Filter arrangement

Discussion in 'Homework Help' started by ali8tor, Apr 15, 2013.

  1. ali8tor

    Thread Starter Member

    May 29, 2012
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    hi i need help understanding the filter arrangement of this circuit
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    C2 and R1 form a High pass filter.
    Fc = 0.16/RC = 0.034Hz
    Next C1 and R7 form a low pass filter Fc = 0.16/RC = 10.6MHz

    Next we have a op-amp non-inverting amp. The C3 and C4 and R2 and R4 form a bandpass type response filter.
    With maximum gain equal to

    Av = 1 + (C3*R3)/(R2*C3 + R4*C4) = 2.4 [V/V]

    At frequency equal to

    Fo = \frac{1}{2 *\pi \sqrt{C3*C4*R2*R4}} = 4.9KHz

    And at the end we have another High pass filter (C6, R6).
    Fc = 0.16/RC = 0.16Hz And that is the end of this simplified analysis.

    Is it those information you were looking for?
     
    Last edited: Apr 15, 2013
    screen1988 likes this.
  3. screen1988

    Member

    Mar 7, 2013
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    Hi Jony,
    I am curious to know why C1 and R7 are separated.
    Why don't we put C1 in parallel with R8?
    And could you explain the role of R8 here?
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    You can place C1 in parallel with R8 if you like. This change don't have any effect on the circuit operation.
    Because C2>>C1 we can treat C2 as a short circuit.
    So C1 time constant is equal to:
    t = C1 * Rth and Fc = 1/(2 * pi *t)

    Rth = R7||R8||R1 ≈ R7

    R8 provide a path to ground for C2 left plate.
    So when we disconnect input source (signal). We don't leave input (C2) floating. So we use R8 as a pull-down resistor.

    R6 at the output is also pull-down resistor.
     
    Last edited: Apr 29, 2013
    screen1988 likes this.
  5. screen1988

    Member

    Mar 7, 2013
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