NEED HELP ON INFRARED LED ARRAY circuit! =(

Discussion in 'The Projects Forum' started by MrWingDings, Jan 26, 2010.

  1. MrWingDings

    Thread Starter New Member

    Jan 26, 2010
    3
    0
    Hello everyone,

    I am working on an infrared LED array and just simply have exhausted ideas on how to fix this problem. This is my second time at this and still have not found out the answer. Here are my specs.

    I have a 9 Volt battery independent source... The positive terminal is then connected to six (6) 940nm wavelength IR LED's then to six(6) one-hundred ohm (100Ω) resistors in parallel then connected to the negative terminal.

    Here is my schematic:

    (+)---->|-->|-->|-->|-->|-->|---|(17.5Ω)|----(Ground (-))

    Here are the spec's to the IR LED's:
    -forward voltage: (1.3 volts)
    -forward current: (100mA)

    I am using 18 gauge wire (AWG rated)

    Here are the spec's to the resistors:
    -100 ohms (Ω)
    -1/2 watt power
    -5% dissipation tolerance

    I have used this website for the corect schematic/wiring diagram but do not know if this is wrong due to it possibly being only for regular LED's:
    -http://led.linear1.org/led.wiz-

    I am using my cell phone camera to test out the array and it picks up 940nm wavelength because I tested it on the Wii's IR sensor bar. (The project I am making this for is correlated to the Wii). Am I not getting the wiring right? (I also used a bread board).

    PLEASE...if I'm doing this wrong correct me because I am a mechanical engineer and I dislike circuits to a great extent.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You wrote:
    "I have a 9 Volt battery independent source"...
    Is that a typical 9v "transistor" battery?

    You may not be aware of this, but those little batteries have a high internal resistance, and a low mAh rating compared to other available batteries.

    Fresh off the shelf with a 24mA load current, you might measure around 8.6v at the terminals. Industrial 9v batteries have an extra internal cell, so they'll measure higher by about 1.5v.

    Basically, your current limiting resistor is calculated by:
    Rlimit >= (Vsupply - Vf_LED_total) / Desired_Current
    So, if your IR LEDs are rated for 1.3v @ 100mA, and you have a rock-solid 9v regulated supply:
    Rlimit >= (9 - (6*1.3))/0.1 = (9-7.8)/.1 = 1.2/.1 = 12 Ohms

    As your battery becomes discharged, it's internal resistance increases. This will happen fairly rapidly with a heavy load like 100mA, particularly a 9v "transistor" battery. A rechargeable NiCD or NiMH will have a rating of 150mAh, but this is when discharged at a 20-hour rate, or 7.5mA. When your load is significantly higher, your battery life becomes much shorter. You'd be lucky to get 20 minutes' use with that kind of load.
     
  3. MrWingDings

    Thread Starter New Member

    Jan 26, 2010
    3
    0
    Thank you very much for your input SgtWookie. I recognize that this forward current load is quite large and I am replacing my 17.5 ohm resistance with a 33.3 ohm resistance to give the resulting forward current to be 36mA. Although this will not be as luminescent it will give a longer life for the 9 volt source. (And to comment on my independent voltage source, it is a common durracel 9 volt battery and my multimeter reading was around 9.22 volts.)

    I am still looking for an answer as to why my circuit is not lighting up. =(
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Ahhh, where did you buy these IR LEDs? Radio Shack, perchance?

    If so, the current rating on the package is not correct; it should be 20mA not 100mA.

    (I bought some of these, and discovered that the first one I tested burned out at 43mA current)

    It is most likely that just one of the LEDs in the string burned out within a few milliseconds of the application of so much current; the remainder may still be usable.

    If you use a voltmeter and measure across each LED, they will all measure less than 1v except for one, which will measure near battery voltage. That one is burned out.
     
  5. MrWingDings

    Thread Starter New Member

    Jan 26, 2010
    3
    0
    PHEW! thank you so much... I switched up the forward current on my bread board to the 33.9mA and it worked like a charm. Luckily none of the LED's, which you guess correctly are radio shack components, burned out. Thank you for your help I appreciate it greatly!

    -Rich-
     
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