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Semiconductor Electronics: Materials, Devices and Simple Circuits – Class 12 Physics notes, important questions and MCQs

NCERT Class 12 Physics, chapter 14. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Semiconductors like silicon have conductivity between conductors and insulators. Doping makes n- and p-type materials, and their junction (diode) is the basis of rectifiers and electronic circuits.

Semiconductors and doping

Energy gap: conductors ~0, semiconductors ~1 eV (Si 1.1 eV), insulators > 3 eV.
n-type: pentavalent dopant (P, As), electrons majority; p-type: trivalent dopant (B, Al), holes majority.
Intrinsic: nₑ = nₕ = nᵢ; extrinsic: nₑnₕ = nᵢ².

p–n junction and uses

Depletion layer and barrier potential form at the junction.
Forward bias: low resistance, current flows; reverse bias: very small current until breakdown.
Half-wave rectifier (one diode), full-wave rectifier (two diodes or bridge); Zener diode as voltage regulator; LEDs and solar cells.

Key terms

Doping: adding small amounts of impurity to a semiconductor
Depletion region: region near the junction without free charge carriers
Forward bias: p-side connected to positive and n-side to negative terminal, reducing the barrier so current flows

Formulas

nₑnₕ = nᵢ²
Full-wave output frequency = 2 × input frequency

Important questions with answers

Why does the resistance of a semiconductor decrease with temperature?2 marks

Heating breaks more covalent bonds and creates more electron–hole pairs, increasing the number of charge carriers, so conductivity rises and resistance falls.

How is the depletion region formed in a p–n junction?3 marks

At the junction, electrons diffuse from n to p and holes from p to n, and they recombine. This leaves uncovered positive donor ions on the n-side and negative acceptor ions on the p-side, forming a layer with no free carriers and a barrier potential that stops further diffusion.

Explain the working of a full-wave rectifier with a centre-tapped transformer and draw (describe) its input and output waveforms. Why is a filter capacitor used?5 marks

The secondary has a centre tap; diodes D₁ and D₂ connect its two ends to the load, with the load returning to the centre tap. In the positive half cycle, D₁ is forward biased and conducts while D₂ is reverse biased; in the negative half cycle, D₂ conducts and D₁ is off. In both halves current flows through the load in the same direction. Input is a sine wave; output is a series of positive half-wave pulses with no gaps (pulsating DC at twice the input frequency). A capacitor across the load charges at the peaks and discharges slowly between them, smoothing the ripple to give nearly steady DC.

Quick revision

n-type: electrons; p-type: holes.
Forward bias conducts.
Zener works in reverse breakdown.

Exam tips

Ripple frequency for 50 Hz mains: 50 Hz (half-wave), 100 Hz (full-wave).
In intrinsic semiconductors nₑ = nₕ; in n-type nₑ ≫ nₕ (donor, group 15); in p-type nₕ ≫ nₑ (acceptor, group 13).

Practice MCQs

Band gap of silicon is about
1. 0 eV
2. 1.1 eV
3. 5 eV
4. 13.6 eV
Show answer
Answer: (b) 1.1 eV. Semiconductor.
Adding phosphorus to silicon makes
1. p-type
2. n-type
3. insulator
4. conductor
Show answer
Answer: (b) n-type. Pentavalent donor.
In forward bias, the depletion layer
1. narrows
2. widens
3. stays same
4. disappears permanently
Show answer
Answer: (a) narrows. Barrier lowers.
A Zener diode is used as a
1. rectifier
2. LED
3. amplifier only
4. voltage regulator
Show answer
Answer: (d) voltage regulator. Reverse breakdown.
Silicon doped with phosphorus becomes
1. p-type
2. insulator
3. intrinsic
4. n-type
Show answer
Answer: (d) n-type. Pentavalent donor.
Output frequency of a full-wave rectifier on 50 Hz mains is
1. 25 Hz
2. 50 Hz
3. 100 Hz
4. 200 Hz
Show answer
Answer: (c) 100 Hz. Both halves rectified.
In an intrinsic semiconductor at room temperature
1. electrons > holes
2. holes > electrons
3. no carriers
4. electrons = holes
Show answer
Answer: (d) electrons = holes. Pairs generated together.
The barrier potential of a silicon p–n junction is about
1. 0.1 V
2. 0.7 V
3. 1.5 V
4. 5 V
Show answer
Answer: (b) 0.7 V. Si ≈ 0.7 V, Ge ≈ 0.3 V.

All chapters of Class 12 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.