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

Conduction angle — where it appears

The fraction of a supply cycle, in degrees, over which a rectifier's diodes conduct and its load draws any current at all. It is twenty-odd degrees for an ordinary reservoir supply, which is why the current is many times the average and the crest factor is large.

Named by 9 essays across 2 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 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 ripple 1000 µF leaves, through the regulator: 60.62 mV rather than 21.11 mV. One settled cycle of the reservoir's output — 1.331 V peak to peak across 1000 µF — split into 1000 Fourier lines, each passed through the solved regulator's rail-to-output response, and summed. The output is 60.62 mV peak to peak. The ripple times the rejection at 100 Hz, -36.00 dB, gives 21.11 mV, which is 2.872 times too little: the rejection worsens at twenty decibels a decade, so each harmonic of the sawtooth arrives at nearly the size of the first. The 100 Hz line carries 23.4% of the output's mean square and the lines above 1 kHz 10.4%; the first three arrive at 7.731 mV at 100 Hz, 7.200 mV at 200 Hz, 6.388 mV at 300 Hz. The dashed curve is the 100 Hz line alone.

The ripple that arrives as a comb

The rejection essay multiplied two numbers: 1.331 volts of reservoir ripple and the regulator's 36.0 decibels of rejection at a hundred hertz, for 21 millivolts at the output. The ripple is a sawtooth, a comb of lines at every multiple of a hundred hertz, and the rejection worsens at twenty decibels a decade — so the second line arrives at nearly the size of the first, and the third too. Summed with their phases, the output is 60.62 millivolts, 2.87 times the estimate, and the hundred-hertz line carries only 23.4 per cent of it. A reservoir twenty-one times larger cuts the rail's ripple 14.8 times and the regulated ripple 5.95.

applied · Regulator A true-RMS reading of a sine: ripple a second filter removes, and a bias it cannot. An explicit converter — square, average through a one-pole of τ = 100 ms, take the root — in steady state on a sine of unit root-mean-square value, integrated exactly over a period at 91 frequencies and by a fourth-order march of its own equation at 6, which agree to 1.9e-8. The upper curve is half the ripple on the reading and the lower one the amount by which its mean is low. The reading is low at every frequency, because the square root is concave; it is 1% low below 1.86 Hz, while the ripple is inside ±1% only above 39.8 Hz. The dashed curve is the small-ripple form, an eighth of the averaged square's ripple power, which the bias approaches as the ripple shrinks.

The average a square root pulls low

A true-RMS converter squares, averages and takes the root, and the root of a quantity that ripples averages below the root of its mean. With a hundred-millisecond averager a sine is read one per cent low below 1.86 hertz, where the ripple is still ±20 per cent — and a second filter that steadies the display takes the ripple away and leaves the reading exactly as low as it was. A square wave is read exactly at any averaging time; a rectifier current conducting for twenty degrees needs 2.41 times the averaging a sine does. The implicit converter is the explicit one at half the time constant, and a reading falls 1.38 times slower than it rises.

power · RMS and average 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 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

Named alongside it

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

Crest factorReservoir capacitorDesign tradeoffForm factorMeasurement conditionModel rangeRectificationTrue-RMSEquivalent series resistanceAveragingTransformerCompanion model

All concepts