# op amp with capacitors Why is Cp here not part of the equation?

• Do you know what the voltage at 'x' is? Mar 7, 2021 at 9:30
• That is not a practical circuit, there is no dc path to ground for the input bias current at the op amp's inverting input.
– user173271
Mar 7, 2021 at 10:40
• The name of this circuit is "charge amplifier". You can see what the idea behind it is in this CD paper. Mar 7, 2021 at 10:46
• @KD9PDP, If the resistor is connected between the inverting input and ground (in parallel to Cp), there will be no DC negative feedback. That is why, it should be connected in parallel to Cf. Then the circuit will work as a DC voltage follower. BTW Cp represents stray capacitances (like the cabel capacitance) in such applicatiobs as charge amplifier. Mar 7, 2021 at 18:31
• @Circuitfantasist Ah yeah, I get what you're saying now - yeah, the DC point isn't stabilized, so there should be that resistor. nice point! Mar 7, 2021 at 21:05

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.

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:=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={{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:

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

• Hi, thanks for the detailed explanation. Just realised i missed out V+=V−=V1=0 because i was only going through the question briefly. Nevertheless, thanks for the reply! Mar 7, 2021 at 10:37
• @Jan, You are a beatiful mind... but I would like to ask you, "Do you have any idea about the role of the op-amp here... and in all inverting circuits?" I am just asking you out of curiosity... Mar 7, 2021 at 18:22
• I would have thought replacing the capacitors with resistors would result in an examination of a fundamentally different circuit.
– K H
Mar 10, 2021 at 1:57