The inductance that limits, and lifts

The essay one rung down read the head of a march everything else in this field reads the tail of. A rectifier’s reservoir capacitor starts empty, so for one conduction the diode is charging a short circuit through the winding resistance alone, and the current is the peak of the mains divided by that resistance — 32 amperes against a repetitive peak of 1.23, twenty-six times, with the worst instant of switching being the peak of the mains and the best a factor of seven below it.

It ended with a result that surprises people who have added a resistor to fix this. The energy the first charge costs, $\int i^2 R\,dt$, is 0.134 joules at a fifth of an ohm and 0.131 at a half — two per cent apart over a factor of four in the resistance — because the energy is $\tfrac{1}{2}CV^2$ and the resistance decides only how long it takes to deliver it. Adding resistance limits the peak, which is what a fuse and a diode’s surge rating care about, and does nothing whatever about the heat.

A real transformer has an inductance as well as a resistance, and it changes both answers.

Two elements, and the transient is no longer an exponential #

With a resistance alone, the first charge is an exponential: the capacitor is a short, the current starts at $\hat{V}/R_s$ and decays as the capacitor fills. That is the picture the rung below measured, and its crest factor and its energy both follow from it.

Put an inductance in series and the same circuit becomes a series resonant one, driven from a rectified sinusoid, with a damping ratio

$$\zeta = \frac{R_s}{2}\sqrt{\frac{C}{L}}$$

At the numbers here — a third of an ohm, a thousand microfarads — twenty microhenries gives $\zeta = 1.24$ and five millihenries gives 0.078. So the range on the figure runs from just-overdamped to a circuit that rings for many cycles, and the two ends behave nothing like each other.

The peak current falls from 36.4 amperes at twenty microhenries to 5.3 at five millihenries, a factor of seven. The bound the surge impedance suggests — $\hat{V}/\sqrt{L/C}$ — is drawn over it and is an over-estimate at every point, by between a factor of two and a factor of three, because the mains is not a step: it is a sinusoid whose own quarter period is comparable with the ring, so the drive falls away before the resonance has delivered its peak.

That is the same shape of statement as the two asymptotic expressions in the reverse-recovery essay: a closed form that is an upper bound rather than an answer, with the range over which it is worth having measured rather than assumed.

The energy falls with the current, which the resistance did not #

The second measurement is the one that separates the two remedies.

$\int i^2\,dt$ over the first conduction falls from 0.370 to 0.073 ampere-squared-seconds across the same range of inductance — a factor of five. Multiplied by the winding resistance, that is the heat the transformer takes, and it has fallen with the current rather than staying where it was.

The reason is the difference between the two elements, stated in one sentence: a resistance limits the current by dissipating the difference, and an inductance limits it by storing it. The energy that goes into the capacitor is $\tfrac{1}{2}CV^2$ either way, but with an inductance in the loop a large part of the energy that used to be dissipated in the resistance is instead cycled into and out of the magnetic field, and what the resistance sees is a smaller current for a longer time — and $\int i^2 dt$ is quadratic in the current and only linear in the time.

Which makes the leakage inductance of a transformer a genuinely free inrush limiter, in the sense that matters: it costs no steady-state dissipation, unlike a series resistor, and no control circuit, unlike a thermistor or a relay-bypassed resistor. Transformers have it whether or not anybody wanted it, and the essay’s first half is a fair description of why mains-frequency linear supplies survive being switched on at all.

The output goes above the peak, and stays there #

The cost is the second curve and it is not a dissipation.

An inductor’s current cannot stop at the instant the two voltages are equal. When the capacitor reaches the peak of the supply the current is still flowing, so it goes on charging — and the capacitor overshoots. At a millihenry the output reaches 21.86 volts from a 17-volt peak, 28.6 per cent above it.

And then it stays there, which is the part that matters. The diodes are reverse biased above the supply, so the charge has no way back out; it can only leak away through the load, at a time constant of a hundred milliseconds. Every steady-state expression in this field — the ripple, the conduction angle, the crest factor, the regulation — is derived from a capacitor that sits just below the peak of its input, and none of them contains a state in which it sits above it.

The consequence for a design is direct and is a stress rather than an error. A supply built from a seventeen-volt peak with a twenty-five-volt reservoir capacitor has forty per cent of margin by the steady-state arithmetic and eleven per cent by this one, on the first cycle after switch-on, at a moment when nothing has warmed up and no load is drawing anything.

