Semiconductor Electronics Class 12 Notes | CBSE Physics Chapter 14

Chapter summary

Semiconductor Electronics explains how the energy band gap sorts solids into conductors, insulators and semiconductors, and how doping pure silicon or germanium creates n-type and p-type material that combine into the p-n junction diode. It builds up to real devices like rectifiers, special diodes, transistors and logic gates, making it a high-yield, formula-and-concept chapter for NEET.

Chapter notes

Key Concepts

1. Energy Bands in Solids

TypeBand GapExamples
ConductorNo gap (bands overlap)Copper, silver, aluminium
SemiconductorSmall gap (~1 eV)Silicon (1.1 eV), Germanium (0.7 eV)
InsulatorLarge gap (> 3 eV)Diamond, rubber, glass

2. Intrinsic and Extrinsic Semiconductors

Intrinsic: Pure semiconductor (equal electrons and holes). Conductivity is low.

Extrinsic: Doped with impurities to increase conductivity.

TypeDopantMajority CarriersExample
n-typePentavalent (P, As, Sb)ElectronsSi doped with As
p-typeTrivalent (B, Al, Ga, In)HolesSi doped with B

3. P-N Junction Diode

When p-type and n-type are joined, a depletion region forms at the junction (thin region with no free carriers).

  • Forward bias: P connected to +, N to −. Current flows (barrier reduced). Diode conducts above ~0.7 V (Si) or ~0.3 V (Ge).
  • Reverse bias: P to −, N to +. Very small reverse current (barrier increased). Diode blocks.

Applications of Diode

  • Rectifier: Converts AC to DC (half-wave or full-wave)
  • Zener diode: Voltage regulator (maintains constant output voltage)
  • LED: Emits light when forward biased
  • Photodiode: Generates current when light falls on it (reverse biased)
  • Solar cell: Converts light to electricity (no external bias needed)

4. Transistor

A transistor has three layers (NPN or PNP) and three terminals: Emitter (E), Base (B), Collector (C).

Current relation: IE = IB + IC (emitter current = base + collector)

Current gain: β = IC/IB (typically 20–200)

Transistor as Amplifier (Common Emitter)

Small change in base current → large change in collector current.

Voltage gain: Av = −βRL/Ri

Input: base-emitter circuit; Output: collector-emitter circuit.

Transistor as Switch

Cutoff region: both junctions reverse biased → transistor OFF (switch open)

Saturation region: both junctions forward biased → transistor ON (switch closed)

5. Logic Gates

GateSymbolBooleanOutput
ANDA · BY = A·B1 only if both inputs are 1
ORA + BY = A+B1 if any input is 1
NOTĀY = ĀInverts input
NAND(A·B)̄Y = (A·B)̄0 only if both inputs are 1
NOR(A+B)̄Y = (A+B)̄1 only if both inputs are 0

NAND and NOR are universal gates - any logic function can be built using only NAND or only NOR gates.


Solved Examples

Example 1

In a transistor, IB = 50 μA and IC = 5 mA. Find IE and β.

Answer: IE = IB + IC = 0.05 + 5 = 5.05 mA. β = IC/IB = 5/0.05 = 100.

Example 2

A Zener diode has breakdown voltage 5 V. Input voltage varies from 7–12 V with 500 Ω series resistance. Find current range.

Answer: I = (V_in − V_z)/R. Min: (7−5)/500 = 4 mA. Max: (12−5)/500 = 14 mA. Output stays at 5 V.


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Quick Revision Points

  • Semiconductor band gap: Si = 1.1 eV, Ge = 0.7 eV
  • n-type: pentavalent dopant (majority: electrons); p-type: trivalent (majority: holes)
  • Diode: forward bias → conducts; reverse bias → blocks
  • Zener: voltage regulator; LED: light emission; Solar cell: light → electricity
  • Transistor: IE = IB + IC; β = IC/IB; amplifier (CE mode): Av = −βRL/Ri
  • Logic gates: AND, OR, NOT, NAND (universal), NOR (universal)

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First Chapter: Ch 1 - Electric Charges and Fields

🃏 Flash Cards: Semiconductor Electronics

Class 12 Physics · Chapter 14 – swipe through all 10 cards to understand the whole chapter.

🔋Start here1/10

Energy Bands of Solids

In a solid, atomic levels split into closely-spaced energy bands, and the gap between two key bands decides if it conducts.

Valence band (filled) → E_g → Conduction band (free)

E_g = forbidden energy gap, measured in eV.

  • Valence band holds bonding electrons; conduction band carries current
  • Free electrons can carry current only after crossing the gap E_g
  • An electron promoted to the CB leaves a hole in the VB
⚖️Core idea2/10

Metals vs Insulators vs Semiconductors

The size of E_g cleanly sorts every solid into conductor, insulator or semiconductor.

