The dc common emitter current gain of an n-p- | vedclass

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Question

(A) Given: Current gain $\beta = 50$,$V_{CE} = 2 \ V$,and $R_{C} = 4 \ k\Omega$.The collector current $I_{C}$ is determined by the output circuit loop

Vedclass · Accepted Answer

$I_{B} = 10 \ \mu A, I_{C} = 0.5 \ mA$

Vedclass · Answer

(A) Given: Current gain $\beta = 50$,$V_{CE} = 2 \ V$,and $R_{C} = 4 \ k\Omega$.The collector current $I_{C}$ is determined by the output circuit loop: $I_{C} = \frac{V_{CE}}{R_{C}} = \frac{2 \ V}{4 	imes 10^{3} \ \Omega} = 0.5 	imes 10^{-3} \ A = 0.5 \ mA$.The base current $I_{B}$ is related to the collector current by the current gain formula: $\beta = \frac{I_{C}}{I_{B}}$.Rearranging for $I_{B}$: $I_{B} = \frac{I_{C}}{\beta} = \frac{0.5 	imes 10^{-3} \ A}{50} = 0.01 	imes 10^{-3} \ A = 10 	imes 10^{-6} \ A = 10 \ \mu A$.Thus,$I_{B} = 10 \ \mu A$ and $I_{C} = 0.5 \ mA$.

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