I am working on a board with a battery that sits around 3.7V. I was using a diode as a reverse polarity protection, but I wanted to switch it to a MOSFET so I'm implementing this circuit.

enter image description here

I know that the MOSFET current depends on VGS, which in my case would be around 3.7V, but that would produce too much current into my board that I am not using. I also know that current is pulled not pushed, but how does that work with the MOSFET? Will it only produce current relative to how much my load needs only?

Do I need to add a Zener diode along with a resistor or is it not needed in this case as VGS is well within the MOSFET capabilities?

  • \$\begingroup\$ What you have shown is a reverse polarity protection mosfet, not a current limiter. \$\endgroup\$
    – Aaron
    Sep 23, 2022 at 3:10
  • \$\begingroup\$ Yes, I know that. My question is about limiting the current going out from the MOSFET that I use as protection in this circuit. Do I even need to? \$\endgroup\$ Sep 23, 2022 at 3:21
  • \$\begingroup\$ Typically you size the mosfet to handle the full load current plus margin. \$\endgroup\$
    – Aaron
    Sep 23, 2022 at 3:29
  • \$\begingroup\$ Yea, so for example, the MOSFET that I am using can handle 4A and my circuit draws max 1A. My question is, regardless of the VGS, will the current coming out of the MOSFET only depend on my load. Like it won't push extra current into my load? \$\endgroup\$ Sep 23, 2022 at 3:33
  • \$\begingroup\$ Yes, it will depend only on the load. Where would it get the extra current from anyway? \$\endgroup\$
    – Hearth
    Sep 23, 2022 at 3:37

1 Answer 1


The MOSFET will not affect the load current. If the maximum load current is 2 A without the MOSFET, then it will be 2 A with the MOSFET. The current will be determined by the load, not the MOSFET. In this case, the MOSFET is just like a switch which is on when the polarity is correct, and off when the voltage polarity is reversed.


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