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

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[–] Vinny_93@lemmy.world 40 points 1 month ago (3 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%.

[–] over_clox@lemmy.world 11 points 1 month ago* (last edited 1 month 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

[–] rhombus@sh.itjust.works 10 points 1 month 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@lemmy.world 1 points 1 month ago (1 children)

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

[–] notarobot@lemm.ee 9 points 1 month ago

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

[–] ArbiterXero@lemmy.world 3 points 1 month 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@lemmy.world 1 points 1 month 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@lemmy.world 1 points 1 month ago

Fair enough. Still nuts.

[–] Flying_Dutch_Rudder@lemmy.world 2 points 1 month ago

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