Fundamentals of Electronics (4311102) - Winter 2024 Solution

Question 1(a) [3 marks]

#

Give the difference between Passive components and Active components

Answer:

Mnemonic: “PAPER-A” - Passive Are Power-free, Energy-storing/Resistive; Active Are Amplifying

Question 1(b) [4 marks]

#

Explain Working of Light dependent resistor with neat diagram.

Answer:

graph LR
    A[Light] --> B[LDR]
    B --> C[Change in Resistance]
    style A fill:#lightblue
    style B fill:#lightgreen
    style C fill:#lightpink

Working of LDR:

• Construction: LDR consists of a semiconductor material (typically cadmium sulfide) with high resistance in darkness
• Photoconductivity: When light falls on the surface, photons transfer energy to electrons, creating free electron-hole pairs
• Resistance variation: Resistance decreases dramatically as light intensity increases - from megaohms in darkness to few hundred ohms in bright light
• Applications: Used in light sensing circuits, automatic street lights, camera exposure control

Mnemonic: “MILD” - More Illumination, Less resistance in Devices

Question 1(c) [7 marks]

#

Define Intrinsic and Extrinsic Semiconductor. Explain P type and N type semiconductors in detail.

Answer:

P-type Semiconductor:

• Doping: Created by adding trivalent impurities (boron, gallium, indium) to pure silicon
• Hole creation: Each impurity atom creates a hole by accepting valence electrons
• Majority carriers: Holes are majority carriers
• Minority carriers: Electrons are minority carriers
• Electrical properties: Positive charge carriers dominate conduction

N-type Semiconductor:

• Doping: Created by adding pentavalent impurities (phosphorus, arsenic, antimony) to pure silicon
• Electron creation: Each impurity atom donates an extra electron
• Majority carriers: Electrons are majority carriers
• Minority carriers: Holes are minority carriers
• Electrical properties: Negative charge carriers dominate conduction

Diagram:

Mnemonic: “PINE” - Positive Impurities make N-type Electrons, Pentavalent donors

Question 1(c) OR [7 marks]

#

What is filter circuit? Give type and necessity of Filter and Explain “PI” Filter circuit in brief.

Answer:

Filter Circuit: Electronic circuit that removes unwanted frequency components from a signal, allowing desired frequencies to pass through.

Necessity of Filters:

• Ripple reduction: Reduces AC ripple from rectifier output
• Clean DC: Provides smoother DC output voltage
• Component protection: Protects downstream components from voltage fluctuations
• Efficiency: Improves overall power supply efficiency

Types of Filters:

Pi (π) Filter:

graph LR
    A[Input] --> B[Capacitor C1]
    B --> C[Inductor L]
    C --> D[Capacitor C2]
    D --> E[Output]
    style A fill:#lightblue
    style B fill:#lightgreen
    style C fill:#lightpink
    style D fill:#lightgreen
    style E fill:#lightblue

• Working: First capacitor (C1) reduces initial ripple, inductor (L) blocks AC components, second capacitor (C2) filters remaining ripples
• Advantage: Provides superior filtering with ripple factor typically below 0.5%
• Applications: Used in high-current power supplies where clean DC is critical

Mnemonic: “PIRO” - Pi filters Input Ripples Out effectively

Question 2(a) [3 marks]

#

Write down different types of capacitors and explain any two.

Answer:

Types of Capacitors:

• Ceramic capacitors
• Electrolytic capacitors
• Tantalum capacitors
• Film capacitors
• Mica capacitors
• Variable capacitors

Ceramic Capacitors:

• Construction: Made from ceramic material as dielectric between metal plates
• Capacity: 1pF to 1μF
• Advantages: Low cost, high stability, non-polarized
• Applications: High-frequency filtering, coupling/decoupling

Electrolytic Capacitors:

• Construction: Aluminum foil with oxide layer as dielectric
• Capacity: 1μF to 10,000μF
• Characteristics: Polarized, higher leakage current
• Applications: Power supply filtering, audio coupling

Mnemonic: “CAPEX” - Ceramics Are Precise, Electrolytics Expand capacity

Question 2(b) [4 marks]

#

Explain air core and toroidal inductor.

