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Elements of Electrical & Electronics Engineering (1313202) - Summer 2024 Solution

23 mins· ·
Study-Material Solutions Electrical-Engineering Electronics-Engineering 1313202 2024 Summer
Milav Dabgar
Author
Milav Dabgar
Experienced lecturer in the electrical and electronic manufacturing industry. Skilled in Embedded Systems, Image Processing, Data Science, MATLAB, Python, STM32. Strong education professional with a Master’s degree in Communication Systems Engineering from L.D. College of Engineering - Ahmedabad.
Table of Contents

Question 1(a) [3 marks]
#

Define: 1. Node, 2. Loop, 3. Branch

Answer:

TermDefinition
NodeA point in a circuit where two or more circuit elements meet or connect
LoopA closed path in a circuit that starts and ends at the same point without passing through any node more than once
BranchA path or element connecting two nodes in a circuit

Mnemonic: “Never Loop Between” - Nodes Link, Loops Bound, Branches Establish connections

Question 1(b) [4 marks]
#

Write statement of Superposition theorem and Maximum power transfer theorem.

Answer:

TheoremStatement
Superposition TheoremIn a linear circuit with multiple sources, the response (voltage or current) in any element equals the algebraic sum of responses caused by each source acting alone, with all other sources replaced by their internal impedances
Maximum Power Transfer TheoremMaximum power is transferred from source to load when the load resistance equals the source’s internal resistance

Diagram:

graph LR
    A[Sources] --> B[Individual Responses]
    B --> C[Sum = Total Response]
    D[Source Rs] --- E[Load RL]
    F[Max Power when Rs = RL]

Mnemonic: “Sum Powers Matched” - Sum individual powers; Match resistance for maximum

Question 1(c) [7 marks]
#

Explain Kirchhoff’s Voltage Law and Kirchhoff’s current Law.

Answer:

LawExplanationMathematical Form
Kirchhoff’s Voltage Law (KVL)The algebraic sum of all voltages around any closed loop in a circuit equals zeroΣ V = 0
Kirchhoff’s Current Law (KCL)The algebraic sum of all currents entering and leaving a node equals zeroΣ I = 0

Diagram:

graph TD
    subgraph "KVL: V1 + V2 + V3 = 0"
        A1[+] --> B1[V1] --> C1[-]
        C1 --> D1[+] --> E1[V2] --> F1[-]
        F1 --> G1[+] --> H1[V3] --> I1[-]
        I1 --> A1
    end
    
    subgraph "KCL: I1 + I2 = I3 + I4"
        A2((Node)) --- I1[I1]
        A2 --- I2[I2]
        A2 --- I3[I3]
        A2 --- I4[I4]
    end

  • Physical interpretation of KVL: Energy is conserved in a circuit loop
  • Physical interpretation of KCL: Charge is conserved at circuit nodes
  • Application of KVL: Finding unknown voltages in circuit loops
  • Application of KCL: Finding unknown currents at circuit junctions

Mnemonic: “Voltages Loop to Zero, Currents Node to Zero”

Question 1(c) OR [7 marks]
#

Explain series and parallel connection of resistors with necessary equations.

Answer:

ConnectionCharacteristicsEquivalent ResistanceCurrent-Voltage Relationship
Series ConnectionSame current flows through all resistorsReq = R1 + R2 + R3 + … + RnI = V/Req
Parallel ConnectionSame voltage appears across all resistors1/Req = 1/R1 + 1/R2 + 1/R3 + … + 1/RnI = I1 + I2 + I3 + … + In

Diagram:

graph LR
    subgraph "Series"
        A1[+] --- R1[R1] --- R2[R2] --- R3[R3] --- B1[-]
    end
    
    subgraph "Parallel"
        A2[+] --- R4[R1]
        A2 --- R5[R2]
        A2 --- R6[R3]
        R4 --- B2[-]
        R5 --- B2
        R6 --- B2
    end

  • Current in series: I = I1 = I2 = I3 = … = In
  • Voltage in series: V = V1 + V2 + V3 + … + Vn
  • Current in parallel: I = I1 + I2 + I3 + … + In
  • Voltage in parallel: V = V1 = V2 = V3 = … = Vn

Mnemonic: “Same Current Series, Same Voltage Parallel”

Question 2(a) [3 marks]
#

State limitations of Ohm’s law.

