simulate this circuit – Schematic created using CircuitLab

I'm trying to build a circuit with 2x NiMH batteries as power source that can output up to 2A when tact switch at mosfet gate is closed, and will stop at battery voltage is at or below 2.0V (to protect batteries from over discharging.)

To achieve this, I'm using a MOSFET with minimum Vgs of 2.0V.

The problem is, I can't find a way to do voltage protection and tact switch protection at the same time.
If I want to prevent my tact switch from burning, I'll have to limit the current passing through by adding resistors. But doing this is of course impractical as it will make gate voltage too low to operate and work as over discharge protection....

I'm really curious if what I'm doing is actually feasible. I have no electrical engineering background so I have very limited knowledge on ways and parts to achieve desired function.

I do did some research and find some circuit called Soft Latching Power Switch, which uses single tact switch to control high current circuit.
It is very close to what I want to do, but I have no idea how to tweak the circuit to realize my thoughts...

  • 2
    \$\begingroup\$ "tact" means what? Where's your circuit? \$\endgroup\$
    – Andy aka
    Commented Jan 20, 2019 at 14:59
  • 1
    \$\begingroup\$ @Andyaka Sorry if I misused any terms or words, I'm not a native English speaker.... Is "tactile switch" more precise? I'm talking about those tiny SMD type button switches. I'll try to draw the circuit and will post here when I'm finished! \$\endgroup\$
    – Jimmy.D
    Commented Jan 20, 2019 at 16:17
  • 1
    \$\begingroup\$ Please draw a schematic. There is a built-in tool here if you press edit and then on the schematic symbol. \$\endgroup\$
    – winny
    Commented Jan 21, 2019 at 15:56
  • \$\begingroup\$ Added original schematic in my mind as requested. Nothing special really, but I hope it would help! \$\endgroup\$
    – Jimmy.D
    Commented Jan 21, 2019 at 16:23
  • \$\begingroup\$ It sure did! @Jimmy.D \$\endgroup\$
    – winny
    Commented Jan 21, 2019 at 16:34

2 Answers 2


The main feature of the MOSFET is very high Gate impedance, which basically means the gate-channel "connection" works like a capacitor. The only time when any significant current flows in the Gate circuit is when the Gate is loading its charge when its connected(happens very fast). You could basically connect the switch between the battery positive terminal and the Gate, (and the positive terminal to drain and 0 to source) and it Will work without burning your switch, but a single resistor in series will limit the initial loading current. This way the current Will flow from drain to source when the battery voltage is over the threshold. Just remember to use some load and not connect the battery directly to the drain and source with the switch closed.

  • \$\begingroup\$ Clear and informative answer! I'll definitely try it when I get the chance \$\endgroup\$
    – Jimmy.D
    Commented Jan 21, 2019 at 16:27

These FET's have too high a threshold voltage from 2 to 4V.

You must use a FET with a Vt or Vgs(th)~ < 1/2 of your batteries lowest voltage of <2V

Thus search for low RdsOn < 1% of load with \$V_{gs(th)} << 1V\$ such as 0.7 to 0.9V

Then you may have to learn to solder SMD chips which are preferred in this range.

Then add gate resistor to source (0V) of any high value (<1Meg) to ensure off when the switch is open.

This will not latch, so you need a low voltage logic chip to make a set/reset latch or a toggle flipflop. (search this site for clues)


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