One cycle at 1.00 Hz, 2% duty, 5 W average
Drawn above at its default parameters, which is almost never how an essay calls it. A placement states the numbers that essay is arguing about, so the figure a reader meets is about that argument rather than about the generator — 98% of the placements on this site pass one, and the phase that raised that number from 12% found eight captions describing a figure the page was not showing.
At those defaults the edge it states is peak 349 K against a steady 49 K — the right-hand slot
of the caption strip, which on this site is never used for anything else, and which is read
back out of the drawing above rather than out of the code that wrote it.
It belongs to Before the steady state, which is to say a change to it is a
change to lib/figures/transients.js.
It takes a slider on switching frequency (Hz) with 5 settings,
and every one of them has passed the same assertions as the frame above — a figure whose
circuit stops doing what its caption says at any setting stops the build.
Called by 5 essays
which is the blast radius of changing it
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.
The degrees a thermocouple cannot see
Every thermal answer in this collection has been one temperature, and a core makes its heat in its volume and loses it from a surface, so it has two. Solved as a conduction problem, a twenty-millimetre core in still air is 2.59 kelvin hotter in the middle than on the outside — 2.9 per cent of a ninety-kelvin rise, which is why the lumped answer has been good enough. Cool the same core on a plate and the gradient does not shrink; it grows to 3.37 kelvin and becomes 78 per cent of what is left.
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.
The loss that depends on what it causes
Every thermal figure in this collection has had the power handed in. A ferrite's has no business being: its saturation flux falls with temperature, its permeability rises, and both move the loss. Closing that loop makes the temperature a fixed point rather than a product — and the fixed point has a stable root at 89 degrees whose loop gain is negative, an ignition root at 191 whose loop gain is 120, and a thermal resistance of 183 kelvin per watt at which the two touch and neither exists.
The boundary that is a starting point
A wound part with a stable operating point at 88.8 degrees and an ignition temperature at 191.1 will never reach the second, because nothing takes it there. Marched in time rather than solved for its steady states, the same equation says what does: a trajectory starting at 189.6 degrees settles back and one starting at 192.6 leaves the material's range in twelve seconds — two starts three kelvin apart. And an overload of four times the normal loss is survivable for ever, while seven times is survivable for seventeen minutes.