Answer:

Air Core Inductor:

• Construction: Wire coiled around non-magnetic material (plastic, air)
• Properties: Lower inductance, no magnetic core saturation
• Applications: High-frequency circuits, RF applications
• Advantages: No core losses, linear operation, no saturation

Toroidal Inductor:

• Construction: Wire wound around a ring-shaped magnetic core
• Properties: Higher inductance, self-shielding magnetic field
• Applications: Power supplies, filters, transformers
• Advantages: Low electromagnetic interference, efficient flux containment

Mnemonic: “TACO” - Toroids Are Contained, Omnidirectional field reduction

Question 2(c) [7 marks]

#

Explain Half wave rectifier and Compare different rectifier circuits.

Answer:

Half Wave Rectifier:

graph LR
    A[AC Input] --> B[Transformer]
    B --> C[Diode]
    C --> D[Load]
    C --> E[Ground]
    style A fill:#lightblue
    style B fill:#lightpink
    style C fill:#lightyellow
    style D fill:#lightgreen
    style E fill:#lightgray

Working Principle:

• During positive half-cycle: Diode conducts, current flows through load
• During negative half-cycle: Diode blocks, no current flows
• Output contains only positive half-cycles of input waveform

Comparison of Rectifiers:

Mnemonic: “BRIEF” - Bridge Rectifiers Improve Efficiency Fundamentally

Question 2(a) OR [3 marks]

#

Write down different capacitor specifications and explain any two in detail.

Answer:

Capacitor Specifications:

• Capacitance value
• Voltage rating
• Tolerance
• Temperature coefficient
• ESR (Equivalent Series Resistance)
• Leakage current
• Dielectric type

Capacitance Value:

• Definition: Amount of electric charge stored per volt
• Units: Measured in farads (F), typically microfarads (μF), nanofarads (nF), or picofarads (pF)
• Importance: Determines application suitability for coupling, filtering, timing
• Marking: Directly printed or color-coded on component

Voltage Rating:

• Definition: Maximum voltage that can be applied without breakdown
• Specification: Working voltage (WVDC) and surge voltage
• Importance: Exceeding rating causes dielectric breakdown and failure
• Safety factor: Typically use capacitors rated 50% higher than circuit voltage

Mnemonic: “CAVERN” - Capacitance And Voltage Ensure Reliable Network

Question 2(b) OR [4 marks]

#

Explain classification of Resistor based on materials.

Answer:

Characteristics of Carbon Film Resistors:

• Temperature coefficient: -250 to 500 ppm/°C
• Tolerance: 5% to 10%
• Noise: Moderate to low

Characteristics of Metal Film Resistors:

• Temperature coefficient: 50 to 100 ppm/°C
• Tolerance: 0.1% to 2%
• Noise: Very low

Mnemonic: “COMFORT” - Carbon Offers Moderate Films, Others Resist Temperature better

Question 2(c) OR [7 marks]

#

Explain full wave bridge and center tapped rectifier with diagram and waveform.

Answer:

Full Wave Bridge Rectifier:

graph LR
    A[AC Input] --> B[Transformer]
    B --> C[Bridge
Rectifier]
    C --> D[D1]
    C --> E[D2]
    C --> F[D3]
    C --> G[D4]
    D & E & F & G --> H[Load]
    H --> I[Ground]
    style A fill:#lightblue
    style B fill:#lightpink
    style C fill:#lightyellow
    style H fill:#lightgreen
    style I fill:#lightgray

Working:

• Positive half-cycle: D1 and D3 conduct, current flows through load
• Negative half-cycle: D2 and D4 conduct, current still flows through load in same direction
• Output: Both half-cycles of input converted to positive output

Center Tapped Full Wave Rectifier:

graph LR
    A[AC Input] --> B[Center-Tapped
Transformer]
    B -->|Upper Half| C[D1]
    B -->|Lower Half| D[D2]
    C & D --> E[Load]
    E --> F[Ground]
    F --> B
    style A fill:#lightblue
    style B fill:#lightpink
    style C fill:#lightyellow
    style D fill:#lightyellow
    style E fill:#lightgreen
    style F fill:#lightgray

