I might have a mental block here... Trying to get my head around this load-sharing circuit from
https://blog.zakkemble.net/a-lithium-battery-charger-with-load-sharing/
The idea is that the battery charger IC U1 (MCP73831) should not see the additional current draw from the load -- to limit the current draw on U1, and to allow the charger to correctly monitor the battery charging curve. So when external USB power is supplied, the load is powered directly from that via diode D1. So far, so good.
But isn't p-channel MOSFET Q1 connected the wrong way round? Unless I am missing something, when there is no external USB power, Q1 is turned on, but will still only conduct via its source-drain diode? (With a less-than-ideal voltage drop.) Might as well just use another Schottky diode instead of Q1, right?
Flipping the source & drain connections for Q1 should get a better result for the no-external-power situation, with Q1 conducting the battery voltage to the load with a lower voltage drop. But with USB power applied, Q1's source-drain diode would then charge the battery directly and in an uncontrolled way -- not good.
So what is the right way of doing this? Is there an accepted standard solution using MOSFETs and/or diodes, which minimizes the voltage drop when supplying the load from the battery? Or am I just misunderstanding the circuit? Thanks for your help!