I'm using the following circuit to drive an electric pulse motor triggered by an optical switch and timing disk, which works very well, but the MOSFET driver (UCC27524P) keeps getting fried.


PCB layout

If the rotor spins, the system can run for a very long time without any of the components getting hot, but then once in a while, usually if the rotor stops in a position where the timing disk happens to turn the driver HIGH, or after I've changed the position of the timing disk, the driver gets fried.


In the traces above, the red trace is the 12V input into the IC, and the blue trace is the drain-source waveform.

I've tried the Zener that @PSTechPaul suggested, which didn't fix it.

Then I suspected too much current from the opto-interrupter board, so I added a 330 ohm resistor and 0.1uF filter capacitor to the PWM line, and for a while that seemed to do the trick, but yesterday my last IC still died.

What is your theory about why this gate driver keeps blowing up? Please let me know, as I'm running out of ideas and ICs.

  • 1
    \$\begingroup\$ DC is fine. Try looking for transients, maybe something's frying it. \$\endgroup\$ Nov 29, 2022 at 15:41
  • 1
    \$\begingroup\$ Perhaps the 12V supply is drifting higher when there is no significant load. You could add a 15V zener to make sure. \$\endgroup\$
    – PStechPaul
    Nov 29, 2022 at 21:04
  • \$\begingroup\$ Thanks! @TimWilliams wouldn't transients happen when things are switching, not when there is a steady voltage? I could try a TVS to catch the transients. PStechPaul I'll try a zener. Could I perhaps need a current limiting resistor on the INA and INB pins? \$\endgroup\$
    – kregus
    Nov 29, 2022 at 21:40
  • \$\begingroup\$ Note point 7 in the layout guidelines in the datasheet of your MOSFET driver, the ground plane must not be a conduction path for the currents. Consider as star grounding as point 6 says. Also make sure to probe the input signal to the IC, as stated in "8.3.6 Low Propagation Delays and Tightly Matched Outputs", the transitions should be faster than 20 V/μs (I would expect an opt-interrupter to give an analog signal if it is partially covered, which means one channel can source an HIGH, while the other sinks a LOW) \$\endgroup\$
    – Ferrybig
    Dec 6, 2022 at 12:24
  • 1
    \$\begingroup\$ @Ferrybig The opto interrupter has a comparator output. It should be sharp, though it will be smeared by the long wires. The pulse may also be short because the module has a low pass filter on the phototransitor. It may also "chatter" if the encoder wheel stops just at the edge of the detector. \$\endgroup\$
    – JRE
    Dec 6, 2022 at 13:20

1 Answer 1


I'm going to go out on a limb here and say that the edges of your "PWM" signal aren't sharp enough.

From page 18 of the UCC27524P datasheet:

Caution must be exercised when directly connecting OUTA and OUTB pins together because there is the possibility that any delay between the two channels during turnon or turnoff may result in shoot-through current conduction as shown in Figure 29. While the two channels are inherently very well matched (4-ns Max propagation delay), note that there may be differences in the input threshold voltage level between the two channels which causes the delay between the two outputs especially when slow dV/dt input signals are employed.

TI recommends the following guidelines whenever the two driver channels are paralleled using direct connections between OUTA and OUTB along with INA and INB:

  • Use very fast dV/dt input signals (20 V/µs or greater) on INA and INB pins to minimize impact of differences in input thresholds causing delays between the channels.
  • INA and INB connections must be made as close to the device pins as possible.
  1. You've got your optical gate connected through a long cable, which will tend to round off the rising and falling edges of the signal. Due to the noted variations in the trigger levels for the inputs, you could easily get one output high when the other is low.
  2. You've got the two outputs connected directly in parallel. Any shoot through would allow full current from the power supply to flow through both output stages.

Here's what I would do:

  1. Put a Schmitt-trigger in between the optical gate and the driver inputs - that's on your PCB and as close to the driver inputs as possible.
  2. Put a low value resistor in series with each output before connecting them together.

The first should greatly reduce the chance of shoot through, while the second should prevent damage to the driver if shoot through occurs.

You've got a two channel scope. Cut the traces going from the driver outputs to R1. Connect a scope probe to each output, then rotate the motor by hand. See if you ever get one high and one low.

You might also consider using just one channel of the driver. The GaN FET you are using claims a low gate charge and fast switching - you may not need as much drive current as you think you do for satisfactory switching.

  • \$\begingroup\$ Ahh this would make a lot of sense, as the driver only breaks when this opto-interrupter board was used, but works perfectly with a function generator. The long wire is only used in this temporary test setup, but now I understand that it could be delaying the rise of the signal, and also that I've been silly not to notice I needed a Rgate per channel.. thanks for those insights! \$\endgroup\$
    – kregus
    Dec 6, 2022 at 20:29

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