Chapter 29: Semiconductor Electronics: Materials, Devices and Simple Circuits

Introduction

Overview of Electronic Devices

• Early electronic devices: vacuum tubes (diode, triode, tetrode, pentode).
• Vacuum tubes: bulky, high power consumption, high voltage (~100 V), limited life, low reliability.
• Modern electronics use solid-state semiconductors for controlled electron flow.
• Semiconductors' charge carriers can be controlled by light, heat, or voltage.
• CRT displays replaced by LCD monitors due to semiconductor technology.

Solved Examples

Ex 1

Identify the main disadvantage of vacuum tubes compared to semiconductors.

• Answer: High power consumption and bulkiness.

Practice Set

• Q1: Which device is bulky and consumes high power?
  A) Transistor
  B) Vacuum tube
  C) Diode
  D) LED
• Q2: What replaces CRT monitors in modern electronics?
  A) Vacuum tubes
  B) LCD monitors
  C) Cathode ray tubes
  D) Pentodes
• Q3: Which excitation can change mobile charge carriers in semiconductors?
  A) Light
  B) Heat
  C) Small voltage
  D) All of the above

Answer Key

• Q1: B - Vacuum tubes are bulky and consume high power.
• Q2: B - LCD monitors replaced CRTs due to semiconductor tech.
• Q3: D - Light, heat, and voltage affect charge carriers in semiconductors.

Memory Aids

• Vacuum tubes = bulky + high power.
• Semiconductors = controlled electron flow.
• LCD replaces CRT in displays.

Classification of Metals, Conductors and Semiconductors

Material Classification by Conductivity

• Conductivity (σ) measures ease of electric current flow.
• Metals: high conductivity, electricity flows easily.
• Semiconductors: medium conductivity, moderate current flow.
• Insulators: very low conductivity, negligible current flow.
• Conductivity and resistivity (ρ) are inversely related: \( \sigma = \frac{1}{\rho} \).

Solved Examples

Ex 1

Classify a material with conductivity \(10^7\) S/m.

• Answer: Metal (high conductivity).

Practice Set

• Q1: Which has the highest conductivity?
  A) Metal
  B) Semiconductor
  C) Insulator
  D) Vacuum
• Q2: Resistivity and conductivity are:
  A) Directly proportional
  B) Inversely proportional
  C) Equal
  D) Unrelated
• Q3: Which material allows moderate current flow?
  A) Metal
  B) Semiconductor
  C) Insulator
  D) Vacuum tube

Answer Key

• Q1: A - Metals have highest conductivity.
• Q2: B - Resistivity and conductivity are inversely proportional.
• Q3: B - Semiconductors allow moderate current flow.

Memory Aids

• Metal = High σ, low ρ.
• Semiconductor = Medium σ.
• Insulator = Low σ, high ρ.

Energy Bands of Solids or Band Theory of Solids

Key Energy Bands

• Valence band: Contains valence electrons; fully or partially filled; electrons do not conduct current.
• Conduction band: Empty or partially filled; contains free electrons that conduct current.
• Forbidden energy gap (band gap): Energy gap between valence and conduction bands; no electrons present here.
• Energy band gap formula: \( E_g = h \nu = \frac{hc}{\lambda} \), where \(h\) = Planck's constant, \(c\) = speed of light, \(\lambda\) = wavelength.
• Band gap varies with temperature.
• Fermi energy: Maximum energy of free electrons at 0 K.

Solved Examples

Ex 1

Calculate band gap if wavelength \(\lambda = 620\) nm. Use \(h = 6.63 \times 10^{-34}\) Js, \(c = 3 \times 10^8\) m/s, \(1 eV = 1.6 \times 10^{-19}\) J.

