The boundary that is a starting point

The loss that depends on what it causes solved T = Tₐ + Rₜₕ·P(T) and found two roots: a stable operating point at 88.8 degrees with a loop gain of −0.192, and an unstable one at 191.1 with a loop gain above eighteen. It said, correctly, that the second is an ignition temperature — a boundary the part must stay below — and that no converging method can find it, which is why it was located by scanning a residual for sign changes.

What it did not say is how a part gets there. And on the arithmetic as it stands, it cannot: a part sitting at 88.8 degrees is at a stable fixed point, every small disturbance decays, and the ignition temperature a hundred kelvin above is a solution of the same equation that the part will never visit.

A steady-state analysis has no way to express “and then something happened”. So the equation needs a clock.

The same equation with a capacitance in it #

A thermal resistance has a thermal capacitance beside it, and with one the steady statement becomes a differential equation:

C dT/dt = P(T) − (T − Tₐ)/Rₜₕ.

Its equilibria are exactly the roots above — that is what an equilibrium is — and what it adds is the direction of motion everywhere else. Below the operating point the right-hand side is positive and the part warms. Between the operating point and the ignition point it is negative and the part cools back. Above the ignition point it is positive again and the part warms without limit.

So the unstable root is not a temperature the part reaches. It is the watershed: the boundary between initial conditions that return and initial conditions that do not.

Two details of the march are load-bearing rather than cosmetic.

The step is adaptive on the temperature change, not fixed in time. dT/dt goes through zero at each fixed point and grows without bound above the second, so a step small enough to resolve an escape is thousands of times finer than the approach to equilibrium needs, and one comfortable for the approach walks straight past the escape.

And the integration is second order. An Euler step walks outwards from an unstable fixed point at a rate that depends on the step size, so the escape time it reports is a property of the arithmetic rather than of the part. Heun’s method costs one extra evaluation and removes that.

The model that had to be turned into a table #

The dissipation here is not a formula. Each evaluation of P(T) builds a hysteretic core at that temperature, marches a settled loop through twenty-four play operators — the object the core the solver has to remember put into a netlist — integrates its area, and adds the copper loss of the magnetising current it implies. It costs about thirteen milliseconds.

The loss that depends on what it causes could afford that because a scan asks for P at a fixed grid of temperatures and a memo answers ninety-nine of every hundred requests. A march is the opposite case. Its temperatures are wherever the trajectory went, so every one is new, and a two-hundred-step march costs two hundred marched loops — which makes a thirty-step bisection over marches six thousand of them.

So the model is sampled once on a grid and interpolated, and the interpolation is on the logarithm. That is not a refinement. This dissipation falls gently to 0.61 watts at 195 degrees and then rises to 888 by 209, as the flux the volt-seconds demand passes what the material can hold: a factor of fourteen hundred over fourteen kelvin. A straight line between two samples of that is 14 per cent out at their midpoint on a one-degree grid, and a straight line between their logarithms is under one — which is measured at the midpoints the grid never sampled and reported, rather than asserted.

Where the capacitance comes from, and why it is the easy number #

Everything above scales with one quantity the rung below did not need, so it is worth saying what it is and how badly it has to be known.

A thermal capacitance is a mass times a specific heat. A 20-millimetre ferrite core is about a hundred grams of material at 750 joules per kilogram per kelvin, which is 75 joules per kelvin; the bobbin, the copper and the potting add to it. Forty joules per kelvin is a small part and a hundred and sixty is a large one, which is the range the slider covers.

It does not move any temperature in the picture. The fixed points are solutions of the steady equation and the capacitance is not in it, so the operating point stays at 88.8 degrees and the ignition point at 191.1 whatever the part is made of. What the capacitance sets is the rate, and therefore every duration: the time constant Rₜₕ·Cₜₕ, the twelve seconds an escape takes, and the seventeen minutes an overload is survivable for. All of them scale in exact proportion.

