Field

Before the steady state

Everything on the frequency axis assumes a settled circuit. These are the figures about getting there — and about the one limit no transfer function contains, where a step becomes large enough that the response stops being a scaled copy of a smaller one.

One step response, computed twice: from the poles, and by walking the network forward. A damping ratio of 0.22, so the overshoot is 49.2%. The two curves are drawn on top of each other; the panel below is the difference between them, which is the trapezoidal rule's error at 500 steps and reaches 1.70e-3 V.

One step, computed twice

A step response from the poles is exact. The same step walked forward in time is not, and the difference between them is the trapezoidal rule's own error rather than anything about the circuit. It falls by a factor of four every time the step is halved, which is a claim about a method and can be watched.

Two poles at ζ = 0.3, recovered from the matrix. The poles are at -477.5 ± j1518 hertz. Their distance from the origin is the natural frequency to six digits; the cosine of their angle from the negative real axis is the damping ratio. The step response beside them follows.

Where the behaviour is written down

Two numbers in the complex plane contain everything a second-order circuit will ever do. Their distance from the origin is the natural frequency, the cosine of their angle is the damping — and the fastest-settling circuit is not the critically damped one, which is the case the textbooks name.

Five steps, each divided by its own size, from an amplifier limited to 0.50 V/µs. A linear circuit would put these five curves exactly on top of each other. The 20.0 mV step is linear; everything above 79.6 mV is not, and the largest step takes 16.0 µs to travel a distance the linear model says takes 0.159 µs.

The step that is too big

A linear circuit scales — double the input and the output doubles, exactly. A real amplifier does not, because its output can only move at a fixed rate, and the amplitude at which the two stop agreeing is about eighty millivolts for an ordinary part. No transfer function contains that number, because no transfer function can.

How long a second-order step takes to arrive inside ±2%. computed by solving, not by drawing from the residue expansion at 260 damping ratios. The fastest is ζ = 0.780 at 3.60/ω₀; critical damping takes 5.83/ω₀, which is 62% longer. Between ζ = 0.775 and 0.780 the time falls by 33% in one step of the sweep, because which excursion is the last one outside the band changes there — the overshoot at the fastest damping is 1.99%, which is the band itself, and one step to the left it is larger. The faint curves are the other bands, each with its own step in a different place.

The cliff before the fastest settling

Settling time against damping is not a smooth curve with a minimum. It falls by a third in one step of a sweep of five thousandths, and the fastest damping sits on the edge of that step — so a design a hundredth of a damping ratio to the left of the optimum settles forty-eight per cent slower, with a waveform that looks no different.

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

The half that never arrives

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

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

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

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

A step on a series RLC at ζ = 0.079, and the two numbers read off H(s). computed by solving, not by drawing. A 50.3 kHz series RLC driven by a one-volt step, with the capacitor voltage and the inductor voltage drawn together. Two limits of the transfer function are two points of the waveform and neither needs the waveform: H(0) = 1.000000 is where the capacitor ends up, and H(∞) across the inductor is 1.0000, which is what it does at the first instant — the expansion gives 1.000000 for it at t = 0. Here the damping ratio is 0.0791, the response is inside ±2% after 7.6 cycles, and 100.0% of the last twenty-four cycles sit there. The poles are at a real part of -7.91e-2 of ω₀, which is the condition the final-value theorem actually has — not a property of H but of where sY(s) has its poles.

Two numbers without solving for the waveform

Where a step response starts and where it ends are two limits of the transfer function, and neither needs the waveform. Both are exact here — 1.000000000 volts at the end and the whole step at the first instant — and one of them is a lie waiting to happen: take the damping to zero and the final-value theorem still returns 1.000000 for a response that swings between 0 and 2 for ever. Its condition is not on the transfer function but on where the poles are, and the practical condition is narrower still: at five ohms the poles are safely in the left half-plane and sixty cycles is not enough time.

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

The cancellation that leaves a tail

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

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

The diode that conducts backwards

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

Nine tenths of the heat is in the switch, and above 1.28 MHz there is no temperature at all. computed by solving, not by drawing. The junction temperature a recovering diode settles at, against switching frequency, with the carrier lifetime rising as the 1.8 power of absolute temperature so that the recovery gets worse as the junction gets hotter. Each event costs 15.43 µJ when cold, of which 90% is dissipated in whatever is pulling the current down rather than in the diode: while the junction is still conducting it holds almost no voltage, and the recovered charge is delivered through the switch at the full 100 V. That energy equals the diode's own conduction loss at 29.2 kHz. The fixed point T = Ta + Rth·P(T) is iterated from the ambient upward; the junction passes its rated 150 °C at 904 kHz, and above 1.28 MHz there is no temperature that satisfies it at all — the loop has gain and the solver reports the refusal rather than the last iterate.

