powering 1 watt luxeon off 12 volts (car battery)

hi , i`m pretty sure i asked this some time ago , but i`ve forgotten, i want to run a red 1 watt luxeon, and a 1 watt amber luxeon off 12 volts, car battery.

what value and wattage resister would i need. or is there a formula for working out the required resistors ? thanks. mark k

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mark krawczuk
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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

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Ken Taylor

hi, whats IIRC? thanks. mark k

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mark krawczuk

IIRC = If I Remember Correctly

Reply to
Pippa Reeves

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

Reply to
Mike

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 :)

Reply to
David L. Jones

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

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Andrew M

On Tue, 12 Apr 2005 03:31:02 GMT, "mark krawczuk" put finger to keyboard and composed:

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

- Franc Zabkar

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