Folks,

I've pulled two LEDs our of a Cellphone board and I want to use them. When I measured one with my multimeter while it was still on the board I got a 0.672 vdrop. My supply voltage is 3.3v, I'd like to put 20mA through the leds. I worked out the resistance from that with this equation:

R = (3.3v-(2*.672v))/.02A R = 97.2 ohms

So I designed my circuit with a 100ohm resistor. Not that I have it put together I find the LEDs to be very dim. I measured the current through the two (connected in series) and I get 3.45mA.

Am I doing the math right if I then calculate the voltage drop from this reading and the 100 ohm resistor?

V = (100 ohms) * (0.00345A) V = 0.345v Vdrop = Vsource -Vcalcuated Vdrop = 3.3v - 0.345v Vdrop = 2.955v

So I'm reading about a 3v drop. I If I calculate the resistance value I want i get 15 ohms

R = (Vsource - Vdrop) / A R = (3.3v - 3v) / .02A R = 15 ohms

So a 15 ohm resistor is what I need? Does is this look correct?

Thanks!