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State Board · Class 11 · Physics

Lesson 14: Semiconductors

7 concepts · 56 practice questions with answers · Free to read

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Conductors, insulators and semiconductors; conductivity

From daily life

The chips in mobiles and computers use silicon — a semiconductor. Link to earlier classes : conductors, insulators and current of classes 9-10 now extend to semiconductors, band theory and diodes.

Simple definition

A semiconductor has resistivity between conductors and insulators; its resistance falls as T rises.

Range of ρ10⁻⁸10⁻⁵110⁶10⁸conductorsemiconductorinsulatorρ (Ω m) → ; σ = 1/ρ

In the figure : Range of ρ (log scale).

Rule 1: Classification — Conductors : ρ ~ 10⁻⁸-10⁻⁵ Ω m; semiconductors : 10⁻⁵-10⁶; insulators : > 10⁸ Ω m. σ = 1/ρ.

  • Easy : Copper ρ = 1.7 × 10⁻⁸ ⇒ conductor.
  • Medium : Silicon ρ ≈ 2300 Ω m ⇒ semiconductor.
  • Exam (CET) : Semiconductor? (1) Si (2) Cu (3) glass (4) Al → answer 1)

Rule 2: Conductivity — σ = e (n_e μ_e + n_h μ_h); μ = mobility (m²/V s).

  • Easy : Si : n = 1.5×10¹⁶, μ_e = 0.135, μ_h = 0.048 ⇒ σ = 4.39 × 10⁻⁴ S/m.
  • Medium : ρ = 1/σ = 2277 Ω m.
  • Exam (CET) : Unit of σ? (1) S/m (2) Ω m (3) A (4) V/m → answer 1)

Rule 3: Effect of temperature — In a semiconductor higher T ⇒ more carriers ⇒ σ rises, ρ falls (α negative); opposite in metals.

  • Easy : Heated Si ⇒ resistance falls.
  • Medium : Heated Cu ⇒ resistance rises.
  • Exam (CET) : Semiconductor R as T rises? (1) falls (2) rises (3) same (4) zero → answer 1) Trick : opposite to metals.

Derivation : σ = e(nμ_e + pμ_h)

  • Step 1 : J = n e v_d (electrons) + p e v_d (holes).
  • Step 2 : v_d = μ E.
  • Step 3 : J = e (n μ_e + p μ_h) E.
  • Step 4 : J = σ E ⇒ σ = e (n μ_e + p μ_h).
Remember

σ = 1/ρ = e(nμ_e + pμ_h).

Common mistake / CET trap

α of a semiconductor is negative — opposite to a metal.

Shortcut

Memory clue : "Heat it, carriers grow".

Spot the difference

Classρ (Ω m)As T rises
Conductor10⁻⁸-10⁻⁵ρ rises
Semiconductor10⁻⁵-10⁶ρ falls
Insulator> 10⁸Very high

Practice questions

Easy
  1. 1σ in terms of ρ?
  2. 2Examples of semiconductors?
  3. 3σ of Si as T rises?
Medium
  1. 4ρ = 0.5 Ω m : σ?
  2. 5n = 10¹⁹, μ_e = 0.4 (electrons only) : σ?
  3. 6σ = 200 S/m : ρ?
Challenging
  1. 7Why is α negative for a semiconductor?
  2. 8Why count both carriers?
Show answers (try them yourself first)
  1. 1σ = 1/ρ
  2. 2Si, Ge
  3. 3Rises
  4. 42 S/m
  5. 50.64 S/m
  6. 65 × 10⁻³ Ω m
  7. 7Carriers increase
  8. 8Both carry current

Band theory of solids

From daily life

Bulbs, solar cells and LEDs work with light or current — thanks to band theory.

Simple definition

Band theory : electron energies in a solid form bands; the gap between them is E_g.

Energy bandsCBVBconductorCBVBsemicond.CBVBinsulatorE_g : Si 1.1 eV, Ge 0.67 eVE_g (band gap)

In the figure : Three kinds : the gap makes the difference.

