Difficulty: Easy
Correct Answer: Correct
Explanation:
Introduction / Context:
Understanding how the depletion region and the built-in (barrier) potential of a P–N junction are related is fundamental to semiconductor physics. This determines current flow under forward and reverse bias and explains why diodes, bipolar junction transistors (BJTs), and junction field-effect transistors (JFETs) behave the way they do.
Given Data / Assumptions:
Concept / Approach:
For an abrupt junction, the depletion width W satisfies W ∝ sqrt((V_bi + V_R) * (1/N_A + 1/N_D)), where V_R is applied reverse bias, and N_A, N_D are acceptor and donor concentrations. As V_bi or reverse bias increases, the electric field strengthens and W widens. Thus, larger depletion width accompanies a larger total junction potential (built-in plus any applied reverse bias).
Step-by-Step Solution:
Verification / Alternative check:
Device I–V curves under reverse bias show reduced capacitance C_j ≈ εA/W; as W increases, C_j decreases. Measured junction capacitance shrinking with reverse voltage confirms widening depletion tied to increased junction potential.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing cause and effect: doping changes can make W smaller even if V_bi is large because W also scales with 1/sqrt(doping). The statement assumes a scenario where an increased potential (built-in or reverse) is what widens W.
Final Answer:
Correct
Discussion & Comments