An N-P-N transistor in a common emitter mode | vedclass

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Question

(A) From the circuit diagram,the potential difference across the load resistance $R_L$ is given by:$V_{R_L} = V_{CC} - V_{CE} = 8 \ V - 4 \ V = 4 \ V$

Vedclass · Accepted Answer

$R_L = 1 \ k\Omega, R_B = 185 \ k\Omega$

Vedclass · Answer

(A) From the circuit diagram,the potential difference across the load resistance $R_L$ is given by:$V_{R_L} = V_{CC} - V_{CE} = 8 \ V - 4 \ V = 4 \ V$Using Ohm's law,$V_{R_L} = I_C R_L$,we have:$R_L = \frac{V_{R_L}}{I_C} = \frac{4 \ V}{4 	imes 10^{-3} \ A} = 10^3 \ \Omega = 1 \ k\Omega$For the base circuit,the potential difference across the base resistance $R_B$ is:$V_{R_B} = V_{CC} - V_{BE} = 8 \ V - 0.6 \ V = 7.4 \ V$The base current $I_B$ is related to the collector current $I_C$ by the amplification factor $\beta$:$I_B = \frac{I_C}{\beta} = \frac{4 	imes 10^{-3} \ A}{100} = 4 	imes 10^{-5} \ A$Now,calculating $R_B$:$R_B = \frac{V_{R_B}}{I_B} = \frac{7.4 \ V}{4 	imes 10^{-5} \ A} = 1.85 	imes 10^5 \ \Omega = 185 \ k\Omega$Thus,$R_L = 1 \ k\Omega$ and $R_B = 185 \ k\Omega$.

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