warm pieces at the start, which emit most, are attenuated most: the warm half of the line supplies 17.99 K, the cold half 46.71 K.

The floor, which bounds from below

The end a lossy line looks like

A line running from room temperature into a 4 K stage is a chain of pads, each at its own temperature, and its noise at the cold end is each piece's emission attenuated by everything after it. Cut into forty pads along a linear gradient, a 20 dB line fed from a 300 K source delivers 67.7 K to the cold end, where the continuous closed form says 67.63; its cold half supplies 46.7 K of that and its warm half 18.0. Read from the warm end, the same line's own noise is 184 K at 10 dB against 89.7 at the cold: a lossy line looks like the end it is read from. And twenty decibels placed as one attenuator at the 4 K stage deliver 6.96 K, where at room temperature they deliver 300.

A 150 nH ground lead on a ten-to-one probe, at a 50 Ω source: the screen shows 49.4% of overshoot ringing at 121 MHz on an edge that has none. computed by solving, not by drawing, twice. A 1 V step from a 50 Ω source, watched through a ten-to-one probe with 11.5 pF at its tip, its ground returned through 150 nH of lead — about fifteen centimetres — and the scope reading the tip against the lead's end (solid, from residues; dots, a trapezoidal march, peak 1.4942). The same probe with its ground at the circuit's ground (dashed). The node itself steps cleanly. With the lead the tip capacitance and the lead inductance resonate at 1/(2π√(LC)) = 121 MHz, damped only by the source: ζ = (Rs/2)√(C/L) = 0.219, and the display overshoots by 49.4%.

Measurement, which is a circuit on a circuit

The ring the ground lead adds

Every earlier measurement of a probe returned it to the circuit's ground directly. A real ten-to-one probe returns through fifteen centimetres of clip lead, about 150 nH, and its 11.5 pF tip and that lead are a series resonance at 121 MHz across the node being watched. The only damping it has is the source's own resistance, ζ = (Rs/2)·√(C/L), so the screen shows 49.4% of overshoot on a clean step from 50 Ω, 87.1% from 10 Ω, and none above 228 Ω — the stiffer the node, the worse the picture. Solved by residues and by a march of the netlist, the overshoot follows the second-order law at every source. A spring tip's 5 nH rings at 664 MHz and not at all from 50 Ω; a 150 Ω resistor in series with the tip brings a clip lead under 2% at the price of a 3.65 ns rise.

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.

Filters, measured not tabulated

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.

A fifth-order Chebyshev ladder built from parts of Q = 100: its passband sags by 0.94 dB near the edge and 0.38 dB at low frequency — 8.686·δ·τg, the group delay times the dissipation. computed by solving, not by drawing. A doubly terminated 0.5 dB Chebyshev ladder, fifth order, 1 kHz edge, 1 kΩ terminations, lossless (dashed) and with every inductor given a series resistance and every capacitor a shunt conductance for a uniform dissipation δ = ωc/Q = 62.8 s⁻¹ (solid), and the lossless response plus 8.686·δ·τg(f), with τg read off the lossless filter's phase (dots). The prediction holds to 3.6% of the extra loss across the passband. The extra is smallest where the delay is smallest, 0.383 dB at 38 Hz, and largest at 953 Hz near the edge, 0.943 dB, where the delay peaks at 1.79 ms: the ripple the design specified is rounded off where it was steepest.

Networks, and how a solve is checked

The derivative that is a delay

A doubly terminated ladder filter is stationary in every element's value at its ripple peaks, and that insensitivity says nothing about the parts' losses. Give every reactance the same dissipation δ — a series resistance δL in each inductor, a shunt conductance δC across each capacitor — and every reactance has the impedance it had at s + δ, so the lossy ladder is exactly H(s + δ). By the Cauchy–Riemann equations the loss then rises by 8.686 times δ times the group delay, in decibels: the derivative of a ladder filter's response with respect to its parts' dissipation is its group delay. Solved with the resistors in, a fifth-order 0.5 dB Chebyshev of Q = 100 parts sags 0.383 dB at low frequency and 0.943 dB near its edge, where the delay peaks at 1.79 ms, the law holding to 3.6%. The sag falls as one over Q, grows with the order — 0.37 dB at third order, 2.74 at ninth — and is set by the mean of the inductors' and capacitors' dissipations.