That is a convenient division of labour. The temperatures come from the material and the thermal path, both of which are known to a few per cent; the times come from a mass, which is known to better than that. A design that gets the capacitance wrong by a factor of two has all its temperatures right and all its durations out by two — which is a very different failure from one where the boundary itself has moved.

What an ignition temperature is for #

A boundary in the initial condition sounds like an academic distinction until it is asked what moves a part across it, and the answer is a list every power designer already has.

A stalled fan. The thermal resistance triples, the removal line flattens, and the two roots walk towards each other. The loss that depends on what it causes bisects the resistance at which they touch — 182.7 kelvin per watt, with the stable point at 187.9 degrees and the ignition at 188.7 — and past it there is no steady state at all.

An ambient excursion, which is the same statement with the line translated rather than tilted.

And an overload, which is the curve moving instead of the line, and which is the one with a duration attached. An inrush thermistor is the same shape of argument run backwards, and the protection that is gone by the second time prices what happens when the part designed around a fixed point has not had time to leave one.

An overload with a stopwatch on it #

Multiply the dissipation by a factor for a stated time, then remove it, and ask whether the temperature came back.

The shape of that result is the useful part.

There is a factor below which duration does not matter at all. At 4.03 times the normal dissipation the removal line still crosses the loss curve, so the part has a hotter operating point and sits at it indefinitely. Nothing about a four-times overload is a race.

Above it, every overload is a race, and the times are minutes. The thermal time constant here is Rₜₕ·Cₜₕ = thirty minutes, and every window drawn is a fraction of it, which is what one would expect: the part is being asked to climb a hundred kelvin and it climbs them at the rate a thermal mass allows.

Which means a fast protection circuit is protecting against something else. A current limit that acts in microseconds is there for the switch, not for the magnetics; the magnetic part’s own protection is a thermal cut-out or a timed shutdown, and the number it needs is a duration in minutes rather than a threshold in amperes — which is also why a junction that measures its own dissipation, as the sensor inside its own answer does, is measuring a different loop from this one.

The gap between the two starts, and what it is not #

The two nearest trajectories drawn start three kelvin apart and do opposite things. It is tempting to read that as a knife edge, and it is not one.

A fixed point is a fixed point: exactly at 191.1 degrees the part stays there for ever, and a thousandth of a degree either side decides which way it goes — eventually. What the three kelvin buys is time. A start at 190 degrees takes tens of minutes to fall back because dT/dt near the watershed is nearly zero; a start at 192.6 takes twelve seconds to leave, because the loss curve above the watershed is nearly vertical.

So the asymmetry is not in the boundary but in what is on each side of it. Below, a shallow slope back to safety. Above, a cliff. That asymmetry is the whole reason the ignition point is worth locating: a part that has crossed it gives no warning time whatever, and a part that is approaching it gives a great deal.

Two roots, and the different questions each of them answers #

It is worth setting the two crossings side by side, because a reader who has met them only as “the solutions of a quadratic-looking equation” will treat them as equally real and they are not equally useful.

The stable root is a measurement. A part in normal operation sits at it, a thermocouple reads it, and it can be compared against a prediction. Its loop gain of −0.192 says the core is a stabilising feedback: a hotter core makes less heat, so a disturbance decays, and this collection’s what is left at crossover is the same statement about an amplifier in a different set of units.

The unstable root is never measured, and it is the specification. Nothing sits at it and no instrument reads it, so it exists only as the output of a model — which is precisely why the model has to be one that can produce it. An iteration converges away from it; a bench measurement never visits it; the only route to it is a residual scanned for sign changes, and the only thing it can be checked against is a march that starts on each side and does opposite things.

That check is what the figure at the top of this page is. Two trajectories, three kelvin apart, and the number the residual scan produced sitting between them.

What the march does not include #

The temperature is one number. The part is treated as isothermal, and a core makes its heat in its volume and loses it from its surface, so there is a gradient inside it. The flux ceiling is decided by the hottest part, which makes every ignition temperature here a slight overestimate. That is measured in the degrees a thermocouple cannot see.