Answer:

Limitations of Ohm’s Law
Non-linear components: Does not apply to components like diodes, transistors
Temperature changes: Not valid when temperature varies significantly
High frequencies: Breaks down at very high frequencies

Mnemonic: “Ohm’s Not Linear Thermal High” - Non-linear, Temperature, High frequency

Question 2(b) [4 marks]
#

Define: 1. Doping, 2. Intrinsic Semiconductor, 3. Extrinsic Semiconductor, 4. Dopant

Answer:

TermDefinition
DopingProcess of adding impurity atoms to pure semiconductor to modify electrical properties
Intrinsic SemiconductorPure semiconductor with equal number of electrons and holes
Extrinsic SemiconductorDoped semiconductor with unequal number of electrons and holes
DopantImpurity element added to semiconductor during doping process

Mnemonic: “Do In-Ex-Do” - Doping Introduces Extrinsic properties through Dopants

Question 2(c) [7 marks]
#

Define Trivalent material and give examples of it. Explain Formation of P-type Semiconductor with the help of proper diagram.

Answer:

Trivalent material: Elements with 3 valence electrons in their outermost shell.

Examples: Boron (B), Aluminum (Al), Gallium (Ga), Indium (In)

P-type Semiconductor Formation:

Diagram:

SilicoSniatom(4valencee-)TrhiovlaelBentatom(3valencee-)
ProcessResult
DopingSilicon doped with trivalent atoms like Boron
Bond formationTrivalent atoms form 3 covalent bonds with 4 neighboring Silicon atoms
Hole creationOne bond remains incomplete, creating a hole (positive charge carrier)
Majority carriersHoles become majority carriers
Minority carriersElectrons become minority carriers

Mnemonic: “Three Makes Positive” - Three valence electrons make a Positive hole

Question 2(a) OR [3 marks]
#

Enlist factors affecting Resistance and explain any one of them.

Answer:

Factors Affecting Resistance
Length of conductor
Cross-sectional area
Material (resistivity)
Temperature

Explanation of Temperature effect: The resistance of most metallic conductors increases with temperature as: R = R₀[1 + α(T - T₀)] where:

  • R = Resistance at temperature T
  • R₀ = Resistance at reference temperature T₀
  • α = Temperature coefficient of resistance

Mnemonic: “LAMT” - Length, Area, Material, Temperature affect resistance

Question 2(b) OR [4 marks]
#

Define: 1. Valance band, 2. Conduction band, 3. Forbidden energy gap, 4. Free electron

Answer:

TermDefinition
Valence bandEnergy band filled with valence electrons that are bound to atoms
Conduction bandHigher energy band where electrons can move freely and conduct electricity
Forbidden energy gapEnergy range between valence and conduction bands where no electron states exist
Free electronElectron that has gained enough energy to escape from valence band to conduction band

Diagram:

graph TD
    A[Conduction Band] --- B[Free Electrons]
    C[Forbidden Energy Gap/Band Gap]
    D[Valence Band] --- E[Bound Electrons]
    
    A --- C --- D

Mnemonic: “Very Clearly Freedom Follows” - Valence, Conduction, Forbidden gap, Free electrons

Question 2(c) OR [7 marks]
#

Define Pentavalent material and give examples of it. Explain Formation of N-type material with the help of proper diagram.

Answer:

Pentavalent material: Elements with 5 valence electrons in their outermost shell.

Examples: Phosphorus (P), Arsenic (As), Antimony (Sb)

N-type Semiconductor Formation:

Diagram:

SilicoSniatom(4valencee-)PefnrteaevPaelleencttraotnom(5valencee-)
ProcessResult
DopingSilicon doped with pentavalent atoms like Phosphorus
Bond formationPentavalent atoms form 4 covalent bonds with 4 neighboring Silicon atoms
Free electronFifth valence electron remains free (negative charge carrier)
Majority carriersElectrons become majority carriers
Minority carriersHoles become minority carriers

Mnemonic: “Five Makes Negative” - Five valence electrons make a Negative carrier

Question 3(a) [3 marks]
#

Define: 1. Depletion region, 2. Knee voltage, 3. Breakdown voltage in accordance of diode.

Answer:

TermDefinition
Depletion regionRegion at P-N junction devoid of mobile charge carriers due to diffusion and recombination
Knee voltageForward voltage at which current begins to increase rapidly (typically 0.7V for silicon, 0.3V for germanium)
Breakdown voltageReverse voltage at which diode rapidly conducts current in reverse direction

Mnemonic: “Depleted Knees Break” - Depletion occurs, Knee begins conduction, Breakdown ends blocking

Question 3(b) [4 marks]
#

Explain V-I characteristics of P-N junction diode with necessary graph.

