this post was submitted on 22 Jan 2025
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Funny: Home of the Haha

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[–] Vinny_93 39 points 15 hours ago (4 children)

Sure there is. A display signal is essentially just current through specific lines. The way the current is routed makes no sense, but there will definitely be current running through the wires. The only thing needed is for the charging pin of the micro-usb to be connected to any vga pin that transfers current. The rest is just the magic of conducting wires.

It won't charge quickly though, I'd expect it'd take hours just to charge like 20%.

[–] Flying_Dutch_Rudder 1 points 2 hours ago

Wrong way though. The source feeds to the display not the other way around. The is no downstream voltage from a vga monitor.

[–] [email protected] 1 points 6 hours ago

I think you still need a minimum amperage of like 1.2 or something like that though. Not sure how the display would be giving that much.

[–] over_clox 12 points 14 hours ago* (last edited 14 hours ago) (1 children)

Page 553 of this document (third page in as it starts at the appendix) says that pin 9 is optional, but if used, is 5V

https://vhdl.us/book/Pedroni_VHDL_3E_AppendixI.pdf

[–] [email protected] 8 points 14 hours ago (1 children)

+5v to the monitor, not from. Even if it was from, I can’t imagine it being rated for enough current to make charging feasible.

[–] Aqarius 2 points 14 hours ago (1 children)

Well, it's hooked up to a laptop, so, yeah, VGA out.

[–] [email protected] 7 points 12 hours ago

Its not hooked up to a laptop. Its hooked up to a display. It says so in the message

[–] ArbiterXero 2 points 15 hours ago (1 children)

I guess the vertical and horizontal sync are +5, but how would that be connected to the +5 on USB? Seems unlikely but possible I guess?

[–] over_clox 2 points 14 hours ago (1 children)

Nah, pin 9 is reserved and not needed, but when implemented, offers 5 volts power.

https://vhdl.us/book/Pedroni_VHDL_3E_AppendixI.pdf

[–] ArbiterXero 1 points 14 hours ago

Fair enough. Still nuts.