The heat a recovery leaves behind

The essay below this one measured how much current a diode conducts backwards and for how long, and stopped there. Both numbers are multiplied by a voltage somewhere, and the surprise is where: while the junction is still conducting it holds almost nothing, so nine tenths of the energy is dissipated in the transistor pulling the current down and not in the diode. Repeat it a hundred thousand times a second and it is 1.5 watts, the lifetime rises with temperature, and above 1.28 megahertz the diode's own loop has no fixed point at all.

Two loops, and one heatsink

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

$The frequency at which a pulse train becomes an average. computed by solving, not by drawing. The same 5 watts of average dissipation at every frequency, delivered 2 per cent at a time. The flat line is the steady-state answer, which does not know about the frequency. The falling curve is the marched peak junction temperature, which does. They meet at 308 Hz, and that frequency is not a property of the converter: it is a fraction of one junction time constant per period — f·τ = 0.738 at this duty, with τ = 2.40 ms, and between 0.78 and 0.56 across the duties on the slider. A hundred-kilohertz converter fits 240 periods inside that time constant, and at the top of the sweep — 10.0 kHz — the steady state is already exact to 0.46 per cent, so the averaged-power fixed point is right and this is the measurement that says why. The march puts 48 steps inside each pulse, which is what the answer is sensitive to: at six it put the boundary 19 per cent too high.$

The pulse the heatsink does not feel

A thermal resistance iterated to a fixed point with a diode or a switch is a statement about a power — so it assumes that a hundred and fifty watts for two per cent of the time is three watts. The die's own heat capacity decides whether that is true, and it decides it at a frequency: above 308 hertz the junction integrates, by a hundred kilohertz the fixed point is exact to five parts in ten thousand, and at one hertz the same average power on the same heatsink puts the junction three hundred kelvin hotter.

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

The capacitor that remembers

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

Three cliffs, not one, and the fastest damping is on the last of them. computed by solving, not by drawing. Settling time against damping for a third-order response — a complex pair at unit natural frequency and a real pole at 3 — with the second-order case behind it. Both are staircases: the settling time is set by the last excursion outside the band, so there is one step for each excursion that stops happening, and there are 3 of them between 0.3 and 0.98. They are at 0.378, 0.522, 0.773, with jumps of 1.24, 1.30, 1.42. The rung below found the last and largest of them and did not look below it. The fastest damping is 0.775, sitting on the edge of the last step, and a design a hundredth to the left of it settles 42 per cent slower.

Three cliffs, and where they are

The rung below sweeps a second-order step's damping, finds the settling time falling by a third in one step of a five-thousandth sweep, and calls it the cliff. There are three of them between 0.3 and 0.98, one for each excursion that stops leaving the band, and adding a third pole moves all three left and makes all three shallower — so the classic 0.78 for fastest two per cent settling is a second-order number, and at a third pole one and a half times the natural frequency the answer is 0.745 and 0.78 is on the wrong side of the step.

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

The gap a derivative needs

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

A staircase costs one Nth, computed rather than quoted. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. The charge is broken into N equal risers, each held for 16 time constants so that it completes. The measured losses are 1.00000, 0.500001, 0.250000, 0.125000, 0.0625000, 0.0312500 of ½CV² — which is 1.000004, 1.000002, 1.000001, 1.000001, 1.000000, 1.000000 times 1/N, so the law is exact to four parts in a million at the worst rather than approximately true. The fitted exponent is -1.00000 and the energy account closes to 2.17e-6 at the worst.

The half a switch keeps

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

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.

The fastest damping is a surface, and the band is worth 5 times the third pole. computed by solving, not by drawing. Each point is the last settling cliff, bisected — the damping at which the first overshoot's peak lands exactly on the band's edge, which is where the fastest settling is. Across the five bands the optimum moves by 0.231 of damping ratio; across a third pole from 1.5 times the natural frequency out to a second-order response it moves by 0.047. The two axes are worth 5.0 to one, and the expensive one is the specification rather than the parasitic. The classic 0.78 for fastest two per cent settling is the second-order curve's value at ±2%, 0.7797; at ±1% the same response wants 0.8261.

The best damping is not the one to build

The fastest settling damping is the right-hand limit at a discontinuity, so two thousandths below it costs 41 per cent and two thousandths above it costs 0.34 — a ratio of 120 in the penalty for the same error. With ±2 per cent on the damping ratio the nominal that minimises the worst case is 0.7927 rather than the optimum's 0.7734, and it guarantees 4.243/ωₙ against 5.943. The band moves the optimum by 0.231 of damping ratio and the third pole by 0.047, and 0.78 is exact at ±2% and 55 per cent slow at ±1%.

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

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.

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

Shorted instead of opened, and the error changes sign

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

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.

