WeMos self-turn-off with N-channel MOSFET

Question

I'm trying to build an internet-connected button. When button is pressed, WeMos D1 turns on, makes a network call and then cuts off the power by pulling the irlb8721 gate to LOW.

Video: https://photos.app.goo.gl/QEpIgPa3gxHatTyG3

Here's the program I currently use for test:

void setup() {
  pinMode(2, OUTPUT);
  digitalWrite(2, LOW); // Turn on built-in LED

  pinMode(5, OUTPUT);
  digitalWrite(5, HIGH); // Keep ourselves powered after button is released
  delay(5000); // Do the work, e.g. make a network call
  digitalWrite(5, LOW); // Cut off the power
}

void loop() {}

The problem is that after the power is turned off, there's still ~20mA current leaking, which is a deal-breaker for a battery-powered button. Disconnecting the pin 5 (D1) gets rid of that for some reason.

Resistor between gate and source is 676kOhm, between D1 and gate - 10kOhm. Thanks for your help!

Answer 1

I think , because you are using the WeMos to turn itself off, what is happening is as the MOSFET turns off, the voltage on the pin driving it goes up.

Or to put it another way, the pin driver is standing on the ground it is trying to pull up.

This is what you are trying to do..

^{simulate this circuit – Schematic created using CircuitLab}

Without studying the specs on the device that internal pull-up may or may not be present and controllable.

At some point it will turn into some form of linear regulator or oscillate.

You may have more success using

pinMode(5, INPUT);

instead of

digitalWrite(5, LOW);

That will allow the pull-down to do it's job, assuming that pull-up does not exist or can be turned off.

If it exists and can not be turned off, you need to significantly reduce the pull-down value and gate resistor and add another one to the switch line and eat the current loss through the divider.

^{simulate this circuit}

However, either way, cutting it's own throat may be problematic since the device may reset itself during the power down or brown out.

Answer 2

This is due to leakage through the circuitry of your MCU. Even when your MOSFET is turned off, the MCU remains connected 3.3V. Some current can flow from the 3.3V input, out through pin D1, through the 10k resistor, through the 676k resistor, to ground. This pulls the gate of the MOSFET up, allowing more current to flow (but due to the forward voltage drop of the diode, the gate voltage remains limited and perhaps the MOSFET does not turn on enough for the MCU to boot up).

A possible solution would be to use a high-side switch instead (using a P-channel MOSFET), whose gate you pull down using a second (N-channel) MOSFET. This is illustrated in an answer to another question on this site.