1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Active low pass filter

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Design this circuit to give a cut off (-3dB) frequency of 1kHz, a low frequency gain of 20, and an input impedance equal to 22kΩ

    2. Relevant equations



    3. The attempt at a solution

    So, I started off with Vout/Vin = [-R2/R1]/[1 + sCR2. I also know that the dc gain will be -R2/R1. I also know that frequency = 1/[2∏R2C]

    Im a little lost as to where to start!

    http://upload.wikimedia.org/wikipedia/commons/thumb/5/59/Active_Lowpass_Filter_RC.svg/300px-Active_Lowpass_Filter_RC.svg.png
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 28, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Hi kimandrew20! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    What is the input impedance of the op-amp filter you show? Start with that.

    Are you going to use just a single stage?
     
    Last edited by a moderator: May 5, 2017
  4. Mar 28, 2012 #3
    Is the input impedance just the value that the problem has specified? The 22kΩ value?

    Yes, just a single stage.
     
  5. Mar 28, 2012 #4
    I tried it again and I got a value for the capacitance and the resistors. Does this look correct?

    R1 will equal the input impedence (The two inputs, V- and V+ are at approximately the same voltage, R1 is at ground voltage, therefore R1 = input impedence = 22 kΩ

    Gain (A) = (Rf/ R1) → Rf/22k = 20
    Therefore, Rf = 440 kΩ.

    f = 1/ [2∏ * R * C] therefore
    C = 1/ [2∏ * R * f] → 1/[2∏ * (22*1000) * (1 * 1000))] = 7.234 x 10-9 F
     
  6. Mar 28, 2012 #5

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You got the right Rf but the wrong C. Reconsider the R you used to compute C ...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Active low pass filter
  1. Low Pass Filter (Replies: 5)

Loading...