Rule 1: Bands — Valence band (VB) filled; conduction band (CB) empty/partly filled; the forbidden gap in between = E_g.

  • Easy : Conductor : VB and CB overlap.
  • Medium : Insulator : E_g > 3 eV (diamond ≈ 5.4).
  • Exam (CET) : Semiconductor E_g? (1) ~ 1 eV (2) 0 (3) 5 eV (4) 10 eV → answer 1)

Rule 2: E_g and wavelength — A photon with E ≥ E_g lifts an electron to CB : λ_max = 1240 ÷ E_g (nm).

  • Easy : Si 1.1 eV ⇒ λ_max = 1127 nm.
  • Medium : Ge 0.67 eV ⇒ 1851 nm.
  • Exam (CET) : Larger E_g ⇒ λ_max? (1) smaller (2) larger (3) same (4) zero → answer 1)

Rule 3: Thermal energy — At 300 K : kT = 0.026 eV ≪ E_g, so only a few electrons cross.

  • Easy : kT = 8.62×10⁻⁵ × 300 = 0.0259 eV.
  • Medium : At 0 K a semiconductor is an insulator.
  • Exam (CET) : Si at 0 K? (1) insulator (2) conductor (3) varies (4) semiconductor → answer 1)

Derivation : λ_max = 1240/E_g

  • Step 1 : E_g = h c / λ.
  • Step 2 : λ = hc / E_g.
  • Step 3 : hc = 1240 eV nm.
  • Step 4 : λ_max = 1240 ÷ E_g (nm).
Remember

λ_max = 1240/E_g (nm); kT = 0.026 eV (300 K).

Common mistake / CET trap

Large gap ⇒ insulator; overlap ⇒ conductor.

Shortcut

Memory clue : "Small gap = semiconductor".

Spot the difference

MaterialE_gBands
Conductor0Overlap
Ge0.67 eVSmall gap
Si1.1 eVSmall gap
Insulator> 3 eVLarge gap

Practice questions

Easy
  1. 1What is VB?
  2. 2What is E_g?
  3. 3Gap in a conductor?
Medium
  1. 4E_g = 2.0 eV : λ_max?
  2. 5Photon E at λ = 800 nm?
  3. 6Does 1000 nm light work for Si?
Challenging
  1. 7How does E_g vary with T?
  2. 8Why is diamond an insulator?
Show answers (try them yourself first)
  1. 1The filled valence band
  2. 2Width of the forbidden gap
  3. 30 (overlap)
  4. 4620 nm
  5. 51.55 eV
  6. 6Yes (1000 < 1127)
  7. 7Nearly constant
  8. 8Large E_g (5.4 eV)

Intrinsic semiconductors: electrons and holes

From daily life

When a seat empties in a row, neighbours shift; the empty seat moves back — that is a 'hole'.

Simple definition

A hole is a vacancy left by an electron, acting like a positive charge.

Electrons and holesholen_e = n_h = n_iI = I_e + I_hT ↑ ⇒ n_i ↑

In the figure : Lattice, hole and electron.

Rule 1: Intrinsic — Pure Si/Ge; electron-hole pairs; n_e = n_h = n_i.

  • Easy : 1 mm³ Si : n_i = 1.5×10¹⁶ m⁻³ × 10⁻⁹ m³ = 1.5×10⁷ electrons.
  • Medium : n_e = 2×10¹⁶ ⇒ n_h = 2×10¹⁶.
  • Exam (CET) : Pure : n_e : n_h? (1) 1 : 1 (2) 2 : 1 (3) 1 : 2 (4) 0 → answer 1)

Rule 2: Hole — A hole is a missing electron; it acts like a positive charge. I = I_e + I_h.

