The current amplification factor for a transistor in a common ... | Technical Ability MCQ | Examoogle

Technical AbilityElectronics Engineering

The current amplification factor for a transistor in a common base configuration is 0.75. If the emitter current is 4A, find the base current.

3 A

1 A

5 A

0 A

Correct Answer

⚙️ TE • Technical Concept & PrincipleTechnical AbilityElectronics Engineering

Option B

1 A

Quick Summary:

In a common base configuration, the current amplification factor ( $\alpha$ ) is the ratio of collector current ( $I_C$ ) to emitter current ( $I_E$ ). Given $\alpha = 0.75$ and $I_E = 4 \text{ A}$ , the collector current is calculated as $I_C = \alpha \times I_E = 0.75 \times 4 = 3 \text{ A}$ . The base current ( $I_B$ ) is then derived using Kirchhoff's Current Law, $I_B = I_E - I_C$ .

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$\alpha = \frac{I_C}{I_E}$ — Current amplification factor in common base configuration

$I_E = I_B + I_C$ — Kirchhoff's current law for a transistor

$I_B = I_E(1 - \alpha)$ — Derived relation for base current

⚙️ Working Principle

The Bipolar Junction Transistor operates based on the principle that the total emitter current must equal the sum of base and collector currents ( $I_E = I_B + I_C$ ). In common base mode, most of the charge carriers from the emitter reach the collector, characterized by the factor $\alpha$ , which is typically slightly less than 1.

📌 Key Points

▸
The current gain $\alpha$ (alpha) is always less than unity for a BJT.
▸
The base current is typically a small fraction of the emitter current.
▸
Common base configuration is often used for high-frequency applications due to low input impedance.

✅ Advantages

▸
High voltage gain
▸
Excellent high-frequency response

❌ Disadvantages / Limitations

▸
Low current gain
▸
Low input impedance

🛠️ Applications / Uses

▸
High-frequency amplifiers
▸
Microphone pre-amplifiers

📄 Additional Information

▸
Calculation: $I_C = 0.75 \times 4 = 3 \text{ A}$ . Thus, $I_B = 4 \text{ A} - 3 \text{ A} = 1 \text{ A}$ .
▸
Option A (3 A) represents the collector current ( $I_C$ ).
▸
Option C (5 A) is physically impossible as $I_B$ cannot exceed $I_E$ in a functional transistor.

📊 Diagram / Illustration

✅

B is correct — Using the relation $I_B = I_E(1 - \alpha)$ , we find $I_B = 4(1 - 0.75) = 4 \times 0.25 = 1 \text{ A}$ .

Core Concepts Used

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BJT Terminal Currents Common Base Configuration Alpha-Beta Relationship

💡 EXAM TIP

Always remember that for any transistor, the sum of base and collector currents equals emitter current; if you calculate $I_C$ correctly, $I_B$ is just the remaining portion.

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