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Class 12 · Physics · Chapter 14

Semiconductor Electronics: Materials, Devices and Simple Circuits: Class 12 Important Questions with Answers

81 important questions for Class 12 Physics Chapter 14, Semiconductor Electronics: Materials, Devices and Simple Circuits, the way they come in the board exam: MCQs, assertion–reason, case-based and short and long answer questions, with answers. Taken from our Class 12 Physics notes for 2026-27.

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Jump to: n_e n_h = n_i² numericalsPractice set: test yourself20 MCQsAssertion–ReasonAnswers to the practice set

n_e n_h = n_i² numericals

Method: (1) majority carriers ≈ dopant concentration (ppm × atoms per m³); (2) minority = n_i²/majority; (3) more electrons → n-type, more holes → p-type.

NCERT Example 14.2 Si (5 × 10²⁸ atoms/m³) doped with 1 ppm of As; n_i = 1.5 × 10¹⁶ m⁻³. n_e and n_h?

n_e ≈ N_D = 5 × 10²⁸ × 10⁻⁶ = 5 × 10²² m⁻³; n_h = (1.5 × 10¹⁶)²/5 × 10²² = 4.5 × 10⁹ m⁻³.

CBSE 2024 · 55/3/1 Same Si doped with 1 ppm of boron. n_h, n_e? Type? [2]

Full answer in the notes pack

CBSE 2026 · 55/5/1 Si doped with 5 × 10²² m⁻³ of arsenic, n_i = 1.5 × 10¹⁶ m⁻³. Majority and minority concentrations? [2]

Majority (electrons) 5 × 10²² m⁻³; minority (holes) 4.5 × 10⁹ m⁻³.

CBSE 2024 · 55/5/1 A pure Si crystal having 5 × 10²⁸ atoms m⁻³ is doped with 1 ppm concentration of antimony. If the concentration of holes in the doped crystal is found to be 4.5 × 10⁹ m⁻³, the concentration (in m⁻³) of intrinsic charge carriers in Si crystal is about: (A) 1.2 × 10¹⁵ (B) 1.5 × 10¹⁶ (C) 3.0 × 10¹⁵ (D) 2.0 × 10¹⁶ [1]

Full answer in the notes pack

CBSE 2025 · 55/4/1 An intrinsic semiconductor has n_i = 5 × 10⁸ m⁻³; after doping n_h = 8 × 10¹² m⁻³. (a) Type of dopant and semiconductor? (b) n_e? [2]

(a) trivalent; p-type. (b) n_e = 25 × 10¹⁶/8 × 10¹² = 3.125 × 10⁴ m⁻³.

CBSE 2026 · 55/4/1 n_h in an intrinsic semiconductor is 5 × 10⁸ m⁻³. After doping n_e = 4 × 10¹² m⁻³. New n_h? Type? [2]

Full answer in the notes pack

CBSE 2025 · 55/1/1 p-type Si: one dopant atom per 5 × 10⁷ Si atoms; Si has 5 × 10²⁸ atoms/m³. Holes per cm³? One such dopant? [2]

5 × 10²⁸/5 × 10⁷ = 10²¹ m⁻³ = 10¹⁵ cm⁻³; dopant: Al, In or Ga.

CBSE 2026 · 55/2/1 Case (ii): n_e = 5 × 10²² m⁻³, n_i = 1.5 × 10¹⁶ m⁻³. n_h? [1]

Full answer in the notes pack

CBSE 2024 · 55/4/1 Case (ii): n_i = 2 × 10¹⁰ cm⁻³; after pentavalent doping holes = 8 × 10³ cm⁻³. n_e? [1]

4 × 10²⁰/8 × 10³ = 5 × 10¹⁶ cm⁻³ = (D) 5 × 10²² m⁻³.

Practice set: test yourself

Try every question before looking at the answers below. 64 or more out of 80 means you are ready for the board.

A. Fill in the blanks (1 mark each)

  1. The energy gap of Si is about ______ eV.
  2. In n-type, the majority carriers are ______.
  3. Boron is a(n) ______ impurity.
  4. n_e n_h = ______.
  5. The barrier potential of Si is about ______ V.
  6. In forward bias the depletion layer becomes ______.
  7. Full-wave output frequency for 50 Hz input: ______.
  8. Reverse current is due to ______ carriers.

B. True or false? (1 mark each)

  1. An intrinsic semiconductor is an insulator at 0 K.
  2. p-type semiconductors are positively charged.
  3. Doping distorts the crystal lattice.
  4. Drift current flows due to the junction field.
  5. In reverse bias the barrier height decreases.
  6. A diode obeys Ohm's law.
  7. Conductivity of a semiconductor rises with temperature.
  8. In a full-wave rectifier both diodes conduct together.

C. Very short answer (1 mark each)

  1. Define a hole.
  2. Name one pentavalent dopant.
  3. What is the depletion region?
  4. What is the knee voltage?
  5. Why is Ge doped only slightly?
  6. What is rectification?
  7. Which current dominates in forward bias?
  8. Value of E_g for diamond?
  9. What is breakdown voltage?
  10. Write the symbol of a diode.

