Question 1(a) [3 marks]

Define Forward and reverse bias of diode.

Answer:

Forward Bias of Diode:

Connection Method: P-type connected to positive terminal and N-type connected to negative terminal of battery
Barrier Width: Barrier width decreases
Resistance: Low resistance (typically 100-1000Ω)
Current Flow: Allows current to flow easily through the diode

Reverse Bias of Diode:

Connection Method: P-type connected to negative terminal and N-type connected to positive terminal
Barrier Width: Barrier width increases
Resistance: Very high resistance (typically several MΩ)
Current Flow: Blocks current flow (only small leakage current flows)

Diagram:

graph TD
    A[Forward Bias] --> B[P connected to +ve<br>N connected to -ve]
    A --> C[Current flows easily]
    D[Reverse Bias] --> E[P connected to -ve<br>N connected to +ve]
    D --> F[Current blocked]

Mnemonic: "PFNR" - "Positive to P Forward, Negative to P Reverse"

Question 1(b) [4 marks]

Explain construction and working of LDR.

Answer:

Construction of LDR:

Material: Made of semiconductor material (Cadmium Sulfide)
Pattern: Zigzag pattern of photosensitive material on ceramic base
Electrodes: Metal electrodes at both ends
Package: Encapsulated in transparent plastic or glass case

Working Principle:

Photoconductivity: Based on photoconductivity principle
Dark Resistance: High resistance (MΩ range) in dark conditions
Light Exposure: When exposed to light, photons release electrons
Resistance Drop: Resistance decreases (kΩ range) in bright light

Diagram:

 +------+
 |      |    Zigzag pattern of
 | +-\/-+ <- semiconductor material
 | |    |
 | +-/\-+
 |      |
 +------+
  |    |
  |    |
  L    D <- Leads

Mnemonic: "MILD" - "More Illumination, Less Dark-resistance"

Question 1(c) [7 marks]

Explain the color band coding method of Resistor. Write color band of 47kΩ ±5% resistance.

Answer:

Color Band Coding Method:

Color	Value	Multiplier	Tolerance
Black	0	10⁰	-
Brown	1	10¹	±1%
Red	2	10²	±2%
Orange	3	10³	-
Yellow	4	10⁴	-
Green	5	10⁵	±0.5%
Blue	6	10⁶	±0.25%
Violet	7	10⁷	±0.1%
Grey	8	10⁸	±0.05%
White	9	10⁹	-
Gold	-	10⁻¹	±5%
Silver	-	10⁻²	±10%
Colorless	-	-	±20%

4-Band Resistor Color Code:

First Band: First significant digit
Second Band: Second significant digit
Third Band: Multiplier
Fourth Band: Tolerance

For 47kΩ ±5%:

First digit: 4 = Yellow
Second digit: 7 = Violet
Multiplier: 10³ = Orange (for kΩ)
Tolerance: ±5% = Gold

Color bands for 47kΩ ±5%: Yellow-Violet-Orange-Gold

Diagram:

+---+---+---+---+-------------+
|   |   |   |   |             |
| Y | V | O | G |             |
|   |   |   |   |             |
+---+---+---+---+-------------+
  |   |   |   |
  |   |   |   +-- Gold (±5%)
  |   |   +------ Orange (10³)
  |   +---------- Violet (7)
  +-------------- Yellow (4)

Mnemonic: "BAND" - "Beginning digits, Amplify with Multiplier, Note tolerance with last band, Decode carefully"

Question 1(c) [7 marks] (OR)

Explain Aluminum Electrolytic wet type capacitor.

Answer:

Aluminum Electrolytic Wet Type Capacitor:

Construction:

Plates: Two aluminum foils (anode and cathode)
Dielectric: Aluminum oxide layer on anode foil
Electrolyte: Liquid electrolyte (boric acid, sodium borate, etc.)
Separator: Paper separator soaked in electrolyte
Enclosure: Aluminum can with rubber seal

Working Principle:

Oxide Layer: Thin aluminum oxide layer acts as dielectric
Electrolyte: Acts as cathode connection to second plate
Polarization: Has defined polarity (+ and -) terminals

Characteristics:

Capacitance Range: 1μF to 47,000μF
Voltage Rating: 6.3V to 450V
Polarity: Polarized (must connect correctly)
Leakage Current: Higher than other capacitor types
ESR: Higher equivalent series resistance

Diagram:

graph TD
    A[Aluminum Electrolytic Capacitor] --> B[Aluminum Can]
    A --> C[Anode Foil]
    A --> D[Cathode Foil]
    A --> E[Electrolyte]
    A --> F[Separator]
    A --> G[Aluminum Oxide Layer]
    A --> H[Terminal Posts]

Mnemonic: "POLE" - "Polarized, Oxide layer, Liquid electrolyte, Enormous capacitance"

Question 2(a) [3 marks]

Draw the symbol of Schottkey diode, LED and Photo-diode.

