Op Amp transfer function

Discussion in 'General Electronics Chat' started by RFeng, Jul 5, 2012.

  1. RFeng

    Thread Starter New Member

    May 30, 2012
    27
    0
    Hi,

    I'm trying to figure out the transfer function of the following Op Amp Intergrator:
    [​IMG]

    Can you please advice on that?

    Thank you!
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,764
    2,534
    It is a integrator. Not sure what the biasing network on the + input is about, it could be replaced with either power supply or a resistor network and a single power supply.
     
  3. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Replace the feedback network with a single impedance that is the parallel combination of the R and C. Then analyze it like any other opamp circuit, noting that the voltage on the inverting input is not at virtual ground, but rather at Vplus.
     
  4. RFeng

    Thread Starter New Member

    May 30, 2012
    27
    0
    Hi guys.
    Thanks!

    I analyzed it and reach its transfer function:
    * R1 - Resistor at (-) Input
    * Zf - Impedance of Feedback
    [​IMG]

    How can you tell what is the pole of this transfer function?

    Thanks again!
     
  5. RFeng

    Thread Starter New Member

    May 30, 2012
    27
    0
    Back to page 1 if I may :)
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Z_f=\frac{R_ f\frac{ 1}{ C_f s}}  {R_f+\frac{ 1}{C_f s } }

    etc.

    No negative sign.
     
  7. AdrianN

    Active Member

    Apr 27, 2009
    97
    1
    Well, I do not see R1 in your schematic so the transfer function must be:

    Vo = -Zf/R2 * Vi + V(+)*(1+Zf/R2)

    Since V(+) is DC, it has no dynamic influence. It just moves the output DC level up or down depending on the combination of VSET and Vdc DC voltage level. The signal rides on top of this DC level. So, the transfer function can be rewritten as:

    Vo = -Zf/R2 * Vi + V(+)*(1+Rf/R2)

    The signal amplitude is given by

    Vo_signal = -Zf/R2 * Vi, and can be written as

    Vo_signal = -1/(1+ s Cf Rf)* Rf/R2 * Vi

    So the pole is at

    1/(2 pi Cf Rf) = 159.15 kHz. It is only determined by Cf Rf time constant. Since both resistors Rf and R2 are 1kohm, the signal amplitude is identical with the input amplitude (inverted), but rides on top of the DC level we discussed. The unity gain starts dropping after 159.15kHz.
     
    Last edited: Jul 7, 2012
Loading...