On Mon, 10 Oct 2005 21:36:11 +0000 (UTC), Ken Smith <kensmith@xxxxxxxxxxxxxxx>
In article <1128979008.34671@ftpsrv1>, Terry Given <my_name@xxxxxxxx> 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
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
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
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.