Working:

• Positive half-cycle: D1 conducts, D2 blocks
• Negative half-cycle: D2 conducts, D1 blocks
• Output: Both half-cycles of input converted to positive output

Waveforms:

Mnemonic: “FOUR-TWO” - FOUr diodes for Bridge, TWO diodes for Center-Tap

Question 3(a) [3 marks]

#

Explain the characteristic of Varactor diode.

Answer:

Varactor Diode Characteristics:

graph LR
    A[Reverse Bias
Voltage] --> B[Depletion
Layer Width]
    B --> C[Junction
Capacitance]
    C --> D[Frequency
Tuning]
    style A fill:#lightblue
    style B fill:#lightpink
    style C fill:#lightgreen
    style D fill:#lightyellow

• Operating principle: Junction capacitance varies with reverse bias voltage
• C-V relationship: Capacitance decreases as reverse voltage increases
• Tuning ratio: Typically 4:1 to 10:1 capacitance variation
• Applications: Voltage-controlled oscillators, FM modulation, tuning circuits

Mnemonic: “VARA” - Voltage Adjusts Reverse-biased capacitance Automatically

Question 3(b) [3 marks]

#

State and explain Faraday’s laws of electromagnetic induction.

Answer:

Faraday’s Laws of Electromagnetic Induction:

First Law:

• Statement: Whenever a conductor cuts magnetic flux, an EMF is induced in the conductor
• Mathematical expression: EMF ∝ Rate of change of magnetic flux
• Application: Basis for generators, transformers, inductors

Second Law:

• Statement: The magnitude of induced EMF equals the rate of change of magnetic flux linkage
• Mathematical expression: EMF = -N × (dΦ/dt)
  • Where: N = number of turns, dΦ/dt = rate of change of flux
• Negative sign: Indicates direction (Lenz’s Law) - induced current opposes the change

Diagram:

Mnemonic: “FACE” - Flux Alteration Creates Electricity

Question 3(c) [7 marks]

#

Compare different Transistor Configurations.

Answer:

Relationship between α, β and γ:

• β = α/(1-α)
• α = β/(1+β)
• γ = β+1

Mnemonic: “BEC” - Base input for Emitter output needs Collector as common terminal

Question 3(a) OR [3 marks]

#

What is forbidden energy gap? Draw the energy band diagram for insulator, conductor and semiconductor.

Answer:

Forbidden Energy Gap: Energy range in a solid where no electron states exist, separating the valence band from the conduction band.

Energy Band Diagrams:

• Insulator: Large forbidden gap (>5eV) prevents electrons from reaching conduction band
• Conductor: Overlapping bands allow free electron movement
• Semiconductor: Small gap (~1eV) allows some electrons to cross at room temperature or when excited

Mnemonic: “IBCS” - Insulators Block, Conductors Share, Semiconductors have gap Between

Question 3(b) OR [4 marks]

#

Explain the function of Zener diode as a voltage regulator

Answer:

graph LR
    A[Unregulated
DC Input] --> B[Series
Resistor]
    B --> C[Load]
    B --> D[Zener
Diode]
    D --> E[Ground]
    style A fill:#lightblue
    style B fill:#lightpink
    style C fill:#lightgreen
    style D fill:#lightyellow
    style E fill:#lightgray

Working Principle:

• Normal operation: Zener diode is reverse biased and conducts when voltage reaches breakdown voltage
• Voltage regulation: When input voltage rises, more current flows through Zener diode, maintaining constant voltage across it
• Load variation: When load draws more current, less current flows through Zener, keeping voltage stable
• Series resistor: Limits current and drops excess voltage

Circuit behavior:

• V_out = V_z (Zener breakdown voltage)
• I_z = (V_in - V_z)/R - I_L

Mnemonic: “SERZ” - Series resistor Enables Regulation with Zener

Question 3(c) OR [7 marks]

#

Explain V-I char of P-N junction diode and give comparison between P-N junction diode and Zener diode.

