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

Switching loss — where it appears

The energy a device dissipates in the transitions between its on and off states, which is paid once per event and becomes a power by repetition. For a diode's recovery most of it is not in the diode: the recovered charge is delivered through the switch at the full supply.

Named by 8 essays across one field — each of them below, with the objects they name alongside it.

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.

transients · Reverse recovery 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.

transients · Reverse recovery 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

Nine tenths of a reverse recovery's energy is dissipated in the transistor and not in the diode, and the obvious calculation of the diode's junction temperature puts all of that energy in it anyway. 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.

transients · Reverse recovery 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.

transients · Reverse recovery A staircase costs one Nth, computed rather than quoted. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. The charge is broken into N equal risers, each held for 16 time constants so that it completes. The measured losses are 1.00000, 0.500001, 0.250000, 0.125000, 0.0625000, 0.0312500 of ½CV² — which is 1.000004, 1.000002, 1.000001, 1.000001, 1.000000, 1.000000 times 1/N, so the law is exact to four parts in a million at the worst rather than approximately true. The fitted exponent is -1.00000 and the energy account closes to 2.17e-6 at the worst.

The half a switch keeps

Charging a capacitor from a step loses half the delivered energy whatever the resistance, and a ramp takes the loss down as 2τ/T with no floor. A staircase of N settled risers costs one Nth of the step, exact to four parts in a million, and the law ends at a dwell of 5.272 time constants. A switch is the other half of the same product and buys nothing at all: with the supply held at five volts and the channel conductance ramped over a thousand time constants, the loss is 1.00000000 of ½CV².

transients · Switching energy A capacitor charged from 5 V through 1 mH, with and without a diode, at a circuit Q of 10. computed by solving, not by drawing, from the netlist's poles. Without a diode the capacitor rings about 5 V and settles there, and the resistance takes 1.00000 of ½CV² — exactly what a resistor alone would take. With an ideal diode in series the current stops at its first zero, 99.5 µs in, and the capacitor is left at 9.272 V, which is V(1 + d) with d = 0.8545 the step's overshoot. That charge cost 0.2699 of ½CV².

The swing a diode keeps

Charging a capacitor through a resistor loses half the energy whatever the resistance, and putting an inductor in the path changes nothing: twenty-four settled charges from 10 µH to 10 mH and Q from 0.3 to 100 lose ½CV² to 1.4 parts in a hundred million. A diode that stops the current at its first zero changes everything. The capacitor keeps the overshoot, V(1 + d), the loss falls to ½CV²(1 − d²), and a charge of duration T costs π²/8 of what a ramp of the same duration costs.

transients · Switching energy A 1 µF capacitor at 5 V joined through 100 Ω to an empty one of the same size. computed by solving, not by drawing, from the netlist's poles. Charge is conserved and both settle at 2.500 V. Energy is not: the pair holds 50.0% of the 12.50 µJ the first one started with, and the resistor has taken 50.00% — C2/(C1 + C2), with no resistance in it.

The energy two capacitors cannot share

Join a charged capacitor to an empty one and the charge is shared exactly while the energy is not: the pair loses C₂/(C₁ + C₂) of what the first one held, through 1 Ω or 10 kΩ, with an inductor ringing for hundreds of cycles or without one. An inductor and a diode turn the sharing into an exchange — equal capacitors swap 96.9% of the energy at a loop Q of 50. And a charge pump is the sharing loss billed once a cycle: a doubler's efficiency is its output voltage over twice its input, to four parts in ten million, at every load and every clock.

transients · Switching energy At 1.00 A forward the silicon-carbide Schottky hands the switch more charge than the silicon PN diode above 245 V. computed by solving, not by drawing. The charge each diode returns through the switch when it is forced to block, against the reverse voltage, at 1.00 A forward and 100 A/µs. The silicon-carbide Schottky's is its depletion capacitance charged to the rail, closed form checked against Simpson's rule: 11.8 nC at 100 V, 25.0 nC at 400 V, 30.9 nC at 600 V, the same at any current. The ultrafast silicon diode's is its stored minority charge, 18.1 nC at this current from the recovery's closed form, plus 1.59 nC of its own capacitance at 400 V. The two are equal at 245.3 V.

The recovery a Schottky does have

A Schottky diode stores no minority charge and is quoted as having no reverse recovery. It has a depletion capacitance instead, and to block it must be charged to the rail through the switch: a 650 V silicon-carbide part holds 25.0 nC at 400 V whatever current it was carrying. An ultrafast silicon diode with a 20 ns lifetime returns 19.7 nC at 1 A, so below 1.42 A forward the diode said to have no recovery hands the switch more charge, and at 125 °C the boundary falls to 632 mA. In energy the capacitive charge costs 64.9 per cent of V·Q rather than all of it, and the boundary is 799 mA.

transients · Reverse recovery

Named alongside it

The objects these essays reach for when they reach for this one.

Reverse recoveryCarrier lifetimeEnergy-storageModel rangeCharging exponentialClosed formDesign tradeoffReal powerStored chargeThermal runawayMarchingOn-resistance

All concepts