A transistor is operated in common emitter configuration at constant collector voltage `V_(c)=1.5 V` such that a change in the base current from `100 muA` to `150 muA` produces a change in the collector current from `5 mA` to `10 mA`. The current gain `(beta)` is

Options

A. 67
B. 75
C. 100
D. 50

Answer

To find the current gain (β) of the transistor in a common emitter configuration, we can follow these steps:

### Step 1: Identify the changes in base current (Ib) and collector current (Ic)
- The base current changes from \( I_{b1} = 100 \, \mu A \) to \( I_{b2} = 150 \, \mu A \).
- The collector current changes from \( I_{c1} = 5 \, mA \) to \( I_{c2} = 10 \, mA \).

### Step 2: Calculate the change in base current (ΔIb)
\[ \Delta I_b = I_{b2} - I_{b1} = 150 \, \mu A - 100 \, \mu A = 50 \, \mu A \]

### Step 3: Calculate the change in collector current (ΔIc)
\[ \Delta I_c = I_{c2} - I_{c1} = 10 \, mA - 5 \, mA = 5 \, mA \]

### Step 4: Convert units if necessary
- Convert \( \Delta I_b \) from microamperes to amperes:
\[ \Delta I_b = 50 \, \mu A = 50 \times 10^{-6} \, A \]
- Convert \( \Delta I_c \) from milliamperes to amperes:
\[ \Delta I_c = 5 \, mA = 5 \times 10^{-3} \, A \]

### Step 5: Calculate the current gain (β)
The current gain \( \beta \) is defined as the ratio of the change in collector current to the change in base current:
\[ \beta = \frac{\Delta I_c}{\Delta I_b} \]
Substituting the values:
\[ \beta = \frac{5 \times 10^{-3}}{50 \times 10^{-6}} = \frac{5}{50} \times 10^{3} = 100 \]

### Final Answer:
The current gain \( \beta \) is \( 100 \).

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