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I am in a situation where my project requires the use of a slip ring so wires can freely rotate. I will be limited to the number of wires that can run through it, which is a total of 12.

I have a USB 3.0 device that I want to connect through the slip ring using USB-A 3.1 and USB-C breakouts on each end. I don't need to have the redundant row/reversible plug option so I will only be working with one row (A1-A12). The device I am connecting will only be using a 3.0 port, never 2.0. Through some reading, I found that the D pins shown below exist for USB 2.0 compatibility:

USB C Connections

It seems that they are only there to transfer data if the plug is connected to a USB 2.0 port. Can I safely omit the USB 2.0 pins in this situation? Doing so would free up two wires on the slip ring which are important for connecting another device.

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    \$\begingroup\$ Slip rings are unlikely to work at the high speeds of the super-speed pins. \$\endgroup\$
    – nanofarad
    Sep 2, 2020 at 3:55
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    \$\begingroup\$ electronics.stackexchange.com/questions/297031/… \$\endgroup\$
    – Jeroen3
    Sep 2, 2020 at 9:01
  • \$\begingroup\$ Why does this site get so many questions of people trying to abuse USB connectors? If you cannot connect all wires needed for reversing the plug, use another connector, e.g. a Type A Super Speed connector or a custom plug altogether. \$\endgroup\$
    – Codo
    Sep 2, 2020 at 13:49
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    \$\begingroup\$ Why do you believe that the slip ring is capable of differentially transmitting 2.5-5 Ghz signals? If not, your arrangement might be useless. \$\endgroup\$ Sep 3, 2020 at 2:48
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    \$\begingroup\$ And you don't need 12 wires, 6 is enough. Maybe two more grounds. \$\endgroup\$ Sep 3, 2020 at 2:56

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If you do not want to support BC1.2 and high speed 480Mbps, you can ignore differential pairs of USB 2.0.

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