Need Help With Cutoff Frequency

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

  1. Mawangs1

    Thread Starter New Member

    Mar 26, 2012
    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
    The correct answer is step 4 to 6.
  3. Youssef.zein86

    New Member

    Apr 24, 2013
    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.