In a message dated 96-07-26 15:33:14 EDT, you write:

>i have a super bright , jumbo LED that reqires 2.6v @ 50ma
>
>how do i figure the resistance of the dropping resistor for a 12v 
>source?

Figuring that your 12V source is really around 13V, heres the scoop:

13V less 2.6V is 10.4V dropped across the resistor.  Since the current you
desire is 50mA (.05 Amps) and R=V/I, you have: Resistance = R=10.4/.05 = 208
ohms.

It's not too critical that you get exactly 208 ohms.  Somewhere in the 200 to
220 ohm range will work o.k.

Good luck,
Jim Davis
87 GL Syncro
88 GL Wolfsburg