Answer:

Symbols:

Schottky Diode      LED                 Photo-diode
   +----|<---+      +---|>|---+         +---|<|---+
   |         |      |    |    |         |    ↓    |
   |         |      |   / \   |         |   / \   |
   +---------+      |  /   \  |         |  \   /  |
                    | /     \ |         |   \ /   |
                    +---------+         +---------+

Key Features:

Schottky Diode: Standard diode symbol with curved bar (represents metal-semiconductor junction)
LED: Standard diode symbol with two arrows pointing away (represents light emission)
Photo-diode: Standard diode symbol with two arrows pointing toward diode (represents light detection)

Mnemonic: "SLP" - "Schottky has curve, LED emits, Photo-diode absorbs"

Question 2(b) [4 marks]

Define Active and Passive Components with example.

Answer:

Passive Components:

Characteristic	Description	Examples
Power	Cannot generate power	Resistors, Capacitors, Inductors
Signal	Cannot amplify signals	Transformers, Diodes
Control	No control over current flow	Connectors, Switches
Energy	Store or dissipate energy	Fuses, Filters

Active Components:

Characteristic	Description	Examples
Power	Can generate power	Transistors, ICs
Signal	Can amplify signals	Op-amps, Amplifiers
Control	Control current flow	SCRs, MOSFETs
Dependency	Require external power	Voltage regulators, Microcontrollers

Diagram:

graph TB
    A[Electronic Components] --> B[Active Components]
    A --> C[Passive Components]
    B --> D[Transistors]
    B --> E[ICs]
    B --> F[Amplifiers]
    C --> G[Resistors]
    C --> H[Capacitors]
    C --> I[Inductors]

Mnemonic: "PASS-ACT" - "Passive stores or dissipates, Active controls or amplifies"

Question 2(c) [7 marks]

Explain working of full wave bridge rectifier.

Answer:

Full Wave Bridge Rectifier:

Circuit Construction:

Diodes: Four diodes arranged in bridge configuration
Input: AC supply from transformer secondary
Output: Pulsating DC across load resistor with filter capacitor

Working Principle:

Positive Half Cycle: D1 and D3 conduct, D2 and D4 block
Negative Half Cycle: D2 and D4 conduct, D1 and D3 block
Current Flow: Always flows through load in same direction

Performance Parameters:

Ripple Frequency: 2× input frequency (100 Hz for 50 Hz input)
Efficiency: 81.2%
PIV: V₀(max) per diode
TUF: 0.812 (Transformer Utilization Factor)

Diagram:

graph LR
    A[AC Input] --> B[Bridge Rectifier]
    B --> C[D1]
    B --> D[D2]
    B --> E[D3]
    B --> F[D4]
    C --> G[Load]
    D --> G
    E --> G
    F --> G
    G --> H[Pulsating DC Output]
    H --> I[Filter Capacitor]
    I --> J[Smooth DC Output]

Mnemonic: "BRIDGE" - "Better Rectification with Improved Diode Geometry Efficiency"

Question 2(a) [3 marks] (OR)

Explain construction and working of LED.

Answer:

Construction of LED:

Material: Semiconductor (GaAs, GaP, AlGaInP, etc.)
Junction: P-N junction with heavily doped semiconductors
Package: Encased in transparent or colored epoxy lens
Cathode: Identified by flat side on package or shorter lead

Working Principle:

Forward Bias: Applied to P-N junction
Recombination: Electrons and holes recombine at junction
Energy Release: Energy released as photons (light)
Wavelength: Determined by band gap of semiconductor material

Diagram:

        +-------+
        |       |
        |   ^   |
        |  / \  | <- Epoxy lens
        | /   \ |
        |/     \|
    ----+-------+----
    |       |       |
    |       |       |
    |       |       |
  Anode   Chip   Cathode

Mnemonic: "LEDS" - "Light Emits During electron-hole recombination in Semiconductor"

Question 2(b) [4 marks] (OR)

Explain composition type resistors.

