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I'm doing some analysis of op-amp circuitry including the high-pass filter shown below.

I was given three questions to try further my understanding on the analysis of such system.

However, I am not quite sure where to start on this one. I have listed the three questions below, the derivation of the transfer function is where I'm currently stuck.

  1. Derive a general expression for the transfer function of the filter shown.

  2. Let R1 = 3.9 kΩ, R2 = 1.8 kΩ, and C = 0.33 μF. What is the filter frequency magnitude response at very low frequencies?

  3. Calculate the frequency at which the magnitude response will be 3 dB lower than it is at high frequencies.

Circuit Diagram

Any help would be greatly appreciated.

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Well, we are trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\text{I}_2\\ \\ \text{I}_3=\text{I}_4 \end{cases}\tag1 $$

When we use and apply KVL, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{p}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_3-\text{V}_2}{\text{R}_4} \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} \frac{\text{V}_\text{p}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}\\ \\ \frac{\text{V}_2}{\text{R}_3}=\frac{\text{V}_3-\text{V}_2}{\text{R}_4} \end{cases}\tag3 $$

Now, when we have an ideal opamp we know that \$\text{V}_x:=\text{V}_+=\text{V}_-=\text{V}_1=\text{V}_2\$. So we can rewrite equation \$(3)\$ as follows:

$$ \begin{cases} \frac{\text{V}_\text{p}-\text{V}_x}{\text{R}_1}=\frac{\text{V}_x}{\text{R}_2}\\ \\ \frac{\text{V}_x}{\text{R}_3}=\frac{\text{V}_3-\text{V}_x}{\text{R}_4} \end{cases}\tag4 $$

Now, we can solve for \$\text{V}_x\$ and \$\text{V}_3\$:

  • $$\text{V}_x=\frac{\text{V}_\text{p}\text{R}_2}{\text{R}_1+\text{R}_2}\tag5$$
  • $$\text{V}_3=\frac{\text{V}_\text{p}\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)}\tag6$$

So, the transfer function is:

$$\mathcal{H}:=\frac{\text{V}_3}{\text{V}_\text{p}}=\frac{1}{\text{V}_\text{p}}\cdot\frac{\text{V}_\text{p}\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)}=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)}\tag7$$


Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

$$\text{R}_1=\frac{1}{\text{sC}}\tag8$$

So, the transfer function will be:

$$\mathcal{H}\left(\text{s}\right)=\frac{\text{v}_3\left(\text{s}\right)}{\text{v}_\text{p}\left(\text{s}\right)}=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\frac{1}{\text{sC}}+\text{R}_2\right)}\tag9$$

So, when we use the transformation \$\text{s}=\text{j}\omega\$ (where \$\text{j}^2=-1\$), we get:

$$\underline{\mathcal{H}}\left(\text{j}\omega\right)=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right)}\tag{10}$$

The magnitude response can be found by solving for the absolute value of equation \$(10)\$:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\left|\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right)}\right|=\frac{\left|\text{R}_2\left(\text{R}_3+\text{R}_4\right)\right|}{\left|\text{R}_3\left(\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right)\right|}=$$ $$\frac{\left|\text{R}_2\left(\text{R}_3+\text{R}_4\right)\right|}{\left|\text{R}_3\right|\cdot\left|\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right|}=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\sqrt{\text{R}_2^2+\left(\frac{1}{\omega\text{C}}\right)^2}}\tag{11}$$

The magnitude response will be at a maximum when \$\omega\to\infty\$, at that point the response is:

$$\lim_{\omega\to\infty}\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\sqrt{\text{R}_2^2+\left(0\right)^2}}=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\text{R}_2}=\frac{\text{R}_3+\text{R}_4}{\text{R}_3}=1+\frac{\text{R}_4}{\text{R}_3}\tag{12}$$

Using the log scale we look at the following transfer function:

$$20\log_{10}\left(\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|\right)\tag{13}$$

Solving for 3-dB lower gives:

$$20\log_{10}\left(\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\sqrt{\text{R}_2^2+\left(\frac{1}{\omega\text{C}}\right)^2}}\right)=20\log_{10}\left(\frac{1}{\sqrt{2}}\cdot\left(1+\frac{\text{R}_4}{\text{R}_3}\right)\right)\space\Longleftrightarrow\space$$ $$\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\sqrt{\text{R}_2^2+\left(\frac{1}{\omega\text{C}}\right)^2}}=\frac{1}{\sqrt{2}}\cdot\left(1+\frac{\text{R}_4}{\text{R}_3}\right)\space\Longrightarrow\space\omega=\frac{1}{\text{CR}_2}\tag{14}$$

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Brandon, the transfer function of the whole circuit is nothing else than the product of the passive C-R2 block multiplied with the gain of the non-inverting opamp stage. Just basics!

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