Photodetector to turn on LED

Discussion in 'The Projects Forum' started by photon1, Oct 11, 2007.

  1. photon1

    Thread Starter New Member

    Oct 11, 2007
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    Hey,
    I am working on a project that will light up an LED when light hits a photodetector, and needs to be run off a battery, preferably a watch battery. I know I need to bias the detector but do I need a load resistor? The real part I am stuck on is how to switch on the LED. The voltage and current off the detector is way low so using an AND wouldn't work. Is there anything else out there I can use or do I need an amplifier?
    Thanks for your help
     
  2. photon1

    Thread Starter New Member

    Oct 11, 2007
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    0
    Forgot one thing, the photodetector is a TO type.
     
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    What might work is a CDS cell with an on resistance low enough to pass current through the LED. That would have to be in the realm of 50 ohms or so. Anything else either won't work well at 1.5 volts or will drain the cell so fast that it won't be worth the efort.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    I've picked up a few of Radio Shack's Cadmium-Sulfide (CDS) photocell assortments over the years (Stock# 276-1657) so I thought I'd give your idea a try with what I had on hand.

    Lowest reading I got with one of them was 90 ohms when held approximately 1" from a 60-watt incandescent bulb. At about 5' distance facing two 60-watt bulbs, average readings were in the 2k-3k ohm range. In dim light conditions (hand covering most of both sides) I was reading between 100k-200k Ohms.

    It appears that using a commonly-available CDS cell, you'd pretty much have to use a semiconductor that was biased ON by the CDS cell. Then there's the problem of "looping" - what if the CDS cell "sees" the light from the LED, causing it to stay in a low resistance state? You'd have to disconnect the battery or move the LED away from the CDS cell to turn the thing off.

    Might take a look at phototransistors instead. I don't happen to have any commonly available phototraonsistors on hand to test, though - just some ancient oddballs. Radio Shack has some I/R phototransistors, which would solve the "loop" problem; LED's output light in a very narrow spectrum, and an I/R phototransistor would not "see" it's output. However, an incandescent lamp puts out light all across the spectrum, including IR and UV. I don't know if they are "switching" type, though.

    The LED needs to be connected up in series with the phototransistor and a load resistor, and the battery across those three. At this point, I have no clue as to what your LED needs as far as voltage drop and current, nor do I have any data for your photodetector.
     
  5. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    You can use a phototransistor and a comparator for this. The detector stage would consist of a voltage divider mede of a phototransistor and a resistor. The ouput of the voltage divider (assuming that you will ground the phototransistor) would be at the "-" input of the comparator. Connected to the "+" input would be a reference voltage divider. Then the comparator would give enough current and an "on-off" signal output.
     
  6. zlazer

    New Member

    Oct 12, 2007
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    0
    I think what this guy wants is the ability and therefore what circutry is needed to turn on an LED when light hits a silicon photodetector....since the detector outputs a very low amount of current it probably is not enough to power the LED. The batter is necessary to power the LED but there must be a circuit where the detector current will allow the battery to pass voltage and turn on the led...right? I have solar cells at home that turn on led's when it gets dark...should be the same idea?
    ideas for this guy?
     
  7. photon1

    Thread Starter New Member

    Oct 11, 2007
    3
    0
    Yes, that is more of what I am looking for. I am unsure what I need between the PD and LED.
     
  8. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    I think my solution solves the problem, since all the current is provided by the comparator.

    You can use a LM393 as a comparator, between the detector circuit (the phototransistor and resistor connected as a voltage divider) and the LED. Just as I said.

    Of course there are other solutions, and some better...but I prefer others to present them instead.
     
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