I am trying to follow the design in this schematic link at page 9.

enter image description here

The schematic above is connecting only RX1 TX1 and D1 data lines on the USB-C receptacle. And there are no pins shown for the other pairs in the USB-C standard (RX2, TX2, D2).

Does this imply leaving the other pairs floating?

I only need my USB-C receptacle to properly transfer my RX, TX and D signals. And couldn't figure out what to do. I spent a lot of time Googling the matter. Using a MUX as TI's HD3SS3220 could be my best chance. But I really want to know if I can just leave these pins floating, voiding extra components.


1 Answer 1


You will want to refer to the USB Type-C Specification. (The USB Implementers Forum has these and other documents.)

One of the main features of the Type-C connector is that it can be flipped. I think section of the linked document may be of most interest to you:

Since the USB Type-C plug can be inserted in either right-side-up or upside-down position, the hosts and devices that support USB data bus functionality must operate on the signal pins that are actually connected end-to-end. In the case of USB 2.0, this is done by shorting together the two D+ signal pins and the two D− signal pins in the host and device receptacles. In the case of USB 3.2 SuperSpeed USB or USB4 TX/RX signals in a single-lane implementation, it requires the functional equivalent of a switch in both the host and device to appropriately route the TX and RX signal pairs to the connected path through the cable. For a USB 3.2 SuperSpeed USB or USB4 dual-lane implementation, the host and/or device resolves the lane ordering.

I'm not quite sure how your project fits into the various examples given, but I think that you'll likely find what you're looking for there.


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