Answer:

V-I Characteristics of P-N Junction Diode:

Diagram:

I/KneeoltaBgroeelat(kadg0oe.w7nV)V
RegionBehavior
Forward Bias (V > 0)Current increases exponentially after knee voltage
Reverse Bias (V < 0)Very small leakage current until breakdown voltage
Breakdown RegionSharp increase in reverse current at breakdown voltage
  • Forward equation: I = Is(e^(qV/nkT) - 1)
  • Knee voltage: ~0.7V for silicon, ~0.3V for germanium

Mnemonic: “Forward Flows, Reverse Restricts, Breakdown Bursts”

Question 3(c) [7 marks]
#

Draw characteristic of Varactor diode. Explain working of Varactor diode with diagram and write its application.

Answer:

Varactor Diode Characteristics:

Diagram:

CVR

Working of Varactor Diode:

Circuit Symbol:

PrincipleExplanation
Basic structureSpecial P-N junction diode optimized for variable capacitance
Reverse bias operationAlways operated in reverse bias condition
Depletion regionWidth varies with applied reverse voltage
Capacitance variationCapacitance decreases as reverse voltage increases
Mathematical relationC ∝ 1/√VR where VR is reverse voltage

Applications of Varactor Diode:

  • Voltage-controlled oscillators (VCOs)
  • Frequency modulators
  • Electronic tuning circuits
  • Automatic frequency control circuits
  • Phase-locked loops (PLLs)

Mnemonic: “Capacitance Varies Reversely” - Capacitance Varies with Reverse voltage

Question 3(a) OR [3 marks]
#

Write application of following diode: 1. Varactor diode, 2. Photo diode, 3. Light Emitting Diode

Answer:

Diode TypeApplications
Varactor DiodeVoltage-controlled oscillators, Frequency modulators, Electronic tuning circuits
Photo DiodeLight sensors, Optical communication, Smoke detectors, Camera light meters
Light Emitting Diode (LED)Display devices, Indicators, Lighting systems, Optical communication

Mnemonic: “Vary Photo Emit” - Varactor varies frequency, Photo detects light, LED emits light

Question 3(b) OR [4 marks]
#

Explain working of P-N junction diode in forward bias and reverse bias.

Answer:

Bias ConditionWorking PrincipleCharacteristics
Forward BiasP-side connected to positive terminal, N-side to negative terminalDepletion region narrows, current flows easily after knee voltage (~0.7V)
Reverse BiasP-side connected to negative terminal, N-side to positive terminalDepletion region widens, only small leakage current flows until breakdown

Diagram:

graph LR
    subgraph "Forward Bias"
        A1["+"] --- P1[P]
        P1 --- J1[Junction]
        J1 --- N1[N]
        N1 --- B1["-"]
        C1[Current flows]
    end
    
    subgraph "Reverse Bias"
        A2["-"] --- P2[P]
        P2 --- J2[Junction]
        J2 --- N2[N]
        N2 --- B2["+"]
        C2[Negligible current]
    end

Mnemonic: “Forward Flows, Reverse Resists”

Question 3(c) OR [7 marks]
#

Draw characteristic of Photo diode. Explain working of Photo diode with diagram and write its application.

Answer:

Photo Diode Characteristics:

Diagram:

I/LiingchrteaisniVtnegnsity

Working of Photo Diode:

Circuit Symbol:

PrincipleExplanation
Basic structureP-N junction diode with transparent window or lens
Reverse bias operationTypically operated in reverse bias condition
Light absorptionPhotons create electron-hole pairs in depletion region
Carrier generationLight intensity proportional to generated carriers
Current generationReverse current increases with light intensity

Applications of Photo Diode:

  • Light detectors in optical communication
  • Photometers and light meters
  • Smoke detectors
  • Barcode readers
  • Medical equipment (pulse oximeters)

Mnemonic: “Light In, Current Out” - Light intensity controls current output

Question 4(a) [3 marks]
#

Explain working of Half wave rectifier with circuit diagram.

Answer:

Half Wave Rectifier:

Circuit Diagram:

AooCD>ROutpoout
Operation PhaseDescription
Positive Half CycleDiode conducts, current flows through load, output follows input
Negative Half CycleDiode blocks, no current flows, output is zero
  • Output frequency: Same as input frequency
  • Form factor: 1.57
  • Ripple factor: 1.21
  • Efficiency: 40.6%
  • PIV of diode: Vmax

Mnemonic: “Half Passes Positive” - Only positive half-cycle passes through

Question 4(b) [4 marks]
#

Explain Zener diode as a voltage regulator.

