Laser diode operation

Discussion in 'General Electronics Chat' started by sharanbr123, Oct 20, 2014.

  1. sharanbr123

    Thread Starter Member

    Sep 29, 2014
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    Hello All,

    I am going through a document explaining how laser diode circuit operates.
    I have attached a image of the circuit along with description snippets from the document.

    The following questions and statements are based on this description. I hope someone can comment ...

    1) Transistor Q1 is in active region normally (with Vbe > threshold value). Is this correct?
    2) It is mentioned that as laser diode output rises, the monitor diode's current increases, reducing the conduction of Q1. I am not able to appreciate this. Since base of Q1 is connected to negative potential of supply, when monitor diode is forward biased, no current flows from Anode to Cathode terminal since base of Q1 which is also Anode of photo diode is connected to negative potential.
    3) Even assuming that when monitor diode's current increases, how come the conduction of Q1 reduces.
    Is this because effective Vbe is reduced?
    4) Similarly, reduction in conductance of Q1 reduces conductance of Q2 also. I am not sure how this is possible?
    Is this because base voltage of Q2 has dropped further due to increased resistance at Q1?

    PS: I am relatively new to the world of circuits and analog world. Probably this is the first time I am trying to analyze a circuit on my own. Please bear with me ...
     
  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    Do you follow this OK.
    E
    BTW: you should be aware that a photo diode will also generate a small voltage when illuminated, look it up.
     
  3. sharanbr123

    Thread Starter Member

    Sep 29, 2014
    49
    0
    Hello Eric,

    Thanks a lot.

    I will have to read explanations more thoroughly to understand better, only because I am new to this subject and take time to grasp.
    Anyway, one more question - increase in monitor's current reduces conduction of Q1 & Q2.
    Does reduced conduction of Q1 have any direct impact on current into laser diode?
    I am assuming, it does not. But as leakage current into photodiode increases, current into laser diode decreases to that extent.
    I assume this is not the real cause of reducing current into laser diode but due to increased resistance of Q2.

    Many thanks for helping ...
     
  4. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Yes, thats the purpose of the circuit, 'closed loop', it maintains a constant current thru the laser diode.
    Without some form of current limit circuit, the laser diode would self destruct

    Look up, 'negative feed back' circuits, thats where a small opposite polarity signal is fed back from the output to the input.
    As the output tends to exceed the desired level, the negative feedback signal from output to input increases, so the circuit is held in balance.
    Dont think in terms of Q1 or Q2 changing 'resistance' , think in terms of collector current flow due to the Vbe [ Base current] of the transistors.

    E
     
  5. sharanbr123

    Thread Starter Member

    Sep 29, 2014
    49
    0
    Hello Eric,

    I am re-starting this thread with several questions. But first with question on your above comment.

    In the above feedback case, I don't see opposite polarity going back to input. Rather, the feedback just forward biases monitor diode.
    Your comments please.

    Instead of having monitor diode in the direction shown, what would be the effect if it was connected in opposite way.
    Anode connected to V+. Still current would flow into base of Q1 and reduce conductance of Q1. Your comments please.

    Another related question is, in the above scheme, instead of having Q2 transistor to control current into laser, could we just remove this and have a
    Q1 and bias it more using monitor diode and hence reduce current into laser diode. This is like having a variable resistor connected in parallel to laser diode and controlling it using monitor diode? If this does not work then would it work if voltage source is replaced by current source.
    Since current is constant, more current would flow into Q1 emitter to collector and hence would equivalently reduce current into laser diode.

    Also, I am not clear in the above scheme, what is the purpose of two resistors connected to emitter and collector of Q1.
    One of them would do if the idea is to limit current flow when Q1 is in active region.

    Looking forward to answers.

    Also, I would appreciate if there are simple circuits that I can work out on my own and see if I am able to get my thoughts clear.
     
  6. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    If you are interested in learning about circuits there are many tutorials online.
    E
     
  7. sharanbr123

    Thread Starter Member

    Sep 29, 2014
    49
    0
    Hello Eric,

    Thanks a lot. Your responses hav gievn me a lot to think. Thanks again ...
     
  8. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    2,504
    380
    hi 123,
    An analogy would be you filling a kettle from a water tap.
    You are the controller [photodiode and Q1] and Q2 is the water tap.[Laser diode]
    The tap is the controlled part of the system.

    E
     
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