Elements of Electrical & Electronics Engineering (1313202) - Winter 2023 Solution

Question 1(a) [3 marks]

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Explain difference between Active and passive network.

Answer:

Mnemonic: “Active Adds Power, Passive Pulls Power”

Question 1(b) [4 marks]

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State and explain Kirchhoff’s voltage law (KVL).

Answer:

Kirchhoff’s Voltage Law (KVL): The algebraic sum of all voltages around any closed path (loop) in a circuit is zero.

Diagram:

graph LR
    A((A)) -- V1 --> B((B))
    B -- V2 --> C((C))
    C -- V3 --> D((D))
    D -- V4 --> A

Mathematical Form: V₁ + V₂ + V₃ + V₄ = 0

• Circuit Application: When moving around a loop, voltage rises (batteries) are positive and voltage drops (components) are negative
• Physical Meaning: Total energy in a closed loop is conserved

Mnemonic: “Voltage Loop Sum Zero”

Question 1(c) [7 marks]

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Define the following terms: (1) Charge (2) Current (3) Potential (4) E.M.F. (5) Inductance (6) Capacitance (7) Frequency.

Answer:

Mnemonic: “Careful Currents Pass Easily Into Circuit Frequently”

Question 1(c) OR [7 marks]

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State Ohm’s law. Write its application and limitation.

Answer:

Ohm’s Law: The current flowing through a conductor is directly proportional to the potential difference across it and inversely proportional to its resistance.

Mathematical Form: I = V/R

Diagram:

Applications of Ohm’s Law:

• Computing current, voltage, resistance in circuits
• Design of electrical networks
• Power calculations (P = VI = I²R = V²/R)
• Voltage division and current division

Limitations of Ohm’s Law:

• Not valid for non-linear elements (diodes, transistors)
• Not applicable at very high frequencies
• Not valid for non-metallic conductors like semiconductors
• Not applicable for vacuum tubes and gaseous devices

Mnemonic: “Voltage Drives, Resistance Restricts”

Question 2(a) [3 marks]

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Draw and explain energy band diagrams for insulator, conductor and Semiconductor.

Answer:

Energy Band Diagrams:

graph TD
    subgraph "Conductor"
    A1[Conduction Band] --- B1[Overlap]
    B1 --- C1[Valence Band]
    end
    
    subgraph "Semiconductor"
    A2[Conduction Band] --- B2[Small Gap]
    B2 --- C2[Valence Band]
    end
    
    subgraph "Insulator"
    A3[Conduction Band] --- B3[Large Gap]
    B3 --- C3[Valence Band]
    end

• Conductor: Valence and conduction bands overlap, allowing easy electron flow
• Semiconductor: Small energy gap (~1eV) between bands; electrons can jump with thermal energy
• Insulator: Large energy gap (>5eV) prevents electron movement between bands

Mnemonic: “Conductors Connect, Semiconductors Sometimes, Insulators Impede”

Question 2(b) [4 marks]

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Write statement of Maximum power transfer theorem and reciprocity theorem.

Answer:

Diagram:

Mnemonic: “Match Resistance to Maximize Power; Switch Source and Sink, Current Stays Same”

Question 2(c) [7 marks]

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Explain the formation and conduction of N-type materials.

Answer:

N-type Semiconductor Formation:

graph TD
    A[Silicon/Germanium Lattice] --> B[Doped with Pentavalent Element]
    B --> C[Extra Electron from Each Dopant Atom]
    C --> D[Free Electrons - Majority Carriers]
    D --> E[Holes - Minority Carriers]
    E --> F[Net Negative Charge]

• Doping Process: Silicon/Germanium (4 valence e⁻) doped with pentavalent elements (P, As, Sb)
• Extra Electron: Each dopant atom provides 1 extra electron after covalent bonding
• Conduction Mechanism:
  • Majority Carriers: Free electrons (negative charge carriers)
  • Minority Carriers: Holes (very few)
• Electrical Properties: Increased conductivity and negative charge carriers

Mnemonic: “Pentavalent Provides Plus one Electron, Negative-type”

Question 2(a) OR [3 marks]

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Define valence band, conduction band and forbidden gap.

