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Thread: Project : Laser By Batteries

  1. #61
    Join Date
    Jan 2006
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    Charleston, SC
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    Quick question here... If you tie in your new power supply at the points on the circuit board where you've cut the traces, is the optical feedback and current control circuitry still in the loop?

    If so, then all you need is a constant voltage source that can deliver a little over 1/2 amp of current. I'd use an LM317 regulator and any convenient battery that will deliver 5 volts or more. (see HERE for the circuit) You can ignore the AC portion of the circuit. Just connect your battery directly to the input of the LM317. (No filter capacitors needed on the input, but I'd keep the one on the output.)

    Just 5 parts (two resistors, a capacitor, a bypass diode to protect against reverse voltage, and the LM317 itself) gives you a stable voltage source capable of delivering up to 2 amps. (Ok, when you add the power switch and the battery you're up to 7 total parts I guess!) But the LM317 has built-in over-current protection so you don't need a fuse. It also has built-in over-temperature protection! All this in a package the size of your thumbnail. (Might want to add a heat sink to it though...)

    Now, if you've actually bypassed the optical feedback and/or current regulator circuits on the old board where you've cut the traces, then you're going to need to build a current regulator into your new power supply. Fortunately, there is a diode test power supply design in Sam's laser FAQ that should work well. (Google the FAQ for Laser Diode Power Supply and you'll find several different circuits.)

    Adam

  2. #62
    Join Date
    Jan 2006
    Location
    PARIS
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    125

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    there's no optical feedback (2 wires to the laser diode)
    and the current regulator is onboard. he need to keep this part.
    the LM317 is the linear régulator solution but the voltage drop is about 2.5v
    3.8v+2.5v =6.3v minimum for the battery voltage.
    the power dissipation of the LM317 is (Vin - Vout * Iout)
    if you take a 12v battery the power dissipation is about (12-3.8 )*0.530 = 4.34W ouch! a big heatsink is needed! and this power is lost! due to the bad efficiency.
    and with a linear regulator you can't draw all the battery capacity due to the voltage drop across the regulator.

    the best solution is to use a switching regulator like the LTC3440.
    thi is a buck boost regulator.
    the feedback is adjusted to 3.8v at the output and the input can vary between 2.5 and 4.5v
    with a Li-ion or Li-polymer battery (like in my producta application) of 1800mA you can use the laser for about 2h30 to 3h before a charge due to the high efficiency (92%) and without heatsink.
    the battery with LTC3440 is big like a little scratch box. but yes! this is more complex.
    LTC3440 datasheet
    http://www.linear.com/pc/productDeta...42,C1116,P2123

    perhaps i have one prototype i can send to Aidan

  3. #63
    Join Date
    Jul 2005
    Location
    UK
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    262

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    That mostly over my head! hehe

    I would be verry gratefull if you could send me a working circuit + acceptable battery which you spoke of for 2h+ usage, so i can just "plug and play". (Solder the laser wires on )

    Would this be something you are willing to do?

    If so could you give me a price for the parts and your trouble?

    Thanks,
    Regards,

    Aidan

  4. #64
    Join Date
    Jan 2006
    Location
    PARIS
    Posts
    125

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    i send to you a PM for a deal

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