If I used this connector to connect a USB-C cable into a circuit (e.g. breadboard), what voltage would I expect to get?
USB power output is always limited to 5VDC (4.75V to 5.25V) unless computers on both ends of the USB connection agree to use a different voltage via the power delivery negotiation protocol (USB PD).
Since that jack has only power wires, power negotiation can not be done so the output will be limited to 5VDC.
Got it. I tested this with various USB power bricks and cheap ones deliver 5V but any smart one (e.g. MacBook supply) doesn’t deliver power. What would I have to do to get the smart power bricks to decide to transmit 5V?
Welcome to the TechForum! In many standard USB connections, there are 4 wires that are used. Much like PaulHutch alluded to earlier, there are VCC and GND wires present in that module but it is missing the D- and D+ connection. The power negotiation he mentioned may be required for certain supplies as a form of “confirmation”. So without those lines, the supply may not provide anything.
Edit: Sorry, I missed the second part the question. To get the smart power bricks to transmit anything (5v or otherwise) it would be necessary to have the additional lines to complete the power negotiation.
This Hackaday article should help.
Comments and article helped a lot. So given my connector with 2 wires, that will never be able to tell a smart power supply to turn on… Okay, if I put this on one end, it has the CC line and the resistor, I just need a female connector with 3 wires to complete it.
Like this one:
FYI, I am trying to extend the USB port on an ESP32 board to the outside of a case, and any cables I have tried so far won’t activate smart power supplies.
I got these connectors with the CC and D+/D- wires, soldered them together and am still not getting 5V from a Smart Power Brick. What am I doing wrong? I have tried so many variations this is really puzzling me. I tried both the White and Blue wires, and the complete parts are shown together in the second photo.
Do you have a datasheet for that 4-pin type-c adapter?
Using an adapter like this will pull 5 Volts from the power-supply : 4090 Adafruit Industries LLC | Prototyping, Fabrication Products | DigiKey
Thank you. I ordered some of those power boards to test. FWIW here is the spec image from aliexpress. Hard to read but it appears there may be a 6 wire version that has the CC where the one I have only has D+/D-. But it’s hard to interpret.
I came across the existence of USB-C Voltage Decoy boards which I think is another way to solve this. I will experiment with those also.
From pictures I could find on this plug adapter it looks like the resistor is a pull-up tied to VBS
I believe that indicates it is spoofing the Downstream Facing Port, specifically the Rp resistor in the below schematic:
I believe what you would be looking for is a cable that spoofs Upstream Facing Port’s Rd resistors in the above schematic.
The Downstream Facing Port in your setup is the power supply. the Upstream Facing Port is your peripheral.
I believe what you are saying is the black adapter is altering the CC line voltage to the microcontroller board (downstream), which has no effect on telling the power supply (upstream) to deliver 5V. i.e. this is essentially backwards and thus doesn’t work. I need the VBS to CC resistor on the female port that the USB-C cable comes in to. Thus why the aforementioned power board and/or decoy board should work. Or I need the opposite cable of the one shown here to make this swap on the inbound lines. If I have it right, this helps a lot. I ordered a bunch of variants of ports and boards to experiment with options. I don’t see a female equivalent of the black adapter that would function on the inbound side.
Question: Why would you want to spoof the downsteam port?
I believe you’re correct on the backwards part, that’s what I was indicating.
I’m not sure why they would want to put that resistor there to spoof the DFP, possibly to override the sourcing device’s advertised capabilities and tell the end device to only expect 500mA.