Answer:

Composition Resistors:

Construction:

Core Material: Carbon particles mixed with insulating material (clay/ceramic)
Binding: Resin binder forms solid cylindrical shape
Terminals: Metal caps with leads attached to ends
Protection: Coated with insulating paint or plastic

Characteristics:

Resistance Range: 1Ω to 22MΩ
Power Rating: 1/8W to 2W
Tolerance: ±5% to ±20%
Temperature Coefficient: -500 to +500 ppm/°C

Advantages & Limitations:

Cost: Low cost
Noise: Higher noise level
Stability: Less stable with temperature
Applications: General purpose, non-critical applications

Diagram:

    +---------------------+
    |                     |
    |  +---------------+  |
    |  | Carbon        |  | <- Insulating
    |  | Composition   |  |    coating
    |  +---------------+  |
    |                     |
    +---------------------+
    |         |
    |         |
Lead         Lead

Mnemonic: "CCRI" - "Carbon Composition Resistors are Inexpensive"

Question 2(c) [7 marks] (OR)

Explain working of full wave rectifier with two diodes.

Answer:

Full Wave Rectifier with Two Diodes (Center-tap):

Circuit Construction:

Transformer: Center-tapped transformer secondary
Diodes: Two diodes connected to opposite ends of secondary
Output: Taken between center tap and diode junction

Working Principle:

Positive Half Cycle: Upper half of secondary positive, D1 conducts, D2 blocks
Negative Half Cycle: Lower half of secondary positive, D2 conducts, D1 blocks
Current Flow: Always flows through load in same direction

Performance Parameters:

Ripple Frequency: 2× input frequency (100 Hz for 50 Hz input)
Efficiency: 81.2%
PIV: 2V₀(max) per diode (twice the center-tap rectifier)
TUF: 0.693 (Transformer Utilization Factor)

Diagram:

graph LR
    A[AC Input] --> B[Center-Tapped Transformer]
    B -->|Upper Half| C[D1]
    B -->|Lower Half| D[D2]
    B -->|Center Tap| E[Ground]
    C --> F[Load]
    D --> F
    F --> E
    F --> G[Pulsating DC Output]
    G --> H[Filter]
    H --> I[Smooth DC Output]

Mnemonic: "CTFWR" - "Center Tap Facilitates Whole-cycle Rectification"

Question 3(a) [3 marks]

Explain working of schhotkey diode.

Answer:

Working of Schottky Diode:

Junction Type: Metal-Semiconductor (M-S) junction instead of P-N
Charge Carriers: Majority carrier device (electrons in N-type)
Barrier: Schottky barrier formed at metal-semiconductor interface
Forward Voltage: Lower forward voltage drop (0.2-0.4V vs 0.7V for Si diode)

Key Characteristics:

Switching Speed: Very fast switching (no minority carrier storage)
Applications: High-frequency circuits, power supplies
Recovery Time: Negligible reverse recovery time

Diagram:

Metal    |    N-type
         |
      +--+--+
      |     |
      | M-S |  <- Schottky Barrier
      |     |
      +-----+

Mnemonic: "SFAM" - "Schottky's Fast And Metal-based"

Question 3(b) [4 marks]

Explain N type semiconductor.

Answer:

N-type Semiconductor:

Formation:

Base Material: Intrinsic semiconductor (Silicon or Germanium)
Doping Element: Pentavalent impurity (P, As, Sb)
Doping Process: Thermal diffusion or ion implantation
Concentration: Typically 1 part impurity to 10⁸ parts silicon

Characteristics:

Majority Carriers: Electrons (negative charge carriers)
Minority Carriers: Holes
Conductivity: Higher than intrinsic semiconductor
Fermi Level: Closer to conduction band

Diagram:

graph TD
    A[N-type Semiconductor] --> B[Silicon Atom]
    A --> C[Pentavalent Impurity Atom]
    C --> D[Extra Free Electron]
    A --> E[Majority Carriers: Electrons]
    A --> F[Minority Carriers: Holes]

Mnemonic: "PENT" - "Pentavalent Element makes N-Type with free electrons"

Question 3(c) [7 marks]

Explain construction and working of PN Junction Diode.

