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I am trying to use a N-channel MOSFET to toggle battery power going to a 5V voltage booster using the circuit shown in the attached image. Battery voltage is 3.7V and the gate voltage is 3.3V. High voltage at the gate is used to turn the power off and low is used to turn the power on (on by default using the 5K pull down resistor). The MOSFET switches as expected and I am able to toggle the boosted 5V on and off until I put a load in the 5V bus which seems to switch the MOSFET (BAT-SW in the schematic becomes grounded turning off power to the 5V booster circuit).

Here is a link for the MOSFET I am using https://statics3.seeedstudio.com/images/opl/datasheet/0440200P1.pdf

Any help or suggestion would b much appreciated!

Switching circuit diagram

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    \$\begingroup\$ You can see that any load you put on BAT-SW has to try to draw current through R2, yes? \$\endgroup\$ – brhans Jan 4 at 15:34
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As noted, your power passes through a resistor which is undesirable.

A high side switch usually uses a P channel device that has negligible resistance when on. Here is one way to do it:

schematic

simulate this circuit – Schematic created using CircuitLab

Taking enable high pulls down the N channel drain, turning on the P channel pass device. This particular pass device has about 6 milliohms of on resistance at these drive levels (3V or so).

I used a N channel device to drive the pass element to get around any logic leakage from logic drive circuits inadvertently turning on the P channel device.

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If I understand, the problem is R2. When you apply a load, you begin to draw current from the battery and it must flow through R2. According to V=IR, this will create a voltage drop across R2. So, at 0A, the voltage at BAT_SW is 3.7v. At 10mA, BAT_SW=3.7-300*0.01=0.7v. So the boost converter is struggling to boost 0.7v up to 5v and the problem gets worse as it tries to draw more current.

The real problem is the configuration of the transistor. You are turning off the boost converter by diverting power around it to ground. So even when it is off, you have current flowing through R2 to ground, effectively burning (3.7^2)/300=0.045W continuously.

I recommend you research “mosfet as switch” to find some tutorials about this. You probably want to use a P-channel instead because N-channel configurations disconnect the switched circuit from GND so you really have 2 grounds. This is fine for motors and stuff but probably not boost converters.

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