# Circuit analysis using an op-amp (non-ideal integrator)

I've the following circuit:

simulate this circuit – Schematic created using CircuitLab

My question is, is my method right to find the output voltage in relation to the input voltage?

My Work: The circuit can be analyzed by applying Kirchhoff's current law at the node $\text{V}_2\left(t\right)$, keeping ideal op-amp behaviour in mind:

$$\text{I}_1\left(t\right)=\text{I}_3\left(t\right)+\text{I}_2\left(t\right)\tag1$$

In an ideal op-amp we know that $\text{I}_3\left(t\right)=0$. So we can write:

$$\text{I}_1\left(t\right)=0+\text{I}_2\left(t\right)=\text{I}_2\left(t\right)\tag2$$

For $\text{I}_2\left(t\right)$ I used Laplace transform:

$$\text{i}_2\left(\text{s}\right)=\frac{\left(\text{v}_2\left(\text{s}\right)-\text{v}_{\space\text{out}}\left(\text{s}\right)\right)\cdot\left(1+\text{R}_2\cdot\text{C}\cdot\text{s}\right)-\text{R}_2\cdot\text{C}\cdot\text{V}_\text{c}\left(0\right)}{\text{R}_2+\text{R}_3+\text{R}_2\cdot\text{R}_3\cdot\text{C}\cdot\text{s}}\tag3$$

And for $\text{I}_1\left(t\right)$ we can write (in the s-plane):

$$\text{i}_1\left(\text{s}\right)=\frac{\text{v}_\text{in}\left(\text{s}\right)-\text{v}_2\left(\text{s}\right)}{\text{R}_1}\tag4$$

We know that $\text{I}_1\left(t\right)=\text{I}_2\left(t\right)$ and $\text{V}_2\left(t\right)=0$, so we can write:

$$\frac{\text{v}_\text{in}\left(\text{s}\right)-0}{\text{R}_1}=\frac{\left(0-\text{v}_{\space\text{out}}\left(\text{s}\right)\right)\cdot\left(1+\text{R}_2\cdot\text{C}\cdot\text{s}\right)-\text{R}_2\cdot\text{C}\cdot\text{V}_\text{c}\left(0\right)}{\text{R}_2+\text{R}_3+\text{R}_2\cdot\text{R}_3\cdot\text{C}\cdot\text{s}}\tag5$$

Which simplifies to:

$$\frac{\text{v}_\text{in}\left(\text{s}\right)}{\text{R}_1}=\frac{\text{R}_2\cdot\text{C}\cdot\text{V}_\text{c}\left(0\right)-\text{v}_{\space\text{out}}\left(\text{s}\right)\cdot\left(1+\text{R}_2\cdot\text{C}\cdot\text{s}\right)}{\text{R}_2+\text{R}_3\cdot\left(1+\text{R}_2\cdot\text{C}\cdot\text{s}\right)}\tag6$$

And I know that $\text{R}_3=\frac{1}{\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}}$, so:

$$\frac{\text{v}_\text{in}\left(\text{s}\right)}{\text{R}_1}=\frac{\text{R}_2\cdot\text{C}\cdot\text{V}_\text{c}\left(0\right)-\text{v}_{\space\text{out}}\left(\text{s}\right)\cdot\left(1+\text{R}_2\cdot\text{C}\cdot\text{s}\right)}{\text{R}_2+\frac{1+\text{R}_2\cdot\text{C}\cdot\text{s}}{\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}}}\tag7$$

Question: Am I right about my relation between the output and input voltage?

• Your method may be correct but it's way more complex that whats required to solve a circuit like that, and it leads you to a wrong answer (units doesn't match) . You forget to apply the virtual short circuit condition(V2=V3) therefore V2=0. See my answer for a much simpler solution. Feb 5, 2018 at 15:02

## 1 Answer

Assuming your opamp is ideal, meaning that it can source or sink infinite current, draws no current at the inputs and has no offset voltage. You can neglect R3 and R4, since they don't have any effect to the output voltage.

Then you can just apply the formula for the inverting amplifier:

$$V_{\mathrm{out}}=V_{\mathrm{in}}\frac{-Z_{\mathrm{f}}}{Z_{\mathrm{in}}}$$ with Zf being the fedback impedance, and Zin the input impedance.
In your case: $$Z_{\mathrm{f}}=R_2||\frac{1}{s C }=\frac{R_2}{R_2Cs+1}$$ Then: $$V_{\mathrm{out}}=V_{\mathrm{in}}\frac{\frac{R_2}{R_2Cs+1}}{R_1}=V_{\mathrm{in}}\frac{R_2}{R_1}\frac{1}{R_2Cs+1}$$

• What about the initial condition? Feb 5, 2018 at 15:11
• Right, basically a low pass filter with gain if R2 does not equal R1 Feb 5, 2018 at 15:31