How Does Changing Capacitor Value Affect OPAMP Output Voltage?

[etf]

Hi!
Here is my task:

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

View attachment 73553

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

View attachment 73554

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

Here is what I have done:
Since there are two sources, one DC and one time dependent pulse waveform, I used superposition method. When time dependent source is active, our circuit is:
View attachment 73555
$$v2(t)=0\rightarrow v1(t)=0$$

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

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

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

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 :(

 

[rude man]

1. you need a feedback resistor to keep from saturation if you're looking at a simulation or a real circuit. With an ideal op amp that's not necessary.
2. you have not described the inpt in words. I'm not familiar with LTSpice.

 

[lazyaditya]

The circuit you have is an integrator.The output for pulse waveform will be triangular wave and the output for the dc signal will be a straight line with a negative slope. When you add the equation of both of your output for ac and dc inputs you will have waveform as given in first figure. Second figure that you have shown has clipping of waveform at the bottom that is since the output waveform has gone below the lower threshold of OP-AMP and thus into saturation (-Vee).
I hope this would help.

 

[etf]

Solved it :)

 

[rezeew]

Hi there,

It looks like you have started off well with using the superposition method to analyze the circuit. However, I believe there may be a mistake in your calculations for the v3(t) output voltage. The correct equation for v3(t) should be:

v3(t) = -50*t for 0 ≤ t ≤ 100*10^-3 s
v3(t) = 50*(t-100*10^-3) for 100*10^-3 s ≤ t ≤ 200*10^-3 s

This is because when the input voltage is positive, the capacitor will charge and the voltage across it will be increasing, while when the input voltage is negative, the capacitor will discharge and the voltage across it will be decreasing. This results in the triangle shape of the output voltage.

As for the change in output voltage when the capacitor's value is changed from 200nF to 33nF, this can be explained by the time constant of the circuit. The time constant is given by RC, where R is the resistance and C is the capacitance. When C is decreased, the time constant decreases as well, meaning that the capacitor will charge and discharge faster. This results in a sharper change in the output voltage, leading to the cut-off bottom and lack of gradual decrease in the output voltage.

I hope this helps and let me know if you have any further questions or need clarification on anything. Keep up the good work with your circuit analysis!