In the schematics of a Firebeetle ESP32 board (first PDF page) we can see this part for the 3.3V power:

enter image description here

This board can be powered in different ways:

  • From 5V USB (5V in the schematic)
  • From a 3.7V battery (VB in the schematic)
  • From a external power supply (Vcc in the schematic)

My question is what is the mode of the P-channel MOSFET SI2301 in case of power from battery? In this scenario, VB = 3.7V, Vcc and 5V are not connected. Thus, gate of MOSFET is at 0V across pull-down resistor RES1A (10k), drain is at 3.7V from battery, but what is at source?

What is the value of Vgs in this scenario?

Another similar circuit is this one (from here):

enter image description here

If first power supply is not-connected and second power supply is +5V, gate is at 0V across pull-down 1k resistor, drain is at +5V, but what is in source (what is Vgs value)?

  • \$\begingroup\$ @V.V.T: in the first circuit the symbol for D10 is for a Schottky of 6.2v and D2 normal diode, in the second circuit D1 symbol is for a Zenner (I expected both D1 and D2 being Schottky and D10 Zenner). \$\endgroup\$ Sep 1, 2021 at 10:54
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    \$\begingroup\$ Note that if VCC (ext PSU) is present and 5V (USB) is not, Vg = 0V and VCC is thus connected to the battery ... boom! \$\endgroup\$
    – user16324
    Sep 1, 2021 at 13:54
  • \$\begingroup\$ @user_1818839: yes, good point. Moreover, using Vcc it is not possible to charge the battery. Better think on Vcc only as output. Something not nice if other power sources than USB wants to be used, because there are not easy pin to connect it. \$\endgroup\$ Sep 1, 2021 at 19:10

1 Answer 1


There is an implicit "body diode" from drain to source. This diode passes current when forward biased regardless of the value of Vgs. This means that the source voltage is close to the drain voltage, making Vgs large enough to switch the transistor on.

  • \$\begingroup\$ Thanks for your answer. Could you add a few words about the kind of diodes in these circuits (see question comments) ?. \$\endgroup\$ Sep 1, 2021 at 10:56
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    \$\begingroup\$ D1 in the second diagram passes current from the first power supply to the load when it is connected. If it is a zener diode, then it also protects the MOSFET from excessive Vgs when that supply is disconnected. \$\endgroup\$
    – Dave Tweed
    Sep 1, 2021 at 11:07
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    \$\begingroup\$ The purpose of D10 in the first diagram is not clear. The symbol for Schottky is used, but it also has a voltage rating, implying that it is actually a zener. Either way, it never conducts, except possibly under fault or surge conditions. Not sure what the designer had in mind. \$\endgroup\$
    – Dave Tweed
    Sep 1, 2021 at 11:11
  • \$\begingroup\$ I suppose D10 is a zener diode to protect the system in case the 5V point (typically from a USB power source) becomes faulty and greater than 6.2V. Wrong symbol (as usual). \$\endgroup\$ Sep 1, 2021 at 19:07

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