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%.
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%.
Wrong way though. The source feeds to the display not the other way around. The is no downstream voltage from a vga monitor.
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
+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.
Well, it’s hooked up to a laptop, so, yeah, VGA out.
Its not hooked up to a laptop. Its hooked up to a display. It says so in the message
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.
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?
Nah, pin 9 is reserved and not needed, but when implemented, offers 5 volts power.
https://vhdl.us/book/Pedroni_VHDL_3E_AppendixI.pdf
Fair enough. Still nuts.