Finding the frequency response of a parallel RLC circuit The parallel RLC circuit in Fig. B.44 has R = 50 Ω, L = 4 mH, and C = 0.15 µF.
(a) Determine the parallel resonant frequency f_{ p } , the damping ratio δ, the cutoff frequencies, the bandwidth BW _{ p } , and the quality factor Q_{ p } of the circuit.
(b) Use PSpice/SPICE to plot the magnitude and phase angle of the output voltage for R = 50 Ω, 100 Ω, and 200 Ω. The frequency f varies from 100 Hz to 100 kHz. Assume I_{ m }=1 A peak AC.
(a) R=50 \Omega, L=4 mH , C=0.15 \mu F \text {, and } I_{ m }=1 A peak AC, so
\omega_{ n }=\frac{1}{\sqrt{L C}}=\frac{10^{5}}{\sqrt{4 \times 1.5}}=40,825 rad / s
The parallel resonant frequency is
f_{ p }=\frac{\omega_{ n }}{2 \pi}=\frac{40,825}{2 \pi}=6497.5 Hz
Since \alpha=1 /(2 R C)=1 /\left(2 \times 50 \times 0.15 \times 10^{-6}\right)=66.667 \times 10^{3}, the damping ratio is
\delta=\frac{\alpha}{\omega_{ n }}=66.667 \times \frac{10^{3}}{40,825}=1.633
For the lower cutoff frequency, Eqs. (B.85) and (B.86) give
u_{1}=-\delta+\sqrt{1+\delta^{2}} (B.85)
\omega_{1}=u_{1} \omega_{n}=\omega_{ n }\left(-\delta+\sqrt{1+\delta^{2}}\right) (B.86)
\begin{aligned}&u_{1}=-\delta+\sqrt{1+\delta^{2}}=-1.633+\sqrt{1+1.633^{2}}=0.28186 \\&\omega_{1}=u_{1} \omega_{ n }=0.28186 \times 40,825=11,507 rad / s\end{aligned}
Thus, f_{1}=11,507 / 2 \pi=1831 Hz For the upper cutoff frequency, Eqs. (B.83) and (B.84) give
u_{2}=\delta+\sqrt{1+\delta^{2}} (B.83)
\omega_{2}=u_{2} \omega_{ n }=\omega_{ n }\left(\delta+\sqrt{1+\delta^{2}}\right) (B.84)
\begin{aligned}&u_{2}=\delta+\sqrt{1+\delta^{2}}=1.633+\sqrt{1+1.633^{2}}=3.54786 \\&\omega_{2}=u_{2} \omega_{ n }=3.54786 \times 40,825=144,841 rad / s\end{aligned}
Thus, f_{2}=144,841 / 2 \pi=23,052 Hz From Eq. (B.98), the bandwidth is
BW_p =2 \frac{1}{2 R} \sqrt{\frac{L}{C}} \times \frac{1}{\sqrt{L C}}=\frac{1}{R C} \quad(\text { in } rad / s ) (B.98)
BW _{ p }=f_{2}-f_{1}=\frac{1}{R C}=\frac{1}{50 \times 0.15 \times 10^{-6}}=133,333.3 rad / s , \text { or } 21,220 Hz
Using Eq. (B.90), we can find the quality factor:
BW _{ p }=\frac{f_{ p }}{Q_{ p }} (B.90)
Q_{ p }=\frac{f_{ p }}{B W_{ p }}=\frac{6497.5}{21,220}=0.3062
(b) The parallel RLC circuit for PSpice simulation is shown in Fig. B.46. The list of the circuit file is as follows.
\begin{aligned}&\text { Example B.17 Frequency Response of a Para11e1 RLC Circuit } \\&\text {.PARAM RVAL = 50 } \\&\text {.STEP PARAM RVAL LIST } \quad 50 \quad 100 \quad 200 \\&\text { IM } \quad 0 \quad 1 \quad \text { AC } \quad 1 \text { A } \quad ; \text { ac input of } 1 \vee \text { peak } \\&\text { L } \quad 1 \quad 0 \quad 4 \text { MH } \\&\text { C } \quad 1 \quad 0 \quad 0.15 \text { UF } \\&\text { R } \quad 1 \quad 0 \quad\{\text { RVAL }\} \\&\text {.AC DEC } 100 \quad 100 \text { HZ } \quad 1 \text { MEGHZ } \\&\text {.PROBE } \\&\text {.END }\end{aligned}The PSpice plots of the magnitude and phase angle (using EXB-17.SCH) are shown in Fig. B.47. The plot for R=50 \Omega \text { gives } f_{1}=1834 Hz , f_{2}=22.56 kHz , f_{ p }=6457 Hz \text {, and } BW _{ p }=f_{2}-f_{1}=20,726 Hz .
