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I read a lot about MOSFET protection and built this scheme.

schematic

simulate this circuit – Schematic created using CircuitLab

I was thinking that gate and DS are protected enough but experience showed that I was wrong. I made 3 PCB's and all blowed. Motor during continous work takes 20A but I'm sure that overcurrent or power loss wasn't reason why MOSFET blown.

1. Why it doesn't work?

Now I know that D2 should be between R2 and SW1 but what else could cause damages?

2. How can I make it possible most failure-free?

I've seen scheme where between drain and gate was added bidirectional TVS diode but whether it is better solution than unidirectional TVS diode between source and drain?

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  • \$\begingroup\$ What MOSFET are you using? What is the maximum Vgs? \$\endgroup\$ – Evan Dec 7 '16 at 23:53
  • \$\begingroup\$ Vgs = 20V, Vds = 30V, Id = 100A \$\endgroup\$ – Piotr Dec 8 '16 at 7:36
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    \$\begingroup\$ Add a flyback diode across your motor! \$\endgroup\$ – winny Dec 8 '16 at 15:39
  • \$\begingroup\$ I can't add flyback diode because project conditions doesn't allow to connect flyback diode cathode to VCC. The only way to do this is connect flyback diode between R1 and SW1 so I'm looking for other solutions. \$\endgroup\$ – Piotr Dec 8 '16 at 16:45
  • \$\begingroup\$ From what you have shown, there is no reason you can't put a flyback diode across the motor, which leads me to think that you haven't explained enough of the problem to get good answers. When the MOSFET switches off, the current flowing through the motor will continue to flow, and it will break the MOSFET unless you provide a return path to for it that keeps Vds < 30 V. \$\endgroup\$ – Evan Dec 8 '16 at 17:38
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First, your motor inductance with 1 uH is totally unrealistic... A more realistic value would be in the milihenry range.

Second, I think you problem is that you have an overvoltage condition in the drain to source voltage. Your gate seems well protected with a TVS diode and a 100 ohms resistor in serie with the gate. However, you need a fast recovery diode across the motor. When the transistor will turn off (or become in open circuit as others prefer to say), the current trapped in the motor inductance must be able to flow somewhere and this is where the fast recovery diode come into play. In your range of voltage, a schottky diode would be perfect to be placed in anti-parallel with the motor.

If you don't do this, the high inductance in your motor will cause a big overvoltage condition than can ring well over your maximum drain to source voltage (30 V in your case). You may think a TVS diode could shunt the overvoltage but, in order to do that, they must conduct a lot of current which TVS diode are not made for. This is why a shottky diode or a fast recovery diode are much more suited for your application.

I hope this answer will solves your problem!

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  • \$\begingroup\$ Thank you for answer! Yes, It was random value of motor inductance. I've found second solution: TVS between gate and drain. Will it work in this situation? \$\endgroup\$ – Piotr Dec 8 '16 at 16:46
  • \$\begingroup\$ I don't think replacing (or augmenting) D1 with a fast recovery diode will help -- the direction is wrong to conduct the positive current when the mosfet switches off. You need either a diode across the motor itself, or a reverse-breakdown style diode at D1, either a TVS or a regular zener diode. \$\endgroup\$ – Evan Dec 8 '16 at 17:30
  • \$\begingroup\$ @Evan oups... yes you are totally right! The fast recovery diode must be in anti-parallel across the motor as you said! I shall edit my post... \$\endgroup\$ – A. Baril Dec 8 '16 at 19:29
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    \$\begingroup\$ @Piotr I don't think a TVS diode across the drain and gate would help.... I never saw a TVS diode connected there and in theory I can't find a reason why this would help... sorry! \$\endgroup\$ – A. Baril Dec 8 '16 at 19:36
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    \$\begingroup\$ @A.Baril When voltage on drain is too high, TVS is conducting, so MOSFET turns on and current from transient is flowing to groud via transistor. On the upper coment it is scheme. \$\endgroup\$ – Piotr Dec 8 '16 at 20:26
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Consider the inductance in SOURCE, given 20 amps is being switched. Assume that inductance is 20nanoHenry (about 1 inch of wire). Assume the 20 amps switches in 20 nanoSeconds. What happens? V = L * dI/dt

Delta_V_Source = 20nH * 20amp/20nS = 20 volts

Your FET is not off. Until that transient is completed.

Also the inductance, plus Cgate, may resonate. Install 100 Ohm, with body of resistor right against the Gate pin. (gate-stopper R).

Also, how far away is the power supply? A meter? 1uH of wire. Put 0.1UF right between top of the Motor and the GND PLANE. Purpose: to minimize the loop area, thus minimize the magnetic-field stored energy.

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