Question

For the shown $\text{n-p-n}$ transistor connected in common emitter configuration, the power gain and $I_B$ is [Internal resistance of transistor is $200\Omega$]

• A. $3.5,1.04\times {10}^{-3}\text{A}$
• B. $3.7,1.04\times {10}^{-3}\text{A}$
• C. $3.7,2.04\times {10}^{-3}\text{A}$
• D. $3.5,2.04\times {10}^{-3}\text{A}$

Answer 1

The correct option is B $3.7,1.04\times {10}^{-3}\text{A}$

Collector current,

$I_C=\frac{\text{Voltage drop across}{R}_L}{R_L}$

$\Rightarrow I_C=\frac{0.8}{800}=1\times {10}^{-3}\text{A}$

Current gain, $\beta =\frac{I_C}{I_B}$

$\Rightarrow \frac{25}{26}=\frac{1\times {10}^{-3}}{I_B}$

$\Rightarrow I_B=1.04\times {10}^{-3}\text{A}$

Further, voltage gain,

$A_v=\beta \frac{R_{output}}{R_{intput}}=\frac{25}{26}\times \frac{800}{200}=3.85$

And, power gain,

$P_r=\beta \times A_v=\frac{25}{26}\times 3.85=3.7$

Hence, $\left(B\right)$ is the correct answer.