this post was submitted on 17 Mar 2026
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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.

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[–] jet@hackertalks.com 2 points 5 days ago (1 children)

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)

[–] evenwicht@lemmy.sdf.org 0 points 5 days ago (1 children)

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.

[–] jet@hackertalks.com 4 points 5 days ago

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.