Once you have the resistor sized then you need to find the wattage you will need, both of these equations work.

where W= power

W = I<squared2> x R
W = V<squared2> / R

Go with a higher wattage resistor so things don't get too hot! You can put two in series to double the resistance or parallel to half the resistance.

Darn, you just lost me. LOL.

Let me find out the voltage first and we can work from there.

:help: :doh: :think: :confused:

nigxdoof,

assuming it is the angel eyes for older models (pre-lci), the stock AE bulbs were rated 12v, 10watts. I measured the voltage of the angel eyes a couple of years ago and don't remember the values being much higher than 12v. The light module controls the power to the angel eyes pretty well.

the LED listed below with the forward current of 25ma will be very very dim. for comparison, the predator ice output is 5 watts. stock bulb use 10 watts. wattage is not directly related to how much light can be seen but it does give us an idea of how bright the object may be. the power dissipation of the LED below = maxcurrent x forward voltage => power = .025A x 3.6V => 0.09W. I think for the angel eyes to be noticeable, the LED should be able to take in at least 1A (1000mA) or higher. This is a ballpark figure since some LED uses a bigger forward voltage but lower current. More light and more current will generate more heat. When dealing with those high power LED, big heatsinks are needed or LED will probably burn up in a matter of less than a minute.