Previous year questions: Semiconductor Electronics

Board pattern for this chapter

The chapter is a reliable source of one-mark MCQs on doping and bias, two-mark explanations of the depletion region or barrier, and three-mark rectifier questions with circuit and waveforms. All ten questions below are rewritten from actual CBSE board papers between 2022 and 2024. The marks in each heading tell you the length, and the year tells you which paper to open.

Q1 (CBSE Board 2022, 2 marks)

Explain the formation of the depletion region in a p-n junction.

Answer: Holes are far more numerous on the p-side and electrons on the n-side, so holes diffuse to the n-side and electrons to the p-side. Each electron that leaves the n-side leaves a fixed positive donor ion behind, and each hole that leaves the p-side leaves a fixed negative acceptor ion behind. The two space-charge layers build up on either side of the junction and together form the depletion region, which is almost free of mobile carriers. The resulting field drives a drift current opposite to the diffusion current until the two balance.

Q2 (CBSE Board 2022, 3 marks)

With the help of a circuit diagram, explain the working of a p-n junction diode as a full-wave rectifier, and draw its input and output waveforms.

Answer: The circuit uses two diodes and a centre-tapped transformer. The p-sides of the two diodes connect to the two ends A and B of the secondary, their n-sides are joined, and the output is taken across the load between that common point and the centre tap. When A is positive with respect to the centre tap, D1 is forward biased and conducts while D2 is reverse biased. In the next half cycle B becomes positive, so D2 conducts and D1 is off. The load therefore receives a pulse in both half cycles, and the output waveform is a series of half-sinusoid pulses with twice the input frequency, while the input is a full sine wave.

Q3 (CBSE Board 2022, 2 marks)

What is meant by the energy band gap in a solid? Draw the energy band diagrams for a conductor, an insulator and a semiconductor.

Answer: The band gap is the energy needed to move an electron from the top of the filled valence band to the bottom of the empty conduction band. In a conductor the conduction and valence bands overlap, or the conduction band is partly filled, so electrons can move freely. In an insulator the gap is large, above about 3 eV, so almost no electron can cross at room temperature. In a semiconductor the gap is small, between about 0.2 and 3 eV, so a few electrons cross at room temperature and a small conductivity appears.

Q4 (CBSE Board 2022, 2 marks)

Define barrier potential. Why does the thickness of the depletion layer in a p-n junction diode vary with an increase in reverse bias?

Answer: The barrier potential is the built-in potential difference across the depletion region, with the n-side positive relative to the p-side, which opposes further diffusion of carriers. Under reverse bias the external voltage adds to the barrier field, so the effective barrier rises to V0 + V. The stronger field pushes more majority carriers away from the junction on both sides, so the depletion layer widens.

Q5 (CBSE Board 2023, 2 marks)

Explain the roles of diffusion current and drift current in the formation of the depletion layer of a p-n junction.

Answer: Diffusion moves electrons from n to p and holes from p to n because of the concentration gradient, and leaves behind fixed ionised donors and acceptors that form the space-charge layers of the depletion region. The space charge creates an electric field from n to p, and this field drives the drift current in the opposite direction, returning electrons to the n-side and holes to the p-side. Diffusion is larger at first and grows smaller in relative terms as the space charge builds, and the junction becomes stable when the two currents are equal.

Q6 (CBSE Board 2023, 2 marks)

How is the width of the depletion layer of a p-n junction diode affected when it is (i) forward biased and (ii) reverse biased? Justify your answers.

Answer: (i) Under forward bias the applied field opposes the barrier field, the effective barrier falls to V0 - V, and the depletion layer becomes narrower. (ii) Under reverse bias the applied field adds to the barrier field, the effective barrier rises to V0 + V, and the depletion layer becomes wider. In both cases the change follows from the direction of the applied voltage relative to the built-in field at the junction.

Q7 (CBSE Board 2023, 2 marks)

Explain the property of a p-n junction that makes it suitable for rectifying alternating voltages, and differentiate between a half-wave and a full-wave rectifier.

Answer: A p-n junction conducts only in forward bias and blocks almost completely in reverse bias, so when an alternating voltage is applied the current flows only in one half of each cycle. This unidirectional conduction is what makes rectification possible. A half-wave rectifier uses one diode, gives one output pulse per input cycle, and its output frequency equals the input frequency. A full-wave rectifier uses two diodes with a centre-tapped transformer, gives one pulse in each half cycle, and has twice the output frequency.

Q8 (CBSE Board 2023, 1 mark)

When an intrinsic semiconductor is doped with a small amount of a trivalent impurity, then

(A) its resistance increases (B) it becomes a p-type semiconductor (C) there will be more free electrons than holes in the semiconductor (D) dopant atoms become donor atoms

Answer: Option (B), it becomes a p-type semiconductor. A trivalent atom has one valence electron fewer than silicon, so it leaves an unsatisfied bond, which is a hole. Holes then outnumber electrons, the material becomes p-type, and its resistance falls rather than rises.

Q9 (CBSE Board 2024, 2 marks)

How does the conductivity of an intrinsic semiconductor vary with temperature? Explain, and show the variation in a plot.

Answer: Conductivity rises with temperature. Heating breaks more covalent bonds, so more electron-hole pairs are generated and the carrier concentration increases rapidly. The plot of conductivity against temperature therefore curves upward. This behaviour is opposite to that of a metal, whose conductivity falls with temperature because the lattice scatters the electrons more strongly.

Q10 (CBSE Board 2024, 2 marks)

Explain the role of diffusion and drift in the formation of a p-n junction.

Answer: Diffusion drives electrons from the n-side to the p-side and holes the other way because of the concentration difference across the junction. The fixed ionised donors left on the n-side and acceptors left on the p-side create a space-charge region and a built-in field. That field produces a drift current of electrons back to the n-side and holes back to the p-side. The junction settles when diffusion and drift are equal and the net current is zero, which leaves the depletion region and barrier potential in place.

How to read the year labels

Every item above is a board-paper question, so the year label tells you the paper to open. When you practise, cover the answer and write only the steps; then compare with the worked answer. If the steps match but a number differs, the error is arithmetic; if the order differs, the error is conceptual and the topic page needs another read. That check is faster than rereading the whole chapter.

What the board rewards

For rectifier questions the marks are for the circuit, the conduction half cycles, the waveform and the output frequency. For bias questions, state the change in barrier height and depletion width before explaining the current. For energy band questions, draw the three bands and label the gap, then link the gap size to conductivity. Numericals on carrier concentration always start with n e n h = n i squared.

Final list for the exam

For semiconductor electronics the minimum equipment is: the three band diagrams, the carrier relations n e = n h = n i and n e n h = n i squared, the donor and acceptor pictures, the depletion region story with diffusion and drift, the diode characteristic with threshold and reverse saturation, and the two rectifier circuits with their frequencies. If all of those can be reproduced on one blank sheet, the chapter is ready; if one is missing, the topic page for it is the one to open.