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

Reverse-recovery — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The reverse peak reaches the forward current at 12.6 A/µs. computed by solving, not by drawing. The peak current a diode conducts backwards while its stored charge is removed, against the rate the external circuit drives its current down. The charge equation is integrated in closed form and marched trapezoidally, and the two agree to a part in a thousand. Drawn over it are the two expressions that describe the limits: √(2·IF·a·τ), which is what a reference gives and which is 16.7% high at this slope, and aτ, which contains no forward current at all and is what the peak approaches when the ramp is slow. At 100 A/µs the junction goes on conducting for 48.3 ns and reaches 3.83 A backwards — 3.83 times the current it was carrying forwards — and the two are equal at 12.6 A/µs.

    The diode that conducts backwards

    Every diode in this collection is an instantaneous function of its own voltage, which is exact for an operating point and has no time in it at all. A conducting junction holds a charge, and until that charge is gone it cannot block: drive its current down at a hundred amperes a microsecond and it conducts 3.83 amperes backwards for 48 nanoseconds, against the one ampere it was carrying forwards. The expression every reference gives for that peak is 17 per cent high there, and is right to a per cent only above sixteen thousand amperes a microsecond.

    part 1 · transients
  2. Nine tenths of the heat is in the switch, and above 1.28 MHz there is no temperature at all. computed by solving, not by drawing. The junction temperature a recovering diode settles at, against switching frequency, with the carrier lifetime rising as the 1.8 power of absolute temperature so that the recovery gets worse as the junction gets hotter. Each event costs 15.43 µJ when cold, of which 90% is dissipated in whatever is pulling the current down rather than in the diode: while the junction is still conducting it holds almost no voltage, and the recovered charge is delivered through the switch at the full 100 V. That energy equals the diode's own conduction loss at 29.2 kHz. The fixed point T = Ta + Rth·P(T) is iterated from the ambient upward; the junction passes its rated 150 °C at 904 kHz, and above 1.28 MHz there is no temperature that satisfies it at all — the loop has gain and the solver reports the refusal rather than the last iterate.

    The heat a recovery leaves behind

    The essay below this one measured how much current a diode conducts backwards and for how long, and stopped there. Both numbers are multiplied by a voltage somewhere, and the surprise is where: while the junction is still conducting it holds almost nothing, so nine tenths of the energy is dissipated in the transistor pulling the current down and not in the diode. Repeat it a hundred thousand times a second and it is 1.5 watts, the lifetime rises with temperature, and above 1.28 megahertz the diode's own loop has no fixed point at all.

    part 2 · transients
  3. Two loops on one heatsink give out at 135 kHz, and it is the switch that goes. computed by solving, not by drawing. The junction temperatures of the diode and the switch against switching frequency, with each device's own thermal resistance to a case they share. Each has a positive temperature loop and they are different loops — the diode's runs through its carrier lifetime and its recovery, the switch's through its on-resistance and its conduction — and the electrical coupling goes one way, since the charge the switch has to take at full supply is the diode's. The pair has no settled temperature above 135 kHz and the component that gives out is the switch, which has no exponential in it and is taking 84 per cent of the heat. The same two devices with the same total thermal resistance and no case in common survive to 485 kHz; the diode on its own to 1.28 MHz.

    Two loops, and one heatsink

    The rung below this one found that nine tenths of a reverse recovery's energy is dissipated in the transistor and not in the diode, and then computed the diode's junction temperature with all of that energy in it. Repaired, the diode alone survives to 1.28 megahertz instead of 128 kilohertz — a factor of exactly the ninety per cent. What replaces the number is the arrangement that exists: two devices with two different positive temperature loops on one piece of aluminium, giving out at 135 kilohertz, and it is the switch that goes.

    part 3 · transients
  4. The frequency at which a pulse train becomes an average. computed by solving, not by drawing. The same 5 watts of average dissipation at every frequency, delivered 2 per cent at a time. The flat line is the steady-state answer, which does not know about the frequency. The falling curve is the marched peak junction temperature, which does. They meet at 308 Hz, and that frequency is not a property of the converter: it is a fraction of one junction time constant per period — f·τ = 0.738 at this duty, with τ = 2.40 ms, and between 0.78 and 0.56 across the duties on the slider. A hundred-kilohertz converter fits 240 periods inside that time constant, and at the top of the sweep — 10.0 kHz — the steady state is already exact to 0.46 per cent, so the averaged-power fixed point is right and this is the measurement that says why. The march puts 48 steps inside each pulse, which is what the answer is sensitive to: at six it put the boundary 19 per cent too high.

    The pulse the heatsink does not feel

    A thermal resistance iterated to a fixed point with a diode or a switch is a statement about a power — so it assumes that a hundred and fifty watts for two per cent of the time is three watts. The die's own heat capacity decides whether that is true, and it decides it at a frequency: above 308 hertz the junction integrates, by a hundred kilohertz the fixed point is exact to five parts in ten thousand, and at one hertz the same average power on the same heatsink puts the junction three hundred kelvin hotter.

    part 4 · transients
  5. The ladder's step response, and the sum of its own stages — 0.95 per cent apart at worst. computed by solving, not by drawing. A step of power into a three-stage thermal ladder, and the junction's rise divided by it. The solid curve is exact: the impedance is a continued fraction in s, its denominator has 3 real negative roots, and the partial-fraction expansion of Z(s)/s is a sum of that many ordinary exponentials — no march, no step size. The dashed curve is the sum every account of a thermal path writes, each stage's own resistance times 1 − exp(−t/RC) with its own local time constant, and it is an approximation because the stages load each other. What that costs is 0.950 per cent, once, at 12.9 ms — between the fastest stage's 2.4 ms and the next one's 200 ms, which is the only place two stages are moving together. It is one-sided: the sum never reads low.

    Two ladders the terminals cannot tell apart

    A thermal path drawn as a ladder and the same path drawn as a sum of exponentials are called different models of one object, and the difference between them has never been priced because pricing it needs an exact answer. Solved in closed form, the sum is 0.950 per cent high at worst and never low; the marched netlist is right to a part in 21,169; and the largest disagreement in the picture was 2.919 per cent that has nothing to do with heat at all, which reading the curve one sample differently removes.

    part 5 · transients

All series