On Mon, 10 Oct 2005 11:50:18 -0700, Jamie
I would like to drive an IGBT from a microcontroller (CMOS).
Is there a driver circuit that can outputs around +15v (on) and a -5v (off)
for switching frequencies around 40Khz.
I think -ve voltage supply is avaliable
What is this type of circuit called?
i really don't understand why you need that?, normally pulling the
gate low will do how ever if you insist.
I am actually playing with IGBTs.
I have Toshiba-MG200Q2YS40 IGBT.
It does NOT turn off itself if I simply remove Gate/Emitter voltage.
I have to apply reverse voltage to turn it off. I actually tried it.
a decent short from gate to emitter would do it.
It would, possibly, switch itself off with a small resistor between
gate and emitter, but it is just a hypothesis of mine.
what you are loking at is a combination of things - miller effect
mostly. There is a non-linear capacitance between gate and collector,
call it Ccg. Whenever the collector voltage changes value (say from
Vcesat to Vdc when turning off) the voltage slews at some rate, dV/dt.
This causes current to flow thru Ccg, Icg = Ccg*dVce/dt. This current
flows into the gate circuit, charging up Cge. If the gate driver is not
very low impedance, Icg will happily charge Cge up to Vth, and beyond.
when turning the IGBT on, Vce falls so Icg flows out of Cge, discharging
it - IOW trying to turn it back off.
This can greatly increase switching time (and hence losses) and, if the
gatedrive is bad enough, can make the device "latch" in the on-state,
with Vge sitting around Vth - but you need a seriously piss-weak gate
driver to do that. It can even over-voltage the gate, poke a hole thru
the thin gate oxide layer, and KABOOM, one dead IGBT.
a negative gate bias means the switch-off miller current has to supply
more charge to reach Vth - instead of supplying Qg = Vth*Cge, it must
supply (Vth + |Vnegative|)*Cge. This allows the use of higher gate
impedance circuitry. Its not just the resistance, its the inductance
too. bad gate drives have high resistance and high inductance; good gate
drives have low resistance and low inductance.
to give you an idea, I've worked on gatedrives for 0.5kW - 2.2kW drives
that have about -3V reverse-bias, and -15V for 400kW drives (6 300A fuji
IGBTs directly paralleled).
modern IGBT designs have much better packaging - far lower inductance,
both Lce and Lge. with some of these IGBTs it is theoretically possible
to keep the gate impedance sufficiently low that negative bias isnt
necessary. Like I said, IR wrote a pretty good paper on the topic - why
you dont, in theory, need -ve bias for little IGBTs.
In practice, if you screw up the gatedriver, your circuit is going to go
BANG. If you really know what you are doing, and are very good at taking
measurements, its fairly easy to dewsign gatedrivers and make them work.
OTOH if you are not too experienced, its very easy to destroy many, many
Generally the gatedrive is a lot cheaper than the IGBTs, so its sensible
to throw -ve bias at it. For little IGBTs, it can be as simple as a
zener in series with the gate resistor, with a 100nF cap across it.