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Homework Help: Band pass filter

  1. Apr 23, 2010 #1
    1. The problem statement, all variables and given/known data


    Given: K=17, R1=R2=6,247 Ohm

    Calculate the value of C1=C2=C to set the center frequency of the filter to 17.3 kHz.

    2. Relevant equations

    w0 = 1/[C*sqr(Rth*KR2)] from lecture notes

    3. The attempt at a solution

    w0 = 1/[C*sqr(Rth*K*R2)]
    C= 1/[w0*sqr(Rth*K*R2)]
    C= 1/[2*pi*17.3k*sqr(6247*17*6247)]
    C= 3.57e-10 F

    The equation above is from a similar circuit that doesn't have R2. I wasn't sure if it still would work. Looking at the circuit above. It seems like R2 does not impact the function of the circuit.

    Any help would be appreciated.

  2. jcsd
  3. Apr 23, 2010 #2


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    Staff: Mentor

    If the equations are for a circuit without an R2, then why is there an R2 in the equations? Maybe I'm being dense.

    Do you understand how to write the KCL equations for an opamp circuit with feedback? That's how you can solve for the transver function of this circuit.

    EDIT -- Welcome to the PF, BTW!
  4. Apr 23, 2010 #3
    The equation uses K*R2 which is equal to R2 in the similar circuit.

    No, I don't know how to write KCL equation for opamp circuit with feedback. It's been a very long time since I used it. Can you help me get started?
  5. Apr 24, 2010 #4


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    Staff: Mentor

    In this case, the equations are a mix of KCL and something else. The theme is to figure out how many unknowns you are dealing with, and to see that you can write the same number of unique equations to help you to solve for those unknowns.

    So in this circuit, Vout is an unknown, and (call it) V1 is an unknown. Call the voltage at the intersection of R1 and C2 "V1". You might initially be inclined to also call the V- input to the opamp a variable, but it's not. Why not?

    So write 2 equations for these two unknown voltages. The equation for V1 is just a KCL equation, but the equation for Vout is a slight variation. Still very solvable. Does that help?
  6. Apr 24, 2010 #5
    V- does not need to be called a variable because the V+ is connected to ground thus V- is a virtual ground too. Right?

    The KCL equation I got for node V1 is:
    KCL1: (V1-Vin)/R1 = (Vout - V1)/(1/sC) + V1/(R2 + (1/sC)).

    For Vout node i got:
    KCL2: Vout/KR2 + (Vout - V1)/(1/sC) = 0

    Are those equation correct? So, now I use KCL2 and solve for Vout? Then substitute it into KCL1 and solve for what?
  7. Apr 26, 2010 #6


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    Staff: Mentor

    Correct about V- = 0V.

    You have a sign error in your first equation. Write KCL equations as the sum of all currents out of a node = 0.
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