Circuits that do a job, and the range they do it over

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.

17 min read 7 figures Measured, not tabulated Two routes to a number Every model has an edge

Assumes: The direct voltage that is a sawtooth · What a meter multiplies by

The direct voltage that is a sawtooth marched a full-wave rectifier and a reservoir capacitor, compared the ripple against the expression $I/2fC$ , found the expression high by the fraction of the cycle the diodes conduct for, and then said something about the peak current that needs correcting. It named a crest factor of 13.4 as the quantity that actually sizes the transformer, the ripple expression having no opinion about it.

It does not. A winding heats as the mean square of the current in it, so the number that sizes it is the ratio of the root-mean-square current to the direct current the load draws, and that ratio is 3.0069 rather than 13.374. Both numbers are in the same trace and neither had been taken out of it.

The same march holds a second unread quantity, and it is the mean output voltage — returned on every call beside the marched answer, never compared against it, and wrong in three separate places whose signs are not the same.

The check that comes first

A rectifier’s current is a short tall pulse and every number below is an integral over it, so the first question is whether the trace’s currents are right at all rather than merely plausible.

They have a test that costs nothing and that nothing in the figures needs. Over a settled cycle the reservoir capacitor gains no net charge, so the mean of the rectifier’s own current must be the direct current the load draws. The first of those is a sum over two thousand samples of the diode law evaluated at each step’s terminal voltages; the second is the mean output voltage divided by the load resistance. They share the solved trace and nothing else — one of them never mentions the load and the other never mentions a diode.

The two agree to two parts in $10^{12}$ at a thousand microfarads and to better than $10^{-11}$ at every capacitance and every load drawn. That is the same pairing one step computed twice requires of a transient, applied to the quantity rather than to the step, and it is what entitles the rest of this essay to quote a current.

It is worth noticing what the check does not cover, because the check is exact and the quantity this essay is about is not the one it tests. A mean is a first moment and it is insensitive to how the charge is distributed inside the cycle; a march that put the whole cycle’s charge into a single step would pass it. What the mean confirms is that the diode currents are rebuilt correctly from the element laws and that no charge is created or lost — which is the failure mode a companion model has — and what it leaves to the discretisation is the shape.

So the shape is checked separately, by halving the step. At 500, 1000, 2000 and 4000 steps a cycle the form factor comes out 3.01331, 3.00813, 3.00694 and 3.00666; consecutive differences fall by 4.35 and 4.13, which is second order and is the trapezoidal rule’s own. The charge balance above, meanwhile, holds to better than $5\times10^{-13}$ at every one of those step counts, including the coarsest — which is the demonstration that it is not a convergence test.

The crest factor does not converge like that, and the reason is the essay’s own argument arriving from the numerical side. Over the same four grids it comes out 14.388, 13.408, 13.374 and 13.361: consecutive differences of 0.980, 0.0342 and 0.0127, which is no clean order at all. A root-mean-square is an integral and averages the grid’s errors away; a peak is a single sample of a pulse two degrees wide, and a coarse grid either lands near the top of it or does not. Quoted at two thousand steps a cycle the form factor is settled to a part in ten thousand and the crest factor to a part in a thousand, and those are the numbers below.

Fig. 1 The march the rest of the essay reads. The upper panel is the output against the rectified sinusoid behind it, at 15.69 volts with 1.33 volts of ripple; the lower is the current in one diode, peaking at 2.098 amperes for 28.8 degrees of each half cycle. Every ratio below is an integral over that lower panel. The slider is the reservoir capacitance.

Three ratios, and only one of them heats

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. — Fig. 2 The crest factor, the form factor and the copper multiplier of the same settled march, against the reservoir. The crest factor runs 6.43 to 24.25; the form factor — root-mean-square current over the load’s direct current — runs 2.114 to 4.077; its square, which is the copper loss against a winding carrying the direct current alone, runs 4.47 to 16.62. At a thousand microfarads each half of the secondary carries 0.3335 amperes root-mean-square.

The three answer different questions and the design uses all of them.

The crest factor is what the diode’s peak rating and the capacitor’s surge rating have to cover, and it is what an average-responding instrument fails on. The form factor is what the winding heats by. Its square is the number a transformer is sized against, because a secondary specified for 157 milliamperes of direct current and asked to deliver it through a rectifier dissipates 9.0417 times the copper loss it was specified for.

That last number is the one worth having in front of a catalogue, and it can be turned into the units a catalogue uses. Each half of the secondary is at 12.021 volts root-mean-square and carries 0.33352 amperes root-mean-square, so the two halves between them present 8.018 volt-amperes while the load takes 2.462 watts. A transformer for this supply has to be rated at 3.26 times the direct power it delivers, and the factor is not a rule of thumb — it is the form factor and the winding geometry, and it moves with the reservoir exactly as the form factor does.

