Fundamentals of Electronics (DI01000051) - Winter 2024 Solution

Question 1(a) [3 marks]

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Define Active and Passive Components with example.

Answer:

Table: Active vs Passive Components

• Active components: Control and amplify electrical signals using external power
• Passive components: Store or dissipate energy without amplification

Mnemonic: “Active Amplifies, Passive Preserves”

Question 1(b) [4 marks]

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Explain construction and working of LDR.

Answer:

Construction:

• Serpentine track of cadmium sulfide on ceramic substrate
• Metal electrodes at both ends for connections
• Protective coating prevents moisture damage

Working Principle:

• Light intensity ↑: Resistance ↓ (conducts more)
• Darkness: Resistance ↑ (conducts less)
• Applications: Street lights, automatic cameras

Mnemonic: “Light Low Resistance”

Question 1(c) [7 marks]

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Define Capacitance and explain Aluminum Electrolytic wet type capacitor.

Answer:

Capacitance Definition: Ability to store electrical charge. C = Q/V (Farads)

Aluminum Electrolytic Capacitor:

Construction:

• Anode: Aluminum foil with oxide layer
• Dielectric: Thin aluminum oxide film
• Cathode: Liquid electrolyte with aluminum foil
• Polarity: Must be connected correctly

Features:

• High capacitance values (1µF to 10,000µF)
• Polarized - has positive and negative terminals
• Applications: Power supply filtering, coupling

Mnemonic: “Aluminum Always Amplifies”

Question 1(c OR) [7 marks]

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Explain the color band coding method of Resistor. Write color band of 32 Ω ±10% resistance.

Answer:

Color Code Table:

For 32 Ω ±10%:

Calculation: 3 × 2 × 0.1 = 3.2 × 10 = 32 Ω

Mnemonic: “Big Boys Race Our Young Girls But Violet Generally Wins”

Question 2(a) [3 marks]

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Define following terms: 1) Rectifier 2) Ripple factor 3) Filter

Answer:

• Rectifier: Uses diodes to allow current in one direction
• Ripple factor: Lower value means better filtering
• Filter: Uses capacitors/inductors to reduce ripples

Mnemonic: “Rectify Ripples, Filter Fixes”

Question 2(b) [4 marks]

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Draw and explain positive clipper circuit with waveform.

Answer:

Circuit Diagram:

Working:

• When Vin > +V: Diode conducts, output = +V
• When Vin < +V: Diode off, output follows input
• Result: Clips positive peaks above +V level

Waveform:

Applications: Signal limiting, protection circuits

Mnemonic: “Positive Peaks Prevented”

Question 2(c) [7 marks]

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Explain working of full wave rectifier with two diodes.

Answer:

Circuit Diagram:

Working:

• Positive half-cycle: D1 conducts, D2 off
• Negative half-cycle: D2 conducts, D1 off
• Both diodes work alternately
• Output frequency = 2 × input frequency

Key Parameters:

Advantages:

• Better efficiency than half-wave
• Lower ripple content
• Higher transformer utilization

Mnemonic: “Two Diodes, Two Halves”

Question 2(a OR) [3 marks]

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Define rectifier and write its applications.

Answer:

Definition: Electronic circuit that converts alternating current (AC) into direct current (DC) using diodes.

Applications:

• Primary function: AC to DC conversion
• Essential component: In all electronic devices

Mnemonic: “Rectify AC, Deliver DC”

Question 2(b OR) [4 marks]

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Explain working of Pi(π) type capacitor filter.

Answer:

Circuit Diagram:

Working:

• C1: Filters initial ripples from rectifier
• Inductor L: Opposes current changes, smooths further
• C2: Final filtering for smooth DC output
• Combined effect: Excellent ripple reduction

Characteristics:

Advantages:

• Excellent filtering performance
• Low ripple content
• Good voltage regulation

Mnemonic: “Pi Provides Perfect”

Question 2(c OR) [7 marks]

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Compare half wave and full wave bridge rectifier.

