The pulse the heatsink does not feel

The rung below this one gave a diode and the switch that drives it a thermal model each, put them on a shared case, and iterated two positive temperature loops to a fixed point — finding that the pair gives out at 135 kilohertz and that it is the switch that goes. Every number in it came from a thermal resistance: so many kelvin per watt, junction to case, case to sink, sink to ambient.

A resistance is a statement about a steady state, and a steady state is a statement about a power. So that rung took the energy each recovery costs, multiplied it by the switching frequency to get watts, and asked what temperature those watts produce. Doing that assumes something it never said: that a hundred and fifty watts for two per cent of the time is the same as three watts all the time.

The junction’s own heat capacity decides whether that is true, and this essay is about where it stops being true. The answer is a frequency, it is 308 hertz, and it is not a property of the converter.

The thermal path is a netlist, because it is one #

Kelvin per watt obeys Ohm’s law and joules per kelvin obeys a capacitor’s law. Temperature is the across quantity, power is the through quantity, and the two constitutive relations are the same two relations this site’s solver has been assembling since its first commit.

So the path from a die to the air is a ladder: 1.2 K/W from junction to case with 2 millijoules per kelvin of silicon behind it, half a kelvin per watt across the grease with 0.4 J/K of copper, eight kelvin per watt from sink to air with 60 J/K of aluminium. A direct-current solve gives the fixed point the rung below iterated to; a march gives everything before it, with Kirchhoff’s current law checked at every step — which, in this network, is conservation of energy.

The three stages differ by three orders of magnitude in time constant and that is the whole of this rung:

A converter switching at a hundred kilohertz delivers ten thousand pulses inside the fastest of those. A start-up surge delivers one.

The march has to start from the steady state #

There is a practical obstacle in the way of the measurement and it is worth stating, because a first version of this walked into it.

The heatsink’s time constant is eight minutes and the switching period is ten microseconds. Marching from cold to the steady state would be fifty million steps to answer a question about one cycle, and a first version that marched twelve cycles from cold reported a mean junction rise of 1.9 kelvin against a steady state of 194 — it was measuring how far the heatsink had got, which was nowhere.

The fix is exact rather than approximate: the steady state is a direct-current solve of the same netlist with the average power in it, and the march starts from there. Same network, one solve, and the periodic answer arrives after five or six cycles. This is the same manoeuvre the flux-walking essay could not use — there the whole point was that there is no steady state to start from — and it is available here because there is one.

At one hertz the fixed point is three hundred kelvin wrong #

Five watts of average dissipation delivered as 250 watts for two per cent of each cycle, at one hertz.

The steady-state answer is 48.5 kelvin above ambient: five watts times 9.7 K/W, and it is the number the rung below would quote. The marched junction reaches 349.0 kelvin above ambient. The mean over the cycle is 48.55 — the average is right to a part in a thousand, exactly as conservation of energy requires — and the peak is seven times it.

The case, one stage down, moves by 11.5 kelvin over the same cycle. The sink moves by 39 millikelvin. So a case-temperature measurement, which is the measurement anybody can actually make, reports essentially the steady-state answer while the junction is three hundred kelvin above it. That is not a small discrepancy in a quantity; it is a measurement that cannot see the failure it is being used to rule out.

Slide the frequency up and the excursion collapses.

The boundary is a fraction of a time constant per period #

The two curves meet — the peak within a tenth of the steady state — at 308 hertz, and the interesting thing about that number is that it contains nothing about the converter.

The junction’s time constant is 2.4 ms. At 308 Hz the period is 3.25 ms, so f·τ = 0.738: the crossover is at about three quarters of a junction time constant per period. The duty cycle moves it, and by a measurable amount rather than a negligible one — f·τ is 0.780 at half a per cent, 0.746 at one, 0.738 at two, 0.710 at five and 0.557 at twenty, with the crossing itself walking from 325 hertz to 232. A sparser pulse delivers its energy in a shorter burst, so the junction gets a little further up its own exponential before the pulse ends and needs a little more integrating to hide it.

What is invariant is not the fraction but what the fraction is of. Across a factor of forty in duty the boundary never leaves the band between a half and one time constant per period, and nothing in it is the converter’s — not the peak power, which changes fortyfold across the same slider, and not the average, which is fixed. That is a boundary stated in the die’s own units, which is the shape this collection prefers — the same shape as the amplitude boundary that is a fixed number of thermal voltages and the frequency boundary that is a fixed fraction of a self-resonance, and it is a weaker statement than “one time constant per period” in exactly the way a measured boundary is weaker than an assumed one.

Above it, the die integrates. At 100 kilohertz the junction’s ripple is 24.5 millikelvin on a 48.5 kelvin rise: the steady-state model is right to five parts in ten thousand.

