Electronics Devices & Circuits (1323202) - Summer 2023 Solution

Table of Contents

Question 1(a) [3 marks]
#

Draw the symbol of (1)SCR (2)Diac(3)Triac

Answer:

Diagram:

SCR Symbol:        DIAC Symbol:       TRIAC Symbol:
   A                   A1                  MT2
   |                    |                   |
   ▼                    ▼                   ▼
  ┌─┐                  ┌─┐                 ┌─┐
  │ │                  │ │                 │ │
──┤ ├──              ──┤ ├──              ──┤ ├──
  │ │                  │ │                 │ │
  └─┘                  └─┘                 └─┘
   ▲                    ▲                   ▲
   |                    |                   |
   K                   A2                  MT1
  /                                        /
 /                                        /
G                                        G

SCR (Silicon Controlled Rectifier): Three-terminal device with Anode, Cathode, and Gate
DIAC (Diode AC switch): Two-terminal bidirectional device with terminals A1 and A2
TRIAC (Triode AC switch): Three-terminal bidirectional device with MT1, MT2, and Gate

Mnemonic: “AGK for SCR, AA for DIAC, MMG for TRIAC”

Question 1(b) [4 marks]
#

Explain the term(1) CMRR (2) Slew rate

Answer:

Table: Op-Amp Parameters

Parameter	Definition	Significance
CMRR (Common Mode Rejection Ratio)	Ratio of differential gain to common mode gain expressed in dB	Higher CMRR means better rejection of common input signals
Slew Rate	Maximum rate of change of output voltage (V/μs)	Determines how fast op-amp responds to rapidly changing inputs

CMRR formula: CMRR = 20 log₁₀(Ad/Acm) dB
Slew Rate importance: Affects high-frequency performance and prevents distortion

Mnemonic: “Common Mode Rejected Rapidly, Slew shows Signal Speed”

Question 1(c) [7 marks]
#

Draw and explain summing amplifier.

Answer:

Diagram:

graph LR
    V1 -- R1 --> A
    V2 -- R2 --> A
    V3 -- R3 --> A
    A -- Rf --> B[Op-Amp]
    B --> Vout
    B -- - --> A
    A -- + --> Ground

Operation of Summing Amplifier:

Circuit function: Adds multiple input voltages with scaling
Output equation: Vout = -(Rf/R1 × V1 + Rf/R2 × V2 + Rf/R3 × V3)
Inverting configuration: Input signals undergo 180° phase shift
Gain control: Rf/Rn determines weight of each input signal
Application: Audio mixing, analog computation, signal processing
Key feature: Virtual ground at inverting input simplifies analysis

Mnemonic: “Sum with Weights: Vout = -Rf(V1/R1 + V2/R2 + V3/R3)”

Question 1(c OR) [7 marks]
#

Draw and explain DA converter

Answer:

Diagram:

graph LR
    D0 -- 2⁰R --> S1
    D1 -- 2¹R --> S2
    D2 -- 2²R --> S3
    D3 -- 2³R --> S4
    S1 & S2 & S3 & S4 --> A[Summing Amp]
    A --> Vout

R-2R Ladder DAC Operation:

Function: Converts digital binary input to analog output voltage
Working principle: Weighted resistor network creates scaled currents
Binary weighting: Each bit contributes voltage proportional to its position (2ⁿ)
Resolution: Determined by number of bits (N) as 1/2ᴺ of full scale
Advantages: Simple design, good accuracy, fast conversion
Applications: Audio equipment, signal generation, control systems

Mnemonic: “Digital Bits to Analog Steps - R-2R makes the magic”

Question 2(a) [3 marks]
#

Describe thermal run away of transistor.

Answer:

Thermal Runaway Process:

graph TD
    A[Increased Temperature] --> B[Increased Collector Current]
    B --> C[More Power Dissipation]
    C --> A

Definition: Self-accelerating process where transistor heats up and draws more current
Cause: Negative temperature coefficient of base-emitter voltage
Prevention: Use proper heat sink and stabilization circuits

Mnemonic: “Heat feeds Current feeds Heat - a dangerous loop”

Question 2(b) [4 marks]
#

Draw and explain voltage series negative feedback.