  • Easy : In a field e moves one way, holes the other; both currents add.
  • Medium : Hole charge +e = 1.6×10⁻¹⁹ C.
  • Exam (CET) : A hole behaves like? (1) +e (2) −e (3) neutron (4) photon → answer 1)

Rule 3: Temperature — T ↑ ⇒ n_i ↑ (exponentially); at 0 K n_i = 0.

  • Easy : T doubled ⇒ n_i far greater.
  • Medium : 0 K ⇒ no current.
  • Exam (CET) : n_i with T? (1) rises (2) falls (3) same (4) zero → answer 1)

Derivation : I = I_e + I_h

  • Step 1 : Electrons : I_e = n_e e A v_e.
  • Step 2 : Holes : I_h = n_h e A v_h.
  • Step 3 : Both flow the same way.
  • Step 4 : I = I_e + I_h.
Remember

n_e = n_h = n_i; I = I_e + I_h.

Common mistake / CET trap

A hole is not a real particle — a vacancy.

Shortcut

Memory clue : "One electron leaves, one hole arrives".

Spot the difference

FeatureElectronHole
Charge−e+e
BandCBVB
MobilityHigherLower

Practice questions

Easy
  1. 1What is an intrinsic semiconductor?
  2. 2What is a hole?
  3. 3What is n_i?
Medium
  1. 4n_e = 5×10¹⁵ : n_h?
  2. 5I_e = 3 mA, I_h = 1 mA : I?
  3. 6n_i as T rises?
Challenging
  1. 7Does a hole move or does an electron move?
  2. 8Pure Si at 0 K?
Show answers (try them yourself first)
  1. 1One without impurities
  2. 2A missing electron
  3. 3Carrier density in a pure semiconductor
  4. 45×10¹⁵
  5. 54 mA
  6. 6Rises
  7. 7Electrons shift; the hole seems to move
  8. 8An insulator

Extrinsic semiconductors: doping, n-type and p-type

From daily life

A pinch of sugar changes tea; a pinch of impurity changes silicon.

Simple definition

Doping is adding a small amount of impurity to a pure semiconductor.

n-type and p-typen-type (P)p-type (B)majority emajority holesn_e n_h = n_i²N_D / N_A (impurity)

In the figure : Majority carriers of n and p.

Rule 1: n-type — Pentavalent impurity (P, As, Sb) = donor; donor level just below CB; n_e ≈ N_D ≫ n_h.

  • Easy : N_D = 10²², n_i = 1.5×10¹⁶ ⇒ n_h = n_i²/N_D = 2.25 × 10¹⁰ m⁻³.
  • Medium : σ ≈ e n_e μ_e = 216 S/m.
  • Exam (CET) : Majority in n-type? (1) electrons (2) holes (3) photons (4) protons → answer 1)

Rule 2: p-type — Trivalent impurity (B, Al, Ga, In) = acceptor; acceptor level just above VB; n_h ≈ N_A ≫ n_e.

  • Easy : N_A = 10²¹ ⇒ n_e = 2.25 × 10¹¹ m⁻³.
  • Medium : p-type is neutral; only holes are the majority.
  • Exam (CET) : Acceptor impurity? (1) B (2) P (3) As (4) Sb → answer 1)

Rule 3: Mass-action law — n_e n_h = n_i² (fixed T); one rises, the other falls.

  • Easy : n_e doubled ⇒ n_h halves.
  • Medium : n_e = 10²⁰ ⇒ n_h = 2.25×10¹² m⁻³.
  • Exam (CET) : n_e n_h = ? (1) n_i² (2) n_i (3) 2n_i (4) 1 → answer 1)

Derivation : n_h = n_i²/N_D

  • Step 1 : n-type : n_e ≈ N_D.
  • Step 2 : n_e n_h = n_i².
  • Step 3 : n_h = n_i² / n_e.
  • Step 4 : n_h = n_i² / N_D (very small).
Remember

n_e n_h = n_i²; n : electrons majority; p : holes majority.

Common mistake / CET trap

n-type is not negatively charged — it is neutral.

