In the following circuit, LEDPWR goes the power lines of some digitally addressable LEDs. But the LEDs are not always on, and even when they are not being lit, they are consuming about half a milliamp each, which will drain the battery over a few days even if it's not in use.

I was under the impression that I could use a MOSFET like a switch. So the micro-controller could simply set the gate to LOW and the LEDs would no longer be connected to power and would not drain the battery when not in active use.

This is the circuit I devised.

enter image description here

But it doesn't work like I expected. When I pull the gate LOW, current appears to continue to flow, and the LEDs continue to draw power. Why?

I tested each prong of the transistor with a multimeter in both the on and off states and found that despite the gate being LOW, I'm getting voltage on the drain. Here's what I measured:


  • Gate: 4.20V
  • Source: 4.36V
  • Drain: 3.73V
  • Current: ~18mA


  • Gate: 0V
  • Source: 4.11V
  • Drain: 3.51V
  • Current: ~5mA

If I disconnect the LED power wire in the OFF state, current plummets to less than 0.1mA, so the MOSFET is clearly not doing what I expect.

So first, how do I get my MOSFET to actually turn disconnect the source and drain?

And second, why am I seeing a 0.6V loss at the drain? With a single cell LiPo (3.7-4.2V) as a power source, that's a lot.

My MOSFET's datasheet: MOSFET datasheet: http://www.nxp.com/documents/data_sheet/2N7002P.pdf

  • 1
    \$\begingroup\$ Your mosfet is upside down. Look at the body diode in the schematic, see ? It's forward biased, that explains the 0.6v drop. \$\endgroup\$
    – Mike
    Commented Jan 21, 2015 at 9:28
  • \$\begingroup\$ Is it? I thought I read that the source should be your power, and the drain should be the load, which matches my footprint and schematic, I think. \$\endgroup\$
    – Alex Wayne
    Commented Jan 21, 2015 at 9:34
  • \$\begingroup\$ source -> power and drain -> load is a configuration of a P-Channel mosfet, what you have is an N-Channel mosfet. brunningsoftware.co.uk/Pictures/U-fig10.jpg \$\endgroup\$
    – Mike
    Commented Jan 21, 2015 at 9:44
  • \$\begingroup\$ Well crap. These pcbs are boned then :( So if I swap the source and the drain, you think this will work like I expected? If so, add an answer as such and I'll accept it. \$\endgroup\$
    – Alex Wayne
    Commented Jan 21, 2015 at 9:47

1 Answer 1


MOSFET is "upside down" as Mike said.


3 "fixes":

  1. Use P Channel FETs and all shall be wellish.

FET will be on when gate is LOW (~= 0 V).
FET will be off when gate is HIGH(~= 4.1V)

  1. IF you can isolate the MOSFET 4.1V and gnd feeds from the rest of the PCB by cutting tracks then

    • Swap MOSFET 4.1V and ground

    • Reverse LED polarity

  1. Depending on pads:


Bend leads UP

Invert MOSFET - solder MOSFET D to old S pad

Solder MOSFET S to old D pad.

Solder MOSFET G to G pad.

Or use wire tacked on leads to make it easier.

  • \$\begingroup\$ Thanks for the help! I just spent some time on solution 3 with a hot rework station and the bastard finally went in, but by now I think I cooked it. Penance for my mistake, surely. :P \$\endgroup\$
    – Alex Wayne
    Commented Jan 21, 2015 at 10:47

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