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I just want to make a "USB C" charging cable from 5V power supply (have cut cable w/ red black green and white pins connected from factory). I just want to use 500 mA max.

How do I connect this? If I do not have CC1/2 physically wired, how to proceed?

NOTE: an identical new (not cut) USB A to USB C cable charges my phone but not the remote control!

Here is a picture of the remote, the cable to be used and a picture of what are the factory connected pins.

enter image description here

enter image description here

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2 Answers 2

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Connect + and - to the power supply. It will work just as “well” as the USB-A to USB-C cable worked before you cut it.

That cable is a piece of junk designed to work with particular devices that are “lenient” and work with no CC1/2 connection.

The cable is not USB-C compliant and many devices will not work with it. By design. They weren’t meant to. Buy a USB-A to USB-C cable that has CC1/2 connected. The terminals on the plug will terminate in resistors inside the plug, so CC won’t go onto the cable, but at least it the plug will present itself as able to deliver 500mA/5V.

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  • \$\begingroup\$ thanks, but that is why a put the 2nd picture if you see carefuly you will see no pin is in the cc1 or cc2 location!! so you are wrong, no terminate in resistors inside \$\endgroup\$ Commented Nov 17, 2023 at 22:50
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The connector does not have CC pins and is not compliant with USB specs and it may not work with devices that detect from CC pins how much current is available.

Buy a better cable.

If it did not allow you to charge your device before cutting it, there is no way to make it charge your device after cutting it either.

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  • \$\begingroup\$ great thanks very clear , rare thing was that charged (slow) my android s9 plus ...that is why i cut it \$\endgroup\$ Commented Nov 17, 2023 at 22:49
  • \$\begingroup\$ @MarceloJSolano If the phone accepts 5V without detection it can do that. It also must charge slowly as it cannot determine or communicate how fast it can charge. Not a rare thing, that is how it works. \$\endgroup\$
    – Justme
    Commented Nov 17, 2023 at 23:13

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