Metal: E_g ≈ 0 (overlap) · Semiconductor: E_g ≈ 1 eV · Insulator: E_g > 3 eV

Si: E_g ≈ 1.1 eV, Ge ≈ 0.7 eV, diamond ≈ 6 eV.

  • Metals: VB and CB overlap, so plenty of free electrons
  • Semiconductor conductivity rises with temperature (resistance falls)
  • This is opposite to a metal, whose resistance rises with temperature
💎Pure crystal3/10

Intrinsic Semiconductor

A perfectly pure Si or Ge crystal where thermal energy breaks bonds to make electron-hole pairs.

n_e = n_h = n_i

n_i = intrinsic carrier concentration; conductivity is tiny.

  • Electrons and holes are always created in equal pairs
  • At 0 K it behaves like an insulator
  • Heating it generates more pairs and raises conductivity
🧪Key process4/10

Doping: n-type & p-type

Adding a controlled impurity (doping) makes an extrinsic semiconductor with far higher conductivity.

Mass-action law: n_e · n_h = n_i2

Doped crystal stays electrically neutral overall.

  • Pentavalent dopant (As, P, Sb) → n-type → electrons majority
  • Trivalent dopant (B, Al, In, Ga) → p-type → holes majority
  • Donor level sits just below CB; acceptor level just above VB
🔗Core device5/10

p-n Junction & Depletion Layer

Joining p and n material lets carriers diffuse and recombine, leaving a carrier-free depletion layer with a barrier potential.

Barrier potential V_b ≈ 0.7 V (Si) · 0.3 V (Ge)

At equilibrium, diffusion current balances drift current.

  • Fixed ions in the depletion region create the barrier potential
  • Diffusion = majority carriers; drift = minority carriers
  • No external voltage applied → net current is zero
🔀Core law6/10

Forward vs Reverse Bias

Biasing changes the barrier width, so a diode conducts strongly one way and blocks the other.

Forward: thin barrier, conducts · Reverse: thick barrier, blocks

Reverse bias passes only a tiny reverse saturation current.

  • Forward (p to +): barrier and width decrease, large current above V_b
  • Reverse (p to −): barrier and width increase, high resistance
  • A diode is a one-way valve; current rises sharply past the knee voltage
〰️Application7/10

Rectifiers (AC → DC)

A diode’s one-way conduction converts alternating current into direct current.

Half-wave: output freq = f · Full-wave: output freq = 2f

A capacitor filter smooths the pulsating DC output.

  • Half-wave: one diode, uses only one half of each AC cycle
  • Full-wave: two diodes (centre-tap) or a 4-diode bridge, more efficient
  • Full-wave output frequency is double the input frequency
💡Special diodes8/10

Zener, LED, Photodiode, Solar Cell

Specially designed diodes regulate voltage, emit light, detect light or generate power.

Zener = regulate · LED = emit · Photodiode = detect · Solar = generate

LED needs E_g ≥ photon energy → GaAs/GaAsP, not Si/Ge.

  • Zener: works in reverse breakdown as a voltage regulator
  • Photodiode: reverse-biased; light increases the reverse current
  • Solar cell: produces emf from light with no external bias
📡Advanced9/10

Transistor (BJT) & Gains

A bipolar junction transistor uses a thin lightly-doped base so emitter carriers reach the collector, enabling amplification and switching.

I_E = I_B + I_C · β = I_C / I_B · α = I_C / I_E · β = α / (1 − α)

Active mode: emitter-base forward, collector-base reverse biased.

  • Common-emitter gain β is large (~50–300); α is slightly less than 1
  • CE amplifier gives a 180° phase shift between input and output
  • As a switch: cutoff = OFF, saturation = ON
🧮Wrap up10/10

Logic Gates

Gates are the digital building blocks that combine binary inputs into one output.

OR: Y = A + B · AND: Y = A·B · NOT: Y = Ā

NAND and NOR are universal gates.

