Active band-pass filter design questions

Question

I am trying to design an active bandpass filter using a uA741 op-amp, and I am having trouble finding where I made a mistake. My math seems to check out, but when I test it in PSpice it doesn't work properly. Any advice would be greatly appreciated.

Answer 1

Well, let's analyze the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

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

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\text{I}_2+\text{I}_4+\text{I}_5\\ \\ 0&=\text{I}_3+\text{I}_4+\text{I}_5 \end{alignat*} \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_\text{i}-\text{V}}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_2}\\ \\ \text{I}_4&=\frac{\displaystyle\text{V}_--\text{V}_\text{o}}{\displaystyle\text{R}_3}\\ \\ \text{I}_4&=\frac{\displaystyle\text{V}-\text{V}_-}{\displaystyle\text{R}_4}\\ \\ \text{I}_5&=\frac{\displaystyle\text{V}-\text{V}_\text{o}}{\displaystyle\text{R}_5} \end{alignat*} \end{cases}\tag2 $$

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

$$ \begin{cases} \begin{alignat*}{1} \frac{\displaystyle\text{V}_\text{i}-\text{V}}{\displaystyle\text{R}_1}&=\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_2}+\frac{\displaystyle\text{V}_--\text{V}_\text{o}}{\displaystyle\text{R}_3}+\frac{\displaystyle\text{V}-\text{V}_\text{o}}{\displaystyle\text{R}_5}\\ \\ 0&=\text{I}_3+\frac{\displaystyle\text{V}_--\text{V}_\text{o}}{\displaystyle\text{R}_3}+\frac{\displaystyle\text{V}-\text{V}_\text{o}}{\displaystyle\text{R}_5}\\ \\ \frac{\displaystyle\text{V}_\text{i}-\text{V}}{\displaystyle\text{R}_1}&=\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_2}+\frac{\displaystyle\text{V}-\text{V}_-}{\displaystyle\text{R}_4}+\frac{\displaystyle\text{V}-\text{V}_\text{o}}{\displaystyle\text{R}_5}\\ \\ 0&=\text{I}_3+\frac{\displaystyle\text{V}-\text{V}_-}{\displaystyle\text{R}_4}+\frac{\displaystyle\text{V}-\text{V}_\text{o}}{\displaystyle\text{R}_5} \end{alignat*} \end{cases}\tag3 $$

When working with an ideal opamp we know that:

$$\text{V}_+=\text{V}_-\tag4$$

In this case we notice that \$\text{V}_+=0\space\text{V}\$, so we can rewrite \$(3)\$:

$$ \begin{cases} \begin{alignat*}{1} \frac{\displaystyle\text{V}_\text{i}-\text{V}}{\displaystyle\text{R}_1}&=\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_2}+\frac{\displaystyle0-\text{V}_\text{o}}{\displaystyle\text{R}_3}+\frac{\displaystyle\text{V}-\text{V}_\text{o}}{\displaystyle\text{R}_5}\\ \\ 0&=\text{I}_3+\frac{\displaystyle0-\text{V}_\text{o}}{\displaystyle\text{R}_3}+\frac{\displaystyle\text{V}-\text{V}_\text{o}}{\displaystyle\text{R}_5}\\ \\ \frac{\displaystyle\text{V}_\text{i}-\text{V}}{\displaystyle\text{R}_1}&=\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_2}+\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_4}+\frac{\displaystyle\text{V}-\text{V}_\text{o}}{\displaystyle\text{R}_5}\\ \\ 0&=\text{I}_3+\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_4}+\frac{\displaystyle\text{V}-\text{V}_\text{o}}{\displaystyle\text{R}_5} \end{alignat*} \end{cases}\tag5 $$

And solve for the transfer function:

$$\mathscr{H}:=\frac{\displaystyle\text{V}_\text{o}}{\displaystyle\text{V}_\text{i}}=-\frac{\displaystyle\text{R}_2\text{R}_3\text{R}_5}{\displaystyle\text{R}_1\text{R}_2\left(\text{R}_3+\text{R}_4+\text{R}_5\right)+\text{R}_4\text{R}_5\left(\text{R}_1+\text{R}_2\right)}\tag6$$

