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I've got a design that I've inherited with a pretty standard n-channel mosfet driving a relay that controls a motor and actuator.

On a recent build we started getting a 50% failure rate on the n-channel mosfet. Previously we had no failures of the mosfet. The only differences I've been able to find so far are different date codes on the relay and mosfet. Otherwise nothing has changed.

The mosfet is an ON Semiconductor 2N7002LT1G

The relay is a Omron Electronics G6RL-1-ASI-DC24

The flyback diode is a ON semiconductor MRA4003T3G

The mosfet was examined by ON semiconductor and it was found that it was most likely destroyed by excessive voltage. But I've not been able see a voltage spike on the mosfet above 30V so far.

Here is the part of the circuit with the mosfet/relay/diode.

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5 Answers 5

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I'm guessing the diode wasn't soldered down properly on your recent build, or maybe you got some bad parts. Take one of the boards that failed, replace the FET, and look at the drain with a fast scope while the relay is being turned off. Then reflow all the solder connections around the diode and maybe even solder wires directly from the diode to the relay and look at the signal again.

You show the schematic, but not the physical layout. Where is the diode relative to the relay and the FET? If it too far away, then the inductance to it partially defeats its purpose.

Another possibility is that this was a bad design all along, and now you got some parts where the difference matters. Try putting a small cap immediately accross the relay. That will slow down voltage changes so that other parts of the circuit can keep up. If the relay is off board, then you have to protect the FET drain separately. This could mean a separate reverse diode on the board an maybe a small cap to ground on the drain. You don't want to put too much there because it will cause a small surge when switching on, but a few 100 pF to a nF or so should slow down the voltage changes.

What voltage is VBATT? Why isn't the diode a Schottky?

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    \$\begingroup\$ Thanks for your feedback. The diode is located right next to the relay. VBATT is 24V from a battery. I'm not sure why the diode is not a Schottky that would make more sense. The part is used other places on the board so I'm guessing they wanted to save on having different part types. \$\endgroup\$ Sep 19, 2012 at 23:17
  • \$\begingroup\$ You need the diode physically closest to the FET, not the relay. If the other end can't be close to the 24 V supply, connect a 0.1 uF from the diode to GND (all close to the FET -- within a few cm) \$\endgroup\$
    – jp314
    May 30, 2021 at 1:18
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  • Changing R38 to 10k MAY help.

  • Adding a zener across gate-source may help

Showing all the relevant circuitry may well help - in this case what is hiding behind ACTCTRL1 may or may not be relevant.

Why it would change between batches is not obvious, but something to check is that gate voltage can never exceed (or closely approach its max rated value (Vgsmax). This depends on the impedance of ACTCTRL1. Miller capacitance will couple turn off voltage from drain to gate and this MUST be clamped by attached gate impedance to less than Vgsmax. Vgsmax may vary between FET batches but this is not overly likely.

If there is any doubt then placing a zener diode of slightly more voltage than V_gate_drive_max from gate to source (cathode to gate so zener usually does not ever conduct).

R38 is probably far higher than necessary at 100k. Odds are that this can be say 10k and this may have been changed between batches without being notices. Miller capacitance energy has to drive this to above Vgsmax to destroy the FET so a 10k makes this 10x harder energy wise. With 5V drive a 10k will require 0.5 mA drive so most drivers will have no problem with this. If ACTCTRL1 is not a direct connection to a drive pin and has series resistance then this may need to be reduced proportionately.

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You may need a faster diode. The data sheet I'm pulling up for that part doesn't list a forward recovery time, which generally means it's long enough that nobody who cares about recovery time would use it. One batch of diodes may have had a faster recovery time, another slower, and now that you've got the slow batch the inductive kick is enough to break your FET before the diode can recover.

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You mention that the failure analysis points to overvoltage, so this may not be relevant, but ensure that the diode wasn't placed backwards. With a 500ma(max) FET and a 1A(max) diode, it is almost a surety that the FET will fail first in the case of a forward-biased diode.

We once had an assembly house do this to us with SMT diodes like yours (the silkscreen was totally obscured by the part). It took a embarrassingly long time to find, but was a simple fix...at a new assembly house.

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  • \$\begingroup\$ If this was the case then the relay would have never come on. That is something they presumably would have noticed immediately. \$\endgroup\$ Sep 20, 2012 at 11:23
  • \$\begingroup\$ @OlinLathrop - For extra points the diode could be destroyed by incorrect polarity and fail open and the relay would operate. .... Drum roll ..... THEN you turn the relay off :-) . Zap. \$\endgroup\$
    – Russell McMahon
    Sep 20, 2012 at 11:35
  • \$\begingroup\$ @Russell: I haven't looked at a the datasheets myself, but DeanB seems to think the diode is more robust than the FET. Still, this is certainly something to check. See if the diode is installed backwards on the failed units. \$\endgroup\$ Sep 20, 2012 at 12:14
  • \$\begingroup\$ Thanks. We will definitely go back and check the diode polarity on some failed units. I'm not sure if that was checked or not but I have had these sorts of parts install backwards before. \$\endgroup\$ Sep 20, 2012 at 13:27
  • \$\begingroup\$ @Olin - yes - I looked at this in my second answer - there is every indication that the FET would die 1st. Murphy may be having a good day and deem otherwise but, mostly, the diode should easily outlast the FET - main reason is that the FET current limits at about 1.5A with consequent very high Rds while diode drops about 1V at this current. Dissipation is almost all in FET and FET Rth is even worse than diode's. . \$\endgroup\$
    – Russell McMahon
    Sep 21, 2012 at 2:24
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I see that this is essentially what DeanB said. This adds a few figures and wanders around the general area a bit.

If D21 is installed with incorrect polarity, the FET will fail almost instantly. :

Failure from over-dissipation is almost certain.
If the diode fails instead, the FET will fail soon after due to inductive spikes.


On FET turn-on, the diode conducts from 24V to ground via FET.
Diode fails open-circuit.
Relay now operates.
On relay release, you now have an inductive spike and no diode ... :-(.

The 7002 is not overly high-current-capable and will probably current-limit at "a few" amps. It may be a race between diode and MOSFET to see which can self destruct first. If the MOSFET dies first, the relay never operates. If the diode dies first, the relay operates at least once, and possibly a number of times.

So:

  • Check diode polarity.
  • Observe Drain with oscilloscope.
  • Observe base with oscilloscope (see my other answer).

The diode datasheet here is rated at 88 C/W with 1 inch square pads so needs not overly-much overcurrent to die thermally.

The MOSFET is rated at 300 mW dissipation and 417 C/W !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! . Datasheet here With all the drive in creation, it's good for about 1.6A and will then drop as much voltage as you want to feed it, whereas the diode is hardly breaking a sweat at 1.6A with Vf of about 1 Volt, so if the diode is reversed, you'll get about P_transistor = V.I ~~~= (24-1) x 1.6 =~ 30 Watts.

Death would be nearly instant.

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