# Current Limiting Formula for LEDs



## KiwiEV (Jul 26, 2007)

*Re: Current Limiting Formula - LED's*

I had this link emailed to me this morning from a budding converter in Russia. He's just playing around with LED's at the moment but created these low power indicators. 
He used 24 red LEDs (8 candell each) and 100 ohms resistors per 4 LEDs. A regular bulb is disassembled and the leds are connected to the metal part of the open bulb.
The page is in Russian but the pictures are self explanatory. I might just do this myself in the future.


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## lazzer408 (May 18, 2008)

If you want to know the voltage across the resistor, wouldn't that be the supply voltage - led forward voltage?

Your example:
_________________________________________________________
_(13.5 - 9.1) / 0.5 = 8.8
13.5 volts was used as the Supply Voltage,
9.1 Volts was used as the Forward Voltage, of the LED module I'm using,
0.5 was used as the LED operating current,
8.8 Ohms was the resulting value of the current limiting resistor._
_________________________________________________________

13.5v-9.1v = 4.4v

4.4v*.5a = 2.2w

Not 4.55w as you've calculated.


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## dmac257 (Jun 30, 2010)

I was looking at the OPTEK website and it shows LED solution for all lights EXCEPT the headlights. Is there a model number for the white light you intend to use for headlight and does it already meet FMVSS for illumination?


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## TexomaEV (Jul 26, 2007)

I ended up just replacing all the lights in the car with LED versions, and used HID lamps for the headlights for the time being. Holding out for some LED Headlights, once the mainstream auto industry starts to use them. I hope one would be able to eventually obtain replacements when parts houses stock repair parts.


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## notailpipe (May 25, 2010)

lazzer408 said:


> If you want to know the voltage across the resistor, wouldn't that be the supply voltage - led forward voltage?
> 
> 13.5v-9.1v = 4.4v
> 
> ...


Lazzer, yes you are right. Perhaps another way to look at is is the power dissipated in a resistor is equal to the current squared times the resistance (play with equations to see equivalence).

P = I^2 * R

So 0.5 * 0.5 * 8.8 = 2.2W as you have correctly observed. This might help others avoid the confusion since the voltages are not needed and the current is the same through both elements since they are in series.


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