NCERT Solutions for Class 12 Physics Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits
All 6 exercise questions of the latest NCERT textbook (2025–26 reprint) answered with full elaborate treatment of every option — n-type and p-type semiconductors, energy bands of C, Si and Ge, the p-n junction under zero and forward bias and half-wave and full-wave rectification — with diagrams linked to the exact NCERT pages, comparison tables, complete revision notes and exam tips.
6 questions6 diagrams9 tablesFull revision notes
Chapter at a glance — complete revision notes
14.1 Introduction
Devices in which a controlled flow of electrons can be obtained are the building blocks of all electronic circuits. Before the transistor (1948) such devices were vacuum tubes (valves): the vacuum diode (anode/plate and cathode), triode (cathode, plate, grid), tetrode and pentode. In them electrons are supplied by a heated cathode and flow in vacuum (so as not to lose energy in collisions with air molecules) only in one direction — hence “valves”. They are bulky, consume high power, operate at high voltages (~100 V) and have a limited life and low reliability.
The seed of solid-state semiconductor electronics goes back to the 1930s: simple excitations (light, heat, a small applied voltage) can change the number of mobile charges in a semiconductor; the charge carriers are supplied and flow within the solid — no heating or evacuated space — so semiconductor devices are small, consume low power, operate at low voltages and have long life and high reliability. CRT displays are being replaced by LCD monitors with solid-state electronics. A naturally occurring crystal of galena (PbS) with a metal point contact had been used as a detector of radio waves long before the theory was understood. The chapter introduces junction diodes (two-electrode) and mentions the bipolar junction transistor (three-electrode).
14.2 Classification of metals, conductors and semiconductors
On the basis of conductivity/resistivity: (i) metals — very low resistivity (10⁻² to 10⁻⁸ Ω m), high conductivity; (ii) semiconductors — resistivity/conductivity intermediate between metals and insulators; (iii) insulators — high resistivity (> 10⁸ Ω m). Other differences also matter (e.g. the temperature dependence and band structure). Elemental semiconductors: Si, Ge; compound semiconductors: inorganic (CdS, GaAs, CdSe, InP …) and organic (anthracene, doped phthalocyanines) and organic polymers (polypyrrole, polyaniline, polythiophene) — devices based on organic semiconductors and semiconducting polymers appeared after 1990 (polymer and molecular electronics). The chapter restricts itself to inorganic Si and Ge.
Energy bands: in a crystal the outer orbits of neighbouring atoms overlap; each electron has a slightly different energy and the closely spaced levels form energy bands. The band containing the valence electrons’ energies is the valence band; the band above it is the conduction band. For Si/Ge (N atoms, 4 valence electrons each): 4N valence electrons and 8N available states; at the interatomic spacing of the crystal the 8N states split into two bands separated by an energy gap E_g (Fig. 14.1): the lower (valence) band is completely filled at 0 K by the 4N electrons, the upper (conduction) band, with 4N states, is completely empty.
Three cases (Fig. 14.2): (a) metals — the conduction band partly filled, or the valence and conduction bands overlap (no gap), so electrons move freely: very low resistance; (b) insulators — a large gap E_g > 3 eV, no electrons in the conduction band, electrons cannot be excited thermally: no conduction; (c) semiconductors — a finite but small gap E_g < 3 eV: at room temperature some valence electrons gain enough energy to cross into the conduction band, leaving vacancies (holes) in the valence band; conduction by both. Example 14.1: C (diamond), Si and Ge have the same lattice structure but their bonding electrons lie in the 2nd, 3rd and 4th orbits so the energy needed to free an electron is greatest for C, then Si, least for Ge; hence the numbers of free electrons are significant in Ge and Si, but negligible in C. Energy gaps: C 5.4 eV, Si 1.1 eV, Ge 0.7 eV (Sn, also group IV, is a metal because its gap is 0 eV).
14.3 Intrinsic semiconductor
Pure Si and Ge have a diamond-like lattice (Fig. 14.3): each atom shares one of its four valence electrons with each of its four neighbours — covalent bonds (Fig. 14.4, idealised at low temperature: no bonds broken; a pure semiconductor behaves as an insulator at 0 K). As the temperature rises thermal energy breaks a few bonds: a free electron (charge −q) and a vacancy with an effective charge +q, a hole (Fig. 14.5a) — a hole behaves as an apparent free particle with positive charge. In an intrinsic semiconductor n_e = n_h = n_i (intrinsic carrier concentration). A hole moves when an electron from a neighbouring bond jumps into it (Fig. 14.5b); under an electric field the electrons give the electron current I_e and the holes the hole current I_h: I = I_e + I_h. Generation of electron–hole pairs is balanced at equilibrium by recombination (an electron colliding with a hole). Conductivity depends on the temperature but is very low at room temperature, so no important devices can be made from intrinsic semiconductors — hence doping.