• \(E_g = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{620 \times 10^{-9}} = 3.21 \times 10^{-19}\) J
• Convert to eV: \(\frac{3.21 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0\) eV

Practice Set

• Q1: Electrons in which band conduct electricity?
  A) Valence band
  B) Conduction band
  C) Forbidden gap
  D) Fermi level
• Q2: Band gap \(E_g\) is energy difference between:
  A) Valence and conduction bands
  B) Two valence bands
  C) Two conduction bands
  D) Electron and hole
• Q3: If \(\lambda = 500\) nm, calculate \(E_g\) in eV. (Use \(h=6.63\times10^{-34}\) Js, \(c=3\times10^8\) m/s, \(1 eV=1.6\times10^{-19}\) J)
  A) 2.48 eV
  B) 1.24 eV
  C) 3.10 eV
  D) 4.00 eV

Answer Key

• Q1: B - Conduction band electrons conduct current.
• Q2: A - Band gap is between valence and conduction bands.
• Q3: A - \(E_g = \frac{6.63\times10^{-34} \times 3\times10^8}{500\times10^{-9}} = 3.98\times10^{-19} J = 2.48 eV\).

Memory Aids

• Band gap formula: \(E_g = \frac{hc}{\lambda}\).
• Conduction band = free electrons.
• Valence band = bound electrons.

Energy Bands Based on Material Type

Metals

• Conduction band partially filled or overlaps valence band.
• No forbidden energy gap.
• Electrons move freely, high conductivity.

Insulators

• Valence band fully filled; conduction band empty.
• Large forbidden energy gap (>5 eV).
• Very low conductivity.

Semiconductors

• Valence band fully filled; conduction band empty at 0 K.
• Small forbidden energy gap (~1 eV): Silicon (1.1 eV), Germanium (0.72 eV).
• Conductivity increases with temperature.

Solved Examples

Ex 1

Identify the material type with band gap 0.7 eV.

• Answer: Semiconductor (e.g., Germanium).

Practice Set

• Q1: Which material has no band gap?
  A) Metal
  B) Semiconductor
  C) Insulator
  D) Vacuum tube
• Q2: Band gap of silicon is approximately:
  A) 0.5 eV
  B) 1.1 eV
  C) 3 eV
  D) 5 eV
• Q3: Which material has a band gap > 5 eV?
  A) Metal
  B) Semiconductor
  C) Insulator
  D) None

Answer Key

• Q1: A - Metals have no band gap.
• Q2: B - Silicon band gap is ~1.1 eV.
• Q3: C - Insulators have band gap > 5 eV.

Memory Aids

• Metal: No band gap.
• Semiconductor: Small band gap (~1 eV).
• Insulator: Large band gap (>5 eV).

Hole in Semiconductors

Definition and Properties

• Hole: Positive charge carrier formed when electron leaves covalent bond.
• Charge of hole = +e (same magnitude as electron charge).
• Hole mobility is less than electron mobility.

Solved Examples

Ex 1

What charge does a hole carry?

• Answer: Positive charge equal to electron charge.

Practice Set

• Q1: A hole is:
  A) Negative charge carrier
  B) Positive charge carrier
  C) Neutral particle
  D) Electron
• Q2: Hole mobility compared to electron mobility is:
  A) Greater
  B) Equal
  C) Smaller
  D) Zero
• Q3: Hole is created when:
  A) Electron breaks covalent bond
  B) Electron recombines
  C) Electron moves to valence band
  D) Electron is created

Answer Key

• Q1: B - Hole carries positive charge.
• Q2: C - Hole mobility is smaller than electron mobility.
• Q3: A - Hole forms when electron breaks covalent bond.

Memory Aids

• Hole = positive charge carrier.
• Mobility: electron > hole.
• Hole forms by electron leaving bond.

Intrinsic Semiconductor

Definition and Examples

• Pure semiconductor without impurities is called intrinsic semiconductor.
• Examples: Silicon (Si), Germanium (Ge).

Carrier Concentrations

• Number density of electrons in conduction band, \( n_e \).
• Number density of holes in valence band, \( n_h \).
• Intrinsic carrier concentration, \( n_i \).
• In intrinsic semiconductor, \( n_e = n_h = n_i \).

