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

Transient response — where it appears

What a circuit does between one steady state and the next, which its frequency response contains and does not display. Reading it requires a march, and the march's own discretisation error is measured against an exact response rather than assumed away.

Named by 17 essays across 4 fields — each of them below, with the objects they name alongside it.

The window a 10 µF capacitor leaves. Above 939 mΩ the loop holds 45° of margin; below it the regulator rings and then oscillates. The droop after a 100 mA step is smallest at 817 mΩ — 129 mV — and by 19.9 Ω it is 1.468 V, because at the first instant of a step the capacitor cannot move and the whole step falls across its series resistance. Two requirements, opposite directions, and the useful values are between them.

Two requirements pulling one capacitor

The output capacitor's series resistance is a stability requirement and a transient requirement at once, and they pull it in opposite directions. Below 939 milliohms this loop has less than 45 degrees of margin; above about 850 the droop after a load step starts to grow, because at the first instant of a step the capacitor cannot move and the whole step falls across that resistance. The droop is smallest 10 per cent inside the unstable region, which means the best transient this design can have is one it must not be built with.

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

The cancellation that leaves a tail

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

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

The inductance that limits, and lifts

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

power · Inrush 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 earlier essay 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

Sweep a second-order step's damping and the settling time falls by a third in one step of a five-thousandth sweep, which is usually called 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.

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

The best damping is not the one to build

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

transients · Damping A capacitor across the upper divider resistor removes the output capacitor's resistance floor. computed by solving, not by drawing. Two series resistances against the capacitance across the upper divider resistor: the smallest the loop tolerates at 45° of margin (lower curve), and the one that gives the smallest droop after a 100 mA load step (upper). With no capacitor they are 939 mΩ and 885 mΩ — the second BELOW the first, which is the conflict this design has: the best transient is one the loop refuses. The floor falls as the capacitor grows and between 500 and 836.5 pF it leaves the sweep altogether, so every series resistance down to a milliohm is stable. Past about 5000 pF the floor climbs back and overtakes the optimum again. The shaded band is where the design a transient wants is one the loop allows.

The floor a second capacitor removes

Two requirements pulling one capacitor found a regulator whose best transient is one it must not be built with: below 939 milliohms of output-capacitor series resistance the loop has under 45 degrees of margin, and the droop is smallest at 885. The capacitor across the upper divider resistor, added for the reference's sake, dissolves that conflict. At 836.5 picofarads the 45-degree floor leaves the sweep entirely — every series resistance down to a milliohm is stable — and the droop falls 40 per cent at the same time. The band of capacitances that do it runs from 100 picofarads to 5 nanofarads, and above it the conflict returns.

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

The slowest one decides it

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

transients · Doublet Losing capacitance to bias raises this loop's margin; the corner that binds is the ceramic at full value. computed by solving, not by drawing. Phase margin of the five-volt regulator with a 5.00 mΩ ceramic output capacitor, against the capacitance that capacitor keeps at the working voltage, from the full 10 µF down to 1.5 µF, for five capacitors across the upper divider resistor. With none, the margin climbs from 18.0° at full value to 43.6° at 15% kept, so the lost capacitance is stabilising rather than the reverse. 630 pF holds at least 49.7° across the whole range. 836.5 pF, the value that removed the resistance floor at the nominal capacitance, gives 48.8° at full value and 44.3° at 15%: the capacitor that has lost the most is the one it fails on.

The capacitance the bias takes away

A ceramic output capacitor has a few milliohms of series resistance, which a capacitor across the upper divider resistor made legal, and a capacitance that falls with the voltage across it — a ten-microfarad part at five volts is two microfarads to a small signal. Swept from the full ten down to that fifth, the regulator's margin does not collapse. It rises: with no divider capacitor from 18.0° to 38.5°, because the output pole climbs away from the amplifier's. The corner that binds is the capacitor at full value. One divider capacitor, 423 to 808 pF, holds 45° at every capacitance between. The 836.5 pF that removed the resistance floor sits just above that band, and the lost capacitance is paid for in droop: 103 mV at full value, 202 mV at a fifth.

applied · Regulator At full capacitance the margin falls with the load, and below a tenth of an ampere no divider capacitor gets it past 36.9°. computed by solving, not by drawing. Phase margin of the five-volt regulator with a 5 mΩ ceramic keeping 10 µF, against the load current from 5 mA to 1 A, for four capacitors across the upper divider resistor. 630 pF, which held 45° at every capacitance at an ampere, gives 49.7° at 1 A and 36.0° at 5 mA. With no capacitor the margin at 5 mA is 0.1°. The dotted level is 36.87°, the most phase a zero and a pole four times apart can ever lead by, and every curve ends under it. The crossover stays between 12.9 kHz and 13.2 kHz over the whole range.

