Question

A transistor is connected in CE configuration. The collector supply is 10 V and the voltage drop across a resistor of 1000 $ \Omega $ in the collector circuit is 0.5 V. If the current gain factor is 0.96, then the base current is

• A. 256 \( \mu A \)
• B. 20.8 \( \mu A \)
• C. 22.5 \( \mu A \)
• D. 15 \( \mu A \)

Hint:

The base current \( I_B \) can be found using the formula \( I_B = \frac{I_C}{\beta} \), where \( I_C \) is the collector current and \( \beta \) is the current gain factor.

Answer 1

The Correct Option is B

In a common emitter (CE) transistor configuration, the collector current \( I_C \) is related to the voltage drop across the resistor \( R \) using Ohm's law: \[ I_C = \frac{V_{\text{drop}}}{R} = \frac{0.5 \, \text{V}}{1000 \, \Omega} = 0.0005 \, \text{A} = 0.5 \, \text{mA} \] The current gain factor \( \beta \) is related to the base current \( I_B \) and collector current \( I_C \) by the equation: \[ I_C = \beta I_B \] Substitute the given values: \[ 0.5 \, \text{mA} = 0.96 \times I_B \] Solving for \( I_B \): \[ I_B = \frac{0.5 \, \text{mA}}{0.96} = 0.5208 \, \text{mA} = 20.8 \, \mu A \]
Thus, the base current is 20.8 \( \mu A \).