Semiconductor principles — relationship between depletion region width and barrier (built-in) potential In a P–N junction inside a transistor or diode, when the depletion region becomes wider (for example, due to higher built-in potential or added reverse bias), the electrostatic barrier potential across the junction becomes larger. Evaluate this statement.
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ACorrect
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BIncorrect
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CTrue only at very high temperatures
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DTrue only for heavily doped junctions
Answer
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:
- An abrupt P–N junction at thermal equilibrium unless bias is explicitly mentioned.
- Standard definitions: depletion region is the ionized, carrier-depleted space-charge region; barrier potential (V_bi) is the electrostatic potential opposing majority-carrier diffusion.
- Temperature and doping levels are within normal device ranges.
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:
At equilibrium, a diffusion of carriers creates a space-charge region and an opposing field that establishes V_bi.The depletion approximation yields W ∝ sqrt(V_bi) for fixed doping.Applying reverse bias increases the total potential across the junction, which increases W further.Therefore, a larger depletion region is associated with a larger barrier (or total) potential.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:
Incorrect: contradicts the standard depletion-region equations.True only at very high temperatures / True only for heavily doped junctions: temperature and doping influence V_bi and W, but the positive correlation between total potential and W remains general.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