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## Re: Driver for IGBT? (-ve off and +ve on from CMOS)

 Subject: Re: Driver for IGBT? -ve off and +ve on from CMOS Terry Given Tue, 11 Oct 2005 12:26:25 +1300 sci.electronics.basics, sci.electronics.design
 ```Ignoramus14838 wrote: ``````On Mon, 10 Oct 2005 21:36:11 +0000 (UTC), Ken Smith wrote: ``````In article <1128979008.34671@ftpsrv1>, Terry Given wrote: ``````Ignoramus14838 wrote: `````` [.. a bunch of good stuff ..] ```Just as a BTW: If you are current sensing by placing a resistance in the emitter leg (current Xformer included), a small capacitance across it, speeds up both the on an off times a bit. ``` Ken, are you talking about a snubber circuit? `````` ```nope. sometime (in little systems) you can bung a resistor between the emitter and 0V, to sense current. plop a cap across this, and the gatedrive can improve. ``` at 200A, its not worth doing. 1mOhm dissipates 40W.... `````` ``````Also: I have used back biasing to speed up the turn off of MOSFETs. `````` just to clarify: ```when discharging the gate of a FET (or an IGBT), it looks like a capacitive discharge - almost. its a bit funny, because the capacitance is nonlinear, but lets pretend its just a cap. ``` ```If the gate voltage is say 12V, and Vth is say 4V, the cap discharge equation is: ``` 4V = 12V*exp(-t/(Rg*Cg) which re-arranges to give: t = -(Rg*Cg)*ln(4V/12V) ```t = 1.099*(Rg*Cg) - it takes 1.1 time constants to discharge from 12V to 4V, at which time the device then begins to turn off. ``` ```actual turn-off time depends on how well the gatedrive can suck out the miller current. ``` ```If Ken used -12V, then the numbers change quite a bit. to make the maths easier, lets just add 12V to all ov the voltages: ``` Von = 12V Vth = 4V Voff = -12V becomes: Von = 24V Vth = 16V Voff = 0V and we can use the same equation, giving t = -(Rg*Cg)*ln(16V/24V) ie t = -0.405*(Rg*Cg) - the fall time has more than halved. ```another way to look at is to ask the question: what is the peak current thru Rg at the instant the gatedriver output switches low? ``` for 12V/0V drive , Cg is charged to +12V, gatedrive Vlow = 0V, so Ig_max = Rg/(12V - 0V) = Rg/12V for +12V/-12V drive , Cg is charged to +12V, gatedrive Vlow = -12V, so Ig_max = Rg/(12V - (-12V)) = Rg/24V ```so the peak current being sucked out of the cap is twice as high. which is why it gets down to Vth a lot faster. ``` *but* there is a penalty. Say Rg = 10R, then Ig_max_unipolar = 12V/10R = 1.2A Pg_max_unipolar = (1.2A^2)*10R = 14.4W but for bipolar drive, Ig_max_bipolar = 24V/10R = 2.4A Pg_max_unipolar = (2.4A^2)*10R = 57.6W so there aint no such thing as a free lunch. ```I also wouldnt use a pissy little 0603 resistor for either of those jobs. the unipolar case certainly exceeds an 0603 peak pulse power rating, which is a few watts, but the bipolar drive is an order of magnitude too high. ``` `````` ```Yes, looks like it is necessary in my case. ```i `````` Cheers Terry ```
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