📌 Snapshot
- Establishes solid-state electronics as the modern replacement for bulky vacuum tubes, with the p-n junction as the "key" to all semiconductor devices.
- Classifies solids as metals, semiconductors, or insulators using both resistivity ranges and energy-band theory (band gap Eg).
- Builds up intrinsic conduction (electron-hole pairs in pure Si/Ge) and then extrinsic conduction via doping with pentavalent (n-type) and trivalent (p-type) impurities; nenh = ni2 always holds.
- Develops the p-n junction (diffusion, drift, depletion region, barrier potential) and the semiconductor diode under forward and reverse bias, including V-I characteristics, cut-in/threshold voltage and breakdown.
- Applies the diode to rectification: half-wave, centre-tap full-wave, and the role of a capacitor filter in producing a steady dc output.
📖 Detailed Notes
2.1 Core concepts
- Before 1948, electron-flow devices were vacuum tubes (diode, triode, tetrode, pentode); they are bulky, need ~100 V, have low reliability — solid-state semiconductor devices replaced them because charge supply happens within the solid (NCERT §14.1, p. 323-324).
- On the basis of conductivity: metals have ρ ~ 10−2 to 10−8 Ω m; semiconductors ρ ~ 10−5 to 106 Ω m; insulators ρ ~ 1011 to 1019 Ω m (NCERT §14.2, p. 324).
- Semiconductors of interest are elemental (Si, Ge) and compound (inorganic like GaAs, CdS, CdSe, InP; organic like anthracene; organic polymers like polyaniline) (NCERT §14.2, p. 324-325).
- Energy-band picture: in a solid, electron energy levels of neighbouring atoms form continuous bands — the valence band (lower, filled by valence electrons) and the conduction band (upper); the gap between top of VB (EV) and bottom of CB (EC) is the energy band gap Eg (NCERT §14.2, p. 325-326).
- Three cases from band theory: (a) Metals — CB and VB overlap, or CB is partially filled; (b) Insulators — Eg > 3 eV, no thermal excitation possible; (c) Semiconductors — small finite Eg < 3 eV, so at room temperature some electrons can cross over to CB (NCERT §14.2, Fig. 14.2, p. 326-327).
- For Si and Ge crystal of N atoms there are 4N valence electrons in 8N available outer-orbit states; at the lattice spacing, these split into two bands separated by Eg (NCERT §14.2, p. 325).
- Intrinsic semiconductor (pure Si, Ge): diamond-like lattice with each atom covalently bonded to 4 neighbours. At T = 0 K behaves as an insulator. At T > 0 K, thermal energy breaks some covalent bonds, producing free electrons in CB and equal-numbered holes in VB (NCERT §14.3, p. 327-329).
- A hole is a vacancy left by an electron in a broken covalent bond and behaves like an apparent free particle of effective positive charge +q (NCERT §14.3, p. 327).
- In an intrinsic semiconductor ne = nh = ni (intrinsic carrier concentration), and the total current I = Ie + Ih (NCERT §14.3, eqns 14.1 and 14.2, p. 327-328).
- Generation–recombination equilibrium: under steady state, rate of thermal generation of carriers equals their rate of recombination (electron colliding with a hole) (NCERT §14.3, p. 328).
- Extrinsic semiconductor: deliberate addition (doping) of a small (~ppm) impurity of nearly the same size as the host atom to increase conductivity manifold. Two dopant types in Si/Ge: pentavalent (As, Sb, P) and trivalent (In, B, Al) (NCERT §14.4, p. 329-330).
- n-type: pentavalent dopant gives one weakly-bound extra electron (ionisation energy ~0.05 eV for Si, ~0.01 eV for Ge — much less than the band gap of 1.1 eV for Si and 0.72 eV for Ge); donor atom donates this electron to CB. Electrons are majority, holes minority; ne >> nh (NCERT §14.4, p. 330-331, eqn 14.3).
- p-type: trivalent dopant accepts an electron, creating a hole in VB; acceptor atom becomes effectively negative. Holes are majority, electrons minority; nh >> ne (NCERT §14.4, p. 331, eqn 14.4).
- Law of mass action (electron–hole concentration product): nenh = ni2 in thermal equilibrium for any (intrinsic or extrinsic) semiconductor; crystal stays overall electrically neutral (NCERT §14.4, eqn 14.5, p. 332).
