Question

A n-p-n transistor is used in common emitter made in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by $2 \mathrm{~mA}\( and \)0.04 \mathrm{~V}$ in input voltage. The current amplification factor is (A) 20 (B) 30 (C) 50 (D) 400

Short answer

The current amplification factor (β) of the given n-p-n transistor in common emitter mode is 50.

Step-by-step solution

Determine the given values

In this problem, we have: - Change in base current (ΔIb): 40 μA - Change in output current (ΔIc): 2 mA - Change in input voltage (ΔV): 0.04 V We need to determine the current amplification factor (β).

Convert the given values to the same unit

To be consistent with our calculations, we need to convert all values to the same unit. We will convert the given current values into Amperes (A). - ΔIb = 40 μA = 40 × 10^(-6) A - ΔIc = 2 mA = 2 × 10^(-3) A

Calculate the current amplification factor (β)

Now, we can calculate the current amplification factor (β) using the formula: β = ΔIc / ΔIb β = (2 × 10^(-3)) / (40 × 10^(-6))

Simplify the expression

By simplifying the expression, we get: β = 50

Choose the correct answer

Since the value of β is 50, we can choose the correct answer to be: (C) 50