14.4 Extrinsic semiconductor
Adding a few parts per million (ppm) of a suitable impurity (doping; the impurity atoms are the dopants) increases the conductivity manifold — an extrinsic (impurity) semiconductor. The dopant should not distort the lattice: it occupies only a very few lattice sites and its size should be nearly the same as that of Si or Ge. Two types of dopants:
(i) n-type: doping Si/Ge with a pentavalent impurity (As, Sb, P) — four of its five valence electrons form covalent bonds, the fifth is loosely bound (needs only ~0.01 eV to be freed, as compared with 0.72 eV for Ge and 1.1 eV for Si); the dopant is a donor (donates one electron, Fig. 14.7). The number of electrons contributed by the dopant depends on the doping level, not on temperature. n_e ≫ n_h: electrons are the majority carriers, holes the minority carriers; n_e n_h = n_i².
(ii) p-type: doping with a trivalent impurity (B, Al, In, Ga) — three covalent bonds are formed and the fourth bond has a vacancy; a neighbouring electron can fill it leaving a hole at its own site; the trivalent atom becomes effectively negatively charged (core of one negative charge with its hole, Fig. 14.8); it is an acceptor; one acceptor atom gives one hole. n_h ≫ n_e: holes are the majority carriers and electrons the minority carriers. The crystal as a whole remains electrically neutral. Because of the abundance of majority carriers, the minority carriers are destroyed by recombination and their concentration is reduced below n_i.
Energy bands (Fig. 14.9): an n-type has a donor level E_D just below the conduction band E_C (electrons enter the conduction band with a very small energy); a p-type has an acceptor level E_A just above the valence band E_V (electrons from the valence band jump into it with a very small energy, leaving holes). At room temperature most donor/acceptor atoms are ionised. Example 14.2: pure Si with 5 × 10²⁸ atoms m⁻³ doped with 1 ppm As: n_e ≈ 5 × 10²² m⁻³, n_h = n_i²/n_e = (1.5 × 10¹⁶)²/5 × 10²² ≈ 4.5 × 10⁹ m⁻³.
14.5 p-n junction
A p-n junction is the basic building block of diodes, transistors etc.: a p-type wafer whose one part is converted to n-type by a pentavalent impurity (a metallurgical junction between the p and n regions). Two processes: diffusion — due to the concentration gradient, holes diffuse p → n and electrons n → p (diffusion current); the electrons leaving the n-side leave behind ionised donors (positive, immobile) and the holes leaving the p-side leave ionised acceptors (negative, immobile): a space-charge (depletion) region about 0.1 μm thick (Fig. 14.10) with a built-in electric field directed from the n-side to the p-side; drift — the field makes electrons move p → n and holes n → p (drift current, opposite to diffusion current). Initially diffusion is large; as the space-charge regions grow, the field and drift grow until diffusion current = drift current: equilibrium, with zero net current. The potential difference across the junction (n-side positive relative to p) is the barrier potential V₀ (Fig. 14.11) opposing further diffusion. Example 14.3: one cannot form a p-n junction by joining two slabs physically — the surface roughness is much larger than the interatomic spacing (2–3 Å), so there is no continuous atomic contact.
14.6 Semiconductor diode
A semiconductor diode (Fig. 14.12) is a p-n junction with metallic contacts at the ends — a two-terminal device; its symbol has the arrow in the direction of conventional current under forward bias. Forward bias (p-side to the positive terminal, n-side to negative; Fig. 14.13): the applied voltage mostly drops across the depletion region (high resistance), opposing V₀: the depletion width decreases and the barrier is lowered to (V₀ − V); large numbers of carriers cross (minority-carrier injection, Fig. 14.14): large forward current (mA). Reverse bias (n-side positive, p-side negative; Fig. 14.15): the barrier increases to (V₀ + V), the depletion region widens, diffusion current is suppressed enormously; only the drift (minority carrier) current of a few μA flows, nearly independent of the voltage — the reverse saturation current, up to the breakdown voltage V_br at which the reverse current rises sharply (the diode may be destroyed by overheating if the current is not limited).
