this post was submitted on 17 Mar 2026
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Lenovo USB-C PSU for laptops powers a Rasberry Pi, but cannot simply charge bicycle lights. WTF? (external-content.duckduckgo.com)
submitted 4 days ago by evenwicht@lemmy.sdf.org to c/askelectronics@discuss.tchncs.de
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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 6 points 4 days ago (1 children)

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.

[–] evenwicht@lemmy.sdf.org 3 points 4 days ago (1 children)

Is the spec ambiguous on that then? Is a 5v default and a PSU without default both compliant?

[–] jet@hackertalks.com 2 points 4 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 4 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 4 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.