Question 11.15: A BJT transistor amplifier shown in Fig. 11.27 has RE = RC =...

\omega _{1p} = 2 \pi f_{L} = 100  \pi   rad/s

We know that                          \omega _{1p}= \frac{1}{C ^{′}_{E}(R_{s}+r_{\pi})^{’}}

Therefore,            C ^{′}_{E}= \frac{1}{\omega _{1p}(R_{s}+r_{\pi})}

C ^{′}_{E}= \frac{1}{100\pi \times 1.6}= 1.99  \mu F

We know that     C_{E} =  (1 + \beta )C ^{′}_{E}

C_{E} = 101 × 1.99 × 10^{–6} = 201   \mu F

We know that      \omega _{1l}= \frac{\omega _{1p}}{10}= \frac{100\pi }{10}= 10\pi = \frac{1}{(R_{s}+R_{\pi })C_{C1}}

C_{C1}= \frac{1}{10\pi \times 1.6}= 19.9   \mu F

We choose            \omega _{12}= \frac{\omega _{1p}}{20}= \frac{100\pi }{20}= 5\pi .

Therefore,            \omega _{12}= 5\pi = \frac{1}{(R_{C}+R_{L })C_{C2}}

Hence,                  C_{C2}= \frac{1}{\omega _{12}(R_{C}+R_{L })}= \frac{1}{5\pi \times 3\times 10^{3} }= 21.23 \mu F

Since \omega_{12} is inversely proportional to C_{C2}, we can choose any value C_{C2} > 21.23   \mu F. Hence, let us choose C_{C2} = 22  \mu F.