▲ | Iulioh 5 days ago | |||||||||||||
It's not the usb c committee problem, the devices you are buying are out of spec This is because the cable is 2 sided so it can't assume polarity So it's a tradeoff for not having to guess how to insert the cable | ||||||||||||||
▲ | rcxdude 5 days ago | parent | next [-] | |||||||||||||
>This is because the cable is 2 sided so it can't assume polarity Not really. The USB-C connection pinout is symmetric about a 180 degree rotation, at least as far as power connections go. It's entirely possible (and common, e.g. when using passive converters) to just put power out of it constantly. The main reason for the signaling resistors is to avoid having power presented on the pins when it's not connected, which is more about avoiding corrosion or wear due to small sparks on connection. | ||||||||||||||
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▲ | Findecanor 5 days ago | parent | prev [-] | |||||||||||||
> This is because the cable is 2 sided so it can't assume polarity To clarify (and to tell my own tale on the topic): The power pins on both sides should be connected in both a plug and in a socket. However, when it comes to the USB 2.0 data pins only the socket end must be double-sided (short A6 to B6 and A7 to B7). Back when "Type C" was new, I wanted to build a project with it, so I got one of the first socket breakout boards available. I built a mechanical keyboard out of aluminium with a slot milled to fit that breakout board. After everything was painted and soldered did I plug it in and it did not work ... It took me a while of troubleshooting before I retried it with the cable plugged in the other orientation. The breakout board had connected only A6/A7. B6/B7 were not available. |