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A transistor has small-signal current gain α_ac = 0.98 in common-base mode. What is the corresponding β_ac in common-emitter mode?

Difficulty: Easy

51
49
47
45
0.98

Correct Answer: 49

Explanation:

Introduction / Context:
Common-base (CB) and common-emitter (CE) current gains (α and β) are related for a bipolar junction transistor. Converting between them is routine in circuit analysis and device modeling.

Given Data / Assumptions:

α_ac = 0.98 (CB current gain).
Small-signal, linear operating region.

Concept / Approach:
The exact relationship between CE and CB current gains is β = α / (1 − α). This follows from current relations in a BJT: IE = IC + IB, with α = IC / IE and β = IC / IB.

Step-by-Step Solution:

Step 1: Use β = α / (1 − α).Step 2: Substitute α = 0.98.Step 3: Compute denominator: 1 − 0.98 = 0.02.Step 4: β = 0.98 / 0.02 = 49.

Verification / Alternative check:

Cross-check with inverse: α = β / (β + 1) = 49 / 50 ≈ 0.98, consistent.

Why Other Options Are Wrong:

51 or 47 or 45: Do not satisfy α = β / (β + 1) near 0.98.0.98: That is α, not β.

Common Pitfalls:

Using β ≈ α / (1 − α) but miscomputing the small denominator; forgetting that small α error produces large β change.

A transistor has small-signal current gain α_ac = 0.98 in common-base mode. What is the corresponding β_ac in common-emitter mode?

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