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Physics · Class 12 · Chapter 14

Semiconductor Electronics: Materials, Devices and Simple Circuits

Every phone and computer is built on the p-n junction. This chapter explains why semiconductors conduct the way they do, how a junction forms, and how a diode turns AC into DC. It is short and scoring if you keep the carriers and bias directions straight.

In this chapter: classification of solids by conductivity and by energy bands, intrinsic semiconductors, electrons and holes, n-type and p-type semiconductors, the relation nenh = ni2, formation of a p-n junction, depletion region and barrier potential, the diode in forward and reverse bias, V-I characteristics, half-wave and full-wave rectifiers.

Classification of solids

On the basis of conductivity (NCERT values):

TypeResistivity ρ (Ω m)Conductivity σ (S m−1)
Metals10−2 to 10−8102 to 108
Semiconductors10−5 to 106105 to 10−6
Insulators1011 to 101910−11 to 10−19

Elemental semiconductors are silicon and germanium. Compound semiconductors include inorganic ones such as CdS, GaAs, CdSe and InP, and organic ones such as anthracene and polymers like polypyrrole.

Energy bands

In a crystal, the energy levels of the atoms spread into bands. The valence band holds the valence electrons; above it lies the conduction band, and the gap between them is the energy gap Eg.

  • Metals: the conduction band is partly filled or overlaps the valence band, so many electrons are free to move.
  • Insulators: the gap is large (Eg > 3 eV); diamond, for example, has about 5.4 eV. Thermal energy cannot lift electrons across.
  • Semiconductors: the gap is small (Eg < 3 eV): about 1.1 eV for silicon and 0.7 eV for germanium. At room temperature a few electrons cross the gap and conduct; at 0 K a semiconductor behaves as an insulator.

Intrinsic semiconductors

In pure Si or Ge, each atom forms four covalent bonds with its neighbours. When thermal energy breaks a bond, a free electron is released and leaves behind a vacancy with an effective charge +e, called a hole. A neighbouring bound electron can jump into the hole, so the hole effectively moves and carries current. In a pure semiconductor

ne = nh = niI = Ie + Ihni is the intrinsic carrier concentration; it rises steeply with temperature

Electrons and holes also recombine continuously, and at equilibrium the rate of generation equals the rate of recombination. Because ni rises with temperature, the resistivity of a semiconductor falls as it gets hotter, the opposite of a metal.

Extrinsic semiconductors

Adding a tiny amount of a suitable impurity (doping, a few parts per million) increases conductivity enormously. The dopant atoms should be about the same size as the host atoms so they fit into the lattice without distorting it.

 n-typep-type
DopantPentavalent (group 15): arsenic, antimony, phosphorusTrivalent (group 13): indium, boron, aluminium
Name of dopantDonor (gives an extra electron)Acceptor (creates a hole)
Majority carriersElectronsHoles
Minority carriersHolesElectrons
Energy levelDonor level just below the conduction bandAcceptor level just above the valence band

An n-type or p-type semiconductor is still electrically neutral: the extra electrons of an n-type crystal are balanced by the positive donor ions. In thermal equilibrium the product of the carrier concentrations is fixed:

nenh = ni2

Doping one type of carrier therefore suppresses the other, since extra electrons recombine with holes.

Worked example: A silicon crystal (5 × 1028 atoms m−3, ni = 1.5 × 1016 m−3) is doped with 1 ppm of arsenic. Find the electron and hole concentrations.
Solution: Donor atoms = 5 × 1028 × 10−6 = 5 × 1022 m−3, so ne ≈ 5 × 1022 m−3 (thermally generated electrons are negligible in comparison). nh = ni2/ne = (2.25 × 1032)/(5 × 1022) = 4.5 × 109 m−3.

The p-n junction

When p-type and n-type regions are formed in one crystal, electrons diffuse from the n side (where they are many) to the p side, and holes diffuse from p to n. This diffusion current flows from p to n. Each electron that leaves the n side uncovers a fixed positive donor ion, and each hole that leaves the p side uncovers a fixed negative acceptor ion. A thin layer on both sides of the junction is thus emptied of free carriers; it is called the depletion region.

The fixed ions set up an electric field from the n side to the p side. This field sweeps minority carriers across the junction, giving a drift current opposite to the diffusion current. At equilibrium the two currents are equal and there is no net current. The potential difference across the depletion region is the barrier potential V0, with the n side at the higher potential.

p-n junction and diode characteristicwww.iitmedicoguide.compndepletion−+−+−+E (n to p)fixed ions, no free carriersV₀potential across the junction(a) p-n junction at equilibriumVI (mA)I (μA)≈ 0.7 Vforward biasreverse biasbreakdownVbrreverse saturationcurrent (μA)(b) Si diode V-I characteristicnote the different scales: mA forward, μA reversewww.iitmedicoguide.com
Left: the depletion region holds only fixed ions, and its field points from n to p. Right: a silicon diode conducts appreciably in forward bias only beyond about 0.7 V, while in reverse bias a tiny current flows until breakdown.

Semiconductor diode

A p-n junction with metal contacts at its ends is a diode. Its symbol is an arrow (the p side) pointing to a bar (the n side); the arrow shows the direction of conventional current in forward bias.

Forward bias

The p side is connected to the positive terminal. The applied voltage opposes the built-in potential, so the effective barrier becomes V0 − V and the depletion region becomes narrower. Majority carriers cross in large numbers and the current rises sharply once the applied voltage exceeds the threshold (cut-in) voltage, about 0.7 V for silicon and 0.2 V for germanium. Forward current is of the order of milliamperes.