Answer:

Construction of PN Junction Diode:

Materials: P-type and N-type semiconductor regions
Junction: Formed by diffusion or epitaxial growth
Depletion Region: Forms at junction interface
Contacts: Metal contacts attached to both regions
Packaging: Sealed in glass, plastic, or metal case

Working Principle:

Depletion Region: Forms due to diffusion of carriers
Barrier Potential: Created across junction (0.7V for Si, 0.3V for Ge)
Forward Bias: Current flows when forward voltage > barrier potential
Reverse Bias: Only small leakage current flows until breakdown

Diagram:

    +-------+-------+
    |       |       |
    |   P   |   N   |
    |       |       |
    +-------+-------+
        |       |
      Anode  Cathode

    Depletion region at junction

Mnemonic: "BIRD" - "Barrier forms at Interface, Rectifies Direct current"

Question 3(a) [3 marks] (OR)

Explain working of photo-diode.

Answer:

Working of Photo-diode:

Operation Mode: Reverse biased P-N junction
Light Absorption: Photons create electron-hole pairs in depletion region
Carrier Generation: Light energy > band gap energy creates free carriers
Current Flow: Photocurrent proportional to light intensity

Key Characteristics:

Sensitivity: Depends on semiconductor material and wavelength
Response Time: Very fast (ns range)
Operating Modes: Photovoltaic mode or photoconductive mode
Applications: Light sensors, optical communication

Diagram:

       Light
         ↓
    +----+----+
    |         |
 ---+         +---
    |    PN   |
    | Junction|
    |         |
 ---+         +---
    |         |
    +---------+

Mnemonic: "PLIP" - "Photons Lead to Increased Photocurrent"

Question 3(b) [4 marks] (OR)

Explain P type Semiconductor.

Answer:

P-type Semiconductor:

Formation:

Base Material: Intrinsic semiconductor (Silicon or Germanium)
Doping Element: Trivalent impurity (B, Al, Ga)
Doping Process: Thermal diffusion or ion implantation
Concentration: Typically 1 part impurity to 10⁸ parts silicon

Characteristics:

Majority Carriers: Holes (positive charge carriers)
Minority Carriers: Electrons
Conductivity: Higher than intrinsic semiconductor
Fermi Level: Closer to valence band

Diagram:

graph TD
    A[P-type Semiconductor] --> B[Silicon Atom]
    A --> C[Trivalent Impurity Atom]
    C --> D[Hole Formation]
    A --> E[Majority Carriers: Holes]
    A --> F[Minority Carriers: Electrons]

Mnemonic: "TRIP" - "TRIvalent impurity Produces holes in P-type"

Question 3(c) [7 marks] (OR)

Compare half wave and full wave rectifier.

Answer:

Comparison between Half Wave and Full Wave Rectifier:

Parameter	Half Wave Rectifier	Full Wave Rectifier
Circuit Complexity	Simple, uses 1 diode	Complex, uses 2 or 4 diodes
Output Waveform	Pulsating DC for half cycle	Pulsating DC for full cycle
Efficiency	40.6%	81.2%
Ripple Factor	1.21	0.48
Ripple Frequency	Same as input (50 Hz)	Twice the input (100 Hz)
PIV of Diode	V_m	2V_m (center-tap), V_m (bridge)
TUF	0.287	0.693 (center-tap), 0.812 (bridge)
DC Output Voltage	0.318V_m	0.636V_m
Form Factor	1.57	1.11
Applications	Low power applications	Power supplies, battery chargers

Diagram:

graph TD
    A[Rectifiers] --> B[Half Wave]
    A --> C[Full Wave]
    C --> D[Center-Tapped]
    C --> E[Bridge]
    B --> F[Uses 1 diode]
    B --> G[Lower efficiency]
    D --> H[Uses 2 diodes]
    E --> I[Uses 4 diodes]
    C --> J[Higher efficiency]

Mnemonic: "HERO" - "Half wave: Efficiency Reduced, One-half cycle only"

Question 4(a) [3 marks]

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

Answer:

Transistor Symbols and Construction:

NPN Symbol         PNP Symbol
    C                  C
    |                  |
    |                  |
    —                  —
   /                  /
  |                  |
  |\                 |>
  | \                |
  |  >               |/
  | /                |
  |/                 |
    —                  —
    |                  |
    |                  |
    B                  B
    |                  |
    |                  |
    —                  —
    |                  |
    |                  |
    E                  E

Construction:

NPN Construction           PNP Construction
    +-------+                 +-------+
    |   N   |                 |   P   | <- Collector
    +-------+                 +-------+
    |   P   |                 |   N   | <- Base
    +-------+                 +-------+
    |   N   |                 |   P   | <- Emitter
    +-------+                 +-------+

Mnemonic: "NIN-PIP" - "N-P-N layers for NPN, P-N-P layers for PNP"

Question 4(b) [4 marks]

Explain working of Transistor amplifier.

