NCERT Solutions for Class 12 Physics Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits

These Class 12 Physics Chapter 14 solutions cover Semiconductor Electronics: Materials, Devices and Simple Circuits from the latest NCERT textbook (session 2026–27). Every NCERT “Exercises” question (14.1–14.6) is reproduced exactly as in the book and solved step by step, followed by key formulas, extra practice, MCQs and assertion–reason questions with answer keys. This is the last chapter of the Class 12 Physics course.

Class: 12 Subject: Physics Chapter: 14 Topic: Semiconductor Electronics Exercises: 14.1 – 14.6 Session: 2026–27
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Chapter Overview

Chapter 14 introduces solid-state semiconductor electronics, the foundation of all modern devices. On the basis of conductivity, solids are classified as metals (very low resistivity), semiconductors (intermediate) and insulators (very high resistivity); on the basis of energy bands, the gap between the valence band and the conduction band (Eg) decides the category — Eg ≈ 0 for metals, Eg < 3 eV for semiconductors and Eg > 3 eV for insulators. A pure semiconductor (Si, Ge) is intrinsic, with equal electrons and holes (ne = nh = ni). Doping creates extrinsic semiconductors: pentavalent donors give n-type (electrons majority), trivalent acceptors give p-type (holes majority), with nenh = ni2 always holding. A p-n junction forms a depletion region and barrier potential; under forward bias the barrier lowers and large current flows, under reverse bias the barrier rises and only a tiny saturation current flows. This one-way conduction is used to build half-wave and full-wave rectifiers that convert ac to dc.

Key Concepts & Definitions

Intrinsic semiconductor: a pure semiconductor (Si, Ge) in which the number of free electrons equals the number of holes, both produced by thermal excitation: ne = nh = ni.

Doping: the deliberate addition of a small amount of suitable impurity (dopant) to a pure semiconductor to increase its conductivity, giving an extrinsic semiconductor.

n-type semiconductor: Si/Ge doped with a pentavalent donor (As, Sb, P); electrons are majority carriers and holes are minority carriers (ne >> nh).

p-type semiconductor: Si/Ge doped with a trivalent acceptor (In, B, Al); holes are majority carriers and electrons are minority carriers (nh >> ne).

Depletion region: a region near the p-n junction depleted of free charge carriers, containing immobile ionised cores, which sets up the barrier potential V0.

Forward bias: p-side to +ve terminal, n-side to −ve terminal — barrier height becomes (V0 − V), depletion width decreases, large current (mA) flows.

Reverse bias: n-side to +ve terminal, p-side to −ve terminal — barrier height becomes (V0 + V), depletion width increases, only a small reverse saturation current (µA) flows.

Rectifier: a circuit using the one-way conduction of a diode to convert ac to dc — half-wave (one diode) or full-wave (centre-tap, two diodes).

Important Formulas

Intrinsic carrier relation: ne = nh = ni

Mass-action law (all cases): nenh = ni2

Total current in a semiconductor: I = Ie + Ih

Effective barrier (forward bias): Vbarrier = V0 − V

Effective barrier (reverse bias): Vbarrier = V0 + V

Dynamic resistance of a diode: rd = ΔV / ΔI

Rectifier output frequency: half-wave → fout = fin; full-wave → fout = 2 fin

NCERT Exercises (14.1–14.6) — Solutions

14.1 In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.

ANSWER Correct option: (c) In n-type silicon the dopant is a pentavalent atom (As, Sb, P), which donates one extra electron. Hence electrons are the majority carriers and holes are the minority carriers. Therefore the only true statement is (c) — holes are minority carriers and pentavalent atoms are the dopants. Option (a) and (d) wrongly use trivalent dopants, and (b) wrongly calls electrons minority carriers.

14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductors.

ANSWER Correct option: (d) In a p-type semiconductor the dopant is a trivalent atom (In, B, Al), which creates a hole. Hence holes are the majority carriers and electrons are the minority carriers. The statement that matches this is (d) — holes are majority carriers and trivalent atoms are the dopants.

14.3 Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge

ANSWER Correct option: (c) The energy band gaps are approximately: carbon (diamond) Eg ≈ 5.4 eV, silicon Eg ≈ 1.1 eV and germanium Eg ≈ 0.7 eV. Therefore (Eg)C > (Eg)Si > (Eg)Ge, which is option (c). This is why carbon behaves as an insulator while Si and Ge behave as semiconductors.

14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above.

ANSWER Correct option: (c) In an unbiased (zero-voltage) p-n junction there is no external potential difference, so motion is driven only by the concentration gradient. The hole concentration is much higher in the p-region than in the n-region, so holes diffuse from high to low concentration, i.e. from p → n. Hence the correct reason is (c).

