# 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. – jagjordi Feb 5 '18 at 15:02

$$V_{\mathrm{out}}=V_{\mathrm{in}}\frac{-Z_{\mathrm{f}}}{Z_{\mathrm{in}}}$$ 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: $$V_{\mathrm{out}}=V_{\mathrm{in}}\frac{\frac{R_2}{R_2Cs+1}}{R_1}=V_{\mathrm{in}}\frac{R_2}{R_1}\frac{1}{R_2Cs+1}$$