JEE Main Physics · Chapter 22

Semiconductor Electronics

Energy bands, doping, p–n junction diode, rectifiers, Zener diode, LED, photodiode and logic gates.

NCERT Class 12JEE Main: 1 Q per shift 9 min read

1. Chapter at a Glance

Semiconductor electronics explains how materials like silicon conduct, how doping controls their conductivity, and how p–n junction devices work. The current JEE Main syllabus covers energy bands, intrinsic and extrinsic semiconductors, the p–n junction diode and its I–V characteristics, rectifiers, special diodes (LED, photodiode, solar cell, Zener diode) and logic gates. It is a scoring chapter with one question almost every year.

What JEE asks most
  • Carrier concentrations: nenh = ni².
  • Forward and reverse bias; current in diode circuits.
  • Half-wave and full-wave rectifiers; Zener regulator calculations.
  • Truth tables and combinations of logic gates.

2. Energy Bands and Semiconductors

  • In solids, atomic levels spread into bands: the valence band (filled) and conduction band, separated by an energy gap Eg.
  • Conductors: bands overlap. Insulators: Eg > 3 eV. Semiconductors: Eg ~ 1 eV (Si 1.1 eV, Ge 0.7 eV).
  • Resistance of a semiconductor decreases as temperature rises (more electrons cross the gap).
  • Intrinsic (pure) semiconductor: ne = nh = ni.
  • n-type: doped with pentavalent atoms (P, As, Sb — donors); majority carriers are electrons. p-type: doped with trivalent atoms (B, Al, In — acceptors); majority carriers are holes. Both remain electrically neutral.
  • Mass-action law: nenh = ni². Conductivity σ = e(neμe + nhμh).

3. The p–n Junction Diode

  • Diffusion of carriers across the junction creates a depletion layer with no free carriers and a barrier potential (≈ 0.7 V for Si, 0.3 V for Ge).
  • Forward bias (p to +): barrier lowers, depletion layer narrows, current rises sharply after the knee voltage.
  • Reverse bias: barrier rises, depletion layer widens, only a tiny reverse saturation current (minority carriers) flows until breakdown.
  • Dynamic resistance rd = ΔV/ΔI.
0.7 V forward currentbreakdown tiny reverse currentVI
Silicon diode: current rises steeply beyond the knee voltage (~0.7 V) in forward bias; in reverse bias only a tiny current flows until breakdown.

4. Rectifiers and Special Diodes

  • Half-wave rectifier: one diode; output frequency = input frequency (50 Hz).
  • Full-wave rectifier (centre-tap or bridge): output frequency = 2 × input (100 Hz). A capacitor filter smooths the output.
  • Zener diode: heavily doped, works in reverse breakdown with a nearly constant voltage VZ — used as a voltage regulator.
  • LED: forward-biased; electron–hole recombination emits light of energy ≈ Eg (λ = 1240/Eg nm).
  • Photodiode: reverse-biased; light of energy > Eg creates carriers and increases reverse current — used as a light detector.
  • Solar cell: unbiased p–n junction that converts light to electricity; needs Eg around 1–1.8 eV (Si, GaAs).

5. Logic Gates

ABAND (A·B)OR (A+B)NANDNORXOR
0000110
0101101
1001101
1111000
  • NOT gate: Y = Ā. NAND and NOR are universal gates — any gate can be built from them alone.
  • De Morgan's laws: (A + B)′ = Ā·B̄ and (A·B)′ = Ā + B̄.
  • NAND with both inputs joined = NOT; NAND followed by NOT = AND; NOT on each input then NAND = OR.