So this rung’s finding is that the rung below is correct, and that is a rarer result than a refutation. The three rungs below computed a runaway frequency of 135 kilohertz from a fixed point, and a fixed point is only the right object if the junction cannot follow the individual pulses. It cannot: at 135 kHz it is three hundred and twenty-four periods inside its own time constant, and the ripple is 18 millikelvin — four parts in ten thousand of the rise it sits on. The model was right and it was right for a reason nobody had checked.

Where the boundary bites, and it is not the converter #

A hundred-kilohertz converter is nowhere near this boundary in normal operation. Four things put a circuit on the wrong side of it.

Mains-frequency switching. A thyristor phase controller, a soft-start relay, a crowbar: these switch at fifty hertz or once, and the junction sees the pulse.

Burst mode. A converter that skips cycles at light load delivers packets of switching at a repetition rate of a few hundred hertz to a few kilohertz. The average power is small, which is why burst mode exists, and the packet rate can be exactly on this boundary.

Fault current. The interesting case, because it is the one the fixed point is being used to rule out. A short circuit puts fifty times the normal current through the switch for a few hundred microseconds before the protection acts — a condition the switching-energy essay prices in joules and this one has to price in kelvin, and the question is not what steady state that implies — there is no steady state, the protection acted — but how hot the junction got. That question is answered by the single-pulse curve rather than by a resistance.

And the inrush. The first cycle of a rectifier’s charging current is one event, at mains frequency, and the diode that survives its steady-state dissipation with eighty per cent margin may not survive its first cycle at all.

One pulse, from cold #

The single-pulse curve is the whole path’s step response, and it is three shoulders rather than one exponential: the die fills in a few milliseconds, the case in a fifth of a second, the sink in eight minutes.

Reading a junction temperature off the 9.7 K/W total is therefore right only for pulses longer than the last of those. At one millisecond the transient impedance is 0.409 K/W — 23.7 times less. So a hundred watts for a millisecond raises the junction by forty-one kelvin, not by nine hundred and seventy, and a part that would be destroyed on the resistance reading survives comfortably.

The curve above passes through 0.488 at the nearest length it reports, and the difference from 0.409 is the grid rather than the heat. Sixty pulse lengths spread logarithmically over six and a half decades sit 29 per cent apart in time, so the one nearest a millisecond is at 1.25 milliseconds, and a value lifted off a plotted curve at a round number is the length after it. On a curve climbing this steeply that is a fifth of an impedance, which is worth more than the difference between the two models drawn in the picture — which is why the number quoted above is computed at a millisecond instead.

That is the direction that matters commercially: the transient impedance is what lets a switch be rated for a fault current forty times its continuous one. It is also the direction that makes the resistance reading useless rather than merely conservative — a design that used it would specify a switch twenty-four times too large.

The two routes in this figure are not equal. The dashed curve is the marched netlist, which agrees with an exact expansion of the same ladder to a part in 6,584 at every length both reach. The solid curve is the Foster sum — each stage’s resistance times $1 - e^{-t/RC}$ with its own local time constant — and it is an approximation, because a Cauer ladder’s stages load each other: the fast one charges into a case that is itself moving. They agree at the two ends and part company by 0.94 per cent where the die and the case are moving together, which is sixty-two times the distance between the march and the truth — the ratio that says which of the two curves is the approximation.

That 0.94 is itself a number read off a grid. The true worst departure of the sum is 0.950 per cent at 12.9 milliseconds; the sixty lengths drawn here straddle it, the nearest sitting at 12.2, and the curve compared against is the march rather than the exact expansion — two grids and one substituted route, for 1.2 per cent of the answer. All three are separated in the rung above, where the disagreements visible in a picture of this kind are priced against a closed form with no step size in it.

A first version of that closed form built each stage’s time constant from every capacitance below it as well. That is right for the slowest stage and it gives the fastest one a time constant of 72 seconds — it reported a thermal impedance of 5×10⁻⁵ K/W at a millisecond, nearly four orders below the marched answer, and looked entirely plausible on a logarithmic axis. It was caught by comparing the two routes, which is the only thing that catches it.

A boundary that is a duration rather than a frequency #

There is a second way to state the same thing and it is the more useful one for a fault.

A pulse train has a frequency; a single event has a duration. The two are the same measurement read against the same 2.4 milliseconds, and the single-pulse form is the one a protection designer needs: how long may a given overload last? The transient impedance answers it directly. A junction that may reach 175 °C from a 60 °C case has 115 kelvin to spend; against the 0.409 K/W of a millisecond that is 281 watts, against the 0.0490 of a hundred microseconds it is 2,348, and against the 0.00499 of ten microseconds it is 23,048 — where the steady-state resistance would have allowed twelve.