Answer:

Working of Transistor Amplifier:

Circuit Configuration:

Common Emitter: Most commonly used
Biasing: Proper DC bias provided to operate in active region
Coupling: Input/output coupling through capacitors
Load: Collector resistor as load

Working Principle:

Input Signal: Applied to base-emitter junction
Base Current: Small base current controls larger collector current
Amplification: Small input voltage variations cause larger output voltage variations
Phase Shift: 180° phase shift between input and output

Key Parameters:

Voltage Gain: A_v = V_out/V_in
Current Gain: β = I_c/I_b
Input Impedance: Typically 1-2kΩ in CE configuration

Diagram:

graph LR
    A[Input Signal] --> B[Base Current]
    B --> C[Controls Collector Current]
    C --> D[Voltage Drop Across R<sub>C</sub>]
    D --> E[Amplified Output Signal]

Mnemonic: "ABCD" - "Amplification through Base Controlled collector Current Dynamics"

Question 4(c) [7 marks]

Explain working of Zener diode.

Answer:

Working of Zener Diode:

Basic Structure:

Junction: Heavily doped P-N junction
Construction: Similar to normal diode but optimized for breakdown
Breakdown: Designed to operate in reverse breakdown region

Working Principle:

Forward Bias: Acts like normal diode
Reverse Bias:
- Below breakdown: Small leakage current
- At breakdown: Sharp increase in current at Zener voltage
- Beyond breakdown: Maintains constant voltage

Breakdown Mechanisms:

Zener Effect: Dominant below 5V (direct tunneling)
Avalanche Effect: Dominant above 5V (impact ionization)

Applications:

Voltage Regulation: Maintains constant output voltage
Reference Voltage: Precise voltage reference
Overvoltage Protection: Protects sensitive components

Diagram:

    I
    ^
    |               /
    |              /
    |             /
    |            /
    |           /
    +----------+------> V
    |         /|
    |        / |
    |       /  |
    |      /   |
    |  Reverse | Forward
    |  Breakdown

Mnemonic: "ZEBRA" - "Zener Effect Breaks at Regulated Avalanche voltage"

Question 4(a) [3 marks] (OR)

Explain transistor as a switch.

Answer:

Transistor as a Switch:

Operating Regions:

Cutoff Region: Transistor OFF (I_B = 0, I_C ≈ 0)
Saturation Region: Transistor ON (I_B > I_C/β, V_CE ≈ 0.2V)

Switching Operation:

OFF State: No base current, high V_CE, acts as open switch
ON State: Sufficient base current, low V_CE, acts as closed switch

Switching Characteristics:

Turn-ON Time: Time to go from cutoff to saturation
Turn-OFF Time: Time to go from saturation to cutoff

Diagram:

graph TD
    A[Transistor Switch] --> B[OFF State: Cutoff Region]
    A --> C[ON State: Saturation Region]
    B --> D[I<sub>B</sub> = 0, I<sub>C</sub> ≈ 0]
    B --> E[High V<sub>CE</sub> ≈ V<sub>CC</sub>]
    C --> F[I<sub>B</sub> > I<sub>C</sub>/β]
    C --> G[Low V<sub>CE</sub> ≈ 0.2V]

Mnemonic: "COST" - "Cutoff Off, Saturation Turns-on"

Question 4(b) [4 marks] (OR)

Draw and Explain characteristics of CE amplifier.

Answer:

CE Amplifier Characteristics:

Input Characteristics:

Plot: I_B vs V_BE at constant V_CE
Behavior: Resembles forward-biased diode curve
Knee Voltage: Approximately 0.7V for silicon
Input Resistance: Slope of curve (ΔV_BE/ΔI_B)

Output Characteristics:

Plot: I_C vs V_CE at constant I_B
Regions:
- Saturation (V_CE < 0.2V)
- Active (V_CE > 0.2V)
- Cutoff (I_B = 0)
Early Effect: Slight increase in I_C with increasing V_CE

Diagram:

   I_C |           I_B3
       |         ,------
       |        /
       |       /
       |      /  I_B2
       |     ,------
       |    /
       |   /
       |  /  I_B1
       | ,------
       |/
       +-------------> V_CE
       |
   
   I_B |
       |        /
       |       /
       |      /
       |     /
       |    /
       |   /
       |  /
       | /
       |/
       +-------------> V_BE
       |   0.7V

Mnemonic: "IAOC" - "Input curves At Origin, Output curves show Current gain"

Question 4(c) [7 marks] (OR)

Explain working of Varactor diode.

