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First, these are the components I'm using on my PCB, click on them to see the datasheets:

U7: Infineon 1EDI60N12AF MOSFET Gate driver

Q6: Infineon IRF250P224 Power MOSFET

D5: Fairchild (ON Semiconductor) RURG5060 Freewheeling diode

U1: Infineon TLI4970-D050T4 Current Sensor

The schematic is attached below. Sorry if I'm not using some right symbols. I took the schematic from the PCB software.

Vbus voltage comes from rectified and filtered mains (110VAC). PWM frequency is 10 kHz.

The gate driver is isolated, so GND1 is ground from the rectifier, and GND2 is the digital ground.

I tested the PCB with two different PMDC motors:

Motor 1:
130VDC
25.5A

Motor 2:
130VDC
27A

These are the steps that I followed before the transistor blew:

  1. I tested the circuit with a light bulb first. Everything ok.

  2. Removed the bulb, then I connected Motor 1 with no load. Everything ok.

  3. Removed Motor 1, I connected motor 2 with a slight load. It started ok at 5% of duty cycle. Then, I increased duty cycle to 10% and the transistor blew up, so the motor went to max speed uncontrollably.

I haven't programmed the fast overcurrent detection from the current sensor. That way I would have probably saved the MOSFET.

The question is... what could be the cause of this? high dv/dt and di/dt? need an snubber?

Update: I've blown up a second MOSFET at startup of the motor. I've replaced the FET with this IGBT IKW50N60DTP and so far it's working well. Anyways, I will capture some waveforms soon.

schematic diagram

Go to this link to see a piece of the PCB where the parts are located: https://i.imgur.com/4UWVIGm.jpg enter image description here

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    \$\begingroup\$ Show some waveforms, they usually blow up because they are in a linear region too long. \$\endgroup\$ – PlasmaHH Sep 15 '18 at 19:28
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    \$\begingroup\$ What is U7? What is U1? At first glance it seems strange that the source of the FET isn't connected to ground. \$\endgroup\$ – Olin Lathrop Sep 15 '18 at 20:56
  • \$\begingroup\$ @Olin U7 is a gate driver with separate source and sink, U1 is a low side current sensor. \$\endgroup\$ – Nick Alexeev Sep 15 '18 at 20:58
  • \$\begingroup\$ I will get some waveforms when I get my scope. In the meantime, I'm trying to guess what could be the cause. \$\endgroup\$ – Xavier Pacheco Paulino Sep 15 '18 at 21:19
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    \$\begingroup\$ Show your layout. Show Vgs and Vds oscillograms. You really can’t guess these things without measurements. \$\endgroup\$ – winny Sep 16 '18 at 8:13
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If the damage isn't caused by overheating, I suspect that inductive voltage transients exceeding the MOSFETs voltage rating may be the issue here. In that case the culprit(s) may be the physical layout, component selection and/or the lack of a snubber network.

Where is the freewheeling diode D5 physically located? Across the motor terminals, right next to the MOSFET Q6 or somewhere else? Even if the inductive transients generated by the motor are clamped by D5 right at the source (motor), the parasitic inductance of the motor wires will still subject the MOSFET to a voltage spike of its own every time the current is interrupted. This is especially true if the motor wires are long and/or follow different routes (large loop area).

Another potential issue could be the freewheeling diode's speed. The gate driver advertises a typical fall time of just 10 ns, while a fall time of 58 ns is advertised for the MOSFET (at 58 A, 125 V and a gate resistance of 3 ohm). This is slightly faster than the 75 ns reverse recovery time of that diode.


I'd add another freewheeling diode as close as possible to Q6, unless already present, in which case a snubber network should cure the problem. While the freewheeling diode would still take care of the majority of the stored energy, even a simple series RC snubber between the drain and source would take the fastest edge off that spike. Yet another modification could be to purposefully slow down the MOSFETs on-off transition by increasing R29, reducing the rate of change of current (and thus induced voltage) and allow more time for the diode to switch on.

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  • \$\begingroup\$ I've added a picture of a piece of the PCB (Top layer) in my original post. M+ is connected to D5 in the bottom layer which can't be appreciated. \$\endgroup\$ – Xavier Pacheco Paulino Sep 15 '18 at 23:12
  • \$\begingroup\$ As we are just guessing because I don't have the waveforms yet, let's say that voltage transients are the problem. I would like to fix the problem in this PCB (if possible) and then, if things improve, I proceed to design a new PCB with the changes. So, for now, do you think I should consider a simple RC network across drain and source? I've also seen RC series networks across the motor directly. \$\endgroup\$ – Xavier Pacheco Paulino Sep 15 '18 at 23:36
  • \$\begingroup\$ Should I try increasing R29 (3.3 ohms) to 10 ohms? \$\endgroup\$ – Xavier Pacheco Paulino Sep 15 '18 at 23:41
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    \$\begingroup\$ @XavierPachecoPaulino I think that an RC network (or just a capacitor without a resistor) right across the motor is primarily intended for reducing EMI from brush noise. If you want to reduce dV/dt at the MOSFET, you should add it across the MOSFET. You could indeed try to increase R29 to 10 ohms. \$\endgroup\$ – jms Sep 15 '18 at 23:42
  • \$\begingroup\$ So far, I will try the RC network across drain and source. Although, it's no easy task to design one. Another flyback diode and increasing R29 are good attempts as well. \$\endgroup\$ – Xavier Pacheco Paulino Sep 16 '18 at 0:35
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It looks like you chose parts with sufficient voltage rating. Something blowing up is therefore due to excessive power dissipation.

