Refer to Figure 9$r_{O}=\infty$ Assume$\left|V_{B E}\right|=0.7 \mathrm{~V}$
(A) For 9.10.1

Question

Refer to Figure 9$r_{O}=\infty$  Assume$\left|V_{B E}\right|=0.7 \mathrm{~V}$  
(A) For 9.10.1

Examlex · Accepted Answer

The Answer is $Figure 9.10.1 (a) \begin{aligned} I_{E 1} & =I_{E 2}=0.25 \mathrm{~mA} \ I_{C 6} & =I_{E 1}+I_{E 2}=0.5 \mathrm{~mA} \ \Rightarrow I_{C 9} & =I_{C 8}=0.5 \mathrm{~mA} \ R_{1} & =\frac{30-0.7-0.7}{0.5}=57.2 \mathrm{k} \Omega \ I_{C 1} & =I_{C 2}=0.25 \mathrm{~mA} \ V_{C} & =V_{D}=+10 \mathrm{~V} \ R_{2} & =R_{3}=\frac{15-10}{0.25}=20 \mathrm{k} \Omega \ I_{C 7} & =2 I_{C 8}=2 imes 0.5=1 \mathrm{~mA} \ I_{C 3} & =I_{C 4}=\frac{1}{2} I_{C 7}=0.5 \mathrm{~mA} \ V_{E} & =0 \mathrm{~V} \ V_{B 5} & =V_{E}+0.7=+0.7 \mathrm{~V} \ R_{4} & =\frac{V_{B 5}-(-15)}{I_{C 4}} \ R_{4} & =\frac{0.7+15}{0.5}=31.4 \mathrm{k} \Omega \ R_{5} & =\frac{V_{E}-(-15)}{I_{E 5}} \ R_{5} & =\frac{0+15}{3}=5 \mathrm{k} \Omega \end{aligned} (b) R_{i d}=2(\beta+1)\left(r_{e 1,2}+R_{e} ight) where \begin{aligned} \beta & =99 \ r_{e 1,2} & =\frac{V_{T}}{I_{E 1,2}}=\frac{25 \mathrm{mV}}{0.25 \mathrm{~mA}}=100 \Omega \end{aligned} Thus, \begin{aligned} 40 & =2(99+1)\left(0.1+R_{e} ight) \ \Rightarrow R_{e} & =0.1 \mathrm{k} \Omega=100 \Omega \end{aligned} (c) A_{d 1}=\alpha \frac{\left(R_{2}+R_{3} ight) \| R_{i 2}}{\left(2 r_{e 1,2}+2 R_{e} ight)} where \begin{aligned} R_{i 2} & =2(\beta+1) r_{e 3,4}=2 r_{\pi 3,4} \ r_{e 3,4} & =\frac{25 \mathrm{mV}}{0.5 \mathrm{~mA}}=50 \Omega \ R_{i 2} & =2 imes 100 imes 0.05=10 \mathrm{k} \Omega \ A_{d 1} & \simeq 1 imes \frac{[(20+20) \| 10]}{(2 imes 0.1+2 imes 0.1)} \ & =20 \mathrm{~V} / \mathrm{V} \end{aligned} (d) A_{d 2}=\alpha \frac{R_{4} \| R_{i 3}}{\left(r_{e 3}+r_{e 4} ight)} where R_{i 3}=(\beta+1)\left(r_{e 5}+R_{5} ight) and \begin{aligned} r_{e 5} & =\frac{V_{T}}{I_{E 5}}=\frac{25 \mathrm{mV}}{3 \mathrm{~mA}}=8.3 \Omega \ R_{i 3} & =(99+1)(0.0083+5) \ & =500.83 \mathrm{k} \Omega \ A_{d 2} & \simeq 1 imes \frac{31.4 \| 500.83}{0.1} \ & =295.5 \mathrm{~V} / \mathrm{V} \end{aligned} (e) A_{3}=\frac{R_{5}}{R_{5}+r_{e 5}}=\frac{5000}{5000+8.3}=0.998 \mathrm{~V} / \mathrm{V} (f) \begin{aligned} A_{v} & =\frac{v_{e}}{v_{b}-v_{a}}=20 imes 295.5 imes 0.998 \ & =5900 \mathrm{~V} / \mathrm{V} \end{aligned}$

