Fundamentals of Electronics (4311102) - Summer 2024 Solution

Question 1 [14 marks]

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Answer any seven out of ten.

Question 1(1) [2 marks]

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Define resistor and give its unit.

Answer: A resistor is an electronic component that opposes the flow of electric current. Its unit is Ohm (Ω).

Table: Resistor Properties

Mnemonic: “Resistors Oppose Current” (ROC)

Question 1(2) [2 marks]

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Give two examples of active and passive components each.

Answer:

Table: Electronic Components Classification

Mnemonic: “TARD” - Transistors And Resistors Differ

Question 1(3) [2 marks]

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Draw symbols of any two semiconductor devices.

Answer:

Diagram:

graph TD
    subgraph Diode
        A[Plus] --> B["|<|"] --> C[Minus]
    end

    subgraph NPN_Transistor
        D[C] --> E
        F[E] --> E
        G[B] --> E
    end

Mnemonic: “Diodes Direct, Transistors Transfer”

Question 1(4) [2 marks]

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Differentiate between intrinsic and extrinsic semiconductor.

Answer:

Table: Intrinsic vs Extrinsic Semiconductors

Mnemonic: “Pure In, Doped Ex”

Question 1(5) [2 marks]

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LED stands for _________________.

Answer: LED stands for Light Emitting Diode.

Diagram:

graph LR
    A[Light] --> B[Emitting] --> C[Diode]
    style A fill:#f96,stroke:#333
    style B fill:#9cf,stroke:#333
    style C fill:#f9f,stroke:#333

Mnemonic: “Light Emitters Dazzle”

Question 1(6) [2 marks]

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State any two applications of Photo-diode.

Answer:

Table: Photo-diode Applications

Mnemonic: “Light Sensing Communication” (LSC)

Question 1(7) [2 marks]

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List the types of transistor and draw their symbols.

Answer:

Types of Transistors:

1. NPN Transistor
2. PNP Transistor

Diagram:

graph TD
    subgraph "NPN"
    A[C] --- B --- C[E]
    D[B] --- B
    end
    subgraph "PNP"
    E[E] --- F --- G[C]
    H[B] --- F
    end

Mnemonic: “Not Pointing iN, Pointing outP”

Question 1(8) [2 marks]

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Give the value of forward voltage drop of Germanium and Silicon diode.

Answer:

Table: Forward Voltage Drop Values

Mnemonic: “Germanium’s Three, Silicon’s Seven” (0.3V, 0.7V)

Question 1(9) [2 marks]

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The _________________ diode can be used as a light detector.

Answer: The Photodiode can be used as a light detector.

Diagram:

graph LR
    A[Light] -->|detected by| B[Photodiode]
    B -->|generates| C[Current]
    style A fill:#ff9,stroke:#333
    style B fill:#9cf,stroke:#333
    style C fill:#f96,stroke:#333

Mnemonic: “Photo Detects Light” (PDL)

Question 1(10) [2 marks]

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Define Q-factor of a coil.

Answer: Q-factor (Quality factor) of a coil is the ratio of its inductive reactance to its resistance, indicating how efficiently it stores energy.

Table: Q-Factor

Mnemonic: “Quality equals Reactance over Resistance” (QRR)

Question 2(a) [3 marks]

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Explain colour coding method of resistor.

Answer:

Resistor color coding uses colored bands to indicate resistance value and tolerance.

Table: Resistor Color Code

For a 4-band resistor:

• First band: First digit
• Second band: Second digit
• Third band: Multiplier
• Fourth band: Tolerance

Mnemonic: “Bad Boys Race Our Young Girls But Violet Generally Wins” (Colors in order: Black, Brown, Red, Orange, Yellow, Green, Blue, Violet, Grey, White)

Question 2(a) OR [3 marks]

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Explain Light Dependent Resistor with its characteristics.

Answer:

LDR is a resistor whose resistance decreases when light intensity increases.