Current and Conductivity

• Total current, \( I = I_e + I_h \), where \( I_e \) is electron current and \( I_h \) is hole current.
• Conductivity, \( \sigma = n_i e (\mu_e + \mu_h) \), where \( e \) is electron charge, \( \mu_e \) and \( \mu_h \) are mobilities.
• Resistivity, \( \rho = \frac{1}{\sigma} = \frac{1}{e n_i (\mu_e + \mu_h)} \).
• Current, \( I = n_i e A (v_e + v_h) \), where \( A \) is cross-sectional area, \( v_e \) and \( v_h \) are drift velocities.

Temperature Effect

• Increasing temperature increases electron-hole pair density \( \Rightarrow \) conductivity increases.
• Resistivity decreases with temperature.
• Semiconductors have negative temperature coefficient of resistance.

Crystal Structure and Hole Motion

• Si or Ge atoms form covalent bonds with four neighbors (+4 core charge).
• At low temperature, all bonds intact.
• Electron movement from one bond to hole site causes apparent hole motion.
• Hole acts as positive charge carrier moving through lattice.

Solved Examples

Ex 1

Calculate conductivity if \( n_i = 1.5 \times 10^{16} \text{ m}^{-3} \), \( \mu_e = 0.14 \text{ m}^2/\text{V·s} \), \( \mu_h = 0.05 \text{ m}^2/\text{V·s} \), and \( e = 1.6 \times 10^{-19} \text{ C} \).

• Use \( \sigma = n_i e (\mu_e + \mu_h) \).
• \( \sigma = 1.5 \times 10^{16} \times 1.6 \times 10^{-19} \times (0.14 + 0.05) = 3.456 \times 10^{-3} \text{ S/m} \).

Practice Set

Q1

In an intrinsic semiconductor, the number density of electrons is:

• A) Greater than holes
• B) Less than holes
• C) Equal to holes
• D) Zero

Q2

Which of the following increases conductivity of intrinsic semiconductor?

• A) Decreasing temperature
• B) Increasing temperature
• C) Adding impurities
• D) Decreasing carrier mobility

Q3

Calculate resistivity if \( n_i = 2 \times 10^{16} \text{ m}^{-3} \), \( \mu_e = 0.15 \text{ m}^2/\text{V·s} \), \( \mu_h = 0.05 \text{ m}^2/\text{V·s} \), and \( e = 1.6 \times 10^{-19} \text{ C} \).

• A) \( 1.56 \times 10^{2} \, \Omega\cdot m \)
• B) \( 3.12 \times 10^{2} \, \Omega\cdot m \)
• C) \( 1.56 \times 10^{-2} \, \Omega\cdot m \)
• D) \( 3.12 \times 10^{-2} \, \Omega\cdot m \)

Answer Key

• Q1: C) Equal to holes — Intrinsic semiconductors have equal electron and hole densities.
• Q2: B) Increasing temperature — Higher temperature generates more electron-hole pairs, increasing conductivity.
• Q3: A) \( 1.56 \times 10^{2} \, \Omega\cdot m \) — \( \rho = \frac{1}{e n_i (\mu_e + \mu_h)} = \frac{1}{1.6 \times 10^{-19} \times 2 \times 10^{16} \times 0.2} = 156 \ \Omega\cdot m \).

Memory Aids

• Intrinsic: \( n_e = n_h = n_i \).
• Conductivity \( \sigma = n_i e (\mu_e + \mu_h) \).
• Resistivity \( \rho = 1/\sigma \).
• Conductivity ↑ with temperature ↑ (negative TCR).
• Hole motion = electron hopping in covalent bonds.