Answer:

Diagram:

graph TD
    A[Conduction Band] --- B[Forbidden Gap]
    B --- C[Valence Band]

Mnemonic: “Valence Holds, Forbidden Blocks, Conduction Flows”

Question 2(b) OR [4 marks]

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Define the terms active power, reactive power and power factor with power triangle.

Answer:

Power Terms in AC Circuits:

Power Triangle:

• Apparent Power (S): Vector sum of active and reactive power
• Power Triangle: Right triangle with P, Q, and S as sides
• Power Factor: cos θ = P/S (0 to 1)

Mnemonic: “Active Power Works, Reactive Power Waits”

Question 2(c) OR [7 marks]

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Explain the structure of atom of trivalent, tetravalent and pentavalent elements.

Answer:

Atomic Structures:

Diagram:

graph TD
    subgraph "Trivalent (B, Al, Ga)"
    A1[Nucleus] --- B1[Inner Shells]
    B1 --- C1[3 Valence Electrons]
    end
    
    subgraph "Tetravalent (C, Si, Ge)"
    A2[Nucleus] --- B2[Inner Shells]
    B2 --- C2[4 Valence Electrons]
    end
    
    subgraph "Pentavalent (P, As, Sb)"
    A3[Nucleus] --- B3[Inner Shells]
    B3 --- C3[5 Valence Electrons]
    end

• Trivalent Elements: Used as p-type dopants in semiconductors
• Tetravalent Elements: Form semiconductor base materials
• Pentavalent Elements: Used as n-type dopants in semiconductors

Mnemonic: “Three Tries to Bond, Four Forms Full bonds, Five Frees an Electron”

Question 3(a) [3 marks]

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Draw the symbol of photodiode and state it’s application.

Answer:

Photodiode Symbol:

Applications of Photodiode:

• Light sensors and detectors
• Optical communication systems
• Camera exposure controls
• Barcode scanners
• Medical instruments
• Solar cells

Mnemonic: “Photons Produce Current”

Question 3(b) [4 marks]

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Write a Short note on LED.

Answer:

LED (Light Emitting Diode):

Advantages:

• High efficiency (low power consumption)
• Long life (50,000+ hours)
• Small size and durability
• Various colors available

Applications:

• Indicators and displays
• Lighting systems
• TV/monitor backlights
• Traffic signals

Mnemonic: “Light Emits when Diode conducts”

Question 3(c) [7 marks]

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Draw and explain VI characteristic of PN junction diode.

Answer:

P-N Junction Diode V-I Characteristic:

Forward Bias Region:

• Knee Voltage: 0.3V (Ge), 0.7V (Si) where current starts flowing
• Current Equation: I = Is(e^(qV/kT) - 1)
• Conductivity: High (low resistance)

Reverse Bias Region:

• Leakage Current: Very small reverse current (micro-amps)
• Breakdown Region: Sharp increase in current at breakdown voltage
• Conductivity: Very low (high resistance)

Key Points:

• Barrier Potential: Decreases in forward bias, increases in reverse bias
• Diode Resistance: Dynamic resistance changes with applied voltage
• Temperature Effect: Voltage drop decreases with temperature increase

Mnemonic: “Forward Flows Freely, Reverse Resists”

Question 3(a) OR [3 marks]

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List the applications of PN junction diode.

Answer:

Applications of PN Junction Diode:

Mnemonic: “Rectify, Process, Protect, Logic, Regulate, Light”

Question 3(b) OR [4 marks]

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Explain the formation of depletion region in unbiased P-N junction.

Answer:

Depletion Region Formation:

graph LR
    subgraph "P-Type"
    A[Holes]
    end
    
    subgraph "Depletion Region"
    B[No Free Carriers]
    end
    
    subgraph "N-Type"
    C[Electrons]
    end
    
    A --Diffusion--> B
    C --Diffusion--> B

Process:

• Diffusion: Electrons from n-side diffuse to p-side; holes from p-side diffuse to n-side
• Recombination: Electrons and holes recombine at the junction
• Immobile Ions: Exposed positive ions in n-region, negative ions in p-region
• Electric Field: Forms between positive and negative ions, opposing further diffusion
• Equilibrium: Diffusion current equals drift current; no net current flows

Properties of Depletion Region:

• No free charge carriers
• Acts as insulator
• Width depends on doping levels
• Contains built-in potential barrier

Mnemonic: “Diffusion Depletes Carriers, Creating Electric barrier”

Question 3(c) OR [7 marks]

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Explain construction, working and applications of PN junction diode.

