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

Crest factor — where it appears

The ratio of a waveform's peak value to its root-mean-square value, which is the square root of two for a sinusoid and larger for anything peakier. It is what decides whether a current of a given root-mean-square value will damage a component, and it is invisible in that value.

Named by 12 essays across 4 fields — each of them below, with the objects they name alongside it.

1000 µF across a 100 Ω load, rectified from 17 V peak. The output sits at 15.69 V with 1.331 V of ripple, against the 1.569 V the expression I/2fC gives — 15.1% high, because the capacitor is being recharged for part of the cycle rather than discharging throughout it. The lower panel is why: the diode conducts for 28.8° of each half cycle and carries 2.10 A at the peak, which is 13.4 times the 157 mA the load draws.

The direct voltage that is a sawtooth

A rectifier and a reservoir capacitor make what everybody calls a direct voltage. Marched with the diodes in the netlist, a thousand microfarads across a hundred ohms gives 15.69 volts with 1.33 volts of ripple on it, against the 1.57 the textbook expression predicts. The expression is high by the fraction of the cycle the diode conducts for — measured at 0.94 to 0.96 of it across two sweeps — and it has no opinion at all about the quantity that actually sizes the transformer, which is a peak diode current 13.4 times the current the load draws.

applied · Unregulated supply Two meters, one current, and neither of them measuring the heat. computed by solving, not by drawing, at 35 conduction angles. The first curve is an average-responding meter: it rectifies, averages and multiplies by 1.1107, which is exactly right for a sinusoid — -7.8e-5% here — and exactly 11.07% high on a square wave, because the error is the ratio of two form factors and contains neither the amplitude nor the frequency. On a rectifier drawing its 100 W in sixty degrees of conduction it is -35.90% low. The second curve is a true-RMS meter that reaches 9 harmonics, which has no shape assumption in it and a bandwidth instead: it returns the root-sum-square of the lines it can see, and is one per cent low below every angle here of conduction. The crest factor at sixty degrees is 1.733, which is inside every instrument's rating — neither meter is failing because the peak is large. One is failing because the shape is not a sinusoid and the other because the spectrum is wider than it is.

What a meter multiplies by

An average-responding meter rectifies, averages and multiplies by 1.1107, which makes it exactly right for a sinusoid and wrong for everything else by the ratio of two form factors — 11.07 per cent high on a square wave and 35.9 per cent low on a rectifier drawing its current in sixty degrees. It is also exactly right at one other waveform, a 145.90 degree conduction angle, which is nobody's sinusoid. Beside it a true-RMS meter that reaches nine harmonics is two per cent low on a square wave and never within one per cent of anything narrower.

power · Rms and average The first conduction carries 26× the repetitive peak, and the factor of 7 is the user's. computed by solving, not by drawing. The largest diode current in the first conduction after switch-on, against the phase of the mains at the instant of switching, marched from an empty capacitor through the same netlist the ripple and the conduction angle are read from. The settled circuit's largest current is 1.230 A; the first one is 32.39 A at the worst instant and 4.64 A at the best, so which of them a design has to survive is decided by nothing in the design. The estimate that treats the capacitor as a short circuit gives 34.00 A, 4.7% high, the difference being the charge the capacitor takes during the pulse itself. Holding the first peak to ten times the repetitive one would need 1.90 Ω of winding resistance.

The first cycle, which no steady state contains

Every number this field computes about a rectifier — the ripple, the crest factor, the conduction angle, the power factor — is read from the settled state, and the march that produces them starts from an empty capacitor and throws the first cycle away. That first conduction carries 32.4 amperes against a repetitive peak of 1.23, it is 26 times larger than anything the circuit ever does again, and how large it is depends on when somebody's hand closed the switch.

power · Inrush 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 The crest factor is 13.4 and the winding is sized by 3.01. computed by solving, not by drawing. Three ratios of the same settled march, against the reservoir. The crest factor — the peak diode current over the load's direct current — runs 6.43 to 24.25. The form factor, which is the root-mean-square current over the same direct current and is what a winding heats by, runs 2.114 to 4.077. Its square is the copper loss against a winding carrying the direct current alone, and that runs 4.47 to 16.62. The first ratio is 3.04 times the second at 220 µF and 5.95 times at 4700, so quoting one of them tells a reader nothing about the other.

The current that sizes the transformer

A reservoir's crest factor is 13.374 at a thousand microfarads and the winding is not sized by it. The root-mean-square of the same marched current is 3.0069 times the load's direct current, so the copper dissipates 9.0417 times what it would carrying the direct current alone — and the two ratios diverge, from 3.04 apart at 220 microfarads to 5.95 apart at 4700. The expression for the mean output is wrong in three places whose signs differ, and at 313.9 microfarads they cancel to six microvolts while the ripple expression inside it is still 32.3 per cent high.

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

The resistance that lowers the ripple

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

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

The resistance that belongs in the winding

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

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

Two capacitors that are one

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

applied · Unregulated supply A switching load's comb in a 10 mV reading: every tooth the same height. computed by solving, not by drawing. The harmonics of a 50-per-cent trapezoid of 100 mA at 100 kHz with a 10 ns edge, each multiplied by the shared conductor's impedance at its own frequency. Below 79.6 kHz the conductor is 5.00 mΩ of resistance and the teeth fall as 1/n. Above it the conductor rises as n while the harmonics still fall as 1/n, so the two cancel and every tooth from 239 kHz to 10.6 MHz is 0.828 mV — flat to 5.4 per cent across 52 of them. The comb stops at 31.8 MHz, which is the edge's own corner, so the total of 12.6 mV is decided by how many teeth there are: fitted over five edge rates it grows as the edge time to the power -0.497, against the −½ that a quadrature sum of equal teeth gives. A faster edge raises no harmonic and adds teeth. What a slow measurement keeps is a different term entirely: the zeroth harmonic, which is the 100 mA average through 5.00 mΩ of resistance — 500.0 µV, with no inductance in it and no edge, against the comb's 12.6 mV.

