How to find the Transfer Function of this circuit?

Question

I am trying to find the transfer function of this circuit, but I can't find how. Does anyone have a clue?

Answer 1

I think you might find what you are looking for in this book: https://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf

It is a reference guide that is invaluable for opamp design. Around the page 400, you might find what you look for!

Answer 2

You can easily determine the transfer function of this circuit by considering a simple \$RC\$ low-pass filter followed by a non-inverting structure described in this Wikipedia link. Because the op-amp does not load the low-pass filter, you can cascade the two transfer functions conveniently:

The final expression shows the presence of a dc gain determined when \$s=0\$, one zero and two poles. The magnitude and phase plots between the three possible expressions are identical.

Answer 3

Well, we are trying to analyze the following circuit:

We know, using the voltage divider formula, that we can write:

• $$\text{V}_+=\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1+\text{R}_2}\cdot\text{V}_\text{i}\tag1$$
• $$\text{V}_-=\frac{\displaystyle\text{R}_3}{\displaystyle\text{R}_3+\text{R}_4}\cdot\text{V}_\text{o}\tag2$$

When using an ideal OPAMP, we know that:

$$\text{V}_+=\text{V}_-\tag3$$

So, we get:

$$\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1+\text{R}_2}\cdot\text{V}_\text{i}=\frac{\displaystyle\text{R}_3}{\displaystyle\text{R}_3+\text{R}_4}\cdot\text{V}_\text{o}\space\Longleftrightarrow\space$$ $$\mathscr{H}:=\frac{\text{V}_\text{o}}{\text{V}_\text{i}}=\frac{\displaystyle\frac{\text{R}_2}{\text{R}_1+\text{R}_2}}{\displaystyle\frac{\text{R}_3}{\text{R}_3+\text{R}_4}}=\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1+\text{R}_2}\cdot\frac{\displaystyle\text{R}_3+\text{R}_4}{\displaystyle\text{R}_3}=\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\displaystyle\text{R}_3\left(\text{R}_1+\text{R}_2\right)}\tag4$$

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Applying this to your circuit we get:

\begin{equation} \begin{split} \mathscr{H}\left(\text{s}\right)&=\frac{\displaystyle\frac{\displaystyle1}{\displaystyle\text{sC}_3}\left(\frac{\displaystyle\text{R}_5\text{R}_6}{\displaystyle\text{R}_5+\text{R}_6}+\frac{\displaystyle\text{R}_7\cdot\frac{1\displaystyle}{\displaystyle\text{sC}_4}}{\displaystyle\text{R}_7+\frac{1\displaystyle}{\displaystyle\text{sC}_4}} \right)}{\displaystyle\frac{\displaystyle\text{R}_5\text{R}_6}{\displaystyle\text{R}_5+\text{R}_6}\left(\frac{\displaystyle\text{R}_7\cdot\frac{1\displaystyle}{\displaystyle\text{sC}_4}}{\displaystyle\text{R}_7+\frac{1\displaystyle}{\displaystyle\text{sC}_4}} +\frac{\displaystyle1}{\displaystyle\text{sC}_3}\right)}\\ \\ &=\frac{\displaystyle\frac{\displaystyle1}{\displaystyle\text{sC}_3}\left(\frac{\displaystyle\text{R}_5\text{R}_6}{\displaystyle\text{R}_5+\text{R}_6}+\frac{\displaystyle\text{R}_7}{\displaystyle\text{sC}_4\text{R}_7+1} \right)}{\displaystyle\frac{\displaystyle\text{R}_5\text{R}_6}{\displaystyle\text{R}_5+\text{R}_6}\left(\frac{\displaystyle\text{R}_7}{\displaystyle\text{sC}_4\text{R}_7+1} +\frac{\displaystyle1}{\displaystyle\text{sC}_3}\right)}\\ \\ &=\frac{\displaystyle\frac{\displaystyle\text{R}_5\text{R}_6}{\displaystyle\text{R}_5+\text{R}_6}+\frac{\displaystyle\text{R}_7}{\displaystyle\text{sC}_4\text{R}_7+1} }{\displaystyle\frac{\displaystyle\text{R}_5\text{R}_6}{\displaystyle\text{R}_5+\text{R}_6}\left(\frac{\displaystyle\text{sC}_3\text{R}_7}{\displaystyle\text{sC}_4\text{R}_7+1} +1\right)}\\ \\ &=\frac{\displaystyle\text{R}_5\text{R}_6+\frac{\displaystyle\text{R}_7\left(\text{R}_5+\text{R}_6\right)}{\displaystyle\text{sC}_4\text{R}_7+1} }{\displaystyle\text{R}_5\text{R}_6\left(\frac{\displaystyle\text{sC}_3\text{R}_7}{\displaystyle\text{sC}_4\text{R}_7+1} +1\right)}\\ \\ &=\frac{\displaystyle\text{R}_5\text{R}_6\left(\text{sC}_4\text{R}_7+1\right)+\text{R}_7\left(\text{R}_5+\text{R}_6\right)}{\displaystyle\text{R}_5\text{R}_6\left(\text{sC}_3\text{R}_7+\text{sC}_4\text{R}_7+1\right)}\\ \\ &=\frac{\displaystyle\text{R}_5\text{R}_6\left(\text{sC}_4\text{R}_7+1\right)+\text{R}_7\left(\text{R}_5+\text{R}_6\right)}{\displaystyle\text{R}_5\text{R}_6\left(\text{sR}_7\left(\text{C}_3+\text{C}_4\right)+1\right)} \end{split}\tag5 \end{equation}