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

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).

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.

Frequency, which is the same solve

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.

A ten-to-one divider with 15 pF of its arm behind a kilohm, stepped through 1 kΩ: one trimmer overshoots 0.00936% where an ideal source shows 13.2%, and the second branch that was exact no longer settles first. computed by solving, not by drawing: the step of the divider of the earlier essay, 9 MΩ and a trimmer over 1 MΩ with 100 pF direct and 15 pF behind a kilohm, through a source resistance of 1 kΩ, by residues. One trimmer at its calibrator setting (solid) overshoots 0.00936%, and the second branch that made the arm a scaled copy (dashed) rises without overshoot on the source's own lag. An ideal source (dotted) shows a 13.2% hook with one trimmer and nothing with two. A source resistance charges the input capacitance, 11.5 pF, so its lag and the hook act on one edge, and when they are comparable the hook's excess goes into shortening the lag.

Measurement, which is a circuit on a circuit

The source the hook hides behind

A compensated divider with a late branch in its lower arm overshoots 13.2 per cent on an ideal step, and every measurement of that hook has been made from a source with no resistance. Through a real source the input capacitance charges first, and the hook and the source's lag act on the same edge. The overshoot is 10.5 per cent from 50 Ω, 4.13 from 300 Ω and gone by a kilohm, where the source's lag equals the hook's. The excess has not gone away: it is in the edge, which the screen shows 15 per cent faster than the source and the probe's capacitance allow. And the second branch that made the divider exact only makes its ratio exact, so from a kilohm the exact divider settles in 65.6 ns and the one with the hook in 44.

A reference 0° behind the excitation at 20 kHz, after a calibration at +1000 µε: matched, the worst error is 5.79 ppm; with a tenth of a per cent of screen mismatch, 0.607%. computed by solving, not by drawing: the error of a synchronous reading of a 350 Ω quarter bridge through ten metres of screened pair, against the strain read, with the reference 0° from the excitation at the bridge and a shunt calibration at +1000 µε through the same reference, for a matched screen (solid) and for 0.01 and 0.1 per cent of mismatch. The calibration takes out cos(φ − θ), because it multiplies the signal at every strain. It cannot take out a voltage that does not scale with the strain, and the mismatch's leak is such a voltage: a hyperbola through zero at the calibration strain. Worst over ±2000 µε: 0%: 5.79 ppm, 0.01%: 606 ppm, 0.1%: 0.607%.

Circuits that do a job, and the range they do it over

The reference that should be late

A carrier amplifier's reference is taken from the excitation because that is what the output is meant to be in phase with, and nothing says that is where it belongs. A screen mismatch's leak arrives late by the common-mode node's lag and the difference's together, 525 ns on ten metres of pair and a 350 Ω bridge, so a reference on the excitation reads sin 3.7° of it at 20 kHz: 0.607 per cent of error at a tenth of a per cent of mismatch, and 72.6 parts per million with the reference 3.7° late. The null is a delay rather than a phase and does not move with the mismatch, and a delayed reference allows about five times the carrier on any mismatched screen. How exactly it must be held narrows as mismatch times carrier.

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.

Frequency, which is the same solve

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.

$The same 15 pF spread along 2 kΩ of line, on a 1 ns edge: with 1 of its branches copied the output is flat to 1.07%, against 11.2% with one trimmer alone. computed by solving, not by drawing: the response to a 1 ns edge of a ten-to-one divider whose lower arm is 100 pF direct and 15 pF along a 2 kΩ resistive line, with one trimmer (dashed) and with the first 1 branch of the line's exact partial-fraction expansion copied into the upper arm (solid). The expansion's time constants are 12.2 ns, 1.35 ns, 486 ps and so on, as 1/(2k − 1)². Inverted numerically by Talbot's method, and the line itself checked against a netlist of sections.$ Measurement, which is a circuit on a circuit

The branches a square root needs

A resistive cable's late charge is a continuum and not one branch, and the partial fractions of its tanh are an exact description of it: 81.1 per cent of the capacitance behind a time constant of 12.2 ns, 9.01 per cent behind a ninth of that, 3.24 behind a twenty-fifth, all through the same kilohm. So a divider's upper arm can copy it one piece at a time. On a 1 ns edge one trimmer alone is 11.2 per cent out, one copied branch 1.07, two 0.252 and four 0.0353. The first branch does most of the work and the uncopied capacitance falls only as one over the count; what saves the rest is that a branch faster than the edge is averaged away. Faster edges ask for more branches, and the tolerance on the first ends the series before the arithmetic does.

$A shorted reading at k = 0.99: within 1% of the leakage from 279 Hz to 112 kHz, too high below through the winding's resistance and above through the leakage's resonance with 100 pF, at 1.13 MHz. computed by solving, not by drawing, in closed form, and by a netlist with a coupling element at three frequencies. The reading Im(Z)/ω at the driven terminals of a 10 mH and 40 mH pair, with the other winding shorted through its own 1 Ω, 0.5 Ω in the driven winding and 100 pF across it, against frequency, as a fractional distance from the leakage L₁(1 − k²) = 199.00 µH. Below the band the short does not cancel the shared flux. Above it the leakage and the capacitance resonate, at 1.13 MHz, and the reading rises to meet them. Band: 279 Hz to 112 kHz, a ratio of 402.1.$ Two windings, and the band between them

The window that moves

A shorted winding reads its own leakage only above the frequency at which the short really cancels the shared flux, and only below the leakage's resonance with whatever capacitance is across it. A tighter coupling raises the first edge as one over the root of 1 − k², and it was expected to pull the second down and close the window. It raises the second as well, by the same factor. So the band is 82 Hz to 36 kHz at k = 0.9, 279 Hz to 112 kHz at 0.99 and 2.81 kHz to 1.12 MHz at 0.9999, about four hundred times wide throughout. Its width is 0.01·L₂/(R₂k√(L₁C)), with the leakage nowhere in it, so what closes the window is the winding and not the coupling. Reading the real part as well is exact on paper and limited by phase.

A 640 kvar bank on a 16 MVA 400 V bus: alone its resonance with the supply is at h = 5.00, and with a 7% reactor in series at 3.01, with the bank's own null at 3.78 between that and everything above it. computed by solving, not by drawing: the impedance per phase at a 400 V bus, with a short-circuit level of 16 MVA and a 640 kvar correction bank, against harmonic order, in closed form and at the fifth by a netlist. The bank alone (dashed) resonates with the supply at h = 5.00, where the bus impedance peaks. With a reactor of 7% of the capacitor's fundamental reactance in series (solid), the bank's own series resonance is at h = 1/√0.07 = 3.78, a deep null, and the parallel peak has moved down to h = 3.01. Above the null the bank is an inductor, in parallel with the supply's inductance, so there is nothing for any higher harmonic to resonate with.

Power, and the part that does no work

The reactor that keeps a bank out of it

A correction bank on a stiff bus resonates with the supply at a harmonic set by the two sizes, and at 640 kvar on 16 MVA that harmonic is the fifth, which it multiplies by 34. A reactor of 7 per cent of the capacitor's reactance in series puts a null at h = 3.78, and above that null the bank is an inductor, in parallel with another inductor, with nothing left to resonate. So the fifth is never raised at any size of bank. It was not free. The capacitor now stands 7.5 per cent above the bus, the resonance has gone below the null and can land on the third, 22.7 times at 654 kvar, and the number that looks like a detail, 7 rather than the 4 that would tune the bank to the fifth, is what keeps an ageing bank on the right side of its null.

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

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.

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