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(B) Given: $\beta_{dc} = 100$,$R_B = 220 	ext{ k}\Omega$,$R_L = 4.7 	ext{ k}\Omega$,$I_C = 1.5 	ext{ mA}$,$V_{CC} = 24 	ext{ V}$.$1$. Calculate ba

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$V_{BE} = +20.7 	ext{ V}, V_{BC} = +3.75 	ext{ V}$ and the amplifier is not working.

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(B) Given: $\beta_{dc} = 100$,$R_B = 220 	ext{ k}\Omega$,$R_L = 4.7 	ext{ k}\Omega$,$I_C = 1.5 	ext{ mA}$,$V_{CC} = 24 	ext{ V}$.$1$. Calculate base current $I_B$:$I_B = \frac{I_C}{\beta_{dc}} = \frac{1.5 	imes 10^{-3} 	ext{ A}}{100} = 15 	imes 10^{-6} 	ext{ A} = 15 	ext{ } \mu	ext{A}$.$2$. Calculate $V_{BE}$ using Kirchhoff's voltage law in the base loop:$V_{CC} = I_B R_B + V_{BE}$$V_{BE} = V_{CC} - I_B R_B = 24 	ext{ V} - (15 	imes 10^{-6} 	ext{ A} 	imes 220 	imes 10^3 	ext{ } \Omega) = 24 	ext{ V} - 3.3 	ext{ V} = 20.7 	ext{ V}$.$3$. Calculate collector voltage $V_C$:$V_C = V_{CC} - I_C R_L = 24 	ext{ V} - (1.5 	imes 10^{-3} 	ext{ A} 	imes 4.7 	imes 10^3 	ext{ } \Omega) = 24 	ext{ V} - 7.05 	ext{ V} = 16.95 	ext{ V}$.$4$. Calculate $V_{BC}$:$V_{BC} = V_B - V_C = V_{BE} - V_C = 20.7 	ext{ V} - 16.95 	ext{ V} = 3.75 	ext{ V}$.$5$. Conclusion:For an $NPN$ transistor to work as an amplifier,the base-emitter junction must be forward-biased $(V_{BE} > 0)$ and the base-collector junction must be reverse-biased $(V_{BC} < 0)$. Here,$V_{BC} = +3.75 	ext{ V}$,which means the base-collector junction is forward-biased. Therefore,the transistor is in saturation and the amplifier is not working.

$A$ common emitter amplifier circuit is shown in the figure. For the transistor used in the circuit,the current amplification factor is $\beta_{dc} = 100$. Other parameters are given in the figure.

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