Semiconductors and Band Theory

Master intrinsic and extrinsic semiconductors, band theory, and doping for JEE Physics with smartphone processor analogy

Prerequisites

Before studying this topic, make sure you understand:

The Hook: What Powers Your Smartphone?

Connect: Your Phone → Semiconductors

Your smartphone has over 10 billion transistors in its processor - all made from semiconductors! A single grain of sand-sized chip contains more transistors than all the people who ever lived on Earth.

Intriguing fact: Pure silicon is actually a poor conductor. But by adding just 1 impurity atom per 100 million silicon atoms, engineers create the processors that run Instagram, YouTube, and yes - even JEE prep apps!

How does adding such tiny amounts of “dirt” transform a poor conductor into the brain of modern technology? Let’s discover the science!

Interactive Demo

Visualize energy bands and electron transitions in semiconductors:


The Core Concept: What Are Semiconductors?

The Big Picture

Materials are classified by their ability to conduct electricity:

Conductors (Metals like Copper):

  • Resistivity: $\rho \sim 10^{-8}$ Ω·m
  • Lots of free electrons
  • Conductivity increases with decreasing temperature

Insulators (Rubber, Glass):

  • Resistivity: $\rho \sim 10^{12}$ - $10^{16}$ Ω·m
  • Almost no free electrons
  • Poor conductors at all temperatures

Semiconductors (Silicon, Germanium):

  • Resistivity: $\rho \sim 10^{-3}$ to $10^{3}$ Ω·m
  • Conductivity increases with temperature (opposite to metals!)
  • Conductivity can be controlled by adding impurities
The Smartphone Connection

Your phone’s processor uses billions of tiny semiconductor switches (transistors) that can turn on and off billions of times per second. Each time you tap your screen, millions of these switches flip their states!

The magic? By controlling the impurity atoms, engineers can make semiconductors conduct in specific patterns - creating the logic gates that form computer processors.


Band Theory: The Foundation

Energy Bands in Solids

When atoms come together to form a solid, individual atomic energy levels split into bands:

  1. Valence Band (VB):

    • Filled with electrons at 0 K
    • Highest occupied energy band
    • Electrons here are bound to atoms
  2. Conduction Band (CB):

    • Empty at 0 K
    • Electrons here are free to move
    • Responsible for electrical conduction
  3. Band Gap ($E_g$):

    • Energy difference between VB and CB
    • Forbidden energy region
    • Determines electrical properties

Classification by Band Gap

MaterialBand Gap ($E_g$)Example
ConductorsNo band gap (CB and VB overlap)Copper, Aluminum
SemiconductorsSmall gap ($E_g < 3$ eV)Si: 1.1 eV, Ge: 0.7 eV
InsulatorsLarge gap ($E_g > 3$ eV)Diamond: 5.4 eV
Why Temperature Matters

At 0 K: All electrons in valence band → semiconductor acts as insulator

At room temperature: Thermal energy $kT \approx 0.026$ eV gives some electrons enough energy to jump the gap!

For Silicon ($E_g = 1.1$ eV):

  • Not all electrons can jump (1.1 eV » 0.026 eV)
  • But enough electrons jump to make it conduct slightly
  • Higher temperature → more electrons jump → better conductivity

This is why semiconductors have NEGATIVE temperature coefficient (unlike metals!)


Intrinsic Semiconductors

Definition

Intrinsic semiconductor = Pure semiconductor with no impurities

Characteristics:

  • Number of free electrons = Number of holes
  • $n_i = $ intrinsic carrier concentration
  • At room temperature: $n_i \approx 10^{10}$ carriers/cm³ (very small!)

How Conduction Happens

  1. Electron generation: Thermal energy breaks covalent bond
  2. Free electron created in conduction band
  3. Hole (vacancy) left behind in valence band
  4. Both electron and hole contribute to current!
$$\boxed{n_i = p_i}$$

where:

  • $n_i$ = intrinsic electron concentration
  • $p_i$ = intrinsic hole concentration

Intrinsic Carrier Concentration

$$\boxed{n_i^2 = A T^3 e^{-E_g/kT}}$$

where:

  • $A$ = constant depending on material
  • $T$ = absolute temperature
  • $E_g$ = band gap energy
  • $k$ = Boltzmann constant ($8.62 \times 10^{-5}$ eV/K)

Key insight: $n_i$ increases exponentially with temperature!

Conductivity of Intrinsic Semiconductor

$$\boxed{\sigma_i = n_i e (\mu_e + \mu_h)}$$

where:

  • $\mu_e$ = electron mobility
  • $\mu_h$ = hole mobility
  • $e$ = electronic charge

For Silicon at 300 K:

  • $\mu_e \approx 0.14$ m²/(V·s)
  • $\mu_h \approx 0.05$ m²/(V·s)
  • Electrons are about 3 times more mobile than holes!