6. Formula Master Table

QuantityFormula
Mass-action lawnenh = ni²
Conductivityσ = e(neμe + nhμh)
Diode current (simple model)I = (V − Vknee)/R
Zener regulatorIs = (Vin − VZ)/Rs; IZ = Is − IL
LED / cut-off wavelengthλ = 1240/Eg(eV) nm
Rectifier ripple frequencyhalf-wave f; full-wave 2f
De Morgan(A + B)′ = Ā·B̄; (A·B)′ = Ā + B̄

7. Mind Map

Semiconductors
Bands
  • Conductor, semiconductor, insulator
  • Intrinsic, n-type, p-type
p–n junction
  • Depletion layer
  • Forward, reverse bias
  • I–V curve
Devices
  • Rectifiers
  • Zener, LED
  • Photodiode, solar cell
Logic
  • AND, OR, NOT
  • NAND, NOR universal
  • De Morgan

8. Tips and Tricks

Shortcuts
  • In doped material, majority carrier ≈ dopant concentration; minority = ni²/majority.
  • In a diode circuit, first decide which diodes are forward biased; treat an ideal forward diode as a wire and a reverse diode as a break.
  • For a gate combination, write the Boolean expression step by step, then simplify with De Morgan.
  • A Zener regulator works only while IZ > 0; this sets the smallest allowed load resistance.
Mistakes students make
  • Thinking doped semiconductors carry a net charge — they are neutral.
  • Using forward bias for a photodiode or Zener regulator (both work in reverse).
  • Forgetting the two diode drops in a bridge rectifier.

9. Important One-liners

  • Majority carriers in n-type semiconductors are electrons.
  • The depletion layer widens under reverse bias.
  • The resistance of a semiconductor decreases with rise in temperature.
  • A Zener diode is used as a voltage regulator.
  • NAND and NOR gates are universal gates.
  • Full-wave rectifier output has twice the input frequency.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working.

Average level

N1Average

Silicon (ni = 1.5 × 10¹⁶ m⁻³) is doped with 10²² donor atoms per m³. Find the electron and hole concentrations.

Show answer & solution

Concept: ne ≈ ND; nh = ni²/ne.

  1. ne ≈ 10²² m⁻³.
  2. nh = (1.5 × 10¹⁶)²/10²² = 2.25 × 10³²/10²².
  3. = 2.25 × 10¹⁰ m⁻³.

Answer: 10²² m⁻³ electrons; 2.25 × 10¹⁰ m⁻³ holes

N2Average

A 50 Hz supply is fed to (a) a half-wave rectifier, (b) a full-wave rectifier. Find the output ripple frequency in each.

Show answer & solution

Concept: half-wave passes one pulse per cycle; full-wave passes two.

  1. (a) 50 Hz.
  2. (b) 2 × 50 = 100 Hz.

Answer: 50 Hz; 100 Hz

N3Average

A silicon diode (knee voltage 0.7 V) is connected in forward bias with a 1 kΩ resistor across a 5 V battery. Find the current.

Show answer & solution

Concept: voltage across R = supply − diode drop.

  1. VR = 5 − 0.7 = 4.3 V.
  2. I = 4.3/1000 = 4.3 mA.

Answer: 4.3 mA

N4Average

A 10 V Zener diode is used with a 500 Ω series resistor on a 15 V supply, feeding a 2 kΩ load. Find the currents through the series resistor, the load and the Zener.

Show answer & solution

Concept: load voltage is held at VZ; IZ = Is − IL.

  1. Is = (15 − 10)/500 = 10 mA.
  2. IL = 10/2000 = 5 mA.
  3. IZ = 10 − 5 = 5 mA.

Answer: 10 mA; 5 mA; 5 mA

N5Average

An LED is made from a semiconductor with band gap 1.9 eV. Estimate the wavelength and colour of its light.

Show answer & solution

Concept: photon energy ≈ Eg; λ = 1240/Eg nm.

  1. λ = 1240/1.9 ≈ 653 nm.
  2. This is in the red part of the spectrum.

Answer: about 653 nm (red)

Advanced level

N6Advanced

Inputs A and B go into a NOR gate, and its output goes into a NOT gate. Write the truth table and identify the equivalent single gate. How would you build the same gate using only NAND gates?

Show answer & solution

Concept: NOT of NOR restores OR; De Morgan: A + B = (Ā·B̄)′.

  1. NOR outputs for (00, 01, 10, 11): 1, 0, 0, 0.
  2. After NOT: 0, 1, 1, 1 — this is the OR gate.
  3. With NANDs: invert A and B using NANDs with joined inputs, then feed Ā and B̄ into a third NAND: (Ā·B̄)′ = A + B.
  4. So three NAND gates make one OR gate.