Answer:

Working of Varactor Diode:

Basic Structure:

Junction: Special P-N junction diode
Operation: Always operated in reverse bias
Property: Junction capacitance varies with reverse voltage

Working Principle:

Depletion Layer: Widens with increasing reverse voltage
Capacitance Effect: Depletion region acts as dielectric between P and N regions
Capacitance Formula: C ∝ 1/√V_R
Tuning Range: Typically 4:1 to 10:1 capacitance

Applications:

Voltage-Controlled Capacitor: In electronic tuning circuits
Frequency Modulation: In voltage-controlled oscillators (VCOs)
Automatic Frequency Control: In receivers
Parametric Amplification: In microwave circuits

Diagram:

graph LR
    A[Varactor Diode] --> B[Reverse Bias Operation]
    B --> C[Depletion Region Width]
    C --> D[Junction Capacitance]
    D --> E[Changes with Applied Voltage]
    E --> F[Electronic Tuning]

Mnemonic: "VCAP" - "Voltage Controls cAPacitance"

Question 5(a) [3 marks]

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

Answer:

Transistor Regions of Operation:

Region	Definition	Biasing Condition	Application
Active Region	Both junctions are properly biased (BE forward, BC reverse)	I_B > 0, V_CE > V_CE(sat)	Amplification
Saturation Region	Both junctions forward biased	I_B > I_C/β, V_CE ≈ 0.2V	Switching (ON state)
Cut-off Region	Both junctions reverse biased	I_B = 0, I_C ≈ 0, V_CE ≈ V_CC	Switching (OFF state)

Diagram:

   I_C |
       |         Active
       |         Region
       |        /|
       |       / |
       |      /  |
       |     /   |
       |    /    |
       |   /     |
       |  /      |
       | /       |
       |/        |
       +---------+------> V_CE
       |Saturation|Cut-off

Mnemonic: "ASC" - "Active for Signals, Saturation & Cutoff for switches"

Question 5(b) [4 marks]

If the value of I_C = 10mA and I_B = 100μA then find the value of current gains α and β.

Answer:

Given:

Collector current (I_C) = 10 mA
Base current (I_B) = 100 μA = 0.1 mA

Calculate β (Common Emitter Current Gain):

β = I_C / I_B
β = 10 mA / 0.1 mA
β = 100

Calculate α (Common Base Current Gain):

I_E = I_C + I_B = 10 mA + 0.1 mA = 10.1 mA
α = I_C / I_E
α = 10 mA / 10.1 mA
α = 0.990 or 0.99

Relation between α and β:

α = β / (β + 1)
α = 100 / (100 + 1) = 100 / 101 = 0.990
β = α / (1 - α)
β = 0.99 / (1 - 0.99) = 0.99 / 0.01 = 99 ≈ 100

Mnemonic: "ABC" - "Alpha equals Beta divided by (Beta plus one) for Current gains"

Question 5(c) [7 marks]

Discuss Strategies of electronic waste management in the small electronics Industries.

Answer:

E-Waste Management Strategies for Small Electronics Industries:

Strategy	Description	Implementation
Segregation	Separate e-waste from general waste	Dedicated collection bins for different components
Reduce	Minimize waste generation	Efficient design, extended product life, repair services
Reuse	Use components again	Refurbish, repurpose working parts
Recycle	Process for material recovery	Partner with authorized recyclers, follow guidelines
Training	Educate employees	Regular workshops on proper handling procedures

Key Implementation Steps:

Inventory Management: Track electronic components throughout lifecycle
Authorized Partnerships: Work only with certified e-waste handlers
Documentation: Maintain records of waste disposal for compliance
Green Design: Design products for easy disassembly and recycling

Regulatory Compliance:

Registration: Register with pollution control board
Authorization: Obtain necessary permits
Annual Returns: Submit regular compliance reports

Diagram:

graph TD
    A[E-Waste Management] --> B[Collection & Segregation]
    A --> C[Storage]
    A --> D[Transportation]
    A --> E[Processing]
    B --> F[Separate bins for different components]
    C --> G[Safe storage in designated areas]
    D --> H[Authorized carriers only]
    E --> I[Authorized recyclers]
    E --> J[Material recovery]
    E --> K[Safe disposal of residues]

Mnemonic: "SRRTA" - "Segregate, Reduce, Reuse, Train, Authorize"

Question 5(a) [3 marks] (OR)

Draw CB, CE and CC transistor configuration circuits.