The overshoot has an interior maximum, and both ends have reasons #

The output curve rises and then falls, which is worth reading carefully because the two sides are different mechanisms rather than one curve with a maximum.

Below the maximum the circuit is too damped to ring. At twenty microhenries $\zeta$ is above one and the current dies before it can carry the capacitor past the supply; the output sits just under the peak exactly as the resistance-only case does, and the inductance is doing nothing but limiting.

Above the maximum the ring is slow compared with the mains. At five millihenries the resonance is at seventy hertz, which is not much above the fifty the supply is being driven at, so the mains reverses before the resonance has finished delivering its energy: the conduction is cut short, the capacitor is left below where the ring would have taken it, and the output falls — at ten millihenries it does not even reach the peak.

That second regime has a name outside this collection. A supply with enough series inductance is a choke-input filter, its output tends towards the mean of the rectified sinusoid rather than its peak, and it is chosen deliberately for exactly the current-limiting reasons this essay measures. The interior maximum is the boundary between the two designs, and it is a resonance frequency compared against a mains frequency rather than an inductance compared against anything.

For this supply the maximum is at about a millihenry, where the ring is at 160 hertz — three times the mains. Which is the useful form of the rule: the overshoot is worst when the leakage rings at a few times the line frequency, and a transformer whose leakage puts the resonance there is the one to check.

Two limiters, priced side by side #

The comparison this essay makes possible is between the two ways of surviving switch-on, and it is worth setting out as a table would if this collection used them.

A series resistance limits the peak to $\hat{V}/R_s$ — over-estimating by four to five per cent, as the rung below measured — and leaves $\int i^2 R\,dt$ where it was. It also stays in the circuit: every ampere the load draws for the rest of the supply’s life goes through it, at $I^2R$ of continuous dissipation and $IR$ of regulation. A third of an ohm on a supply delivering an amp is a third of a volt and a third of a watt, for ever, to solve a problem that lasted ten milliseconds.

A series inductance limits the peak by a factor this figure measures, takes the energy down with it, dissipates nothing, and is invisible at direct current — the load current sees $j\omega L$ at a hundred hertz, which for a millihenry is 0.6 ohms of reactance to the ripple current and nothing at all to the mean.

On those terms the inductance wins every comparison, and the reason it is not the universal answer is the overshoot this essay measures and the size of the component. A millihenry that carries the full load current without saturating is a physical object; a transformer’s leakage is free but is whatever the winding geometry gave it, typically one to three per cent of the primary inductance, and is not a design variable unless the transformer is being specified rather than bought.

Which is the practical reading of the figure: a designer does not usually choose this inductance, but does need to know where on the curve the transformer has put the design — and the answer is a resonance frequency, computable from the leakage and the reservoir with no march at all.

What the march reads, and why it is read at the capacitor #

One measurement decision in this essay is worth stating because the obvious alternative gives eight orders of nonsense.

The other essays in this field rebuild each diode’s current from the junction’s own exponential at the solved node voltages, which is this site’s standing second route and is the right one nearly everywhere. With an inductance in the loop it stops being one at a single instant per cycle. The diode has to go from conducting to fifteen volts of reverse bias inside one time step; the Newton iteration’s guess is clamped to two tenths of a volt per pass and cannot get there in sixty; the step ends with a device voltage that is stale by about a volt, and an exponential turns a stale volt into a hundred megaamperes.

The march itself is fine — the node voltages are consistent, the residual is $5\times10^{-12}$ — and it is the reading that fails. So the current here is read at the capacitor instead: Kirchhoff’s law at the reservoir node says that whatever the diodes deliver goes into the capacitor or into the load, and both of those are computed from the one node voltage the march solves for directly. It shares no arithmetic with the junction’s exponential and it cannot manufacture amperes from a rounding error.

The grid is stated with the result, for the same reason. Below about twenty microhenries the ring is fast enough that seven hundred steps per mains cycle resolve it to ten points a period, and the peak read from such a march moves by ten per cent when the step is halved. The sweep therefore starts at twenty, and the figure says so.

What a soft start is, in these terms #

The arrangement that supplies without either problem is worth naming, because it is what this measurement argues for.

A soft start puts a resistance in the loop for the first few cycles and then shorts it out — with a relay, a triac, or a thermistor whose own resistance falls as it heats. During the first conduction the circuit is the resistance-limited one this essay’s rung below measured: a peak of $\hat{V}/R_s$, an energy of $\tfrac{1}{2}CV^2$ that the resistor takes, and no overshoot at all, because a resistance cannot carry the capacitor past the supply. After a few cycles the resistance leaves and the steady state is the undissipative one.

