OPamp integrator.

Thread Starter

Mathishard

Joined Apr 22, 2016
2
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...
 

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Jony130

Joined Feb 17, 2009
5,183
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
 

crutschow

Joined Mar 14, 2008
25,421
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.
 
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