D. Short answer I (2 marks each)

  1. Distinguish intrinsic and extrinsic semiconductors.
  2. Si (5 × 10²⁸) doped with 2 ppm of P; n_i = 1.5 × 10¹⁶ m⁻³. Find n_e and n_h.
  3. Draw band diagrams of a metal and an insulator.
  4. Why can't we join two slabs to make a p-n junction?
  5. Draw the circuit for reverse-bias characteristics.
  6. Explain why the reverse current is almost independent of voltage.
  7. Draw input and output waveforms of a half-wave rectifier.
  8. Find the dynamic resistance if ΔV = 0.1 V gives ΔI = 5 mA.
My score (fill in after checking the answers)
SectionA /8B /8C /10D /16E /18F /20G (bonus)Total /80
My marks

Rough work

E. Short answer II (3 marks each)

  1. Explain the formation of the depletion region and the barrier potential.
  2. Explain n-type and p-type semiconductors with band diagrams.
  3. Explain forward and reverse bias with circuits and the effect on the barrier.
  4. Draw and explain the V–I characteristic of a Si diode.
  5. Explain a full-wave rectifier with circuit and waveforms.
  6. Distinguish metals, semiconductors and insulators on the basis of bands.

F. Long answer (5 marks each)

  1. (a) How are holes created in an intrinsic semiconductor? (b) Explain p-type doping with a band diagram. (c) Si (5 × 10²⁸) doped with 1 ppm of In; n_i = 1.5 × 10¹⁶ m⁻³: find n_e and n_h and the type.
  2. (a) Explain the formation of a p-n junction. (b) Draw the V–I characteristic and explain it. (c) Explain the full-wave rectifier with waveforms.

G. HOTS (higher-order thinking)

  1. Why is Si preferred over Ge for diodes, though Ge has a smaller gap?
  2. Why does a reverse-biased junction behave like a capacitor?
  3. What happens to the output of a full-wave rectifier if one diode is open?
  4. Why is the conductivity of n-type higher than p-type for the same doping?

20 MCQs

  1. Energy gap of an insulator
    (a) < 1 eV (b) 1.1 eV (c) > 3 eV (d) 0.7 eV
  2. Donor impurity
    (a) arsenic (b) boron (c) indium (d) aluminium
  3. Majority carriers in p-type
    (a) electrons (b) protons (c) ions (d) holes
  4. n_e n_h equals
    (a) n_i (b) n_i² (c) 2n_i (d) √n_i
  5. Forward bias: barrier
    (a) increases (b) decreases (c) unchanged (d) vanishes always
  6. Reverse current is of the order of
    (a) A (b) mA (c) kA (d) μA
  7. Depletion region contains
    (a) immobile ions (b) holes (c) free electrons (d) nothing
  8. Full-wave rectifier output frequency (60 Hz in)
    (a) 30 Hz (b) 60 Hz (c) 120 Hz (d) 240 Hz
  9. At 0 K, pure Si is
    (a) conductor (b) n-type (c) superconductor (d) insulator
  10. Barrier potential of Si
    (a) 0.3 V (b) 0.7 V (c) 1.1 V (d) 3 V
  11. Drift current is due to
    (a) concentration gradient (b) magnetic field (c) junction field (d) heat
  12. In an unbiased junction at equilibrium
    (a) diffusion = drift (b) only drift (c) only diffusion (d) no carriers
  13. Doping with In gives
    (a) n-type (b) p-type (c) intrinsic (d) insulator
  14. Energy to free a donor electron in Si
    (a) 1.1 eV (b) 5 eV (c) 0.05 eV (d) 0.7 eV
  15. A diode is used as
    (a) amplifier (b) oscillator (c) transformer (d) rectifier
  16. Reverse bias: depletion width
    (a) increases (b) decreases (c) zero (d) unchanged
  17. In a full-wave rectifier each diode conducts for
    (a) half cycle (b) full cycle (c) quarter cycle (d) never
  18. Minority carrier injection occurs in
    (a) breakdown only (b) reverse bias (c) no bias (d) forward bias
  19. E_g order
    (a) Ge > Si > C (b) C > Si > Ge (c) Si > C > Ge (d) all equal
  20. n-type semiconductor is
    (a) negative (b) positive (c) neutral (d) depends on doping

1 (c) · 2 (a) · 3 (d) · 4 (b) · 5 (b) · 6 (d) · 7 (a) · 8 (c) · 9 (d) · 10 (b) · 11 (c) · 12 (a) · 13 (b) · 14 (c) · 15 (d) · 16 (a) · 17 (a) · 18 (d) · 19 (b) · 20 (c)

Assertion–Reason (6)

(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

  1. A: The reverse current of a diode is very small. R: It is due to minority carriers.
  2. A: Si is a semiconductor while C is an insulator. R: The energy gap of C is much larger.
  3. A: A p-type semiconductor is positively charged. R: Its majority carriers are holes.
  4. A: Full-wave output has twice the input frequency. R: Output comes in both half-cycles.
  5. A: In forward bias the depletion layer widens. R: The applied voltage opposes the barrier.
  6. A: Doping increases conductivity. R: It increases the number of charge carriers.

Key: 1 (a) · 2 (a) · 3 (d) · 4 (a) · 5 (d) · 6 (a)

Answers to the practice set

A. 1. 1.1 2. electrons 3. acceptor (trivalent) 4. n_i² 5. 0.7 6. thinner 7. 100 Hz 8. minority

B. 1. T 2. F 3. F 4. T 5. F 6. F 7. T 8. F

C. 1. a vacancy in a bond acting as +e 2. As / P / Sb 3. a region near the junction with immobile ions and no free carriers 4. the forward voltage after which current rises sharply 5. heavy doping would distort the lattice and destroy semiconducting behaviour 6. ac → dc 7. diffusion 8. 5.4 eV 9. reverse voltage at which current rises suddenly 10. ▷| (arrow = conventional current)

D. 1. p. 3 2. n_e = 10²³ m⁻³, n_h = 2.25 × 10⁹ m⁻³ 3. p. 2 4. roughness breaks continuity 5. p. 8 (μA meter, p to −) 6. minority carriers limited 7. p. 11 8. 20 Ω

Answers to E, F, G, case study (short, long and HOTS answers, step by step) are in the notes pack

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