Energy Band Diagram

Band Structure at Different Temperatures

• At \( T=0 \text{ K} \), valence band (VB) fully filled, conduction band (CB) empty.
• Intrinsic semiconductor behaves as insulator at 0 K.
• At \( T>0 \text{ K} \), electrons gain thermal energy and jump from VB to CB.
• Equal number of holes created in VB.
• Electrons occupy lowest energy states in CB.
• Number of electrons in CB equals number of holes in VB.

Solved Examples

Ex 1

Explain conduction at 0 K and room temperature.

• At 0 K: No electrons in CB, no conduction.
• At room temperature: Thermal excitation creates electron-hole pairs, enabling conduction.

Practice Set

Q1

At absolute zero, an intrinsic semiconductor is:

• A) Conductor
• B) Insulator
• C) Semiconductor
• D) Superconductor

Q2

Electron-hole pairs in intrinsic semiconductor are created by:

• A) Impurities
• B) Thermal energy
• C) Electric field only
• D) Magnetic field

Q3

At room temperature, the number of electrons in conduction band is:

• A) Less than holes in valence band
• B) Equal to holes in valence band
• C) Greater than holes in valence band
• D) Zero

Answer Key

• Q1: B) Insulator — At 0 K, no free carriers, so no conduction.
• Q2: B) Thermal energy — Heat excites electrons to conduction band.
• Q3: B) Equal to holes in valence band — Electron-hole pairs are created in equal numbers.

Memory Aids

• 0 K: VB full, CB empty, no conduction.
• \( T>0 \): Thermal excitation creates equal electron-hole pairs.
• Electron-hole pairs enable conduction.

Extrinsic Semiconductor

Concepts

• Impure semiconductor with added impurities is called extrinsic semiconductor.
• Doping: Adding small amounts of impurities with different valency to pure semiconductor.
• Doping changes number of mobile charge carriers (electrons or holes), increasing conductivity.
• Two types: n-type (pentavalent impurities) and p-type (trivalent impurities).
• Energy band diagrams show donor and acceptor levels affecting electron movement.

Solved Examples

Ex 1: Identify majority carriers in n-type semiconductor doped with phosphorus.

• Phosphorus is pentavalent impurity (group V).
• Donates extra electrons.
• Majority carriers are electrons (negative charge).

Practice Set

• Q1: What is doping in semiconductors?
  A) Removing impurities
  B) Adding impurities to increase conductivity
  C) Heating semiconductor
  D) Cooling semiconductor
• Q2: In n-type semiconductor, majority carriers are:
  A) Holes
  B) Electrons
  C) Ions
  D) Neutrons
• Q3: Which impurity is used for p-type doping?
  A) Phosphorus
  B) Arsenic
  C) Boron
  D) Antimony

Answer Key

• Q1: B - Doping means adding impurities to increase conductivity.
• Q2: B - Electrons are majority carriers in n-type semiconductor.
• Q3: C - Boron (trivalent) is used for p-type doping.

Memory Aids

• Doping = Adding impurities to increase conductivity.
• n-type: Donor atoms (group V), electrons majority.
• p-type: Acceptor atoms (group III), holes majority.

n-type Semiconductor

Concepts

• Pure Si or Ge (tetravalent) doped with pentavalent impurities (As, Sb, P).
• Donor atoms donate one free electron each for conduction.
• Conduction due to electrons (negative charge carriers).
• Electrons are majority carriers; holes are minority carriers.
• Charge neutrality maintained.
• Carrier concentration: \( n_e \approx N_d \gg n_h \), where \( N_d \) = donor density.

Solved Examples

Ex 1: Calculate majority carrier concentration if donor density \( N_d = 10^{16} \) cm^-3.

• Majority carrier concentration \( n_e \approx N_d = 10^{16} \) cm^-3.

Practice Set

• Q1: Donor atoms in n-type semiconductor are from group:
  A) III
  B) IV
  C) V
  D) VI
• Q2: Majority carriers in n-type semiconductor are:
  A) Holes
  B) Electrons
  C) Ions
  D) Protons
• Q3: If donor density is \( 5 \times 10^{15} \) cm^-3, approximate electron concentration is:
  A) \( 5 \times 10^{15} \) cm^-3
  B) \( 5 \times 10^{14} \) cm^-3
  C) \( 10^{16} \) cm^-3
  D) \( 10^{14} \) cm^-3

Answer Key

• Q1: C - Donor atoms are pentavalent (group V).
• Q2: B - Electrons are majority carriers in n-type.
• Q3: A - Electron concentration \( \approx N_d = 5 \times 10^{15} \) cm^-3.

Memory Aids

• n-type = Negative charge carriers (electrons).
• Donor atoms = group V elements.
• Majority carriers \( n_e \approx N_d \).

p-type Semiconductor

Concepts

• Pure Si or Ge doped with trivalent impurities (Al, B, In).
• Acceptor atoms create holes by accepting electrons.
• Conduction due to holes (positive charge carriers).
• Holes are majority carriers; electrons are minority carriers.
• Charge neutrality maintained.
• Carrier concentration: \( n_h \approx N_a \gg n_e \), where \( N_a \) = acceptor density.

Solved Examples

Ex 1: Find majority carrier concentration if acceptor density \( N_a = 2 \times 10^{16} \) cm^-3.

• Majority carrier concentration \( n_h \approx N_a = 2 \times 10^{16} \) cm^-3.

Practice Set

• Q1: Acceptor atoms in p-type semiconductor belong to group:
  A) III
  B) IV
  C) V
  D) VI
• Q2: Majority carriers in p-type semiconductor are:
  A) Electrons
  B) Holes
  C) Ions
  D) Neutrons
• Q3: If acceptor density is \( 1 \times 10^{15} \) cm^-3, hole concentration is approximately:
  A) \( 1 \times 10^{15} \) cm^-3
  B) \( 1 \times 10^{14} \) cm^-3
  C) \( 10^{16} \) cm^-3
  D) \( 10^{14} \) cm^-3

Answer Key

• Q1: A - Acceptor atoms are trivalent (group III).
• Q2: B - Holes are majority carriers in p-type.
• Q3: A - Hole concentration \( \approx N_a = 1 \times 10^{15} \) cm^-3.

Memory Aids

• p-type = Positive charge carriers (holes).
• Acceptor atoms = group III elements.
• Majority carriers \( n_h \approx N_a \).

p-n Junction

Formation and Charge Distribution

• p-n junction formed by joining p-type (holes) and n-type (electrons) semiconductors.
• Electrons diffuse from n-side to p-side, filling holes; holes diffuse from p-side to n-side.
• Creates positive charge layer on n-side and negative charge layer on p-side near junction.
• These immobile charges form the depletion layer, preventing further charge carrier movement.
• Potential barrier (V₀) of 0.1–0.3 V (0.3 V for Ge, 0.7 V for Si) develops across junction.
• Electric field intensity across depletion region: E = V/d, where V = potential barrier, d = depletion width.

Current Types in p-n Junction

• Diffusion current: Due to concentration gradient, holes move p → n and electrons n → p.
• Drift current: Minority carriers move across junction due to barrier potential.

Solved Examples

• Ex 1: Calculate electric field intensity if potential barrier V = 0.7 V and depletion width d = 2 \(\mu\)m.
  • Use E = V/d
  • E = 0.7 V / 2 × 10^-6 m = 3.5 × 10⁵ V/m

Practice Set

• Q1: What is the typical potential barrier for silicon p-n junction?
  • A) 0.1 V
  • B) 0.3 V
  • C) 0.7 V
  • D) 1.0 V
• Q2: Which current is caused by minority carriers moving due to barrier potential?
  • A) Diffusion current
  • B) Drift current
  • C) Conduction current
  • D) Displacement current
• Q3: If the depletion width is 1 \(\mu\)m and potential barrier is 0.3 V, what is the electric field intensity?
  • A) 3 × 10⁵ V/m
  • B) 3 × 10⁴ V/m
  • C) 3 × 10⁶ V/m
  • D) 3 × 10³ V/m

Answer Key

• Q1: C) 0.7 V — Silicon p-n junction has a potential barrier around 0.7 V.
• Q2: B) Drift current — Minority carriers move due to barrier potential causing drift current.
• Q3: A) 3 × 10⁵ V/m — E = V/d = 0.3 V / 1 × 10^-6 m = 3 × 10⁵ V/m.

