MOSFET WTF

Sep 23, 2017 21 Replies

See how it's angled in during the onset of the plateau?

Suppose Cdg were massively smaller at that point; what would happen? Well, Vds would drop a whole lot faster, and Vgs would rise faster, too. Though not quite as fast as below threshold.

Now reflect upon my waveform:

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The first 'break' in the rising gate voltage, where Vds first drops, is exactly where that happens. Cdg is extremely small, perhaps 10pF.

Well, let's see. The initial slope is about 3:1, or 6V per 5.2nC, or 0.87nF (which is Ciss, close to the 1nF datasheet figure).

When Vds starts to drop, it gets more shallow, about 1:1 slope, or 2V/5nC, or 2.5nF equivalent. We could calculate the voltage gain and Miller effect, but better to use the totals: delta Vgs = 1.2V (4.0 to 5.2V) delta T = 3.3us --> delta Q = 3.45nC delta Vds = 340V

So delta Qg = 1.044nC (still assuming 0.87nF), so delta Qgd = 1.4nC. 1.4nC / 340V = 4.11pF average.

Which is pretty astonishingly low!

It's not fully "on" until 20us after trigger -- even though Vds swings 90% in the first 7us. That's the magic of these SuperJunction parts. The capacitance acts like a snubber network built into the MOSFET, massively saving on switching losses, even for hard switched loads!

Tim

Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website: http://seventransistorlabs.com

Right, if the measurement is made at a high current density for the MOSFET, it'll show a slope.

Thanks, - Win

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