V–I characteristics (Fig. 14.16): in forward bias the current is negligible until the threshold (cut-in) voltage (~0.2 V for Ge, ~0.7 V for Si), after which it rises almost exponentially; a milliammeter is used for the forward measurement and a microammeter for reverse bias. The diode conducts easily in one direction only — forward-bias resistance ≪ reverse-bias resistance. Dynamic resistance r_d = ΔV/ΔI. Example 14.4 (V–I of a Si diode, Fig. 14.17): between 10 mA and 20 mA, V changes from 0.7 V to 0.8 V, a dynamic resistance of ≈ 10 Ω.
14.7 Application of junction diode as a rectifier
Since a diode conducts only when forward-biased, an ac input is converted to a unidirectional (pulsating dc) output: a rectifier. Half-wave rectifier (Fig. 14.18): a diode in series with the load; during the positive half-cycle (end A positive) the diode is forward biased and conducts; during the negative half-cycle it is reverse biased and does not conduct (the reverse saturation current is negligible): output only for half of the input wave; output frequency = input frequency. Full-wave rectifier (Fig. 14.19): two diodes with a centre-tap transformer: the p-sides of the diodes are connected to the ends of the secondary and the n-sides joined together; the output is taken between this common point and the centre tap; each diode rectifies for alternate half-cycles and each handles only half of the secondary voltage; the output is rectified for both half-cycles, so output frequency = 2 × input frequency. (Another circuit with four diodes does not need a centre tap.)
Filtering (Fig. 14.20): a capacitor input filter across the load charges to the peak voltage and discharges slowly through the load; the rate of fall depends on the time constant C R_L (large C gives a small ripple), so the output approaches the peak voltage of the rectified voltage — widely used.
Summary & key points (NCERT)
Semiconductors: intrinsic (n_e = n_h) and extrinsic (doped: n-type with pentavalent donors, n_e ≫ n_h; p-type with trivalent acceptors, n_h ≫ n_e); energy bands (E_g > 3 eV insulators; < 3 eV semiconductors; overlapping/partly filled: metals); in extrinsic semiconductors n_e n_h = n_i² and the crystal is electrically neutral; p-n junction: depletion region and barrier potential; forward bias lowers, reverse bias raises the barrier; diode characteristics; half-wave and full-wave rectifiers with a filter.
Q1
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
Concept and context. Pure (intrinsic) silicon has four valence electrons per atom and a diamond-like lattice in which each atom shares one electron with each of its four neighbours through covalent bonds. At room temperature it has very few free charge carriers (n_e = n_h = n_i ≈ 10¹⁶ m⁻³) and hence a very low conductivity. To increase the conductivity enormously, a small quantity (a few parts per million) of a suitable impurity is added in a process called doping; the impurity atoms are called dopants, and the doped material is an extrinsic (impurity) semiconductor. There are two kinds of dopants: pentavalent atoms (valency 5: arsenic As, antimony Sb, phosphorus P) which make n-type semiconductors, and trivalent atoms (valency 3: boron B, aluminium Al, indium In) which make p-type semiconductors. In the n-type semiconductor “n” stands for the negative charge carriers (electrons) which are the majority carriers.
Fig. — a pentavalent donor atom in the silicon lattice: four electrons complete the bonds, the fifth is free to conduct, making an n-type semiconductor (compare NCERT Fig. 14.7, a pentavalent donor atom in Si) (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.7, page 8 ↗
How n-type silicon is formed (step by step)
Doping with a pentavalent atom: a phosphorus (or As, Sb) atom replaces a Si atom in the lattice. It has five valence electrons; four of them form covalent bonds with the four neighbouring Si atoms exactly as a Si atom would.
The fifth electron: it is not needed for bonding and is only very loosely bound to the impurity atom (the energy required to set it free is only about 0.01 eV, compared with the 1.1 eV needed to break a Si–Si bond). It is therefore free to move in the crystal at room temperature and contributes to the conduction. Such a dopant is called a donor (it donates one electron). The donor atom, having lost an electron, becomes a fixed, immobile positive ion — the crystal as a whole remains electrically neutral.
