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IR Illuminator

Discussion in 'Electronic Design' started by Rob Graham, Nov 6, 2003.

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  1. Rob Graham

    Rob Graham Guest

    Can anyone help me out? I'm interested in creating a simle IR LED
    Illuminator. I'm interested to see if I can increase the range on a
    camera with IR sensitivity by creating an alternate/higher power IR
    source than the one mounted to the camera. I've searched the WEB, and
    all links seem to lead back to the same one, which is now dead.

    Thanks for any info-

    Robert Graham
     
  2. Illuminators are just big bunches of LEDs with a bit of resistance in
    series to stabilize the current, powered from a DC supply. The most
    important detail is to get good, high output LEDs. My favorite, right
    now are the Vishay TSAL6100, 100 ma, 10 degree output beam, 950 nm.
    Mouser sells then for $0.21 each.
    http://www.mouser.com//index.cfm?handler=search.processsearchb&supplierpage= 782-TSAL6100

    A typical arrangement might be to connect strings of 6 of them in
    series with 33 ohms 1/2 watt resistor. Connect as many strings in
    parallel as you wish, and power from a 12 volt supply.
     
  3. pkh

    pkh Guest

    Just wire a battery, a resistor, and an IR LED in series (with the
    proper polarities, of course)! The hardest part is figuring out the
    resistor value & watt rating required

    Most LEDs have current ratings, the trick is to figure out the
    on-voltage across the LED at the rated current, if it is not spec'd
    (I've found it usually isn't). Best to assume a low on-voltage initially
    for the LED (say, 0.5-0.7), which would give you a higher resistor value
    (and lower current) and measure the actual on-voltage. Then recalculate
    the resistor value based on the measured on voltage (which should be a
    lower resistor value). Using the higher resistor value initially give
    you a lower than rated current for the LED and keeps you from frying it!

    Example:

    100mA LED
    1.5V battery
    Assume 0.5V LED on-voltage
    Resistor must drop 1V with 100mA through it: R=V/I=10 Ohms

    Wire it up and try it. Measure LED voltage. If measured LED voltage is,
    say, 1.0 Volts, calculate a new resistor value...

    Resistor must drop 0.5V at 100mA, so new R value is 5 Ohms.

    Wire it up and try it. Measure LED voltage. If measured LED voltage is
    higher still, recalc the resistor and try it again. (As you increase the
    current through the LED, the on-voltage will increase slightly, so you
    may have to try a couple of iterations).

    Min resistor power rating is R*I^2 (resistor value * rated current
    squared) = R*(0.010) in this case.

    Hope that helps!

    Paul
     
  4. (snip)

    The data sheets for LEDs specify the typical forward voltage drop at
    rated current. No need to start down at .7 volts.
     
  5. pkh

    pkh Guest

    Like I said, if it's spec'd, great, otherwise you'll have to experiment.
    If you get them from Radio Shack, the "data sheet" is just the info
    printed on the back of the packaging card, and it rarely has the forward
    voltage drop listed for LEDs.

    Regards,

    Paul
     
  6. If you have access to AC power, there's a really simple way--

    Go buy one of those cheap 300W halogen outdoor lights.
    Put a 3-amp 400V diode in series with it so it's not so white.

    Put a couple layers of theatrical "congo blue" and "deep red" filter gels on
    the front.
    Very little visible light will get thru, but many watts of IR will.

    You should be able to "see" many hundreds of feet with this.
     
  7. Good point. However, if you are using essentially unknown parts, you
    also probably can't assume the rated current (or that the listed spec
    is right). I have found that most LEDs driven with a current source
    (or a dropping resistor that has several times more voltage drop than
    the LED) will show little change in forward voltage with time till
    they approach rated current, then the voltage drops a few percent.
    This is the temperature effect on the LED die showing up in the
    forward voltage. So, if you measure the forward voltage, and it sags
    only 15 mv or so in the first few seconds of operation, you are
    probably not overheating the die with that current.
     
  8. This works well with two bulbs connected in series so they each run at
    half voltage.
     
  9. Considered IR laser diodes with appropriate expanding optics?

    Mark L. Fergerson
     
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