Now, we can transform the transfer function using Laplace transform. This gives \$\displaystyle\text{R}_4=\frac{\displaystyle1}{\displaystyle\text{sC}_1}\$ and \$\displaystyle\text{R}_5=\frac{\displaystyle1}{\displaystyle\text{sC}_2}\$. So, we get:

$$ \begin{alignat*}{1} \mathscr{H}\left(\text{s}\right)&=-\frac{\displaystyle\text{R}_2\text{R}_3\cdot\frac{\displaystyle1}{\displaystyle\text{sC}_2}}{\displaystyle\text{R}_1\text{R}_2\left(\text{R}_3+\frac{\displaystyle1}{\displaystyle\text{sC}_1}+\frac{\displaystyle1}{\displaystyle\text{sC}_2}\right)+\frac{\displaystyle1}{\displaystyle\text{sC}_1}\cdot\frac{\displaystyle1}{\displaystyle\text{sC}_2}\cdot\left(\text{R}_1+\text{R}_2\right)}\\ \\ &=-\frac{\displaystyle\text{R}_2\text{R}_3\cdot\frac{\displaystyle1}{\displaystyle\text{sC}_2}}{\displaystyle\text{R}_1\text{R}_2\left(\text{R}_3+\frac{\displaystyle1}{\displaystyle\text{sC}_1}+\frac{\displaystyle1}{\displaystyle\text{sC}_2}\right)+\frac{\displaystyle1}{\displaystyle\text{sC}_1}\cdot\frac{\displaystyle1}{\displaystyle\text{sC}_2}\cdot\left(\text{R}_1+\text{R}_2\right)}\\ \\ &=-\frac{\displaystyle\text{R}_2\text{R}_3\cdot\frac{\displaystyle\text{sC}_2}{\displaystyle\text{sC}_2}}{\displaystyle\text{R}_1\text{R}_2\left(\text{sC}_2\text{R}_3+\frac{\displaystyle\text{sC}_2}{\displaystyle\text{sC}_1}+\frac{\displaystyle\text{sC}_2}{\displaystyle\text{sC}_2}\right)+\frac{\displaystyle1}{\displaystyle\text{sC}_1}\cdot\frac{\displaystyle\text{sC}_2}{\displaystyle\text{sC}_2}\cdot\left(\text{R}_1+\text{R}_2\right)}\\ \\ &=-\frac{\displaystyle\text{R}_2\text{R}_3}{\displaystyle\text{R}_1\text{R}_2\left(\text{sC}_2\text{R}_3+\frac{\displaystyle\text{C}_2}{\displaystyle\text{C}_1}+1\right)+\frac{\displaystyle1}{\displaystyle\text{sC}_1}\cdot\left(\text{R}_1+\text{R}_2\right)}\\ \\ &=-\frac{\displaystyle\text{sC}_1\text{R}_2\text{R}_3}{\displaystyle\text{R}_1\text{R}_2\left(\text{sC}_1\text{sC}_2\text{R}_3+\frac{\displaystyle\text{sC}_1\text{C}_2}{\displaystyle\text{C}_1}+\text{sC}_1\right)+\frac{\displaystyle\text{sC}_1}{\displaystyle\text{sC}_1}\cdot\left(\text{R}_1+\text{R}_2\right)}\\ \\ &=-\frac{\displaystyle\text{sC}_1\text{R}_2\text{R}_3}{\displaystyle\text{R}_1\text{R}_2\left(\text{sC}_1\text{sC}_2\text{R}_3+\text{sC}_2+\text{sC}_1\right)+\text{R}_1+\text{R}_2}\\ \\ &=-\frac{\displaystyle\text{C}_1\text{R}_2\text{R}_3\text{s}}{\displaystyle\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_3\text{s}^2+\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\text{s}+\text{R}_1+\text{R}_2} \end{alignat*} \tag7 $$

