Op-amp transfer function

Discussion in 'Homework Help' started by deki, Sep 20, 2012.

  1. deki

    Thread Starter New Member

    Sep 6, 2011

    Ok I know how to find the transfer function but what has me confused is the resistor on the positive terminal. I thought I'd treat it as an inverting op-amp, hence the gain= -Z2/Z1 and I did that and ended up getting two poles. Since we've been told to assume that it's an ideal op-amp, wouldn't that make the voltage at the positive terminal 0? Since the input currents are equal to zero, hence the voltages are equal?

    Also for part b, I'm not exactly sure how to go about it. There is a formula in the lecture notes: ft = G * f3db, but ft is the gain BW product. So is our f1 the ft in the formula?

  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Each pole sets the frequency at which one sees a 3dB drop in transfer gain. You make the poles the same value in each case - per the problem statement.

    The equivalent time constant for each pole will be of the form


    and the pole frequency would be

    ω=1/T radians per second.

    Presumably you were assigned a specific frequency in Hz. So you would convert that to radians per second in the first instance.

    And yes - R3 is just a distraction to confuse you if the op-amp is ideal.
  3. deki

    Thread Starter New Member

    Sep 6, 2011
    So my poles are: (1 + sC1R1) (1 + sC2R2)
    So I simply equate 1/R1C1 = 2PI*f1 and 1/R2C2 = 2PI*f1

  4. WTP Pepper

    New Member

    Aug 1, 2012
    Ignore the resistor on the +ve input. It's there to settle the output offset voltage which in an AC coupled circuit is not an issue. It is on a DC high gain circuit though.
  5. druff1989

    New Member

    Dec 9, 2010
    Sorry I can't help but I have a question relating to this. How can you find the transfer function in the frequency domain? Thanks