Is TVS diode fast enough to protect MCU from MOSFET malfunction?

Question

I am using standart N-channel MOSFET (Fairchild FDS6676) switch like in this picture:

Load is actually 10A resistive and 27V power source is fused with 15A non-recovery fuse. I am driving it directly from 3,3V logic - ARM MCU.

I know that MOSFET will be quite stressed because it is designed for 14,5A continuous current and 10A will be drained for sure. There might me some peaks to 12-13A for few miliseconds. As I calculated, there will be about 3-4W of power loss on this MOSFET.

I have heatsink mounted, but still concerned about reliability (always with thermally stressed parts). In case of MOSFET break/malfunction (overcurrent peak above 14,5A or over-temperature failure), MOSFETS often shorts all terminals, so I could get 27V to gate and directly to MCU port. In case that S-D burned together will blow my fuse, but in case only S-G not...

My question is - is TVS protection diode fast enought to protect MCU in such scenario? Or is there any better approach? I am considering Littelfuse AQ4020-01FTG-C.

Thank you

Answer 1

Yes a TVS Diode, unidirectional, can protect your MCU if the fuse blows fast enough before the TVS diode gets burned itself. So you must use a fast blowing fuse with the smallest amperage possible. (Your symbol shows a bidirectional and it's not necessary.)

You didn't explain what type of signal will come from the MCU. Is it on/off, low frequency or high frequency? Depending on this, you can chose stronger protection with varistors. They are more robust than TVS diodes but have a capacitance of 1000 to 10 000 pF, compared to single digit pF or sub pF capacitance of a TVS Diode.

Also add a resistor, between the TVS Diode or the varistor if you choose one, and the MCU. you can go as high as 22K but here again, the higher the value, the lower the frequency.

But there is another problem, not part of the question: 3.3V is a bit low, the bare minimum in fact, for the MOSFET gate, For better MOSFET efficiency, less Rds, it's better to use at least 5V and possibly more, up to 16V. The more the better. So I would suggest to add a circuit to bring more voltage to this gate. With a second MOSFET for example. For 10A, that's what I would do.

Answer 2

You could split Rin in half and add a clamp to the middle node, for example a cheap jellybean BJT.

Speaking of cheap- it sounds like you're going a little cheap on the MOSFET, why not use a better MOSFET that has a higher rating and lower Rds(on) so you're not throwing away watts of power? It also does not have guaranteed Rds(on) with 3.3V drive so you're gambling. For example, this one, which would have perhaps 300mW dissipation. It's also about the same price.