Answer:

Zener Diode Voltage Regulator:

Circuit Diagram:

oVoinRsZe>nerRLooVout
ComponentFunction
Series resistor RsLimits current and drops excess voltage
Zener diodeMaintains constant voltage across load
Load resistor RLRepresents the circuit being powered

Working Principle:

  • Zener operates in reverse breakdown region
  • Maintains constant voltage regardless of input changes
  • Excess current flows through Zener diode
  • Voltage regulation equation: Vout = Vz (Zener voltage)

Mnemonic: “Zener Zeros Voltage Variations”

Question 4(c) [7 marks]
#

Write need of Rectifier. Explain Bridge wave rectifier with circuit diagram and draw its input and output waveform.

Answer:

Need of Rectifier:

  • To convert AC voltage to DC voltage
  • Most electronic devices require DC for operation
  • Power supply systems need DC output from AC mains

Bridge Wave Rectifier:

Circuit Diagram:

AoC><DD12<>DD34RLoOutput

Input and Output Waveform:

IOnuptuptuttt
Working in Positive Half CycleWorking in Negative Half Cycle
D1 and D4 conductD2 and D3 conduct
Current flows through load in same directionCurrent flows through load in same direction
  • Output frequency: Twice the input frequency
  • Form factor: 1.11
  • Ripple factor: 0.48
  • Efficiency: 81.2%
  • PIV of diode: Vmax

Mnemonic: “Bridge Both Better” - Bridge rectifier uses both half cycles

Question 4(a) OR [3 marks]
#

Explain working of Shunt capacitor filter.

Answer:

Shunt Capacitor Filter:

Circuit Diagram:

AooC>DCRLOouotput
OperationDescription
ChargingCapacitor charges during peak of rectified output
DischargingCapacitor discharges slowly through load when voltage drops
Smoothing effectProvides almost constant DC output by filling gaps
  • Ripple reduction: Significant reduction in ripple voltage
  • Time constant: RC must be much larger than period of input
  • Discharge equation: V = V₀e^(-t/RC)

Mnemonic: “Capacitor Catches Peaks” - Capacitor stores peak voltage

Question 4(b) OR [4 marks]
#

Compare Center tap full wave rectifier and Bridge wave rectifier

Answer:

ParameterCenter Tap Full Wave RectifierBridge Wave Rectifier
Number of diodes24
TransformerCenter-tapped transformer requiredSimple transformer sufficient
PIV of diode2VmaxVmax
Efficiency81.2%81.2%
Output frequencyTwice input frequencyTwice input frequency
CostHigher due to center-tapped transformerLower, simpler transformer but more diodes
SizeLargerSmaller

Mnemonic: “Center Taps Transformer, Bridge Bypasses Tapping”

Question 4(c) OR [7 marks]
#

Write need of Filter circuit in rectifier. Explain π filter with circuit diagram and draw its input and output waveform.

Answer:

Need of Filter Circuit in Rectifier:

  • Reduces ripple in rectified output
  • Provides steady DC voltage required by electronic circuits
  • Improves efficiency of power supply
  • Prevents damage to sensitive electronic components

π Filter:

Circuit Diagram:

AooC>DC1LC2RLOouotput

Input and Output Waveform:

IOnuptuptut(Rectified)tt
ComponentFunction
Input capacitor (C1)Initial filtering of rectified output
Choke (L)Blocks AC ripple and allows DC to pass
Output capacitor (C2)Further filtering for smoother output
  • Superior filtering: Better ripple reduction than simple capacitor filter
  • Ripple factor: Much lower than capacitor filter alone
  • Voltage regulation: Better voltage regulation under load variations

Mnemonic: “Capacitor-Inductor-Capacitor Perfectly Irons” (π shape resembling CIC filter)

Question 5(a) [3 marks]
#

Explain Working of PNP Transistor with the necessary diagram.

Answer:

PNP Transistor Working:

Diagram:

BaseCNPEomliltePtcetror
BiasingWorking
Base-Emitter junctionForward biased
Base-Collector junctionReverse biased
Majority carriersHoles
Current flowEmitter to Collector
  • Emitter: Heavily doped P-region that emits holes
  • Base: Thin, lightly doped N-region that controls current flow
  • Collector: Moderately doped P-region that collects holes

Mnemonic: “Positive-Negative-Positive” - PNP structure

Question 5(b) [4 marks]
#

Explain working of N-channel JFET with diagram.

