I want to add reverse polarity protection to a device which is powered by a single AA battery, so its voltage might go down to 0.8 V. Forward diode is too huge of a drop. I've already thought about adding P-MOSFET but what bothers me is a diode's forward voltage. It's usually around 1 V. What it means is that if battery is inserted the right way, MOSFET won't be open because Vgs is not high enough. As I understand, if after diode voltage is high enough, MOSFET will be opened and diode won't matter anymore. Am I wrong or should I consider other ways to implement reverse polarity protection?

Just to be clear what circuit I'm talking about:

enter image description here

I was considering Si2329DS which has a maximum Vgs of 0.8 V but diode voltage leaves it unusable.

edit: So, I've been doing some research and I've found ideal diode controllers that enable me to use NMOS. I'm going to tell if I find something suitable

  • \$\begingroup\$ 0.8V is pretty low for Vgs. The FET may then still have tens of Ohms between drain and source. What current do you need? \$\endgroup\$
    – JimmyB
    Commented Jul 27, 2016 at 17:57
  • \$\begingroup\$ Not much, 200mA should suffice \$\endgroup\$ Commented Jul 27, 2016 at 18:00
  • 1
    \$\begingroup\$ Silly question maybe: what will happen if the device is powered backwards? \$\endgroup\$
    – Transistor
    Commented Jul 27, 2016 at 18:00
  • \$\begingroup\$ Diode would be connected backwards, and Vgs would not be right to open FET, am I right? That way gate is high and source should be low, and it would mean positive voltage though negative one is needed \$\endgroup\$ Commented Jul 27, 2016 at 18:02
  • 5
    \$\begingroup\$ How about just shunting it away with a Schottky diode and resettable fuse? \$\endgroup\$ Commented Jul 27, 2016 at 18:18

3 Answers 3


Did you see (and understand) Spehro Pefhany's comment? I take the liberty of turning it into an answer because that's also what I'd suggest.

Add a diode in front of your circuit that shorts the voltage if the battery is reversed (=V2). In that case the (resettable) fuse will break the current flow. If the battery is not reversed (=V1) there is almost no voltage drop (only a small one across the fuse because it won't have 0 resistance):


simulate this circuit – Schematic created using CircuitLab

The diode must pass enough current at reversed polarity to deactivate the fuse. The fuse, of course, must be able to sustain enough current for supplying the circuit in normal operation but must blow/deactivate if the battery is shorted via the diode.

  • \$\begingroup\$ Yes, I've already seen it and see it as a good answer not only for me but also for future users. If you added what how diode and fuse properties should be selected that will be an almost ideal answer to what have asked \$\endgroup\$ Commented Jul 27, 2016 at 20:49
  • \$\begingroup\$ @Artūras Jonkus: what about a wire fuse? Another option would be to do completely without fuse (i.e. the battery will stay shorted via diode in case of reverse polarity) if that is safe and acceptable. \$\endgroup\$
    – Curd
    Commented Jul 28, 2016 at 7:51

You dismiss diodes very quickly. Have you looked for Schottky diodes or specialized diodes with very low voltage drop? Look for the TI SM74611. It has a Vf of 26mV and is available for less than $5 on digikey.

  • 1
    \$\begingroup\$ This is not an actual schottky, but interesting active device though. I'm wondering what the supply current for the device itself will be (as it is only specified from 2A onwards). \$\endgroup\$
    – Douwe66
    Commented Jul 27, 2016 at 18:37
  • \$\begingroup\$ One downside of it is its huge package - it's an overkill for device sinking only 200 mA \$\endgroup\$ Commented Jul 27, 2016 at 18:42
  • 3
    \$\begingroup\$ This device is probably not suitable for low-power operation. Note that the forward voltage jumps up to over 500 mV whenever the internal charge pump needs to run to charge the internal capacitor. The forward current at this time must also be enough to run the charge pump. \$\endgroup\$
    – Dave Tweed
    Commented Jul 27, 2016 at 19:41

Ok, I will add one more answer myself for more clarity for future. Usually when trying to block reverse current diode seems like a nice way, nevermind if it's Schottky or regular Silicon diode. The problem with it is a fairly high voltage drop which might be unacceptable in most of cases where device is battery powered. The good thing is that you can use devices called Ideal Diode Controllers. It drives N-MOSFET's gate so you can drive voltages as low as 0.6-0.8 V. Some of these controllers tend to work from 4 V or something onwards but there's LM74610 which is suitable for such an application


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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