P-Channel Mosfet Gate Drive

[smssk]

Mosfets are fairly new to me. Will it be ok driving a DMP3125L-7 directly from 12v? I am trying to automatically switch +MCU_V between the buck converters 3.2v output and a lipo when +12v is connected. I will be drawing less than 100ma from +MCU_V.

Also does anyone see any other issues with my circuit? Is 100K a reasonable value as a gate pulldown?

The TPS560200DBVR datasheet does specify to leave EN floating to Enable.

DMP3125L-7 Max\Min RDS ON According to digikey: 4.5V, 10V

https://www.diodes.com/assets/Datasheets/DMP3125L.pdf

[inse]

Flip drain and source of the FET, don’t exceed 20V gate/source and everything should be fine

[eTobey]

Here is something to play around with to learn:
http://www.falstad.com/circuit/

[smssk]

Flip drain and source of the FET, don’t exceed 20V gate/source and everything should be fine

Thanks. Swapping the source and drain makes sense to me but I may still be misunderstanding some fundamentals.

I keep finding examples where the source and drain appear to be swapped. Is there an error in the attached schematic or am I missing something? Is the battery connected to drain in this example?

[liaifat85]

I can see that the Gate-Source Voltage is ±20 V. So, 12V direct input shouldn't be a problem. 
https://www.diodes.com/assets/Datasheets/DMP3125L.pdf

[RFDx]

I keep finding examples where the source and drain appear to be swapped. Is there an error in the attached schematic or am I missing something? Is the battery connected to drain in this example?

Yes, the battery is connected to the drain of the PMOSFET because, contrary to your circuit, the voltage of the USB power supply is higher than the battery voltage.
Without 5V from USB, the transistor gate is grounded, the PMOS is switched on (ideal diode) and the load is supplied from the battery. With USB connected, the PMOS transistor is off and the load is supplied with ~4.6V (5V minus the voltage drop of a Schottky diode). The body diode of the PMOSFET is not conducting because the output voltage (~4.6V) exceeds the voltage of the (fully charged) battery.

[Peabody]

+1.  If the mosfet were oriented "normally", the high external voltage could pass through the body diode back to the battery, which you defintely don't want.  The orientation used in load sharing prevents that.  But the mosfet still functions properly as a switch because, as @RFDx says, the body diode is reverse biased when the high voltage is present, so no current flows through it.

Here's an app note:

http://ww1.microchip.com/downloads/en/AppNotes/01149c.pdf