Two loops, and one heatsink

The rung below this one is the essay that found the recovery loss in the wrong component. While the junction is still conducting it holds almost no voltage, so the diode dissipates almost nothing during the recovery itself; the charge it hands back is delivered through the transistor pulling the current down, which at that instant is standing across the whole supply. Nine tenths of the energy is the switch’s.

It then computed the diode’s junction temperature — and gave the diode all of it.

That is the first thing this rung had to do, and the correction is a factor of ten in every frequency the rung below quotes. What replaces those numbers is not a corrected single-body model but a different model, because the switch is not a heatsink with a fixed temperature either. It is a second device with a positive temperature loop of its own, on the same piece of aluminium.

The correction, and why it is exactly a factor of ten #

The thermal loop in the rung below is $T = T_a + R_\mathrm{th}P(T)$, iterated from the ambient upward, with $P$ the power dissipated in the diode. What was passed as $P$ was pTotal — the recovery’s whole energy times the frequency, plus the conduction loss — and the recovery’s whole energy is $V\!\cdot\!Q_{rr}$ plus the diode’s own sixth of $I_{RM}Vt_b$, of which the first term is the switch’s.

So the diode’s junction was being heated by the transistor’s dissipation, on the rung whose entire finding is that the transistor has it.

Corrected — the diode charged with $f\!\cdot\!E_\mathrm{diode}$ and its conduction, and nothing else — the same device on the same 40 K/W of thermal resistance settles at:

The whole curve has moved out by a decade, and it has moved by a decade rather than by some other amount because the recovery term dominates the conduction term and the recovery term has been divided by ten. The frequency at which the loop stops converging goes from 128 kilohertz to 1.28 megahertz; the rated 150 °C is passed at 904 kilohertz instead of 90.4.

Two things worth saying about a repair of that size.

Nothing failed. Every assertion in the figure and in the site’s gate went on passing: the switch’s share was asserted, the monotonicity was asserted, the existence of a runaway frequency was asserted, and the number was in prose. A quantity that only appears in prose is a quantity nothing is checking.

The rung below’s headline is unaffected. Ninety per cent of the energy is still not in the diode; that was measured correctly and separately. What was wrong is what happened to the ninety per cent afterwards, which is the subject of this rung.

Two loops, and they are not the same loop #

The switch has a positive temperature loop of its own and it runs through a different quantity.

A silicon field-effect transistor’s on-resistance rises with temperature — roughly as the 2.3 power of absolute temperature, because carrier mobility falls — so its conduction loss climbs with its own junction temperature. That is a loop with gain in it exactly as the diode’s is, and it has nothing to do with stored charge.

The diode’s loop runs through its carrier lifetime and its recovery. Hotter lattice, slower recombination, longer lifetime, more stored charge, a larger reverse peak, more energy per event.

The two are coupled, and the coupling is asymmetric in a way that decides the answer. The energy the switch takes at each recovery is $V\!\cdot\!Q_{rr}$, and $Q_{rr}$ is the diode’s stored charge — so the diode’s temperature is an input to the switch’s power. The switch’s temperature is an input to nothing of the diode’s. Electrically the heat runs one way.

Thermally it runs both ways, through the case they share:

$$T_\mathrm{case} = T_a + R_\mathrm{case}(P_d + P_s), \qquad T_d = T_\mathrm{case} + R_{jd}P_d, \qquad T_s = T_\mathrm{case} + R_{js}P_s$$

which is the ordinary two-body model and is written out because the shared term is the whole point. At $R_\mathrm{case} = 0$ they are two devices that happen to be in the same circuit. At 35 K/W they are two dice on one tab, and each one’s heat is most of the other’s ambient.

What the pair does #

Iterating both temperatures together from the ambient upward:

The share rising from 38 to 80 per cent is the recovery loss growing with frequency while the conduction losses do not. Below about thirty kilohertz the diode is the hotter of the two and the heat is mostly conduction; above it the switch is, and the heat is mostly recovery.

And the pair has no settled temperature above 135 kilohertz, where the component that gives out is the switch.

That is the finding. The device with the exponential in it, the one whose stored charge is the subject of the whole anchor, is not the one that fails. The one that fails is the one taking the energy — and it has no exponential, no stored charge, and no part in the mechanism except that it is where the charge is delivered.

What sharing a case costs #

The comparison that makes the coupling a measurement rather than a modelling choice is against the same two devices with the same total thermal resistance to ambient and no case in common — 40 K/W for the diode, 37 for the switch, exactly what each has now.

