Unknown circuit

Discussion in 'Homework Help' started by wave runner, Jun 1, 2009.

  1. wave runner

    Thread Starter New Member

    May 21, 2009
    4
    0
    Hello to everyone.
    Im new in the forum and i need your help.
    Can you explain me the operation of the circuit below ?
    I tried but failed to give explanation.
    Is it a peak detector??
    How do I calculate the output,the capacitors kai the resistors?

    Thanks,
    waverunner
     
  2. bertus

    Administrator

    Apr 5, 2008
    15,646
    2,345
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    The diode will have some effect on the output. The capacitors will have some reactance at 4 KHz that will also affect circuit operation.

    Use Xc = 1/2PI*F*C
     
  4. wave runner

    Thread Starter New Member

    May 21, 2009
    4
    0
    Thanks for your answers.
    I did that:
    R3\\Xc4=9,2KΩ , R4+Xc3=497,88Ω , Vpeak=0,54V

    Α=(9,2ΚΩ/497,88Ω)+1=19,47

    Vdc=Vpeak/PI=0.172V

    Vout=(0,172V*19,47)+2,5V=5,84V ... but, that is not true.

    Where i do wrong??
    Thanks
     
  5. wave runner

    Thread Starter New Member

    May 21, 2009
    4
    0
    If : Vin=0,212Vrms --> Vout=2,8Vdc
    Vin=0,382Vrms --> Vout=3,03Vdc
    Vin=0,743Vrms --> Vout=3,528Vdc

    If I do : Vout = Vpeak+Voffset
    for example Vout=0,382/0,707 + 2,5V=3.04V
    or 0,212/0,707 + 2,5 = 2,799V

    Is that way true??
     
  6. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Yes the circuit is a peak detector.
    The opamp has an internal gain of about 200,000 that reduces the 0.7V forward voltage of the diode to almost nothing since the diode is inside the negative feedback loop.
    Then the output voltage is the 2.5V input bias voltage plus the peak of the input sine-wave which is 1.414 times its RMS value.
     
  7. wave runner

    Thread Starter New Member

    May 21, 2009
    4
    0
    Thanks all the members for their answers..
     
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