Linear Integrated Circuit (4341105) - Summer 2024 Solution

Table of Contents

Question 1(a) [3 marks]
#

State and explain the difference between positive and negative feedback with diagram.

Answer:

Parameter	Negative Feedback	Positive Feedback
Signal	Output signal is fed back to input with opposite phase	Output signal is fed back to input with same phase
Gain	Decreases	Increases
Stability	Improves	Reduces
Applications	Amplifiers	Oscillators

Diagram:

graph LR
    A[Input] --> B[Amplifier]
    B --> C[Output]
    C --> D{Feedback Network}
    
    %% Negative Feedback
    subgraph Negative Feedback
    D -->|180° Phase Shift| E[Subtractor]
    E --> B
    end
    
    %% Positive Feedback
    subgraph Positive Feedback
    D -->|0° Phase Shift| F[Adder]
    F --> B
    end

Phase relationship: In negative feedback, signal is 180° out of phase while in positive feedback, signal is in phase
Purpose: Negative feedback stabilizes system while positive feedback creates oscillations

Mnemonic: “Negative Needs Stability, Positive Produces Oscillations”

Question 1(b) [4 marks]
#

Explain the effect of negative feedback on input impedance of the Amplifier.

Answer:

Type of Feedback	Effect on Input Impedance	Formula
Voltage Series	Increases	Z(in-f) = Z(in)(1+Aβ)
Current Series	Increases	Z(in-f) = Z(in)(1+Aβ)
Voltage Shunt	Decreases	Z(in-f) = Z(in)/(1+Aβ)
Current Shunt	Decreases	Z(in-f) = Z(in)/(1+Aβ)

Diagram:

graph LR
    A[Input Signal] --> B[Input Impedance]
    B --> C[Amplifier]
    C --> D[Output]
    D --> E[Feedback Network]
    E --> F[Summing Point]
    F --> B
    style B fill:#f9f,stroke:#333,stroke-width:2px

Series feedback: When feedback signal is in series with input, input impedance increases
Shunt feedback: When feedback signal is in parallel with input, input impedance decreases
Magnitude: Change is proportional to (1+Aβ) where A is gain and β is feedback factor

Mnemonic: “Series Soars, Shunt Shrinks”

Question 1(c) [7 marks]
#

List the advantages and Disadvantages of negative feedback.

Answer:

Advantages	Disadvantages
Stabilizes gain	Reduces overall gain
Increases bandwidth	Requires additional components
Reduces distortion	May cause oscillation if improperly designed
Reduces noise	Requires careful phase compensation
Improves input/output impedance	Increases power consumption
Reduces temperature sensitivity	Makes circuit more complex
Controls frequency response	May reduce signal-to-noise ratio in some cases

Diagram:

graph TD
    A[Negative Feedback] --> B[Advantages]
    A --> C[Disadvantages]
    
    B --> D[Stable Gain]
    B --> E[Wider Bandwidth]
    B --> F[Lower Distortion]
    B --> G[Better Impedance]
    
    C --> H[Reduced Gain]
    C --> I[More Components]
    C --> J[Complex Design]

Performance tradeoff: Sacrifices gain to achieve better stability and linearity
Frequency considerations: May require compensation to prevent oscillations at high frequencies
Design complexity: More complex to design properly but offers better long-term performance

Mnemonic: “Stability Grows As Gain Drops”

Question 1(c) OR [7 marks]
#

Explain Voltage series feedback amplifier in detail with block diagram and draw the Practical voltage series feedback circuit.

Answer:

Parameter	Effect in Voltage Series Feedback
Input signal	Voltage
Feedback signal	Voltage
Input impedance	Increases
Output impedance	Decreases
Gain stability	Improves
Bandwidth	Increases

Diagram:

graph TD
    A[Input Vi] --> B[+]
    B --> C[Amplifier A]
    C --> D[Output Vo]
    D --> E[Feedback Network β]
    E --> F[-]
    F --> B
    
    style C fill:#bbf,stroke:#333,stroke-width:1px
    style E fill:#fbb,stroke:#333,stroke-width:1px

Practical Circuit:

Sampling method: Output voltage is sampled and fed back to input
Mixing method: Feedback signal is mixed in series with input signal
Working principle: Reduces gain for improved stability and linearity
Applications: Audio amplifiers, instrumentation amplifiers

Mnemonic: “Voltage Series - Impedance In Up, Out Down”

Question 2(a) [3 marks]
#

Write short note on Colpitts oscillator circuit.

Answer:

Component	Function
LC Tank	Determines oscillation frequency
Capacitive Voltage Divider	Provides feedback
Active Device	Provides gain to sustain oscillations

Diagram:

Frequency formula: f = 1/(2π√(L×(C1×C2)/(C1+C2)))
Feedback: Provided by capacitive voltage divider (C1 and C2)
Applications: RF oscillators, communication circuits

Mnemonic: “Colpitts Contains Capacitive divider”

Question 2(b) [4 marks]
#

Explain requirement of oscillator. i) Barkhausen Criterion. ii) Tank circuit. iii) Amplifier.

