The first cycle, which no steady state contains

This field’s account of an unregulated supply is a settled one. A transformer, four diodes, a reservoir capacitor and a load; a ripple whose textbook expression turns out to be a first-order approximation with a stated error; a conduction angle of twenty-nine degrees measured rather than assumed; a crest factor of 13.4, which is what makes the power factor of such a supply so poor. Every one of those numbers is read from a march after it has settled, and the march always starts from an empty capacitor.

So the first cycle is computed on every run and discarded on every run. It is the largest current the circuit will ever carry — larger than the repetitive peak by more than an order of magnitude — and it is the number that decides the diode, the fuse, the switch and the transformer.

It is also, uniquely among the quantities in this field, not a property of the design. It depends on the phase of the mains at the instant of switching, which nothing in the circuit chooses, and the difference between the best instant and the worst is a factor of seven.

Why it is so large, in one sentence #

An empty capacitor is a short circuit. At the instant of switching there is nothing in the circuit to limit the current except the transformer’s winding resistance and the diode’s forward drop, so the current is approximately the instantaneous supply voltage divided by the winding resistance — and the instantaneous supply voltage can be anything up to the peak.

That is the whole mechanism, and it explains both the size and the phase dependence. Close the switch at a zero crossing and the voltage rises from nothing while the capacitor is already charging, so the two effects fight each other and the peak is modest: 4.64 amperes here. Close it at the peak and the full 17 volts appears across half an ohm with an empty capacitor on the other side: 32.4 amperes.

In the settled state none of that applies, because the capacitor is nearly charged. The diode conducts only while the supply exceeds the capacitor’s voltage, which for these values is twenty-nine degrees of each half cycle, and the current that flows is what replaces the charge the load took: 1.23 amperes at the peak.

26.3 times. That is the ratio the first conduction has to the largest current the settled circuit ever draws, and no expression in the settled analysis contains it.

The estimate that is four per cent high, at every resistance #

The obvious estimate is the one the mechanism suggests: the capacitor is a short, so the peak is $\hat V/R_s$. With 17 volts and half an ohm that is 34.0 amperes against a marched 32.4.

The interesting part is not that it is close. It is that it is close by the same amount at every resistance: 4.87 per cent high at 0.2 ohms, 4.78 at 0.35, 4.73 at 0.5, 4.66 at 0.8, 4.60 at 1.2, 4.52 at 2 and 4.44 at 3.5. A factor of seventeen in the resistance and a factor of seventeen in the current, and the estimate’s error moves by less than half a per cent.

What the estimate leaves out is the charge the capacitor takes during the pulse, which lifts its voltage while the current is flowing and so reduces the peak. That correction is a fraction of the supply, and the fraction depends on how much charge the pulse moves relative to the capacitor’s size rather than on the resistance — which is why it is nearly constant here and why it would not be if the capacitor were smaller.

So the estimate is usable with its error stated, which is the most this collection asks of an approximation: four to five per cent optimistic, always in the same direction, over an order of magnitude in the quantity that matters.

The energy does not care about the resistance #

There is a second quantity in the first cycle, and it behaves in a way the first one does not.

Integrating $i^2R_s$ over the first conduction gives the energy the winding resistance dissipates while the capacitor charges. At a fifth of an ohm it is 0.134 joules. At half an ohm — where the peak current is two and a half times smaller — it is 0.131 joules. Two per cent apart, for a factor of four in the resistance.

That is the charging-energy result, arriving in a place it was not being looked for. Charging a capacitance $C$ to a voltage $V$ through any resistance dissipates $\tfrac12CV^2$ in the resistance, regardless of its value, because a smaller resistance passes a larger current for a proportionally shorter time. Here $\tfrac12CV^2$ is 0.1445 joules and the measured figures are eight per cent below it, the difference being the load current drawn during the charge and the diodes’ forward drops.

The practical consequence is worth stating because it cuts against the obvious reading. Adding resistance to limit the inrush does not reduce the energy the first charge dissipates; it only spreads it over a longer time and moves some of it from the diodes to the added resistor. What it reduces is the peak, which is what the semiconductor and the fuse care about, and the two are different specifications with different limits.