The first thing to check is whether the FET is being switched properly, and whether it is switching properly as a result. Any significant time spent in the transition region will quickly increase the power dissipation.

However, even with the FET switching instantly, you have to consider the power dissipation. With RDSON of 12 mΩ and 27 A, the FET will dissipate nearly 9 W when on. That FET can handle 9 W, but only with a proper heat sink or some kind of cooling. Add to that the inevitable additional dissipation during switching transitions. If you had just the bare package sticking up in the air, then it's no surprise it fried.

Response to comments

at 10% duty cycle the average conduction loss is only 0.9W

True. I was just trying to point out that dissipation at full on can't be ignored.

I expect the dissipation during switching transitions is the real culprit, especially since we haven't seen any waveforms to the contrary. 110 V AC rectified makes about 150 V DC. The worst case is when half that is across the FET and half across the motor. (75 V)(13.5 A) = 1013 W. Even short times in the transition region add up to significant dissipation.

27A are at full load, and I don't intend to reach that point.

You don't have a choice. The motor will draw the full current when starting. Actually it's not clear whether 27 A is the expected running current or the stall current. If that's only the maximum expected running current, then the numbers I show above get a lot worse.

The inductance of the coils do help some, but at only 10 kHz it's hard to guess how much. With high inductance, the current changes little between the on and off phases of the PWM, and the current during the on time approaches the average. However, when switching slowly relative to the inductive time constant, you get much higher than the average current by the end of the PWM on phase. Since on-state dissipation goes with the square of the current, the latter results in more dissipation.

Too slow switching also results in more dissipation in the motor, something you likely want to minimize at this power level. To convince yourself of that, consider the current thru the motor as a composite of the DC average and the AC ripple due to the pulses. Only the DC average moves the motor. The AC ripple does no useful work, and only causes heating due to I2R losses in the resistance of the motor coils.

Again, we need to see waveforms to determine how well the FET is switching, and what the ripple component of the motor current is.

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    \$\begingroup\$ Good point, although at 10% duty cycle the average conduction loss is only 0.9W (a little higher due to the on-resistance rising with temperature). That might or might not need a heatsink, depending on airflow. \$\endgroup\$ – pericynthion Sep 15 '18 at 22:53
  • \$\begingroup\$ 27A are at full load, and I don't intend to reach that point. But I chose parts far above that value. I've added a picture of a piece of the PCB (Top layer) in my original post. M+ is connected to D5 in the bottom layer which can't be appreciated. \$\endgroup\$ – Xavier Pacheco Paulino Sep 15 '18 at 23:14
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Hay, your MOSFET looks good with Vds (2 times Vbus), Rds (12 mOhm) and Ids (128A).

Can you provide additional info about your DC motor like full load starting current ?

5% duty cycle is too low, I have seen geared DC motor that don't even start up before 60 % duty cycle and keeps humming with low duty cycle.

Are you taking proper ESD protection handling MOSFETs ? Sometimes MOSFET blow up simply because of damaged Gate insulation ( the SiO2 layer which have +/-20 V spec ). Try again, sometimes components are faulty to begin with - blow at least 3 MOSFETs and then come back !

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  • \$\begingroup\$ That MOSFET is insanely expensive. I want to avoid blowing many of them. I will also upload the motors' nameplates. \$\endgroup\$ – Xavier Pacheco Paulino Sep 16 '18 at 0:31
  • \$\begingroup\$ Blowing MOSFET was a joke, man ! Is it essential to use DC motors for your solution, why don't use AC with VFD for speed control ? \$\endgroup\$ – user198591 Sep 16 '18 at 2:10
  • \$\begingroup\$ I was doing some research on MOSFET blow up reasons Here is a good article on this matter: powerelectronictips.com/how-and-when-mosfets-blow-up \$\endgroup\$ – user198591 Sep 16 '18 at 2:12
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The switching of 50 amps in 10 nanoseconds, across 5cm of PCB trace with NO plane underneath, thus about 50nanoHenries inductance (1nH/millimeter), produces

V = L * dI/dT

V = 50 nH * 50 amps/ 10 nanoseconds, and the nano cancel the nano

V = 50 * 50/10 = 250 volts

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