Apart, they run away at 485 kilohertz. Together, at 135.

A factor of 3.6, from an arrangement in which nothing about either device has changed and no thermal resistance has been added. What has changed is that each one’s heat now arrives at the other’s ambient, so each positive loop is driving the other. Two loops that are each stable become one loop that is not.

The factor is not a coincidence of one design either. Across the shared resistance on the slider:

Between 2.8 and 3.8 times, at every setting, and always in the same direction.

Which component to fix, which is not the one that fails #

The natural reading of “the switch gives out first” is that the switch is the part to improve, and the measurement says otherwise. Perturbing one parameter at a time from the design above and re-bisecting the runaway frequency:

Halving the diode’s stored charge doubles the frequency the pair survives to. Halving the switch’s own on-resistance — which is the whole of the switch’s own loop, and what a designer buying a better transistor is buying — moves it by four per cent.

The reason is in the asymmetry established above. The switch’s power is mostly the diode’s charge delivered at the supply, and only a little of it is the switch’s own conduction. So the loop that actually has the gain runs: diode hotter, more charge, more energy into the switch, switch hotter, almost nothing back. The switch is where the failure appears and the diode is where the gain is.

The two junction-to-case resistances buy nothing at all, and that is not a surprise once the numbers are looked at: 5 and 2 K/W against 35 shared. The thermal design is entirely in the part they have in common, which is the same statement as the previous section’s factor of 3.6 read from the other end.

Whether the iteration is entitled to its answer #

Two coupled fixed points is a stronger claim than one, and the site’s habit is to say why the arithmetic is allowed to report what it reports.

The map is $(T_d, T_s) \mapsto (f_d(T_d), f_s(T_s, T_d))$ and both components are increasing in every argument: more temperature is more lifetime is more charge is more power, and more temperature is more on-resistance is more power. A monotone map iterated from below the lowest fixed point converges to the lowest fixed point if one exists, and increases without bound if one does not — which is exactly the dichotomy wanted, and is why the iteration starts at the ambient rather than at a guess near the answer.

Starting from a high temperature instead would find the upper fixed point where there are two, and the upper one is the unstable one: a device that reached it would leave it in either direction. A device switched on cold reaches the lower one. The iteration models the device rather than the equation.

And the refusal is the same refusal the rung below makes for one body, extended: when either temperature passes the ceiling the solver returns which of the two did it, not a temperature. A frequency alone does not tell a designer what to change, and this rung’s whole content is that the two answers to which differ.

The number the rung below reported, and why it was nearly right #

There is a coincidence here worth naming rather than leaving for a reader to notice.

The rung below reported 128 kilohertz. The correct answer for the pair on a shared heatsink is 135. The two are within six per cent, and they are within six per cent for a reason that is not a reason at all: charging one device with two devices’ heat is arithmetically similar to putting two devices on one heatsink, because in both cases one thermal resistance is carrying all the power.

It is the right size and it is the wrong statement. It attributes the failure to the diode, it puts the limit at the diode’s carrier lifetime, and a designer acting on it would buy a faster diode — which moves the runaway by very little, because the switch’s conduction loop is untouched by anything the diode does and the switch is the component that goes.

A number can be right to six per cent and point at the wrong component, and the only way to tell is to model both.

Where the model stops being a model of anything #

Three boundaries are worth naming, because the arrangement above is a steady state and most switching converters are not in one.

There is no thermal capacitance here. A die has a heat capacity and a package has more, so the temperatures above are what a device reaches after seconds to minutes of a fixed switching frequency. A converter that starts up into a short, or that runs a burst at a high frequency and then idles, has a transient thermal problem this arithmetic says nothing about — and in the direction that matters, since the peak junction temperature in a burst can exceed the steady value the same average power would give.

The switch’s overlap loss is absent. Everything charged to the switch here is the recovered charge at full supply plus its own conduction. A real transistor also dissipates while its voltage and current cross during each transition, and that loss rises with frequency exactly as the recovery loss does. Its omission makes every frequency in this essay optimistic, which is the same direction the rung below’s omission ran in and is stated for the same reason.

And the case-to-ambient resistance is treated as a constant. For a heatsink in still air it is not: natural convection carries more heat per kelvin as the surface gets hotter, so $R_\mathrm{case}$ falls with temperature — which is a negative feedback term sitting underneath two positive ones, and would push every runaway frequency up. With a fan it is nearly constant, which is the case this arithmetic is right for.