The charge is the same however it arrives #

One quantity in the first cycle does not move at all, and it is worth putting beside the two that do.

The capacitor has to acquire $CV$ of charge before it can hold the peak, and $CV$ here is 17 millicoulombs. Integrating the marched diode current over the first conduction gives 17.9 millicoulombs at a fifth of an ohm and 17.3 at a half — the excess over $CV$ being what the load drew while the charging was going on.

So there are three quantities in the first cycle and they behave in three different ways as the winding resistance moves. The peak is inversely proportional to it. The energy is nearly independent of it. The charge is exactly fixed by the capacitor and the voltage and has nothing to do with it at all.

A designer who thinks in any one of the three and reasons about the others from it will get two of them wrong. That is not a subtle trap: “limiting the inrush” sounds like it should reduce all three, and it reduces one.

Where the model stops #

Three things in this march are worth naming as absent, because each would move the numbers and none of them is in the netlist.

The transformer is a source and a resistance. A real one has leakage inductance, which limits the rate the current can rise and therefore reduces the peak by an amount this model does not compute; the magnetics field has the machinery for it and this circuit does not use it.

The diodes have no reverse recovery and no thermal model. Their forward drop is an exponential solved at every step, which is right for the current, and their junction temperature is not modelled at all — so the surge argument above is about a rating rather than about a computed temperature.

And the switch is ideal. A real one bounces, which means the first conduction happens several times over a few milliseconds with the capacitor part-charged at each attempt, and the worst of those is not the first.

Each of the three makes the real answer smaller than the marched one except the last, which makes it harder to reason about. That is the honest direction to leave an estimate in.

What a fuse and a diode are specified in #

Neither a fuse nor a rectifier is specified in amperes for this purpose, and the reason is exactly the energy argument above.

A fuse’s element melts when enough energy has gone into it, and for a pulse short compared with its thermal time constant nothing is conducted away, so what matters is $\int i^2dt$ — a quantity in ampere-squared-seconds that the industry writes as $I^2t$ and that has nothing in it about the shape of the pulse. For the worst instant here that integral is 0.266 ampere-squared-seconds, which is what a fuse has to survive and a load current of 0.16 amperes has to blow.

A rectifier’s surge rating is the same idea with a different constant: a single half-cycle at some tens of times its continuous rating, because the junction’s own thermal mass carries the heat for the ten milliseconds it takes. That is why a one-ampere diode is untroubled by 32 amperes for four milliseconds and destroyed by 5 amperes for a second.

So the design question is not “is the peak under the rating” but “is the integral under the rating for a pulse of this length”, and the second question needs the marched waveform rather than a peak.

What is actually done about it #

Three repairs are used and each is a different answer to which quantity is being bounded.

More winding resistance, which is what the slider on the figure moves. Holding the first peak to ten times the repetitive one needs 1.90 ohms here, against the 0.5 a transformer of this size actually has. Those 1.90 ohms would drop 2.3 volts at the repetitive peak and dissipate continuously, so this is a repair that costs regulation and efficiency for the whole life of the supply to bound one event.

A negative-temperature-coefficient thermistor in series, which is the same repair with the cost removed after the first few seconds: cold it is several ohms, and the inrush itself heats it to a fraction of one. Its failure mode is the interesting one — it does not cool down between a switch-off and a switch-on a second later, so the repair is absent exactly when somebody power-cycles a supply, which is the most common way of testing one.

A relay across a resistor, closed a hundred milliseconds after the supply comes up, which bounds the peak and costs nothing continuously and needs a circuit to decide when to close it.

The first is measured here; the second and third are named and not built, because both need a temperature or a timer that this collection’s netlist has no element for.

Why the first cycle was there all along #

The last thing worth saying is about the measurement rather than the circuit.

