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I would like to use these exact female USB-C ports (labeled JRC-B008):

JRC-B008

to power a microcontroller at 5 V. Soldering the V -> 5Vin, G -> GND works if the power source is a 5 V phone charger.

When I plug in the USB-C charger from my Macbook, however, there is no power negotiation, so there is no output to the device. Is it possible to somehow add the power negotiation here?

Unfortunately the question "https://electronics.stackexchange.com/questions/512557/usb-c-as-power-source-and-negotiation-with-usb-2-0-and-3-0" does not address it.

I have read about soldering a 56 kΩ or 5.1 kΩ resistor from the CC pin to the +5V, but the CC pads are not exposed, just the V, -D, +D and G pads (essentially USB 2.0). Since this is for an ESP32-WROOM-32E microcontroller or similar, <500 mA should suffice; higher would be great but isn't necessary.

I also see pads for R1, but I am unsure what soldering something there would achieve.

ocrdu
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Sriram
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  • Have you got a manual for the connector board? You can also look or measure what R1 does. – Justme Nov 18 '21 at 10:56
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    Does this answer your question? [USB-C as power source and negotiation with USB 2.0 and 3.0](https://electronics.stackexchange.com/questions/512557/usb-c-as-power-source-and-negotiation-with-usb-2-0-and-3-0) – Elliot Alderson Nov 18 '21 at 11:22
  • @Justme unfortunately not. It's available all over [AliExpress](https://nl.aliexpress.com/i/4001173873247.html) but without much documentation on usage. I'll keep looking to see if the R1 is indeed what I need to add an smd resistor to. – Sriram Nov 18 '21 at 13:03
  • @ElliotAlderson I read through that but it seems to indicate soldering resistor to CC, but there's no exposed pad for the CC pin. I could try some solder-fu to identify and solder to the CC pin directly, but I'm not entirely sure if that's the only solution here. – Sriram Nov 18 '21 at 13:05
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    @Sriram If I had to guess, they are just selling a faulty or prototype batch of manufactured items that would otherwise have to be disposed properly as electronic waste. Or maybe they are just selling random items cheaply without documentation, so if you buy those you are on your own. – Justme Nov 18 '21 at 13:42
  • That clearly says "BCC8", not "B008". – Hearth Dec 20 '21 at 02:14
  • Its B008 but the via cuts into the silk screen, you can see on the image in the answer below (where it gets disturbed in a different place) – Miron Sep 09 '22 at 14:29

3 Answers3

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I have the same connector. The R1 pads expose the CC pins. Both CC pins are tied to the same pad so you just need a single resistor. I put a 5.1K on there and I get 5.0v with a smart charger.

Connector with 5.1k resistor on R1 pads

Jerther
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    That is not good, that's the RPi4 mistake basically https://www.scorpia.co.uk/2019/06/28/pi4-not-working-with-some-chargers-or-why-you-need-two-cc-resistors/ (one must use two separate resistors!) – myfreeweb Jan 30 '22 at 16:58
  • Yes but this specific board has both hard wired :( Your comment is something to keep in mind though. – Jerther Jan 30 '22 at 17:03
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Christopher Biggs above is right.

In addition, the large pad (left of JRC, below R1) is also connected to the CC pins. So if you don't have an 0402 sized resistor, you can use an 0805 across the left pad of R1 and the large pad.

See photo.

JRC-B008 board with two 0805 5.1K resistors

Greenonline
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AlexisAV
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Cut the track just above the hyphen on the silkscreen, then use another 5k1 resistor (but in size 0805) to join the pin above the B with the ground pin at the right hand side).

Picture of the bodge

Christopher Biggs
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