Muon power circuitry question

[jpeytonj]

Hello there,

I was looking at the power circuitry of the Muon, and the pattern shown in the screenshot below shows up in multiple places.

I think the PNP transistor circuit behaves such that the P-channel MOSFET behaves like an ideal diode (in parallel with its own flyback diode), thereby eliminating the voltage drop that would be seen if a simple passive diode was used instead. This makes sense because it allows multiple potential power sources to be connected to a common point (eg. PMIC_VIN in this case) without the risk of contention. Can anyone confirm this, or correct me as necessary?

By the same token, can anyone explain why two conventional passive diodes are used (in parallel presumably to handle large currents), instead of another copy of this 'ideal diode' circuit, when connecting the DC_IN to PMIC_VIN (see lower part of the image) – and why is there a 10K resistor connected directly across the DC_IN power input?

Thank you!

[rickkas7]

I don't have answers to all of your questions, but the Muon VIN minimum is 6V. Thus the voltage drop across the input diode is not an issue and the diodes are less expensive.

The SBR3U40P1 diode is 1A, so I believe you are correct that there are two in parallel to handle the required current.

I think but am not positive that the pull-down is to bleed voltage off when the VIN is a wall adapter and the high voltage side is unplugged.

Since the USB voltage could be 5V, the voltage drop with a normal rectifier or Schottky diode could drop the voltage too low.

Here's a version without the weird mirrored text that occurs in the schematic viewer: