powering 1 watt luxeon off 12 volts (car battery)

If they are the same as the ones in the RS catalog, the Vf is 2V at 20mA, so
for a 12V battery you'd need a 500 Ohm resistor. If it's actually 13.8V then
you'd need a 690Ohm. A 680 Ohm may do the trick; I'd be hesitant to
over-drive one of these as I seem to recall people saying they don't like it
(heatsink the blighter!). A proper regulator circuit is recommended IIRC.

Ken

hi, whats  IIRC?
thanks.
mark k

IIRC =
If I Remember Correctly

On Tue, 12 Apr 2005 03:31:02 GMT, "mark krawczuk"



It's The Uppercase Version Of iirc. Like When You Use The Shift Key.


- Franc Zabkar

Ken clearly doesn't remember correctly, because 2V @ 20mA is only 40mW -
nowhere NEAR 1Watt!  His formula is for a bog ordinary red LED.

   i

volts,
formula for

You need Ohms Law and the Luxeon LED specs.
At a maximum battery voltage of 13.8V, with a drop of around 2.9V
across the LED (same for red and amber) and a maximum current of 350mA
you'd be looking at around a 31ohm resistor at almost 4W.
Round that up to the nearest prefered value of 33ohms. A 5W resistor
would work but it will get HOT. Better to use say a 10W.

Dave :)

Use an LM317 configured as a constant current generator. Then you're not
dependant on battery voltage. You can also run both LEDs in series if
necessary, which results in the LEAST dissipation in the LM317.

There is a formula - it's Ohms law.

R=E/I. E=voltage across the resistor, I= current through the resistor


-Andrew M