Question 1(a) [3 marks]

Explain difference between Active and passive network.

Answer:

Active Network	Passive Network
Contains at least one energy source	Contains no energy source
Can deliver power to other elements	Cannot deliver power to other elements
Examples: Transistors, Op-amps, Batteries	Examples: Resistors, Capacitors, Inductors

Mnemonic: "Active Adds Power, Passive Pulls Power"

Question 1(b) [4 marks]

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]

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

Answer:

Term	Definition
Charge	The basic electrical quantity measured in coulombs (C); flow of electrons creates electricity
Current	The rate of flow of electric charge, measured in amperes (A); I = dQ/dt
Potential	Electric potential energy per unit charge, measured in volts (V)
E.M.F.	Electromotive force, energy supplied by source per unit charge, measured in volts (V)
Inductance	Property of a conductor to oppose change in current, measured in henry (H)
Capacitance	Ability of a component to store electric charge, measured in farad (F)
Frequency	Number of cycles per second of an alternating quantity, measured in hertz (Hz)

Mnemonic: "Careful Currents Pass Easily Into Circuit Frequently"

Question 1(c) OR [7 marks]

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:

            +
            |
            V
            |
  A---------/\/\/\---------B
            R
            |
            |
            -

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]

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]

Write statement of Maximum power transfer theorem and reciprocity theorem.

Answer:

Theorem	Statement
Maximum Power Transfer Theorem	Maximum power is transferred from source to load when load resistance equals the source internal resistance (RL = RS)
Reciprocity Theorem	In a linear passive network with a single source, if the source is moved from position A to B, the current at A due to source at B will equal the current at B when source was at A

Diagram:

Maximum Power Transfer:
        +---[Source]---+
        |              |
        R(source)      R(load)
        |              |
        +------+-------+

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

Question 2(c) [7 marks]

Explain the formation and conduction of N-type materials.

Answer:

N-type Semiconductor Formation:

graph LR
    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]

Define valence band, conduction band and forbidden gap.

Answer:

Term	Definition
Valence Band	The highest energy band filled with electrons, where electrons are bound to atoms
Conduction Band	The band above valence band where electrons move freely and contribute to electrical conduction
Forbidden Gap	The energy range between valence and conduction bands where no electron states exist

Diagram:

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

Mnemonic: "Valence Holds, Forbidden Blocks, Conduction Flows"

Question 2(b) OR [4 marks]

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

Answer:

Power Terms in AC Circuits:

Term	Definition
Active Power (P)	Actual power consumed, measured in watts (W); P = VI cosθ
Reactive Power (Q)	Power oscillating between source and load, measured in VAR; Q = VI sinθ
Power Factor (PF)	Ratio of active power to apparent power; PF = cosθ

Power Triangle:

                S (VA)
               /|
              / |
             /  |
            /   |
           /θ   |
          /_____|
         P(W)   Q(VAR)

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]

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

Answer:

Atomic Structures:

Element Type	Valence Electrons	Examples	Electronic Configuration
Trivalent	3	Boron, Aluminum, Gallium	3 electrons in outermost shell
Tetravalent	4	Carbon, Silicon, Germanium	4 electrons in outermost shell
Pentavalent	5	Nitrogen, Phosphorus, Arsenic	5 electrons in outermost shell

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]

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]

Write a Short note on LED.

Answer:

LED (Light Emitting Diode):

Parameter	Description
Structure	p-n junction with special doping materials
Working	Electrons recombine with holes, releasing energy as photons
Materials	GaAs (red), GaP (green), GaN (blue), etc.
Voltage	Forward voltage typically 1.8V to 3.3V depending on color

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]

Draw and explain VI characteristic of PN junction diode.

Answer:

P-N Junction Diode V-I Characteristic:

                        |
                        |         /
                        |        /
                        |       /
                        |      /
                        |     /
                        |    /
-----------+------------+---+----------
           |            |  /|
           |            | / |
           |            |/  |
           |            |   |
           |            |   |
           +            +   +
       Reverse       Origin Forward
       Region               Region

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]

List the applications of PN junction diode.

Answer:

Applications of PN Junction Diode:

Application Category	Examples
Rectification	Half-wave rectifier, Full-wave rectifier, Bridge rectifier
Signal Processing	Signal demodulation, Clipping circuits, Clamping circuits
Protection	Voltage spike protection, Reverse polarity protection
Logic Gates	Diode logic circuits, Switching applications
Voltage Regulation	Zener diodes for voltage references
Light Applications	LEDs, Photodiodes, Solar cells

Mnemonic: "Rectify, Process, Protect, Logic, Regulate, Light"

Question 3(b) OR [4 marks]

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]

Explain construction, working and applications of PN junction diode.