Stepped at 20 of its time constant, a 1 µs pole rings between 1.818 and 0.331 V, and needs 23 steps to settle. Marched with the trapezoidal rule at a step of 20.0 µs. A 1 µs pole (1 kΩ, 1 nF) drives, through a unity buffer, a 1 ms pole (1 kΩ, 1 µF). The fast node's exact response reaches its final volt within a few microseconds; the march's first values are 1.8182, 0.3306, 1.5477, 0.5519, 1.3666 V. Its distance from its final volt is multiplied by (1 − h/2τ)/(1 + h/2τ) = −0.8182 every step, measured and checked against that form, so it changes sign every step and takes 23 steps to fall below 1% — 460 µs. The slow node it drives is 1.23e-5 V from exact at 1 ms, because a 1 ms pole averages an alternation at half the stepping rate to nothing.

The ringing that belongs to the rule

The trapezoidal rule is stable for every stable circuit and every step size, and it is not damping. March a one-microsecond pole with twenty-microsecond steps and its node reads 1.818, 0.331, 1.548, 0.552 volts — an oscillation at half the stepping rate, its distance from the final volt multiplied by exactly −0.8182 every step, taking twenty-three steps to fall below one per cent. The slow node that pole drives is right to 1.2 × 10⁻⁵ V at a millisecond. One backward-Euler step at the discontinuity cuts the first swing from 0.818 V to 0.048 and two to 0.0023, because backward Euler multiplies the same error by 1/(1 + h/τ) and the trapezoidal rule by (1 − h/2τ)/(1 + h/2τ), which approaches −1.

At 20 steps a cycle, ten cycles of an undamped LC: the trapezoidal rule keeps the amplitude and falls 29.2° behind; backward Euler keeps 0.0082% of it. Marched, both rules, against 1 − cos ωt for a 1 kHz inductor–capacitor pair stepped with no resistance at all. At 20 steps a cycle the trapezoidal march's amplitude stays at 1.00000 a cycle and its frequency is slow: it loses 2.918° a cycle, measured from the march's own recurrence, against 2π − 2N·atan(π/N) = 2.918°, so after ten cycles it is 29.2° behind. Backward Euler keeps 0.3901 of its amplitude a cycle, against (1 + (2π/N)²)^(−N/2) = 0.3901, so 0.0082% is left after ten, and it loses 11.19° a cycle. No resistance is in the circuit; every loss is the rule's.

The phase the rule loses

An inductor and a capacitor with no resistance ring for ever, and two ways of marching them disagree about how. The trapezoidal rule keeps the amplitude exactly — its factor per step has a magnitude of one — and loses phase instead: 2π − 2N·atan(π/N) a cycle, 2.918° at twenty steps a cycle, so ten cycles later it is 29.2° behind the circuit. Backward Euler keeps 0.3901 of the amplitude a cycle at the same step, and after ten cycles 0.0082 per cent of the ringing is left, in a circuit that has no loss. The two errors fall at different rates: the trapezoidal rule's phase as the square of the steps a cycle, backward Euler's amplitude as the first power. A hundred cycles to within one per cent needs 182 steps a cycle of one and 196,404 of the other.

$A step through r sections starts as (t/τ)^r: it reaches 1% at 10.1 µs, 105 µs, 243 µs, 380 µs, 508 µs for r = 1 to 5. Solved, and expanded two ways. The step response of buffered RC sections of time constants τ, τ/2, … τ/r, with τ = 1 ms, on logarithmic axes. The relative degree of the recovered transfer function is r, so the first r − 1 derivatives of the step are zero at the start and the r-th is lim s^r·H(s) = r!/τ^r, read off the network solved far above its poles and off the expansion of H about infinity; the step therefore starts as (t/τ)^r, a straight line of slope r. The expansion about infinity and the residue expansion agree to a part in a million where both are well conditioned. The output reaches 1% at 10.1 µs (r = 1), 105 µs (r = 2), 243 µs (r = 3), 380 µs (r = 4), 508 µs (r = 5), and half its final value at 693 µs, 1.23 ms, 1.58 ms, 1.84 ms, 2.04 ms. For these time constants the whole step is (1 − e^(−t/τ))^r, checked against both expansions, so the time to a fraction ε is −τ·ln(1 − ε^(1/r)).$

The start a step takes from infinity

The initial-value theorem reads where a step starts off H at infinite frequency. Apply it again to s·H, s²·H and on, and it reads how the step starts: the first r − 1 derivatives are zero for a network r degrees more poles than zeros, and the r-th is the ratio of the leading coefficients. So a step through r sections begins as a power of time — for sections of τ, τ/2, … τ/r, exactly (t/τ) to the r — and reaches one per cent at 10.1 µs through one section, 105 µs through two and 508 µs through five. Put a zero anywhere, even a thousand times above every pole, and the step starts linearly instead, with a slope of twice the zero's time constant over τ² that is the larger term for the first two of them.

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