For the network of Fig. 17.64 , if VBB = 12 V, R = 20 kΩ, C = 1 mF, RK = 100 Ω, RB1 a. V P . b. R max and R min . c. T and frequency of oscillation. d. The waveforms of vA, vG, and v K . = 10 kΩ, RB2 = 5 kΩ, IP = 100 µA, VV = 1 V, and IV = 5.5 mA, determine: a. VP . b. Rmax and Rmin . c. T and

Question

For the network of Fig. 17.64 , if [latex]V_{B B}=12 V , R=20 k \Omega, C=1 \mu F, R_{K}=100 \Omega, R_{B_{1}}=10 k \Omega, R_{B_{2}}=5 k \Omega, I_{P}=100 \mu A , V_{V}=1 V , ext { and } I_{V}=5.5 mA ,[/latex] determine:

a. [latex]V_P [/latex] .
b. [latex]R_{max} [/latex] and [latex]R_{min} [/latex] .
c. T and frequency of oscillation.
d. The waveforms of [latex]v_{A}, v_{G}[/latex], and [latex]v_{K}[/latex] .

Accepted Answer

a. Eq. (17.20): [latex]V_{P}=\eta V_{B B}+V_{D}[/latex]

[latex]=\frac{R_{B_{1}}}{R_{B_{1}}+R_{B_{2}}} V_{B B}+0.7 V[/latex]

[latex]=\frac{10 k \Omega}{10 k \Omega+5 k \Omega}(12 V )+0.7 V[/latex]

=(0.67)(12 V )+0.7 V = 8 . 7 V

b. From Eq. (17.25): [latex]R_{\max }=\frac{V_{B B}-V_{P}}{I_{P}}[/latex]

[latex]=\frac{12 V -8.7 V }{100 \mu A }= 3 3 k \Omega[/latex]

From Eq. (17.26): [latex]R_{\min }=\frac{V_{B B}-V_{V}}{I_{V}}[/latex]

[latex]=\frac{12 V -1 V }{5.5 mA }= 2 k \Omega[/latex]

R: 2 k Ω < 20 k Ω < 33 k Ω

c. Eq. (17.23): [latex]T=R C \log _{e} \frac{V_{B B}}{V_{B B}-V_{P}}[/latex]

[latex]=(20 k \Omega)(1 \mu F ) \log _{e} \frac{12 V }{12 V -8.7 V }[/latex]

[latex]=20 imes 10^{-3} \log _{e}(3.64)[/latex]

[latex]=20 imes 10^{-3}(1.29)[/latex]

= 25.8 ms

[latex]f=\frac{1}{T}=\frac{1}{25.8 ms }= 3 8 . 8 ~ H z[/latex]

d. Indicated in Fig. 17.67 .

Question 17.3: For the network of Fig. 17.64 , if VBB = 12 V, R = 20 kΩ, C ...

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