Characteristics of LDR:

Table: LDR Properties

Diagram:

graph TD
    A[Increase Light] -->|Causes| B[Decrease Resistance]
    C[Decrease Light] -->|Causes| D[Increase Resistance]
    style A fill:#ff9,stroke:#333
    style B fill:#9cf,stroke:#333
    style C fill:#999,stroke:#333
    style D fill:#f96,stroke:#333

Mnemonic: “Light Up, Resistance Down” (LURD)

Question 2(b) [3 marks]

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Explain classification of capacitors in detail.

Answer:

Capacitors are classified based on dielectric material and construction.

Table: Capacitor Classifications

Diagram:

graph TD
    A[Capacitors] --> B[Fixed]
    A --> C[Variable]
    B --> D[Ceramic]
    B --> E[Electrolytic]
    B --> F[Polyester/Film]
    C --> G[Air Gang]
    C --> H[Trimmer]
    style A fill:#f96,stroke:#333

Mnemonic: “CEPT” (Ceramic, Electrolytic, Polyester, Tantalum)

Question 2(b) OR [3 marks]

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Explain classification of inductor in detail.

Answer:

Inductors are classified based on core material and construction.

Table: Inductor Classifications

Diagram:

graph TD
    A[Inductors] --> B[Air Core]
    A --> C[Iron Core]
    A --> D[Ferrite Core]
    A --> E[Toroidal]
    style A fill:#9cf,stroke:#333

Mnemonic: “Air Iron Ferrite Toroid” (AIFT)

Question 2(c) [4 marks]

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State and explain Faraday’s laws of Electromagnetic Induction.

Answer:

Faraday’s laws explain how electromagnetic induction works.

Faraday’s First Law: When a magnetic field linked with a conductor changes, an EMF is induced in the conductor.

Faraday’s Second Law: The magnitude of induced EMF is proportional to the rate of change of magnetic flux.

Table: Faraday’s Laws Summary

Diagram:

graph LR
    A[Moving Magnet] -->|Creates| B[Changing Magnetic Field]
    B -->|Induces| C[EMF in Conductor]
    style A fill:#f96,stroke:#333
    style B fill:#9cf,stroke:#333
    style C fill:#ff9,stroke:#333

Mnemonic: “Change Magnetic Field, Create Electric Current” (CMFCEC)

Question 2(c) OR [4 marks]

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Enlist specifications of capacitors and explain two in detail.

Answer:

Specifications of Capacitors:

1. Capacitance value
2. Voltage rating
3. Tolerance
4. Leakage current
5. Temperature coefficient

Detailed Explanation:

Capacitance Value: The amount of charge a capacitor can store per volt, measured in Farads (F).

Voltage Rating: The maximum voltage that can be applied without damaging the capacitor.

Table: Capacitor Specifications

Diagram:

graph TD
    A[Capacitor Specifications] --> B[Capacitance Value]
    A --> C[Voltage Rating]
    A --> D[Tolerance]
    A --> E[Leakage Current]
    A --> F[Temperature Coefficient]
    style A fill:#9cf,stroke:#333

Mnemonic: “Capacitors Very Tolerant of Low Temperatures” (CVTLT)

Question 2(d) [4 marks]

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Write colour band of 47Ω±5% resistance.

Answer:

For 47Ω±5% resistor, the color bands are:

Table: Color Bands for 47Ω±5%

Diagram:

graph LR
    A[Yellow] -->|4| B[Violet] -->|7| C[Black] -->|×10⁰| D[Gold] -->|±5%| E[47Ω±5%]
    style A fill:#ff9,stroke:#333
    style B fill:#f0f,stroke:#333
    style C fill:#000,stroke:#fff
    style D fill:#fd0,stroke:#333
    style E fill:#fff,stroke:#333

Mnemonic: “Yellow Violets Bring Gold” (The colors of the bands)

Question 2(d) OR [4 marks]

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Calculate value of resistor and tolerance for a given colour code: Brown, Black, yellow.

Answer:

Table: Interpretation of Brown, Black, Yellow

Calculation: 1st digit: 1 2nd digit: 0 Multiplier: 10⁴

Value = 10 × 10⁴ = 100,000Ω = 100kΩ

No 4th band means ±20% tolerance

Diagram:

graph LR
    A[Brown] -->|1| B[Black] -->|0| C[Yellow] -->|×10⁴| D[100kΩ ±20%]
    style A fill:#a52a2a,stroke:#333
    style B fill:#000,stroke:#fff
    style C fill:#ff0,stroke:#333
    style D fill:#fff,stroke:#333

Mnemonic: “Big Black Yield” (Brown-Black-Yellow)

Question 3(a) [3 marks]

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Define doping. Give the name of semiconductor materials fabricated by doping with an example of each.

Answer:

Doping is the process of adding impurities to a pure semiconductor to modify its electrical properties.

Table: Doped Semiconductors

Diagram:

graph LR
    A[Pure Semiconductor] --> B[Add Trivalent Impurity] --> C[P-type]
    A --> D[Add Pentavalent Impurity] --> E[N-type]
    style A fill:#9cf,stroke:#333
    style C fill:#f96,stroke:#333
    style E fill:#99f,stroke:#333

Mnemonic: “Positive has Plus Holes, Negative has Numerous Electrons” (PHNE)

Question 3(a) OR [3 marks]

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Define Ripple factor, Peak Inverse Voltage (PIV), Rectification efficiency.

Answer:

Table: Rectifier Terms

Diagram:

graph TD
    A[Rectifier Parameters] --> B[Ripple Factor]
    A --> C[Peak Inverse Voltage]
    A --> D[Rectification Efficiency]
    style A fill:#9cf,stroke:#333

Mnemonic: “Ripples Peak Efficiently” (RPE)

Question 3(b) [3 marks]

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Explain working of Crystal diode.

Answer:

Crystal diode is a point-contact diode made with a semiconductor crystal.

Table: Crystal Diode Properties

Diagram:

graph LR
    A[RF Signal] --> B[Crystal Diode] --> C[Rectified Signal]
    style A fill:#9cf,stroke:#333
    style B fill:#f96,stroke:#333
    style C fill:#9f9,stroke:#333

Mnemonic: “Crystal Detects Radio Frequencies” (CDRF)

Question 3(b) OR [3 marks]

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Explain working of photodiode.

Answer:

Photodiode converts light energy into electrical current when operated in reverse bias.

Table: Photodiode Characteristics

Diagram:

graph LR
    A[Light] -->|Strikes| B[PN Junction]
    B -->|Creates| C[Electron-Hole Pairs]
    C -->|Produces| D[Current Flow]
    style A fill:#ff9,stroke:#333
    style D fill:#9cf,stroke:#333

Mnemonic: “Light In, Current Out” (LICO)

Question 3(c) [4 marks]

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Explain half-wave rectifier with circuit diagram and waveforms.

Answer:

Half-wave rectifier converts AC to pulsating DC by allowing current flow only during positive half cycles.

Circuit Diagram:

graph LR
    A[AC Input] --- B[Transformer] --- C[Diode] --- D[Load Resistor] --- E[Ground]
    E --- A
    style A fill:#9cf,stroke:#333
    style D fill:#f96,stroke:#333

Waveforms:

graph TD
    subgraph "Input AC"
    A[+Vp] --- B[(0)] --- C[-Vp]
    end
    subgraph "Output DC"
    D[+Vp] --- E[(0)] --- F[(0)]
    end
    style A fill:#9cf,stroke:#333
    style C fill:#9cf,stroke:#333
    style D fill:#f96,stroke:#333

Table: Half-Wave Rectifier Properties

Mnemonic: “Half Wave Passes Half” (HWPH)

Question 3(c) OR [4 marks]

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Explain full-wave rectifier with circuit diagram and waveforms.

Answer:

Full-wave rectifier converts both halves of AC input to pulsating DC output.

Circuit Diagram (Bridge type):

graph LR
    A[AC Input] --- B[D1]
    A --- C[D3]
    B --- D[D2] --- E[+Output]
    C --- F[D4] --- G[-Output]
    E --- H[Load] --- G
    style A fill:#9cf,stroke:#333
    style H fill:#f96,stroke:#333

Waveforms:

graph TD
    subgraph "Input AC"
    A[+Vp] --- B[(0)] --- C[-Vp] --- B
    end
    subgraph "Output DC"
    D[+Vp] --- E[(0)] --- D
    end
    style A fill:#9cf,stroke:#333
    style C fill:#9cf,stroke:#333
    style D fill:#f96,stroke:#333

Table: Full-Wave Rectifier Properties

Mnemonic: “Full Wave Makes Full Use” (FWMFU)

Question 3(d) [4 marks]

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Draw and explain VI characteristics of PN junction diode.

Answer:

VI Characteristics:

graph TD
    subgraph "Forward Bias"
    A[Vf] --> B[If]
    end
    subgraph "Reverse Bias"
    C[Vr] --> D[Ir]
    E[Breakdown] --> F[Reverse Current Increases]
    end
    style A fill:#9cf,stroke:#333
    style C fill:#f96,stroke:#333
    style E fill:#f00,stroke:#333

Table: PN Junction Diode Characteristics

Forward Bias: Positive voltage to P-side, current flows easily after threshold. Reverse Bias: Positive voltage to N-side, only small leakage current flows.

Mnemonic: “Forward Flows, Reverse Restricts” (FFRR)

Question 3(d) OR [4 marks]

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Write difference between P-type and N-type semiconductor.

Answer:

Table: P-type vs N-type Semiconductor

Diagram:

graph LR
    subgraph "P-type"
    A[Silicon] --- B[Boron]
    C[Holes] --- D[+]
    end
    subgraph "N-type"
    E[Silicon] --- F[Phosphorus]
    G[Electrons] --- H[-]
    end
    style C fill:#f96,stroke:#333
    style G fill:#9cf,stroke:#333

Mnemonic: “Positive has Plus Holes, Negative has Numerous Electrons” (PHNE)

Question 4(a) [3 marks]

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Explain the principle of operation of LED.

Answer:

LED (Light Emitting Diode) emits light when forward biased due to electron-hole recombination.

Principle of Operation: When forward biased, electrons from N-side move to P-side and recombine with holes, releasing energy as photons (light).

Table: LED Operation

Diagram:

graph LR
    A[Forward Bias] -->|Causes| B[Current Flow]
    B -->|Creates| C[Electron-Hole Recombination]
    C -->|Releases| D[Photons or Light]

Mnemonic: “Forward Current Emits Light” (FCEL)

Question 4(a) OR [3 marks]

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State applications of LED.

Answer:

Table: LED Applications

Diagram:

graph TD
    A[LED Applications] --> B[Indicators]
    A --> C[Displays]
    A --> D[Lighting]
    A --> E[Communication]
    A --> F[Signals]
    style A fill:#9cf,stroke:#333

Mnemonic: “Display Lights In Clever Signals” (DLICS)

Question 4(b) [4 marks]

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Explain Zener diode as voltage regulator.

Answer:

Zener diode maintains constant output voltage despite input voltage fluctuations when operated in reverse breakdown region.

Circuit:

graph LR
    A[Unregulated DC] --- B[Series Resistor] --- C[Output]
    C --- D[Zener Diode] --- E[Ground]
    C --- F[Load] --- E
    style A fill:#9cf,stroke:#333
    style C fill:#9f9,stroke:#333
    style D fill:#f96,stroke:#333

Working:

• Series resistor limits current
• Zener operates in breakdown region
• Maintains constant voltage across load

Table: Zener Regulator Characteristics

Mnemonic: “Zeners Break to Regulate” (ZBR)

Question 4(b) OR [4 marks]

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Give limitations of zener voltage regulator.

Answer:

Table: Limitations of Zener Voltage Regulator

Diagram:

graph TD
    A[Zener Limitations] --> B[Power Limits]
    A --> C[Current Limits]
    A --> D[Temperature Effects]
    A --> E[Efficiency Issues]
    A --> F[Noise Generation]
    style A fill:#f96,stroke:#333

Mnemonic: “Power Current Temperature Efficiency Noise” (PCTEN)

Question 4(c) [7 marks]

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Discuss the necessity of filter circuit in rectifier. List various types of filter circuits used in rectifier and explain any one with neat diagram.

