Luxeon star driver

Sep 28, 2005 3 Replies

Is there a diagram out there to drive a Luxeon III star from 12V without having to use a huge limiting resistor?


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+12 ----| 7805 |---[2R]--- Vout to Luxeon, 650 mA ------ | Gnd --------+

Figuring Vf at 3.7 volts, 2 ohms will limit current to

650 mA, so a 2 watt resistor will do. The 7805 can go up to 1 amp, and needs a heat sink. It will dissipate up to ~7 watts. If you want to go "full bore" (>1A), use a different circuit:

------- 2W

+12 ----| LM317 |---+---[1R]--- Vout to Luxeon ------- | | [240R]1/2W | | +-------+---[2.2K]------+ | 1/2W | [810R] 1/2W [POT] 5K | 1/2W | Gnd --------+-----------[120R]------+

The above will allow you to vary the current to the Luxeon from about 670 mA to 1.34 amps.

You could use a 2 watt one ohm limiting resistor, but it will dissipate about 1.4 watts, so a little higher wattage gives a wider margin and is desireable. The other resistors are 1/2 watt. The LM317 will need to dissipate ~ 10 watts, so a heat sink is needed.

Ed

I would get a $0.40 MC34063, a a well chosen inductor, and make a low dissipation current source. ~600 mA without external boost fet / transistor is achievable.

Beats huge heatsinks etc. Including the ~ 50 uH inductor it would be cheaper too I guess. My 0.02

- René

yep MC30463, some links that seem OK

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martin

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