Extrinsic Semiconductors: The Game Changer

What is Doping?

Doping = Deliberately adding impurity atoms to control conductivity

Key numbers:

  • Pure Si: $10^{22}$ atoms/cm³
  • Typical doping: 1 impurity per $10^8$ Si atoms
  • But conductivity increases by factor of 1000!

This is like adding 1 drop of ink to 10 million drops of water - tiny amount, huge effect!

n-Type Semiconductors (Electron Rich)

Dopant: Group V elements (Phosphorus, Arsenic, Antimony)

  • 5 valence electrons
  • 4 bond with Si, 1 extra electron becomes free

Result:

  • Majority carriers: Electrons ($n_n$)
  • Minority carriers: Holes ($p_n$)
  • $n_n >> p_n$

Key relation:

$$\boxed{n_n \approx N_D}$$

where $N_D$ = donor concentration

Mass action law still holds:

$$\boxed{n_n \times p_n = n_i^2}$$

So if $n_n$ increases, $p_n$ decreases!

p-Type Semiconductors (Hole Rich)

Dopant: Group III elements (Boron, Aluminum, Gallium)

  • 3 valence electrons
  • Creates hole (electron vacancy)

Result:

  • Majority carriers: Holes ($p_p$)
  • Minority carriers: Electrons ($n_p$)
  • $p_p >> n_p$

Key relation:

$$\boxed{p_p \approx N_A}$$

where $N_A$ = acceptor concentration

Mass action law:

$$\boxed{n_p \times p_p = n_i^2}$$
The Processor Connection

Modern processors use billions of n-type and p-type regions placed next to each other. The boundary between n-type and p-type forms a p-n junction - the basis of diodes and transistors!

By arranging these junctions cleverly, engineers create:

  • Diodes (one p-n junction) - Allow current in one direction
  • Transistors (two junctions) - Act as electronic switches
  • Integrated circuits - Billions of transistors on a chip!

Important Formulas Summary

For Intrinsic Semiconductor

$$\boxed{n_i = p_i}$$ $$\boxed{\sigma_i = n_i e (\mu_e + \mu_h)}$$

For n-Type Semiconductor

$$\boxed{n_n \approx N_D \text{ (majority)}}$$ $$\boxed{p_n = \frac{n_i^2}{N_D} \text{ (minority)}}$$ $$\boxed{\sigma_n = n_n e \mu_e + p_n e \mu_h \approx N_D e \mu_e}$$

For p-Type Semiconductor

$$\boxed{p_p \approx N_A \text{ (majority)}}$$ $$\boxed{n_p = \frac{n_i^2}{N_A} \text{ (minority)}}$$ $$\boxed{\sigma_p = p_p e \mu_h + n_p e \mu_e \approx N_A e \mu_h}$$

Universal Relation (Mass Action Law)

$$\boxed{n \times p = n_i^2}$$

This holds for ALL semiconductors in thermal equilibrium!


Memory Tricks & Patterns

Mnemonic for Semiconductor Types

“n-type has Negative charge carriers” → Electrons (negative)

“p-type has Positive holes” → Holes (absence of electrons)

Dopant Groups Memory

“5 gives electrons, 3 takes electrons”

  • Group V (5 valence) → Donor → n-type
  • Group III (3 valence) → Acceptor → p-type

Pattern Recognition

  1. Temperature coefficient:

    • Metals: Resistance increases with temperature (+ve coefficient)
    • Semiconductors: Resistance decreases with temperature (-ve coefficient)
  2. Typical values to memorize:

    • Si band gap: $E_g = 1.1$ eV
    • Ge band gap: $E_g = 0.7$ eV
    • Room temperature $kT \approx 0.026$ eV
    • Si intrinsic concentration: $n_i \approx 1.5 \times 10^{10}$ cm⁻³ at 300 K
  3. Doping ratio:

    • Heavy doping: 1 impurity per $10^6$ atoms
    • Medium doping: 1 impurity per $10^8$ atoms
    • Light doping: 1 impurity per $10^{10}$ atoms

When to Use This

Decision Tree

Use intrinsic semiconductor formulas when:

  • Problem states “pure silicon” or “intrinsic”
  • Asked about temperature dependence of pure semiconductor
  • Need to find $n_i$ for use in mass action law

Use n-type formulas when:

  • Dopant is Group V (P, As, Sb)
  • Problem gives donor concentration $N_D$
  • Majority carriers are electrons

Use p-type formulas when:

  • Dopant is Group III (B, Al, Ga)
  • Problem gives acceptor concentration $N_A$
  • Majority carriers are holes

Use mass action law ($np = n_i^2$) when:

  • Given majority carrier concentration, find minority
  • Comparing intrinsic and extrinsic semiconductors
  • ANY problem involving carrier concentrations!

