RESISTORS FOR LEDS

[windrules]

Do you need to have a resistor in a string of leds.I have 3 x 4 volt leds to be run off of 12 volt and using led calculator it tells me you can't do it or won't give me a resistor value.My guess is that without a resistor in that string of leds that they would blow.I was going to do strings of 2 leds with 130 ohm resistors but would you believe I can't get those resistors from the electronics shop.I would go to another shop but I live on an Island so its not that easy.

Thanks,

Mos

[T2B53]

to turn on leds off 12volt line why not just

use something like la7805 ic,5 volt regultor.

simple circuit.might help

[iFred]

each LED has a 1.7 volt drop. I generally use 330-480 ohm  1/4 watt resistors for pull up for almost all applications upto 12 volts. One resistor will do nicely. Put it in series, if the resistor gets a little hot increase the resistor to 1K, if it's still gets hot then increase it again. a good place to start is 1K 1/4 watt for 12 volts and work from there. put a current meter in series and measure the current, you should be around 10-15 Milliamps per diode at about 1.3 volts for turn on and 1.7 volts for a little brighter. Good Luck!

[TrotFox]

If you've got 4V LEDs I assume they are white or blues?  You can run those 3 at 12V but it's not the ideal.  You'd be better off using 2&1 or maybe even running 4 in a 2&2 configuration.  That way any voltage fluxuation won't harm the LEDs and you'll have a bit of overhead to ensure that they stay lit also.  Having 3 in a single circuit will get you the least power consumption but won't be the best situation for the LEDs.  However, if you want to go that route it'll likely work so long as you include some low value resistor to protect against over-voltage situations.  I'd guess something like 10-33 Ohms would be fine.

Again, giving yourself some headroom on the voltage would stabilise the circuit better...

Trot, the electronically trained, fox...

[finnsawyer]

Another way to protect the LEDs is to use a Zener diode.  Pick a Zener which has a  voltage VZ greater than that of the stacked LEDs and less than lowest battery voltage.  Determine the LED current ID and chose a resistor R1 such that R1xId equals the voltage difference between the Zener diode and the LED stack.  Place R1 in series with the stack and that combination in parallel with the Zener.  Then place a resistor R2 between the battery and this new combination such that the current through R2 is somewhat greater than ID for lowest battery voltage.  When maximum battery voltage is reached the increase in current is dumped through the Zener diode.  Zener current will reach a value IZmax.  The Zener must then be able to safely dissipate the power VZxIzmax.   Its also useful to allow for the effects of temperature.  

[(unknown)]

Sorry! I tried to give this answer a good rating... But I think I did it wrong.

This answer was a to-the-point, practical, way of going about it. (Without a lot of theory, which can put off some folks that "Just want to jump in and do it") I would only change one thing... I would add that one could also simply use a low value, linear taper pot to arrive at the proper resistance much easier and more quickly. Once the pot is set, just measure with meter what it's set at, and bingo, you have your answer! Just pick the nearest 'standard' value resistor out of the 'ole junkbox. (If, say, the pot indicated that 325 OHMS was ideal, use a real world resistor of 330 ohms... It will be close enough for the real world... remember, even a 5% unit will vary by 330 +-5%=16.5 ohms, more OR less than 330, or 313.5 ohms to 346.5 ohms. If you really want or need to be more precise, keep the pot, or buy 1% resistors.)

Junkman