Answer:

Diagram:

graph LR
    Vin --> A[Amplifier]
    A --> Vout
    Vout -- Feedback Network --> B[Subtractor]
    B --> A

Voltage Series Negative Feedback:

Parameter	Effect of Negative Feedback
Gain stability	Improved, less dependent on amplifier parameters
Bandwidth	Increased proportional to feedback factor
Distortion	Reduced significantly
Input impedance	Increased

Working principle: Output voltage is sampled and fed back to input
Gain formula: Closed-loop gain = Open-loop gain/(1 + βA)

Mnemonic: “Series says Sample Voltage, Stabilize Gain”

Question 2(c) [7 marks]
#

Draw and explain DC load line for common emitter amplifier.

Answer:

Diagram:

graph TD
    subgraph DC Load Line
    A[Point A: IC=0, VCE=VCC] --> B[Operating Point Q]
    B --> C[Point B: IC=VCC/RC, VCE=0]
    end

DC Load Line Characteristics:

Definition: Graphical representation of all possible operating points
Equation: IC = VCC/RC - VCE/RC
Key points:
- Saturation point (VCE ≈ 0V, IC = VCC/RC)
- Cutoff point (IC ≈ 0mA, VCE = VCC)
- Q-point (selected operating point for amplification)
Significance: Determines biasing stability and output signal limits
Relationship: DC load line is fixed by circuit components (VCC and RC)

Mnemonic: “Connect Cutoff to Saturation for DC Load Line”

Question 2(a OR) [3 marks]
#

Explain operating point(Q-point) in transistor

Answer:

Q-Point (Operating Point):

      |
  Ic  |      DC Load Line
      |          /
      |         /
      |        /
      |       * Q-Point
      |      /
      |     /
      |    /
      |___/____________
          Vce

Definition: Specific DC bias point where transistor operates in active region
Importance: Determines output signal range without distortion
Selection criteria: Center of load line for maximum swing

Mnemonic: “Quality amplification needs Quiet bias at Q-point”

Question 2(b OR) [4 marks]
#

Draw and explain hartley oscillator.

Answer:

Diagram:

graph LR
    A[Transistor] -- Feedback --> B[LC Tank Circuit]
    B --> A
    B -- L1, L2, C --> Output

Hartley Oscillator:

Configuration: Common emitter with tapped inductor feedback
Frequency formula: f = 1/[2π√(C×(L1+L2))]
Phase shift: Ensures 360° total phase shift for oscillation
Feedback: Inductive voltage divider provides positive feedback

Mnemonic: “Hartley Has two coils with inductance for LC oscillation”

Question 2(c OR) [7 marks]
#

Draw and explain AC load line for common emitter amplifier.

Answer:

Diagram:

graph TD
    subgraph AC and DC Load Lines
    A[DC Load Line] --> B[Q-Point]
    B --> C[AC Load Line - Steeper]
    end

AC Load Line Characteristics:

Definition: Represents dynamic operation during signal amplification
Equation: ic = (VCC-VCEQ)/R’c - vce/R’c where R’c = RC||RL
Comparison with DC load line:
- AC load line is steeper than DC load line
- Passes through Q-point
- Determines voltage and current signal swings
Significance: Defines maximum undistorted output signal
Limiting factor: Avoiding saturation and cutoff regions

Mnemonic: “AC Amplitude Controlled by Load line Angle”

Question 3(a) [3 marks]
#

Draw the fixed bias circuit and explain working of it

Answer:

Diagram:

      Vcc
       |
       R
       |
       |C
       |----Output
       |
      /|
     / |
    /--|
   /   |
  |    |
  B    E
  |    |
  Rb   |
  |    |
  |____|
  |
  Vin

Structure: Base resistor connected to VCC, collector resistor for load
Operation: Fixed base current biases transistor
Disadvantage: Poor stability against temperature changes

Mnemonic: “Fixed Bias Feeds Base from power supply”

Question 3(b) [4 marks]
#

In hartley oscillator L1=5mH, L2=10mH, C=0.01µF. Calculate frequency of oscillations.

Answer:

Solution:

Given: L1=5mH, L2=10mH, C=0.01µF
Frequency formula: f = 1/[2π√(C×(L1+L2))]
Calculation:
- Total inductance LT = L1 + L2 = 5mH + 10mH = 15mH = 15×10⁻³ H
- C = 0.01µF = 1×10⁻⁸ F
- f = 1/[2π√(15×10⁻³ × 1×10⁻⁸)]
- f = 1/[2π√(15×10⁻¹¹)]
- f = 1/[2π×3.873×10⁻⁶]
- f = 1/[24.33×10⁻⁶]
- f = 41,101 Hz ≈ 41.1 kHz

Mnemonic: “For Hartley’s frequency, add coils then take square root”

Question 3(c) [7 marks]
#

Draw and explain the frequency response curve of two stage RC coupled amplifier.

Answer:

Diagram:

graph TD
    subgraph Frequency Response
    A[Low Frequency] --> B[Mid Frequency]
    B --> C[High Frequency]
    end

Two-Stage RC Coupled Amplifier Frequency Response:

Low-frequency region: Gain rises with frequency (< 50Hz)
- Limited by coupling and bypass capacitors
Mid-frequency region: Constant maximum gain (50Hz-20kHz)
- Flat response, ideal operating region
High-frequency region: Gain drops with frequency (> 20kHz)
- Limited by transistor capacitances and Miller effect
Bandwidth: Range of frequencies with gain ≥ 70.7% of maximum gain
Cutoff frequencies: Points where gain drops by 3dB (0.707 times max gain)

Mnemonic: “Low-flat-high: capacitors block, amplify well, then roll off”

Question 3(a OR) [3 marks]
#

Explain in detail barkhausen criterion for oscillation.

Answer:

Barkhausen Criterion:

Condition	Requirement
Loop Gain	Must equal exactly 1 (Aβ = 1)
Phase Shift	Must be 0° or 360° around loop

Purpose: Ensures sustained oscillations without damping
Consequences:
- If Aβ < 1: Oscillations die out
- If Aβ > 1: Oscillations grow until limited by nonlinearity
- If Aβ = 1: Stable oscillations maintained

Mnemonic: “Barkhausen’s Balance: Loop Gain=1, Phase=360°”

Question 3(b OR) [4 marks]
#

Explain the effect of negative feedback on the gain of amplifier

Answer:

Effect of Negative Feedback on Amplifier Gain:

Parameter	Without Feedback	With Feedback
Voltage Gain	A	A/(1+Aβ)
Stability	Less stable	More stable
Bandwidth	Lower	Higher
Distortion	Higher	Lower

Gain reduction: Gain decreases by factor (1+Aβ)
Gain-bandwidth tradeoff: Bandwidth increases as gain decreases
Gain stabilization: Less affected by temperature and component variations

Mnemonic: “Negative Feedback: Less Gain, More Stability”

Question 3(c OR) [7 marks]
#

Draw fan regulator circuit and explain how it will control the speed of fan.

Answer:

Diagram:

graph LR
    A[AC Supply] --> B[DIAC]
    B --> C[TRIAC]
    C --> D[Fan]
    E[Variable Resistor] --> F[RC Network]
    F --> B

Fan Regulator Operation:

Control method: Phase angle control using TRIAC and DIAC
Working principle: RC network creates variable phase shift
Speed control: Variable resistor adjusts RC time constant
Operation sequence:
- RC network delays DIAC firing
- DIAC triggers TRIAC at adjustable point in AC cycle
- TRIAC conducts for remaining portion of AC half-cycle
- Less conduction time = lower power to fan = slower speed
Advantages: Simple design, smooth control, energy efficient
Applications: Ceiling fans, exhaust fans, cooling systems

Mnemonic: “Delay the TRIAC firing, control fan’s speed”

Question 4(a) [3 marks]
#

Write short note on natural commutation

Answer:

Natural Commutation:

Definition: SCR turns off automatically when current falls below holding current
Process: Occurs in AC circuits at each zero-crossing point
Requirements: No external components needed, inherent to AC operation

Mnemonic: “Natural Commutation: Zero Current Crossings Turn Off Thyristors”

Question 4(b) [4 marks]
#

Explain the parameters gain and bandwidth of amplifier.

