op amp with capacitors

Question

Why is Cp here not part of the equation?

Answer 1

Quick answer: because the op amp makes sure node x is always zero volts. It's a virtual ground. The voltage across Cp is always 0, so no current flows. No current, no voltage means no contribution to the transfer function.

Answer 2

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following opamp-circuit:

^{simulate this circuit – Schematic created using CircuitLab}

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_2=\text{I}_1+\text{I}_3\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2-\text{V}_1}{\text{R}_3} \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$\frac{\text{V}_1}{\text{R}_2}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_2-\text{V}_1}{\text{R}_3}\tag3$$

Now, using an ideal opamp, we know that:

$$\text{V}_+=\text{V}_-=\text{V}_1=0\tag4$$

So we can rewrite equation \$(3)\$ as follows:

$$\frac{0}{\text{R}_2}=\frac{\text{V}_\text{i}-0}{\text{R}_1}+\frac{\text{V}_2-0}{\text{R}_3}\tag3$$

Now, we can solve for the transfer function:

$$\mathcal{H}:=\frac{\text{V}_2}{\text{V}_\text{i}}=-\frac{\text{R}_3}{\text{R}_1}\tag6$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global`*"];
V1 = 0;
FullSimplify[
 Solve[{I2 == I1 + I3, I1 == (Vi - V1)/R1, I2 == V1/R2, 
   I3 == (V2 - V1)/R3}, {I1, I2, I3, V2}]]

Out[1]={{I1 -> Vi/R1, I2 -> 0, I3 -> -(Vi/R1), V2 -> -((R3 Vi)/R1)}}

My equation was also confirmed using LTspice.

---

When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so \$\text{y}\left(\text{s}\right)\$ is the Laplace transform of the function \$\text{Y}\left(t\right)\$):

• $$\text{R}_1=\frac{1}{\text{sC}_1}\tag7$$
• $$\text{R}_2=\frac{1}{\text{sC}_2}\tag8$$
• $$\text{R}_3=\frac{1}{\text{sC}_3}\tag9$$

So, we can rewrite the transfer function as:

$$\mathscr{H}\left(\text{s}\right)=-\frac{\frac{1}{\text{sC}_3}}{\frac{1}{\text{sC}_1}}=-\frac{\text{C}_1}{\text{C}_3}\tag{10}$$