For the network of Fig. 5.48. determine:

a. $r_e$ .
b. $Z_i$ .
c. $Z_o$ .
d. $A_{v}$ .
e. Repeat parts (b) through (d) with $r_o = 20 kΩ$ and compare results.

Question

For the network of Fig. 5.48. determine:

a. [latex]r_e [/latex] .
b. [latex]Z_i [/latex] .
c. [latex]Z_o [/latex] .
d. [latex]A_{v}[/latex] .
e. Repeat parts (b) through (d) with [latex]r_o = 20 kΩ [/latex] and compare results.

Accepted Answer

a. [latex]I_{B}=\frac{V_{C C}-V_{B E}}{R_{F}+\beta R_{C}}=\frac{9 V -0.7 V }{180 k \Omega+(200) 2.7 k \Omega}[/latex]

= 11.53 µA

[latex]I_{E}=(\beta+1) I_{B}=(201)(11.53 \mu A )=2.32 mA[/latex]

[latex]r_{e}=\frac{26 mV }{I_{E}}=\frac{26 mV }{2.32 mA }= 1 1 . 2 1 \Omega[/latex]

b. [latex]Z_{i}=\frac{r_{e}}{\frac{1}{\beta}+\frac{R_{C}}{R_{C}+R_{F}}}=\frac{11.21 \Omega}{\frac{1}{200}+\frac{2.7 k \Omega}{182.7 k \Omega}}=\frac{11.21 \Omega}{0.005+0.0148}[/latex]

[latex]=\frac{11.21 \Omega}{0.0198}= 5 6 6 . 1 6 \Omega[/latex]

c. [latex]Z_{o}=R_{C}\left\|R_{F}=2.7 k \Omega ight\| 180 k \Omega=2.66 k \Omega[/latex]

d. [latex]A_{v}=-\frac{R_{C}}{r_{e}}=-\frac{2.7 k \Omega}{11.21 \Omega}=- 2 4 0 . 8 6[/latex]

e. [latex]Z_i[/latex]: The condition [latex]r_{o} \geq 10 R_{C}[/latex] is not satisfied. Therefore,

[latex]Z_{i}=\frac{1+\frac{R_{C} \| r_{o}}{R_{F}}}{\frac{1}{\beta r_{e}}+\frac{1}{R_{F}}+\frac{R_{C} \| r_{o}}{\beta r_{e} R_{F}}+\frac{R_{C} \| r_{o}}{R_{F} r_{e}}}=\frac{1+\frac{2.7 k \Omega \| 20 k \Omega}{180 k \Omega}}{\frac{1}{(200)(11.21)}+\frac{1}{180 k \Omega}+\frac{2.7 k \Omega \| 20 k \Omega}{(200)(11.21 \Omega)(180 k \Omega)}+\frac{2.7 k \Omega \| 20 k \Omega}{(180 k \Omega)(11.21 \Omega)}}[/latex]

[latex]=\frac{1+\frac{2.38 k \Omega}{180 k \Omega}}{0.45 imes 10^{-3}+0.006 imes 10^{-3}+5.91 imes 10^{-6}+1.18 imes 10^{-3}}=\frac{1+0.013}{1.64 imes 10^{-3}}[/latex]

= 617.7 Ω vs. 566.16 Ω above

[latex]Z_{0}:[/latex]

[latex]Z_{o}=r_{o}\left\|R_{C} ight\| R_{F}=20 k \Omega\|2.7 k \Omega\| 180 k \Omega[/latex]

= 2.35 k Ω vs. 2.66 kΩ above

[latex]A _{ v }:[/latex]

[latex]=-\left(\frac{R_{F}}{R_{C} \| r_{o}+R_{F}} ight) \frac{R_{C} \| r_{o}}{r_{e}}=-\left[\frac{180 k \Omega}{2.38 k \Omega+180 k \Omega} ight] \frac{2.38 k \Omega}{11.21}[/latex]

= - [0.987] 212.3

= - 209.54

Question 5.9: For the network of Fig. 5.48. determine: a. r e . b. Z i . c...

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