Shortcut

Memory clue : "n = negative carriers; p = positive carriers".

Spot the difference

Featuren-typep-type
ImpurityPentavalent (donor)Trivalent (acceptor)
MajorityElectronsHoles
MinorityHolesElectrons
ExampleP, AsB, Al

Practice questions

Easy
  1. 1What is doping?
  2. 2Impurity for n-type?
  3. 3Is p-type charged?
Medium
  1. 4N_D = 10²¹ : n_h?
  2. 5N_A = 10²² : n_e?
  3. 6n_e = 3×10²⁰ : n_h?
Challenging
  1. 7Where is the donor level?
  2. 8Why is p-type neutral?
Show answers (try them yourself first)
  1. 1Adding impurity
  2. 2Pentavalent (P)
  3. 3No
  4. 42.25 × 10¹¹ m⁻³
  5. 52.25 × 10¹⁰ m⁻³
  6. 67.50 × 10¹¹ m⁻³
  7. 7Just below CB
  8. 8Atoms are neutral

p-n junction, depletion region and biasing

From daily life

A p-n junction is a one-way door for current.

Simple definition

A p-n junction is the boundary formed on joining p and n; it has a depletion region.

p-n junctionpnV_Bdepletion regionE_iforward : p → +reverse : p → −

In the figure : p, n and the depletion region.

Live figure

p-n junction, depletion region and biasing

In the app you can move this figure with your finger.

Open in the app

Rule 1: Depletion region — On joining, diffusion carries e and holes across; a layer of fixed ions forms the depletion region with built-in field E_i and barrier V_B (Si 0.7 V, Ge 0.3 V).

  • Easy : V_B = 0.7 V, width 0.5 µm ⇒ E = 1.4 × 10⁶ V/m.
  • Medium : V_B = 0.3 V for Ge.
  • Exam (CET) : V_B of Si? (1) 0.7 V (2) 0.3 V (3) 5 V (4) 0 → answer 1)

Rule 2: Forward bias — p → battery +; barrier falls, depletion region narrows; mA current flows.

  • Easy : Battery 1.5 V, R = 800 Ω, V_B = 0.7 ⇒ I = (1.5 − 0.7)/800 = 1 mA.
  • Medium : Battery 0.5 V (< 0.7) ⇒ I ≈ 0.
  • Exam (CET) : Depletion region in forward bias? (1) narrow (2) wide (3) same (4) vanishes → answer 1)

Rule 3: Reverse bias — p → battery −; barrier rises, region widens; only a µA leakage of minority carriers.

  • Easy : Reverse bias 5 V ⇒ I ≈ µA.
  • Medium : In reverse bias the diode ≈ open.
  • Exam (CET) : Reverse bias : current? (1) tiny (2) large (3) always zero (4) keeps rising → answer 1)

Derivation : Forming the depletion region

  • Step 1 : More e in n, more holes in p ⇒ diffusion.
  • Step 2 : e cross, recombine with holes; +ions remain on n side.
  • Step 3 : −ions on p side ⇒ built-in E_i (n → p).
  • Step 4 : E_i stops diffusion ⇒ equilibrium, V_B.
Remember

V_B : Si 0.7 V, Ge 0.3 V; forward = narrow, reverse = wide.

Common mistake / CET trap

Forward current flows only after V_B; before that ≈ 0.

Shortcut

Memory clue : "+ to p = forward".

Spot the difference

Biasp joined toDepletionI
Forward+NarrowmA
Reverse−WideµA

Practice questions

Easy
  1. 1What is the depletion region?
  2. 2How to forward bias?
  3. 3V_B of Ge?
Medium
  1. 43 V, R = 460 Ω (Si) : I?
  2. 5V_B = 0.3, width 1 µm : E?
  3. 60.4 V forward on Si : I?
Challenging
  1. 7Why leakage in reverse bias?
  2. 8Can V_B be measured with a voltmeter?
Show answers (try them yourself first)
  1. 1An ion layer free of carriers
  2. 2p to +
  3. 30.3 V
  4. 45 mA
  5. 53 × 10⁵ V/m
  6. 6≈ 0
  7. 7Minority carriers
  8. 8No

Junction diode and its characteristics

From daily life

A phone charger turns wall AC into DC — thanks to diodes.

