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Luxeon star driver

Discussion in 'Electronic Design' started by Mike, Sep 28, 2005.

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  1. Mike

    Mike Guest

    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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  2. ehsjr

    ehsjr Guest

    ------
    +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
     
  3. René

    René Guest

    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
     
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