# OPAMP Circuit analysis

• Engineering
Hi!

OPAMP circuit and output voltag are 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).

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

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

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

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

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 :(

NascentOxygen
Staff Emeritus
The output should comprise a triangle wave produced by integrating the squarewave, plus a slow ramp produced by integrating a small fixed current. Does your mathematical derivation exhibit these two features?

The waveform from your simulation is what I'd expect to see.

Here is how I have done it:
I found Fourier series of $$vin(t)$$, $$vin(t)=\sum_{n=1}^{\infty}\frac{2(1-(-1)^{n})}{n\pi }\sin{(nw0t)}$$. Then I used superposition method to find $$viz(t)$$. When $$vin(t)$$ is active, our circuit is (I didn't show OPAMP power supply): Using KVL I got:
$$viz(t)+\frac{1}{C}\int i(t)+Ri(t)=vin(t)$$

$$viz(t)+\frac{1}{C}\int\frac{vin(t)}{R1}dt+Ri(t)=vin(t)$$

$$viz(t)+\frac{1}{R1C}\int vin(t)+R1\frac{vin(t)}{R1}=vin(t)$$
$$viz(t)=\frac{-1}{R1C}\int (\sum_{1}^{\infty}2\frac{1-(-1)^{n}}{n\pi }\sin{(nw0t)})dt$$
$$viz(t)=\frac{-2}{R1C}\sum_{n=1}^{\infty}\frac{1-(-1)^{n}}{n\pi }\int \sin(nw0t)dt=\frac{2}{R1C}\sum_{n=1}^{\infty}\frac{1-(-1)^{n}}{n^{2}\pi w0 }\cos{(nw0t)}$$

Here is plot of viz(t) Now it is necessary to do DC analysis and add viz(t) and Viz and (hopefully) we will get output as in simulator. I didn't do DC analysis yet.

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NascentOxygen
Staff Emeritus
With referece to your infinite series... Uh huh. That's all well and good, but what does but tell you? I think you are expected to come up with an analysis and explanation of what you see in the waveforms. Along the lines of what I outlined.

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

I don't see much call for Fourier here. Not explicitly, anyway.

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Well, I thought I must have explicit expression of input voltage for analysis. How would I express output voltage (viz(t)) if I hadn't expression for input voltage?

NascentOxygen
Staff Emeritus
Well, I thought I must have explicit expression of input voltage for analysis. How would I express output voltage (viz(t)) if I hadn't expression for input voltage?
I think you should express each input as a function of time, then integrate and sum the result.

What I think you are doing is expressing the input as a Fourier series and integrating that, then using a computer program to convert that back to a function of time and plot it. That is a roundabout process I doubt you would have time to pursue in any exam.

Have you memorised the Fourier series of a triangle wave?

Now I'm completely lost. I did analysis in Multisim and I got same shape for output voltage but different peak to peak value of signal. I give up. Here is code in Matlab I used to plot viz(t) I got by hand:
clear
clc
R1=100000;
C=200*10^(-9);
t=linspace(0,1,10000);
w0=2.*pi./0.2;
suma=0;
for n=1:10000
suma = suma + ((1-(-1).^n).*(cos(n.*w0.*t))./(n.*n.*pi.*w0));
end
red=(2./(R1.*C)).*suma;
plot(t,red)
grid

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NascentOxygen
Staff Emeritus
In my view, you've shown all the mathematics that's needed back in your first post: the integral of a step. Now, apply that to one cycle and show how the squarewave input becomes a triangular wave at the output, and how the other input leads to a ramp at the output. Then combine them.

It might be instructive if you were to extend the duration of your simulation, so you capture some "interesting output" when non-linear things start to appear.

Before you dive into complicated maths you should look back at what you are specifically asked to explain, in post #1. I interpret it to be they want you to demonstrate a practical understanding of that circuit's behaviour under the two different conditions.

• etf
NascentOxygen
Staff Emeritus
When using the 33nF capacitor, could you also please record the waveform at the junction of R1 and R2. It might contain some answers ....

I think I solved it (without Fourier series :) )
Firstly I analysed circuit for t between 0 and 0.1s (VCC=15V, vin(t)=1V). I got vout(t)=-57.5t. Then I analysed circuit for t between 0.1s and 0.2s (VCC=15V, vin(t)=-1V). I got vout(t)=42.5t. On time interval between 0s and 0.2s our output voltage vout(t) is What do you think? I didn't plot it on some wider interval since it's a little bit tricky to connect those lines and plot :)

Here is code I used for plot:
>> t1=linspace(0,0.1,100000);
>> t2=linspace(0.1,0.2,100000);
>> vout1=-57.5.*t1;
>> vout2=-5.75+42.5.*(t2-0.1);
>> plot(t1,vout1,t2,vout2)
>> grid
>>
It is necessary to plot decreasing line, then on it's end increasing one etc. :)

NascentOxygen
Staff Emeritus