14.5 When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lowers the potential barrier. (d) None of the above.

ANSWER Correct option: (c) Under forward bias the applied voltage V opposes the built-in potential V0, so the effective barrier height becomes (V0 − V). The depletion width decreases and the barrier is lowered, allowing a large majority-carrier current to flow. Hence (c) is correct (option (a) is opposite, and (b) is wrong because majority current actually increases).

14.6 In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.

ANSWER Given: input frequency fin = 50 Hz. Half-wave rectifier: it conducts during only one half of each input cycle, so one output pulse is produced per input cycle. Hence output frequency = input frequency: fout (half-wave) = fin = 50 Hz. Full-wave rectifier: it conducts during both halves of each input cycle, so two output pulses are produced per input cycle. Hence output frequency = twice the input frequency: fout (full-wave) = 2 × fin = 2 × 50 = 100 Hz. Result: output is 50 Hz for a half-wave rectifier and 100 Hz for a full-wave rectifier.

Extra Practice Questions

Short Answer Type Questions

Q1. State the mass-action law for a semiconductor in thermal equilibrium and explain one consequence.

ANSWERThe mass-action law states nenh = ni2, where ni is the intrinsic carrier concentration. A consequence is that increasing the majority-carrier concentration by doping automatically reduces the minority-carrier concentration, because their product must stay constant at a given temperature.

Q2. Why does the conductivity of an intrinsic semiconductor increase with temperature?

ANSWERAs temperature rises, more thermal energy becomes available to break covalent bonds, generating additional electron–hole pairs. More charge carriers means higher conductivity (lower resistivity). At T = 0 K an intrinsic semiconductor behaves like an insulator because no carriers are excited across the gap.

Q3. Distinguish between diffusion current and drift current in a p-n junction.

ANSWERDiffusion current is caused by the concentration gradient: majority carriers diffuse across the junction (holes p→n, electrons n→p). Drift current is caused by the junction electric field acting on minority carriers, and flows opposite to the diffusion current. At equilibrium the two are equal and opposite, so net current is zero.

Q4. Define threshold (cut-in) voltage and give its approximate value for Si and Ge diodes.

ANSWERThe threshold or cut-in voltage is the forward voltage beyond which the diode current rises sharply (almost exponentially). It is about 0.7 V for a silicon diode and about 0.2 V for a germanium diode.

Q5. Why is a capacitor connected across the load in a rectifier circuit?

ANSWERThe rectified output is unidirectional but pulsating. A capacitor connected in parallel with the load acts as a filter: it charges to the peak voltage and slowly discharges through the load between pulses, smoothing the ripples to give a steadier (nearly dc) output. A large capacitance gives a large time constant (RLC) and a smoother output.

Long Answer Type Questions

Q1. Explain the formation of the depletion region and barrier potential in a p-n junction.

ANSWERWhen a p-n junction is formed, the large concentration gradient causes holes to diffuse from p→n and electrons from n→p. As electrons leave the n-side, they uncover immobile positive donor ions; as holes leave the p-side, they uncover immobile negative acceptor ions. This creates a region near the junction depleted of free carriers, the depletion region (about a tenth of a micrometre thick). The exposed ion charges set up an electric field directed from the n-side to the p-side. This field opposes further diffusion and is the source of the barrier potential V0, which makes the n-side positive relative to the p-side. Equilibrium is reached when the diffusion current is exactly balanced by the drift current, giving zero net current.

Q2. With the help of a circuit description, explain the working of a full-wave rectifier and compare it with a half-wave rectifier.

ANSWERA full-wave rectifier uses a centre-tap transformer and two diodes D1 and D2. The p-sides of the diodes connect to the two ends (A and B) of the secondary, and their n-sides join to give the output across the load RL taken between this common point and the centre tap. The voltages at A and B are always out of phase. During the half cycle when A is positive, D1 is forward biased and conducts while D2 is reverse biased; during the next half cycle B is positive, so D2 conducts and D1 is off. In both halves current flows through RL in the same direction, so output is obtained for the full ac cycle. A half-wave rectifier uses only one diode and conducts in only one half of each cycle, wasting the other half. Hence the full-wave rectifier is more efficient and gives an output frequency of 2f (against f for half-wave), with less ripple and easier filtering.

Q3. Compare the behaviour of a p-n junction diode under forward bias and reverse bias.

ANSWERUnder forward bias (p to +ve, n to −ve), the applied voltage opposes V0, so the effective barrier (V0 − V) decreases and the depletion width shrinks. Many majority carriers gain enough energy to cross the junction (minority-carrier injection), giving a large current of the order of milliamperes; the forward resistance is low. Under reverse bias (n to +ve, p to −ve), the applied voltage adds to V0, so the effective barrier (V0 + V) increases and the depletion width widens. The diffusion current is suppressed and only a small, nearly voltage-independent reverse saturation current of a few microamperes flows due to minority-carrier drift; the reverse resistance is very high. Beyond a critical breakdown voltage Vbr, the reverse current increases sharply. This asymmetry makes the diode a one-way conductor, used for rectification.

