lokki However if the PocketBeagle already is being supplied with power from elsewhere (e.g. the USB client connector) then USB HOST VIN will have power on it, and could be tied to VBUS.
we did exactly this, so USB VIN P1.07 is tied to USB VBUS P1.05. The trace that connects them cannot be cut because it is underneath the socket header.
lokki but you suggested P1.05 which is not the same as P1.07, right?
... which is why I talk about P1.05 instead of P1.07: they are connected together, so they are electrically equivalent. However VIN (P1.07) is supposed to be for voltage coming IN from either USB port, while VBUS (P1.05) is supposed to be for the voltage that goes OUT to power the USB device. So it is more accurate to talk about P1.05 when talking about the power for the external device.
lokki and here a solution for the usb power "issue":
Yup, involves having an extra power management IC.
Try the Schottky solution, I think it will be fine. That would provide a trouble-free solution where regardless of whether you connect the external PSU, the USB micro, or both, the USB device will always be powered.
A less preferred alternative if this does not work, which is also "safe", is to cut the leg of P1.07 between the PB and the BelaMini cape, and then connect P1.05 to P1.01 directly (without diode). This will require an external PSU to be connected to P1.01 in order for the USB device to be powered.
Last, a potentially dangerous solution would be tie P1.01 to P1.05 directly (without diode) and not cutting the pin between P1.07. The danger here is that if you connect both the USB micro and the external PSU, then there will be current flowing from one to the other which would cause the flow of unneeded/unwanted current between the two, and could possibly to the damage of either 5V source.