Answer:

Amplifier Parameters:

Parameter	Definition	Formula
Gain (A)	Ratio of output to input signal	A = Vout/Vin
Bandwidth (BW)	Frequency range with gain ≥ 70.7% of maximum	BW = fH - fL

Gain-bandwidth product: Remains constant (GBP = Gain × Bandwidth)
Cutoff frequencies: Lower (fL) and higher (fH) frequencies where gain drops by 3dB
Significance: Determines amplifier’s ability to handle different frequencies

Mnemonic: “Good Amplifiers Balance Width and Magnitude”

Question 4(c) [7 marks]
#

Draw the construction and characteristics of triac and describe working of it, also write the application of triac.

Answer:

TRIAC Construction and Characteristics:

           MT2
            |
      ------+------
     /      |      \
    /  P    |    N  \
   +--------+--------+
   |        |        |
   |    N   |    P   |
   +--------+--------+
   |        |        |
   |    P   |    N   |
   +--------+--------+
    \       |       /
     \      |      /
      ------+------
            |
           MT1
            |
            G

I-V Characteristics:

    I
    ^
    |      /|
    |     / |
    |    /  |
    |---+---|----> V
    |   /   |
    |  /    |
    | /     |

TRIAC Operation:

Structure: Five-layer PNPN bidirectional device
Switching: Conducts in both directions when triggered
Triggering modes: Four quadrant operation possible
Turn-off: Natural commutation at current zero-crossing

Applications:

Light dimmers
Fan speed controllers
Heater controls
Motor speed regulation
AC power switching

Mnemonic: “TRIAC Takes AC Control in Both Directions”

Question 4(a OR) [3 marks]
#

Write any three application of SCR.

Answer:

Applications of SCR:

Application	Function
DC Motor Speed Control	Provides variable DC to motors
Battery Chargers	Regulates charging current
Power Inverters	Converts DC to AC efficiently

Advantages: High power handling, efficient control, robust operation
Limitations: Requires forced commutation in DC circuits

Mnemonic: “SCR Controls DC - Motors, Batteries, Inverters”

Question 4(b OR) [4 marks]
#

Explain holding current and latching current with reference to SCR

Answer:

SCR Current Parameters:

Parameter	Definition	Typical Values
Holding Current (IH)	Minimum current to maintain conduction	5-40 mA
Latching Current (IL)	Minimum current to establish conduction	10-100 mA

Latching current: Must be exceeded briefly after triggering for SCR to latch
Holding current: Must be maintained to keep SCR in conduction
Relationship: Usually IL > IH
Significance: Critical for reliable switching operation

Mnemonic: “Latch with more, Hold with less, both keep SCR conducting”

Question 4(c OR) [7 marks]
#

Draw and explain in detail block diagram of operational amplifier.

Answer:

Operational Amplifier Block Diagram:

graph LR
    A[Input Differential Stage] --> B[Intermediate Stage]
    B --> C[Output Stage]
    D[Bias Circuit] --> A & B & C
    E[Frequency Compensation] --> B

Op-Amp Blocks and Functions:

Input differential stage:
- High input impedance
- Rejects common-mode signals
- Provides differential voltage gain
Intermediate stage:
- Additional voltage gain
- Level shifting
- Frequency compensation
Output stage:
- Low output impedance
- Current amplification
- Power capability for driving loads
Bias circuit:
- Establishes proper operating points
- Temperature stability
Frequency compensation:
- Prevents oscillation
- Controls frequency response

Mnemonic: “Differential Input, Gain in Middle, Power at Output”

Question 5(a) [3 marks]
#

Draw and explain in brief inverting amplifier.