Answer: OR gate; three NAND gates

N7Advanced

A Zener regulator has Vin = 20 V, VZ = 12 V and series resistance 400 Ω. Find the smallest load resistance for which regulation is maintained.

Show answer & solution

Concept: regulation needs IZ ≥ 0, so IL ≤ Is.

  1. Is = (20 − 12)/400 = 20 mA.
  2. Maximum load current = 20 mA.
  3. RL,min = 12/0.020 = 600 Ω.

Answer: 600 Ω

N8Advanced

An intrinsic semiconductor has ni = 2.5 × 10¹⁹ m⁻³, electron mobility 0.39 m² V⁻¹ s⁻¹ and hole mobility 0.19 m² V⁻¹ s⁻¹. Find its conductivity and resistivity.

Show answer & solution

Concept: σ = nie(μe + μh) for an intrinsic semiconductor.

  1. μe + μh = 0.58 m² V⁻¹ s⁻¹.
  2. σ = 2.5 × 10¹⁹ × 1.6 × 10⁻¹⁹ × 0.58 = 2.32 S m⁻¹.
  3. ρ = 1/σ ≈ 0.43 Ω m.

Answer: 2.32 S m⁻¹; about 0.43 Ω m

N9Advanced

A bridge rectifier made of silicon diodes (0.7 V each) is fed with 12 V rms AC. Find the peak output voltage.

Show answer & solution

Concept: in a bridge, two diodes conduct in each half-cycle, so two drops are subtracted from the peak.

  1. Input peak = 12√2 ≈ 16.97 V.
  2. Drop = 2 × 0.7 = 1.4 V.
  3. Output peak ≈ 16.97 − 1.4 ≈ 15.6 V.

Answer: about 15.6 V

N10Advanced

Silicon has a band gap of 1.1 eV. Find the longest wavelength a silicon photodiode can detect. Can it detect light of 1300 nm?

Show answer & solution

Concept: a photon must have energy ≥ Eg to create an electron–hole pair: λmax = 1240/Eg.

  1. λmax = 1240/1.1 ≈ 1127 nm.
  2. Light of 1300 nm has energy 1240/1300 ≈ 0.95 eV < 1.1 eV.
  3. So silicon cannot detect it (a smaller-gap material like Ge would).

Answer: about 1127 nm; no

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

In an n-type semiconductor, the majority charge carriers are:

  1. Holes
  2. Electrons
  3. Protons
  4. Positive ions
Show solution

Pentavalent donors add free electrons.

Answer: (B)

Q2Easy

Under reverse bias, the depletion layer of a p–n junction:

  1. Narrows
  2. Widens
  3. Disappears
  4. Stays the same
Show solution

The external field adds to the barrier field, pulling carriers away from the junction.

Answer: (B)

Q3Numerical

Find the output of a NAND gate when both inputs are 1.

Show solution

AND gives 1; NOT of that is 0.

Answer: 0

Q4Medium

A Zener diode is mainly used as a:

  1. Rectifier
  2. Amplifier
  3. Voltage regulator
  4. Light source
Show solution

In reverse breakdown its voltage stays almost constant over a wide current range.

Answer: (C)

Q5Hard

The correct order of energy gaps is:

  1. Conductor > semiconductor > insulator
  2. Insulator > semiconductor > conductor
  3. Semiconductor > insulator > conductor
  4. All equal
Show solution

Insulators have large gaps (> 3 eV), semiconductors about 1 eV, conductors none (overlapping bands).

Answer: (B)

Q6Numerical

A full-wave rectifier is fed from a 60 Hz supply. Find the ripple frequency of the output in Hz.

Show solution

2 × 60 = 120 Hz.

Answer: 120

12. Quick Revision Summary

Remember before the exam
  • Eg: insulator > semiconductor > conductor; nenh = ni².
  • n-type: donors, electrons majority; p-type: acceptors, holes majority.
  • Forward bias narrows, reverse bias widens the depletion layer.
  • Half-wave f, full-wave 2f; Zener regulates in reverse breakdown.
  • LED λ = 1240/Eg; NAND and NOR are universal.
All Physics chaptersJEE Main Mock Tests