Reverse bias

The n side is connected to the positive terminal. The barrier becomes V0 + V and the depletion region widens. Diffusion of majority carriers almost stops; only minority carriers drift across, giving a small reverse current of a few microamperes that hardly changes with voltage (the reverse saturation current). At a large enough reverse voltage, the breakdown voltage Vbr, the reverse current rises suddenly; an ordinary diode can be damaged if operated beyond it.

The dynamic resistance of a diode is rd = ΔV/ΔI at a given point on the characteristic. It is low in forward bias and very high in reverse bias, which is the one-way behaviour that makes rectification possible.

Worked example: On the forward characteristic of a silicon diode, the current rises from 10 mA to 30 mA when the voltage rises from 0.7 V to 0.8 V. Find the dynamic resistance in this range.
Solution: rd = ΔV/ΔI = 0.1 V/(20 × 10−3 A) = 5 Ω.

The diode as a rectifier

A rectifier converts AC into DC (in one direction, though not steady).

  • Half-wave rectifier: a single diode in series with the load. It conducts only during the half cycle in which it is forward biased, so the output has one hump per cycle and the other half cycle is lost. The output frequency equals the input frequency (50 Hz for mains).
  • Full-wave rectifier: two diodes and a centre-tap transformer. The two ends of the secondary are always in opposite phase with respect to the centre tap, so D1 conducts in one half cycle and D2 in the other. Current through the load is in the same direction in both halves, and the output has two humps per input cycle, so the ripple frequency is 100 Hz for a 50 Hz input.
Full-wave rectifier with centre-tap transformerwww.iitmedicoguide.comAC inputD₁D₂RLcentre tap+−current in RL same wayin both half cyclesABInput voltage (50 Hz)Output across RL (100 Hz ripple)teal: D₁ conducts; amber: D₂Full-wave rectifier with a centre-tap transformer; a capacitor across RL smooths the outputwww.iitmedicoguide.com
In a full-wave rectifier D1 and D2 conduct in alternate half cycles, but both send current through RL in the same direction. The output is a series of positive pulses at twice the input frequency.

The pulsating output is smoothed by a filter, usually a capacitor connected across the load. It charges to the peak voltage while a diode conducts and discharges slowly through the load in between, so the output stays close to the peak. A larger capacitor (larger CRL) gives a smoother output.

Zener diodes, optoelectronic devices (photodiode, LED, solar cell), transistors and logic gates were removed from the rationalised NCERT book. Some of them still appear in the JEE Main syllabus, so check the current syllabus of your exam.

Common mistakes: (1) Saying an n-type semiconductor is negatively charged; it is neutral. (2) Mixing up donors and acceptors: pentavalent atoms give n-type, trivalent atoms give p-type. (3) Saying the depletion region widens in forward bias; it narrows in forward bias and widens in reverse bias. (4) Thinking reverse current is due to majority carriers; it is due to minority carriers. (5) Giving 50 Hz as the output frequency of a full-wave rectifier on 50 Hz mains; it is 100 Hz.

JEE and NEET focus

  • Energy band pictures of metals, semiconductors and insulators, with typical band gaps.
  • n-type and p-type doping, majority and minority carriers, and nenh = ni2 numericals.
  • Formation of the depletion layer, barrier potential, diffusion and drift currents.
  • Forward and reverse bias behaviour, the V-I characteristic and dynamic resistance.
  • Half-wave and full-wave rectifier circuits, output waveforms and ripple frequency; circuits with ideal diodes.

Practice questions

In an n-type silicon crystal, the majority carriers and the type of dopant are:

  1. holes, trivalent
  2. electrons, trivalent
  3. electrons, pentavalent
  4. holes, pentavalent
Show answer
C. A pentavalent donor gives an extra electron.

The energy gap of an insulator is typically:

  1. zero
  2. less than 1 eV
  3. about 1 eV
  4. more than 3 eV
Show answer
D. Diamond, for instance, has about 5.4 eV.

When a p-n junction is forward biased, the width of the depletion region:

  1. increases
  2. decreases
  3. stays the same
  4. becomes infinite
Show answer
B. The applied voltage opposes the barrier.

The small current in a reverse-biased diode below breakdown is due to:

  1. majority carriers
  2. minority carriers
  3. donor ions
  4. acceptor ions
Show answer
B. The junction field sweeps minority carriers across.

The input to a full-wave rectifier is 50 Hz AC. The fundamental frequency of the output ripple is:

  1. 25 Hz
  2. 50 Hz
  3. 100 Hz
  4. 200 Hz
Show answer
C. There are two output pulses per input cycle.

In a p-type semiconductor, ni = 1.5 × 1016 m−3 and nh = 4.5 × 1022 m−3. The electron concentration is:

  1. 5 × 109 m−3
  2. 1.5 × 1016 m−3
  3. 3 × 106 m−3
  4. 4.5 × 1022 m−3
Show answer
A. ne = ni2/nh = 2.25 × 1032/4.5 × 1022.

As the temperature of a semiconductor rises, its resistivity:

  1. increases
  2. decreases
  3. stays constant
  4. first decreases then increases
Show answer
B. More electron-hole pairs are generated.

To make p-type silicon, the dopant used could be:

  1. phosphorus
  2. arsenic
  3. boron
  4. antimony
Show answer
C. Boron is trivalent and acts as an acceptor.
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