- In the band diagram of doped material the donor level ED lies just below EC; the acceptor level EA lies just above EV (NCERT §14.4, Fig. 14.9, p. 331-332).
- Energy gaps reported: C (diamond) 5.4 eV, Si 1.1 eV, Ge 0.7 eV; Sn 0 eV (a metal). This explains why C is an insulator while Si and Ge are intrinsic semiconductors despite having the same lattice (NCERT §14.4, p. 332; Example 14.1, p. 329).
- p-n junction formation: in the same wafer one region is doped p-type and another n-type. Two processes occur — diffusion (holes p → n, electrons n → p, due to concentration gradient) and drift (carriers driven by the electric field of the depletion region). At equilibrium diffusion current = drift current and net current is zero (NCERT §14.5, p. 333).
- Depletion region: as carriers diffuse across, immobile ionised donors leave a positive space-charge on the n-side and ionised acceptors leave a negative space-charge on the p-side; thickness ~ one-tenth of a micrometre. This sets up a barrier potential V0 opposing further diffusion (NCERT §14.5, Fig. 14.10-14.11, p. 333-334).
- Forward bias (p to +, n to −): applied V opposes V0, so depletion width and effective barrier (V0 − V) decrease; majority carriers cross the junction (minority carrier injection); current is in mA (NCERT §14.6.1, p. 334-335).
- Reverse bias (n to +, p to −): applied V adds to V0, so depletion width and effective barrier (V0 + V) increase; only a small drift current (~μA) due to minority carriers flows; nearly voltage-independent up to breakdown (NCERT §14.6.2, p. 335-336).
- V-I characteristics: forward current is negligible until the threshold/cut-in voltage (~0.2 V for Ge, ~0.7 V for Si), after which it rises sharply (exponentially). In reverse bias, current is a small reverse saturation current (~μA); at breakdown voltage Vbr it suddenly increases (NCERT §14.6.2, Fig. 14.16, p. 336-337).
- Dynamic resistance rd = ΔV/ΔI; in forward bias rd ~ a few Ω, in reverse bias ~107 Ω (NCERT §14.6.2, eqn 14.6 and Example 14.4, p. 337).
- Rectification: since a diode conducts only when forward biased, it converts ac into pulsating dc. Half-wave rectifier uses one diode and gives output only during one half-cycle (NCERT §14.7, Fig. 14.18, p. 338).
- Full-wave rectifier uses two diodes with a centre-tap transformer; D1 and D2 conduct in alternate half-cycles, giving an output for both halves of the input ac (NCERT §14.7, Fig. 14.19, p. 338-339).
- A capacitor in parallel with RL acts as a filter: it charges to peak rectified voltage and discharges slowly through RL (time constant ∝ CRL). Large C gives a steady dc nearly equal to peak voltage; widely used in power supplies (NCERT §14.7, p. 339-340; Fig. 14.20).
- Output frequency depends on rectifier type: a half-wave rectifier reproduces the input frequency (one pulse per ac cycle, so 50 Hz → 50 Hz output), whereas a full-wave rectifier delivers two pulses per cycle so the output ripple frequency is twice the input (50 Hz → 100 Hz). This is a recurring distinction tested by NCERT exercises (NCERT §14.7, p. 339; Exercise 14.6).
- The depletion-region width in a typical silicon p-n junction is of order 0.1 µm; despite being so thin it sets up an internal field strong enough (~10⁵ V cm⁻¹) to balance the chemical-potential difference between p- and n-regions at equilibrium (NCERT §14.5, p. 333).
- In reverse bias the small reverse saturation current arises from minority carriers — electrons in the p-region and holes in the n-region — which the depletion-region field happily sweeps across; this is why the reverse current is almost independent of the applied voltage but depends strongly on temperature (NCERT §14.6.2, p. 336).
