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

Closed form — where it appears

An expression that gives an answer directly rather than by iteration, and which is worth having for its shape as much as for its value. Every one here is checked against a solve or a march that shares none of its arithmetic, and the range over which it holds is measured rather than assumed.

Named by 33 essays across 13 fields — each of them below, with the objects they name alongside it.

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

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

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

networks · Conditioning 5 sections, equal ripple, and the band that is 134% rather than 97. computed by solving, not by drawing. The repaired five-section equal-ripple design over the band it was designed for. The horizontal rule is the 0.1 the specification allows and the 4 interior peaks sit on it, level to 8.5e-4 per cent — which is the condition for a minimax solution and is now checked rather than assumed. The dots are the eighty-one frequencies the objective used to be evaluated at: the worst of them is 0.09999998 and the worst of the design over the whole band is 0.10013858, so an optimiser shown only the dots drove them down to the specification and left the true peaks 13.9 parts in ten thousand above it. That is nothing until something downstream is a threshold, and the band measurement was one: it reported 97.34 per cent for a design that holds 134.04.

The number that was wrong

An earlier figure printed 97.3 per cent of band for a five-section transformer where the answer is 134, said in its own text that the figure was wrong, and blamed a search that had converged to eight digits. The search was fine. The objective was the worst of a grid rather than the worst of a band, the band was then measured by bisecting a function that crosses its threshold five times, and the check guarding all of it passed — because 97.3 is still more than 92.6.

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

What the fourth order says about the third

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

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

The order that stops helping

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

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

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

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

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

Ten seconds, and fifteen minutes

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

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

The factor the expression leaves out

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

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

The dither that is a decision

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

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

Two ladders the terminals cannot tell apart

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

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

Where the estimate stops being a bound

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

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

The wire that is not a foil

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

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

The gap that is bigger than it is

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

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

The inductance that is a shape

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

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

The optimum that does not move

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

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

The count a core survives

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

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

The part that decides the match

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

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

The voltage at which the slope is zero

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

semiconductors · Temperature A capacitor of 67% of the load's reactive power, moved along a feeder whose load is spread evenly over 10 km. computed by solving, not by drawing at each of 30 positions: one phase of an 11 kV feeder of 0.3 + j0.35 Ω/km, carrying 2 MW at 0.8 lagging spread over 30 equal loads. Uncorrected it loses 16.8 kW a phase; with every load corrected to unity at its own terminals, 11.1 kW — so 5.66 kW is the reactive current's share. This capacitor does best at 67% of the way out, where the loss is 11.8 kW and 87.6% of the reactive share is gone. At the substation end it removes nothing; at the far end it overshoots.

The capacitor two-thirds of the way out

A feeder's reactive current costs loss in every section it passes through, and on a feeder whose load is spread along its length that current is largest at the substation and falls to nothing at the far end. Correcting it at the substation removes no loss at all. One phase of an 11 kV feeder carrying 2 MW at 0.8 lagging over ten kilometres loses 16.8 kW, of which 5.66 kW is the reactive current's; the best single capacitor is about two-thirds of the load's reactive power placed about two-thirds of the way out, and it removes 87.6% of that share — against the eight-ninths the textbook rule gives for an idealised line. Two capacitors at 2/5 and 4/5 remove 95.6%. The same capacitor left in at light load costs more than it saves below half load, and the rule for a continuous load is a statement about many taps.

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

The bow a trim leaves

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

semiconductors · Temperature A 10 mH winding read at 1 kHz while a 40 mH winding coupled to it at k = 0.99 is loaded from a short to an open. computed by solving, not by drawing at 61 loads for each coupling. Open, the first winding reads 10.0000 mH, its own inductance. Shorted, it reads 199.00 µH — L₁(1 − k²), the leakage seen from this side, because the second winding's current cancels all of the flux the two share. The change happens around a load equal to the second winding's reactance, 251 Ω. At k = 0.5, 0.9, 0.99 and 0.999 the shorted readings are 7500 µH, 1900 µH, 199.00 µH, 19.99 µH.

The leakage a series reading cannot see

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

magnetics · Mutual inductance A capacitor charged from 5 V through 1 mH, with and without a diode, at a circuit Q of 10. computed by solving, not by drawing, from the netlist's poles. Without a diode the capacitor rings about 5 V and settles there, and the resistance takes 1.00000 of ½CV² — exactly what a resistor alone would take. With an ideal diode in series the current stops at its first zero, 99.5 µs in, and the capacitor is left at 9.272 V, which is V(1 + d) with d = 0.8545 the step's overshoot. That charge cost 0.2699 of ½CV².

