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for the series circuit with 1 βR , C , and L respectively, we obtain for the parallel circuit
Ο 1 = β 1 2 R C + ( 1 2 R C ) 2 + 1 L C \omega_1 = -\frac{1}{2RC} + \sqrt{\left(\frac{1}{2RC}\right)^2 + \frac{1}{LC}} Ο 1 β = β 2 R C 1 β + ( 2 R C 1 β ) 2 + L C 1 β β \n
Ο 2 = 1 2 R C + ( 1 2 R C ) 2 + 1 L C \omega_2 = \frac{1}{2RC} + \sqrt{\left(\frac{1}{2RC}\right)^2 + \frac{1}{LC}} Ο 2 β = 2 R C 1 β + ( 2 R C 1 β ) 2 + L C 1 β β \n(14.45)
B = Ο 2 β Ο 1 = 1 R C B = \omega_2 - \omega_1 = \frac{1}{RC} B = Ο 2 β β Ο 1 β = R C 1 β (14.46)
Q = Ο 0 B = Ο 0 R C = R Ο 0 L Q = \frac{\omega_0}{B} = \omega_0 RC = \frac{R}{\omega_0 L} Q = B Ο 0 β β = Ο 0 β R C = Ο 0 β L R β (14.47)
It should be noted that Eqs. (14.45) to (14.47) apply only to a parallel RLC circuit. Using Eqs. (14.45) and (14.47), we can e xpress the halfpower frequencies in terms of the quality factor. The result is
Ο 1 = Ο 0 1 + ( 1 2 Q ) 2 β Ο 0 2 Q , Ο 2 = Ο 0 1 + ( 1 2 Q ) 2 + Ο 0 2 Q \omega_1 = \omega_0 \sqrt{1 + \left(\frac{1}{2Q}\right)^2} - \frac{\omega_0}{2Q}, \qquad \omega_2 = \omega_0 \sqrt{1 + \left(\frac{1}{2Q}\right)^2} + \frac{\omega_0}{2Q} Ο 1 β = Ο 0 β 1 + ( 2 Q 1 β ) 2 β β 2 Q Ο 0 β β , Ο 2 β = Ο 0 β 1 + ( 2 Q 1 β ) 2 β + 2 Q Ο 0 β β \n(14.48)
Again, for high-Q circuits (Q β₯ 10)
Ο 1 β Ο 0 β B 2 , Ο 2 β Ο 0 + B 2 \omega_1 \simeq \omega_0 - \frac{B}{2}, \qquad \omega_2 \simeq \omega_0 + \frac{B}{2} Ο 1 β β Ο 0 β β 2 B β , Ο 2 β β Ο 0 β + 2 B β (14.49)
Table 14.4 presents a summary of the characteristics of the series and parallel resonant circuits. Besides the series and parallel RLC considered here, other resonant circuits exist. Example 14.9 treats a typical example.
TABLE 14.4 Summary of the characteristics of resonant RLC circuits.
Characteristic Series circuit Parallel circuit Resonant frequency, Ο0 ____ 1 ____ 1 Quality factor, Q Ο0L ____ ____ R Bandwidth, B Ο0 Ο0 Half-power frequencies, Ο1, Ο2 Ο0 Ο0 For Q β₯ 10, Ο1, Ο2 B B
For Example 14.8.
Example 14.8 In the parallel RLC circuit of Fig. 14.27, let R = 8 kΞ©, L = 0.2 mH, and C = 8 ΞΌ F. (a) Calculate Ο 0, Q , and B . (b) Find Ο 1 and Ο 2. (c) Determine the power dissipated at Ο 0, Ο 1, and Ο 2.