Nothing new was computed for this essay’s central number. The march that produces the ripple and the conduction angle starts at $t = 0$ with an empty capacitor and runs for twelve cycles; every figure in this field reads its answers from the last of those; and the first was in the trace the whole time. What was needed was a winding resistance — without one the first conduction is limited by nothing but an exponential and the model has no opinion worth having — and the decision to read the head of the array rather than its tail.

That is the same shape as the power field’s own origin: the energy check every solve on this site performs had been computing real power on every solve since the first commit and discarding it, and the field exists because somebody read it. A number that a machine computes and throws away is cheaper to find than a number that has to be derived, and it is much easier to miss.

What the first cycle costs elsewhere #

An inrush is the one measurement in this field that a steady-state model cannot contain at all, and it decides three things measured elsewhere. The half that never arrives is the energy the reservoir takes whatever the resistance, which is what the first cycle delivers all at once. The direct voltage that is a sawtooth is the steady state that follows, with a conduction angle no ripple specification contains. The inductance that limits, and lifts is the repair that costs nothing in dissipation and lifts the output above the peak. The protection that is gone by the second time is the limiter that survives one switch-on and not two, and The current that does no work is the steady-state current the same transformer carries.

What is checked #

Five assertions, and the second is the one that says the phase dependence is real rather than an artefact of where the march happens to start.

That the first conduction carries many times the largest current the settled circuit draws, at every winding resistance the slider offers.

That the worst instant is the peak of the mains, at every one of them — swept over nineteen phases rather than assumed, because a march that started at a fixed phase would have reported one number and called it the answer.

That the best and worst instants differ by more than half again, so that the phase is a variable rather than a detail.

That the capacitor-as-a-short estimate is an over-estimate by between four and five per cent, at every resistance — a bracket, because the whole point is that the error barely moves.

And that holding the first peak to ten times the repetitive one is a stated resistance, solved on the same march, so that the cost of the repair is a number beside the number it repairs.

The three repairs and what each keeps #

That resistance is one of three ways of holding the first cycle down, and the rest of this field measures the other two — with the result that none of them is simply better.

A series resistance is what this essay computes, and the inductance that limits, and lifts is where its limitation is measured: it limits the first peak and does not reduce the energy at all, with $\int i^2 dt$ two per cent apart over a factor of four in the resistance. So the semiconductor’s own survival — which is an energy question rather than a current one — is barely improved by it.

A thermistor is the same resistance arranged to disappear once the supply has started, and the protection that is gone by the second time measures both halves: ten ohms cold and 1.94 ohms once the load current has warmed it, which is the design working, and 198 seconds to recover half its cold resistance against a reservoir that empties in tens of milliseconds — which is the design absent for exactly the event it was bought for.

Leakage inductance divides the peak by seven and the energy by five and dissipates nothing to do it, which makes it the only one of the three that improves the quantity the parts are chosen on. What it charges 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 — so the repair changes the circuit’s ordinary operating point rather than only its first cycle.

Which is the useful way to hold the three. The first cycle is a current, an energy and a duration at once, and each repair addresses a different one of the three.

Thrown away, in the essay that threw it away #

Every number the steady-state analysis of this circuit produces — the ripple, the crest factor, the conduction angle, the power factor — comes from a march that starts with an empty capacitor and discards its first cycles. The direct voltage that is a sawtooth says so explicitly: twelve cycles marched at fifty hertz, with the last of them reported. That is the correct procedure for measuring a settled state and it is a decision to discard the largest current in the record.

The discarding is not incidental to the method; it is the method. The amplitude nothing linear predicts records the opposite failure on a different circuit — a run stopped in the middle of its growth reporting 51 millivolts against a settled 703 — and the repair there was a run length computed from the growth rate and a flag on every result that had not settled. A settled-state measurement has to run long enough to settle, and the price of running long enough is that the transient scrolls off the front of the record unless somebody asks for it.

Which is what makes this rung’s result the kind worth writing down. It is not a correction to the steady-state numbers, all of which stand; it is a quantity that the procedure producing them is constructed to remove, and that decides the rating of every part in the circuit.