Electronic – How does this Op Amp affect voltage


I have a circuit as follows:

enter image description here

Note that \$V^+\$ is a voltage source that has both an AC and DC component. I want to see how the OpAmp affects both components. My guess is that we can write:

$$V_{out}(AC)=(1 + \frac{R2}{R1} – \frac{X_c}{\sqrt{X_c^2+ R_1^2}})V^+(AC) $$

Where I assumed the role of the RC part as a low pass filter as seen by \$V^+\$. Then, since low pass filters let all DC pass through unaffected, I would say that as long as C1 is not very large, $$V_{out}(DC)=(1 + \frac{R2}{R1})V^+(DC) $$

If I was told C1 is large enough to be considered as a short for AC and an open for DC, I would say $$V_{out}(AC)=(1 + \frac{R2}{R1})V^+(AC)$$ and $$V_{out}(DC)=V^+(DC)$$

Can someone tell me if this is correct or provide alternative explanations?

Best Answer

This diagram explains everything

enter image description here

At DC the capacitor reactance is very high (open circuit)\$X_C = \infty \$.

So, op-amp work as a voltage follower (gain one).

As the input signal frequency increases the capacitor reactance decreases.

And when \$X_C = (R_1 + R_2) \$ at \$F_1 = \frac{1}{2 \pi C_1 (R_1 + R2)}\$

The op-amp voltage gain start to increases and will reach \$ \large (1 +\frac{R_2}{R_1})\$ for signal frequency larger than :

\$\large F_2 = \frac{1}{2 \pi C_1 R_1}\$ When \$X_C < R_1\$