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Need help in opamp oscillator

  1. Sep 9, 2013 #1
    Hi everyone,
    ImageUploadedByPhysics Forums1378782185.797644.jpg
    I was trying to solve this problem. Here at calculate 3 db frequency the gain should me 1/sqrt(2) times of the maximum voltage gain.
    So I calculated maximum gain which is 1+6/3=3 ( capacitor will be open for maximum gain). At 3db gain will be 3/1.414
    3/1.414=(1+6k/(3k||(1/jwc))). Then I solved for the frequency. But I am getting the wrong answer. Can somebody please tell me that wether I am making a mathematical mistake or applying wrong concept.
     
  2. jcsd
  3. Sep 9, 2013 #2

    meBigGuy

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    What happens at high frequencies when the capacitor is a short.
     
  4. Sep 9, 2013 #3
    At heigh frequency the inverting terminal is connected to ground. And since it is a negative feed back circuit so the output will also be 0.
     
  5. Sep 10, 2013 #4

    meBigGuy

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    Nope. The op amp wants the voltage between + and - to be zero.
     
  6. Sep 10, 2013 #5
    Yes it does. But at heigh frequency the capacitor is short and 100% voltage is fended back to the opamp. That's why output will be zero
     
  7. Sep 10, 2013 #6

    NascentOxygen

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    Fixed. :wink:
     
  8. Sep 10, 2013 #7

    rude man

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    Have you figured out what kind of circuit this is? What is meant by "cutoff frequency" for this circuit? The term 'cutoff frequency' is a very poor one, since the gain is finite (=3) at 0 Hz and approaches infinity for very high frequencies.

    So the 'cutoff frequency' is 3db above the dc gain of 3.
    That frequency is not formed by the 3K resistor only.
     
  9. Sep 10, 2013 #8
    I think that the gain is o at very heigh frequency and as gain vary from 3 to 0 the cutoff frequency will be where the gain will be 3/1.414
     
  10. Sep 10, 2013 #9

    NascentOxygen

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  11. Sep 10, 2013 #10

    rude man

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    I think that the gain is very high at very high frequency.
     
  12. Sep 11, 2013 #11
    How? Would you please explain.
     
  13. Sep 11, 2013 #12
    At very heigh frequency the inverting terminal is connected to ground through short capacitor and now if we apply vi then there is a potential difference between the input terminals of opamp so it will produce heigh output. But here at the same time we are getting feedback through 6k resistor Wouldn't that effect the gain of the opamp.
    Please explain this.
     
  14. Sep 11, 2013 #13

    rude man

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    Write the equation for the gain of the circuit and you will see.

    If you remember the gain for non-inverting input = 1 + Zf/Zi it is obvious that at high frequencies Zi → 0.
     
  15. Sep 11, 2013 #14

    meBigGuy

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    please explain how one gets feedback through the 6K resistor if the - pin side is connected to ground? What voltage can the opamp output that will raise ground to 1V?
     
  16. Sep 11, 2013 #15
    Ok I get that gain become very heigh at heigh frequency. Sorry I misunderstood the concept. But if gain is varying between 3 to very heigh value then what's the meaning of 3db frequency?
     
  17. Sep 12, 2013 #16

    meBigGuy

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    I'm going to let you chew on that one. Reread the responses now that you understand the basics.
     
  18. Sep 12, 2013 #17

    rude man

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    Please read my post #7.
    The phrase "cutoff frequency" is extremely badly chosen for this circuit. There is really no 'cutoff' frequency.

    There is however a frequency at which the gain is 3dB above the dc gain of 3.
     
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