Question

In an n-p-n common emitter (CE) transistor the collector current changes from 5 m A to 16 mA for the change in base current from 100 μ A and 200 μA , respectively. The current gain of transistor is Options 1. 110 2. 0.9 3. 210 4. 9

• A. 0.9
• B. 9
• C. 110
• D. 210

Answer 1

The Correct Option is C

To find the current gain (\beta) of the n-p-n common emitter (CE) transistor, we use the relationship between the change in collector current and the change in base current. The formula for the current gain, \beta, is given by:

\beta = \frac{\Delta I_{C}}{\Delta I_{B}}

Where:

• \Delta I_{C} is the change in collector current.
• \Delta I_{B} is the change in base current.

From the given data, the change in collector current, \Delta I_{C}, is:

\Delta I_{C} = 16 \, \text{mA} - 5 \, \text{mA} = 11 \, \text{mA}

Similarly, the change in base current, \Delta I_{B}, is:

\Delta I_{B} = 200 \, \mu\text{A} - 100 \, \mu\text{A} = 100 \, \mu\text{A}

Converting \Delta I_{B} to milliamperes for consistency:

\Delta I_{B} = 100 \, \mu\text{A} = 0.1 \, \text{mA}

Now, substituting these values into the formula:

\beta = \frac{11 \, \text{mA}}{0.1 \, \text{mA}} = 110

Thus, the current gain of the transistor is 110.

Therefore, the correct answer is 110.