Answer:

Construction of PN Junction Diode:

• P-Type Region: Silicon/Germanium doped with trivalent impurities (boron, aluminum)
• N-Type Region: Silicon/Germanium doped with pentavalent impurities (phosphorus, arsenic)
• Junction: Interface between p and n regions with depletion layer
• Terminals: Anode (p-side) and Cathode (n-side)

Working Principle:

Applications:

• Rectification in power supplies
• Signal demodulation in radios
• Voltage regulation (Zener)
• Signal clipping and clamping
• Logic gates and switching
• Light emission and detection

Mnemonic: “Forward Flow, Reverse Restrict, Convert AC to DC”

Question 4(a) [3 marks]

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Define: (1) Ripple frequency (2) Ripple factor (3) PIV of a diode.

Answer:

Mnemonic: “Ripples Per second, Ripple Proportion, Reverse Peak Voltage”

Question 4(b) [4 marks]

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Give comparison between full wave rectifier with two diodes and full wave bridge rectifier.

Answer:

Diagram:

graph TD
    subgraph "Center-Tapped"
    A[Transformer with center-tap] --> B[2 Diodes]
    end
    
    subgraph "Bridge"
    C[Transformer] --> D[4 Diodes in Bridge]
    end

Mnemonic: “Bridge Beats Tap with Lower PIV but Needs More Diodes”

Question 4(c) [7 marks]

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Explain zener diode as voltage regulator.

Answer:

Zener Diode Voltage Regulator:

Working Principle:

• Reverse Biased: Zener operates in breakdown region
• Constant Voltage: Maintains fixed voltage (Vz) across its terminals
• Current Regulation: Series resistor (Rs) limits current
• Load Changes: When load current changes, Zener current changes to maintain constant output voltage

Design Equations:

• Rs = (Vin - Vz) / (IL + Iz)
• Power rating of Zener: Pz = Vz × Iz(max)

Advantages:

• Simple circuit
• Low cost
• Good regulation for small loads
• Fast response to load changes

Limitations:

• Power wastage in Rs and Zener
• Limited output current capability
• Temperature dependence of Vz

Mnemonic: “Zener Stays at breakdown Voltage despite Current changes”

Question 4(a) OR [3 marks]

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What is rectifier? Explain full wave rectifier with waveforms.

Answer:

Rectifier: A circuit that converts AC voltage to pulsating DC voltage by allowing current flow in one direction only.

Full Wave Rectifier:

Waveforms:

• Operation: Both half cycles of AC input are converted to same polarity
• Frequency: Output ripple frequency is twice the input frequency
• Voltage: Vdc = 0.637Vm (where Vm is peak input voltage)

Mnemonic: “Full Wave Forms Full Output”

Question 4(b) OR [4 marks]

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Why filter is required in rectifier? State the different types of filter and explain any one type of filter.

Answer:

Need for Filters: Rectifiers produce pulsating DC with large ripples; filters smooth this output to provide steady DC voltage.

Types of Filters:

• Capacitor (C) filter
• Inductor (L) filter
• LC filter
• π (Pi) filter
• RC filter

Capacitor Filter:

Working:

• Capacitor charges during voltage rise
• Discharges slowly through load during voltage fall
• Acts as temporary storage element
• Time constant RC determines discharge rate
• Reduces ripple by providing discharge path

Advantages:

• Simple and economical
• Good smoothing for light loads
• Increases DC output voltage

Mnemonic: “Capacitor Catches Charge and Releases Slowly”

Question 4(c) OR [7 marks]

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Write the need of rectifier. Explain bridge rectifier with circuit diagram and draw its input and output waveforms.

Answer:

Need for Rectifiers:

• Convert AC to DC for electronic devices
• Power supplies for DC-operated equipment
• Battery charging circuits
• DC power for industrial drives
• Signal demodulation in communication

Bridge Rectifier Circuit:

Working Principle:

• Positive Half Cycle: D1 and D4 conduct, D2 and D3 block
• Negative Half Cycle: D2 and D3 conduct, D1 and D4 block
• Both Half Cycles: Current flows in same direction through load

Input-Output Waveforms:

Characteristics:

• Vdc = 0.637Vm (Vm: peak input voltage)
• PIV of each diode = Vm
• Ripple factor = 0.48
• Efficiency = 81.2%
• TUF = 0.693

Mnemonic: “Bridge Brings Both halves to Direct Current”

Question 5(a) [3 marks]

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Explain causes of electronic waste.

