Need Help With Cutoff Frequency

Discussion in 'Homework Help' started by Mawangs1, Apr 23, 2013.

  1. Mawangs1

    Thread Starter New Member

    Mar 26, 2012
    23
    0
    I'm getting confused. I'm trying to make a sallen key low pass filter where it's been established that R1 = R2 and C1 = C2.

    ω = 1/sqrt(R*C)

    Let's say I want to make an active lowpass filter that cuts out all frequencies above 400Hz.

    With this in mind, I would do:

    (1) 400 = 1/sqrt(R*C)

    (2) plug in a random C value

    (3) solve for the R.

    I'm being told I'm off though. I hear that I actually need to do:

    (4) ω = 2*∏*f

    (5) put in 400 for the f and calculate ω

    (6) do steps (2) and (3) as before


    Steps (4) to (6) make no sense to me though. If I do this, wouldn't it would make my signal take longer to attenuate?:confused:
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    The correct answer is step 4 to 6.
     
  3. Youssef.zein86

    New Member

    Apr 24, 2013
    5
    1
    Indeed the correct answer is step 4; however you should be logical in choosing the values of R and C.
    The salen key design assums that the imput resistance of the op-amp is infinity, so you have to choose R relatively small to the imput resistance of the op-amp.
    Typical voltage controlled op-amps have imput impedances that range 1 - 2 Mohms, so choosing R to be 50 times smaller will give good results, so R should not be higher of about 40 K, and you will find C as the last unknown of the equation.
     
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