Answer:

N-channel JFET Working:

Diagram:

GatePDSroaNuirncePGate
TerminalFunction
SourceSource of charge carriers (electrons)
DrainCollects charge carriers
GateControls width of the channel

Working Principle:

  • Channel formed by N-type material between source and drain
  • P-type gate regions form PN junctions with channel
  • Gate-to-source junction always reverse biased
  • Increasing negative gate voltage widens depletion region
  • Narrower channel increases resistance between source and drain
  • FET operates as voltage-controlled resistor

Mnemonic: “Negative Channel Junction Effect” - N-channel JFET

Question 5(c) [7 marks]
#

Compare BJT and JFET

Answer:

ParameterBJT (Bipolar Junction Transistor)JFET (Junction Field Effect Transistor)
StructureThree-layer structure (NPN or PNP)Single channel with gate junctions
Control mechanismCurrent-controlled deviceVoltage-controlled device
CarriersBoth majority and minority carriers (bipolar)Only majority carriers (unipolar)
Input impedanceLow to medium (1-10 kΩ)Very high (10⁸-10¹² Ω)
NoiseHigher noiseLower noise
Power consumptionHigherLower
Switching speedSlower due to charge storageFaster due to absence of charge storage
Temperature stabilityLess stableMore stable

Diagram:

graph TD
    subgraph "BJT"
        A1[Current-Controlled] --> B1[Bipolar Carriers]
        B1 --> C1[Low Input Impedance]
        C1 --> D1[Higher Noise]
    end
    
    subgraph "JFET"
        A2[Voltage-Controlled] --> B2[Unipolar Carriers]
        B2 --> C2[High Input Impedance]
        C2 --> D2[Lower Noise]
    end

Mnemonic: “Current Bipolar Low, Voltage Unipolar High” - BJT vs JFET key differences

Question 5(a) OR [3 marks]
#

Enlist methods to dispose E-waste and explain any one method of them.

Answer:

E-waste Disposal Methods
Recycling
Reuse
Incineration
Landfilling
Take-back systems

Explanation of Recycling: E-waste recycling involves collecting, dismantling, and separating electronic waste into recoverable materials. Components are shredded and sorted into raw materials like plastic, glass, and metals (including precious metals like gold, silver, copper). These materials are then processed and can be used to manufacture new products. Recycling reduces environmental impact, conserves resources, and recovers valuable materials.

Mnemonic: “RRIL-T” - Recycling, Reuse, Incineration, Landfill, Take-back

Question 5(b) OR [4 marks]
#

Compare PNP and NPN Transistor.

Answer:

ParameterPNP TransistorNPN Transistor
SymbolArrow points inward to baseArrow points outward from base
StructureP-type, N-type, P-type layersN-type, P-type, N-type layers
Majority carriersHolesElectrons
Biasing voltageBase negative with respect to emitterBase positive with respect to emitter
Current directionEmitter to collectorCollector to emitter
SpeedSlower (holes mobility is less)Faster (electrons mobility is more)

Diagram:

graph TD
    subgraph "PNP"
        A1[P-N-P Layers] --> B1[Hole Carriers]
        B1 --> C1[Negative Base Bias]
        C1 --> D1[E to C Current]
    end
    
    subgraph "NPN"
        A2[N-P-N Layers] --> B2[Electron Carriers]
        B2 --> C2[Positive Base Bias]
        C2 --> D2[C to E Current]
    end

Mnemonic: “Positive-Negative-Positive (Holes), Negative-Positive-Negative (Electrons)”

Question 5(c) OR [7 marks]
#

Draw and explain Input and Output Characteristics of CE configuration.

Answer:

Input Characteristics of CE Configuration:

Diagram:

Ib(μA)/VCE=/0VVCE=V5CVE=V1B0EV(V)

Output Characteristics of CE Configuration:

Diagram:

Ic(ARmceAtg)Iiibvo=enI0b=I1b0=μI2Ab0SR=μaeI3Atgb0uiI=μrob4Aan=0t5μi0AoμnAVCE(V)
CharacteristicDescription
Input CharacteristicsRelationship between base current (IB) and base-emitter voltage (VBE) at constant collector-emitter voltage (VCE)
Output CharacteristicsRelationship between collector current (IC) and collector-emitter voltage (VCE) at constant base current (IB)

Regions in Output Characteristics:

RegionDescription
Saturation RegionBoth junctions forward biased, VCE is small, IC is almost constant regardless of VCE
Active RegionBase-emitter junction forward biased, base-collector junction reverse biased, IC proportional to IB
Cutoff RegionBoth junctions reverse biased, negligible current flows

Important Parameters:

  • Current gain (β): Ratio of collector current to base current (IC/IB)
  • Input resistance: Ratio of change in VBE to change in IB at constant VCE
  • Output resistance: Ratio of change in VCE to change in IC at constant IB

Mnemonic: “Input Shows Voltage Effects, Output Shows Current Control”

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