# OP AMP Integrator Circuit

I have a HW question with a simple op amp circuit and was wondering if I could get some help. Here is a quick glance at the problem: and here is my attempt to solve it: I am not sure if my signs are correct when solving it. Also should I solve for Vo(t) for each case or did I do it correctly by solving for the formula for a general Vi and then graphing each result separately. Could I also receive some insight on part B. I am not sure how to go about it. the only thing I am thinking is that I multiply Vo(t) by tau and the RC's cancel out but this doesn't seem correct.

Any help appreciated. Thank you!

• I think the graphs would be OK if they all started at 0V (t=0) as opposed to some arbitrary positive voltage that can be mistaken for the input voltage at t=0. – Andy aka Apr 30 '14 at 16:51
• @Andyaka should I state that at (t=0) V0 = V (the V in the graphs? – user39571 May 1 '14 at 22:42
• I think you should commence the graphs at 0 volts. This doesn't make any assumptions about what the signal is before t=0. – Andy aka May 1 '14 at 22:51

a)

1. I think you missed a minus sign in (If) current equation , according to the current direction.

2. Don't know why you used dVout = dQ/C and then again replaced dQ=dVout*C. No effect.

3. Didn't understand what you tried to do line #4. you should just do an integration in line #3 with limiting value on Vout. e.g at t=0 Vout=0 and t=t Vout=Vout. Final equation is right though.

4. The waveforms are correct .

b) T=RC is time constant. It acts as a gain constant for the integrator. If you provide a square signal with period T, the RC value should be greater than or equal to T/2 to get a triangular waveform. Otherwise the capacitor won't get enough time to charge or discharge.

• so, when Vin = +V, current Iin/If goes through the capacitor and charges it. This charging rate depends on Capacitance C & Rin that is the time constant Tau. If it charges too quickly you wouldn't get a ramp-shaped output for square inputs. See more explanation here: electronics-tutorials.ws/opamp/opamp_6.html – hassansin Apr 30 '14 at 18:25

A) That looks about right. Except you might want to think about why your value for f(0) is what it is. (This is not a problem at this point, but you'll later need to think about the question of power-on behavior.)

B) Think "slope".