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I'm designing a circuit which has a bluefruit nrf52832 micro controller built into it. When referencing the schematic provided by Adafruit I noticed that they had +VBAT going to the drain of the p-channel FET. Does anyone know why this is? Is there a reason or does it not matter/just a mistake on the schematic?

Link to bluefruit: https://www.adafruit.com/product/3406

Schematic provided by Adafruit: Adafruit Schematic of bluefruit nrf52382

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4 Answers 4

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The FET comes with a "free" body diode (due to its construction) so it has to be connected this way. If it was the other way around, then current would flow through the body diode from VBUS (presumably +5V) into VBAT (it's a LiPo, so 4.2V max), which would overcharge the battery in an uncontrolled and probably spectacular way (ie, boom).

In fact, the FET could be replaced with a diode, but that would cause more losses.

The FET itself doesn't care much whether it is wired in the "proper" way or not. In order to turn on a PMOS, the condition is not "Gate voltage has to be a threshold below source" but rather "Gate voltage has to be a threshold below source or drain". In this case, source and drain are functionally swapped but it will still work fine. TrenchFETs aren't symmetrical though, so most other properties like capacitance etc will not be swapped, which does not matter here.

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The FET is turned on when VBUS is not present. If you had only a diode, current could flow only from the battery through the diode and into your circuit. Using this approach, current can flow in both directions. You can attach a battery charging circuit of the correct voltage to the "JST LIPO INPUT" and run the board and charge the battery without disconnecting the battery from the board. With a diode alone, the battery could not be charged in this way.

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It looks correct to me.
The source of that mosfet is connected to Vbus, which is higher than Vbat I suppose.
So Vbat is disconnected when Vbus is present. Otherwise Vbus is connected and mosfet is turned on.

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Yes, the source/drain orientation of \$Q_3\$ matters.

If you were to swap the terminals, you would have a stiff path from VBUS to the battery through the body diode of \$Q_3\$. With the orientation used in the schematic, when VBUS is applied, \$Q_3\$ is turned off via a positive \$V_{GS}\$ and the body diode is reverse biased.

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