Is there a single answer to the following question?
Design a band-pass filter using a parallel resonant circuit to meet the following specifications: \$BW=500Hz, Q=40, I_{C_{max}}=20mA, V_{C_{max}}=2.5V\$.
Full disclosure: this is question 35 on page 655 of Pearson's Electronics Fundamentals 8th Ed.
I posit the following circuit:
simulate this circuit – Schematic created using CircuitLab
I deduce that: \$f_r = 20kHz\$ because \$f_r = Q . BW\$, \$V_{in} = 2.5V\$ since at resonance the \$LC\$ tank will have infinite impedance and therefore all the input voltage, and therefore \$R = \frac{V_{in}}{20mA} = 125 \Omega\$.
Next I conclude that \$LC = \frac{1}{(2 \pi f_r)^2}\$ from the standard formula \$f_r = \frac{1}{2\pi\sqrt{LC}}\$.
The book gives \$L = 989\mu H, C = 0.064\mu F\$, which satisfies this, but I don't see how they got specific values of \$L\$ and \$C\$.
Best Answer
Well, first of all we know that:
$$\text{Q}=\frac{\text{f}_\text{r}}{\Delta\text{f}_\text{r}}=\frac{\text{f}_\text{r}}{\text{BW}}\space\Longleftrightarrow\space\text{f}_\text{r}=\text{Q}\cdot\text{BW}\tag1$$
Now, the resonance frequency is given by:
$$2\pi\text{f}_\text{r}=\frac{1}{\sqrt{\text{CL}}}\space\Longleftrightarrow\space\text{f}_\text{r}=\frac{1}{2\pi\sqrt{\text{CL}}}\tag2$$
So:
$$\text{Q}\cdot\text{BW}=\frac{1}{2\pi\sqrt{\text{CL}}}\space\Longleftrightarrow\space\text{CL}=\frac{1}{\left(2\pi\text{Q}\cdot\text{BW}\right)^2}\tag3$$
Now, the voltage across the capacitor is given by:
$$\text{V}_\text{C}=\left|\frac{\text{j}\omega\text{L}}{\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}\cdot\frac{\hat{\text{V}}_\text{in}\exp\left(\varphi\text{j}\right)}{\text{R}+\frac{\frac{\text{j}\omega\text{L}}{\text{j}\omega\text{C}}}{\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}}\cdot\frac{1}{\text{j}\omega\text{C}}\right|=\frac{\hat{\text{V}}_\text{in}\text{L}\omega}{\sqrt{\left(\text{L}\omega\right)^2+\left(\text{R}\left(\text{CL}\omega^2-1\right)\right)^2}}\tag4$$
The maximum occurs when:
$$\omega=\frac{1}{\sqrt{\text{CL}}}\tag5$$
So:
$$\hat{\text{V}}_\text{C}=\lim_{\omega\to\frac{1}{\sqrt{\text{CL}}}}\text{V}_\text{C}=\hat{\text{V}}_\text{in}\tag6$$
The current through the capacitor is given by:
$$\text{I}_\text{C}=\left|\frac{\text{j}\omega\text{L}}{\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}\cdot\frac{\hat{\text{V}}_\text{in}\exp\left(\varphi\text{j}\right)}{\text{R}+\frac{\frac{\text{j}\omega\text{L}}{\text{j}\omega\text{C}}}{\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}}\right|=\frac{\hat{\text{V}}_\text{in}\text{CL}\omega^2}{\sqrt{\left(\text{L}\omega\right)^2+\left(\text{R}\left(\text{CL}\omega^2-1\right)\right)^2}}\tag7$$
The maximum occurs when:
$$\omega=\frac{\text{R}\sqrt{2}}{\sqrt{\text{L}\left(2\text{CR}^2-\text{L}\right)}}\tag8$$
With the condition that \$\text{L}<2\text{CR}^2\$.
So:
$$\hat{\text{I}}_\text{C}=\lim_{\omega\to\frac{\text{R}\sqrt{2}}{\sqrt{\text{L}\left(2\text{CR}^2-\text{L}\right)}}}\text{I}_\text{C}=\frac{2\text{CR}\hat{\text{V}}_\text{in}}{\sqrt{\text{L}\left(4\text{CR}^2-\text{L}\right)}}\tag9$$