I was wondering if a power MOSFET fails due to overheating, will it usually burn as an open-circuit or short-circuit?
Also, does over-temp damage usually cause the gate oxide to blow through as a low resistance ( < 100 Ohms)?
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Assuming my answer is correct, my intuition was wrong about this. The short answer is I would expect a MOSFET to fail as an open circuit due to over-temperature conditions.
This wikipedia article suggests that:
Increase in drain-to-source resistance. It is observed in high-temperature devices, and is caused by metal-semiconductor interactions, gate sinking and ohmic contact degradation.
... at least in monolithic microwave integrated circuits, but the terminology seems consistent with MOSFETs...
This other article also suggests it will fail open, but for different (fundamentally mechanical) reasons:
Exactly what happens depends on how excess the power is. It may be a sustained cooking. In this case, the MOSFET gets hot enough to literally unsolder itself. Much of the MOSFET heating at high currents is in the leads - which can quite easily unsolder themselves without the MOSFET failing! If the heat is generated in the chip, then it will get hot - but its maximum temperature is usually not silicon-restricted, but restricted by the fabrication. The silicon chip is bonded to the substrate by soft solder and it is quite easy to melt this and have it ooze between the epoxy and the metal of the body, forming solder droplets. This may well not destroy the chip!
Usually, a MOSFET will fail short first. This is because excessive heat will, by diffusion, mix the dopants enough to create a good conductor instead of the p-n or n-p barriers that were there originally. Often, the gate oxide will be taken into the diffusion, too, causing a short betweem all three terminals.
Only if the short circuit current after this first mode of failure is high enough to blow the bond wires or the entire transistor, there is an open circuit.