I am trying to design a board in M.2 key B form factor to split the USB and PCI-E signals to serve two devices.
However, M.2 key B only provides 2.5 A current (5 power pins, and 0.5 A each), but the total power requirement for the devices is 4 A. The motherboard I am working on has a standby 5 V pin rated at 2 A. I wonder whether there is a way to satisfy the power requirement by sharing current between the two sources.
Some of my "ideas":
Motherboard 5 V->buck converter to 3.3 V->ideal diode->load
M.2 3.3 V->ideal diode->load
The main problem I see in this idea is the step down of 5 V may not really match the 3.3 V generated by the motherboard, and the current sharing does not happen until the load voltage drops below the minimum of the two. I am not sure whether the over-current protection could be triggered before current sharing even starts.
Active current sharing like UCC39002. May work, but seems quite complex and I am not familiar with the relevant field.
Would really appreciate evaluations on my "ideas" and/or alternative solutions to the problem.