MCQs & Assertion–Reason

1. The energy band gap is largest for:

(a) germanium    (b) silicon    (c) carbon (diamond)    (d) tin

2. In a p-type semiconductor, the majority carriers are:

(a) electrons    (b) holes    (c) protons    (d) neutrons

3. The dopant used to obtain an n-type semiconductor is:

(a) boron    (b) indium    (c) aluminium    (d) arsenic

4. For an intrinsic semiconductor at thermal equilibrium:

(a) ne > nh    (b) ne < nh    (c) ne = nh    (d) ne = 0

5. Under forward bias, the width of the depletion region:

(a) increases    (b) decreases    (c) stays the same    (d) becomes infinite

6. The reverse saturation current in a diode is of the order of:

(a) amperes    (b) milliamperes    (c) microamperes    (d) kiloamperes

7. The cut-in (threshold) voltage of a silicon diode is approximately:

(a) 0.2 V    (b) 0.7 V    (c) 1.1 V    (d) 5.4 V

8. If the input frequency to a full-wave rectifier is 60 Hz, the output frequency is:

(a) 30 Hz    (b) 60 Hz    (c) 120 Hz    (d) 240 Hz

9. In all semiconductors at equilibrium, the relation nenh equals:

(a) ni    (b) ni2    (c) 2ni    (d) zero

10. A capacitor connected across the output of a rectifier acts as a:

(a) rectifier    (b) amplifier    (c) filter    (d) oscillator

Answer key: 1-(c), 2-(b), 3-(d), 4-(c), 5-(b), 6-(c), 7-(b), 8-(c), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: In an n-type semiconductor, electrons are the majority carriers.

Reason: An n-type semiconductor is doped with a pentavalent donor impurity that supplies extra electrons.

A-R 2. Assertion: A forward-biased p-n junction offers low resistance.

Reason: Forward bias raises the potential barrier and widens the depletion region.

A-R 3. Assertion: A full-wave rectifier gives an output frequency twice that of the input.

Reason: It produces an output pulse during both halves of each input ac cycle.

A-R 4. Assertion: Carbon (diamond) behaves as an insulator while silicon behaves as a semiconductor.

Reason: The energy band gap of carbon is much larger than that of silicon.

A-R 5. Assertion: In an unbiased p-n junction, holes diffuse from the p-region to the n-region.

Reason: The hole concentration is higher in the p-region than in the n-region.

Answer key: 1-(A), 2-(C), 3-(A), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Confusing the dopants — pentavalent (As, Sb, P) gives n-type; trivalent (In, B, Al) gives p-type. Do not swap them.
  • Thinking n-type material is negatively charged or p-type is positively charged — the crystal stays overall electrically neutral.
  • Writing that forward bias raises the barrier — it lowers it to (V0 − V); reverse bias raises it to (V0 + V).
  • Forgetting that nenh = ni2 holds in all cases, not only for intrinsic semiconductors.
  • Mixing up the rectifier frequencies: half-wave output = fin; full-wave output = 2fin.
  • Mixing diffusion current (concentration-driven, majority carriers) with drift current (field-driven, minority carriers).

Exam tips for this chapter

Memorise the three approximate band gaps (C ≈ 5.4 eV, Si ≈ 1.1 eV, Ge ≈ 0.7 eV) and the two threshold voltages (Si ≈ 0.7 V, Ge ≈ 0.2 V) — these are favourite one-mark questions. For diode questions, always state the effective barrier as (V0 − V) for forward and (V0 + V) for reverse bias. When asked about rectifiers, mention the number of diodes, which half of the cycle each conducts, the output frequency relation, and the role of the filter capacitor. This is the last chapter of the syllabus, so quick, accurate recall of these standard facts secures easy marks.

Frequently Asked Questions

What is Class 12 Physics Chapter 14 about?

Chapter 14, Semiconductor Electronics: Materials, Devices and Simple Circuits, explains the classification of solids by conductivity and energy bands, intrinsic and extrinsic (n-type and p-type) semiconductors, the p-n junction and its behaviour under forward and reverse bias, and the use of diodes as half-wave and full-wave rectifiers.

What is the output frequency of half-wave and full-wave rectifiers for a 50 Hz input?

A half-wave rectifier gives an output frequency equal to the input, 50 Hz, while a full-wave rectifier gives twice the input, 100 Hz, because it produces an output pulse in both halves of each ac cycle.

Why is carbon an insulator while silicon and germanium are semiconductors?

The energy band gap of carbon (diamond) is about 5.4 eV, far larger than silicon (1.1 eV) and germanium (0.7 eV). The large gap prevents thermal excitation of electrons into the conduction band, so carbon behaves as an insulator while Si and Ge conduct slightly as semiconductors.

Are these Class 12 Physics Chapter 14 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Physics are free and follow the official NCERT textbook for session 2026-27, with every exercise solved step by step.

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