Answer:

V-I Characteristics of P-N Junction Diode:

Key Points:

• Forward bias: Conducts easily after exceeding knee voltage (~0.7V for silicon)
• Reverse bias: Very small leakage current until breakdown voltage
• Breakdown region: Occurs at high reverse voltage, causes damage in normal diodes

P-N Junction Diode vs. Zener Diode:

Mnemonic: “FORD” - Forward Operation for Rectifiers, Diodes; reverse operation for Zeners

Question 4(a) [3 marks]

#

Describe working principle of Photodiode.

Answer:

Working Principle of Photodiode:

graph LR
    A[Light] --> B[P-N Junction]
    B --> C[Electron-Hole
Pairs]
    C --> D[Photocurrent]
    style A fill:#lightyellow
    style B fill:#lightpink
    style C fill:#lightblue
    style D fill:#lightgreen

• Construction: P-N junction diode with transparent window or lens
• Operation: Reverse biased operation for light detection
• Photon absorption: Incoming photons create electron-hole pairs in depletion region
• Current generation: Electric field sweeps carriers to respective terminals, creating photocurrent
• Light sensitivity: Current proportional to light intensity

Mnemonic: “LIGER” - Light Induces Generation of Electrons in Reverse-bias

Question 4(b) [4 marks]

#

Explain the characteristic of Schottky barrier diode.

Answer:

Schottky Barrier Diode Characteristics:

• Low forward voltage drop: 0.2-0.3V compared to 0.7V for silicon PN junction
• Fast switching: No minority carrier storage, minimal reverse recovery time
• Construction: Metal-semiconductor junction instead of P-N junction
• No reverse recovery time: Majority carrier device (no stored charge)
• Applications: High-frequency applications, rectifiers in power supplies

Mnemonic: “FAST” - Forward voltage low, Allows Switching Timely

Question 4(c) [7 marks]

#

Explain working principle of PNP and NPN transistor.

Answer:

NPN Transistor Structure and Working:

• Biasing: Emitter-base junction forward biased, collector-base junction reverse biased
• Current flow: Electrons from emitter to collector through thin base region
• Amplification principle: Small base current controls larger collector current
• Current relationship: I_E = I_B + I_C
• Majority carriers: Electrons

PNP Transistor Structure and Working:

• Biasing: Emitter-base junction forward biased, collector-base junction reverse biased
• Current flow: Holes from emitter to collector through thin base region
• Amplification principle: Small base current controls larger collector current
• Current relationship: I_E = I_B + I_C
• Majority carriers: Holes
• Current direction: Opposite to NPN (conventional current from emitter to collector)

Mnemonic: “NPNP” - Negative carriers in NPN, Positive carriers in PNP

Question 4(a) OR [3 marks]

#

Describe working principle of LED.

Answer:

Working Principle of LED:

graph LR
    A[Forward Bias] --> B[Electron-Hole
Recombination]
    B --> C[Energy Release
as Photons]
    C --> D[Light Emission]
    style A fill:#lightblue
    style B fill:#lightpink
    style C fill:#lightyellow
    style D fill:#lightgreen

• Construction: P-N junction made from direct bandgap semiconductor materials
• Forward biasing: Electrons from n-region and holes from p-region recombine at junction
• Recombination: Electrons fall from conduction band to valence band
• Energy emission: Energy released during recombination emits photons (light)
• Color determination: Bandgap energy determines wavelength (color) of emitted light

Mnemonic: “REBEL” - Recombination of Electrons and holes By Energetic Light emission

Question 4(b) OR [4 marks]

#

Explain function of transistor as switch in cut off and application of saturation region.