Memory Aids

• Potential barrier: Si ~0.7 V, Ge ~0.3 V.
• Depletion layer = immobile charges, no free carriers.
• Diffusion current: majority carriers move due to concentration gradient.
• Drift current: minority carriers move due to electric field.
• Electric field intensity formula: E = V/d.

Semiconductor Diode

Basic Structure and Function

• Formed by joining p-type and n-type semiconductors creating a p-n junction.
• Metallic contacts are attached at ends for external voltage application.
• Depletion region forms at junction, acting as a barrier to current.
• Allows current flow mainly in one direction (p-side to n-side) under forward bias.
• Symbol: Arrow indicates conventional current direction in forward bias.

Solved Examples

• Ex 1: Identify the direction of current flow in a forward biased diode.
  • Current flows from p-side to n-side (arrow direction).

Practice Set

• Q1: What forms the depletion region in a semiconductor diode?
  • A) Metallic contacts
  • B) Junction of p-type and n-type materials
  • C) External battery
  • D) Insulating layer
• Q2: The arrow in the diode symbol indicates:
  • A) Electron flow direction
  • B) Conventional current direction under forward bias
  • C) Direction of applied voltage
  • D) Direction of reverse bias
• Q3: What happens to current flow when a diode is forward biased?
  • A) Current flows easily from p to n side
  • B) Current is blocked
  • C) Current flows from n to p side
  • D) No current flows

Answer Key

• Q1: B - Depletion region forms at the junction of p-type and n-type materials.
• Q2: B - Arrow shows conventional current direction when diode is forward biased.
• Q3: A - Forward bias reduces barrier, allowing current from p to n side.

Memory Aids

• Diode = p-n junction + metal contacts.
• Arrow in symbol = forward current direction.
• Depletion region blocks current unless forward biased.

Biasing of p-n Junction

Unbiased Junction

• No external voltage applied.
• Diffusion current (i_df) balances drift current (i_dr).
• Net current (i_net) = 0.
• Built-in potential barrier exists across depletion region.

Forward Bias

• Positive terminal connected to p-side, negative to n-side.
• External voltage reduces barrier potential.
• Depletion width decreases.
• Low resistance at junction; high current flows.
• Net current flows in forward direction.

Reverse Bias

• Positive terminal connected to n-side, negative to p-side.
• Barrier potential increases.
• Depletion width increases.
• High resistance at junction; very low leakage current flows.
• Net current is almost zero.

Solved Examples

• Ex 1: What happens to depletion width when a diode is forward biased?
  • Depletion width decreases.
• Ex 2: Describe current flow in reverse bias.
  • Very small leakage current; net current nearly zero.

Practice Set

• Q1: In forward bias, the barrier potential:
  • A) Increases
  • B) Decreases
  • C) Remains same
  • D) Becomes zero
• Q2: Reverse bias causes:
  • A) Decrease in depletion width
  • B) Increase in depletion width
  • C) No change in depletion width
  • D) Current flow from p to n side
• Q3: Which of the following is true for reverse biased diode?
  • A) High current flows
  • B) Low resistance at junction
  • C) Net current is almost zero
  • D) Barrier potential decreases

Answer Key

• Q1: B - Forward bias reduces barrier potential.
• Q2: B - Reverse bias increases depletion width.
• Q3: C - Net current is nearly zero in reverse bias.

Memory Aids

• Forward bias: Barrier down, width down, current up.
• Reverse bias: Barrier up, width up, current negligible.
• Depletion region controls current flow.