The dissipation is a steady-state one at each temperature. P(T) is the loss of a settled hysteresis loop, so the march is valid while the electrical settling is fast compared with the thermal — which at a hundred kilohertz against a thirty-minute time constant is nine orders of magnitude, and is the one approximation here that costs nothing.

And the thermal capacitance is a single number. A real part has a core, a winding, a bobbin and a heatsink, each with its own mass and its own path, which is a ladder rather than a lump — the pulse the heatsink does not feel measures what that distinction is worth when the times are short. At the minutes this essay works in, the ladder has mostly equalised and a lump is the right model; at the seconds an overload transient occupies, it is not.

The number worth carrying #

Three kelvin between coming back and not, and 4.03 times the normal loss between “for ever” and “seventeen minutes”.

The habit is about what a steady-state solution is and is not. Solving for equilibria gives every temperature at which a part could sit; it says nothing about which of them it will be found at, and nothing at all about the ones it can reach only by being pushed. Both of those are properties of the trajectory, and a trajectory needs a capacitance — which is one extra number, and it turns an unstable root from a curiosity into a duration a design can be specified against.

Three kelvin, and what can supply it #

A trajectory starting at 189.6 degrees settling back and one starting at 192.6 leaving in twelve seconds makes three kelvin the quantity this whole result turns on, and it is worth naming what in an ordinary design is that size.

The degrees a thermocouple cannot see is the first and it is exactly three kelvin: a twenty-millimetre core is 2.59 kelvin hotter in the middle than on the outside in still air and 3.37 on a cold plate. So the equilibrium computed for a lumped core and the temperature at its centre differ by the whole of this essay’s margin, and the ignition threshold — which is a property of the material and therefore of the hottest material present — should be compared against the centre.

The pulse the heatsink does not feel is the second and is much larger: above 308 hertz a die’s own heat capacity integrates, and by a hundred kilohertz a fixed point computed from an average power is exact to five parts in ten thousand, while at one hertz the same average power on the same heatsink puts the junction three hundred kelvin hotter. A core driven by a converter that bursts rather than switching continuously is in that regime.

And an ambient excursion supplies it trivially. A part whose stable root is at 88.8 degrees in a twenty-five degree room has its whole margin to the ignition root reduced kelvin for kelvin by a warmer one — so the twelve seconds measured here is a property of a starting temperature that the room can move most of the way to.

An overload that is survivable for ever, and one that is not #

Four times the normal loss survivable indefinitely and seven times survivable for seventeen minutes is the result a specification can be written from, and it has a shape worth naming: a fault current that is either harmless or fatal with almost nothing between.

That is what a positive feedback loop does to a rating. Without the loop, dissipation and temperature are proportional and every overload has a steady temperature — larger overloads simply give hotter ones. With it, the loop gain rises with temperature, so below some overload the fixed point still exists and above it there is none, and the transition is not gradual. The seventeen minutes is not a derating curve; it is the time taken to cross a region where no equilibrium exists.

The pulse the heatsink does not feel is where the same structure meets a duty cycle, and its finding is the complementary one: above 308 hertz the die’s own heat capacity integrates, and a hundred kilohertz is far enough above that boundary for a fixed point computed from an average power to be exact to five parts in ten thousand, while at one hertz the same average power puts the junction three hundred kelvin hotter. So an overload’s shape decides whether the average is the right thing to compute — and an intermittent fault at a few hertz is a case where a four-times overload arrives as a much larger instantaneous one.

So the specification this rung produces — a survivable overload, a survivable duration, and a starting temperature — is complete only for a steady overload on a part with a fast thermal path. Any of the three assumptions failing moves the numbers, and the machinery that would say by how much is a march with two capacitances in it rather than one — one for the core’s bulk and one for the surface it loses heat from, which is the arrangement the degrees a thermocouple cannot see solves as a conduction problem and this rung treats as a single node.