Ok on my bodykits front bumper i have no provisions for the stock turnsignals. I could cut out the holes for the stockers but they are ugly and huge. So i have decided to take 20 (10 on each side) superbright orange leds and drill some 5mm holes in the bumper spaced 1" apart. Then epoxy the leds in. I know i have to solder resistors in and cover everything in heat shrink tubing. MY question is what size resistors do i use to get the voltage down frpm 13.5 to 2.35 volts. and also how do i get them to flash with the stock flasher? I know that if i dont modify it a certian way then they will kick the blown bulb indicator circut in the flasher. I have heard that you can modify a Tridon Stant EL-12 flasher to work with the leds and not trip the blown bulb indicator circut. I found a writeup on how to modify it here. http://www.stu-offroad.com/electrical/led1/led-3.htmany ideas on how to do this or make this flasher work with our cars. Btw the car is in my sig.

 

there should be a maximum forward current rating supplied with your leds, the rating should be in mA's. to find the required resistor you need divide the supply voltage by the maximum forward current. the led's i deal with are generally around 40mA.

so we use ohms law.
voltage = current x resistance or V = I x R

flop that around and get R = V / I

R = 13.5V / .040A
R = 337.5omh's

 

theres the info on the leds.

Emitted color : ORANGE
Size : 5mm T1 3/4
Lens color : water clear
Peak wavelength (nm) : 615 ~ 620
Forward voltage (V) : 1.8 ~ 2.2
Reverse current : <= 30μA
Luminous intensity : 5000mcd average
Life rating : 100000 hours
Viewing angle : ±10°

Absolute maximum ratings (Ta = 25°C)
Maximum power dissipation : 80mw
Maximum continuous forward current : 30mA
Maximum peak forward current : 75mA
Reverse voltage : 5 ~ 6V
Lead soldering temperature : 240°C (< 5 seconds)
Operating temperature range : -25°C ~ +85°C
Preservative temperature range : -30°C ~ +100°C



so if i use ohms law then 13.5 / .030a = 450 ohm's. so i would just put ever how many resistors it takes to get 450 and then solder them to the long leg of the led right

now that we have that figured out what do i do about the flasher?

 

you're probably actually want a little more resistor than the formula works out to as the way i have it stated would be the maximum continous voltage that the led can take, and depending on the charging system it may be closer to 14 volts.

as far as the flasher goes, try a search in the apearance section, led's won't usually function properly as a turn signal without a stand alone flasher to make them flash, and i've never used them in that application.

ps. don't forget the way that effective resistance "stacks up" when used in parralell or series if you need more than one, it can have quite the effect on the final output reached!

 

ok so what number resistors would i have to use when stacked in paralell to achive the required resistance? electronics isnt my strongest point.

 

See where it says 30 mA & forward voltage 1.8-2.2. Those are the numbers you got to work with. Subtract the voltage drop for the LED from the supply voltage (alternator), then divide by the current to get the resistor value, then make it a little bigger like WS6T3RROR says to make it last. You might just be able to put a bunch of LEDs (maybe 7) in a series and let their equivalent internal resistance control the current & not need a resistor, but I've never done it that way. There are also some LEDs with built-in resistors.

 

I wouldn't recommend stringing LED's in series without a resistor.

Once those babies are forward biased, their internal resistance is very low.
Aside from that, the voltage drop will be ( x * number of diodes).

If anything, tie the cathode of all LED's together and then use a single resistor
to limit current.

Voltage is common since they will be in parallel, but all of the current will flow
through one resistor.

anywhere around 400 ohms is a good starting point as everyone is suggesting. If that is too dim, drop down the resistance by 50 ohms.