So the lone LED in the middle, with two resistors, is going to be on all the time, as a night light to the night light.
If the rest of the LEDs are on a switch, will I have to run two completely separate wires for the single LED, isolating it on its own circuit?
I’m tentatively planning on doing that, using heat shrink or something like that to tidy up the wires, then use two DC barrel jacks to connect each set of wires to the board. Are there any potential problems with this plan?
Something like this?
Also, are there any glaring issues with doing it like this?
This circuit does work but the always on LED can just be connected right to positive it doesn’t need to also be connected to the switch. It does depend on which switch type you’re using, I’m assuming it’s spdt like in your diagram which means that with your current circuit when the switch is in one position the main LEDs will be on and the indicator one is off and with it in the other position the indicator one will be on and the main ones off. Which also works fine if that’s your intention.
Shouldn’t see any issues, this is how it’s usually done. The only thing to watch out for is the longer your wires to the switch the more resistance your adding but it should be negligible unless it’s crazy long.
What the other poster said about running three wires up also works, and is essentially the same circuit but It depends how it’s being set up, I assumed it was going in an enclosure of some kind with the switch on the outside of the enclosure. If the switch is going to be in-between and outlet and your LEDs it would make more sense to run 3 wires up instead of running wires back down for the switch.
I’m the middle of reading your comment I realized what I did wrong when making it in the breadboard.
So I got the switch and parallel LEDs working, but the single LED connected directly to positive wouldn’t light. The resistor and LED made a loop back onto the same positive side of the breadboard, so the current had no reason to flow through them.