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enter image description here

I have built this circuit to drive a 2 W variable solenoid on connector J5. The signal PWM_V is a 5 V, 1400 Hz PWM signal from a microcontroller that varies the duty cycle via potentiometer input. The MOSFET is a DMG3402L N-channel.

The problem is that the MOSFET seems to be always on; the voltage I measured with a multimeter across the two connector pins (J5) varies between 23 and 24 V depending on the pot input.

The signal on PWM_V measures with the right duty cycle and voltage.

I can’t find the problem. Is the MOSFET fried? Wrong component? Any ideas?

EDIT: Thanks a lot for your answers. So, the MOSFET is probably fried. I changed the schematic and will test as soon as the missing parts arrive:

enter image description here

ocrdu
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VLKRT
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  • `voltage I measured with a multimeter` with what kind of parallel load? Try at least a 10kΩ resistor. – greybeard Jan 07 '23 at 09:51
  • @greybeard Thanks for your reply! I measured with the solenoid attached, but I just tried it with a 10k resistor and without the solenoid. Now the voltage is stable 23.3V and doesnt vary with the PWM anymore.. – VLKRT Jan 07 '23 at 09:58
  • While [snubber](https://electronics.stackexchange.com/search?q=%5Bsnubber%5D+is%3Aanswer+score%3A1) belongs as close to a switched inductor as possible, the cautious would have a somewhat fast diode from drain to +24 V just in case (reverse biased…). It would offer some protection to the MOSFET in case an inductive load without snubber gets connected. If that already happened, chances are the MOSFET is fried. (Measure drain to source in 'meter diode setting in both directions.) – greybeard Jan 07 '23 at 10:06
  • Please add a diode across your solenoid. Measure if your MOSFET is intact. – winny Jan 07 '23 at 10:12
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    (Can you add a picture of how the MOSFET is connected? One explanation of voltage varying with a higher load is source&drain transposed.) – greybeard Jan 07 '23 at 10:54
  • Revised circuit looks good. Mounting diode physically close to solenoid terminals helps. Mounting Zener physically & electrically close to FET gate & source leads helps. || If using without a driver (direct from controller) R401 should pobably be more like 10 Ohms and FET switching times MAY be slow enough to cause more FET heating than desirable. If it doesn't get hot it's OK. If a gate driver is needed an extremely simple circuit will work. Details if wanted. – Russell McMahon Jan 11 '23 at 12:33

2 Answers2

8

Your MOSFET is probably dead, assuming that your circuit is as shown.

Test:

  • Connect +5V to input. If MOSFET does not turn on it is dead, if your circuit is correct.

A solenoid MUST have a reverse diode across it or your FET will almost always die at the first turn off due to inductive spike.

Adding a 6v3 or somewhat higher zener diode connected gate to source, cathode to gate (so non conducting with gate at +5V will greatly improve robustness.

FET datasheet here

Update:

Your revised circuit with solenoid connected reverse diode and gate to source zener looks good.

Mounting the diode physically close to the solenoid terminals helps.
Mounting Zener physically & electrically close to the FET gate & source leads helps.

If using without a driver (direct from controller) R401 should probably be more like 10 Ohms and FET switching times MAY be slow enough to cause more FET heating than desirable.
If it doesn't get too hot it's probably OK.

If a gate driver is needed the extremely simple circuit below will work. I've used that circuit for 20+ years, but I copied that diagram from my prior answer here , and the original source is this 2011 article.

enter image description here

Russell McMahon
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4

The problem is that the MOSFET seems to be always on

That's a common failure mode (possibly the most common) when the drain is "hit" with an over-voltage.

The drain-source rating is only 30 volts and, the energy built-up inside the solenoid (when it is first activated) could be several milli joules and this would be the last time the MOSFET was still operational because, to release that energy (without a flyback diode) forces the stored energy into the deactivating MOSFET.

During MOSFET deactivation (rapidly going open-circuit), the voltage produced by the solenoid will be whatever-is-needed to keep forcing current into the only place current is likely to flow (the MOSFET). This could easily generate hundreds of volts and puncture the internals of the 30 volt rated MOSFET.

Andy aka
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