Carrier concentrations: the number of electrons contributed by the donors is fixed by the doping level and does not depend on temperature; the intrinsic generation of electron–hole pairs adds a small number of both. The total electron concentration n_e is thus much larger than the hole concentration n_h (n_e ≫ n_h); moreover the extra electrons increase the recombination rate so that the hole concentration of the doped crystal falls below n_i (the law of mass action n_e n_h = n_i²). Example 14.2 of NCERT: 1 ppm of As in Si (5 × 10²⁸ atoms m⁻³) gives n_e ≈ 5 × 10²² m⁻³ and n_h ≈ 4.5 × 10⁹ m⁻³.
Majority and minority carriers: electrons are the majority carriers and holes the minority carriers in an n-type semiconductor.
Examining each option
Option
Statement
Verdict
Reason
(a)
Electrons are majority carriers and trivalent atoms are the dopants
False
the first half is right but a trivalent dopant has only 3 valence electrons: it leaves a vacancy (hole) in a bond and produces a p-type semiconductor, not n-type
(b)
Electrons are minority carriers and pentavalent atoms are the dopants
False
pentavalent dopants are right but electrons are the majority carriers in n-type silicon (n_e ≫ n_h)
(c)
Holes are minority carriers and pentavalent atoms are the dopants
True
pentavalent dopants add free electrons (majority); the holes, produced only by thermal generation and reduced by recombination, are the minority carriers
(d)
Holes are majority carriers and trivalent atoms are the dopants
False
this describes a p-type semiconductor
Property
n-type Si
p-type Si
Dopant
pentavalent (As, Sb, P) — donor
trivalent (B, Al, In) — acceptor
Majority carriers
electrons
holes
Minority carriers
holes
electrons
Relation
n_e ≫ n_h
n_h ≫ n_e
Charge of the crystal
neutral
neutral
Energy-band picture (NCERT Fig. 14.9a): the donor energy level E_D lies just below the bottom E_C of the conduction band; at room temperature almost all the donor electrons are lifted into the conduction band with a very small energy, so the conduction band is populated mainly by donor electrons (only very few ~10¹² Si atoms per m³ are ionised intrinsically).
Is the n-type crystal negatively charged? No — the free electrons are balanced by the positively charged immobile donor ions.
Exam tip: Memorise: n-type ⇒ pentavalent donors ⇒ electrons majority, holes minority. The right answer is the only option consistent on both clauses (dopant and carriers): (c).
Conclusion: statement (c) is true: in n-type silicon holes are the minority carriers and pentavalent atoms are the dopants (electrons are the majority carriers).
Q2
Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
Answer
Concept and context. A p-type semiconductor is obtained when silicon or germanium is doped with a small amount of a trivalent impurity such as boron (B), aluminium (Al) or indium (In). The “p” stands for positive: the principal charge carriers are the holes — vacancies in the covalent bonds that behave as free particles with an effective positive charge. Each trivalent atom (the acceptor) has only three valence electrons: it forms covalent bonds with three of its four neighbouring Si atoms, but cannot complete the fourth bond; the bond contains a vacancy (a hole).
Fig. — a trivalent acceptor atom in the silicon lattice: one bond is incomplete — a hole — making a p-type semiconductor (compare NCERT Fig. 14.8, a trivalent acceptor atom in Si) (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.8, page 9 ↗
How p-type silicon conducts
Hole formation: an electron from a neighbouring Si–Si bond can jump into the vacancy of the acceptor atom (it needs only a very small energy), leaving a hole at its own site; in this way the hole moves through the crystal (a hole at site 1 → site 2 → site 3 …) — the apparent motion of a positive charge. The acceptor atom, having accepted an electron, becomes an immobile negative ion (the core with one negative charge and its associated hole, as in NCERT Fig. 14.8b). One acceptor atom gives one hole.
Carrier concentrations: the holes from the acceptors add to those generated thermally, while the source of conduction electrons is only intrinsic generation. Hence n_h ≫ n_e, and the thermally generated electrons are reduced by recombination with the abundant holes (n_e n_h = n_i²).
Majority and minority carriers:holes are the majority carriers and electrons the minority carriers in p-type silicon. The crystal remains electrically neutral (the positive holes are balanced by the negative acceptor ions).
