FET Gate Driver w/ BJTs

[Styx]

That's what I was originally talking about, but styx went off on a tangent about gate resistor dissipation: Then he posted an equation, but as far as I can tell, he never said what it applied to. Maybe I missed that part. I thought the equation was for gate resistor dissipation, since that was the topic of his previous post.

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

 

[dknguyen]

Sorry equation wrong (looked at the wrong cell in my excel sheet). I use

**broken link removed**

P = (1/3)*(V*I)*ts*F
Where P == power dissipated in that one switching event
V = peak voltage switching to/from
I = peak voltage switching into/out of
ts = switching time (in this case 50ns)
F = switching freq (in Hz)

its 1/3 due to the waveform of the voltage and the current changing at the same time (see assumptions)

v(t) = (-V/ts)*t
i(t) = (I/ts)*t

so integrate V*I by t over ts:


int( (-VI)/(ts^2)t^2,t,0,ts)
gives E = (1/3)*I*V*ts
so multipying by F will give you the power for all turn-ON (if that ts was the turn-ON time)

The assumptions made in this are
1) that the voltage falls/rises linearly (in practice it doesn't)
2) that the current rises/falls linearly (in practive it doesn't)
3) it neglects reverse-recovery of any freewheel diodes at turn-on
4) neglects any voltage overshoot due to stray inductance at turn off
4) neglects the delay in voltage falling w.r.t. current rising at turn-on & the delay with current falling w.r.t. voltage rising at turn-off

These assumptions are valid for low amp/voltage operation (you should be ok). Even at higher voltages and currents these assumptions can lead to a switching loss that, while not spot-on, isn't too far out.



mmm getting interesting, capturing the backEMF can be tricky and must be sync with your PWM and modulation depth (so you sample during periods of no switching). Still there is only two instances when you would go to the effort of trying to measure the BackEMF
1) space-vector modulation of a BLAC machine
2) a means to sense speed

#1 is bad, there are better ways to derive it indirectly
#2 alot of hassle, might as well strap an encoder disk and count pulses

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