The lowside MOSFET would be getting turned on soon after the body diode conducts. Does anyone know how much this would affect reverse recovery?
Again, it acts in parallel. If the channel voltage drop is higher than a diode drop, then the diode part will carry some current. If the voltage drop is less, it will be shorted out and little diode current will flow.
In an inductive load condition, the diode may be reverse-biased at first, then load current reverses (and the transistor switches on anywhere during this reverse period), recovery occurs seamlessly and the channel carries load current. With Vds > 0 by the time it's switching off again, there's no recovery loss.
I've done this before with 1200V FETs, rated for... max 300ns recovery it looks, but I seem to remember more, maybe I'm looking at the wrong series, maybe they're obsolete now, dunno. Anyway, switching was up to 400kHz, inductive load, and yeah, no problem with recovery.
It's hard switching that's the problem, a general class D amplifier will find this problematic, or in general a capacitive load condition -- anything where one transistor is carrying reverse (body diode) current when the opposing side switches on.
In such a case, it can be better to set dead time to zero, or even slightly negative -- while intentionally increasing loop inductance to limit commutation dI/dt, and snubbing it to keep peak voltages and EMI down.
And yeah, the advance since then has been this: old transistors scaled as Rds ~ Vds^2, i.e., as voltage goes up, performance is not only worse but quadratically worse. So MOSFETs of 600V rating say already performed so much worse than 400V. Performance being a figure of gate charge versus Rds(on), lower (in both) being better.
Over the last decade, the SuperJunction process has been developed, reducing the scaling factor to Rds ~ Vds, i.e., performance is basically independent of voltage, it's plain old proportional, if you want more current or voltage, just scale up the die area proportionally. Or put another way: Qg is proportional to switching area, Id(max) * Vds(max).
Now, the body diode drop doesn't scale nearly so fast (more like log(Vds) probably? Hmm, I never did look at it in detail, I wonder), so you are still disadvantaged, at high voltages, to achieve synchronous rectification -- out of a 12V supply, it's a hell of a lot easier (indeed just about mandatory) to get a 0.1V or even lower Vds(on), than it is to get less than 1V out of a 400V supply. For something of the same power level, say, a 1 HP motor speed controller or something -- in the first case the voltage drop must necessarily be low just as a consequence of keeping power dissipation manageable, in the other the diode drop isn't problematic at all (1V out of 400 is only a 0.25% efficiency hit in conduction loss!) and you have to work a lot harder (spend say $20 on transistors, and use somewhat stronger gate drivers) to eke that little bit more performance out of it.
Put another way, the 400V inverter with synchronous switching, needs whatever, say 4 times more inverter current capacity, just to avoid a quirk of the transistor, whereas the 12V inverter just needs whatever it does.
So it is kind of cruel I guess, that body diode recovery sucks worse at higher voltages, and is that much harder to avoid, through synchronous rectification at least. But we have other ways to solve it, and if nothing else, can employ the OP re-diodeing circuit to prevent it outright (at obvious expense to conduction loss, of course). So it's not a huge problem.
Tim
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