Applying the options of Exercise 14.1 to a p-type semiconductor
Option
Statement
Verdict for p-type
Reason
(a)
Electrons are majority carriers and trivalent atoms are the dopants
False
trivalent dopants are correct but the majority carriers are holes
(b)
Electrons are minority carriers and pentavalent atoms are the dopants
False
electrons are minority carriers (correct) but pentavalent dopants make n-type material
(c)
Holes are minority carriers and pentavalent atoms are the dopants
False
true for n-type, not p-type
(d)
Holes are majority carriers and trivalent atoms are the dopants
True
both clauses describe a p-type semiconductor: acceptors (B, Al, In) and holes in the majority
Energy-band picture (NCERT Fig. 14.9b): the acceptor level E_A lies just above the top E_V of the valence band; with a very small energy an electron from the valence band jumps to E_A and ionises the acceptor negatively, leaving a hole in the valence band; at room temperature most acceptor atoms are ionised, so the density of holes in the valence band is predominantly due to the impurity.
Which carrier carries the current? In a p-type semiconductor the current is mainly the motion of holes in the direction of the electric field (equivalent to electrons hopping in the opposite direction from bond to bond).
Exam tip: p-type = trivalent acceptor = holes majority = electrons minority. Do not confuse with n-type: the answer is option (d).
Conclusion: for p-type semiconductors statement (d) is true: holes are the majority carriers and trivalent atoms are the dopants.
Q3
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 (E_g)_C, (E_g)_Si and (E_g)_Ge. Which of the following statements 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
Answer
Concept and context. In a crystal the energy levels of the valence electrons of neighbouring atoms broaden into energy bands: the valence band (filled with the bonding electrons at 0 K) and the conduction band above it (empty at 0 K), separated by a forbidden region, the energy gap E_g (NCERT Fig. 14.1). The electrical behaviour depends on the size of this gap (Fig. 14.2): a large gap (E_g > 3 eV) — an insulator; a small gap (E_g < 3 eV) — a semiconductor; overlapping bands — a metal. Carbon (diamond), silicon and germanium belong to the same group (group IV) of the periodic table with four valence electrons and the same crystal structure, but their gaps differ because of the different orbits from which the bonding electrons come.
Fig. — energy bands of carbon (diamond), silicon and germanium: the band gap is 5.4 eV, 1.1 eV and 0.7 eV — decreasing from C to Si to Ge (compare NCERT Fig. 14.2, energy bands of metals, insulators and semiconductors) (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.2, page 4 ↗
Position of the bonding electrons: the four bonding electrons of C, Si and Ge lie in the 2nd, 3rd and 4th orbit respectively (Example 14.1 of NCERT).
Binding: an electron in the nearer (n = 2) orbit of carbon is bound much more strongly to the nucleus than one in the n = 3 orbit of silicon or the n = 4 orbit of germanium, so the energy needed to free it (the band gap E_g) is highest for C, then Si, and lowest for Ge.
Values: (E_g)_C = 5.4 eV, (E_g)_Si = 1.1 eV, (E_g)_Ge = 0.7 eV (so (E_g)_C > (E_g)_Si > (E_g)_Ge). Tin (Sn), another group-IV element, has E_g = 0 and is a metal.
Consequence: carbon in the form of diamond (5.4 eV > 3 eV) is an insulator — at room temperature thermal energy (kT ≈ 0.026 eV) cannot lift electrons across such a gap — while Si and Ge are intrinsic semiconductors with a significant number of thermally generated electrons and holes (smaller number in Si than in Ge).
Option
Statement
Verdict
Reason
(a)
(E_g)_Si < (E_g)_Ge < (E_g)_C
False
Ge has a smaller gap than Si (0.7 eV < 1.1 eV)
(b)
(E_g)_C < (E_g)_Ge > (E_g)_Si
False
the gap of C is the largest, not smaller than Ge
(c)
(E_g)_C > (E_g)_Si > (E_g)_Ge
True
5.4 eV > 1.1 eV > 0.7 eV
(d)
(E_g)_C = (E_g)_Si = (E_g)_Ge
False
the gaps are very different — which is why their conductivities differ widely
Element
Bonding electrons in orbit
Band gap E_g
Class
Carbon (diamond)
n = 2
5.4 eV
insulator
Silicon
n = 3
1.1 eV
semiconductor
Germanium
n = 4
0.7 eV
semiconductor
Tin (grey)
n = 5
≈ 0
metal
Exam tip: The gap decreases down the group: C > Si > Ge > Sn. Link it to the principal quantum number of the bonding electrons.