Figure 9.10.1
(a)

\begin{aligned}I_{E 1} & =I_{E 2}=0.25 \mathrm{~mA} \I_{C 6} & =I_{E 1}+I_{E 2}=0.5 \mathrm{~mA} \\Rightarrow I_{C 9} & =I_{C 8}=0.5 \mathrm{~mA} \R_{1} & =\frac{30-0.7-0.7}{0.5}=57.2 \mathrm{k} \Omega \I_{C 1} & =I_{C 2}=0.25 \mathrm{~mA} \V_{C} & =V_{D}=+10 \mathrm{~V} \R_{2} & =R_{3}=\frac{15-10}{0.25}=20 \mathrm{k} \Omega \I_{C 7} & =2 I_{C 8}=2 imes 0.5=1 \mathrm{~mA} \I_{C 3} & =I_{C 4}=\frac{1}{2} I_{C 7}=0.5 \mathrm{~mA} \V_{E} & =0 \mathrm{~V} \V_{B 5} & =V_{E}+0.7=+0.7 \mathrm{~V} \R_{4} & =\frac{V_{B 5}-(-15)}{I_{C 4}} \R_{4} & =\frac{0.7+15}{0.5}=31.4 \mathrm{k} \Omega \R_{5} & =\frac{V_{E}-(-15)}{I_{E 5}} \R_{5} & =\frac{0+15}{3}=5 \mathrm{k} \Omega\end{aligned}

(b)

R_{i d}=2(\beta+1)\left(r_{e 1,2}+R_{e} ight)

where

\begin{aligned}\beta & =99 \r_{e 1,2} & =\frac{V_{T}}{I_{E 1,2}}=\frac{25 \mathrm{mV}}{0.25 \mathrm{~mA}}=100 \Omega\end{aligned}

Thus,

\begin{aligned}40 & =2(99+1)\left(0.1+R_{e} ight) \\Rightarrow R_{e} & =0.1 \mathrm{k} \Omega=100 \Omega\end{aligned}

(c)

A_{d 1}=\alpha \frac{\left(R_{2}+R_{3} ight) \| R_{i 2}}{\left(2 r_{e 1,2}+2 R_{e} ight)}

where

\begin{aligned}R_{i 2} & =2(\beta+1) r_{e 3,4}=2 r_{\pi 3,4} \r_{e 3,4} & =\frac{25 \mathrm{mV}}{0.5 \mathrm{~mA}}=50 \Omega \R_{i 2} & =2 imes 100 imes 0.05=10 \mathrm{k} \Omega \A_{d 1} & \simeq 1 imes \frac{[(20+20) \| 10]}{(2 imes 0.1+2 imes 0.1)} \& =20 \mathrm{~V} / \mathrm{V}\end{aligned}

(d)

A_{d 2}=\alpha \frac{R_{4} \| R_{i 3}}{\left(r_{e 3}+r_{e 4} ight)}

where

R_{i 3}=(\beta+1)\left(r_{e 5}+R_{5} ight)

and

\begin{aligned}r_{e 5} & =\frac{V_{T}}{I_{E 5}}=\frac{25 \mathrm{mV}}{3 \mathrm{~mA}}=8.3 \Omega \R_{i 3} & =(99+1)(0.0083+5) \& =500.83 \mathrm{k} \Omega \A_{d 2} & \simeq 1 imes \frac{31.4 \| 500.83}{0.1} \& =295.5 \mathrm{~V} / \mathrm{V}\end{aligned}

(e)

A_{3}=\frac{R_{5}}{R_{5}+r_{e 5}}=\frac{5000}{5000+8.3}=0.998 \mathrm{~V} / \mathrm{V}

(f)

\begin{aligned}A_{v} & =\frac{v_{e}}{v_{b}-v_{a}}=20 imes 295.5 imes 0.998 \& =5900 \mathrm{~V} / \mathrm{V}\end{aligned}

Refer to Figure 9 $r_{O}=\infty$ Assume $\left|V_{B E}\right|=0.7 \mathrm{~V}$
(A) For 9.10.1

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Refer to Figure 9 rO=∞r_{O}=\inftyrO​=∞ Assume ∣VBE∣=0.7 V\left|V_{B E}\right|=0.7 \mathrm{~V}∣VBE​∣=0.7 V (A) For 9.10.1

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Refer to Figure 9 $r_{O}=\infty$ Assume $\left|V_{B E}\right|=0.7 \mathrm{~V}$
(A) For 9.10.1

Correct Answer:
Verified