The lead a divider can give

At an ampere one capacitor across the upper divider resistor holds this ceramic regulator above 45° at every capacitance the part keeps. At 5 mA nothing does. The crossover stays at 13 kHz while the output pole falls from 3.20 kHz to 31.8 Hz, so the loop arrives at unity with every lag it has and the only phase left is the lead the divider capacitor makes — and that lead has a ceiling, arcsin((G − 1)/(G + 1)), set by the divider's ratio alone. For a five-volt output from a 1.25 V reference it is 36.87°, and the best capacitor reaches 36.83°. Over the rectangle from 5 mA to an ampere and ten microfarads to two, the best single capacitor gives 32.4°. Putting 416 mΩ back in series with the ceramic restores 45° over all of it, and lowers the droop rather than raising it.

applied · Regulator Driven through 1kΩ, the absorbing capacitor's tail is 2.6e-5 of the step, not 1%. computed by solving, not by drawing. The error at a capacitor's terminal after a step applied through 1k ohms, where the capacitor is 1 nF with eight absorption branches spread over two decades from 0.1 ms, together holding one per cent of its capacitance. The poles are found by bisection and the residues in closed form, and a trapezoidal march of the same netlist agrees. The tail the branches leave is 2.58e-5 of the step — the source resistance over the branch resistances in parallel — and settling to 1e-5 takes 173µs against 11.5µs with no branches; the model that weights each branch by its share of the capacitance says 48.4ms.

The tail the source decides

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

transients · Doublet Swept at η = 1, a resonance reads 93.6% of its height, 0.45 bandwidths late and 1.26 bandwidths wide. computed by solving, not by drawing. The envelope of a series resonator's capacitor voltage, Q 50 at 1591.5 Hz, while its drive sweeps linearly upwards through resonance, plotted against the frequency the drive has reached, for sweep rates η = r/B² of 0.1, 0.3, 1 and 3; dashed, the steady-state response the sweep is chasing. At η = 1 — 1013 Hz a second, Q cycles for every bandwidth crossed when η is one — the reading peaks at 93.6 per cent of the true height, 0.445 bandwidths above resonance, and is 1.263 bandwidths wide. Swept at η = 0.03 it is the steady curve to 0.2 per cent.

A sweep that never stops

Sweep a drive continuously through a resonance and read the envelope against the frequency the drive has reached, and the reading depends on one number: the rate in hertz a second divided by the square of the bandwidth. At that rate equal to one — Q cycles for every bandwidth crossed, the dwell a stepped sweep needed — a Q-50 resonance reads 93.6 per cent of its height, 0.445 bandwidths late and 1.263 bandwidths wide, and at Q 10 or 200 the same within a per cent. A tenth of that rate gets within 0.3 per cent of the height and costs ten times the time. The peak always lands beyond resonance in the direction of travel, so an upward and a downward sweep together find it to 0.0075 bandwidths where either alone is off by nearly half.

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

The tail with no slowest relaxation

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

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

The instant a transformer is worst at

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

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

The harmonic that names an inrush

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

power · Inrush The droop of the ceramic regulator against the rise time of a 100 mA load, with the capacitor at full value, half and a fifth. computed by solving, not by drawing, marched from the operating point: 10 µF, 5 mΩ of series resistance, 630 pF across the upper divider resistor, the 100 mA arriving as a straight ramp. A step droops 103 mV at 100%, 137 mV at 50%, 203 mV at 20% of the capacitance. A ramp of a millisecond droops 2.98 mV, 2.56 mV, 2.56 mV, and the slowest ramps approach the loop's inductance, 25.2 µH, times the ramp's rate (dashed) — a quantity with no capacitance in it — times 1.18, 1.03, 1.03, the share the loop's ringing adds. The full-value and fifth-value curves cross at 23 µs; slower than that, the larger capacitor droops more.

The load that rises slower than the loop

Every droop a regulator is sized against is a load step with an infinitely fast edge, and a ceramic regulator droops 103 millivolts on its full 10 µF and 203 on a fifth of it. Give the same 100 mA a rise time and the ordering does not survive: past 23 microseconds the full capacitor droops more than the fifth, and a slow ramp's droop is the loop's own 25.2 µH times the load's rate of change — a number with no capacitance in it, raised 18 per cent on the full capacitor by the impedance peak its slower loop leaves. For a 10 mV budget the load must take 298 µs to arrive on the full capacitor and 253 on the fifth.

applied · Regulator A resonator of Q 10 driven at 1.2 times its frequency, the drive switched off at the capacitor's crest and at its zero. computed by solving, not by drawing: the series circuit's exact free response from its settled state, held to a fourth-order march of the same circuit. Switched off at the capacitor's crest the ring starts at 1.0013 of the settled amplitude and never passes it; switched off at its zero it starts at 1.2015 — ω/ωd — and its first crest reaches 1.1120. The settled arrow turning at the drive's frequency is gone at the switch; what is left turns at the circuit's own, and its length is set by where in its cycle the settled arrow was.

The ring a drive leaves behind

Stopping a resonator's drive removes the arrow that turned at the drive's frequency and leaves the one that turns at the circuit's own, and how long that arrow is depends on the instant the drive stopped. Its length is the energy the circuit held at that instant, and off resonance the energy is not constant through a cycle: it breathes between the capacitor's peak and the inductor's, which differ by (ω/ω₀)². So above resonance a drive stopped as the capacitor's voltage passes through zero leaves a ring 1.2015 times the settled amplitude at 1.2 times resonance, and its first crest passes the settled value; below resonance no instant does.

frequency · Phasors 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

Named alongside it

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

Design tradeoffModel rangeSettling timeResiduesEquivalent series resistancePhase marginDamping ratioLoop gainMarchingOvershootPolesDielectric absorption

All concepts