Conclusion: statement (c) is true: (E_g)_C > (E_g)_Si > (E_g)_Ge (5.4 eV, 1.1 eV and 0.7 eV respectively).
Q4
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
Concept and context. A p-n junction is formed in a single crystal of semiconductor (for example a p-type Si wafer part of which is converted to n-type by adding a pentavalent impurity). In the p-region the majority carriers are holes (n_h ≫ n_e), in the n-region the majority carriers are electrons. “Unbiased” means that no external voltage is applied across the junction. During the formation of the junction two processes occur: diffusion — the movement of carriers from a region of higher concentration to one of lower concentration, caused by the concentration gradient — and drift — the motion of carriers under an electric field (NCERT Section 14.5).
Fig. — a p-n junction at equilibrium: holes diffuse p → n and electrons n → p because of the concentration gradient, leaving the depletion region of ionised acceptors and donors with a built-in field that causes opposing drift (compare NCERT Fig. 14.10, p-n junction formation process) (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.10, page 11 ↗
Why holes diffuse from the p-region to the n-region (explanation)
Concentration gradient: the hole concentration is very high in the p-region (majority carriers) and very low in the n-region (minority carriers, produced only thermally). Holes therefore tend to spread from the region of high concentration to the region of low concentration — just like a drop of ink spreading in water or a perfume spreading in a room. Likewise the electron concentration is high in the n-region, so electrons diffuse from n to p.
No external potential: in an unbiased junction there is no applied voltage; at the instant of joining there is also no potential difference across the junction to push the carriers. So the initial motion is purely diffusion, driven by the concentration gradient.
Space-charge region: a hole crossing to the n-side leaves behind an immobile negative ionised acceptor; an electron crossing to the p-side leaves behind an immobile positive ionised donor. These ions build the depletion region (thickness ≈ 0.1 μm) and an electric field directed from the n-side to the p-side; this field causes a drift of carriers in the opposite sense (holes n → p, electrons p → n). Diffusion continues until the drift current equals the diffusion current: at equilibrium the net current is zero and a barrier potential V₀ is established across the junction.
Examining each option
Option
Statement
Verdict
Reason
(a)
free electrons in the n-region attract them
False
the electrons do not “pull” the holes across the junction; the n-region is electrically neutral as a whole (free electrons are balanced by the positive donor ions), so there is no long-range attraction before the junction forms; recombination at the junction happens only after carriers have diffused
(b)
they move across the junction by the potential difference
False
in the unbiased junction there is no applied potential difference; the built-in potential barrier V₀ opposes (does not cause) the diffusion of holes from p to n — it is created as a result of the diffusion and drives the opposite drift
(c)
hole concentration in the p-region is more as compared to the n-region
True
the concentration gradient is the cause of diffusion
(d)
all the above
False
(a) and (b) are not correct
Drift vs diffusion: drift current (due to the electric field) and diffusion current (due to the concentration gradient) flow in opposite directions at equilibrium and cancel.
Barrier potential: the n-side is positive relative to the p-side (the barrier is a few tenths of a volt in Si and Ge).
Exam tip: Say “diffusion is caused by a concentration gradient, not by an electric field or attraction”. Eliminate (a) and (b) with their reasons.
Conclusion: option (c) is correct: in an unbiased p-n junction holes diffuse from the p-region to the n-region because the hole concentration in the p-region is greater than in the n-region (a concentration gradient).
Q5
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
Concept and context. A semiconductor diode is a p-n junction with metallic contacts for applying an external voltage. At equilibrium a depletion region and a barrier potential V₀ exist across the junction (n-side positive relative to the p-side). When an external battery is connected, the junction is forward biased if the p-side is connected to the positive terminal and the n-side to the negative terminal (NCERT Fig. 14.13); it is reverse biased if the connection is the other way round (Fig. 14.15).
Fig. — potential barrier of a p-n junction: forward bias lowers the barrier from V₀ to (V₀ − V) and narrows the depletion region; reverse bias raises it to (V₀ + V) (compare NCERT Fig. 14.13, p-n junction under forward bias) (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.13, page 13 ↗
What happens under forward bias
The applied voltage opposes V₀: because the depletion region has very high resistance (no free charges) compared with the n- and p-regions, the applied voltage drops almost entirely across the depletion region. The external field is opposite to the built-in field, so the effective barrier height is reduced to (V₀ − V) and the depletion region becomes narrower.
