I'm designing a breakout board for a USB2.0 Type-C connector.

The final board, which will contain the connector breakout circuit, should work as "device", and so as Upstream Facing Port, or UFP. It will also be equipped with a self-powered IC (no need for Vbus) that provides dp and dn lines to the connector.

At the moment, I know how to treat the CC lines (via 5.1kOhm pull-down resistors) but I still don't understand how to connect (or how to use) the VBUS lines.

In this kind of device, can I leave the VBUS lines unconnected or I should follow some specific procedure to connect them?

  • \$\begingroup\$ Don't see any need to do anything with them. I might be tempted to put an LED on the board to show they're powered up (which would signify the host has seen the pull-down resistors and activated VBUS power). \$\endgroup\$
    – Finbarr
    Jun 24, 2020 at 11:20
  • \$\begingroup\$ So, do you mean to leave them Simply unconnected? \$\endgroup\$ Jun 24, 2020 at 14:11

1 Answer 1


Yes, you should follow a specific procedure to connect VBUS to your device. As you said, your "self-powered IC" will provide Dp and Dm (as UFP). So, if you want to follow USB specifications, operations of Dp/Dm lines should be conditional with the presence of VBUS. For more details and references, see this answer.


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