Updated schematic

I am using this solution to drive a large (huge) relay on a boat.

The system works well for months on end, but eventually the MOSFET (T2) dies suddenly.

I have modified the schematic to include the modification suggested by Jack Creasey and others (circled in RED, not part of original design), as well as a further explanation as to how the system is wired to battery ground to illustrate why the typical snubber diode may be inadequate (it dissipates back-EMF to battery ground through a 4-ohm wire).

Simulations suggest that with this cable resistance back-EMF may push LOCAL_GND down to -20 V with respect to BATTERY_GND, the controller would see 34 V and fry.

I am hoping to find suggestions on the correct way to protect the MOSFET in this situation.

I am not able to make modifications to anything outside of the gray box and no strong GND connection is available in this part of the system.

12VDC is the battery voltage and may vary depending on load, generator output, other loads switching, etc. Typically 11-14.4 volts.

The relay draws approximately 6 A from PULSE_OUT, but I have never had access to the actual relay to perform measurements on its inductance or behavior when switched on or off.

Because the relay is bistable and requires only a pulse to turn on, PULSE_TTL is driven with 200 ms pulses at 3.3 V. These pulses are typically very infrequent, but they can occur every 5 seconds worst case. Another identical circuit drives the turn off coil in a similar fashion.


The answer is simple, you need a diode across the output wire to the relay.

When you turn the current to the relay OFF you will get a large negative back EMF spike on the drain of the FET (IRFR5305). The spike will likely far exceed the 55 V rating for the FET, and although the FET will avalanche this may not provide the required protection depending on the energy stored in the relay coil.

Connect the diode like this:


simulate this circuit – Schematic created using CircuitLab

While I've shown a BAT54 here, almost any power diode will do. Even an 1N4001 would be more than adequate.


Your update to the question posed what problems long thin (small gauge) wire to the really would pose. I would suggest that there is no problem.

If the wiring is sufficient to carry a peak current of 6 A to drive the relay, then current at turn OFF cannot be higher than 6 A, and will fall exponentially after the OFF drive transition.

If you are really concerned about your configuration then I'd suggest the following configuration where the FET is kept on to dissipate the inductive energy.


simulate this circuit

Enter image description here

The waveforms would look like the above, but I just used a dummy L limiting to 6 A to demonstrate.

In the schematic above, the components in the box result in the FET being turned on at about 34 V back EMF from the relay. This is all reference to the +12 V supply, so is not impacted in any way by the wiring. Allowing the back EMF to develop to 34 V means the FET never sees anything above 44 V (12 V + 34 V) so comfortably within its 55 V rating. And before anyone asks ….D2 does not clamp during the back EMF, the voltage on the gate will be just enough (V(GS)) to cause M1 to conduct sufficiently to clamp the drain voltage.

NOTE: Your base resistor to Q1 is too high. You should lower this to about 2700 ohms.

| improve this answer | |
  • \$\begingroup\$ Thank you so much for answering! I understand how back-EMF works (and I typically do put in this diode) but in this case I did not for two reasons: 1) i thought the MOSFET avalanche would take care of the problem better than a diode - is this completely wrong in this case? 2) I do not have access to a strong GND connection at the MOSFET device side (0.5mm²/AWG20 for GND vs 2.5/AWG14 for 12V) Is there other ways of safely dissipating back-EMF into +12V such as a zener diode gate to drain, or should I not worry about the wire guage? \$\endgroup\$ – Ronald McFüglethorn Jul 13 '19 at 16:01
  • \$\begingroup\$ Forget wire gauge ….put the diode right at your FET or if it's easier right across the relay. There is some technical difference in placement but it will not impact you. \$\endgroup\$ – Jack Creasey Jul 13 '19 at 16:15
  • \$\begingroup\$ I am not able to modify anything outside of the device, so putting it on the relay side is out of the question. Of course I would put it in right next to the FET, but this means all of the back-emf energy will be dissipated to GND through a rather thin and long wire. If the resistance/inductance of this wire is too high, the diode will not be very effective is what I'm thinking. It would be as if D1 (your schematic) was connected to GND through about 4 ohm worst case - will this really be enough to add significant protection? \$\endgroup\$ – Ronald McFüglethorn Jul 13 '19 at 16:23
  • \$\begingroup\$ The long wire makes no real difference. \$\endgroup\$ – Jack Creasey Jul 13 '19 at 16:31
  • \$\begingroup\$ I have updated my original question to reflect exactly where the weak GND connection is present and how it affects the snubber diode. Perhaps you are saying that 4 ohms is simply insignificant in this case? \$\endgroup\$ – Ronald McFüglethorn Jul 13 '19 at 17:10

When you switch off an inductive load (your large relay) you may well get a large voltage spike. This will eventually damage the transistor if not immediately. You do not appear to have included a snubber network across the MOSFET (12VDC to PULSE_OUT) which will protect it. I suggest you read the appropriate Wikipedia article as a start, then google "Snubber network" for more detailed articles and design notes.

| improve this answer | |
  • \$\begingroup\$ Thank you for answering! I understand back-EMF and snubber diodes and I typically include these. I had ignorantly thought/hoped that in this case the MOSFET would avalanche and adequately handle the back-EMF without the addition of an external diode, but I think you are correct that I need stronger protection. I have updated my question to include your suggestion as well an an explanation on how GND is tied to the system, which I believe presents a problem for the typical snubber solution. I would love your take on this. \$\endgroup\$ – Ronald McFüglethorn Jul 13 '19 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.