If I understand correctly, you have determined that there is no connection between Vbus_3.0 and Vbus_C. This means that the flash controller is not being powered directly from either USB connector.
The simplest way to implement power "OR-ing" is via two diodes:
Code:
Vbus-C o---->|--.
| .-----------.
+---| controller|
| '-----------'
Vbus-3.0 o---->|--'
I don't see any diodes on your PCB, so this would not appear to be how it's done.
An alternative method might involve a power selector switch using a single IC:
Code:
.--------.
Vbus-C --| -o | .-----------.
| \--o-|---| controller|
Vbus-3.0--|--o | '-----------'
'--------'
This method leaves me wondering what the other IC does.
Here is a complicated power selector IC, but I think it should be possible to implement a simpler version with 6-pins:
https://www.mouser.com/ds/2/308/on%20semiconductor_cat6500-d-1194834.pdfThe third method could involve separate ICs for each power source:
Code:
.------.
Vbus-C --|IC #1 |---.
'------' | .-----------.
+--| controller|
.------. | '-----------'
Vbus-3.0--|IC #2 |---'
'------'
The problem with this method is that it doesn't explain why the two ICs are different.
I would test whether any pins of the two 6-pin ICs are connected to each other. This junction would then be the power output to the flash controller.
Maybe there is an error in my logic, but that's how it looks to me.
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