Every tooth the same height

A shared return conductor's impedance rises a decade per decade above 79.6 kilohertz and a switching load's harmonics fall a decade per decade, so the two cancel exactly: every harmonic of a hundred-milliamp square wave puts 0.828 millivolts into a ten-millivolt reading, from the conductor's own corner up to the edge's. The comb is flat rather than falling, so the total is decided by how many teeth there are — and since they add in quadrature it grows as the square root of the edge rate, 39.7 millivolts at a nanosecond against 4.03 at a hundred.

instruments · Common-impedance One rms value, and a peak that moves 41 per cent with a phase. A 10 V fundamental and a 3.33 V 3th harmonic, summed, with the harmonic's relative phase swept. The two are orthogonal over a period, so the root-mean-square value is the quadrature sum 7.45356 V at every phase — to 6.2e-15, which is the one case superposition allows and is a theorem rather than an approximation. The peak is not a sum of anything: it runs from 9.428 V to 13.33 V, a factor of 1.414, against a bound of 13.33 V that it reaches exactly at 180° — where this harmonic's own crest lands on the fundamental's, which for n ≡ 3 (mod 4) is that end of the sweep and not the other. At the far end it flattens the crest to 9.428 V, and there the peak has no closed form at all. So one signal has one heating and a headroom requirement that varies by 41 per cent, decided by a phase that no magnitude spectrum carries.

The peak that only has a bound

Power does not superpose and the gap has a closed form — two times the real part of one current times the conjugate of the other, exact at every setting. A peak has nothing of the kind. A ten-volt fundamental with a third harmonic a third its size has one root-mean-square value, 7.454 volts at every relative phase because the two are orthogonal, and a peak running from 9.428 to 13.333 volts across the same sweep — reaching the sum of the two peaks exactly at 180 degrees, where the harmonic's own crest lands on the fundamental's, and having no closed form at all at the other end.

networks · Superposition Summed over whole periods 1.3% too long, a sine is read to ±0.65% for up to 38 periods, whatever their number. Integrated exactly over each window, the worst over every starting phase. The error in a root-mean-square summed over N assumed periods 1.3% too long, against N, for a sine and a 60° rectifier current, beside an explicit converter averaging over a comparable time, τ of N/2 periods. For the sine the worst error is 0.643% at one period and stays near δ/2 until N approaches 1/(2δ) = 38; it vanishes where Nδ is a whole number of half-periods of the square, and beyond it is bounded by 1/(4πN). The rectifier current's is 1.274% at one period, near δ(CF² − 1)/2 with a crest factor of 1.732. The converter at τ = N/2 periods is low by 15.8 ppm on the sine at N = 10, with a ripple of ±0.796%; on the rectifier current, low by 46.3 ppm with ±1.665%.

The cycle a converter has to know

Summing a waveform's square over a whole number of periods reads its root-mean-square exactly: no averager, no ripple, no bias. It needs the period, and a period known one per cent long puts a hundredth of a period too much into the window. Wherever that extra piece falls, the reading moves — on a sine by up to 0.50 per cent over one period, and by 0.46 per cent over ten, because the extra piece grows with the window as fast as the window does. The worst error is δ(CF² − 1)/2, set by the crest factor and the period error and not by how many periods are summed, until the excess reaches half a period. A square wave is read exactly from any window, and a 60° rectifier current twice as badly as a sine.

power · Rms and average With the period 1.3% long, eight periods read a sine 0.599% wrong through a rectangle, 79.4 ppm through a triangle, 22.8 ppm through a Hann window. Integrated over each weighted window, the worst over every starting phase. The error in a sine's root-mean-square summed over N assumed periods 1.3% too long, against N, for four weights across the window. The rectangle is the plain sum: 0.640% at two periods and flat in N, δ/2 as the earlier measurement found. The triangle is flat in N too but 75 times smaller at eight periods. The Hann window falls with N, as the −2.67 power: 414 ppm at two periods, 0.293 ppm at thirty-two. The four-term Blackman–Harris is below 4.85 ppm at every count from two. A weight that falls to zero at the window's ends counts the extra sliver of period at almost nothing.

The window that forgives the period

A converter that sums a waveform's square over whole periods reads its RMS exactly, and one that sums over periods 1.3% too long reads a sine 0.6% wrong — δ/2, for any number of periods — because the extra sliver of period counts at full weight. Weight the window down to zero at its ends and the sliver barely counts. Over eight periods a triangular weight reads the same sine 79 ppm wrong and a Hann weight 23 ppm; and they are different kinds of cure. The triangle is the only one that changes the error's order, from first to second in δ, so it wins when the period is nearly right; the Hann weight keeps first order but falls with the number of periods, about as its inverse square, so it wins over long windows. Both cost nothing with the period exact — from two periods up. Over one, a Blackman–Harris weight reads even a sine 10% wrong. The price is noise: a Hann-weighted reading of noise scatters √1.5 as widely.

power · Rms and average

Named alongside it

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

Conduction angleModel rangeReservoir capacitorDesign tradeoffRectificationTrue-rmsEquivalent series resistanceForm factorHarmonic contentInrush currentMeasurement conditionMeasurement error

All concepts