Common Mistakes to Avoid

Trap #1: Confusing n-type with Negative Charge

Wrong thinking: “n-type semiconductor is negatively charged”

Reality:

  • n-type is electrically neutral overall
  • More free electrons, but also more positive ions
  • Extra electrons exactly balance positive donor ions

JEE Trap: Questions asking “net charge on n-type semiconductor?” Answer: ZERO (unless stated otherwise)!

Trap #2: Forgetting Minority Carriers

Wrong: “In n-type, only electrons conduct”

Correct:

  • Majority: electrons ($n_n \approx N_D$)
  • Minority: holes ($p_n = n_i^2/N_D$)
  • Both contribute to current (but electrons dominate)

Why it matters: Minority carriers are crucial in p-n junctions and transistors!

Trap #3: Temperature Dependence Confusion

Wrong: “Doped semiconductor conductivity is independent of temperature”

Correct: At low doping or high temperature:

  • Intrinsic carriers can become significant
  • Temperature coefficient can change sign!

Rule of thumb:

  • Extrinsic region (low T): $\sigma \approx$ constant
  • Intrinsic region (high T): $\sigma$ increases exponentially with T

Practice Problems

Level 1: Foundation (NCERT/Basic)

Problem 1.1

Silicon has 5 × 10²² atoms/cm³. If it’s doped with phosphorus (1 atom per 10⁸ Si atoms), find the electron concentration in the doped semiconductor.

Solution:

Phosphorus (Group V) is a donor → creates n-type semiconductor

Donor concentration:

$$N_D = \frac{5 \times 10^{22}}{10^8} = 5 \times 10^{14} \text{ atoms/cm}^3$$

For n-type: $n_n \approx N_D$

$$n_n = 5 \times 10^{14} \text{ electrons/cm}^3$$

Answer: $5 \times 10^{14}$ electrons/cm³

Insight: This is $10^4$ times more than intrinsic Si ($n_i \approx 10^{10}$ cm⁻³)!

Problem 1.2

For the above n-type silicon, find the hole concentration if $n_i = 1.5 \times 10^{10}$ cm⁻³.

Solution:

Using mass action law: $n_n \times p_n = n_i^2$

$$p_n = \frac{n_i^2}{n_n} = \frac{(1.5 \times 10^{10})^2}{5 \times 10^{14}}$$ $$p_n = \frac{2.25 \times 10^{20}}{5 \times 10^{14}} = 4.5 \times 10^5 \text{ holes/cm}^3$$

Answer: $4.5 \times 10^5$ holes/cm³

Key insight: Minority carriers ($p_n$) are $10^9$ times less than majority carriers ($n_n$)!

Level 2: JEE Main

Problem 2.1

A semiconductor has electron and hole mobilities of 0.36 m²/(V·s) and 0.14 m²/(V·s) respectively. If it’s doped with 10²¹ donors/m³, find its conductivity. Given: $n_i = 1.5 \times 10^{16}$ m⁻³.

Solution:

For n-type: $n_n \approx N_D = 10^{21}$ m⁻³

Minority carriers: $p_n = \frac{n_i^2}{N_D} = \frac{(1.5 \times 10^{16})^2}{10^{21}} = 2.25 \times 10^{11}$ m⁻³

Conductivity:

$$\sigma = n_n e \mu_e + p_n e \mu_h$$

Since $n_n >> p_n$:

$$\sigma \approx n_n e \mu_e$$ $$\sigma = 10^{21} \times 1.6 \times 10^{-19} \times 0.36$$ $$\sigma = 57.6 \text{ S/m}$$

Answer: σ = 57.6 S/m

JEE Insight: Minority carrier contribution is negligible when doping is heavy!

Problem 2.2

Pure germanium has resistivity 0.6 Ω·m at 27°C. When doped, its resistivity drops to 0.003 Ω·m. Find the ratio of conductivity (doped/pure).

Solution:

$$\sigma = \frac{1}{\rho}$$ $$\frac{\sigma_{doped}}{\sigma_{intrinsic}} = \frac{\rho_{intrinsic}}{\rho_{doped}}$$ $$= \frac{0.6}{0.003} = 200$$

Answer: Conductivity increases by factor of 200

Key concept: Even small doping dramatically increases conductivity!

Problem 2.3

The energy gap of silicon is 1.1 eV. What fraction of electrons can jump from valence to conduction band at 300 K? (Given: $kT = 0.026$ eV)

Solution:

Fraction of electrons with energy > $E_g$:

$$f \propto e^{-E_g/kT}$$ $$f \propto e^{-1.1/0.026} = e^{-42.3}$$ $$f \approx 10^{-18}$$

Answer: About 1 in 10¹⁸ electrons

Insight: Very few electrons can jump the gap thermally - that’s why pure Si is poor conductor!

