Operated in reverse breakdown to give constant V_z — voltage regulator.
Photodiode & LED
Photodiode: light → current (reverse bias). LED: current → light (forward). λ = hc/E_g.
Solar cell
Unbiased illuminated junction generates EMF. E_g ~ 1-1.8 eV optimum.
Master Reference Tables
Quantity / Relationship
Formula / Value
Mass-action law
n_e · n_h = n_i²
Effective barrier (forward bias)
V_eff = V₀ − V
Effective barrier (reverse bias)
V_eff = V₀ + V
Dynamic resistance
r_d = ΔV / ΔI
Cut-in voltage V_γ
~0.2 V (Ge), ~0.7 V (Si)
Output frequency (half-wave)
f_out = f_in
Output frequency (full-wave)
f_out = 2 f_in
LED wavelength
λ = hc/E_g ≈ 1240 nm·eV / E_g(eV)
Zener regulator condition
V_in − I R_s = V_z
NCERT Exercises — Worked Solutions
Q 14.1 — In an n-type silicon, which of the following statements is true?
(a) Electrons are majority carriers and trivalent atoms are dopants.
(b) Electrons are minority carriers and pentavalent atoms are dopants.
(c) Holes are minority carriers and pentavalent atoms are dopants.
(d) Holes are majority carriers and trivalent atoms are dopants.
n-type ⇒ electrons are majority, holes are minority. Dopants are pentavalent (group V atoms like P, As, Sb).
Correct option: (c) — holes are minority carriers and pentavalent atoms are the dopants.
Q 14.2 — Which of the statements given in Exercise 14.1 is true for p-type semiconductors?
For p-type: holes are majority, electrons are minority. Dopants are trivalent (group III atoms like B, Al, In).
Correct option: (d) — holes are majority carriers and trivalent atoms are the dopants.
Q 14.3 — Carbon, silicon and germanium each have four valence electrons. Their valence and conduction bands are separated by energy band gaps (E_g)_C, (E_g)_Si, (E_g)_Ge. Which of the following is true?
(a) (E_g)_Si < (E_g)_Ge < (E_g)_C
(b) (E_g)_C < (E_g)_Ge > (E_g)_Si
(c) (E_g)_C > (E_g)_Si > (E_g)_Ge
(d) (E_g)_C = (E_g)_Si = (E_g)_Ge
This explains why C is an insulator while Si and Ge are semiconductors.
Q 14.4 — In an unbiased p-n junction, holes diffuse from the p-region to n-region because:
(a) free electrons in n-region attract them.
(b) they move across the junction by potential difference.
(c) hole concentration in p-region is more compared to n-region.
(d) all the above.
Diffusion is driven by concentration gradient, not by the field (the built-in field actually opposes hole motion p→n at equilibrium).
Correct option: (c) — hole concentration in p-region is more than in n-region.
Q 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.
Forward bias opposes the built-in potential V₀, reducing the effective barrier to V₀ − V. Majority carriers can now climb across more easily, giving a large forward current.
Correct option: (c) — lowers the potential barrier.
Q 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?
Half-wave rectifier: only one half of each input cycle produces output → one pulse per cycle.
\[ f_{out, HW} = f_{in} = \mathbf{50\ Hz} \]
Full-wave rectifier: both halves of each input cycle produce output → two pulses per cycle.
Q1 (MCQ). The depletion region width in a p-n junction:
(a) increases under forward bias
(b) decreases under forward bias
(c) is unaffected by bias
(d) becomes infinite at breakdown
(b) Forward bias narrows the depletion region.
Q2 (MCQ). At T = 0 K, an intrinsic semiconductor behaves like:
(a) Conductor
(b) Semiconductor
(c) Insulator
(d) Superconductor
(c) At T = 0 K, all valence electrons are in the VB, the CB is empty, and there are no thermally excited carriers — pure intrinsic Si or Ge behaves like an insulator.
Q3 (Short Answer). Why is the reverse current of a p-n junction diode nearly independent of the applied reverse voltage (until breakdown)?
The reverse current consists of minority carriers (holes in n-region, electrons in p-region) that wander into the depletion region and are swept across by the field. Even a small reverse voltage is sufficient to sweep them all. The current is therefore limited by the supply of minority carriers (set by temperature), not by the applied voltage. Hence reverse-saturation current is essentially V-independent.
Q4 (Numerical). A pure Si crystal at T = 300 K has n_i = 1.5 × 10¹⁶ /m³. It is doped with 10²² /m³ donor atoms. Find n_e and n_h.
N_D = 10²² >> n_i, so n_e ≈ N_D = 10²² /m³. From mass action: n_h = n_i²/n_e = (1.5×10¹⁶)²/10²² = 2.25 × 10¹⁰ /m³.
Q5 (HOTS). Why is GaAs (E_g = 1.43 eV) preferred over Si (E_g = 1.12 eV) for high-efficiency LEDs?
(i) GaAs is a direct band-gap semiconductor: an electron in the CB minimum and a hole in the VB maximum have the same crystal momentum, so they can recombine by emitting a photon directly (radiative recombination is efficient). Si is an indirect band-gap material — recombination requires a phonon to balance momentum, making it slow and mostly non-radiative. (ii) GaAs band gap of 1.43 eV gives near-IR (~870 nm) photons; alloying with phosphorus (GaAsP) shifts emission into the visible. Si emits at 1100 nm with very low efficiency.
Final Assertion–Reason Round
Options: (A) Both true, R correct explanation. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.
Assertion: The conductivity of a semiconductor increases with temperature.
Reason: The number of charge carriers in a semiconductor increases exponentially with T because more electrons gain enough thermal energy to cross E_g.
(A) Both correct, and the reason explains the assertion. n_i ∝ T^(3/2) e^(−E_g/2kT) — the exponential growth in carriers dominates the slight reduction in mobility from phonon scattering.
Assertion: A solar cell can power a load directly without a battery.
Reason: The built-in junction field separates photo-generated carriers and creates an EMF.
(A) Both correct, and the reason explains the assertion.
Assertion: Avalanche breakdown in a Zener diode destroys it.
Reason: The current at breakdown rises sharply and the diode dissipates excessive heat.
(D) Assertion is false: Zener diodes are designed to operate safely at breakdown, provided the current is limited externally by a series resistor. Reason is true (uncontrolled current would damage any diode by overheating).
Interactive — Combined Concept Check
Pick a question type
Mix-and-match practice — see the answer immediately.
Where to next? Chapter 14 ends with diodes and rectifiers — the simplest two-terminal devices. The transistor (three-terminal device) and integrated circuits build on the same p-n junction physics to deliver amplification, switching and the digital logic that powers every computer. Welcome to the foundations of modern electronics.
Frequently Asked Questions - NCERT Exercises and Solutions: Semiconductor Electronics
What are the key NCERT exercise types in Chapter 14 Semiconductor Electronics?
NCERT Class 12 Physics Chapter 14 Semiconductor Electronics exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Semiconductor Electronics?
For numerical problems in NCERT Class 12 Physics Chapter 14 Semiconductor Electronics: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 14?
From NCERT Class 12 Physics Chapter 14 (Semiconductor Electronics), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 14 Semiconductor Electronics problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 14 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 14 Semiconductor Electronics exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Physics Chapter 14 Semiconductor Electronics solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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