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Led illuminator question

Discussion in 'Electronic Design' started by Ripley, Aug 2, 2003.

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  1. Ripley

    Ripley Guest


    I want to make a infrared led illuminator with about 100 leds. For "best
    performance" a led needs about 75-80mA. I have a little problem with the
    led/resistor configuration.

    Lets take only 10 leds and 12V.
    If I try in parallel, I need 800mA to feed my 10 leds. A little to much.
    If I try in serie (5 leds), I need 80mA. That's good. I also need a 150Ohm.
    But when I mesure the current, I get under 25mA. Leds "burn" about
    1.15-1.30V so I can't put as much as I want in serie. They also have a
    resistance so I can't realy calculate the base resistor.

    Since I'm just beginning with electronic, I don't know all the tricks to
    solve this kind of problem.

    So, how can I put all my leds? Ideally, I'd like to use a 9V 1.2A power
    supply. I can do another power supply if I need to, I have a couple transfo
    here, but I preffer to use a already "professional" made one.

  2. Take your LED's an put them in a series string with a smaller series
    resistor in each string.
    Let's make some assumptions...and then you can calculate what you really
    need later on...
    The LED's need exactly 80mA, they burn (actually it's called Vf, forward
    voltage drop) at 1.25v, and you want to use all 100 LEDs.
    First of all, the 9v 1.2a power supply isn't going to be enough, so we'll
    try a 12v power supply, since you can get those just as easily as anything
    We've got 12v. Each LED can drop 1.25v. This means that at most you be
    able to put 9 LED's in series along with a current limiting resistor. 9
    LEDs drop 11.25v, leaving .75v to get dropped somewhere, which is where the
    resistor comes in. .75v/.080A = 9.375ohms, go with 10ohms and you'll be on
    the safe side, 75mA instead of 80mA, close enough. So this means that you
    have 9 LEDs in series per string. Since you want 100 LEDs, you'll need 11
    strings (and 1 LED left over). 11 strings * 75mA = .825amps total. 12v *
    ..825a = 9.9watts, with each resistor dissipating .5625watts. So go with a
    1watt 10 ohm resistor per string, 9 LEDs per string, a 12v 1amp supply.
    If you wanted to use the 9v 1.2a power supply, you'd be cutting it pretty
    close to it's rated max, but it could be done using 7 LEDs per string and a
    3.125ohm 1/2W resistor.
    If the LEDs drop a bit more than 1.25v, you'll need less of a resistor, and
    conversely, if they drop a bit less, you'll need more of a resistor.
    Basically the formula works something like this:
    Input voltage - (individual LED voltage drops multiplied by the number of
    LEDs in series) / desired current = current limiting resistor value.
    desired current * ( input voltage - total series LED voltage drop ) =
    wattage of that resistor

    Hope that helps...probably the only thing I know enough about to actually
    help somebody on here...
  3. N. Thornton

    N. Thornton Guest


    Ditch the LEDs and use a light bulb. If you need it dark run at around half V_rated.

    Regards, NT
  4. Ripley

    Ripley Guest

    Are you telling me I can do an infrared spot just by taking a 100W/125V bulb
    and feed it with 60-75V???
  5. To achieve uniform luminosity across the strings then I suggest using a
    constant current source to drive the leds. Power dissipation, and therefore
    heat, will also be reduced due to the IR drop of the resistors.

    There are LED string driving chips available off the shelf, or you can
    construct one using an mosfet and opamp.
    Hope this helps

    Dana Raymond
  6. Trouble is that he didn't say how well the 12V supply is regulated.
    If it's a car battery, the voltage will vary from 12V to well over 14V
    when charging. So your circuit with a lot of LEDs and a low value
    resistor will allow the current to vary widely over that range, much
    more than the 80 mA he was shooting for. If it's held close to 12V by
    a regulator, then the circuit you recommended would work okay.

    Here's an inexpensive and simple regulator that can be used in place
    of the resistor. It needs about 2V minimum drop.

    +--------------------+------- Positive
    | | Supply V.
    | |
    | --- LED(s)
    4.7k \ \ / =====>
    to 47k / ===
    ohms \ |
    depends / |
    on current | / Q1
    | | / Gen'l
    | | Purp NPN
    +----------------| 2N4401 or
    | | BC337
    | | \ E Must
    Q2 \ \ dissipate
    Gen'l \ | | required
    Purp NPN | 470 ohms | power.
    2N3904 or |------/\/\/\----+
    BC547 | |
    E / | | 30 ohms for 20 mA
    / \
    | / 10 ohms for 60 mA,
    | \ etc.
    | /
    | |
    | |
    | |
    Negative Supply V.

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  7. N. Thornton

    N. Thornton Guest

    Ditch the LEDs and use a light bulb. If you need it dark run at around
    Of course. I thought you wanted something lower power than that
    though. IR is where filament bulbs transmit most of their output.

    Regards, NT
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