Q90 / 100 · BITSAT 2022 · Physics
A transistor is connected in common-emitter (CE) configuration. The collector supply is $8 \mathrm{~V}$ and the voltage drop across a resistor is $500 \Omega$ in the collector circuit is $0.6 \mathrm{~V}$. If the current gain factor $\alpha$ is $0.96$, find the base current
- A$25 \mu \mathrm{A}$
- Correct answerB$50 \mu \mathrm{A}$
- C$20 \mu \mathrm{A}$
- D$35 \mu \mathrm{A}$
Solution
Given, $V_{C C}=8 \mathrm{~V}$
Voltage drop, $V_0=\Delta I_C R_L=0.6 \mathrm{~V}$
Load Resistor, $R_L=500 \Omega$
Gain factor, $\alpha=0.96$
$\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.96}{1-0.96}=24$
Now, $\quad V_0=\Delta I_C R_L$
$0.6=\Delta I_C \times 500$
$\Delta I_C=\frac{0.6}{500}=1.2 \times 10^{-3} \mathrm{~A}$
Now,
$\beta=\frac{\Delta I_C}{\Delta I_B}$
$\therefore$ Base current $\Delta I_B=\frac{\Delta I_C}{\beta}=\frac{1.2 \times 10^{-3}}{24}=5 \times 10^{-5}$
$=50 \times 10^{-6}$
$\Delta I_B=50 \mu \mathrm{A}$
Hence, the correct answer is option: 'B'.