- Forward-bias current in a diode follows the exponential Shockley-like behaviour above the cut-in voltage — that is why textbook V–I curves show a knee at ~0.7 V (Si) followed by a near-vertical rise; below the knee, current is negligibly small (NCERT §14.6.2, Fig. 14.16, p. 337).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Energy band gap (Eg) | Energy gap between the top of the valence band (EV) and the bottom of the conduction band (EC) | 326 |
| Hole | A vacancy in a covalent bond left by an electron; behaves as an apparent free particle of effective positive charge +q | 327 |
| Intrinsic carrier concentration (ni) | Common value of ne = nh in a pure (intrinsic) semiconductor due to thermal generation | 327 |
| Doping | Deliberate addition of a small amount of a suitable impurity (dopant) to a pure semiconductor to alter its conductivity | 329 |
| Donor (n-type) | Pentavalent impurity (P, As, Sb) that donates one electron to the conduction band; gives ne >> nh | 330-331 |
| Acceptor (p-type) | Trivalent impurity (B, Al, In) that accepts an electron from a Si/Ge bond, creating a hole; gives nh >> ne | 331 |
| Law of mass action | nenh = ni2 at thermal equilibrium for any semiconductor | 332 |
| Depletion region | Region around a p-n junction depleted of mobile carriers, containing only immobile ionised donor/acceptor cores | 333 |
| Barrier potential (V0) | Built-in potential across the depletion region at equilibrium that opposes further diffusion | 334 |
| Threshold (cut-in) voltage | Forward voltage above which diode current rises sharply (~0.2 V for Ge, ~0.7 V for Si) | 337 |
| Reverse saturation current | Small (~μA), nearly voltage-independent current in a reverse-biased diode | 337 |
| Breakdown voltage (Vbr) | Reverse voltage at which diode reverse current rises sharply | 336 |
| Dynamic resistance (rd) | Ratio ΔV/ΔI of a small change in voltage to the corresponding small change in current | 337 |
| Rectifier | Circuit using a diode to convert ac into pulsating dc | 338 |
| Valence band (VB) | Band of energy levels occupied by valence electrons at 0 K | 326 |
| Conduction band (CB) | Higher band above EV; electrons here are free to conduct | 326 |
| Intrinsic semiconductor | Pure semiconductor with no significant impurity (ne = nh = ni) | 327 |
| Extrinsic semiconductor | Semiconductor whose conductivity is altered by doping | 329 |
| Forward bias | p-side connected to + and n-side to −; effective barrier (V0 − V) | 334 |
| Reverse bias | p-side connected to − and n-side to +; effective barrier (V0 + V) | 335 |
| Half-wave rectifier | Single-diode circuit conducting during one half of the input ac | 338 |
| Full-wave rectifier | Two-diode centre-tap circuit conducting during both halves of ac | 339 |
| Filter | Capacitor (or other circuit) used to smooth pulsating dc into steady dc | 339 |
2.3 Diagrams / processes to remember
- Fig. 14.1 — energy band positions of a semiconductor at 0 K with EC, EV and Eg labelled (p. 326).
- Fig. 14.2 — band-gap pictures for (a) metal (overlap), (b) insulator (Eg > 3 eV), (c) semiconductor (small Eg) (p. 326).
- Fig. 14.4 / 14.5 — 2-D lattice of Si/Ge showing covalent bonds and the hole-electron generation and hole-motion mechanism (p. 328).
- Fig. 14.7 / 14.8 — pentavalent (donor) and trivalent (acceptor) atoms substituted in the Si lattice; commonly used schematic with +ve core and electron, and −ve core and hole (p. 330-331).
- Fig. 14.9 — band diagram showing the donor level ED just below EC for n-type and the acceptor level EA just above EV for p-type (p. 332).
- Fig. 14.10 / 14.11 — formation of the p-n junction: diffusion and drift, and the resulting depletion region with barrier potential V0 (p. 333-334).
- Fig. 14.13 / 14.15 — diode under forward bias (barrier reduced to V0 − V) and reverse bias (barrier raised to V0 + V) (p. 335-336).
- Fig. 14.16 — circuit for studying V-I characteristics and the typical V-I curve of a silicon diode showing cut-in at ~0.7 V and a reverse saturation current (p. 336).
- Fig. 14.18 — half-wave rectifier circuit with transformer secondary and load RL and output waveform (p. 338).
- Fig. 14.19 — centre-tap full-wave rectifier with diodes D1 and D2 and output waveform across RL (p. 339).
- Fig. 14.20 — full-wave rectifier with a capacitor in parallel with RL acting as a filter; smooth dc output (p. 340).
2.4 Common confusions / NTA trap points
- Eg ranges: metals Eg ≈ 0, semiconductors 0.2-3 eV (Si 1.1, Ge 0.7), insulators > 3 eV (C ~5.4 eV). Students mix up which class has the smallest/largest gap — remember the order (Eg)C > (Eg)Si > (Eg)Ge.
- n-type majority/minority: in n-type, electrons are MAJORITY but the dopant is PENTAVALENT (P/As/Sb). NTA frequently swaps "pentavalent" with "trivalent" to trap students.
- Hole charge: a hole is the absence of an electron in a bond — its effective charge is +q, but the material as a whole remains neutral because the acceptor/donor cores carry the opposite charge.
- Cut-in voltage values: 0.2 V for Ge, 0.7 V for Si — easy to swap.
- Reverse current is voltage-independent (in μA) only up to breakdown; at breakdown it rises sharply. Students wrongly assume it is always negligible.
- Output frequency: half-wave rectifier output frequency equals input frequency (50 Hz → 50 Hz), but a full-wave rectifier output frequency is twice the input (50 Hz → 100 Hz).
- Depletion-region direction of internal field: the built-in field points from the n-side to the p-side (i.e. from + to − space charge), opposing further diffusion of majority carriers. Many students draw the arrow the wrong way.
- Conduction in metal vs semiconductor with temperature: metals' resistivity rises with T (more lattice scattering), but semiconductors' resistivity falls with T (more thermally generated carriers). NTA exploits this contrast.
- nₑnₕ = nᵢ² always — even in heavily doped material. If nₑ is raised by donor doping, nₕ falls accordingly so that the product is fixed at any given temperature.
- Diode is not Ohmic: V–I curve is non-linear, so dynamic resistance rd = ΔV/ΔI varies along the curve. Treating Ohm's law I = V/R as applicable to a diode is wrong.
- The capacitor filter does NOT change input frequency; it merely smooths the pulsating dc. Full-wave output already has frequency 2ν; the filter does not double it further.
2.5 Key formulas table
| Quantity | Symbol / Formula | NCERT reference |
|---|---|---|
| Resistivity ranges (metal) | ρ ~ 10⁻² to 10⁻⁸ Ω m | §14.2, p. 324 |
| Resistivity ranges (semiconductor) | ρ ~ 10⁻⁵ to 10⁶ Ω m | §14.2, p. 324 |
| Resistivity ranges (insulator) | ρ ~ 10¹¹ to 10¹⁹ Ω m | §14.2, p. 324 |
| Energy gap (insulator) | Eg > 3 eV | §14.2, p. 326 |
| Energy gap (semiconductor) | Eg < 3 eV (Si 1.1 eV; Ge 0.7 eV) | §14.4, p. 332 |
| Total current (intrinsic) | I = Iₑ + Iₕ | §14.3, Eq. 14.2, p. 328 |
| Intrinsic concentration | nₑ = nₕ = nᵢ | §14.3, Eq. 14.1, p. 327 |
| n-type carrier inequality | nₑ ≫ nₕ | §14.4, Eq. 14.3, p. 330 |
| p-type carrier inequality | nₕ ≫ nₑ | §14.4, Eq. 14.4, p. 331 |
| Law of mass action | nₑnₕ = nᵢ² | §14.4, Eq. 14.5, p. 332 |
| Ionisation energy of donor in Si | ~0.05 eV | §14.4, p. 330 |
| Forward-bias effective barrier | V0 − V | §14.6.1, p. 335 |
| Reverse-bias effective barrier | V0 + V | §14.6.2, p. 335 |
| Cut-in voltage (Si / Ge) | ~0.7 V / ~0.2 V | §14.6.2, p. 337 |
| Dynamic resistance | rd = ΔV/ΔI | §14.6.2, Eq. 14.6, p. 337 |
| Output frequency (half-wave) | ν_out = ν_in | §14.7, p. 338 |
| Output frequency (full-wave) | ν_out = 2 ν_in | §14.7, p. 339 |
| Filter time-constant | τ = CRL | §14.7, p. 340 |
| Reverse saturation current | I_s ~ µA, voltage-independent below breakdown | §14.6.2, p. 337 |
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. According to the band theory of solids, a material is classified as an insulator when its energy band gap Eg is
▸ Show answer & explanation
Answer: C
For an insulator Eg > 3 eV, so thermal excitation across the gap is not possible. Option (B) describes a semiconductor; option (A) describes a metal.
Q2. Match the materials in List-I with their approximate energy band gap in List-II: List-I (Material) — List-II (Eg) P. C (diamond) — 1. 0 eV Q. Si — 2. 0.7 eV R. Ge — 3. 1.1 eV S. Sn — 4. 5.4 eV
▸ Show answer & explanation
Answer: A
Matching the four band-gap values gives C → 5.4 eV, Si → 1.1 eV, Ge → 0.7 eV, Sn → 0 eV (a metal).
Q3. In an intrinsic semiconductor at T > 0 K, which of the following relations is correct?
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Answer: C
Each broken covalent bond creates one electron and one hole, so the electron and hole densities are equal and both equal the intrinsic carrier concentration ni.
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Q4. A pure Si crystal is doped with a small amount of phosphorus. The resulting semiconductor is
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Answer: B
Phosphorus is pentavalent; the fifth electron, weakly bound (~0.05 eV), is donated to the conduction band, making electrons the majority carriers. P is not trivalent, so (D) is wrong.
Q5. The law of mass action for a semiconductor in thermal equilibrium states that
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Answer: C
The product of electron and hole concentrations at thermal equilibrium equals the square of the intrinsic carrier concentration, irrespective of doping.
Q6. Statement I: In a p-n junction at equilibrium, the depletion region contains only immobile ionised donor and acceptor cores. Statement II: At equilibrium the diffusion current of carriers across the junction is exactly cancelled by the drift current, so the net current is zero.
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Answer: A
Both statements are textbook facts from the formation of a p-n junction.
Q7. The cut-in (threshold) voltage of a silicon p-n junction diode is approximately
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Answer: C
~0.7 V for Si. Option (A) is the Ge value; option (D) is the band gap of Si, not its diode cut-in voltage.
Q8. The V-I characteristic of a p-n junction diode shows that, in reverse bias, the current
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Answer: B
Reverse current saturates at a small value set by minority-carrier concentration and is voltage-independent until breakdown is reached.
Q9. In a half-wave rectifier the input ac frequency is 50 Hz. The output frequency is
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Answer: B
A half-wave rectifier produces one pulse per input cycle, so the output frequency equals the input frequency (50 Hz). A full-wave rectifier would give 100 Hz.
Q10. In the centre-tap full-wave rectifier circuit (Fig. 14.19) at the instant when the voltage at end A of the secondary is positive with respect to the centre tap,
▸ Show answer & explanation
Answer: B
In a centre-tap rectifier the two diodes conduct in alternate half-cycles. When A is positive, D1 conducts; when A goes negative (so B is positive), D2 conducts.
Q11. In a full-wave rectifier circuit a capacitor is connected in parallel with the load resistor RL. Its role is to
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Answer: B
The capacitor charges to the peak rectified voltage and discharges slowly through RL (time constant CRL), smoothing the pulsating output into a near-constant dc.
Q12. Assertion (A): When a p-n junction diode is forward biased, the width of the depletion region decreases. Reason (R): Under forward bias the applied voltage V is in the same direction as the built-in barrier potential V0, so the effective barrier becomes V0 + V.
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Answer: C
The assertion is correct — forward bias does reduce the depletion width. The reason is wrong: under forward bias V opposes V0 and the effective barrier is V0 − V, not V0 + V (that is the reverse-bias case).
Q13. The output frequency of a full-wave rectifier fed by a 50 Hz ac mains is
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Answer: C
A full-wave rectifier delivers two pulses per input cycle, so output frequency is twice the input frequency: 2 × 50 = 100 Hz. Half-wave would give 50 Hz.
Q14. In a p-n junction diode under reverse bias, the width of the depletion region
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Answer: B
Under reverse bias the applied V supports the built-in V0, sweeping more carriers out of the junction region and widening the depletion layer. Under forward bias it narrows.
Q15. Assertion (A): In an n-type silicon semiconductor, the conductivity at room temperature is much higher than that of pure (intrinsic) silicon. Reason (R): A pentavalent dopant such as P or As provides an extra weakly-bound electron (ionisation energy ~0.05 eV) that is easily promoted to the conduction band.
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Answer: A
The donor's fifth electron is bound with only ~0.05 eV (compared to the 1.1 eV band gap of Si), so at room temperature kT is enough to ionise it; nₑ ≫ nᵢ and conductivity rises sharply. R correctly explains A.
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