The swing a diode keeps

Charging a capacitor through a resistor loses half the energy whatever the resistance, and putting an inductor in the path changes nothing: twenty-four settled charges from 10 µH to 10 mH and Q from 0.3 to 100 lose ½CV² to 1.4 parts in a hundred million. A diode that stops the current at its first zero changes everything. The capacitor keeps the overshoot, V(1 + d), the loss falls to ½CV²(1 − d²), and a charge of duration T costs π²/8 of what a ramp of the same duration costs.

transients · Switching energy A chebyshev low-pass of order 5: where its impulse response balances, and where its step crosses half. computed by solving, not by drawing, the step built from the network's poles and the impulse response its derivative. The impulse response (dark, scaled to its peak) balances at 0.7090 ms; the group delay at a thousandth of the corner is 0.7090 ms — the same number by a different route. The step (pale) crosses half its final value at 0.7238 ms, 2.1% after it.

The delay that is a centre of mass

A filter's group delay at zero frequency is not an approximation to when its step arrives; it is exactly where its impulse response balances. Across three families and orders one to seven the two agree to 5.7 parts in a million by routes that share only the netlist. The step's half-way point is a different thing: ln 2 of the delay for one pole, approaching it from below for a Bessel filter, and passing it — by up to 16.9 per cent — for Butterworth and Chebyshev filters whose impulse responses swing negative. Elmore's RC-tree delay is the same centre of mass, and a uniform ladder network crosses half at 0.757 of it.

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

One step of dither per moment

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

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

The current a capacitor adds up

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

instruments · Input bias current A 1 µF capacitor at 5 V joined through 100 Ω to an empty one of the same size. computed by solving, not by drawing, from the netlist's poles. Charge is conserved and both settle at 2.500 V. Energy is not: the pair holds 50.0% of the 12.50 µJ the first one started with, and the resistor has taken 50.00% — C2/(C1 + C2), with no resistance in it.

The energy two capacitors cannot share

Join a charged capacitor to an empty one and the charge is shared exactly while the energy is not: the pair loses C₂/(C₁ + C₂) of what the first one held, through 1 Ω or 10 kΩ, with an inductor ringing for hundreds of cycles or without one. An inductor and a diode turn the sharing into an exchange — equal capacitors swap 96.9% of the energy at a loop Q of 50. And a charge pump is the sharing loss billed once a cycle: a doubler's efficiency is its output voltage over twice its input, to four parts in ten million, at every load and every clock.

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

The Q that peaks before resonance

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

magnetics · Real inductor Averaging an input that drifts 1 µV a cycle through a 1 pF switched-capacitor averager: too short a memory keeps the noise, too long a lag, and the best is 16.5 cycles at 27.2 µV. computed by solving, not by drawing, and seeded. A switched-capacitor exponential averager — 1 pF sampling capacitor, an equal damping capacitor for unity gain, memory set by the feedback capacitor — fed a ramp of 1 µV a cycle. The rms error against the input (solid) is the lag (dashed), (a/(1 − a))·r, and the switch noise (dotted), 2kT(Cs + Cd)/(Cf²(1 − a²)), together; seeded marches of sixty thousand cycles (dots) agree to 5%. The best memory is 16.55 cycles against (kT/(Cs·r²))^(1/3) = 15.88, with 27.2 µV of error: 15.5 µV of lag and 22.3 µV of noise.

The leak worth building in

An integrator that never forgets cannot follow an input that moves. Give a switched-capacitor integrator a deliberate leak — a damping capacitor that samples the output each cycle and takes its charge away — and it becomes an exponential average with a memory of Cf/Cd cycles. Fed an input drifting 1 µV a cycle, through a 1 pF sampling capacitor, too short a memory passes the switch noise and too long a memory lags the drift; the best, on the exact closed form and on seeded marches of sixty thousand cycles, is 16.5 cycles, for 27.2 µV of error against 89.5 µV in a single sample. The best memory follows a cube-root law, the cube root of kT/(Cs·r²), ten times the drift for a memory 4.64 times shorter, and at it the noise variance is twice the squared lag. And the leak is not free: the damping capacitor is switched too, and at unity gain it samples half of the averager's noise.

noise · kT/C noise A zero at 1ωₙ moves the fastest-settling damping from 0.782 to 0.873 and the fastest settling from 3.62 to 2.38/ωₙ. computed by solving, not by drawing. The 2% settling time of a second-order step against its damping ratio, with no zero (grey), a zero at 2ωₙ (dashed) and at 1ωₙ (solid), each located by residues on the recovered polynomial and refined by bisection; the step by residues equals the plain step plus (1/b) times its derivative to 2e-16. zero none: fastest at ζ = 0.7825, 3.623/ωₙ; zero 2ωₙ: fastest at ζ = 0.7975, 3.028/ωₙ; zero 1ωₙ: fastest at ζ = 0.8725, 2.376/ωₙ. The zero adds the step's own derivative, which adds overshoot, so more damping is needed to keep the last excursion inside the band — and it speeds the rise, so the best settling is faster than with no zero at all. The cliffs in the curve are still there, moved.

