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Thread: Dual loc series or parallel

  1. #1
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    Default Dual loc series or parallel

    I'm getting ready to set up the flexmod for my dual LOC red setup and I'm not sure if I should wire them in series or parallel.

    I have the silpads insulating the aixiz modules as I've seen people mention when wiring them in series, and I think I have some 4.7ohm 2 watt resistors and some 2.2 half watt resistors coming.

    Why not parallel? I want to drive them at 400ma each so could I set the flexmod to 800ma and if so, resistor or no? I don't know if there is a "right" way or if it's six of one, half dozen of the other.

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    Its really hard to get diodes to track in parallel without ballast resistors. If you have 12V around, go series. No sense wasting energy making heat in resistors.

    If you only have 5V, or only have one Lasorb, you need to go parallel, and you need to make sure that you use a pair of ballast resistors, and make sure one diode is not hogging much more current then the other. The voltage across the series resistor is proportional to current, so you can at least see whats going on. 1-2 ohms should be a start.

    Your really concerned with the "headroom" available to the driver, if a diode has a 2.4 volt drop, and you have two in series, that is 4.8V volts, leaving only .2 volts with a 5V supply, which is not enough headroom in most cases.

    I'm not familiar with the LOCs forward voltage, so that was a made up example.

    Each ohm of series resistance burns up 1 volt of headroom, so at 250 mA, 1 ohm in series burns 250 mV of headroom.
    Since P=IE, or P = I squared R, you can calculate how many watts of resistor capacity you need.

    Example, 250 mA diode current, 1 ohm ballast, (.250 x .250) * 1 = .0625 watts * 1.5 safety factor = 0.9 watts, so choose a 1 watt resistor.

    Take the number you calculate and multiply by 1.5, so if you need a 1 watt resistor in the math, multiply that by 1.5, and the nearest standard resistor is a two watt, so that is what you buy or build. By build, wattage in parallel adds, so two one watt resistors in parallel is a two watt resistor.

    Steve
    Last edited by mixedgas; 09-29-2010 at 16:53.
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    Thanks Steve,

    I appreciate your help.
    Last edited by Howie; 09-29-2010 at 17:00. Reason: Only caught part of response

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    Quote Originally Posted by king4quarter View Post
    Yo Steve

    I'm not that good with small electronic stuff...Can you just show me were to buy the stuff to connect 2 445nm diodes to one flexmod?

    What exactly do they look like??

    Cheers
    If you have a 12V or higher PS at an amp n half, just wire them in series..
    Last edited by polishedball; 11-26-2010 at 14:35. Reason: added current
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    If you are trying to max out the 445's 1flexmod will not be enough.

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    Quote Originally Posted by chipdouglas View Post
    If you are trying to max out the 445's 1flexmod will not be enough.
    explain please, a single is good for upto 4amps plenty for 2 445's as long as your supply can deliver.
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    Quote Originally Posted by king4quarter View Post
    So what if I run the driver at 2amps for 2x1W will it work without lasorbs or anything else just on its own?

    No you would run it at 1amp, you will have voltage drop not current drop if you wire them in series.

    Thats why you PS need voltage twice that of the vdrop of diode and some headroom.
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    Quote Originally Posted by king4quarter View Post
    I understand better now.....do you reckon 12V is enough to get the 445nm diodes running at full power?
    iirc the Vdrop of a 445 is 4.1v so 12V to run 2 should be fine. As long as your 12v supply has a high enough current rating.
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    I have found that when pushing an A140 445 to around 1500mA, the Flexmod P3 runs best at >8.0V. The Vf of these diodes is a little higher that 4.1V at this current, more like 4.5V.
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    Quote Originally Posted by dnar View Post
    1500mA,
    Thats steaming it what output power are you getting at that amperage?
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