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99 LEDs....

Discussion in 'Electronic Design' started by [email protected], May 2, 2006.

  1. Guest

    Hello!

    I'm wiring 99 LEDs together and want to power it by battery....

    I have a very basic knowledge of circuitry and was wondering if anyone
    could help me get started?

    The LEDs that I have are:

    Kingbright description: L-53HD
    Forward current max.: 25mA
    Reverse voltage max.: 5V
    Wavelength @ peak : 700nm
    Power dissipation PT: 120mW
    Light output [email protected] 10mA: 2mcd
    Light output [email protected] 10mA: 5mcd

    and I have been told that I could power them without a resistor from a
    12v power pack.

    Thanks, any help is greatly appreciated!

    Alex
     
  2. Are the LED's carrefully factory matched?
    If not, you do not have a prayer.


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    Don Lancaster voice phone: (928)428-4073
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  3. Guest

    http://www.dotlight.de/datasheets/L-53HD.pdf

    You've had bad advice, as you will be albe to see from the voltage
    versus current curves in the data sheet.

    The worst case forward votlage at room temperature at 20mA is 2,5V.
    2.25V is typical, so a 12V power pack probably can't drive a string of
    more than 5V diodes.

    You could probably set up a current mirror to put a well-defined 10mA
    through a string of four LEDs running from a 12V supply, but this is a
    long way short of 99, which you'd have to manange with 25 parallel
    strings, controlled by 25 current mirrors.
     
  4. Guest

    Here's what I would do, not knowing exactly what your requirements are.
    These are red, so about 1.7V forward voltage drop. Connect 6 or 7 in
    series, along with a resistor 20-80 ohms. I'm just making wild
    guesses; you'd have to measure the current draw to get the appropriate
    value. Yeah, it could be calculated but it's easier to experiment.

    You'll end up with about 15 of these strings. Just connect them all to
    your battery. One string will have a different amount of LEDs, just
    use a different value resistor.

    It's not precise, you could use about 15 current regulators, but it's
    quick and dirty and all you probably need.
     
  5. Ban

    Ban Guest


    You can make a current regulator from a 2N2222 transistor with a single
    resistor to the base. For 12V
    you need between 100k and 200k for 20mA, but you have to choose it for each
    transistor individually. This regulator only takes 0.4V forward voltage, so
    you can wire 6 LEDs in a row. Usually when they get warm, the forward
    voltage also lowers and they will pull much more current, which might
    shorten lifetime. This is prevented with the transistor, which driven in
    this way will limit the increase to 10% or so.
     
  6. GregS

    GregS Guest

    Crank up the voltage to at least 18. You can probably get one at 24 volds DC also.

    greg
     
  7. Fred Bloggs

    Fred Bloggs Guest

    That's true but the informant failed to mention that means a Royer
    Converter. Now some nut is going to come on here and rage about Peter
    Baxandall, but it is the Royer that you really want.
     
  8. Tam/WB2TT

    Tam/WB2TT Guest

    Assuming some kind of hobby project, this makes the most sense. Although 6
    in a row might be pushing it. L-53HD does not come up at the Kingsbright web
    site; so, no way to tell for sure what the forward drop is. Also, it makes a
    difference based on what kind of "12 Volt" supply there is. They tend to
    come 12.0, 12.6, and 13.8 V. A car battery with the engine running will be
    at least 13.8. Depending on how the car is wired, the cigarette lighter
    socket could go as high as 15V.

    If you want to do something fancier, build a DC/DC converter with an output
    of about 180(?)V, and set the current limit to 20 ma. All LEDs in series.
    Tam
     
  9. Paul Mathews

    Paul Mathews Guest

    This used to be true. However, the resistive component of many high
    efficiency LEDs is large enough, and their matching is often good
    enough 'right out of the bag', that parallel combinations work out just
    fine, particularly if you don't really need brightness matching. It's
    as if they're self-ballasted to a degree. If the battery pack has some
    significant amount of source impedance, you can sometimes get away
    without any resistors. Try it.
    Paul Mathews
     
  10. GregS

    GregS Guest

    The 1 watt and 5 watt Phillips Luxeons are very well matched out of the bag.
    A group of American Bright I used were very well matched.

    greg
     
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