FET Gate Driver w/ BJTs

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Not really a tangent

TOPIC title: FET Gate Driver w/ BJTs

Post #1
Hi. I am trying to put together a PNP/NPN totem pole FET gate driver since I added an isolator and theh bootstrap IC doesn't work anymore (or is more than needed).
Post #3
Yeah. 252nC gate charge and even if I switch at 50ns which is what 3A gets me, I still get 3W of switching losses. The other 4W comes from I^2R.
Problem seems to of arrisen here, the discussion kinda started to tangent here to switching losses but I failed to pick up on it in time (concidering I deal with 2-3W worth of gate resistance losses I didn't see a differenence)

BUT since switching speeds and thus losses can only really be influenced (with some degree of predictivity) by the gate-drive it seemed logical to probe from that way



That equation, as following the discusion is todo with the switching losses of a FET (if prev posts are followed AND that post is followed).



Seems the thread got lost due to what is needed and what info is/was provided differed
To me everything to the initial title and posts after were todo with gate-drive
 
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I hate to revive a dead thread but this thread always comes up as the first hit whenever I google "BJT MOSFET Gate driver".

The power dissipation for the MOSFET due switching is actually what Styx initially said it was:
W = (1/6)*V*I*ts

He was right the first time and then thought he was wrong. It is NOT W = (1/3)*V*I*ts as he has derived in my quote mainly because he missed the positive offset for current (for turn-off) or voltage (for turn-on) at time zero. That is, he didn't include the 'b' in the y=mx+b equation for a line.

Once this is included it adds another term inside the integral. So for the turn-on case:

Energy Dissipated = ∫{ [ (-V/ts)*t +V ] * [ (I/ts)*t] }dt

For the turn off-case it's just the V and I change places because at the start of turn-on, voltage across the FET is non-zero while current is zero. But at the start of turn-off, current is non-zero while voltage is ideally zero:

Energy Dissipated = ∫{ [ (-I/ts)*t +I ] * [ (V/ts)*t] }dt

If you multiply out the equation so you can integrate it you get the same result whether you do it for the turn-on or turn-off case:

Energy Dissipated = ∫{ (-VI/ts²)*t² + (VI/ts)*t }dt

You can then easily integrate it over the time interval 0 to ts, and it becomes the same answer whether or not you did it for the turn-on or turn-off case:

Energy Dissipated = (-1/3*V*I*ts) + 1/2*V*I*ts

The bolded terms are what Styx was missing in his solution of it.

Now the power dissipated due to switching losses in one transition is :
P = E/T = E*f = [ (-1/3*V*I*ts) + 1/2*V*I*ts ] * f = (1/6)*V*I*ts*f

For on and off switching losses it's just simply the sum of the on and off loss:
P = (1/6)*V*I*(ts_on+ts_off)*f
 
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