I've recently wired up some large 12V fan PWM controller with a 11A rated n type MOSFET and a gate driver IC. Testing the MOSFET without its heatsink (probably a bad idea…), I noticed that it would respond to overtemperature in a way that I have difficulty explaining: by entering a partially-on state, where even with 0V at the gate, about 700mA of current were still going through the fan. I found that surprising; testing with different loads and scenarios showed that
1) 10W LED array load shows the same effect
2) 2W LED array load does not show the effect
3) an open circuit (no load) does not show the effect
4) A hot MOSFET will stay off if the circuit is interrupted externally.
My question is now why. It should be noted that the MOSFET is not damaged; after cooling down, it switches the fan just fine. Also, the fan did always have an adequate snubber diode in place, so I don't think the back emf is the expanstion here. I've also tested different MOSFETs that consistently show the same effect.
My question is now: What causes the MOSFET to behave that way? If it were simply free carriers generated by thermal excitation, wouldnt those continue to be there after interrupting and reconnecting the load circuit (#4)?
(Note this is purely academic — I understand that the MOSFET is running out of spec without its heatsink and should be expected to fail one way or the other, but I want to understand the physics behind this particular failure mode.)
Best Answer
We build high-temperature circuits, and find we have to test for this phenomena in an oven, although we generally work at higher voltages. IDSS increases with temperature on all MOSFETS and will eventually reach a point where the leakage is enough to keep it hot or it runs away; that is, the heat generated by the leakage current (in your case .7 * 12 or 8 Watts) heats up the junction enough to sustain the large leakage or allow it to continue to increase. Not an academic response, but maybe there is a solid-state physicist out there who can explain why this occurs.