How long will it take to kill a MOSFET with heat

high voltagemosfetopto-isolator

Example; http://www.aosmd.com/res/data_sheets/AOK42S60.pdf

With this; http://www.irf.com/product-info/datasheets/data/pvin.pdf

As I was taught in another question, the 40nC gate charge being hit with 10uA at 10V gives you 40/10 = 4mS of switching time.

I'm gonna be sticking 3-4A~ of 300VAC@150kHz into the drain/source.

If I turn this FET on manually, i.e. 100% duty cycle, will it just die instantly because it's in the linear region for 4uS?

Or do you need to turn the gate on and off really fast for it to kill your FETs?

According to this;

$$ P = (I^2R) $$

It should dissipate;

$$ P = (4^2*0.099) = 1.584$$

Do I get the I(ds) graph per unit of V(gs) and integrate over it, changing \$ (I)^2 \$ each time for each change in V(gs) over a period of 4uS in the graph? So like, the power dissipation at \$ 0V + … + 10V = total Power Dissipation \$

I'm really confused as to why this MOSFET would burn up instantly like a guy said in the other question.

Best Answer

Well, ~1.6 Watts with a thermal resistance (junction to ambient) of say 40 Degrees Celsius per Watt makes it go from 25 degrees ambient to ~90 degrees which won't really kill it, but that assumes your calculations are perfect. The heat you see here is actually ALWAYS being dissipated while operating in your 4A load condition.

The absolute max ratings is 150 degrees, so during the 4 microsecond switching period, if it manages to generate a lot of heat, enough to bring the junction up to 150+, then it may begin to fail. Seems a bit of a fair stretch, especially if you only switch it a few times. If you are switching many times a second, this may build up average heat enough for it to fail fairly quickly.