More carriers can cross: with the lower barrier many more majority carriers (electrons from the n-side, holes from the p-side) have enough energy to cross the junction; the injected carriers become minority carriers on the other side (minority carrier injection), diffuse away from the junction and recombine; a continuous current flows in the circuit — the forward current, of the order of mA, which rises almost exponentially once V exceeds the threshold (cut-in) voltage (~0.2 V for Ge, ~0.7 V for Si).
Conclusion: forward bias lowers the potential barrier (and does not raise it, nor reduce the majority carrier current to zero — it greatly increases the majority carrier current).
Option
Statement
Verdict
Reason
(a)
raises the potential barrier
False
that is what reverse bias does: the barrier becomes (V₀ + V) and the depletion region widens
(b)
reduces the majority carrier current to zero
False
the majority carrier current increases enormously under forward bias; it is under reverse bias that the diffusion (majority carrier) current is suppressed to almost zero
(c)
lowers the potential barrier
True
V₀ → (V₀ − V)
(d)
none of the above
False
(c) is correct
Feature
Forward bias
Reverse bias
Connection
p to +, n to −
p to −, n to +
Barrier height
V₀ − V (lowered)
V₀ + V (raised)
Depletion width
decreases
increases
Current
large (mA), majority carriers
very small (μA), minority carriers; saturates
Resistance
low
very high
Rectifier action: the diode conducts easily in the forward direction and hardly at all in the reverse direction — the property used for rectification (Exercise 14.6).
Dynamic resistance r_d = ΔV/ΔI is small in forward bias (about 10 Ω for the Si diode of NCERT Example 14.4).
Exam tip: Forward bias ⇒ barrier lowered, depletion layer narrower, large current. Remember the (V₀ − V) expression.
Conclusion: option (c) is correct: when a forward bias is applied to a p-n junction it lowers the potential barrier (to V₀ − V), narrowing the depletion region and allowing a large current.
Q6
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
Concept and context. A rectifier converts an alternating voltage into a unidirectional (pulsating dc) voltage by using the one-way conduction of a p-n junction diode: it conducts only when forward-biased. In a half-wave rectifier (NCERT Fig. 14.18) a single diode in series with a load passes only one half of every input cycle; in a full-wave rectifier (Fig. 14.19) two diodes with a centre-tap transformer (or four diodes in a bridge) pass both halves of the input cycle in the same direction through the load. The frequency of the output is the number of output pulses (ripple cycles) per second.
Fig. — input ac wave, half-wave rectified output (one pulse per input cycle) and full-wave rectified output (two pulses per input cycle) (compare NCERT Fig. 14.18 and Fig. 14.19, half-wave and full-wave rectifiers) (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.19, page 17 ↗
(a) Half-wave rectifier
During the positive half-cycle of the input (end A positive) the diode is forward biased and conducts, producing a current pulse through the load; during the negative half-cycle the diode is reverse biased and does not conduct (the reverse saturation current is negligible), so the output is zero.
Hence one output pulse appears for every input cycle (the repetition rate of the pulses equals that of the input): f_out = f_in = 50 Hz.
(b) Full-wave rectifier
In the centre-tap circuit the p-sides of the two diodes are connected to the ends A and B of the secondary of the transformer and the n-sides are joined together; the output is taken between this common point and the centre tap. When A is positive (and B negative) diode D₁ conducts; in the next half-cycle B is positive and diode D₂ conducts. In both cases the current through the load flows in the same direction.
So the output has two pulses for every input cycle: f_out = 2 f_in = 2 × 50 = 100 Hz.
Why full-wave is better: its output has a higher ripple frequency (100 Hz) and a larger average value, so it is easier to smooth with a smaller capacitor filter (time constant C R_L) — the output of the filtered full-wave rectifier is nearer to the peak value (NCERT Fig. 14.20).
The average (dc) output of a half-wave rectifier is V_m/π ≈ 0.318 V_m while that of a full-wave rectifier is 2V_m/π ≈ 0.636 V_m (V_m = peak voltage); each diode of a centre-tap rectifier handles half the secondary voltage.
Exam tip: Half-wave: f_out = f_in; full-wave: f_out = 2f_in. Support with “one pulse per cycle” and “two pulses per cycle”.
Conclusion: the output frequency of the half-wave rectifier is 50 Hz (equal to the input frequency) and that of the full-wave rectifier is 100 Hz (twice the input frequency).
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