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Class XII · Chapter 14

Semiconductor Electronics — Exam Practice

Practice questions in CBSE, NEET, and JEE Main style covering p-n junctions, diodes, transistors, and logic gates.

📝 10 questions 🎯 CBSE · NEET · JEE Main
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.

1CBSE Board Exam Style

CBSE · 1 Mark
Q1. Why is silicon preferred over germanium for making most semiconductor devices?
Answer Silicon has a larger band gap (≈1.1 eV vs. germanium's ≈0.7 eV), giving devices better thermal stability — silicon-based devices work reliably over a wider temperature range without excessive leakage current. Silicon is also far more abundant and easier to purify at scale.
CBSE · 3 Marks
Q2. With the help of a labelled diagram, explain forward and reverse biasing of a p-n junction diode, and describe how the depletion region behaves in each case.
Answer Forward bias: p-side connected to positive terminal, n-side to negative. The external field opposes the junction's built-in field, shrinking the depletion region; once the barrier potential is overcome, current flows easily and grows rapidly with voltage.

Reverse bias: p-side connected to negative terminal, n-side to positive. The external field reinforces the built-in field, widening the depletion region and allowing only a tiny reverse leakage current to flow.
CBSE · 5 Marks
Q3. (a) Explain, with a circuit diagram, how a transistor works as a common-emitter amplifier. (b) In a transistor circuit, the base current IB = 25 µA and collector current IC = 2.5 mA. Calculate the current gain β and the emitter current IE.
Answer (a) In the common-emitter configuration, a small AC input signal applied between base and emitter produces a much larger, amplified variation in collector current (since β is large), which develops an amplified output voltage across a load resistor in the collector circuit. The output is also phase-inverted relative to the input.

(b) β = IC/IB = 2500/25 = 100.
IE = IB + IC = 25 + 2500 = 2525 µA = 2.525 mA.

2NEET Style (Single-Correct MCQ)

NEET Style
Q4. In a p-type semiconductor, the majority charge carriers are:
Electrons
✓ Holes
Protons
Equal electrons and holes
Solution p-type semiconductors are doped with trivalent (acceptor) impurities, creating an excess of holes as the majority carrier.
NEET Style
Q5. A Zener diode is primarily used for:
Rectification
✓ Voltage regulation
Amplification
Oscillation
Solution Operated deliberately in reverse breakdown, where voltage across it stays essentially constant over a wide current range — making it ideal as a voltage regulator.
NEET Style
Q6. A logic gate that gives output 1 only when both of its inputs are 1 is a/an:
OR gate
✓ AND gate
NOT gate
NOR gate
Solution By definition — AND requires every input to be 1 for the output to be 1; any 0 input forces the output to 0.
NEET Style
Q7. In an intrinsic (pure) semiconductor, the number of free electrons is:
Much greater than the number of holes
✓ Equal to the number of holes
Much less than the number of holes
Zero
Solution Every thermally freed electron leaves exactly one hole behind — ne = nh in an intrinsic semiconductor, with no doping to favour one carrier type over the other.

3JEE Main Style

JEE Main Style · Numerical
Q8. A transistor has a common-base current gain α = 0.98. Find its common-emitter current gain β.
Solution β = α/(1−α) = 0.98/(1−0.98) = 0.98/0.02 = 49.
JEE Main Style · MCQ
Q9. A NAND gate is logically equivalent to:
OR gate followed by NOT
✓ AND gate followed by NOT
NOT gate alone
AND gate alone
Solution NAND = "NOT AND" — output is 0 only when both inputs are 1, the exact inverse of a plain AND gate's truth table.
JEE Main Style · Numerical
Q10. Silicon has a band gap of 1.1 eV. Find the maximum wavelength of light that can generate an electron-hole pair in silicon.
Solution λmax = hc/Eg = (6.63×10⁻³⁴ × 3×10⁸) / (1.1 × 1.6×10⁻¹⁹) ≈ 1.13×10⁻⁶ m ≈ 1130 nm (near-infrared).
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