The equivalent network for the transistor configuration in Fig. 21.37 is provided in Fig. 21.38.
a. Find $f_p$ .
b. Determine $Q_p$ .
c. Calculate the BW.
d. Determine $V_p$ at resonance.
e. Sketch the curve of $V_C$ versus frequency.

Question

The equivalent network for the transistor configuration in Fig. 21.37 is provided in Fig. 21.38.
a. Find [latex]f_p[/latex].
b. Determine [latex]Q_p[/latex].
c. Calculate the BW.
d. Determine [latex]V_p[/latex] at resonance.
e. Sketch the curve of [latex]V_C[/latex] versus frequency.

Accepted Answer

a. [latex]f_s=\frac{1}{2\pi\sqrt{LC} }=\frac{1}{2\pi\sqrt{(5mH)(50pF)} }=318.31 kHz\[0.5cm] X_L=2\pi f_sL=2\pi(318.31 kHz)(5 mH) = 10 kΩ\[0.5cm] Q_l=\frac{X_L}{R_l}=\frac{10 kΩ}{100 kΩ}=100\gt 10\[0.5cm] Therefore,\enspace f_p = f_s = 318.31 kHz.\enspace Using\enspace Eq. (21.32) \enspace results\enspace in ≅ 318.5 kHz.\[0.5cm] f_p=f_s\sqrt{1-\frac{R^2_lC}{L} }\hspace{30 pt} (21.32)[/latex]

b. [latex]Q_p=\frac{R_s\parallel R_p}{X_L} \[0.5cm] R_p=Q^2_lR_l=(100)^2100 Ω=1 MΩ\[0.5cm] Q_p=\frac{50 kΩ\parallel 1 MΩ}{10 kΩ}=\frac{47.62 kΩ}{10 kΩ}=4.76 [/latex]

Note the drop in Q from [latex]Q_l = 100 to Q_p = 4.76 due to R_s[/latex].

c. [latex]BW=\frac{f_p}{Q_p}=\frac{318.31 kHz}{4.76}=66.87 kHz [/latex]

On the other hand,

[latex]BW=\frac{1}{2\pi}\left(\frac{R_l}{L}+\frac{1}{R_sC} \right)=\frac{1}{2\pi}\left[\frac{100 Ω}{5mH}+\frac{1}{(50 kΩ)(50 pF)} \right] =66.85 kHz [/latex]

compares very favorably with the above solution.

d. [latex]V_p=IZ_{T_p}=(2 mA)(Rs\parallel R_p)=(2 mA)(47.62 kΩ) = 95.24 V[/latex]

e. See Fig. 21.39.

Question 21.9: The equivalent network for the transistor configuration in F...

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