# Transfer Function - Opamp

My work:

$$\text{I}_{\text{in}}(t)+\text{I}_{\text{p}}(t)=0\Longleftrightarrow$$ $$\text{I}_{\text{p}}(t)=-\text{I}_{\text{in}}(t)\Longleftrightarrow$$ $$\text{I}_{\text{p}}(t)=-\frac{\text{V}_{\text{in}}(t)}{\text{R}_1}$$

$$\text{I}_{\text{p}}(t)=\text{I}_{\text{C}}(t)+\text{I}_{\text{R}_2}(t)\Longleftrightarrow$$ $$\text{I}_{\text{p}}(t)=\text{C}\text{V}'_{\text{out}}(t)+\frac{\text{V}_{\text{out}}(t)}{\text{R}_2}$$

$$-\frac{\text{V}_{\text{in}}(t)}{\text{R}_1}=\text{C}\text{V}'_{\text{out}}(t)+\frac{\text{V}_{\text{out}}(t)}{\text{R}_2}\Longleftrightarrow$$ $$\mathcal{L}_t\left[-\frac{\text{V}_{\text{in}}(t)}{\text{R}_1}\right]_{(s)}=\mathcal{L}_t\left[\text{C}\text{V}'_{\text{out}}(t)+\frac{\text{V}_{\text{out}}(t)}{\text{R}_2}\right]_{(s)}\Longleftrightarrow$$ $$-\frac{\text{V}_{\text{in}}(s)}{\text{R}_1}=\text{C}s\text{V}_{\text{out}}(s)+\frac{\text{V}_{\text{out}}(s)}{\text{R}_2}\Longleftrightarrow$$ $$\color{red}{\frac{\text{V}_{\text{out}}(s)}{\text{V}_{\text{in}}(s)}=-\frac{\text{R}_2}{\text{R}_1\left(1+\text{C}\text{R}_2s\right)}}$$

• Yes - looks good. However, there is a simpler (and quicker) way for arriving at the transfer function: Calculate from the beginning in the s-domain (without the necessity to perform the Laplace transformation).
– LvW
Feb 20, 2016 at 16:34
• @LvW Thanks for your response, can you show me that method? Feb 20, 2016 at 17:10
• @JanEerland Here's a brief introduction s-domain circuit analysis. Basically, you can transform R, L, and C into complex impedances in terms of s and then all the math becomes much easier. Feb 20, 2016 at 23:00

The solution you have arrived at is correct. The circuit is a practical integrator. The resistor in parallel with capacitor limits low frequency gain and minimizes variations in output. Here is a simpler and quicker solution:

Since the opamp is in inverting configuration, the transfer function is:

$$Av = \frac{-Z2(s)}{Z1(s)}$$ Note that all impedances are in s-domain. Z2(s) happens to be the parallel combination of R2 and 1/sC $$Z2(s) =\frac{R2\cdot\frac{1}{sC}}{R2+\frac{1}{sC}}$$ $$Z1(s) =R1$$ $$\frac{vo(s)}{vin(s)} = -\frac{\frac{R2\cdot\frac{1}{sC}}{R2+\frac{1}{sC}}}{R1}$$

Which on simplification reduces to: $$\frac{vo(s)}{vin(s)} = -\frac{R2}{R1\cdot(1+R2Cs)} = -\frac{\frac{R2}{R1}}{1+R2Cs}$$

So the gain at low frequencies is -R2/R1 which without R2 would have quickly rolled off.