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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:

schematic

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\$.

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  • \$\begingroup\$ Why did you equate I(R) with I(Cmax)? I think I(Cmax) at resonance and 2.5V implies an Xc and therefore a C (therefore an L), and then you can arrive at R from the given Q. \$\endgroup\$
    – user16324
    May 17, 2020 at 19:04
  • \$\begingroup\$ I'm thinking that as frequency tends to infinity \$X_L \rightarrow \infty, X_C \rightarrow 0\$ so \$Z \rightarrow R\$, and the circuit becomes equivalent to R and C in series, which implies that they have the same current, \$V_{in}/{R}\$. No? \$\endgroup\$
    – kikazaru
    May 17, 2020 at 19:54
  • \$\begingroup\$ Try this calculator for checking results. I get R to be 5000 ohms if L and C are as per the 1st answer: stades.co.uk/RLC%20filters/RCL%20BPF/RCL%20BPF.html \$\endgroup\$
    – Andy aka
    May 17, 2020 at 20:49

1 Answer 1

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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:

$$\text{Q}\cdot\text{BW}=\frac{1}{2\pi\sqrt{\text{CL}}}\space\Longleftrightarrow\space\text{CL}=\frac{1}{\left(2\pi\text{Q}\cdot\text{BW}\right)^2}\tag3$$

This implies that:

$$\text{CL}=\frac{1}{\left(2\pi\text{Q}\cdot\text{BW}\right)^2}=\frac{1}{\left(2\pi\cdot40\cdot500\right)^2}=\frac{1}{1600000000\pi^2}\space\left[\text{Second}^2\right]$$

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:

$$\hat{\text{V}}_\text{C}=\lim_{\omega\to\frac{1}{\sqrt{\text{CL}}}}\text{V}_\text{C}=\hat{\text{V}}_\text{in}\tag6$$

This implies that:

$$\hat{\text{V}}_\text{in}=\frac{5}{2}=2.5\space\left[\text{Volt}\right]$$

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:

$$\hat{\text{I}}_\text{C}=\lim_{\omega\to\frac{\text{R}\sqrt{2}}{\sqrt{\text{L}\left(2\text{CR}^2-\text{L}\right)}}}\text{I}_\text{C}=\frac{2\text{CR}\hat{\text{V}}_\text{in}}{\sqrt{\text{L}\left(4\text{CR}^2-\text{L}\right)}}\tag9$$

So, now we can solve this problem and we get:

  • For \$\text{R}>125\$: $$\text{L}=\frac{\text{R}}{160\sqrt{2}\pi\sqrt{\text{R}\left(\sqrt{\text{R}^2-15625}+\text{R}\right)}}$$
  • For \$\text{R}>125\$: $$\text{C}=\frac{\sqrt{\frac{\sqrt{\text{R}^2-15625}+\text{R}}{\text{R}}}}{5000000\sqrt{2}\pi}$$
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  • \$\begingroup\$ Thanks! Step (5) is the key for me. But I thought \$I_{C_{max}}\$ would occur as the frequency tends to infinity and \$Z \rightarrow R\$. Is this incorrect? \$\endgroup\$
    – kikazaru
    May 17, 2020 at 20:01
  • \$\begingroup\$ @kikazaru notice that when \$\omega\to\infty\$ that implies that \$\hat{\text{I}}_\text{C}\to\infty\$ which is practially impossible. \$\endgroup\$ May 17, 2020 at 21:36
  • \$\begingroup\$ Doesn't \$R\$ limit the current that can flow through C? \$\endgroup\$
    – kikazaru
    May 18, 2020 at 12:01
  • \$\begingroup\$ @kikazaru Look at my edit. \$\endgroup\$ May 19, 2020 at 9:31

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