For the parallel resonant circuit in Fig. 21.35 with $R_s = \infty Ω$ :

a. Determine $f_s, f_m, and f_p$ , and compare their levels.
b. Calculate the maximum impedance and the magnitude of the voltage $V_C at f_m$ .
c. Determine the quality factor $Q_p$ .
d. Calculate the bandwidth.
e. Compare the above results with those obtained using the equations associated with $Q_l \geq 10$ .

Question

For the parallel resonant circuit in Fig. 21.35 with [latex]R_s = ∞ Ω[/latex]:

a. Determine [latex]f_s, f_m, and f_p[/latex], and compare their levels.
b. Calculate the maximum impedance and the magnitude of the voltage [latex]V_C at f_m[/latex].
c. Determine the quality factor [latex]Q_p[/latex].
d. Calculate the bandwidth.
e. Compare the above results with those obtained using the equations associated with [latex]Q_l ≥ 10[/latex].

Accepted Answer

a. [latex]f_s=\frac{1}{2\pi\sqrt{LC} }=\frac{1}{2\pi\sqrt{(0.3 mH)(100 nF)} } =29.06 kHz\[0.5cm] f_m=f_s\sqrt{1-\frac{1}{4 }\left[\frac{R^{2}_{l}C}{L} \right] }\[0.5cm] =(29.06 kHz) \sqrt{1-\frac{1}{4}\left[\frac{(20\Omega)^2(100nF)}{0.3mH} \right] }=25.58 kHz\[0.5cm] f_p=f_s\sqrt{1-\frac{R^{2}_{l}C}{L} }=(29.06 kHz)\sqrt{1-\left[\frac{(20\Omega)^2(100nF)}{0.3mH} \right] }=27.06 kHz[/latex]

Both [latex]f_m and f_p[/latex] are less than [latex]f_s[/latex], as predicted. In addition, [latex]f_m[/latex] is closer to [latex]f_s than f_p[/latex], as forecast. [latex]f_m[/latex] is about 0.5 kHz less than [latex]f_s, whereas f_p[/latex] is about 2 kHz less. The differences among [latex]f_s, f_m, and f_p[/latex] suggest a low Q network.

b. [latex]Z_{T_m}=(R_l+jX_L)\parallel -jX_C\quad at\quad f=f_m\[0.5cm] X_L=2\pi f_ml=2\pi (28.58 kHz)(0.3 mH) = 53.87 Ω\[0.5cm] X_C=\frac{1}{2\pi f_mC}=\frac{1}{2\pi(28.58 kHz)(100 nF)}=55.69 Ω\[0.5cm] R_l+jX_L= 20 Ω + j 53.87 Ω = 57.46 Ω ∠69.63°\[0.5cm] Z_{T_m}=\frac{(57.46 Ω ∠69.63°)(55.69 Ω ∠-90°)}{20 Ω + j 53.87 Ω - j 55.69 Ω}=159.34 Ω \angle −15.17° \[0.5cm] V_{C_{max}}=IZ_{T_m}=(2 mA)(159.34 Ω) = 318.68 mV [/latex]

c. [latex]R_s=∞ Ω;\quad therefore,\[0.5cm] Q_p=\frac{R_s\parallel R_p}{X_{L_p}}=\frac{R_p}{X_{L_p}}=Q_l=\frac{X_L}{R_l}\[0.5cm] =\frac{2\pi(27.06 kHz)(0.3 mH)}{20 Ω}=\frac{51 Ω}{20 Ω}=2.55 [/latex]

The low Q confirms the conclusion of part (a). The differences among [latex]f_s, f_m, and f_p[/latex] are significantly less for higher Q networks.

d. [latex]BW=\frac{f_p}{Q_p}=\frac{27.06 kHz}{2.55}=10.61 kHz [/latex]

e. [latex]For Q_l \geq 10, f_m = f_p = f_s = 29.06 kHz\[0.5cm] Q_p=Q_l=\frac{2\pi f_sL}{R_l}=\frac{2\pi(29.06 kHz)(0.3 mH)}{20 Ω}=2.74\hspace{30 pt} (versus\enspace 2.55\enspace above) \[0.5cm] Z_{T_p}=Q^{2}_{l}R_{l}=(2.74)^2\cdot 20 Ω = 150.15 Ω \angle 0°\qquad (versus\; 159.34 Ω ∠-15.17° above)\[0.5cm] V_{C_{max}}=IZ_{T_p}=(2 mA)(150.15 Ω) = 300.3 mV\qquad (versus\; 318.68 mV\enspace above)\[0.5cm] BW=\frac{f_p}{Q_p}=\frac{29.06 kHz}{2.74}=10.61 kHz\qquad (versus\; 10.61 kHz \; above) [/latex]

The results reveal that, even for a relatively low Q system, the approximate solutions are still close compared to those obtained using the full equations. The primary difference is between [latex]f_s and f_p[/latex] (about 7%), with the difference between [latex]f_s and f_m[/latex] at less than 2%. For the future, using [latex]f_s[/latex] to determine [latex]Q_l[/latex] will certainly provide a measure of [latex]Q_l[/latex] that can be used to determine whether the approximate approach is appropriate.

Question 21.7: For the parallel resonant circuit in Fig. 21.35 with Rs = ∞ ...

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