Answer:

Construction of PN Junction Diode:

    +--------+--------+
    |        |        |
    |  P-Type|N-Type  |
    |        |        |
    +--------+--------+
       |     |     |
       |Depletion|
       |  Region |

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:

Bias Condition	Behavior
Forward Bias	Depletion region narrows, current flows when V > 0.7V (Si)
Reverse Bias	Depletion region widens, only small leakage current flows

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]

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

Answer:

Term	Definition
Ripple Frequency	The frequency of AC component present in rectified DC output; for half-wave f = supply frequency, for full-wave f = 2 × supply frequency
Ripple Factor (γ)	Ratio of RMS value of AC component to DC component in rectifier output; γ = Vac(rms)/Vdc
PIV of Diode	Peak Inverse Voltage - maximum reverse voltage a diode can withstand without breakdown

Mnemonic: "Ripples Per second, Ripple Proportion, Reverse Peak Voltage"

Question 4(b) [4 marks]

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

Answer:

Parameter	Center-Tapped Full Wave	Bridge Rectifier
Diodes Used	2 diodes	4 diodes
Transformer	Center-tapped required	No center tap needed
PIV of Diode	2Vm	Vm
Output Voltage	Vdc = 0.637Vm	Vdc = 0.637Vm
Ripple Factor	0.48	0.48
Efficiency	81.2%	81.2%
TUF	0.693	0.693

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]

Explain zener diode as voltage regulator.

Answer:

Zener Diode Voltage Regulator:

    Vin     Rs          
    +------|\/\/\|------+--------+ Vout
    |                   |        |
    |                   Z        RL
    |                   Z Zener  |
    |                   Z        |
    +-------------------+--------+
                        |
                       GND

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]

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:

                    D1
   AC     +--------->|-------+
   Input  |                  |
   o------+                  +-----o
          |                  |     DC
          |                  |     Output
   o------+                  +-----o
          |                  |
          +--------|<--------+
                   D2

Waveforms:

Input:    ^     ^     ^
          |     |     |
   0 -----+-----+-----+----
          |     |     |
          v     v     v

Output:   ^     ^     ^
          |     |     |
   0 -----+-----+-----+----

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]

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:

    +-------+-----+
    |       |     |
    |       C     RL
    |       |     |
    +-------+-----+

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]

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:

           D1       D3
     +----->|----+--|>----+
     |             |      |
AC   |             |      | DC
Input|             |      | Output
     |             |      |
     +------|<----+--|<---+
            D2       D4

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:

Input:     ^      ^      ^
           |      |      |
    0 -----+------+------+-----
           |      |      |
           v      v      v

Output:    ^      ^      ^      ^      ^
           |      |      |      |      |
    0 ------+----+------+------+------+----

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]

Explain causes of electronic waste.

Answer:

Causes of Electronic Waste:

Cause	Description
Rapid Technology Change	Frequent upgrades and obsolescence of electronics
Short Lifecycle	Devices designed with limited useful life
Consumer Behavior	Preference for new gadgets over repair
Manufacturing Issues	Poor quality leading to early failures
Economic Factors	Sometimes cheaper to replace than repair
Marketing Strategies	Promoting new models through planned obsolescence

Mnemonic: "Upgrade, Use, Throw, Repeat"

Question 5(b) [4 marks]

Compare PNP and NPN transistors.

Answer:

Parameter	PNP Transistor	NPN Transistor
Symbol	E \| \|\ \| \ \| >C \| / \|/ B	C \| \|/ \| / \| >E \| \ \|\ B
Current Flow	Emitter to Collector	Collector to Emitter
Majority Carriers	Holes	Electrons
Biasing	Emitter positive, Collector negative	Collector positive, Emitter negative
Switching Speed	Slower	Faster
Usage	Less common	More common

Mnemonic: "PNP: Positive to Negative to Positive; NPN: Negative to Positive to Negative"

Question 5(c) [7 marks]

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

Answer:

MOSFET Symbol (N-Channel Enhancement):

Construction:

graph LR
    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]

Explain methods to handle electronic waste.

Answer:

Methods to Handle Electronic Waste:

Method	Description
Reduce	Designing products with longer lifecycle and upgradability
Reuse	Refurbishing and donating electronics for secondary use
Recycle	Systematic disassembly to recover valuable materials
Responsible Disposal	Proper collection and processing by certified facilities
Extended Producer Responsibility	Manufacturers take back used products
Urban Mining	Recovering precious metals from discarded electronics

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]

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:

α Value	β Value
0.9	9
0.95	19
0.99	99

Mnemonic: "Alpha approaches One as Beta approaches Infinity"

Question 5(c) OR [7 marks]

Explain common collector configuration with its input and output characteristics.

Answer:

Common Collector (Emitter Follower) Configuration:

                   +Vcc
                    |
                    |
                    R
                    |
                    |
    +------+--------+-------+
    |      |                |
    |    B |    C           E
    +-----|       |-------+-+
           |      |       |
         --+      +--     R
           |                |
           |                |
         +-+-+              |
         |   |              |
         GND GND            GND

Input Characteristics:

   Ib
   ^
   |      -------
   |     /
   |    /
   |   /
   |  /
   | /
   |/
   +-----------------> Vbe

Output Characteristics:

   Ie
   ^
   |       --------
   |      /
   |     /
   |    /
   |   /
   |  /
   | /
   |/
   +-----------------> Vce

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.