When working with sinusoidal inputs we know that we can write \$\text{s}:=\text{j}\omega\$ where \$\text{j}^2=-1\$. So we get:

$$ \begin{alignat*}{1} \left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|&=\displaystyle\left|-\frac{\displaystyle\text{C}_1\text{R}_2\text{R}_3\text{j}\omega}{\displaystyle\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_3\left(\text{j}\omega\right)^2+\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\text{j}\omega+\text{R}_1+\text{R}_2}\right|\\ \\ &=\displaystyle\left|-1\right|\cdot\frac{\displaystyle\left|\text{C}_1\text{R}_2\text{R}_3\omega\text{j}\right|}{\displaystyle\left|\text{R}_1+\text{R}_2-\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_3\omega^2+\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\text{j}\right|}\\ \\ &=\displaystyle1\cdot\frac{\displaystyle\text{C}_1\text{R}_2\text{R}_3\omega}{\displaystyle\left|\text{R}_1+\text{R}_2-\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_3\omega^2+\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\text{j}\right|}\\ \\ &=\displaystyle\frac{\displaystyle\text{C}_1\text{R}_2\text{R}_3\omega}{\displaystyle\sqrt{\left(\text{R}_1+\text{R}_2-\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_3\omega^2\right)^2+\left(\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\right)^2}}\\ \\ &=\displaystyle\frac{\displaystyle\text{C}_1\text{R}_2\text{R}_3\omega}{\displaystyle\sqrt{\left(\text{R}_1+\text{R}_2\left(1-\text{C}_1\text{C}_2\text{R}_1\text{R}_3\omega^2\right)\right)^2+\left(\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\right)^2}} \end{alignat*} \tag8 $$

Now, let's analyze a few cases:

1. \$\omega\space\to\space0\$: $$\lim_{\omega\space\to\space0}\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=0\tag9$$
2. \$\omega\space\to\space\infty\$: $$\lim_{\omega\space\to\space\infty}\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=0\tag{10}$$
3. When is \$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|\$ at a maximum: $$\frac{\partial\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|}{\partial\hat{\omega}}=0\space\Longrightarrow\space\hat{\omega}=\sqrt{\frac{\displaystyle\text{R}_1+\text{R}_2}{\displaystyle\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_3}}\tag{11}$$

So, we get:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|=\frac{\displaystyle\text{C}_1\text{R}_3}{\displaystyle\text{R}_1\left(\text{C}_1+\text{C}_2\right)}\tag{12}$$

So, the cut-off frequency is:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{2}}\cdot\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|\space\Longrightarrow\space$$ $$\omega_\pm=\frac{\displaystyle\sqrt{\left(\text{C}_1+\text{C}_2\right)^2+\frac{\displaystyle4\text{C}_1\text{C}_2\text{R}_3\left(\text{R}_1+\text{R}_2\right)}{\displaystyle\text{R}_1\text{R}_2}}\pm\left(\text{C}_1+\text{C}_2\right)}{\displaystyle2\text{C}_1\text{C}_2\text{R}_3}\tag{13}$$

So, the bandwidth is given by:

$$\mathcal{B}:=\left|\omega_+-\omega_-\right|=\frac{\displaystyle\text{C}_1+\text{C}_2}{\displaystyle\text{C}_1\text{C}_2\text{R}_3}\tag{14}$$

And the quality factor is given by:

$$\mathcal{Q}:=\frac{\hat{\omega}}{\mathcal{B}}=\frac{\displaystyle\sqrt{\frac{\displaystyle\text{R}_1+\text{R}_2}{\displaystyle\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_3}}}{\displaystyle\frac{\displaystyle\text{C}_1+\text{C}_2}{\displaystyle\text{C}_1\text{C}_2\text{R}_3}}=\frac{\displaystyle1}{\displaystyle\text{C}_1+\text{C}_2}\cdot\sqrt{\frac{\displaystyle\text{C}_1\text{C}_2\text{R}_3\left(\text{R}_1+\text{R}_2\right)}{\displaystyle\text{R}_1\text{R}_2}}\tag{15}$$