Ironsraven,

Cool idea, making a low intensity red - thanks. There's an unused Solitaire in the log base here somewhere... anyway, your idea gives me another idea along those lines (low intensity red).

I can offer you a tip about the smoked yellow LED: You need a resistor in the circuit. Coin cell batteries like you got from the little 12v MN21 work OK without a limiting resistor because as soon as the LED tries to draw enough current to fry itself, the internal resistance of the battery shoots up - they cannot deliver much current. But AA batteries have the capability to deliver far more current than it takes to fry a 5mm LED. Check the specs on that yellow LED and you'll see it probably has a rated Vf of 2.xxx volts. Two AA in series give you ~3 volts. And the LED has a max rated current - probably about 20ma or maybe 30ma if it's a super bright.

Really really simple math to figure out what size resistor to stick in there and if you do it using an insulating disk to hold the resistor between the batteries (or between the battery stack and either end), it's trivial to pull the resistor and LED and then drop the bulb back in. A resistor automatically means that your circuit efficiency is reduced because you're bleeding some power into heat, but over-all it's a tiny amount of power anyway, so run-time should be really generous - I'm guessing well over 100 hours on two AA - maybe far more than that. Rather than re-hash all the how-to, there's plenty of help on this on the 'net (or surf CPF). Even on and off-line calculators for this.

You might get away with a green in the 2 x AA because IIRC, they want something in the neighborhood of 3.6 volts - anyway, over 3 volts. So it might light up VERY faintly, with extremely low current draw, direct-connecting it to two AA in series - have not tried it myself, but it may work. A 5mm would give a little more light, but you would have to mod either the reflector or the LED package to fit it in a minimag.

Thanks for the low-intensity red idea!

Tom