Answer:

Transistor as a Switch:

Cut-off Region (Switch OFF):

• Base voltage: Below 0.7V (for silicon)
• Base current: Approximately zero
• Collector current: Approximately zero
• Collector-emitter voltage: Equal to supply voltage
• Applications: Logic gates, digital circuits, relay drivers

Saturation Region (Switch ON):

• Base voltage: Well above 0.7V
• Base current: Sufficient to ensure minimum V_CE
• Collector current: Maximum (limited by collector resistor)
• Collector-emitter voltage: Very low (0.2V - 0.3V)
• Applications: Digital switches, motor drivers, LED drivers

Mnemonic: “COSI” - Cutoff Opens Switch, Input saturates to close

Question 4(c) OR [7 marks]

#

Explain common emitter (CE) configuration of Transistor. Derive relation between α and β for transistor amplifier.

Answer:

Common Emitter Configuration:

graph TB
    A[Input Signal] --> B[Base]
    C[Output Signal] --> D[Collector]
    E[Ground] --> F[Emitter]
    style A fill:#lightblue
    style B fill:#lightpink
    style C fill:#lightgreen
    style D fill:#lightyellow
    style E fill:#lightgray
    style F fill:#lightcyan

Characteristics of Common Emitter Configuration:

• Input terminal: Base
• Output terminal: Collector
• Common terminal: Emitter (grounded)
• Current gain (β): High (20-500)
• Voltage gain: High (250-1000)
• Input impedance: Medium (1-2kΩ)
• Output impedance: High (30-50kΩ)
• Phase shift: 180° (output inverted from input)

Relationship between α and β:

By definition:

• α = I_C/I_E (Common Base current gain)
• β = I_C/I_B (Common Emitter current gain)

From Kirchhoff’s Current Law:

• I_E = I_B + I_C

Dividing both sides by I_E:

• 1 = I_B/I_E + I_C/I_E
• 1 = I_B/I_E + α

Therefore:

• I_B/I_E = 1 - α

Now, β = I_C/I_B = (I_C/I_E)/(I_B/I_E) = α/(1-α)

And conversely:

• α = β/(1+β)

Mnemonic: “BEAR” - Beta Equals Alpha divided by (1-alpha) Relation

Question 5(a) [3 marks]

#

What do you mean by E-waste? What are the different methods of E-waste disposal?

Answer:

E-waste (Electronic Waste): Discarded electronic devices and components that have reached end of life or are no longer useful.

Methods of E-waste Disposal:

Mnemonic: “RIPER” - Recycling Is Preferable to Environmentally-harmful Remedies

Question 5(b) [4 marks]

#

Explain methods of handling electronic waste with examples.

Answer:

Methods of Handling Electronic Waste:

graph TD
    A[E-waste
Collection] --> B[Sorting]
    B --> C[Dismantling]
    C --> D[Material
Recovery]
    D --> E[Safe
Disposal]
    style A fill:#lightblue
    style B fill:#lightpink
    style C fill:#lightyellow
    style D fill:#lightgreen
    style E fill:#lightgray

Collection and Segregation:

• Example: Dedicated e-waste bins in public places, e-waste collection drives
• Benefit: Prevents mixing with general waste, enables proper processing

Dismantling and Resource Recovery:

• Example: Recovering gold, silver, copper from circuit boards and connectors
• Benefit: Recovers valuable metals, reduces mining demands

Refurbishment and Reuse:

• Example: Repairing old computers for educational institutions
• Benefit: Extends product lifecycle, reduces waste generation

Proper Disposal of Hazardous Components:

• Example: Specialized treatment for mercury-containing components
• Benefit: Prevents toxic substances from entering environment

Mnemonic: “CREED” - Collect, Recover, Extract, Extend, Dispose safely

Question 5(c) [7 marks]

#

What is ripple factor? Derive the equation of the ripple factor for rectifier.

Answer:

Ripple Factor: Measure of effectiveness of a rectifier’s filtering - the ratio of AC component (ripple) to DC component in the output.