Answer:

Comparison Table:

Circuit Diagrams:

Half Wave:

Full Wave Bridge:

Key Differences:

• Full wave: Better efficiency and lower ripple
• Half wave: Simpler but poor performance
• Bridge: No center-tap transformer required

Mnemonic: “Half Wastes, Full Works”

Question 3(a) [3 marks]

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Draw the symbols of following: 1) Zener diode 2) LED 3) Varactor diode

Answer:

Electronic Symbols:

Symbol Details:

• Zener: Z indicates zener characteristics
• LED: Arrows show light output direction
• Varactor: Lines represent variable capacitance

Mnemonic: “Zener Zigs, LED Lights, Varactor Varies”

Question 3(b) [4 marks]

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Explain construction and working of LED.

Answer:

Construction:

Materials:

• P-type: Boron-doped semiconductor
• N-type: Phosphorus-doped semiconductor
• Common materials: GaAs, GaP, GaN

Working Principle:

• Forward bias: Electrons recombine with holes
• Energy release: In form of photons (light)
• Color: Depends on semiconductor material and bandgap
• Efficiency: High light output with low power

Applications:

• Indicators: Status lights, displays
• Lighting: LED bulbs, strips
• Electronics: Seven-segment displays

Mnemonic: “Light Emitting, Energy Efficient”

Question 3(c) [7 marks]

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Explain working characteristics of Zener diode.

Answer:

V-I Characteristics:

Key Regions:

Important Parameters:

• Zener Voltage (Vz): Breakdown voltage
• Zener Current (Iz): Current in breakdown region
• Maximum Power: Vz × Iz(max)
• Temperature coefficient: Voltage variation with temperature

Applications:

• Voltage regulation: Maintains constant output
• Reference voltage: Precise voltage source
• Overvoltage protection: Protects circuits

Advantages:

• Sharp breakdown: Well-defined voltage
• Low dynamic resistance: Good regulation
• Wide range: Available in many voltages

Mnemonic: “Zener Zones Zero variation”

Question 3(a OR) [3 marks]

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Enlist the applications of varactor diode.

Answer:

Applications Table:

Key Features:

• Voltage variable: Capacitance changes with reverse voltage
• No mechanical parts: Electronic tuning only
• Fast response: Quick frequency changes

Mnemonic: “Voltage Varies Capacitance”

Question 3(b OR) [4 marks]

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Explain working of photo diode.

Answer:

Construction & Symbol:

Working Principle:

• Light absorption: Creates electron-hole pairs
• Reverse bias: Widens depletion region
• Photocurrent: Proportional to light intensity
• Fast response: Quick detection capability

Characteristics:

Applications:

• Light sensors: Automatic lighting systems
• Optical communication: Fiber optic receivers
• Safety systems: Smoke detectors
• Solar panels: Light to electrical energy

Mnemonic: “Photo Produces Proportional current”

Question 3(c OR) [7 marks]

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Explain Zener diode as a voltage regulator.

Answer:

Voltage Regulator Circuit:

Working Principle:

• Zener operates in breakdown region
• Output voltage remains constant at Vz
• Series resistor Rs limits current
• Load changes don’t affect output voltage

Design Equations:

Regulation Characteristics:

• Line regulation: Output change with input variation
• Load regulation: Output change with load variation
• Efficiency: Generally low due to Zener power loss

Advantages:

• Simple circuit: Few components required
• Good regulation: Stable output voltage
• Fast response: Quick voltage correction

Limitations:

• Poor efficiency: Power wasted in Zener
• Limited current: Cannot supply high currents
• Temperature sensitivity: Voltage varies with temperature

Applications:

• Reference voltage: Precise voltage source
• Simple regulators: Low current applications
• Protection circuits: Overvoltage protection

Mnemonic: “Zener Zones provide Zero variation”

Question 4(a) [3 marks]

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Draw the symbol and construction of PNP and NPN transistor with proper notation.

Answer:

Transistor Symbols:

Construction Diagrams:

Terminal Identification:

• Emitter: Heavily doped, arrow shows current direction
• Base: Thin, lightly doped middle region
• Collector: Moderately doped, collects charge carriers

Current Direction:

• NPN: Arrow points outward (emitter to base)
• PNP: Arrow points inward (base to emitter)

Mnemonic: “NPN: Not Pointing iN, PNP: Pointing iN Please”

Question 4(b) [4 marks]

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Draw and Explain characteristics of CE amplifier.

Answer:

CE Amplifier Circuit:

Input Characteristics (IB vs VBE):

Output Characteristics (IC vs VCE):

Key Features:

Regions of Operation:

• Cut-off: Both junctions reverse biased
• Active: BE forward, BC reverse biased
• Saturation: Both junctions forward biased

Mnemonic: “Common Emitter, Current Enlarged”

Question 4(c) [7 marks]

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Derive relation between current gains α, β and γ.

Answer:

Current Gain Definitions:

Derivation:

Step 1: Basic Current Relation IE = IB + IC … (Kirchhoff’s Current Law)

Step 2: Express IC in terms of IE α = IC/IE Therefore: IC = α × IE … (1)

Step 3: Substitute in current equation IE = IB + α × IE IE - α × IE = IB IE(1 - α) = IB IE = IB/(1 - α) … (2)

Step 4: Find β β = IC/IB From (1): IC = α × IE From (2): IE = IB/(1 - α) Therefore: IC = α × IB/(1 - α)

Step 5: Final relation for β β = IC/IB = α/(1 - α) … (3)

Step 6: Express α in terms of β From equation (3): β(1 - α) = α β - βα = α β = α + βα = α(1 + β) Therefore: α = β/(1 + β) … (4)

Step 7: Find γ γ = IE/IB From (2): γ = 1/(1 - α) Substituting α from (4): γ = 1/(1 - β/(1 + β)) γ = (1 + β)/(1 + β - β) γ = 1 + β … (5)

Final Relations:

Typical Values:

• α ≈ 0.98 to 0.995
• β ≈ 50 to 200
• γ ≈ 51 to 201

Mnemonic: “Alpha Beta Gamma, Always Better Gains”

Question 4(a OR) [3 marks]

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Define Active, Saturation and Cut-off region for transistor amplifier.

Answer:

Operating Regions:

Detailed Description:

Active Region:

• Normal amplification mode
• IC = β × IB relationship holds
• Linear operation for small signals

Saturation Region:

• Both junctions forward biased
• Maximum collector current flows
• VCE ≈ 0.2V (very low)
• Used in switching applications

Cut-off Region:

• No base current (IB = 0)
• No collector current (IC = 0)
• Transistor acts like open switch

Mnemonic: “Active Amplifies, Saturated Switches, Cut-off Cuts”

Question 4(b OR) [4 marks]

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Explain working of Transistor as an amplifier.

Answer:

Amplifier Circuit:

Working Principle:

• Small input signal applied to base-emitter
• Input resistance is low (few kΩ)
• Small base current controls large collector current
• Output taken from collector-emitter
• Current amplification: IC = β × IB

Amplification Process:

Key Features:

• Current gain: β (50-200 typical)
• Voltage gain: Depends on load resistance
• Power gain: Product of current and voltage gains
• Phase inversion: 180° in CE configuration

Applications:

• Audio amplifiers: Music systems
• RF amplifiers: Radio transmitters
• Op-amp stages: Integrated circuits

Mnemonic: “Tiny signal Triggers Tremendous output”

Question 4(c OR) [7 marks]

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Compare CB, CC, and CE amplifier configuration.

Answer:

Comprehensive Comparison:

Circuit Diagrams:

Common Base:

Key Characteristics:

Common Base (CB):

• High frequency performance
• No current gain but high voltage gain
• Input-output isolation excellent
• Used in: RF amplifiers, high-frequency circuits

Common Emitter (CE):

• Most popular configuration
• High current and voltage gain
• Good compromise of all parameters
• Used in: Audio amplifiers, general amplification

Common Collector (CC):

• Unity voltage gain (voltage follower)
• High current gain
• Impedance transformation (high to low)
• Used in: Buffer amplifiers, impedance matching

Selection Criteria:

Mnemonic: “CB for Communication, CE for Common use, CC for Coupling”

Question 5(a) [3 marks]

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Draw the pin diagram of IC 555.

Answer:

IC 555 Pin Diagram:

Pin Functions:

Key Points:

• Dual-in-line 8-pin package
• Power supply: 5V to 18V DC
• Output current: Up to 200mA
• Reset pin: Normally connected to Vcc

Mnemonic: “Great Timer, Great Pins”

Question 5(b) [4 marks]

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List out Features of 555 Timer IC.

Answer:

Key Features:

Technical Features:

• CMOS/TTL compatible output levels
• High current output capability
• Wide supply voltage range
• Temperature stable operation

Functional Features:

• Three operating modes available
• External timing components
• Reset capability for control
• Low power consumption design

Advantages:

• Versatile timer for multiple applications
• Easy to use with minimal external components
• Reliable operation in various conditions

Mnemonic: “Fantastic Features, Flexible Functions”

Question 5(c) [7 marks]

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Explain Mono stable multivibrator using 555 timer IC.

Answer:

Monostable Circuit:

Working Principle:

Stable State:

• Output LOW (approximately 0V)
• Capacitor discharged through pin 7
• Threshold voltage below Vcc/3

Triggered State:

• Negative pulse applied to trigger (pin 2)
• Output goes HIGH immediately
• Discharge transistor turns OFF
• Capacitor starts charging through R

Timing Period:

• Duration: T = 1.1 × R × C
• Output remains HIGH for calculated time
• Automatic return to stable state

Return to Stable:

• Capacitor voltage reaches 2Vcc/3
• Threshold triggered (pin 6)
• Output returns to LOW
• Discharge begins again

Key Characteristics:

Applications:

• Pulse generation: Fixed width pulses
• Time delays: Switch-on delays
• Missing pulse detection: Watchdog timers
• Debouncing circuits: Switch contact cleaning

Design Example: For T = 1ms: If C = 0.1µF, then R = 9.1kΩ

Mnemonic: “Mono means One pulse Only”

Question 5(a OR) [3 marks]

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List out applications of IC 555.

Answer:

Timer Applications:

Mode-wise Applications:

Monostable Mode:

• Time delays in circuits
• Pulse width generation
• Debouncing switches

Astable Mode:

• LED flashers and blinkers
• Clock signals generation
• Tone generation for buzzers

Bistable Mode:

• Flip-flop circuits
• Memory elements
• Latch circuits

Common Projects:

• Electronic dice using LEDs
• Traffic light controllers
• Digital clocks and timers

Mnemonic: “Timer for Tremendous Tasks”

Question 5(b OR) [4 marks]

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Draw and explain the internal block diagram of IC 555.

Answer:

Internal Block Diagram:

Block Functions:

Working:

• Comparators set and reset flip-flop
• Output buffer amplifies flip-flop output
• Discharge transistor controlled by flip-flop
• Reference voltages set trigger levels

Mnemonic: “Internal Intelligence, Integrated Implementation”

Question 5(c OR) [7 marks]

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Explain astable multivibrator using 555 timer IC.

Answer:

Astable Circuit:

Working Principle:

Charging Phase:

• Capacitor charges through R1 + R2
• Output HIGH during charging
• Charging time: T1 = 0.693(R1 + R2)C
• Voltage rises from Vcc/3 to 2Vcc/3

Discharging Phase:

• Capacitor discharges through R2 only
• Output LOW during discharging
• Discharging time: T2 = 0.693 × R2 × C
• Voltage falls from 2Vcc/3 to Vcc/3

Frequency Calculations:

Waveforms:

Design Example: For f = 1kHz, D = 60%:

• Choose C = 0.1µF
• Calculate R1 = 7.2kΩ, R2 = 3.6kΩ

Key Features:

• Continuous oscillation without external trigger
• Frequency adjustable by R and C values
• Duty cycle always > 50% in basic circuit
• Stable operation over wide temperature range

Applications:

• LED flashers and blinkers
• Clock generators for digital circuits
• Tone generators for alarms
• PWM signal generation

Modifications for 50% Duty Cycle:

• Add diode in parallel with R2
• Separate paths for charge and discharge
• Equal charge/discharge times possible

Mnemonic: “Astable Always Alternates Automatically”