# Is it meaningful to describe an IGBT as desaturating?

I've heard IGBTs described as desaturating if too much current passes through them, particularly in the context of drivers with desat protection. My understanding of an NPN BJT is that it is in saturation mode as long as the B-E voltage and B-C voltage are positive. Obviously this definition does not apply to IGBTs. The curves on every IGBT data sheet I've seen show a typical junction drop relationship between voltage and current collector-emitter when the device is gated on. What happens off the edge of the sheet, though, is not clear to me.

I see two possibilities.

1. The curve doesn't change off the edge of the sheet. The only disadvantages to running a given device at higher currents than spec'd are thermal, and the general rule that running off the data sheet is not recommended behavior.

2. The curve does change off the edge of the sheet; put too much current through the device and it goes into an analog mode. The losses start going up even faster than extrapolations from the given curve would indicate, leading to rapid catastrophic failure.

Which is the deal? Is there physical meaning to describing an IGBT as desaturating, like a BJT? Or is it just a way of saying "instantaneous overcurrent"?

If you were to define "saturated" as $V_{CE} > V_{GE}$ then for an 'on' gate voltage of, say, 12V it would be desaturated iff $V_{CE}$ > 12V. That would be an indication of something bad happening and it would be time to switch it off fast (but not too fast, or that itself could cause damage).
In reality, that definition does not seem to be widely used, and $V_{CE(SAT)}$is used loosely to describe the 'on' $V_{CE}$ voltage with reasonable collector current and gate voltage. Thus "desaturated" means a reasonable 'on' gate voltage and an unreasonably high $V_{CE}$, indicating an unreasonably high collector current.