I foolishly burned out 2 green LEDs already since I provided too much voltage. The 2 green LEDs I need to light up have a forward voltage of 3.2 volts each. The battery pack has 3 AA rechargeable batteries, for a total of 3.6 V(1.2 V per battery). If I remove one battery, the LEDs won't light up at all. How can I reduce the voltage so it doesn't burn out my LEDs, do I need to use a resistor or can I use something else? If I need resistors, what kind of resistors should I use?
The other posters gave you the math - a good thing But this LED calculator may come handy as well metku /?sectviewn
You don't, you use a resistor to limit the current going to the LED, the voltage taking care of itself.
OK, hopefully you have some basic electronics knowledge, 'cause I'm gonna throw you some specifics here. First, make sure that the LED's anodes are wired in parallel, and connect the positive wire to either the power source, or an on/off switch. Next, you will wire a 22Ω resistor to each cathode, so you'll need 2x 22Ω resistors. Then, connect the other end of the resistors to the negative terminal on your battery.
Your LED needs a minimum of 3.2v to light up. That's why removing one battery fails to work. You're using two LEDs, so they're apparently in parallel. (In series, you'd need 6.4v to make them work.) In any case, it's not excessive voltage that's damaging the LEDs, but excessive current. You need a current-limiting resistor between the battery and the LEDs. Be careful, though. When the resistor limits the current, it will also drop the voltage that gets to the LEDs. So it takes a little math. If you were using a single LED, odds are that its maximum current rating is around 20ma. You're providing 3.6v, but only need 3.2v, so you can afford to drop 0.4v across the resistor. The same current that flows through the resistor will flow through the LED. So the resistor calculation is E/I, or 0.4v divided by .02a, for 20 ohms. Standard resistor values are 22 ohms, so that's good. Since you're using two LEDs in parallel, connect a 22 ohm resistor to each LED, then connect the LED/resistor combination in parallel.
