I have an older electrical device which I would like to switch to charging with a USB C cable and a standard USB power supply. The device's power supply unit supplies 14V and 400 mA, so 5.6 W. The device documentation certifies it for 12V with 2 Ah.
I found a useful answer on stackexchange about the Source Power Rule for USB 3.1. This implies to me that my device is not compatible with 3.1 because at a power draw below 15W, PD 3.1 defaults to 5V.
According to Wikipedia, a USB4 DFP is required to supply at least 7.5W Type-C current, which is above the power requirements for my device.
Based on this, it seems to me that the power supply of my device is too non-standard to be supplied by a standard USB power supply. Have I understood this correctly?
Thank you for your help.
jeinzi@discuss.tchncs.de 3 days ago
I’m not sure about the standard - but in my experience, if a PD trigger IC requests 12V, that will be supplied, no matter the current draw. I have used this to power several low power devices, for example a 12V computer fan, and I have also verified this with no load at all. There is also PD PPS, which could be used to deliver 14V or any other voltage.
cmnybo@discuss.tchncs.de 3 days ago
12V support is optional in USB PD. If the power supply doesn’t support it, you will get 9V instead. PPS support is even less common.
czardestructo@lemmy.world 1 day ago
That is not correct, you won’t get a random lower voltage. You are correct that thw PD supply needs to support the voltage but PD always starts with 5V for backwards comparability then digitally its negotiates the PD voltage it wants, if it asks for 12v and its not available it with just stay at 5V. It won’t spit out the closest value unless the device asks for it. Its a very strict and wonderful standard.
SorryImLate@piefed.social 3 days ago
So, if I added this board to my device, I must be sure to only charge it with a power supply that supports PPS in the 12v range - is that correct?
SorryImLate@piefed.social 3 days ago
Thank you, this is a very elegant solution.