Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can the value Q not be sensitive to the circuit components?

  1. Nov 27, 2012 #1


    User Avatar
    Gold Member

    I have attached a circuit I am trying to analyze,

    Is it possible that Q is not sensitive to the value of C?

    I solved my transfer function Vo/Vi, and I get (2/c^2R^2)/(S^2 + S(1/RC) + 1/(R^2C^2))

    Which looks plausible at the present time, however when I solve for Q, get 1 here? It is not sensitive to any circuit parameters... I'm sure that couldn't be possible.?

    Thanks for looking.

    (Just confirming, because Ra = Rb, K = 2 correct)

    Attached Files:

  2. jcsd
  3. Nov 29, 2012 #2


    User Avatar

    Staff: Mentor

    If you were to re-analyze the circuit, but using dis-similar resistors (e.g., R and m.R instead of both equal to R), and (similarly) capacitors of C and n.C, I expect you'll find that you can set Q to a wide range of values.

    m and n being fractions.
  4. Dec 1, 2012 #3
    This cell does NOT work because at any intersting Q it is exceedingly sensitive to Ra and Rb.

    By the way, when analyzing the sensitivity of a filter cell or any design, you can't keep R=R. You have to give an individual value to each component, because the drifts and tolerances don't match an other.
  5. Dec 2, 2012 #4
    I am not the expert of filter. I thought Q is control by ratio of Ra and Rb. k=2 is cutting a little close, I saw oscillation around this number. I usually try to keep it no more than 1.5.
  6. Dec 2, 2012 #5


    User Avatar

    Staff: Mentor

    With this being a low-pass filter, there generally is no requirement for Q>1. As an easy straight-forward design, the circuit should suit many applications.
  7. Dec 2, 2012 #6
    With Q<1 you don't need RA and RB at all. Just use different capacitors.

    If the filter has several biquads then the needed Q climbs and immediately, this cell gets unusable, because a finite RA/RB gives an infinite Q (the cell oscillates) and before that, Q is ridiculously sensitive to RA/RB.

    I know that so many books and courses recommend this cell, but they didn't try it by themselves.

    It's just like the biquadratic cells that pretend to compensate the limited bandwidth of the op amps by inserting one more identical op amp in the feedback: all these cells oscillate.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook