Q136 / 180 · COMEDK 2023 (Morning Shift) · Physics
A transistor is connected in CE configuration. The collector supply is $10 \mathrm{~V}$ and the voltage drop across a resistor of $1000 \Omega$ in the collector circuit is $0.5 \mathrm{~V}$. If the current gain factor is 0.96 , then the base current is
- A$15 \mu \mathrm{A}$
- B$25_6 \mu \mathrm{A}$
- Correct answerC$20.8 \mu \mathrm{A}$
- D$22.5 \mu \mathrm{A}$
Solution
The collector current $I_C$ is determined by the voltage drop across the collector resistor $R_C = 1000 \Omega$. Given the voltage drop $V_R = 0.5 \mathrm{~V}$, we have $I_C = \dfrac{V_R}{R_C} = \dfrac{0.5}{1000} = 0.5 \times 10^{-3} \mathrm{~A} = 500 \mu \mathrm{A}$.
The current gain factor in common base configuration is given as $\alpha = 0.96$. The relationship between collector current $I_C$ and base current $I_B$ in common emitter configuration is $I_C = \beta I_B$, where $\beta = \dfrac{\alpha}{1 - \alpha}$.
Calculating $\beta$: $\beta = \dfrac{0.96}{1 - 0.96} = \dfrac{0.96}{0.04} = 24$.
Now, calculating the base current $I_B$: $I_B = \dfrac{I_C}{\beta} = \dfrac{500 \mu \mathrm{A}}{24} = 20.833 \mu \mathrm{A}$.
Rounding to the nearest value, we get $I_B \approx 20.8 \mu \mathrm{A}$.
Answer: $20.8 \mu \mathrm{A}$
The current gain factor in common base configuration is given as $\alpha = 0.96$. The relationship between collector current $I_C$ and base current $I_B$ in common emitter configuration is $I_C = \beta I_B$, where $\beta = \dfrac{\alpha}{1 - \alpha}$.
Calculating $\beta$: $\beta = \dfrac{0.96}{1 - 0.96} = \dfrac{0.96}{0.04} = 24$.
Now, calculating the base current $I_B$: $I_B = \dfrac{I_C}{\beta} = \dfrac{500 \mu \mathrm{A}}{24} = 20.833 \mu \mathrm{A}$.
Rounding to the nearest value, we get $I_B \approx 20.8 \mu \mathrm{A}$.
Answer: $20.8 \mu \mathrm{A}$
Hence, the correct answer is option: 'C'.