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.


1 Answer 1


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.

  • \$\begingroup\$ But it'll only be like that during the 'ramp-up', right? Then it'll be in saturation. Either way, if it's heatsinked that's fine for like 50Hz or so. Plus, I can just shut down the power input, then switch the gates to full, then turn the power back on; no more ramp-up. Am I right or is this a terrible idea? (Fed from H-bridge) \$\endgroup\$
    Commented Apr 25, 2015 at 4:32
  • \$\begingroup\$ @ARMATAV see my edit/changes. Also, turning off the high voltage input, and pre-opening the switch, and then turning on the input - this will work. But then you cant keep switching, that defeats the purpose. \$\endgroup\$
    – KyranF
    Commented Apr 25, 2015 at 4:37
  • \$\begingroup\$ Ah, gotcha. See the thing is, I'm only gonna switch it like once every couple minutes. Manually. I'm literally gonna type a command and it'll send a pulse to a bit shift register to open a transistor and therefore open a PVI1050N. Then the PVI1050N will open the MOSFETs. I kinda used 50Hz as a 'maximum' example. I don't think I can manually send commands that fast anyway. Maybe 3-4Hz if I loop it for whatever reason. Also; these calculations aren't including a heatsink, right? Because I'm heatsinking the hell out of the whole setup. So I think I might be good then. \$\endgroup\$
    Commented Apr 25, 2015 at 4:40
  • \$\begingroup\$ @ARMATAV with a heat sink it will be fine - but just think - heat sinks will not help if there is such an insane amount of current that the junction heats up way before it can be transferred to the heat sink. With only 4A continuous or switched at a slow rate like 3-4Hz it's fine. \$\endgroup\$
    – KyranF
    Commented Apr 25, 2015 at 4:42
  • \$\begingroup\$ Gotcha. One last thing, does it matter if the 4A is, as mentioned above, 150kHz and 300VAC through source/drain? Gate is still 3-4Hz. \$\endgroup\$
    Commented Apr 25, 2015 at 4:44

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