Solution: (a)
Ο 0 = 1 L C = 1 0.2 Γ 10 β 3 Γ 8 Γ 10 β 6 = 10 5 4 = 25 Β krad/s \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.2 \times 10^{-3} \times 8 \times 10^{-6}}} = \frac{10^5}{4} = 25 \text{ krad/s} Ο 0 β = L C β 1 β = 0.2 Γ 1 0 β 3 Γ 8 Γ 1 0 β 6 β 1 β = 4 1 0 5 β = 25 Β krad/s Q = R Ο 0 L = 8 Γ 10 3 25 Γ 10 3 Γ 0.2 Γ 10 β 3 = 1 , 600 Q = \frac{R}{\omega_0 L} = \frac{8 \times 10^3}{25 \times 10^3 \times 0.2 \times 10^{-3}} = 1,600 Q = Ο 0 β L R β = 25 Γ 1 0 3 Γ 0.2 Γ 1 0 β 3 8 Γ 1 0 3 β = 1 , 600 B = Ο 0 Q = 15.625 Β rad/s B = \frac{\omega_0}{Q} = 15.625 \text{ rad/s} B = Q Ο 0 β β = 15.625 Β rad/s (b) Due to the high value of Q , we can regard this as a high- Q circuit, Hence,
Ο 1 = Ο 0 β B 2 = 25 , 000 β 7.812 = 24 , 992 Β rad/s \omega_1 = \omega_0 - \frac{B}{2} = 25,000 - 7.812 = 24,992 \text{ rad/s} Ο 1 β = Ο 0 β β 2 B β = 25 , 000 β 7.812 = 24 , 992 Β rad/s Ο 2 = Ο 0 + B 2 = 25 , 000 + 7.812 = 25 , 008 Β rad/s \omega_2 = \omega_0 + \frac{B}{2} = 25,000 + 7.812 = 25,008 \text{ rad/s} Ο 2 β = Ο 0 β + 2 B β = 25 , 000 + 7.812 = 25 , 008 Β rad/s (c) At Ο = Ο 0, Y = 1βR or Z = R = 8 kΞ©. Then
I o = V Z = 10 / β 90 β 8 , 000 = 1.25 / β 90 β Β mA I_o = \frac{V}{Z} = \frac{10/-90^{\circ}}{8,000} = 1.25/-90^{\circ} \text{ mA} I o β = Z V β = 8 , 000 10/ β 9 0 β β = 1.25/ β 9 0 β Β mA Since the entire current flows through R at resonance, the average power dissipated at Ο = Ο 0 is
P = 1 2 β£ I o β£ 2 R = 1 2 ( 1.25 Γ 10 β 3 ) 2 ( 8 Γ 10 3 ) = 6.25 Β mW P = \frac{1}{2} |\mathbf{I}_o|^2 R = \frac{1}{2} (1.25 \times 10^{-3})^2 (8 \times 10^3) = 6.25 \text{ mW} P = 2 1 β β£ I o β β£ 2 R = 2 1 β ( 1.25 Γ 1 0 β 3 ) 2 ( 8 Γ 1 0 3 ) = 6.25 Β mW or
P = V m 2 2 R = 100 2 Γ 8 Γ 10 3 = 6.25 P = \frac{V_m^2}{2R} = \frac{100}{2 \times 8 \times 10^3} = 6.25 P = 2 R V m 2 β β = 2 Γ 8 Γ 1 0 3 100 β = 6.25 mW
At Ο = Ο 1, Ο 2,
P = V m 2 4 R = 3.125 Β mW P = \frac{V_m^2}{4R} = 3.125 \text{ mW} P = 4 R V m 2 β β = 3.125 Β mW Practice Problem 14.8 A parallel resonant circuit has R = 100 kΞ©, L = 50 mH, and C = 2 nF. Calculate Ο 0, Ο 1, Ο 2, Q , and B .
Answer: 100 krad/s, 97.5 krad/s,102.5 krad/s, 20, 5 krad/s.
Solution: The input admittance is
Y = j Ο 0.1 + 1 10 + 1 2 + j Ο 2 = 0.1 + j Ο 0.1 + 2 β j Ο 2 4 + 4 Ο 2 \mathbf{Y} = j\omega 0.1 + \frac{1}{10} + \frac{1}{2 + j\omega 2} = 0.1 + j\omega 0.1 + \frac{2 - j\omega 2}{4 + 4\omega^2} Y = j Ο 0.1 + 10 1 β + 2 + j Ο 2 1 β = 0.1 + j Ο 0.1 + 4 + 4 Ο 2 2 β j Ο 2 β At resonance, Im(Y ) = 0 and
Ο 0 0.1 β 2 Ο 0 4 + 4 Ο 0 2 = 0 β Ο 0 = 2 Β rad/s \omega_0 0.1 - \frac{2\omega_0}{4 + 4\omega_0^2} = 0 \qquad \Rightarrow \qquad \omega_0 = 2 \text{ rad/s} Ο 0 β 0.1 β 4 + 4 Ο 0 2 β 2 Ο 0 β β = 0 β Ο 0 β = 2 Β rad/s Calculate the resonant frequency of the circuit in Fig. 14.29. Practice Problem 14.9
Answer: 173.21 rad/s.