Read through this essay, that arrangement is buying the absence of the overshoot as much as it is buying the current limit — and it is the only one of the three schemes that gets both. The inductance limits the current and lifts the output; the resistance limits the current and dissipates; the soft start limits the current, dissipates once, and then stops being in the circuit.

The thermistor version has an edge of its own that belongs to the same family of measurements as this essay’s: it works because it is hot, so a supply switched off and on again inside a few seconds meets a thermistor that has not cooled, at a small fraction of its cold resistance, and the first conduction is back to the unlimited one. That is a boundary in time between switch-ons — a quantity nothing else in this collection bounds a model with — and it is measured nowhere here.

What the leakage does to the rest of the supply #

A leakage inductance limits the inrush without dissipating anything and lifts the output above the rectified peak, and both consequences reach further into the field. The first cycle, which no steady state contains is the peak it limits. The direct voltage that is a sawtooth is the steady state it changes, because a lifted output changes the conduction angle as well as the level. The band a turns ratio holds over is where the same leakage is measured as an upper band edge rather than as a current limit. The protection that is gone by the second time is the alternative repair, which does dissipate and does not recover quickly. And The cliff before the fastest settling is the damping argument the lift is an instance of.

What is checked #

The peak is asserted to fall by more than three times across the range of inductance drawn, at every winding resistance the slider offers. The energy is asserted to fall by more than half over the same range, which is the comparison against the rung below — where four times the resistance moved it by two per cent. The output is asserted to have an interior maximum in the inductance rather than a monotone dependence, and that maximum is asserted to be above what the same rectifier reaches with no inductance at all. And for a winding under about four tenths of an ohm the maximum is asserted to be above the peak of the supply, which is the claim the essay is named for.

What is not modelled: the transformer’s magnetising inductance and its own core saturation, which at switch-on can carry a first-cycle current of their own that has nothing to do with the reservoir; the diode’s forward recovery, which lets the first millisecond of current through at a higher voltage than the steady drop; and any load current at all, since the march starts with the capacitor empty and the load resistive. Each is named in the field it belongs to and none is in this netlist.

The twenty-nine per cent, and what it is a symptom of #

The output sitting 29 per cent above the peak of its own supply is the result that most needs explaining, because every steady-state expression in this field says it cannot happen — and the reason it does is the same reason the repair works at all.

An inductance in series with a rectifier makes the conduction interval a resonance rather than a charging event. The inductor and the reservoir capacitor form a series circuit driven by the supply’s half-sinusoid, and a series resonant circuit driven for part of a cycle delivers its capacitor more voltage than the drive has, for the same reason a step into an undamped second-order circuit overshoots to twice its final value — where the behaviour is written down is where that shape is established, with two numbers in the complex plane containing everything a second-order circuit will ever do.

So the boundary this essay crosses is not a modelling error in the steady-state expressions; it is the condition under which they apply. They assume the only reactance in the loop is the reservoir, which makes the conduction interval first order and monotone. The direct voltage that is a sawtooth states that assumption without needing it, because every row of its tables has a peak that is the transformer’s peak less a diode. Add one inductance and the peak stops being an upper bound.

Which is why the repair has to be designed rather than fitted. The peak divided by seven and the energy by five are both good, and the twenty-nine per cent is a permanent change to what the regulator downstream sees — a higher input voltage, more dissipation in the pass device, and a rating on the reservoir capacitor that the unrepaired circuit did not need. The last of those is the one most likely to be missed, because a capacitor’s voltage rating is chosen from the transformer’s peak on a schematic that does not show where the leakage inductance is. The inductance that is a shape is where that quantity is computed rather than assumed — twice the magnetic energy in the window under equal and opposite ampere-turns, a geometry rather than a material, with interleaving worth 3.11 times and not the four it is quoted as — so the repair measured here is sized by a winding decision made for reasons of bandwidth and safety.

And the alternative repair is measured on the rung below. The protection that is gone by the second time prices a thermistor: ten ohms cold, 1.94 ohms once the load current has warmed it, and 198 seconds to recover half of what it started with — against a reservoir that empties in tens of milliseconds. The inductance has no such memory, which is the property that recommends it and the one no efficiency figure reports.