So, the amplitude of this transfer function is given by:

\begin{equation} \begin{split} \left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|&=\left|\frac{\displaystyle\text{R}_5\text{R}_6\left(\text{j}\omega\text{C}_4\text{R}_7+1\right)+\text{R}_7\left(\text{R}_5+\text{R}_6\right)}{\displaystyle\text{R}_5\text{R}_6\left(\text{j}\omega\text{R}_7\left(\text{C}_3+\text{C}_4\right)+1\right)}\right|\\ \\ &=\frac{\displaystyle\left|\text{j}\omega\text{C}_4\text{R}_5\text{R}_6\text{R}_7+\text{R}_5\text{R}_6+\text{R}_7\left(\text{R}_5+\text{R}_6\right)\right|}{\displaystyle\left|\text{j}\omega\text{R}_5\text{R}_6\text{R}_7\left(\text{C}_3+\text{C}_4\right)+\text{R}_5\text{R}_6\right|}\\ \\ &=\frac{\displaystyle\sqrt{\left(\text{R}_5\text{R}_6+\text{R}_7\left(\text{R}_5+\text{R}_6\right)\right)^2+\left(\omega\text{C}_4\text{R}_5\text{R}_6\text{R}_7\right)^2}}{\displaystyle\sqrt{\left(\text{R}_5\text{R}_6\right)^2+\left(\omega\text{R}_5\text{R}_6\text{R}_7\left(\text{C}_3+\text{C}_4\right)\right)^2}} \end{split}\tag6 \end{equation}

And the argument is given by:

\begin{equation} \begin{split} \arg\left(\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)&=\arg\left(\frac{\displaystyle\text{R}_5\text{R}_6\left(\text{j}\omega\text{C}_4\text{R}_7+1\right)+\text{R}_7\left(\text{R}_5+\text{R}_6\right)}{\displaystyle\text{R}_5\text{R}_6\left(\text{j}\omega\text{R}_7\left(\text{C}_3+\text{C}_4\right)+1\right)}\right)\\ \\ &=\arctan\left(\frac{\displaystyle\omega\text{C}_4\text{R}_5\text{R}_6\text{R}_7}{\displaystyle\text{R}_5\text{R}_6+\text{R}_7\left(\text{R}_5+\text{R}_6\right)}\right)-\arctan\left(\frac{\displaystyle\omega\text{R}_5\text{R}_6\text{R}_7\left(\text{C}_3+\text{C}_4\right)}{\displaystyle\text{R}_5\text{R}_6}\right)\\ \\ &=\arctan\left(\frac{\displaystyle\omega\text{C}_4\text{R}_5\text{R}_6\text{R}_7}{\displaystyle\text{R}_5\text{R}_6+\text{R}_7\left(\text{R}_5+\text{R}_6\right)}\right)-\arctan\left(\omega\text{R}_7\left(\text{C}_3+\text{C}_4\right)\right) \end{split}\tag7 \end{equation}

Using your values, we get:

• $$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\sqrt{730458729\omega^2+749664400000000000000}}{11 \sqrt{19882681 \omega ^2+1000000000000000000}}\tag8$$
• $$\arg\left(\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)=\arctan\left(\frac{27027\omega}{27380000000}\right)- \arctan\left(\frac{4459\omega}{1000000000}\right)\tag9$$