The damping a zero asks for

The fastest-settling damping of 0.78 belongs to a step with no zero in it. Put a zero at the natural frequency and the step becomes the plain one plus its own derivative: the best damping moves to 0.873 and the best settling falls from 3.62/ωₙ to 2.38/ωₙ. A zero five times further out moves nothing, and a zero at three times moves the optimum by only 0.005, yet keeping 0.78 there costs 52 per cent, because the optimum sits on a cliff. A zero in the right half-plane moves the best damping by 0.015 and costs 20 per cent in settling that no damping can buy back.

transients · Damping Three matches of one length, half a wavelength at f₀: two quarter-wave sections are bad again at 2f₀ and 4f₀; an exponential and a Klopfenstein taper are not bad again at all. computed by solving, not by drawing, as cascades of exact line input impedances. 50 Ω into 200 Ω through three matching sections each half a wavelength long at f₀: a two-section binomial transformer (dashed), an exponential taper (solid) and a Klopfenstein taper designed for 5% ripple (dotted), the tapers as four hundred thin slices. The stepped transformer is a band-pass: matched at f₀, and at 2f₀, where each section is half a wavelength and transparent, |Γ| is the bare mismatch, 0.600, again — and at 4f₀, and every even multiple. The tapers are high-pass: the exponential is 0.0194 at f₀ and under 0.131 everywhere above 1.6f₀; the Klopfenstein holds 0.050 from 1.1f₀ up.

The match with no upper edge

A stepped quarter-wave transformer is a band-pass in disguise: at twice its design frequency every section is half a wavelength and transparent, and a 50-to-200 Ω match reflects the bare 0.600 again. Let the steps multiply over a fixed length and the staircase of n steps returns the mismatch at n·f₀; in the limit the upper edge has gone, and what is left is a taper — a high-pass match. An exponential taper half a wavelength long is 0.019 at f₀ and never above 0.131 past 1.6f₀. The Klopfenstein profile, the Chebyshev design's continuous limit, holds 0.050 from 1.1f₀ up, needs half the exponential's length for |Γ| ≤ 0.1 and an eighth of it for 0.02, and is not continuous: it starts with a step from 50 to 52.6 Ω. And the small-reflection formula that defines all of them as a Fourier transform overstates a 10:1 taper's reflection by 41% — the line gives tanh of it.

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

The derivative that is a delay

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

networks · Sensitivity At 1.00 A forward the silicon-carbide Schottky hands the switch more charge than the silicon PN diode above 245 V. computed by solving, not by drawing. The charge each diode returns through the switch when it is forced to block, against the reverse voltage, at 1.00 A forward and 100 A/µs. The silicon-carbide Schottky's is its depletion capacitance charged to the rail, closed form checked against Simpson's rule: 11.8 nC at 100 V, 25.0 nC at 400 V, 30.9 nC at 600 V, the same at any current. The ultrafast silicon diode's is its stored minority charge, 18.1 nC at this current from the recovery's closed form, plus 1.59 nC of its own capacitance at 400 V. The two are equal at 245.3 V.

The recovery a Schottky does have

A Schottky diode stores no minority charge and is quoted as having no reverse recovery. It has a depletion capacitance instead, and to block it must be charged to the rail through the switch: a 650 V silicon-carbide part holds 25.0 nC at 400 V whatever current it was carrying. An ultrafast silicon diode with a 20 ns lifetime returns 19.7 nC at 1 A, so below 1.42 A forward the diode said to have no recovery hands the switch more charge, and at 125 °C the boundary falls to 632 mA. In energy the capacitive charge costs 64.9 per cent of V·Q rather than all of it, and the boundary is 799 mA.

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

The step that returns the digits

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

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

The window that moves

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

magnetics · Mutual inductance

Named alongside it

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

Model rangeDesign tradeoffVerificationEnergy-storageResiduesBisectionChebyshevField solutionNumerical errorQuality factorSeeded generatorTemperature coefficient

All concepts