I am looking for a way to use a MOSFET as a normally closed switch such that when a battery charger is attached it opens the switch to disconnect the battery from the rest of the circuit. I want the load to be between the MOSFET source and ground, instead of between the battery and the MOSFET drain.

EDIT: I want the load of the battery to be disconnected. Thus the battery is only connected to the charger. The battery is a single cell LiPo and the charger will be built into the board so just charging power is needed to charge the battery.

  • 3
    \$\begingroup\$ Just go ahead and do it. \$\endgroup\$
    – PlasmaHH
    Commented Oct 27, 2015 at 21:09
  • 1
    \$\begingroup\$ If you want the battery to be disconnected when the battery charger attaches, how on earth is it going to charge the battery. Now c'mon, what do you really want!! \$\endgroup\$
    – Andy aka
    Commented Oct 27, 2015 at 21:21
  • \$\begingroup\$ I assume that he means he wants to disconnect the battery from the circuit that it would otherwise be powering the circuit... Maybe? \$\endgroup\$
    – Matt Ruwe
    Commented Oct 27, 2015 at 21:22
  • \$\begingroup\$ There have been questions about this type of thing before, just browse through them and you will most probably find what you need. Maybe even with a couple of diodes. \$\endgroup\$ Commented Oct 27, 2015 at 21:22
  • \$\begingroup\$ I'd be hesitant to put diodes between a battery and its charger. Simple chargers won't care, but they'd have to be specially designed to account for the diode drop. Complex chargers would get really confused as the battery disappears and reappears depending on what part of the algorithm they're in. \$\endgroup\$
    – AaronD
    Commented Oct 27, 2015 at 21:25

2 Answers 2


Look at depletion-mode FET's. They're on by default and require some gate voltage to turn off.

Where an enhancement N-channel requires positive voltage to turn on, a depletion N-channel requires negative voltage to turn off. Similar relation between P-channels.

I'll leave the construction details to the student.

  • 1
    \$\begingroup\$ But you will have a hard time finding a depletion type mosfet that will operate like a decent switch. Uhm, I'd be surprised if you can find any. All switching mosfets are enhancement fets. \$\endgroup\$ Commented Oct 27, 2015 at 21:20
  • \$\begingroup\$ True, but I gave him what he asked for: default on, forced off. He should read the datasheets anyway. \$\endgroup\$
    – AaronD
    Commented Oct 27, 2015 at 21:23
  • \$\begingroup\$ @FakeMoustache -- there are depletion N-channel FETs out there that can be pressed into switching service in a pinch, see the Microchip/Supertex LND01 for an example. \$\endgroup\$ Commented Oct 27, 2015 at 22:45
  • \$\begingroup\$ @ThreePhaseEel OK, then I'll have to eat my hat ;-) Not a very low Rdson though (1.4 ohms) :-( \$\endgroup\$ Commented Oct 28, 2015 at 10:07
  • 1
    \$\begingroup\$ @matteo That would be a P-channel. Less performance than N-ch for the same size die (and often package), or physically bigger for the same performance, but it does have the control law that you asked for. Increasing voltage turns it off. \$\endgroup\$
    – AaronD
    Commented Dec 9, 2019 at 14:55

You could try to use some Schimmit Trigger CI to invert the drain input signal, so when its input is low, the output will put high, and vice-versa. This basically "invert" your transistor gate. Look at some CD40106/74LS14 datasheet and see if they fit in your application. `

https://www.fairchildsemi.com/datasheets/CD/CD40106BC.pdf https://www.egr.msu.edu/eceshop/Parts_Inventory/datasheets/74ls14.pdf


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.