Photodetector to LED

Discussion in 'General Electronics Chat' started by robby991, Apr 8, 2008.

  1. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    Hello. I have a photodetector operating in photoconductive mode with a red LED as the light source. This LED is powered by a freq. generator set to 1Hz pulses. To make sure that my photodetector is generating these input pulses, I attached the output of the detector to an oscilliscope, which did varify I am getting squarewave pulses with 5V peaks. Now when I input this to another LED, I the LED does not pulse. The LED works fine, because when I attached the freq. generator directly to it, the LED pulses perfectly. Does anyone know why this is? The LED is getting more than enough voltage from the photodetector, yet it still doesn't work.
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Can you sketch your circuit to assist us in understanding the way you are connecting up the LED in the circuit where it does not work?

    hgmjr
     
  3. MusicTech

    Active Member

    Apr 4, 2008
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    I am sorry, this is a bit off topic, by why is it a square wave?
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    What type of photodetector?
    If it is a reverse biased photo-diode then it needs an amplifier.

    Pulses are used to blink the transmitter LED. Pulses are usually from a square-wave.
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
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  6. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    It is a planar diffused silicon photodiode (pin-10D), in reverse bias mode (5-7V). I didn't think I needed an amplifier because the pulse peaks were 5V. I even increased the reverse voltage to get higher peak pulses. I just covered the diode and LED with a box to make sure no light got in, then used that output to drive the other LED. 5V is plenty of voltage to drive the LED.

    Beenthere, the signal generator pulses the LED just fine when I connect it directly. It is when it goes through the photodetector when nothing happens.

    I will draw up a schematic today, but the general idea is:

    signal generator-->photodetector in reverse bias mode-->produces 5V pulses-->LED
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Robby991,
    I don't doubt that you're measuring 5v pulse peaks on the output, but it's low current.

    As Audioguru already mentioned, you need an amplifier to multiply the output current in order to supply enough to be able to light the LED.
     
  8. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    Would an opamp be sufficient to amplify current? What is the best method?
     
  9. nomurphy

    AAC Fanatic!

    Aug 8, 2005
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    Connect the detector output to an input of a 74HC14 inverter, connect the final LED to +5V and set its resistor for 10-15mA going to the respective inverter output.

    When the signal from the detector is high (+5V), the LED will light. If you want the opposite, add another inverter in series.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    Attached are a couple of variations on a similar theme you can try.

    The transistors can be any small-signal NPN transistors you happen to have around, like 2N3904, etc. They're used in a Darlington configuration so that the gain of one transistor is multiplied by the gain of the 2nd transistor. This means that very little current is required on the base of the 1st transistor to control a large current on the 2nd transistor.

    The 170 Ohm resistors will limit current through a standard red LED to around 15mA. If you're using something different, select accordingly.

    The 5k resistors limit current through the photodiode and the base of the 1st transistor. You should check the value to ensure the current doesn't exceed the capability of your photodiode. As it is, current will be limited to 1mA.

    Just realized I have the photodiode in backwards - you can make that change yourself.
     
  11. sndpgr

    Member

    Jun 22, 2006
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    Sorry for being off topic.
    'Sgtwookie' why have you used 2 bjt's , would not 1 be enough.
    is it because that would increase the Vbe(twice the Vbe of a single transistor )
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    One might be enough. I wanted a pretty sure bet. ;)

    The gain of the two transistors in a Darlington configuration is multiplied. As any transistor approaches saturation, the gain is dramatically reduced. So, in the schematic I previously posted, the current through Q1 or Q3 is limited by the 5k resistor to less than 1mA, but even that may be too much for the photodiode - I don't have the specs for the one he's using. He may need to increase that 5k resistor to 50k or more, which would leave less than 0.1mA to drive the base.

    A 2N2222 has a minimum hFE of somewhere around 80 when Ice~=150mA and Vce is 5, but it drops off rapidly to 50 as Vce is lowered, which is what will happen in this circuit. So, 50 x 1mA = 50mA, which would be enough to light the LED (but not be in saturation), but 50 x 0.1mA = 5mA, which is not sufficient to provide full LED brightness.

    But with the Darlington configuration, you'd get at gain of at least 50 x 50, or 2500. This ensures that Q2/Q4 will indeed be in saturation.
     
  13. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    Ok thank you I am going to breadboard these tomorrow. I am just curious though, why wouldn't a LM741 opamp work to amplify this signal?
     
  14. Audioguru

    New Member

    Dec 20, 2007
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    The inputs of a 741 opamp don't work when their voltage is within about 2V from the positive supply or 0V in this circuit so it would need a positive supply and a negative supply.
     
  15. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    I breadboarded that amplifier circuit with the LED and it works fine, using AMP05 transistors. I removed the LED and scoped the output because I need it to be an input to another circuit and I wanted to see the pulses. However it was only producing pulses of 500mV peaks. I need about 2V to drive my logic gates. How can I amplify this more? I tried adding another transistor but it was producing the same peaks.
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    The LED with its current-limiting resistor is the load that pulls the output of the transistor high (to a positive voltage). You removed the load so the output ogf the transistor does not go high.
    Add a resistor from the output of the transistor to the +5V supply then the output of the transistor will go to +5v if the resistance of what it feeds is higher than about 20k ohms. Reduce the value of the resistor to 100 ohms if the resistance of what it feeds is 2k ohms.
     
  17. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    Ah I see. Thanks alot. I am now inputting these pulses into a state machine. The state machine works fine with the function generator being used as the clock input, but when I run the function generator into the photodetector then use the output from thi photodetector funny things happen. The state machine doesn't go in order like it should, they are out of order. Is there anything I should do in between photodector amplifier output and clock input?
     
  18. baseball07

    Active Member

    Apr 24, 2007
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    I think the problem is that the input pulses don't get pulled all the way to 0V. It peaks at around 5 to 6V then its minimum is around 500mv to 1V. How can I modify that amplifier circuit to ensure that the pulses get pulled down to 0V? I need to add some sort of pull down network?
     
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