OPamp integrator.

Discussion in 'Homework Help' started by Mathishard, Apr 22, 2016.

  1. Mathishard

    Thread Starter New Member

    Apr 22, 2016
    Hello everybody, I am trying to simulate an opamp integrator with a square wave input and triangle wave output.
    Right now I am puzzled with the capacitor value. My job is to fit the triangle wave in between the voltage swings, by calculating the correct capacitor value.
    Some base stats:
    Generator f=300 Hz
    Generator single peak amplitude=10 V
    Generator duty cycle=50%
    Input resistance R=1000 ohm
    Opamp positive and negative swing +15 to -15 volts.
    To get the best fit into the voltage swings, my RMS output value needs to be 8,666 V
    Ive tried to calculate the capacitor value, and came up with 0,612 uF, but the capacitor value 0,555uF fits properly to the 8,666 V.
    Could somebody please explain to me what I did wrong ?
    Thank you for reading. Cant math for my life...
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    First you need to have a good understand of how this circuit work.
    Next we can solve for C quite easy. From capacitance definition we have
    C = Q/V = (I*t)/V
    I = 10V/1K = 10mA
    t = 1.667ms
    V = 30V

    C = (10mA * 1.666ms)/30V ≈ 0.556μF ≈ 555.6nF
  3. crutschow


    Mar 14, 2008
    A handy thing to remember is that, for a step voltage change at the input, the integrator output voltage will change by a voltage equal to that step voltage in one time-constant (RC) time.