Two things follow that are not obvious from the curve.

The first is that the allowance is not proportional to the pulse length, because the impedance is not proportional to it either. Over the decade from a hundred microseconds to a millisecond the impedance rises by 8.35 rather than by ten, so shortening a fault by a factor of ten buys 8.35 times the power and not ten — which is the practical content of the die’s heat capacity being finite rather than infinite.

The second is that repeating the pulse changes the question. A single event of a millisecond is governed by the fast stage alone; the same event once a second is governed by the fast stage sitting on whatever average the slow stages have reached, which is the picture the first figure in this essay draws. Those are different calculations, and a data sheet that prints a single-pulse curve and a continuous rating has given the two ends of a family and none of the middle.

What the peak-to-average ratio does not do #

There is a tempting shortcut here and it is wrong in an instructive way.

If the junction followed the pulses perfectly, the peak temperature would be the peak power times the thermal resistance — 250 watts times 9.7 K/W, or 2,425 kelvin. If it followed them not at all, the peak would be the steady state, 48.5. The marched answer at one hertz is 349.0, which is neither, and the reason is that the three stages respond differently: the die follows the pulse almost completely, the case follows the cycle average, and the sink follows nothing.

So the peak temperature is the pulse acting on the fast stage, sitting on top of the average acting on the slow ones. There is no single number between the two limits that a peak-to-average ratio can be multiplied by, and a duty-cycle correction applied to the total resistance gets it wrong in both directions depending on where the frequency sits relative to each stage. The right object is the whole curve, which is why data sheets print one.

Two things this does not model #

Nothing here is nonlinear. A real thermal path is: the sink-to-ambient resistance under natural convection falls with temperature, and the die’s own heat capacity rises with it. Both corrections are of order ten per cent over the range these numbers cover, both point the same way — a hot part is slightly better at getting rid of heat than this model says — and neither changes a boundary that is a ratio of two time constants.

And the dissipation is a rectangle. A real recovery event is a current and a voltage overlapping for a few tens of nanoseconds with a shape the rung two below computes in detail. At any frequency in this essay’s range that shape is far shorter than the junction’s own time constant, so the die sees its integral and not its profile — which is the same argument as the one above, applied at the other end, and is why a rectangle is the right idealisation here and would not be for a die with a microsecond time constant.

There is a real limit hiding in that. The 2.4-millisecond figure is for a whole die to a whole case. The hot spot on a die — the few square microns where the current crowds — has a time constant of microseconds, and at that scale a recovery pulse is not integrated at all. A one-stage model of the junction is exactly the mistake this essay is about, made one level down, and measuring it needs a model of the die’s interior that this collection does not have.

What the gate checks #

The current-law residual is asserted at every step of every march, which for this network is conservation of energy and is what would catch a heat capacity stamped with the wrong sign.

The mean of the marched cycle is required to equal the direct-current solve’s steady state at every frequency on the sweep. That is not a tautology — it is what says the march is periodic rather than still settling — and it is the check that would fail if the run were too short.

The peak is asserted to fall monotonically with frequency, and to be within two per cent of the steady state at the top of the range. Both, rather than one: a model with no heat capacity at all would pass the second and a model with the stages transposed would pass the first.

The crossing is asserted against a half to one junction time constant per period rather than against 308 hertz, so the claim moves with the die and holds at every duty cycle on the slider rather than at the one it was written at. It is a band rather than a value because the duty moves the crossing inside it, and stating it as a value is how the earlier version of this essay came to assert that the duty moved it hardly at all.

And the march is required to put at least twenty steps inside the pulse itself, which is the check that would have caught what the earlier version reported. Marching uniformly across a period puts six steps inside a two-per-cent pulse, and the peak is reached at the pulse’s end — so the only part of the cycle the figure is about was the part being resolved worst. Every temperature it printed looked plausible and the boundary came out 19 per cent high, at f·τ = 0.91 rather than 0.74, which is close enough to one to have read as a result. Nothing failed, because the march was solving accurately the problem it had been handed.

And the single-pulse curve is required to rise monotonically and to be still climbing at five minutes, which is the heatsink filling and is the statement that no single resistance describes the path.

What this changes about the rung below #

Nothing, and that is the result.

The two-body thermal model and its runaway stand exactly as computed, at every frequency at which anybody switches. The 135-kilohertz limit it found, the finding that it is the switch rather than the diode that fails, and the factor of two the shared case costs are all statements about a steady state, and the steady state is what the junction is in.

What has changed is that the assumption is now a measurement with a boundary on it, and the boundary turns out to be somewhere no converter goes — but somewhere a fault, an inrush and a burst-mode controller all do. The model was right; it was right for a reason, and the reason is a 2.4-millisecond time constant that appears in none of the three rungs below.