Ok so this is a noob question about LED lighting. I have an amp I'm building (project updates in the general DIY section) and I have a LED pilot lamp. My heater primaries are 6.3 volts at 2.5A.. So If I assume a 2v drop across the LED, and I want around .003 amps,
I should be able to do E=I*R (with the voltage drop calculation you showed there) and come up 4.3/.003 = 1433 Ohms.
So I guess now i have a question of wiring this. I hope its not considered trolling to post a layout/schematic..
So, if anyone looks, I'm running my heater wires off the 269EX up to the Legs of the LED, the + which will have a 1500K resistor on it, (side note.. should I use a high wattage resistor?) and then back down to the board there, where they join on opposite ends of R2 and R3, and then split off and go heat my tubes. I would need to have the wires that leave the LED and head to the board, join up before the resistor wouldn't I? As not to change my amperage delivered to the board/heaters?
Anyway, I know that's a big question, but you guys seem to have the LED stuff figured out, so I thought I'd ask!

Thanks.
-Brett