I-V Characteristic of Diode

Key Points

• Current through diode: \( I = I_0 \left[ e^{\frac{eV}{nkT}} - 1 \right] \)
• Where:
  • \( I_0 \) = reverse saturation current
  • \( e \) = electronic charge
  • \( V \) = applied voltage
  • \( n \) = ideality factor (1 for Ge, 2 for Si)
  • \( k \) = Boltzmann constant
  • \( T \) = absolute temperature (K)
• Dynamic resistance: \( r_d = \frac{\Delta V}{\Delta I} \)
• Static (dc) resistance: \( r_{dc} = \frac{V}{I} \)
• Forward bias: Current increases exponentially after threshold voltage.
• Reverse bias: Current is very small (reverse saturation current) until breakdown.

Zener Diode Characteristics

• Forward bias similar to normal diode.
• Reverse bias shows breakdown at Zener voltage \( V_Z \).
• At \( V_Z \), current increases sharply without damage.
• Used for voltage regulation.

Solved Examples

• Ex 1: Calculate current if \( V = 0.7V \), \( I_0 = 10^{-12} A \), \( n=1 \), \( T=300K \).
  • Use formula \( I = I_0 [e^{eV/nkT} -1] \).

Practice Set

• Q1: In the diode equation, the ideality factor \( n \) is:
  • A) 1 for Si, 2 for Ge
  • B) 2 for Si, 1 for Ge
  • C) Always 1
  • D) Always 2
• Q2: Dynamic resistance \( r_d \) is defined as:
  • A) \( V/I \)
  • B) \( \Delta V / \Delta I \)
  • C) \( I/V \)
  • D) \( V \times I \)
• Q3: At Zener breakdown voltage, the diode:
  • A) Stops conducting
  • B) Allows large reverse current without damage
  • C) Burns out
  • D) Acts as open circuit

Answer Key

• Q1: B - \( n=2 \) for Si, \( n=1 \) for Ge.
• Q2: B - Dynamic resistance is \( \Delta V / \Delta I \).
• Q3: B - Zener diode conducts large reverse current safely at breakdown.

Memory Aids

• Diode current: Exponential with voltage.
• \( n=1 \) Ge, \( n=2 \) Si.
• Zener diode: Reverse breakdown for voltage regulation.

Application of Junction Diode as a Rectifier

Rectifier

• Converts AC voltage to DC voltage.
• Uses diode's property: forward bias conducts, reverse bias blocks.

Half Wave Rectifier

• Diode conducts during positive half cycle only.
• Output is unidirectional pulsating DC.
• Output frequency = input frequency.

Full Wave Rectifier

• Rectifies both positive and negative half cycles.
• Uses two diodes and transformer with center tap.
• Output frequency = 2 × input frequency.
• Output is smoother pulsating DC.

Solved Examples

Ex 1

Calculate output frequency of full wave rectifier if input AC frequency is 50 Hz.

• Output frequency = 2 × 50 Hz = 100 Hz.

Practice Set

Q1: What is the output frequency of a half wave rectifier if input frequency is 60 Hz?

• A) 30 Hz
• B) 60 Hz
• C) 120 Hz
• D) 0 Hz

Q2: In a full wave rectifier, the output frequency is:

• A) Same as input frequency
• B) Twice the input frequency
• C) Half the input frequency
• D) Zero

Q3: Which component is essential for full wave rectification?

• A) One diode
• B) Two diodes and center-tapped transformer
• C) Capacitor only
• D) Resistor only

Answer Key

Q1: B) 60 Hz - Half wave rectifier output frequency equals input frequency.

Q2: B) Twice the input frequency - Full wave rectifier doubles frequency.

Q3: B) Two diodes and center-tapped transformer - Required for full wave rectification.

Memory Aids

• Half wave: Output frequency = input frequency.
• Full wave: Output frequency = 2 × input frequency.
• Full wave needs two diodes + center tap.