Level 3: JEE Advanced

Problem 3.1

An intrinsic semiconductor has intrinsic concentration $n_i$. It’s doped such that majority carrier concentration is $10^4 n_i$. Find the factor by which conductivity increases if $\mu_e = 2\mu_h$.

Solution:

Intrinsic conductivity:

$$\sigma_i = n_i e (\mu_e + \mu_h) = n_i e (2\mu_h + \mu_h) = 3 n_i e \mu_h$$

Extrinsic (assume n-type):

$$n_n = 10^4 n_i$$ $$p_n = \frac{n_i^2}{10^4 n_i} = \frac{n_i}{10^4}$$ $$\sigma_e = n_n e \mu_e + p_n e \mu_h$$ $$= 10^4 n_i e (2\mu_h) + \frac{n_i}{10^4} e \mu_h$$ $$\approx 2 \times 10^4 n_i e \mu_h$$

Ratio:

$$\frac{\sigma_e}{\sigma_i} = \frac{2 \times 10^4 n_i e \mu_h}{3 n_i e \mu_h} = \frac{2 \times 10^4}{3} \approx 6667$$

Answer: Conductivity increases by factor of ~6700

Advanced concept: Effect of doping is amplified by higher electron mobility!

Problem 3.2

A semiconductor has equal concentration of donors and acceptors ($N_D = N_A = 10^{16}$ cm⁻³). Find electron and hole concentrations if $n_i = 10^{10}$ cm⁻³.

Solution:

This is compensated semiconductor!

Net doping: $N_D - N_A = 0$

When donors and acceptors are equal, they cancel out:

  • Each donor electron fills an acceptor hole
  • Result: Behaves like intrinsic!
$$n = p = n_i = 10^{10} \text{ cm}^{-3}$$

Answer: $n = p = 10^{10}$ cm⁻³ (intrinsic behavior)

Advanced concept: Complete compensation makes extrinsic semiconductor behave intrinsic!


Quick Revision Box

TypeDopantMajorityMinority$\sigma$ formula
IntrinsicNone$n_i$ (e⁻)$n_i$ (h⁺)$n_i e(\mu_e + \mu_h)$
n-typeGroup V$N_D$ (e⁻)$n_i^2/N_D$ (h⁺)$N_D e \mu_e$
p-typeGroup III$N_A$ (h⁺)$n_i^2/N_A$ (e⁻)$N_A e \mu_h$

Universal: $n \times p = n_i^2$ (always!)


JEE Strategy: High-Yield Points

What JEE Loves to Test
  1. Mass action law ($np = n_i^2$) - appears in 60% of semiconductor problems!

    • Given majority, find minority
  2. Identifying dopant type:

    • Group V (P, As, Sb) → n-type
    • Group III (B, Al, Ga) → p-type
  3. Temperature dependence:

    • Semiconductors: resistance ↓ with temperature ↑
    • Opposite to metals!
  4. Numerical values to memorize:

    • Si: $E_g = 1.1$ eV
    • Ge: $E_g = 0.7$ eV
    • $kT$ at 300K = 0.026 eV
    • Si: $n_i = 1.5 \times 10^{10}$ cm⁻³
  5. Conductivity calculation:

    • For heavy doping: minority carriers negligible
    • $\sigma \approx N_D e \mu_e$ (n-type) or $N_A e \mu_h$ (p-type)

Time-saving trick: In most JEE problems, minority carriers can be ignored when calculating conductivity if doping is » $n_i$


Within Electronic Devices

  • p-n Junction - What happens when p and n regions meet
  • Diode - Using p-n junction for rectification
  • Transistor - Two p-n junctions working together

Connected Chapters

Real-world Applications

  • Smartphone processors (billions of transistors)
  • LED lights (light-emitting semiconductors)
  • Solar cells (converting light to electricity)
  • Computer memory (storing data in semiconductor circuits)

Teacher’s Summary

Key Takeaways
  1. Semiconductors have moderate conductivity that increases with temperature - opposite to metals, making them unique and useful

  2. Band gap determines semiconductor behavior - Si (1.1 eV) and Ge (0.7 eV) have perfect gaps for electronics at room temperature

  3. Intrinsic semiconductors have equal electrons and holes ($n_i = p_i$) - but very few carriers overall, so poor conductors

  4. Doping transforms semiconductors - adding 1 impurity per 10⁸ atoms increases conductivity 1000×!

    • n-type: Group V donors → excess electrons
    • p-type: Group III acceptors → excess holes
  5. Mass action law $np = n_i^2$ is universal - works for intrinsic and extrinsic, the most important formula in semiconductor physics!

  6. Your smartphone’s brain contains >10 billion semiconductor switches - all based on controlling these tiny impurity concentrations!

“From sand to smartphone: By adding just 1 impurity atom per 100 million, we transformed silicon into the most important material of the digital age!”