View Full Version : What resistor for new angel eye led bulbs?
goober5904
07-13-2005, 10:50 PM
OK i'm replacing the leds in my angel eye fog lights with some new brighter leds (18,000mcd). here are the specs
http://superbrightleds.com/specs/w18030_specs.htm
I was just wondering what size resistor I will need for these and how they should be wired (series/parallel). I believe there are 2 leds in each housing. Thanks.
Dave C
07-13-2005, 11:24 PM
you'd have to give me the specs for the LED that was in it before... if I am guessing correctly (never seen an angeleye lamp in person), then there is some built in resistance. if you have an ohmmeter you can measure the resistence between the positive and negative leads to the LED. this will give you the resistence it already has. go to LED calculator through the link in my sticky, type the specs in for a single LED, and that will give you the resistence you need.
if this needed resistence is less than whats already though the existing LED, then you'll need to add the difference by putting the correct sized LED in line.
goober5904
07-13-2005, 11:34 PM
i was told that the stock resistor is 240 ohm and there are 2 leds in each housing. 4 total for both fog lamps.
Dave C
07-14-2005, 07:35 AM
http://www.metku.net/index.html?sect=view&n=1&path=mods/ledcalc/index_eng
ok, check out this site... you now have everything you need to figure it out.
1. I'd assume you'd wire the LED's in series within each lamp. In this case, you'll use the LED's in series place on the above link.
2. Your supply voltage is 14.4
3. Voltage drop across the LED is listed in the specs at superbrightleds.com
4. current is most likely 20 mA, check the specs as well (peak current is probably 50 mA, but that's not the normal operating current)
5. 2 LED's in a row..
6. Take the ohms that this gives you and then add resisters in series with your LEDs until it = the number given by the calculator (ie added resistence + 240 = needed resistence). If the needed resistence is under 240, then you'll be fine.
My guess is that the resistences will probably be pretty close, and you might not have to add much... you might be able to get away with not adding any at all.
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