Difficulty: Medium
Correct Answer: The minority carrier density becomes 4 times the original value
Explanation:
Introduction / Context:In doped semiconductors, temperature raises intrinsic carrier concentration n_i, which influences minority carriers strongly through the mass-action law. Understanding this interplay is critical when predicting leakage currents and designing devices for high-temperature environments.
Given Data / Assumptions:
Concept / Approach:The mass-action law states n * p = n_i^2 at equilibrium. In a strongly n-type sample, the majority concentration n ≈ N_D (weakly affected by modest changes in n_i), while the minority concentration p ≈ n_i^2 / n ≈ n_i^2 / N_D. If n_i doubles, p scales with n_i^2 and therefore quadruples. Majority carriers remain essentially set by ionized donors until very high temperatures push the material toward intrinsic behavior.
Step-by-Step Solution:
Initial: p_1 ≈ n_i^2 / N_D.After heating: n_i → 2 n_i ⇒ p_2 ≈ (2 n_i)^2 / N_D = 4 n_i^2 / N_D.Therefore p_2 / p_1 = 4; the minority concentration quadruples.Majority concentration remains ≈ N_D for moderate temperature rise.Verification / Alternative check:
Check limiting case: if n begins to deviate from N_D, the material tends to intrinsic and both carriers approach n_i; the stated result holds before that limit.Why Other Options Are Wrong:
Doubling either density ignores the n_i^2 dependence of the minority carriers.Majority carriers do not double unless the material becomes intrinsic.Common Pitfalls:
Forgetting the square dependence p ∝ n_i^2 in extrinsic regimes; assuming symmetric changes for majority and minority carriers.Final Answer:
The minority carrier density becomes 4 times the original value
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