None of the three changes which component gives out first, because that is decided by where the energy is delivered rather than by how it leaves.

The three measurements this one is built on #

Two positive loops on one thermal node is a statement about three things that were measured one at a time. The diode that conducts backwards establishes how much charge comes back and over how long, at a fixed lifetime. The heat a recovery leaves behind makes that lifetime a function of temperature and finds the frequency above which the iteration has no fixed point. The half that never arrives supplies the other loop’s energy — the half of every charging step that is lost whatever the resistance — and Two millivolts a kelvin, and the wrong sign supplies the coefficient that makes both of them temperature-dependent. Neither loop is unstable alone; what this page measures is what they do sharing a heatsink.

The same thermal node carries the switch’s own loss, which The pulse the heatsink does not feel measures as a transient rather than a steady state — and that is the third contribution neither of the two loops above contains.

And the switch on the other side of the same node has a transient of its own that no steady-state thermal resistance contains, which is what The pulse the heatsink does not feel measures and this page inherits as an initial condition.

What is checked #

The repaired single-body loop is checked against the arrangement it generalises: coupledThermal with no shared case, no switch dissipation and the diode’s 40 K/W must reproduce recoveryThermal exactly, so the two-body model is a generalisation rather than a replacement.

Both temperatures are asserted to rise with frequency and the switch’s share of the heat to exceed sixty per cent at the top of the range, which is the rung below’s finding arriving as a temperature rather than as an energy. The runaway is asserted to exist, to be found by bisection on the solver’s own refusal rather than by extrapolating a curve, and — the assertion that carries this rung’s argument — to be the switch.

The coupling is asserted against the uncoupled pair with the same total thermal resistances, and required to cost more than a factor of two in frequency. The diode alone, correctly charged, is asserted to survive to several times either, which is the repaired number kept as a live check rather than as a note.

What is not modelled: the switch’s own turn-on and turn-off overlap loss, which is real and would make the switch hotter still — so every number here is optimistic in the direction that matters; the gate drive, which decides the overlap; the thermal capacitance of either die, so all of this is a steady state and a burst is a different question; and the case-to-ambient path being a function of temperature itself, which it is for natural convection.

Two positive loops on one thermal node #

The arrangement this rung ends on — two devices with two different positive temperature coefficients sharing one piece of aluminium — is the most coupled thermal problem in the collection, and it is worth setting beside the two single-loop cases to say what the coupling adds.

The loss that depends on what it causes is one device closing one loop: a ferrite whose saturation flux falls and whose permeability rises with temperature, giving a fixed point with a stable root at 89 degrees, an ignition root at 191, and a thermal resistance of 183 kelvin per watt at which the two touch and neither exists. The boundary that is a starting point marches the same equation and finds the ignition root is not reached from below — a trajectory starting at 189.6 degrees settles back and one at 192.6 leaves in twelve seconds.

What two loops on one node add is that the fixed point is now a solution of two coupled equations, so neither device’s stability margin is a property of that device. The switch giving out at 135 kilohertz rather than the diode at 1.28 megahertz is the whole content of that: the component that fails is the one whose own loop is steeper, not the one whose dissipation is larger, and swapping either part changes the temperature the other one runs at.

Which is why the repair that halved the diode’s problem — attributing nine tenths of the recovery energy to the transistor — moved the failure rather than removing it. On a shared heatsink an improvement to one device is an improvement to both, and a reattribution between them is an improvement to one and a worsening of the other.

Which is the argument for solving the pair rather than each device in turn. Two coupled fixed-point equations have solutions that neither equation has alone, and the one that decides the design is the smaller of two ignition thresholds rather than either device’s own — a quantity that appears on no data sheet and that cannot be computed from the two parts separately.

The practical remedy is the one the arrangement suggests rather than a better calculation: give the two devices separate thermal paths. Two loops on two heatsinks are two independent problems with two independent margins, and the coupling that makes this rung’s answer different from the sum of its parts is a piece of aluminium rather than anything electrical. Which is not usually available — the two devices are in the same package on many parts, and where they are not, one heatsink is cheaper than two — so the coupled solve is the ordinary case and the decoupled one is the luxury.

Which is the argument for computing the pair rather than for rearranging the hardware. Two fixed-point equations in two temperatures is arithmetic; two heatsinks is a mechanical decision made months earlier by somebody costing an enclosure. The measurement is the cheap half, and the number it produces — which device gives out first, and at what frequency — is one nothing on either data sheet contains.