Here's my stage driver circuit:

stage driver schematic

The circuit worked as expected at around 100v. But at 330, both mosfets burned out at the first firing, and the entire cap bank discharged through the coil. I thought it might have been an issue with my firing circuit so I replaced the mosfets and tried again. Again, burned out immediately.

I can't figure out why. The mosfets are rated for 650V, and a pulse current of 520A, which I don't think I've exceeded. The gate drivers are very strong, so I don't think switching speed is the issue.

HV is a 330v 5400uF bank of flash capacitors.

The MCU is programmed to turn on both fets for 100uS, and then turn them off. I checked, and it does. I also checked the mosfet gates (without the cap bank connected), and they pulse up to 12v as expected. The rise time is very quick, <100nS.

The coil is approximately 686uH, with a DC resistance of 1.8 ohms.

The mosfets are normally 20 bucks each on digikey, but I was able to find 10 of them on ebay for $8 each. Maybe they're counterfeit.

Update: Here's my layout: circuit board picture front and back


Datasheet Links:

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Jun 9, 2020 at 18:45

3 Answers 3


That's most probably inductive kickback killing your MOSFETs by overvoltage.

Current slope is as little as you reckon to 300V/600uH=0.5A/us in the inductor only.

From the power supply point of view you see peak current, say 50A, changing direction from source to sink during MOSFETs switching time. That's the slope you have to cope with.

I'd go for a good low ESL/ESR snubber grade capacitor very close to your half bridge supply.

Also take care of poorly damped ringing that could arise, some kind of snubber may be required too.

A second option would be to slow down switching, that's not an inverter working at 10's kHz rate. In virtually one-off zero duty cycle application switching losses are no concern as long as you don't exit RBSOA.

So a relatively high gate resistance would help a lot keeping EMI low too.

I believe both actions should be taken, together with some layout improvement to reduce high di/dt loops area.

  • \$\begingroup\$ Will try this, thanks. \$\endgroup\$
    – Drew
    May 25, 2020 at 1:31
  • \$\begingroup\$ @Drew going over your post I realised you could also saturate the main coil which is not air-cored. Could you post a couple of scope Vds screenshots while working at say 50V and 100V? \$\endgroup\$
    – carloc
    May 25, 2020 at 5:25
  • \$\begingroup\$ will do once more (cheaper) mosfets come in \$\endgroup\$
    – Drew
    May 25, 2020 at 6:08

Only guesses are possible without having your circuit.

Guess1: When the mosfets are turned off the coil current bulldozes its way through D1 and D2 as you have planned. That leads to an accelerated mosfet Vds jump. Too high d(Vds)/dt can trigger parasitic parts inside the mosfets and that current route maybe doesn't stand what's available. Of course, I cannot prove that's the case. The phenomena is described in this paper: https://www.mouser.com/pdfdocs/Impacts_of_dv-dt_Rate.pdf You very likely could use the same ideas how thyristors are protected against too high d(Uak)/dt.

Guess2: The parts this cheap can be fake and do not fulfill the specs as you said in the question. But you hang at the edge of the genuine parts, too. See the safe operating area curve:

enter image description here

Your coil can have a substantial capacitance. Charging it too fast needs a current which can be too high because Id and Vds are both high at the same time.


So you want to switch 500 amps in about 50 nanoseconds, in a system with 100 nanoHenries of inductance. What will be the spike?

  • V = L * dI/dT

  • V = 100nH * 500 amps / 50 nanoSeconds

The "nano" cancel, and we have 1,000 volts.


OK So what? Given no info on your ground planes, you should assume the worst.

But ... lets look at coupling from high current wiring into low-voltage control traces.

Assume the high current flows 1cm away from 1cm by 10cm loop.

What is the induced voltage? (the high current is parallel, for 10cm)

  • V = [MUo * MUr * Area / ( 2 * pi * Distance)] * dI/dT

which for MUo = 4 * pi * 1e-7, MUr = 1 for air/copper/aluminum/FR-4 becomes

  • Vinduce = (2e-7 * Area / Distance) * dI/dT

and we substitute [I don't know how bad this will be; any more than 1 volt will upset control signals]

  • Vinduce = 2e-7 * (1cm * 10cm)/1cm * 10,000,000,000 amps/second

  • Vinduce = 2e-7 Henry/meter * 0.1 meter * 1e+10 amp/second

and the prediction (remove the card from the mayonnaise jar, please)

  • Vinduce = 2e-8 * 1e+10 ==== 200 volts


You do have a Ground Plane, right? or two.

And you understand the 10,000,000,000 amp/second transient will cause currents to spread over ALL possible paths, as mother nature desperately attempts to solve the many simultaneous differential equations so as to give you the minimum energy solution.

  • \$\begingroup\$ I don't think this analysis is accurate. 1. The low voltage circuit is decently isolated from the H bridge. And 2. The current spike is nowhere near 500A in 50nS. The load is inductive, so the current ramps up, then the mosfets switch off, and the current ramps down, regenerating back into the caps. \$\endgroup\$
    – Drew
    May 23, 2020 at 22:23
  • \$\begingroup\$ (didn't downvote btw) \$\endgroup\$
    – Drew
    Aug 13, 2023 at 19:30

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