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.