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

Integrator Circuit Transfer Function (voltage divide)

  1. Oct 1, 2012 #1
    I have attached a picture of the circuit in which I am attempting to build a transfer function for, however, as you can see it is not a normal integrator. Normally it is evident which resistor is the resistor that is part of the series connection to the op amp.

    When I do the KVL on the circuit to get the values I get the following function

    [itex]

    v_{in} = C\frac{dv_{o}}{dt} + 2.5\times R_1 (\frac{1}{R1} + \frac{1}{R_2})[/itex]

    assuming that I voltage divide to [itex] 2.5V [/itex] using values for [itex] R_1 = 5k\Omega R_2 = 910\Omega [/itex] and the input is 16.3V.

    [itex]
    v_{in} = C\frac{dv_{o}}{dt} + 16.23[/itex]

    From here I do not know how to get the transfer function. I am correct in assuming that the transfer function is no longer a linear line on the Bode plot due to that extra resistor that goes from the input to ground (thus reducing the current).

    Or am I way off base with my assumptions.

    Thank-you for your help
    Lance

    PS: I am an undergrad electrical engineer so the explanation does not have to be overly simplified.
     

    Attached Files:

  2. jcsd
  3. Oct 1, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    You could solve the DE, then for sinusoids let vin = A.sinѠt

    Or you could start again, and let vcap = iC/(jѠc)
    where iC is the capacitor current
     
  4. Oct 1, 2012 #3
    That does make it more difficult to determine system dynamic parameters, ie rise time settling time and the such?
     
  5. Oct 1, 2012 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    Your first post only mentioned Bode plot. If you are going to use Laplace, then let Vcap(s) = Icap(s)/(s.C)
     
  6. Oct 1, 2012 #5
    Yes however, normally I use the Laplace form of the equation to make the transfer function to then recreate the Bode Plot.

    The problem is that the Laplace transform is

    [itex]

    sV_i = A s^2 V_o + B [/itex]

    As you can see the Constant B makes creating the transfer function very difficult!
     
  7. Oct 1, 2012 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    What's that DC offset on op-amp's (+) input in aid ofhttp://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif [Broken] You won't get anywhere while that's there, it will just send your op-amp output directly to the + rail! If that offset is to remain, then you'll need to exactly compensate for it by feeding DC into the (-) input, then superimpose your AC (for the Bode plot) on top of that DC. Capacitor coupling will achieve this.

    Is this a circuit you invented as an exercise? :confused:
     
    Last edited by a moderator: May 6, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integrator Circuit Transfer Function (voltage divide)
Loading...