Answer:

Inverting Amplifier Circuit:

          Rf
          ___
    Vin---| |-----+
          ---     |
                  |
                 _|_
    +------+    /   \
    |      |---+     +---Vout
    |      |    \___/
Vin-+      |      |
    |Op-Amp|      |
    +------+      |
                  |
                 ---
                 ///

Gain formula: Vout = -(Rf/Rin) × Vin
Operation: Input signal inverted with amplification
Virtual ground: Inverting input maintained at 0V

Mnemonic: “Inverting means Negative Gain equals -Rf/Rin”

Question 5(b) [4 marks]
#

Draw and explain the block diagram of regulated power supply.

Answer:

Regulated Power Supply Block Diagram:

graph LR
    A[Transformer] --> B[Rectifier]
    B --> C[Filter]
    C --> D[Regulator]
    D --> E[Output]
    F[Reference] --> D
    G[Feedback] --> D

Regulated Power Supply Stages:

Transformer: Steps down AC voltage to required level
Rectifier: Converts AC to pulsating DC (diode bridge)
Filter: Smooths pulsating DC (capacitors)
Regulator: Maintains constant output despite variations
Reference: Provides stable comparison voltage
Feedback: Monitors output and adjusts regulation

Mnemonic: “Transform, Rectify, Filter, Regulate for Stable DC”

Question 5(c) [7 marks]
#

Draw and explain astable multivibrator.

Answer:

Astable Multivibrator Using 555 Timer:

graph TD
    subgraph 555 Timer
    A[Threshold] --> B[Flip-Flop]
    C[Trigger] --> B
    B --> D[Output]
    end
    E[R1] & F[R2] & G[C] --> A & C

Operation of Astable Multivibrator:

Configuration: Free-running oscillator with no stable states
Timing components: External R1, R2, and C
Oscillation process:
- Capacitor charges through R1+R2
- Capacitor discharges through R2
- Continuous charging/discharging cycle
Output waveform: Rectangular with duty cycle based on R1/R2 ratio
Frequency formula: f = 1.44/((R1+2R2)×C)
Applications: Clock generation, LED flashers, tone generators
Advantages: Simple design, stable frequency, adjustable duty cycle

Mnemonic: “Always Switching, Time set by RC, Both states Least stable”

Question 5(a OR) [3 marks]
#

In an op amp non-inverting amplifier R1=2kΩ and Rf=200kΩ. Find the voltage gain of non-inverting amplifier.

Answer:

Solution:

Given: R1 = 2kΩ, Rf = 200kΩ
Non-inverting amplifier gain formula: A = 1 + (Rf/R1)
Calculation:
- A = 1 + (200kΩ/2kΩ)
- A = 1 + 100
- A = 101
Result: Voltage gain of non-inverting amplifier is 101
Significance: Output voltage will be 101 times the input voltage

Mnemonic: “Non-inverting amplifier gain: One plus Feedback over Ground”

Question 5(b OR) [4 marks]
#

Draw and explain in brief circuit to get -5V regulated dc output voltage.

Answer:

Negative Voltage Regulator Circuit:

     +--------+
     |        |
Vin--+        +---Vout (-5V)
     | 7905   |
     |        |
     +--------+
         |
        ---
        ///

Circuit Operation:

Key component: 7905 negative voltage regulator IC
Input requirement: Negative DC voltage (typically -7V to -25V)
Filtering: Input and output capacitors for stability
Regulation method: Series pass element with feedback control
Output characteristics: Fixed -5V with current up to 1A

Mnemonic: “79XX for Negative, 78XX for Positive regulated voltage”

Question 5(c OR) [7 marks]
#

Draw and explain the block diagram of SMPS.

Answer:

SMPS Block Diagram:

graph LR
    A[AC Input] --> B[EMI Filter]
    B --> C[Rectifier & Filter]
    C --> D[High-Frequency Inverter]
    D --> E[Transformer]
    E --> F[Output Rectifier]
    F --> G[Output Filter]
    G --> H[DC Output]
    I[Feedback & Control] --> D
    H --> I

SMPS Operation: