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

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

Every model has an edge

Four assumptions this collection runs 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

Every model this collection has drawn is right below a number or above one. A transformer is the first that 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.

 The millivolts in the wire, which are nobody's signal. computed by solving, not by drawing at 145 frequencies. A 10 mm run of one-ounce copper carries the return of a 100 mA load and the reference of a 10 mV sensor. Its 5.00 mΩ and 10.0 nH put 500.0 µV in series with the sensor at low frequency — 5.00% of the reading — rising a decade per decade above 79.6 kHz until at 1.59 MHz the error is the whole signal. The full solve and the interfering current times the shared impedance agree to 4.4e-7. The slider is the length of the shared run: both the resistance and the inductance are proportional to it, so every point on the curve moves down together and the corner stays at 79.6 kHz. Measurement, which is a circuit on a circuit

The millivolts in the wire

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

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

The Q the components allow

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

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

The edges that are lengths

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

 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.

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

The energy is in the gap

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

 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 the rung below measured, 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.

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

The resistance that grows with frequency

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

 A 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.

 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 as the rung below drew it, 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

The rung below this one connects two capacitors to a load through nothing, and says so. 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.

 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 the rungs below 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 rungs of this argument 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.

 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 was computed three ways on this site and the agreement was 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 first rung's r_s/(β+1) + 1/g_m. Between 110 MHz and 301 MHz it is negative: r_π and C_π 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.

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

The optimum that hands back a bandwidth

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

 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.

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

The headroom that is the line's own charge

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

 The 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 ladder in this collection 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.

 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.

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

The gap that is three gaps

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

 The 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.

 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.

 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 boundary this ladder's second rung measured. 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 rung 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.

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