I'm working on a project to update a device to use a Li-Ion battery rather than non-rechargables. I would like it to be charged via USB-C.

I have previously used IP2721 to handle the power delivery negotiation but have recently found this chip (MAX77751) which looks like it can handle the PD negotiation and the battery charging in a single IC. It also doesn't need to be configured by a microprocessor which is very useful for my project.

My issue is with the USB-C connector. The IC requires CC1, CC2, D+ and D- in order to negotiate the current draw. I am struggling to find a USB-C connector with these signals available. The PCBs will be soldered by hand so the fewer pins on the connector the better (Not sure I have a steady enough hand to solder a 24-pin connector).

I have found a connector which fits my requirement (Datasheet) but it only has CC1 and SBU1. Would this be able to correctly negotiate PD with only one CC signal?

Alternatively, I found another connector (Datasheet) which has both CC1/CC2 signals but no D+/D-. What would the limitations be if I used this connector? I assume I would lose BC1.2 detection. With a Type C power brick I'm guessing this would be fine but if I used a legacy USB-A power brick with a "Type A-Type C" cable would it work at all?

It's quite a large battery (10000 mAh) so I'd rather not be limited to 500 mA charging if the negotiation fails.

  • \$\begingroup\$ Not directly related to the question, but suggest you get a type "K" tip for your iron and look up "Drag soldering". You'll find that soldering fine-pitch connectors by hand is not very difficult, with some practice. \$\endgroup\$ Commented Feb 28 at 16:22

3 Answers 3


USB PD on Type-C does not use the DP/DN pins, only the CC pins. Or rather, only one pin at a time, and the pin depends on which orientation you plug in the cable to connector.

The first connector with only one CC pin will only work in one orientation, as there is only one CC wire in a cable.

The second connector with no DP/DN pins cannot negotiate BC and won't work with BC supplies with Type-A supplies. As it cannot enumerate with a host, it cannot draw even 500mA, because to draw more than 100mA, it must be requested from host if it is OK to draw more than 100mA.


24-pin connectors are clearly overkill, but 12-pin "USB2-only" connectors do exist and are quite easy to source and solder. Looking for them on famous resellers gives me tons of results.

You will clearly need D+/D- for BC-1.2 compatibility through "legacy" cables with something else than Type-C on the other end.


I am struggling to find a USB-C connector with these signals available.

Complete USB-C connectors have contacts on the mating face for all 24 lines. Here is a list on Digikey.

The PCBs will be soldered by hand so the fewer pins on the connector the better

So, you want a connector with fewer PCB terminals.

Better yet, instead of a connector by itself, consider a USB-C breakout adapter. For example this breakout board with D and CC lines.

USB-C breakout

Source: Amazon.

  • 2
    \$\begingroup\$ Type-C compliant devices may be open on unused pins (SSs, SBUs). One may have a Type-C compliant device with less than 24 physical connections on the connector. But both CC lines are definitely required. \$\endgroup\$
    – Nipo
    Commented Feb 27 at 13:27
  • \$\begingroup\$ Correct. You're discussing "Type-C compliant devices". I am discussing "fully-compliant USB-C connectors". If I am misunderstanding, I would appreciate a correction. \$\endgroup\$ Commented Feb 27 at 13:54
  • 3
    \$\begingroup\$ Type-C devices that do not use some pins shall leave them open. There is no minimal required stub length, so it may be no conductor at all. Such connector is not full featured, but complies with the term "open". \$\endgroup\$
    – Nipo
    Commented Feb 27 at 14:11
  • 1
    \$\begingroup\$ @DavideAndrea From Type-C specs, power connectors do not need to have data pins present, they are optional. Page 41 in r2.0 from 2019 cable and connector specs. Page 56 also has more details. \$\endgroup\$
    – Justme
    Commented Feb 27 at 14:19
  • \$\begingroup\$ Thank you both for your help. I think I understand now. \$\endgroup\$ Commented Feb 27 at 19:18

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