I'm routing a USB-C receptacle on a carrier board for a System on Module. The location and orientation of the receptacle and the SoM headers are dictated by the design of the device.

I noticed that if I would swap the signals running from the SoM to the A and B rows of the USB-C receptacle, routing would be substantially easier. As the USB-C port design is mirrored, one would assume that would works just as well.

But I'm afraid that somehow the bult-in logic of USB-C where it resolves cable orientation would somehow cause trouble in my swapped layout. Could I trick the system by swapping ALL signals except CC1 and CC2? Would DisplayPort still work?

If I just leave it as it is, I will have to have differential signals cross-over, as well as have P/N swaps, wheras in the swapped situation, I can simply route almost everything in more or less straight lines.


Below is an image of the supposed routing if I make an exact copy of the schematics of the development carrier board that my SoM supplier has made avilable to me, shown within the physical constraints of my carrier board layout. I think everybody would agree that if I rotate the pins on rows A and B of the type-c connector, so (SS)TX1 becomes (SS)TX2 and vice versa and (SS)RX1 becomes (SS)RX2 vv., routing becomes much easier. And then I haven't even shown the amount of P/N reversal that would have to be applied in the below case.

Any way i guess the comments that have already been made reassure me I can go ahead and rotate A and B and don't do anything weird with the CC1 and CC2 pins, because I understand now that I can just see this rotation as exactly the same as a normal connector rotation.

enter image description here

  • 1
    \$\begingroup\$ USB C is rotationally symmetrical, not mirror symmetrical. If you just swap adjacent A-B pins, then you would completely change the pinout (e.g. SS Tx would swap with SS Rx). A diagram of what you intend to do would help. \$\endgroup\$ May 26, 2020 at 0:12
  • \$\begingroup\$ Why do you think you need to "trick the system"? The A/B "side" of receptacle does not mean anything until the plug is inserted. If you rotate all signals including CC then electrically there will be no difference. \$\endgroup\$
    – Maple
    May 26, 2020 at 0:58
  • \$\begingroup\$ @Maple I thought it is only reversible because of the CC pins. In alt mode (Displayport) SSTX and SSRX become 4 one way DP lanes. I figured if I swap all pins, the system would observe a swapped cable situation and swap the DP lane order back, while I had already swapped them physically. Note that I have 4 SS pairs running from SoM to receptacle, not 2. \$\endgroup\$
    – Genoil
    May 26, 2020 at 5:47
  • \$\begingroup\$ As @TomCarpenter said, the diagram will help a lot. Without it all I can say is this - the typical 24 pin USB-C receptacle is simple device routing connector pins to PCB pads. Many datasheets do not even name pads on a footprint, it is entirely up to you to decide which row is A and which is B. If you swap all socket signals a1 <-> b1, a2 <-> b2 and so on, and then insert same plug rotated 180° the electrical connection will be exactly the same as before. \$\endgroup\$
    – Maple
    May 26, 2020 at 6:21
  • \$\begingroup\$ thanks @Maple. For clarity, I have updated the question with a diagram. Please let me know if this new information would change your answer. \$\endgroup\$
    – Genoil
    May 26, 2020 at 7:32

1 Answer 1


Let's say you connected the receptacle exactly as it is shown on your diagram. Then you attached the cable and marked corresponding pins on both plug and receptacle, A1..A12 in top row, B12..B1 in the second row.

Now, if you disconnect the cable, rotate it 180° and plug it back, pin marked A1 on the plug will be connected to B1 of the receptacle. Pin marked A2 to B2 and so on.

So, if you now swap the signal traces the same way, i.e. A1 <-> B1, A2 <-> B2 ... then the end result will be electrically identical to the original connection. Your SoM will not see any difference, but the routing will be more convenient.

Note, that the same result can be achieved by rotating SoM in place, but since you said it is fixed the swapping of the traces is a way to go.


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