The two ratios are not one statement in two units

The crest factor is 3.04 times the form factor at 220 microfarads and 5.95 times at 4700. The gap grows monotonically across the whole slider.

That is what separates them as design quantities. Both rise as the reservoir grows — the rung below established the direction, and its reason is unchanged: the charge that has to be replaced each cycle is fixed by the load, the conduction interval shortens from 54.18 degrees to 17.28, and the same charge moving in less time is more current. But the peak rises faster than the root-mean-square, because the pulse is getting taller and narrower rather than simply taller, and a mean square weights the width where a peak does not.

So a designer who reads a crest factor of 24 off a simulation and sizes the winding by it overspecifies the copper by a factor of six, and one who reads a form factor of 4 and sizes the diode by it underspecifies the peak by the same. Neither ratio can be inferred from the other without the conduction angle, and the conduction angle is not on any specification.

4700 µF across a 100 Ω load, rectified from 17 V peak. The output sits at 16.15 V with 312 mV of ripple, against the 344 mV the expression I/2fC gives — 9.2% 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 17.3° of each half cycle and carries 3.92 A at the peak, which is 24.3 times the 161 mA the load draws. — Fig. 3 The largest reservoir on the slider: 312 millivolts of ripple, 17.3 degrees of conduction, and a peak diode current of 3.92 amperes against 161 milliamperes drawn by the load. The lower panel is the pulse whose two integrals the ratios above come from.

Set beside the smallest reservoir on the same slider, what has changed is the pulse’s shape rather than the work it does. Twenty-one times less capacitance divides the peak current by 4.30 and multiplies the conduction interval by 3.14, while the charge each pulse carries — the load’s own current divided by twice the line frequency — changes by only 14.0 per cent, since the load draws 141.7 milliamperes at the smaller reservoir and 161.5 at the larger. Nearly the same charge, moved in a third of the time, at four times the height. The mean square is the only one of the three ratios that reads how that charge is distributed rather than an endpoint of it, and it is the one the copper answers to.

220 µF across a 100 Ω load, rectified from 17 V peak. The output sits at 14.17 V with 4.607 V of ripple, against the 6.441 V the expression I/2fC gives — 28.5% 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 54.2° of each half cycle and carries 0.911 A at the peak, which is 6.43 times the 142 mA the load draws. — Fig. 4 The smallest: 4.61 volts of ripple, 54.2 degrees of conduction, and a peak of 911 milliamperes. This is the shape a form factor of 2.114 belongs to, and the difference between the two figures is the whole of why one ratio cannot be read off the other.

The two halves of the secondary carry the pulses alternately, so each carries a root-mean-square current smaller by exactly $\sqrt{2}$ — 0.33352 amperes each against 0.47166 for the pair at a thousand microfarads. That split is a claim rather than a convention, and the trace tests it: the two halves are equal to a part in $10^9$ , and the sum of their squares is the whole to a part in $10^{12}$ , which is the statement that no instant has both diodes conducting.

And a lighter load makes it worse

The reservoir is the obvious knob and it is not the only one. Holding the capacitor at a thousand microfarads and varying the load moves both ratios the same way, and it moves them further:

load	direct current	crest factor	form factor	copper multiplier
50 Ω	303.2 mA	9.703	2.5680	6.595
100 Ω	156.9 mA	13.374	3.0069	9.042
200 Ω	80.0 mA	17.941	3.4863	12.154
400 Ω	40.4 mA	23.055	3.9696	15.758
1000 Ω	16.3 mA	28.841	4.4804	20.074

The mechanism is the one the rung below identified for the capacitance, arriving through the other variable: a lighter load takes less charge per cycle, the trough falls less far, the conduction interval shortens from 38.2 degrees to 14.9, and the current the diode passes in that interval is a larger multiple of a smaller mean.

The practical reading is unpleasant and worth stating. A supply designed at full load and then run at a tenth of it does not merely become inefficient; its winding’s copper multiplier rises by a factor of three and its diode’s peak current is a larger multiple of a smaller current. The absolute copper loss falls, because the mean current fell faster — 20.07 times 16.3 milliamperes is a smaller number than 9.04 times 156.9 — so the part is safe. What is not safe is the habit of quoting either ratio without the load it was measured at, since neither is a property of the circuit.

The instrument that cannot see it

The consequence for measurement is severe enough to be worth naming, and the instruments field has already priced it.

Fig. 5 Two meters on a rectifier’s current, against the conduction angle. An average-responding meter rectifies, averages and multiplies by 1.1107 — exactly right for a sinusoid, 11.07 per cent high on a square wave, and 35.90 per cent low on a rectifier drawing its power in sixty degrees. The crest factor there is 1.733, well inside every instrument’s rating, so neither meter is failing because the peak is large.

Sixty degrees is a gentler case than anything in this essay. The march above conducts for 28.8 degrees at a thousand microfarads and 17.3 at 4700, so an average-responding meter clipped across the transformer’s secondary reports a current that is not merely inaccurate but wrong in the direction that matters — low, on the quantity that decides whether the winding survives. The reading is a property of the meter’s shape assumption, and the shape is exactly what a reservoir capacitor changes.

The mean, which is right for the wrong reason

The other unread number is the mean output voltage. The expression is the transformer’s peak less half the textbook ripple, $V_\text{pk} - I/4fC$ , and it is returned on every call beside the marched mean.

It is wrong in three places, and they can be separated exactly because each is a difference between two quantities the march reports:

it uses the transformer’s peak rather than the output’s, so it ignores the rectifier’s forward drop — 676 millivolts at a thousand microfarads, and this term always makes it high;
it subtracts half of $I/2fC$ , and $I/2fC$ is itself too large by the conduction fraction — 119 millivolts too much at a thousand microfarads and 917 at 220, and this term makes it low;
and the sawtooth’s mean is not its peak less half its ripple, because a sawtooth whose recharge is faster than its discharge is not symmetric — 27.2 millivolts at a thousand microfarads, also low.

The three add to the total error to the last bit at every capacitance drawn.

At 314 µF the mean is right to 6 µV and every term in it is wrong. computed by solving, not by drawing. The expression for a reservoir's mean output is the transformer's peak less half the textbook ripple, and it is wrong in three places. It ignores the rectifier's own forward drop, which is 676 mV and makes it high. It uses a ripple that is itself too large, which subtracts 119 mV too much at 1000 µF and far more at 220. And the sawtooth's mean is not its peak less half its ripple, which is a further 27.2 mV. The three add to the error exactly, they do not have the same sign, and they cancel at 314 µF — where the ripple expression inside the answer is still 32.3% high. — Fig. 6 The expression’s error and the three terms it is made of, against the reservoir. The total is the curve carrying circles: it runs from −392 millivolts at 220 microfarads to +679 at 4700 and crosses zero at 313.9, where the expression is right to six microvolts. The dashed curve along the top is the diode’s drop, the only term that makes the expression high. The other two both make it low and both shrink as the capacitor grows — the large one starting at −917 millivolts is half the ripple expression’s own error, and the small one that converges onto the zero line is the sawtooth’s asymmetry, −143 millivolts at the smallest reservoir and −3.6 at the largest.

At 313.9 microfarads the expression for the mean output is exact and every term inside it is wrong, and the largest of them by a long way is the ripple: the $I/2fC$ that the expression is built on is 32.3 per cent high at that capacitance. The location was found by a secant on the logarithm of the capacitance rather than interpolated between drawn points, because each evaluation is a twelve-cycle march and a root worth stating is a root worth locating.

This is not a curiosity about one capacitance. It is the reason the expression has survived: at the capacitances a small mains supply actually uses — a few hundred microfarads for a hundred-milliampere load — it is right to a few tens of millivolts, and the agreement is a cancellation rather than a derivation. Move to 4700 microfarads, which any modern design does because two requirements pull the same capacitor, and the expression is 679 millivolts high on a sixteen-volt rail with no warning that anything has changed.

The winding this march does not have

Every ratio above is computed with an ideal transformer, and that is the model’s edge rather than an oversight. Winding resistance is what actually limits the charging pulse, and putting it in changes the answers in the direction that makes them smaller.

Half an ohm of winding takes the crest factor from 13.4 to 8.0. computed by solving, not by drawing. The same 1000 µF reservoir on the same 100 Ω load, with a resistance put in each half of the secondary. The crest factor falls from 13.374 to 5.329 and the copper multiplier from 9.042 to 4.234, because the conduction lengthens from 28.8° to 52.0° and the same charge moves in a lower current. It is paid for in output: the mean falls from 15.69 V to 14.65 V. So every ratio measured with an ideal transformer is an upper bound, and the part that makes it smaller is the part a designer is trying to remove. — Fig. 7 The same reservoir and load with a resistance in each half of the secondary. Half an ohm takes the crest factor from 13.374 to 7.966 and the copper multiplier from 9.042 to 6.240, because the conduction lengthens from 28.8 degrees to 36.5 and the same charge moves in a lower current. It is paid for in output: the mean falls from 15.69 V to 15.44.

So the numbers above are upper bounds, and the thing that lowers them is the thing a designer spends money to remove. A transformer wound with heavier copper to hold its regulation delivers a higher mean voltage and a worse crest factor at the same time; two ohms of secondary resistance brings the crest factor to 5.329 and the copper multiplier to 4.234, at the cost of a volt of output. That is a genuine trade rather than a defect, and it is invisible in a data sheet that quotes regulation and says nothing about conduction.

It also means the ideal march is the right instrument for a specification and the wrong one for a prediction. The peak current a diode has to survive should be quoted from the ideal case, because a transformer with less resistance than assumed is a transformer that has been improved; the copper loss should be quoted from the real one.

What this opens

Two things follow that are computable with what is already here.

The first is the harmonics. A current drawn in 28.8 degrees out of 180 is a comb of them, and the supply’s own wiring carries every one — which is what a power factor is a summary of, and what the current that does no work prices for a reactive load that draws a sinusoid. A rectifier’s is a different object with the same name: nothing is stored and returned, and the whole of the excess current is distortion rather than reactance. The form factor above is one number from that spectrum and the harmonic amplitudes are the rest of it, available from the same trace.

The second is the ripple’s own consequence rather than its size. The rung below measured the ripple and this rung measured the current; what neither has measured is what the ripple does downstream, where a source below a frequency is holding a rail against it with a finite rejection. The two hundred and fifty milliohms of secondary resistance that improves the crest factor above also improves the ripple, by 3.0 per cent, and the essays on either side of that boundary have not yet been made to agree on which of the two the copper is bought for.

That join is a shorter piece of arithmetic than it sounds, because the quantity a regulator’s rejection is quoted against is a frequency and a reservoir’s ripple is a sawtooth. A sawtooth at a hundred hertz has harmonics at every multiple of it, and what gets through from the rail measures a rejection that falls with frequency — so the ripple that reaches the output is not the first harmonic attenuated by the rejection at a hundred hertz, and a design that treats it as one is optimistic in a direction that grows as the conduction angle shrinks. The same short pulse that raises the copper multiplier puts more of the ripple’s energy where the regulator is worse.

What it does not say

It does not say the crest factor is the wrong number. It is the right number for the diode’s surge rating and for the capacitor’s, and it is the number the inductance that limits and lifts is about at switch-on, where the first pulse is larger again. It is the wrong number for the copper, and the error is not small: sizing a winding by a crest factor of 13.374 rather than a form factor of 3.0069 is a factor of 4.45 in current and 19.8 in the loss it is being asked to allow for.

It does not say the mean-voltage expression should be discarded. It is a line of arithmetic against a twelve-cycle march of a nonlinear netlist, and inside a few hundred microfarads it is worth tens of millivolts. What it should carry is a range, which is what this collection asks of every model, and the range here is a capacitance rather than a frequency: exact at 314 microfarads, 280 millivolts out at 470, and 679 out at 4700.

And it does not say anything about what happens before the settled state. Every integral here is taken over the last cycle of a twelve-cycle march, and the first cycle of the same march is a different circuit — an empty capacitor is a short, and the first cycle, which no steady state contains measures a peak there that this one has no term for. The two questions want opposite instruments: the settled ratios above are integrals over a repeating cycle and the first cycle’s peak is a single event whose size depends on when the supply was switched on. A part has to survive both, and only one of them is in a form factor.

There is a boundary between them worth naming, because it decides which number to use. A ratio is a ratio of a repeating quantity, so every number in this essay assumes the load has been constant long enough for the reservoir to reach its cycle — twelve cycles here, which at fifty hertz is 240 milliseconds. A load that changes faster than that, which is most digital loads, is never in the state these ratios describe, and the copper multiplier for it is somewhere between the one measured at its mean current and the one measured at its peak. That is a range rather than a number, and saying which end of it applies is a boundary is a model and a tolerance.

The number worth carrying

Form factor 3.0069 at a thousand microfarads against a crest factor of 13.374, with the copper multiplier 9.0417 and the two ratios diverging from 3.04 apart to 5.95 apart across the slider. The mean-output expression exact at 313.9 microfarads and 679 millivolts high at 4700, with a 32.3 per cent error inside it at the point where it is right.

The habit that goes with both is the same one. Every model has an edge, and a model that agrees at one setting has not been tested — it has been evaluated at the point where its errors cancel. Two of the three terms above shrink with capacitance and one grows, so there had to be a crossing somewhere, and a design verified at that crossing would have reported the expression as excellent. What separates an agreement from a verification is knowing how many independent errors were in the sum, and the only way to find that out is to compute each of them.

Part 2 on unregulated supply

One argument about Unregulated supply, and one of 5 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it: