Hi!
Here is my task:

OPAMP circuit and output voltag eare shown in image below. Assuming that OPAMP is ideal, come up with a solution by hand.



Then change capacitor's value from 200nF to 33nF and output voltage will be:



Explain why output voltages are different ( first has triangle shape and gradually decreases and second has cut off bottom and doesn't decrease).
Any idea here?

 

The circuit is an integrator with no way to discharge C1. Since one of the inputs is a +DC voltage, the integrator capacitor will slowly charge until the output goes to the negative rail of the op amp. If you extend the simulation time you will see that you eventually end with the same result with both capacitor values. It just takes 6 times longer with the larger capacitor.

 

Well the answer is very simple, smaller capacitor will charge faster. And notice that in this circuit the capacitor is charge/discharge via constant current.
And this is why Vout voltage rump up or down depend on whether the capacitor is charged or discharged.
C = Q/V = (I*t)/V --> V = I/C *t
So 33nF will be charge/discharge in rate equal to 10μA/33nF = 303V per second and 200nF in rate 50V per second

 

Here is what I have done:
Since there are two sources, one DC and one time dependent, I used superposition method. When time dependent source is active, our circuit is:



\(v2(t)=0\rightarrow v1(t)=0\)

\(i2(t)=0\rightarrow iin(t)=ic(t)\)

\(iin(t)=\frac{vin(t)}{R1}\)

\(v3(t)+\frac{1}{C}\int ic(t)dt+R1iin(t)-vin(t)=0\)

For positive input voltage \(vin(t)=1V\) we have:

\(v3(t)=-\frac{1}{RC1}\int vin(t)dt=-50\int 1dt=-50t,\,\, 0\leq t\leq 100*10^{-3}s\)

For negative input voltage \(vin(t)=-1V\) we have:
\(v3(t)=-\frac{1}{RC1}\int vin(t)dt=-50\int (-1)dt=50t,\,\,100*10^{-3}s\leq t\leq 200*10^{-3}s\)
But I don't know what to do next

 

Solved it
If someone is interested in solution, let me know