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Question

(A) Given,collector resistance $R_C = 800 \Omega$ and voltage drop across it $V_R = 0.5 V$.Collector current $I_C = \frac{V_R}{R_C} = \frac{0.5}{800}

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$2.6 	imes 10^{-5} A$

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(A) Given,collector resistance $R_C = 800 \Omega$ and voltage drop across it $V_R = 0.5 V$.Collector current $I_C = \frac{V_R}{R_C} = \frac{0.5}{800} = 6.25 	imes 10^{-4} A = 0.625 	imes 10^{-3} A$.We know that the current gain $\alpha = \frac{I_C}{I_E}$,so emitter current $I_E = \frac{I_C}{\alpha} = \frac{0.625 	imes 10^{-3}}{0.96} \approx 6.51 	imes 10^{-4} A$.The base current $I_B = I_E - I_C = I_C \left( \frac{1}{\alpha} - 1 ight) = I_C \left( \frac{1 - \alpha}{\alpha} ight)$.Substituting the values: $I_B = 0.625 	imes 10^{-3} 	imes \left( \frac{1 - 0.96}{0.96} ight) = 0.625 	imes 10^{-3} 	imes \left( \frac{0.04}{0.96} ight) = 0.625 	imes 10^{-3} 	imes \frac{1}{24} \approx 0.02604 	imes 10^{-3} A = 2.6 	imes 10^{-5} A$.

$A$ transistor is connected in common emitter configuration. The collector supply is $8 V$ and the voltage drop across a resistor of $800 \Omega$ in the collector circuit is $0.5 V$. If the current gain factor $\alpha$ is $0.96$,the base current is

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