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I am currently working on a drive for a DC motor and I want to add a reverse voltage protection on my circuit so it doesn't fry in the case of an inverted power supply. I've read many tutorials on that subject and I chose a protection using an enhancement mode N-channel MOSFET.

My circuit will be powered with a 24V source and most of the N-channel MOSFETs only have a max Vgs of +/- 20V which is a problem because in a case of a reverse voltage, Vgs will be -24V, over the absolute maximum characteristics for most MOSFETs.

My question is this: Can I use a resistor voltage divider to create a lower voltage at the gate, thus having a Vgs that is within the specifications?

Something like this:

Circuit protection with resistor divider

I have no background whatsoever with MOSFETs so I am wondering if that will cause any problems.

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You've got that FET drawn backwards, and most of the time gate protection like this is done with a zener diode.

You should make the 9k on your drawing into a 15K , and the 15k resistor into a 15V zener (assuming that 15V is enough to fully enhance the gate of your particular MOSFET). If your MOSFET needs 20V to fully enhance it, then use a 20V zener)

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  • \$\begingroup\$ I did not choose a MOSFET yet because of my problem. I would assume that I don't need 20V to fully enhance the MOSFET as it is it's absolute maximum rating. But I get the concept here. Thank you for your answer. \$\endgroup\$ Commented Feb 6, 2014 at 19:31
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Aside from the S-D swap (the MOSFET is actually used in reverse in such a polarity protection circuit), you should consider pathological cases of partial supply with the correct polarity.

Since this is explicitly a power application, if you have something like a battery or other kind of supply that can experience "brown-out" conditions you may wish to include an "undervoltage lockout" (UVLO) circuit to disable the power stage(s) when the input voltage is out of spec.

If this is not done, you could have the MOSFET operating partially "on" (since the gate voltage might not be enough to turn it fully on) and it could dissipate a great deal more power than would occur during normal operation or (reverse polarity).

To be a bit more explicit on the reason the UVLO is important, the gate voltage is important (to avoid burning up the MOSFET), and avoiding the zener in favor of a voltage divider means that the MOSFET might be less enhanced with a given input voltage. You need to work out the number to make sure that the MOSFET is guaranteed to be in an appropriate state for any possible input voltage to make your design bulletproof. I can't tell you that, but I can say that using a divider in place of a resistor + zener will not be better.

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  • \$\begingroup\$ The raw 24V input is used to create multiple rails (12V to 1.8V) depending on the motors requirements. Each of these rails have an ULVO, is that enough? \$\endgroup\$ Commented Feb 6, 2014 at 20:15
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    \$\begingroup\$ @JephGagnon did you know that with PWM, you can use the motor as a buck converter, and get the same effect of having all those rails, with probably less complexity? \$\endgroup\$
    – Phil Frost
    Commented Feb 6, 2014 at 20:56
  • \$\begingroup\$ A good point about the UVLO, but I'd point out you didn't actually answer the question: "Can I use a resistor voltage divider to create a lower voltage at the gate?" \$\endgroup\$
    – Phil Frost
    Commented Feb 6, 2014 at 21:01
  • \$\begingroup\$ @PhilFrost This is an interesting concept that I might explore later but right now each board is customizable with a feedback of 2 resistors and each board will be dedicated to a motor. So I only need to switch 2 resistors for a different output. It's not really complex as it is (2 chips, 1 rail and 1 driver) edit : and I save complexity for the control of the motor: I can set the decay and I just need 2 pins for the direction and the chips has many protection that I don't need to implement. \$\endgroup\$ Commented Feb 6, 2014 at 21:12
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    \$\begingroup\$ @PhilFrost Ok, I made the connection a bit more explicit, I hope. \$\endgroup\$ Commented Feb 6, 2014 at 21:39

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