Q79 / 90 · JEE Main 2023 (08 Apr Shift 2) · Physics

For a given transistor amplifier circuit in CE configuration $V_{C C} = 1 V, R_{C} = 1 kΩ, R_{b} = 100 kΩ$ and $β = 100$ . Value of base current $I_{b}$ is

A
$I_{b} = 1.0 μA$
B
$I_{b} = 0.1 μA$
C
$I_{b} = 100 μA$
Correct answer
D
$I_{b} = 10 μA$

Solution

Considering the transistor in saturation mode, $V_{C E} = 0$
Using Kirchhoff's voltage rule,
$- I_{C} R_{C} + V_{C C} = 0 \Rightarrow I_{C} = \frac{V_{C C}}{R_{C}} \Rightarrow I_{C} = \frac{1 V}{10^{3} Ω} = 10^{- 3} A$
So,
$β = \frac{I_{C}}{I_{b}} = \frac{10^{- 3} A}{I_{b}} \Rightarrow I_{b} = \frac{10^{- 3}}{10^{2}} = 10^{- 5} A = 10 μA$

Hence, the correct answer is option: 'D'.

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Solution

Relevant Questions

For a given transistor amplifier circuit in CE configuration $V_{C C} = 1 V, R_{C} = 1 kΩ, R_{b} = 100 kΩ$ and $β = 100$ . Value of base current $I_{b}$ is