Answer:

Transistor Configuration Circuits:

Common Base (CB)         Common Emitter (CE)        Common Collector (CC)
                                                     (Emitter Follower)
    +---+                     +---+                      +---+
    |   |                     |   |                      |   |
    | RC|                     | RC|                      |   |
    |   |                     |   |                      |   |
    +---+                     +---+                      +---+
      |                         |                          |
      |                         |                          |
      +-----+             +-----+                    +-----+
      |     |             |     |                    |     |
 Cout +     +---+    Cout +     +---+          +---->+     +---> Vout
      |     |             |     |              |     |     |
      +-----+             +-----+              |     +-----+
        |                   |                  |       |
        |                   |                  |       |
    +---+                   |                  |     +---+
    |   |                   |                  |     |   |
    | RE|               +---+---+              |     | RE|
    |   |               |       |              |     |   |
    +---+               |       |              |     +---+
      |                 +-------+              |       |
      |                     |                  |       |
     GND                   GND                 +-------+
                                                   |
 Input to Emitter      Input to Base           Input to Base
 Output from Collector Output from Collector   Output from Emitter

Key Characteristics:

CB: High stability, low input impedance, high output impedance
CE: Medium stability, medium input impedance, medium output impedance
CC: Low stability, high input impedance, low output impedance

Mnemonic: "EBC" - "Emitter input for CB, Base input for CE/CC, Collector output for CB/CE"

Question 5(b) [4 marks] (OR)

Derive relation between current gains α and β.

Answer:

Relation Between Current Gains α and β:

Given definitions:

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

Step 1: Use current relation in transistor

I_E = I_C + I_B

Step 2: Express α in terms of β

α = I_C/I_E
α = I_C/(I_C + I_B)

Step 3: Substitute I_B = I_C/β

α = I_C/(I_C + I_C/β)
α = I_C/(I_C(1 + 1/β))
α = I_C/(I_C(β + 1)/β)
α = β/(β + 1)

Step 4: Express β in terms of α

β = α/(1 - α)

Diagram:

      I_C
     ↗   ↘
    /     \
   /       \
  I_B       I_E

  α = I_C/I_E
  β = I_C/I_B
  I_E = I_C + I_B

Mnemonic: "ABR" - "Alpha = Beta divided by (Beta plus one) Reciprocally"

Question 5(c) [7 marks] (OR)

Define E-Waste and Explain disposal of electronic waste.

Answer:

E-Waste Definition: Electronic waste (e-waste) refers to discarded electrical or electronic devices that have reached end-of-life or become obsolete, including computers, televisions, mobile phones, printers, and other electronic equipment containing hazardous components like lead, mercury, cadmium, PCBs, and brominated flame retardants.

Disposal Methods of E-Waste:

Method	Description	Environmental Impact
Collection & Segregation	Gathering and separating by type	Reduces contamination
Dismantling	Manual disassembly of components	Enables targeted recycling
Material Recovery	Extracting valuable materials	Conserves natural resources
Refurbishment	Repairing for reuse	Extends product lifecycle
Authorized Recycling	Processing by certified facilities	Ensures proper handling

Disposal Process Flow:

Initial Assessment: Determine if device can be repaired/reused
Data Sanitization: Secure erasure of personal/business data
Disassembly: Separation into component categories
Resource Recovery: Extraction of valuable materials
Hazardous Waste: Special handling of toxic components

Diagram:

graph TD
    A[E-Waste Disposal] --> B[Collection]
    B --> C[Sorting & Segregation]
    C --> D[Recycling]
    C --> E[Recovery]
    C --> F[Safe Disposal]
    D --> G[Disassembly]
    G --> H[Material Sorting]
    H --> I[Crushing & Shredding]
    I --> J[Material Separation]
    J --> K[Refinement]
    K --> L[New Products]
    F --> M[Landfill for Inert Material]
    F --> N[Incineration with Pollution Control]

Mnemonic: "CRESD" - "Collect, Recycle, Extract, Separate, Dispose"