Definition:

• Ripple factor (γ) = RMS value of AC component / DC value
• Lower ripple factor indicates better filtering

Derivation for Half Wave Rectifier:

Let’s assume sinusoidal input: v = V_msinωt

For half wave rectifier:

• Output is v = V_msinωt for 0 ≤ ωt ≤ π
• Output is v = 0 for π ≤ ωt ≤ 2π

Step 1: Find DC component (average value)

• V_DC = (1/2π) ∫₀^2π v(ωt) d(ωt)
• V_DC = (1/2π) ∫₀^π V_msinωt d(ωt)
• V_DC = V_m/π

Step 2: Find RMS value

• V_RMS = √[(1/2π) ∫₀^2π v²(ωt) d(ωt)]
• V_RMS = √[(1/2π) ∫₀^π V_m²sin²ωt d(ωt)]
• V_RMS = V_m/2

Step 3: Find AC component

• V_AC² = V_RMS² - V_DC²
• V_AC² = (V_m/2)² - (V_m/π)²
• V_AC² = V_m²(1/4 - 1/π²)

Step 4: Calculate ripple factor

• γ = V_AC/V_DC
• γ = √(V_m²(1/4 - 1/π²))/(V_m/π)
• γ = π√(1/4 - 1/π²)
• γ = 1.21 (for half wave rectifier)

For Full Wave Rectifier: Following similar steps leads to γ = 0.48

Mnemonic: “ROAD” - Ripple is Output’s AC Divided by DC component

Question 5(a) OR [3 marks]

#

Which are the toxic substances present in e-waste?

Answer:

Toxic Substances in E-waste:

Mnemonic: “LMBCHB” - Lead, Mercury, and Beryllium Cause Harmful Bodily effects

Question 5(b) OR [4 marks]

#

Write important parameters for selecting the right transistor for your application and explain any two.

Answer:

Important Transistor Selection Parameters:

• Maximum collector current (I_C)
• Maximum collector-emitter voltage (V_CEO)
• Maximum collector-base voltage (V_CBO)
• Current gain (hFE or β)
• Frequency response (f_T)
• Power dissipation (P_tot)
• Package type (TO-3, SMT, etc.)
• Temperature range

Maximum Collector Current (I_C):

• Definition: Maximum current that can flow through collector without damage
• Importance: Must exceed application’s peak current requirements with safety margin
• Typical values: 100mA to 100A depending on transistor type
• Application consideration: Select 50% higher rating than maximum required current

Current Gain (hFE or β):

• Definition: Ratio of collector current to base current
• Importance: Determines amplification capability and required base drive
• Typical values: 20-500 for general-purpose transistors
• Application consideration: For switching, high gain reduces base current requirement; for amplifiers, consistent gain across operating range is important

Mnemonic: “GIVE” - Gain and Ic are Very Essential parameters

Question 5(c) OR [7 marks]

#

What is rectifier efficiency? Find out efficiency of the full wave rectifier.

Answer:

Rectifier Efficiency: The ratio of DC output power to the AC input power, expressed as a percentage.

Definition:

• Efficiency (η) = (P_DC/P_AC) × 100%
• Higher efficiency means better conversion of AC to DC power

Derivation for Full Wave Rectifier:

Step 1: Calculate DC output power

• I_DC = V_DC/R_L
• P_DC = I_DC² × R_L = V_DC²/R_L
• For full wave, V_DC = 2V_m/π
• P_DC = (2V_m/π)²/R_L = 4V_m²/(π²R_L)

Step 2: Calculate AC input power

• I_RMS = V_RMS/R_L
• P_AC = I_RMS² × R_L = V_RMS²/R_L
• For sine wave, V_RMS = V_m/√2
• P_AC = (V_m/√2)²/R_L = V_m²/(2R_L)

Step 3: Calculate efficiency

• η = (P_DC/P_AC) × 100%
• η = [4V_m²/(π²R_L)] / [V_m²/(2R_L)] × 100%
• η = [4/(π²)] × 2 × 100%
• η = 8/(π²) × 100%
• η = 8/9.87 × 100%
• η = 81.2%

Full Wave Rectifier Efficiency = 81.2%

For comparison:

• Half Wave Rectifier Efficiency = 40.6%
• Bridge Rectifier Efficiency = 81.2%

Mnemonic: “PIDE” - Power Input Determines Efficiency