I bought a 2nd-hand Lenovo USB-C PSU (ADLX65YLC3D) which indicates a range of voltages (20v, 15v, 9v, 5v) on the label. Tried to charge a few different bicycle lights but the charging indicators did not light up on any of them. I almost tossed it because the 2nd-hand market I bought from is definately dodgy. But then I tried to power a Rasberry Pi and it seems to work on that. So wtf? An a/c adapter either works or it doesn’t. What would cause this: works on some devices but not others? The Rasberry Pi needs 5v just as the bicycle lights. That is the default voltage for USB-c.
The lights most likely do not have the extra circuitry to talk to the charger to negotiate voltages. Since it’s a charger that can change voltage as you stated then the device must be able to say “hey give me 5v”. You will need to use a dumber 5v only charger for those devices.
What would be the meaning of a default voltage then? My understanding of USB PD is that 5v is a default, which I took to mean it would deliver 5v in the absence of a handshake.
Yeah, but some power bricks want to be safe and wont give any power without power delivery negotiation. It’s not unreasonable, and it is safe, it wont burn anything out.
Is the spec ambiguous on that then? Is a 5v default and a PSU without default both compliant?
The spec is very clear, the source does not need to provide any amperage, just voltage. PE_SRC_Disabled (see my other comment in this thread)
voltage = current × resistance, IIRC my high school physics correctly. If current is zero, then voltage must also be zero, no? I don’t understand how voltage can be positive if amperage is zero.
Your right, but it only needs a tiny amount to signal 5V.
The power brick engineers can choose to fail safe (just 5V only minimal amperage), or fail dangerous (5W power delivery) - for this lenovo power brick they decided to fail safe.