I have two of these breakout boards, using them for PD based power passthrough.

I need to do passthrough for all pins except power. Power (vbus) will do breakout into a DC filter, then back to the other connector.
Input is PD capable usb-c PSU, the connected device on other port is PD capable.

So to get true PD passthrough, is it A1 to A1, B1 to B1, or some other schedule? The bottom side has smd resistor on CC1 and CC2, should these just be removed?

 

hi DC,
This link indicates that the cc1 and cc2 resistors should be fitted to ensure general compatibility.
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https://www.google.com/search?q=usi...qQvCBwcwLjkuOC4yyAdEgAgB&sclient=gws-wiz-serp

 

I assume my PD device (the upstream port from a PD psu) already has the needed CC1 CC2 pull-down resistors.

Neither board that I have will be an "upstream" port, I just need to be able to peel away VBUS, pass that through my filter, then back to VBUS on the exiting port.

I think just remove the pull-downs from boards. If I leave them, there will be three pull-downs in parallel per CC line, which is not right.



Gool AI tells me this:

For a simple passive passthrough between two identical USB-C breakout boards, you should connect them in a 1:1 mapping (A1 to A1, A2 to A2, etc.), but you
must flip the pinout of the second connector relative to the first if they are mounted back-to-back, or keep it consistent if they are facing the same direction. For USB-C, ensuring the CC1/CC2 lines are not shorted together, but passed through directly, is crucial for device negotiation.

The flipping over mention is a bit misleading. If the schedule is 1:1 (A1 A1, B1 B1, CC1 CC1, etc) then the orientation of the port makes no difference, because C cable can itself be flipped over.

I also learn that USB-C cables port (spec) have SHLD and GND tied together at each end of the connection. This is an odd specification for USB-C as the shield should not be a current carrying conductor. The shield should be a chassis ground decoupled from signal GND with a cap to short AC noise.