Answer:

Necessity of Filter Circuit: Rectifier output contains AC ripple that must be removed for smoother DC. Filters reduce these ripples to provide steady DC output.

Types of Filter Circuits:

1. Capacitor filter (Shunt capacitor)
2. LC filter
3. π-filter (Pi-filter)
4. RC filter

Explanation of Capacitor Filter:

Circuit Diagram:

graph LR
    A[Rectifier Output] --- B[+]
    B --- C[Load]
    B --- D[Capacitor]
    C --- E[Ground]
    D --- E
    style A fill:#9cf,stroke:#333
    style C fill:#f96,stroke:#333
    style D fill:#9f9,stroke:#333

Working:

• Capacitor charges during voltage peaks
• Discharges slowly during voltage drops
• Maintains output voltage between peaks
• Reduces ripple voltage

Table: Capacitor Filter Characteristics

Waveforms:

graph TD
    subgraph "Rectifier Output"
    A[Pulsating DC]
    end
    subgraph "Filter Output"
    B[Smoother DC]
    end
    style A fill:#f96,stroke:#333
    style B fill:#9f9,stroke:#333

Mnemonic: “Capacitors Hold Voltage During Drops” (CHVDD)

Question 5(a) [3 marks]

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Define e-waste. List common e-waste items.

Answer:

E-waste refers to discarded electronic devices and components that have reached the end of their useful life.

Table: Common E-waste Items

Diagram:

graph TD
    A[E-waste] --> B[Computing]
    A --> C[Communication]
    A --> D[Home Appliances]
    A --> E[Components]
    A --> F[Office Equipment]
    style A fill:#f96,stroke:#333

Mnemonic: “Computers, Communication, Components, Home Appliances” (CCCHA)

Question 5(b) [3 marks]

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State and explain various strategies of e-waste management.

Answer:

Table: E-waste Management Strategies

Diagram:

graph TD
    A[E-waste Management] --> B[Reduce]
    A --> C[Reuse]
    A --> D[Recycle]
    A --> E[Responsible Disposal]
    A --> F[Extended Producer Responsibility]
    style A fill:#9cf,stroke:#333

Mnemonic: “3R’s

Question 5(c) [4 marks]

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Explain transistor as switch.

Answer:

Transistor can function as an electronic switch by operating in either cutoff (OFF) or saturation (ON) region.

Table: Transistor Switch Operation

Circuit Diagram:

graph LR
    A[+Vcc] --- B[Rc] --- C[Collector]
    C --- D[Emitter] --- E[Ground]
    F[Vin] --- G[Rb] --- H[Base]
    H --- D
    style F fill:#9cf,stroke:#333
    style A fill:#f96,stroke:#333

Working:

• When input is HIGH: Transistor saturates, acts like closed switch
• When input is LOW: Transistor cuts off, acts like open switch

Mnemonic: “No Base No Current, Apply Base Connect Circuit” (NBNC-ABC)

Question 5(d) [4 marks]

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Derive relation between α and β for CE configuration of transistor.

Answer:

In transistors, α (alpha) and β (beta) are current gain parameters.

Definitions:

• α = IC/IE (Common Base current gain)
• β = IC/IB (Common Emitter current gain)

Derivation: Since IE = IC + IB, we can write: α = IC/IE = IC/(IC + IB)

Dividing numerator and denominator by IB: α = (IC/IB)/[(IC/IB) + 1] = β/(β + 1)

Therefore: β = α/(1-α)

Table: Relationship between α and β

Diagram:

graph TD
    A[alpha = IC divided by IE] --- B[beta = IC divided by IB]
    C[beta = alpha divided by 1 minus alpha] --- D[alpha = beta divided by beta plus 1]

    style A fill:#9cf,stroke:#333
    style B fill:#f96,stroke:#333

Mnemonic: “Beta equals Alpha divided by One minus Alpha” (BAOA)