  • OR outputs 1 if any input is 1; AND outputs 1 only if all inputs are 1
  • NOT (inverter) flips the input
  • Any logic circuit can be built using only NAND or only NOR gates
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Q1NEET 2021
The electron concentration in an n-type semiconductor is the same as the hole concentration in a p-type semiconductor. An external electric field is applied across each of them. Comparing the resulting currents:
Correct answer: C. With equal majority-carrier concentrations, the current is decided by carrier mobility. Electron mobility (μₑ) exceeds hole mobility (μₕ) in a semiconductor, so for the same field and concentration the drift current I = n e μ E is larger in the n-type sample.
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Q2NEET 2020
Solids which have a negative temperature coefficient of resistance are:
Correct answer: C. A negative temperature coefficient means resistance falls as temperature rises. In both insulators and semiconductors, heating promotes more electrons across the gap, increasing carrier number and conductivity. Metals have a positive coefficient (more lattice scattering on heating).
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Q3NEET 2020
The Boolean output of a logic combination is Y = A · B. This is equivalent to a single:
Correct answer: A. By De Morgan’s theorem AcdotB = A+B, which is the NOR operation. So the output equals that of a single NOR gate.
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Q4NEET 2020
An intrinsic semiconductor is converted into an n-type extrinsic semiconductor by doping it with:
Correct answer: A. n-type material needs a pentavalent donor. Phosphorus (group 15) has five valence electrons; four bond with Si/Ge and the fifth is donated as a free electron. Aluminium is trivalent (gives p-type), silver is not a standard dopant, and germanium is tetravalent (no extra carrier).
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Q5NEET 2020
An n-p-n transistor is connected in common-emitter configuration in which the collector voltage drop across a load resistance of 800 Ω is 0.8 V. The collector current is:
Correct answer: A. The collector current flows through the load: I_C = V_(RC)/R_C = 0.8/800 = 1×10⁻³ A = 1 mA.
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Q6NEET 2019
An LED is constructed from a p-n junction diode using GaAsP with energy gap 1.9 eV. The wavelength of the light emitted is approximately (take h = 6.63×10⁻³⁴ J·s, c = 3×10⁸ m/s):
Correct answer: A. The emitted photon energy equals the gap: E_g = hc/λ, so λ = hc/E_g. With E_g = 1.9×1.6×10⁻¹⁹ = 3.04×10⁻¹⁹ J, λ = (6.63×10⁻³⁴×3×10⁸)/(3.04×10⁻¹⁹) = 6.54×10⁻⁷ m = 654 nm.
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Q7NEET 2018
In a p-n junction diode, the change due to heating (rise in temperature):
Correct answer: D. Heating generates more thermal electron-hole pairs, changing carrier concentrations on both sides. This alters both forward and reverse behaviour, so the entire V–I characteristic (and hence both forward and reverse resistance) is affected.
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Q8NEET 2016
An ideal p-n junction diode in series with a 1 kΩ resistor is operated such that the p-side is at +4 V and the n-side is at −6 V. The current through the diode is:
Correct answer: A. The p-side (+4 V) is at higher potential than the n-side (−6 V), so the ideal diode is forward biased with zero drop. The full 4-(-6) = 10 V appears across the 1 kΩ resistor: I = 10/1000 = 10⁻² A.
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Q9NEET 2012
Carbon (C) and silicon (Si) both have the same lattice structure with 4 bonding electrons each. However, C is an insulator while Si is an intrinsic semiconductor. This is because:
Correct answer: A. C’s bonding electrons sit in the second (n=2) shell, held tightly, giving a large band gap (~6 eV) → insulator. Si’s are in the third (n=3) shell, more loosely bound, giving a small gap (~1.1 eV) → semiconductor. The shell location sets the gap size.
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Q10NEET 2009
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength (take hc = 1240 eV·nm):
Correct answer: D. To be detected, the photon energy must be at least E_g = 2.5 eV, i.e. wavelength ≤ hc/E_g = 1240/2.5 = 496 nm = 4960 Å. Of the options, only 4000 Å (= 400 nm) is shorter than this cutoff, so it carries enough energy to be detected.
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Frequently Asked Questions

What is the difference between intrinsic and extrinsic semiconductors?

An intrinsic semiconductor is a perfectly pure crystal of silicon or germanium where electrons and holes are created in equal pairs by heat, so n_e equals n_h. An extrinsic semiconductor is one that has been doped with impurity atoms to greatly increase conductivity, making it either n-type or p-type.

How does doping create n-type and p-type semiconductors?

Doping with a pentavalent impurity such as arsenic, phosphorus or antimony gives an n-type semiconductor where electrons are the majority carriers. Doping with a trivalent impurity such as boron, aluminium, indium or gallium gives a p-type semiconductor where holes are the majority carriers.

What is the barrier potential of a p-n junction for silicon and germanium?

At the junction, diffusion and recombination leave a carrier-free depletion layer with a built-in barrier potential. This barrier is about 0.7 volts for a silicon diode and about 0.3 volts for a germanium diode.

What is the difference between a half-wave and a full-wave rectifier?

A half-wave rectifier uses one diode and conducts during only one half of each AC cycle, so its output frequency equals the input frequency f. A full-wave rectifier uses two diodes with a centre tap or a four-diode bridge to use both halves, giving an output frequency of 2f and higher efficiency.

Is Semiconductor Electronics important for the NEET physics syllabus?

Yes. It is a Class 12 chapter in the NEET syllabus and is considered high-yield because questions on band gaps, diodes, rectifiers, transistors and logic gates are largely conceptual and quick to solve, making it a reliable scoring chapter.

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