Simple definition

A diode is a p-n junction that conducts mainly one way.

Diode I-VV_kVI (mA)forwardreversebreakdown

In the figure : Forward current rises after V_k.

Rule 1: I-V curve — Forward : after V_k (Si 0.7 V) I rises fast; reverse : µA, at breakdown V_BR I jumps.

  • Easy : V_k = 0.7 V; I ≈ 0 at 0.5 V.
  • Medium : Breakdown at reverse 20 V ⇒ danger.
  • Exam (CET) : Is a diode linear? (1) no (2) yes (3) forward only (4) reverse only → answer 1)

Rule 2: Dynamic resistance — r_d = ΔV / ΔI; small in forward bias; very large in reverse.

  • Easy : ΔV = 0.1 V, ΔI = 20 mA ⇒ r_d = 5 Ω.
  • Medium : Reverse : ΔV = 5 V, ΔI = 1 µA ⇒ 5 MΩ.
  • Exam (CET) : r_d in forward? (1) small (2) large (3) infinite (4) zero → answer 1)

Rule 3: Rectifier — Half-wave : V_dc = V_m/π, ripple f = f; full-wave : V_dc = 2V_m/π, ripple f = 2f.

  • Easy : V_m = 10 V ⇒ half V_dc = 3.18 V; full = 6.37 V.
  • Medium : AC 50 Hz ⇒ half ripple 50 Hz, full 100 Hz.
  • Exam (CET) : Full-wave ripple f? (1) 2f (2) f (3) f/2 (4) 4f → answer 1) Trick : both half-cycles used ⇒ double.

Derivation : V_dc = V_m/π

  • Step 1 : Half-wave : v = V_m sin ωt (0 to π), then 0.
  • Step 2 : Average = (1/2π) ∫₀^π V_m sin θ dθ.
  • Step 3 : = (V_m/2π) × 2.
  • Step 4 : V_dc = V_m/π ≈ 0.318 V_m.
Remember

r_d = ΔV/ΔI; V_dc (HW) = V_m/π; (FW) = 2V_m/π.

Common mistake / CET trap

r_d is ΔV/ΔI, not V/I.

Shortcut

Memory clue : "Full = double".

Spot the difference

FeatureHalf-waveFull-wave
Diodes12 / 4
V_dcV_m/π2V_m/π
Ripple ff2f

Practice questions

Easy
  1. 1What is V_k?
  2. 2What is a rectifier?
  3. 3Ripple f of full-wave?
Medium
  1. 4V_m = 20 V : half-wave V_dc?
  2. 5Same : full-wave V_dc?
  3. 6ΔV = 0.2, ΔI = 10 mA : r_d?
Challenging
  1. 7Half-wave ripple 60 Hz ⇒ source f?
  2. 8Why is reverse r_d large?
Show answers (try them yourself first)
  1. 1Knee voltage where forward current begins
  2. 2AC → DC
  3. 32f
  4. 46.37 V
  5. 512.73 V
  6. 620 Ω
  7. 760 Hz
  8. 8Tiny I

Semiconductor devices, applications and thermistors

From daily life

A phone screen, a red charging light and a solar panel — all semiconductor devices.

Simple definition

Special diodes : Zener, LED, photodiode, solar cell; a thermistor : T-dependent resistance.

Special diodesZenerconstant VLEDlightPhotolight → ISolarlight → VNTCT ↑ ⇒ R ↓λ = 1240/E_gI_Z = (V − V_Z)/R

In the figure : Each device and its job.

Rule 1: Zener — Constant V_Z at reverse breakdown; voltage regulator : I = (V_in − V_Z) / R_s.