Special Purpose p-n Junction Diodes

Zener Diode

• Heavily doped p-n junction diode.
• Operates in reverse bias.
• Zener voltage \(V_Z\) remains constant over wide current range.
• Used as voltage regulator.
• Voltage across Zener diode remains constant despite input voltage changes.

Photodiode

• Special p-n junction diode with transparent window.
• Operated under reverse bias.
• Generates photocurrent proportional to light intensity.
• Used in light sensors and cameras.

Light Emitting Diode (LED)

• Converts electrical energy into light energy.
• Heavily doped p-n junction diode.
• Emits light under forward bias.
• Threshold voltage higher than Si diode; varies with color.
• Reverse breakdown voltage ~5 V.
• Made from GaAsP (visible colors) or GaAs (infrared).

Solar Cell

• Converts solar energy into electrical energy.
• Large area p-n junction generating emf on illumination.
• Works on photovoltaic effect (no external bias).

Solved Examples

Ex 1

What happens to Zener diode voltage when input voltage increases?

• Voltage across Zener diode remains constant.

Practice Set

Q1: Zener diode is used as a:

• A) Rectifier
• B) Voltage regulator
• C) Light emitter
• D) Photodetector

Q2: Photodiode operates under:

• A) Forward bias
• B) Reverse bias
• C) No bias
• D) Both biases

Q3: LED emits light when:

• A) Reverse biased
• B) Forward biased
• C) No bias
• D) Under illumination

Answer Key

Q1: B) Voltage regulator - Zener diode maintains constant voltage.

Q2: B) Reverse bias - Photodiode works in reverse bias.

Q3: B) Forward biased - LED emits light under forward bias.

Memory Aids

• Zener diode: Reverse bias, constant voltage \(V_Z\).
• Photodiode: Reverse bias, current ∝ light intensity.
• LED: Forward bias, electrical to light energy.
• Solar cell: Photovoltaic effect, no external bias.

Digital Electronics and Logic Gates

Signals

• Analog signal: Continuous time-varying voltage.
• Digital signal: Two discrete levels (0 and 1).

OR Gate

• Two or more inputs, one output.
• Output is high if any input is high.
• Boolean expression: \(Y = A + B\).

AND Gate

• Two or more inputs, one output.
• Output is high only if all inputs are high.
• Boolean expression: \(Y = A \cdot B\).

NOT Gate

• One input, one output.
• Output is inversion of input.
• Boolean expression: \(Y = \overline{A}\).

NAND Gate

• AND gate followed by NOT gate.
• Output is low only if all inputs are high.
• Boolean expression: \(Y = \overline{A \cdot B}\).

NOR Gate

• OR gate followed by NOT gate.
• Output is high only if all inputs are low.
• Boolean expression: \(Y = \overline{A + B}\).

Solved Examples

Ex 1

Find output of OR gate when inputs A=0, B=1.

• Output \(Y = 0 + 1 = 1\).

Practice Set

Q1: Output of AND gate when A=1, B=0 is:

• A) 0
• B) 1
• C) Undefined
• D) 2

Q2: Boolean expression for NAND gate is:

• A) \(Y = A + B\)
• B) \(Y = \overline{A \cdot B}\)
• C) \(Y = A \cdot B\)
• D) \(Y = \overline{A + B}\)

Q3: Output of NOT gate when input is 0 is:

• A) 0
• B) 1
• C) Undefined
• D) 2

Answer Key

Q1: A) 0 - AND gate output is 1 only if both inputs are 1.

Q2: B) \(Y = \overline{A \cdot B}\) - NAND is NOT of AND.

Q3: B) 1 - NOT gate inverts input.

Memory Aids

• OR: Output high if any input high.
• AND: Output high only if all inputs high.
• NOT: Output is inverse of input.
• NAND: AND followed by NOT.
• NOR: OR followed by NOT.