Answer:

Causes of Electronic Waste:

Mnemonic: “Upgrade, Use, Throw, Repeat”

Question 5(b) [4 marks]

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Compare PNP and NPN transistors.

Answer:

    E
    |
    |\
    |  \
    |    >C
    |  /
    |/
    B

    C
    |
    |/
    |  /
    |    >E
    |  \
    |\
    B

Mnemonic: “PNP: Positive to Negative to Positive; NPN: Negative to Positive to Negative”

Question 5(c) [7 marks]

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Draw the symbol, explain the construction and working of MOSFET.

Answer:

MOSFET Symbol (N-Channel Enhancement):

Construction:

graph TD
    A[Source - n+] --- B[Channel region - p]
    B --- C[Drain - n+]
    D[Gate] --- E[Silicon Dioxide Insulator]
    E --- B

Components:

• Substrate: P-type semiconductor body
• Source/Drain: Heavily doped n+ regions
• Gate: Metal electrode separated by insulator (SiO2)
• Channel: Forms between source and drain when biased

Working Principle:

• Enhancement Mode: No channel exists initially; gate voltage creates channel
• Threshold Voltage (VT): Minimum gate voltage needed to form channel
• Conducting State: When VGS > VT, electrons form channel, allowing current flow
• Saturation Region: Current remains constant despite increase in VDS
• Linear Region: Current proportional to VDS at low drain voltages

Applications:

• Digital circuits (logic gates)
• Power amplifiers
• Switching applications
• Memory devices

Mnemonic: “Gate Voltage Controls Electron Channel”

Question 5(a) OR [3 marks]

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Explain methods to handle electronic waste.

Answer:

Methods to Handle Electronic Waste:

Diagram:

graph LR
    A[E-Waste] --> B[Collection]
    B --> C[Sorting]
    C --> D[Dismantling]
    D --> E[Material Recovery]
    E --> F[Remanufacturing]

Mnemonic: “Reduce, Reuse, Recycle, Recover Resources”

Question 5(b) OR [4 marks]

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Derive the relationship between αdc and βdc.

Answer:

Relationship between α and β:

Given:

• αdc = IC/IE (Common base current gain)
• βdc = IC/IB (Common emitter current gain)

Derivation: From Kirchhoff’s current law: IE = IC + IB

Dividing both sides by IC: IE/IC = 1 + IB/IC

Since αdc = IC/IE: 1/αdc = 1 + IB/IC

Since βdc = IC/IB: 1/αdc = 1 + 1/βdc

Final Relations:

• αdc = βdc/(1 + βdc)
• βdc = αdc/(1 - αdc)

Table:

Mnemonic: “Alpha approaches One as Beta approaches Infinity”

Question 5(c) OR [7 marks]

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Explain common collector configuration with its input and output characteristics.

Answer:

Common Collector (Emitter Follower) Configuration:

Input Characteristics:

Output Characteristics:

Key Features:

• Voltage Gain (Av): Approximately 1 (unity)
• Current Gain (Ai): High (β + 1)
• Input Impedance: High (β × RE)
• Output Impedance: Low (1/gm) where gm is transconductance
• Phase Relationship: No phase inversion between input and output
• Applications: Impedance matching, buffers, voltage regulators

Characteristics:

• Input Resistance: Ri = β × (re + RL)
• Output Resistance: Ro = (rs + re)/(β + 1)
• Voltage Gain: Av = RL/(RL + re) ≈ 1
• Current Gain: Ai = (β + 1)

Advantages:

• Very high input impedance
• Low output impedance
• Good impedance matching properties
• No phase inversion

Limitations:

• No voltage gain (slightly less than 1)
• Used only for impedance matching

Mnemonic: “Collector Common, Current amplifies, Voltage follows”

This completes the full solutions for the Elements of Electrical & Electronics Engineering (1313202) Winter 2023 examination.