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My goal is to programmatically switch between supply voltages (3.3V and ~2.6V) for a load. The 2.6V level doesn't need to be exact, +/- 0.2V is OK.

I created the following circuit, which uses my MCU's GPIO to control an NPN BJT / P-channel MOSFET pair that drives a load (tens of milliamps). The parallel diode drops the voltage down to ~2.6V by default. When the MOSFET is on, it conducts between source and drain and essentially sets the supply voltage to 3.3V.

Are there any red flags in this circuit or any other minor issues I may have overlooked?

Schematic 1

Additionally, if the first circuit works, could I remove the the parallel diode and use the body diode of the MOSFET instead? This would involve letting current flow from drain to source when the MOSFET is on though so I'm not sure.

Schematic 2

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    \$\begingroup\$ You can't just flip the PMOS to use the body diode. The do not behave symmetrically because the gate specifically references it's voltage to the source terminal. The gate-source voltage has no control over what the body diode does. A MOSFET can only block current in one direction. \$\endgroup\$
    – DKNguyen
    Commented Aug 26, 2021 at 23:32
  • \$\begingroup\$ @DKNguyen I see what you're saying, thanks for the input. Would moving the 10k pull-up resistor from the drain to the source work then? Presumably the body diode would drop the voltage on the source to ~2.6v, which would mean V_GS = 0, and when the BJT turns on V_GS = -2.6V, which is enough to turn on the MOSFET. \$\endgroup\$ Commented Aug 27, 2021 at 0:00
  • \$\begingroup\$ Also as clarification, I don't want the MOSFET to block current, I just want it to short the diode (whether it's the body diode or an external one) to alter the output voltage. \$\endgroup\$ Commented Aug 27, 2021 at 0:03
  • \$\begingroup\$ @kirtan-shah Oh, yes In that case that would work since you select between a body diode and ideal diode but that doesn't change the fact that to make it work you need to modify the gate-source control scheme. \$\endgroup\$
    – DKNguyen
    Commented Aug 27, 2021 at 3:03

2 Answers 2

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As DKNguyen points out, your second circuit will not work, because you are operating the P-channel MOSFET as a source follower. In this configuration its source potential is always lower than its gate by an amount approximately equal to its gate-source threshold voltage (\$ V_{GS(TH)} \$).

Your original (top) design will work quite well, for low load currents. Your claim that the MOSFET will effectively short-circuit the diode is correct. However, the bipolar junction transistor's base cannot exceed 0.7V, so you must protect it from the GPIO's higher potential, and limit current there with another resistor:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

Be aware that Q1's \$ V_{GS(TH)} \$ is critical here. You must use a MOSFET with \$ V_{GS(TH)} \$ lower than 3.3V. As current through the load increases, so does the gate-source potential difference required to support that current. Also, Q2's saturation \$ V_{CE} \$ could be as high as 0.2V, somewhat limiting the best \$ V_{GS}\$ you can achieve. All things considered, \$ V_{GS(TH)} \$ will have to be significantly lower than 3.3V.

Also, please label the components in your schematics properly, it makes answering the question so much easier.

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  • \$\begingroup\$ Good catch on the BJT base voltage, and thanks for the feedback on labeling (this is my first question). \$\endgroup\$ Commented Aug 27, 2021 at 5:54
  • \$\begingroup\$ FYI I'm planning on using AO3401A for Q1, whose V_GS(TH) is -0.9V. \$\endgroup\$ Commented Aug 27, 2021 at 5:55
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Additionally, if the first circuit works, could I remove the the parallel diode and use the body diode of the MOSFET instead? This would involve letting current flow from drain to source when the MOSFET is on though so I'm not sure.

Provided the MOSFET has a sufficiently low Gate drive voltage specification it will work fine.

Initially the voltage between Gate and Source will be lower due to voltage drop across the body diode. Assuming a typical value of 0.8 V at practical body diode current (limited by power dissipation), subtracting this from 3.3 V leaves 2.5 V between the Gate and Source. Once the FET turns on the Gate drive voltage will increase to almost 3.3 V, which will turn it on harder.

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