  • Easy : V_in = 12, V_Z = 6, R_s = 300 Ω ⇒ I = 20 mA.
  • Medium : V_in 14 ⇒ I = 26.7 mA; output still 6 V.
  • Exam (CET) : Zener is connected? (1) reverse biased (2) forward (3) AC (4) series only → answer 1)

Rule 2: LED, photodiode, solar cell — LED : forward bias, emits light; photodiode : reverse bias, light ⇒ I; solar cell : light ⇒ V (no bias). λ = 1240 / E_g.

  • Easy : E_g = 1.9 eV ⇒ λ = 653 nm (red).
  • Medium : E_g = 2.6 eV ⇒ λ = 477 nm (blue).
  • Exam (CET) : Photodiode bias? (1) reverse (2) forward (3) none (4) AC → answer 1)

Rule 3: Thermistor — NTC : T ↑ ⇒ R ↓ (semiconductor); PTC : T ↑ ⇒ R ↑. Uses : thermometers, protection.

  • Easy : Hot NTC ⇒ lower resistance.
  • Medium : PTC works like a protective fuse.
  • Exam (CET) : NTC : T ↑ ⇒ R? (1) ↓ (2) ↑ (3) same (4) 0 → answer 1)

Derivation : Zener regulator

  • Step 1 : R_s in series; Zener in parallel with the load.
  • Step 2 : Voltage across the Zener is V_Z.
  • Step 3 : R_s drops V_in − V_Z.
  • Step 4 : I = (V_in − V_Z) / R_s.
Remember

I = (V_in − V_Z)/R_s; λ = 1240/E_g.

Common mistake / CET trap

Photodiode reverse; LED forward — do not swap.

Shortcut

Memory clue : "LED forward, photo reverse, Zener reverse".

Spot the difference

DeviceBiasJob
ZenerReverseConstant V
LEDForwardEmits light
PhotodiodeReverseLight → I
Solar cellNoneLight → V

Practice questions

Easy
  1. 1Use of a Zener?
  2. 2Which bias for an LED?
  3. 3What is an NTC?
Medium
  1. 4V_in = 10, V_Z = 5.6, R_s = 220 : I?
  2. 5E_g = 1.77 eV : λ?
  3. 6V_in 9 → 15 : Zener V?
Challenging
  1. 7Why no bias for a solar cell?
  2. 8How does a PTC protect?
Show answers (try them yourself first)
  1. 1Regulate voltage
  2. 2Forward
  3. 3T ↑ ⇒ R ↓
  4. 420 mA
  5. 5701 nm
  6. 6V_Z (constant)
  7. 7It generates its own V
  8. 8Hot ⇒ R ↑ ⇒ I falls

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
Conductivityσ = e(n_eμ_e + n_hμ_h)S/m
λ_max1240 / E_gnm
Intrinsicn_e = n_h = n_im⁻³
Extrinsicn_e n_h = n_i²m⁻⁶
V_BSi 0.7, Ge 0.3V
Dynamic Rr_d = ΔV/ΔIΩ
V_dcV_m/π ; 2V_m/πV
ZenerI = (V_in − V_Z)/R_sA
LEDλ = 1240/E_gnm

🎯 Most important points for the exam

  • ρ : conductor 10⁻⁸, semiconductor in between, insulator > 10⁸; T ↑ ⇒ semiconductor R ↓.
  • E_g : Si 1.1, Ge 0.67 eV; λ_max = 1240/E_g.
  • Intrinsic n_e = n_h; I = I_e + I_h.
  • n : pentavalent; p : trivalent; n_e n_h = n_i².
  • Forward bias : p → +, barrier falls; V_B of Si 0.7 V.
  • Full-wave V_dc = 2V_m/π, ripple 2f.
  • Zener reverse; LED forward; photodiode reverse; NTC T ↑ ⇒ R ↓.
Study mantra

ρ → bands → e/hole → n/p → junction → diode → Zener/LED.