14
\$\begingroup\$

So I am working on a project for which I am using a USB Type C Receptacle for the main hardware interface.

I am familiar with the previous USB types and how their pins are laid out, however, upon looking up the pinout for USB Type C, I cannot figure out which pin corresponds to traditional, pins for USB Type A. Does D+ / D- correspond to the traditional Rx / Tx, or does it have to do with the Rx+/- and Tx +/-? Furthermore, which pin corresponds to the ID pin in microUSB B.

Any clarification is appreciated.

enter image description here

\$\endgroup\$
7
  • 10
    \$\begingroup\$ USB 2 and earlier has never had Tx/Rx! It's a single differential pair D+/-. Note that the key at the bottom of your diagram shows purple as "USB High Speed 480Mbps" and those are the D+/- so there's your answer! \$\endgroup\$
    – DoxyLover
    Commented Mar 10, 2017 at 0:47
  • 2
    \$\begingroup\$ Are you implementing a USB2 port, or a full-blown USB 3.1 communication? \$\endgroup\$
    – Ale..chenski
    Commented Mar 10, 2017 at 0:53
  • \$\begingroup\$ @DoxyLover, thank you. I was looking at some old pinout diagrams. \$\endgroup\$
    – Ignatius_Gim
    Commented Mar 10, 2017 at 1:09
  • 1
    \$\begingroup\$ Again, is this for a host, of for a device? \$\endgroup\$
    – Ale..chenski
    Commented Mar 10, 2017 at 2:41
  • 2
    \$\begingroup\$ Your terminology is confusing. So, you are designing a DEVICE. If your other connected device (microprocessor) has OTG port, it will select the host mode automatically as soon as it will see 5.1K pull-down from your DEVICE \$\endgroup\$
    – Ale..chenski
    Commented Mar 11, 2017 at 5:41

2 Answers 2

Reset to default
19
\$\begingroup\$

It's the D+ and D- like normal. There is two of each because USB C is supposed reversible. You can insert the plug both ways and it will still be a valid USB 2 connection, as usb 2.0 does not negotiate the pinning. Flip the plug 180 degrees and the same pins will connect in the same order. Your board should connect both together for maximum connectivity.

There is no ID pin, as that's only implemented on plugs. In USB C, the CC pins handle this, and pulling them to ground with a 5K resistor will initiate OTG HOST mode on the other side of the link.

\$\endgroup\$
6
  • \$\begingroup\$ @alichen plurals. Nice catch. \$\endgroup\$
    – Passerby
    Commented Mar 10, 2017 at 4:41
  • \$\begingroup\$ So how would you initiate OTG mode? \$\endgroup\$
    – Ignatius_Gim
    Commented Mar 10, 2017 at 16:04
  • 3
    \$\begingroup\$ Pulling both CC pins to ground with a 5K resistor places the device in OTG mode if the device supports OTG mode. \$\endgroup\$
    – Passerby
    Commented Mar 10, 2017 at 16:11
  • \$\begingroup\$ You probably need to clarify then which "device" you are talking about, the OP's device, or the "partner microprocessor OTG device" \$\endgroup\$
    – Ale..chenski
    Commented Mar 11, 2017 at 5:43
  • \$\begingroup\$ I'm reading here: synopsys.com/designware-ip/technical-bulletin/… : "o convert an existing USB 2.0 device to USB Type-C, the designer can short the two CC pins, add one Pull-Down resistor and route the USB D+/D- signals to both positions on the USB Type-C receptacle" Should I short the cc pins? Or just connected them to ground with their own 5K pull-down resistors? \$\endgroup\$
    – John Evans
    Commented Feb 3, 2019 at 17:03
14
\$\begingroup\$

If you are fitting an old style USB 2.0 device with new Type-C connector, you need:

(a) connect A6 with B6;

(b) connect A7 with B7;

(c) connect EACH CC1 and CC2 to ground with 5.1k resistor.

All legacy Type-C cables have only one D+/D- signal pair in the cable, so you need (a) and (b). [as a matter of fact, contacts B6 and B7 are absent in the Type-C plug. So the receptacle must have A and B connected]. There is also only one CC wire in the cable, so pull-downs and sensors must be on both CC1 and CC2 in the device.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.