Semiconductors Checkpoint Questions
Five written checkpoints on semiconductor energy bands, intrinsic carriers, doping, depletion regions and diode bias, with worked reasoning.
Learning objectives
- Explain energy bands, intrinsic conduction and how temperature and light change semiconductor resistance.
- Explain n-type and p-type doping, donor and acceptor levels and majority carriers.
- Explain p-n junction formation and bias, and analyse diode, LED, Zener, rectifier-smoothing, sensor and transistor-switching circuits.
Answer each checkpoint before opening the reasoning. These questions diagnose models and causal explanations; use the dedicated quiz for a longer scored set.
Exam question 1: Comparing band modelsCore
Explain why a semiconductor can conduct more strongly when heated while a metal commonly becomes more resistive.
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In a semiconductor, heating generates many more electron–hole pairs, so the rise in carrier number density can dominate the reduction in mobility. In a metal, carrier density changes little; stronger lattice vibrations increase scattering and reduce mobility, so resistivity rises.
Exam question 2: Electron and hole motionCore
An electric field points to the right through an intrinsic semiconductor. State the directions of electron drift, hole drift and conventional current.
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Electrons drift to the left because their charge is negative. Holes behave as positive carriers and drift to the right. Both contributions produce conventional current to the right.
Exam question 3: Why n-type material is neutralCore
A donor atom releases an electron into the conduction band. Explain why the n-type sample does not acquire a net negative charge.
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After donating the electron, the dopant remains as a fixed positive ion in the lattice. Its positive charge balances the mobile electron, so the bulk sample remains approximately neutral even though electrons are its majority carriers.
Exam question 4: Forming the depletion regionCore
Describe how joining p-type and n-type material produces a built-in electric field.
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Electrons diffuse from n to p and holes effectively diffuse from p to n. They recombine near the boundary, exposing fixed positive donor ions on the n-side and fixed negative acceptor ions on the p-side. These ion layers create an electric field from n to p, opposing further majority-carrier diffusion.
Exam question 5: Forward and reverse biasCore
Compare the depletion width and current when the p-side is connected first to the positive terminal and then to the negative terminal of a supply.
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With p connected to positive, the junction is forward biased: the applied field opposes the built-in field, the barrier and depletion width decrease, and majority-carrier current can become substantial. With p connected to negative, reverse bias reinforces the built-in field, widens the depletion region and leaves only a small leakage current before breakdown.