Answer:

Requirement	Function	Explanation
Barkhausen Criterion	Ensures sustained oscillation	Loop gain = 1, Phase shift = 0° or 360°
Tank Circuit	Determines frequency	Resonant LC circuit that stores energy
Amplifier	Provides gain	Compensates for circuit losses

Diagram:

graph TD
    A[Oscillator] --> B[Barkhausen Criterion]
    A --> C[Tank Circuit]
    A --> D[Amplifier]
    
    B --> E[Loop Gain = 1]
    B --> F[Phase Shift = 0° or 360°]
    
    C --> G[Energy Storage]
    C --> H[Frequency Determination]
    
    D --> I[Overcome Losses]
    D --> J[Maintain Amplitude]

Barkhausen Criterion: Mathematical condition for sustained oscillations without damping
Tank Circuit: LC circuit that determines frequency of oscillations
Amplifier: Active device that provides energy to maintain oscillations

Mnemonic: “BAT - Barkhausen Amplifies Tank”

Question 2(c) [7 marks]
#

Explain construction, working and V-I characteristics of UJT.

Answer:

Parameter	Description
Construction	Silicon bar with two base connections and one emitter
Symbol	Triangle with emitter on one side and two bases
Equivalent Circuit	Voltage divider with diode
Key Parameter	Intrinsic standoff ratio (η)

Diagram:

V-I Characteristic Curve:

Construction: N-type silicon bar with P-type emitter junction
Working principle: When emitter voltage > (η×VBB), device conducts
Regions of operation: Cut-off, negative resistance, and saturation
Applications: Relaxation oscillators, timing circuits, triggering devices

Mnemonic: “UJT Peaks Then Valleys - Negative Resistance Rules”

Question 2(a) OR [3 marks]
#

State the advantages, disadvantages and applications of Hartley oscillator.

Answer:

Advantages	Disadvantages	Applications
Easy tuning	Bulky inductors	RF generators
Wide frequency range	Mutual inductance issues	Radio receivers
Simple design	Difficult at high frequencies	Amateur radio
Good frequency stability	Requires center-tapped coil	Communication equipment

Diagram:

Key feature: Uses tapped inductor for feedback
Frequency formula: f = 1/(2π√(C×(L1+L2)))
Distinguishing characteristic: Inductive voltage divider for feedback

Mnemonic: “Hartley Has tapped Inductor”

Question 2(b) OR [4 marks]
#

Explain UJT as relaxation oscillator.

Answer:

Component	Function
UJT	Provides switching action
Capacitor	Timing element
Resistor	Controls charging rate
Output	Sawtooth waveform

Diagram:

Waveforms:

Operating principle: Capacitor charges until UJT firing voltage, then rapidly discharges
Frequency formula: f ≈ 1/(RC×ln(1/(1-η)))
Applications: Timing circuits, pulse generators, control systems

Mnemonic: “Charge-Fire-Repeat - Sawtooth’s Beat”

Question 2(c) OR [7 marks]
#

Explain working of weinbridge oscillator with neat diagram also state the advantage, disadvantage and application for the same.

Answer:

Parameter	Description
Configuration	RC feedback network in bridge formation
Frequency Formula	f = 1/(2πRC) when R1=R3 and C2=C4
Feedback	Positive feedback through RC network
Phase Shift	0° at resonant frequency

Diagram:

graph TD
    A[Amplifier] --> B[RC Bridge]
    B --> A
    
    subgraph "Wien Bridge Network"
    C[R1] --- D[C1]
    D --- E[R2]
    E --- F[C2]
    F --- C
    end

Circuit:

Advantages:

High frequency stability
Low distortion output
Simple RC components
Easy to tune

Disadvantages:

Limited frequency range
Amplitude stabilization needed
Sensitive to component variations
Difficult to start oscillations

Applications:

Audio test equipment
Function generators
Musical instruments
Laboratory signal sources

Mnemonic: “Wien Works at R1C1=R2C2 frequency”

Question 3(a) [3 marks]
#

Give classification of power Amplifier.

Answer:

Classification Basis	Types
Based on Conduction Angle	Class A, B, AB, C
Based on Configuration	Single-ended, Push-pull, Complementary
Based on Coupling	RC coupled, Transformer coupled, Direct coupled
Based on Operation	Linear, Switching

Diagram:

graph TD
    A[Power Amplifiers]
    A --> B[Class A - 360°]
    A --> C[Class B - 180°]
    A --> D[Class AB - 180°-360°]
    A --> E[Class C < 180°]
    
    style B fill:#d4f0f0,stroke:#333
    style C fill:#d4f0f0,stroke:#333
    style D fill:#d4f0f0,stroke:#333
    style E fill:#d4f0f0,stroke:#333

Class A: Conducts for full 360° cycle, highest linearity, lowest efficiency
Class B: Conducts for 180° cycle, medium distortion, medium efficiency
Class AB: Conducts for 180°-360° cycle, good linearity, good efficiency
Class C: Conducts for <180° cycle, highest distortion, highest efficiency

Mnemonic: “A All-time, B Bisects, AB Almost-Bisects, C Cuts-more”

Question 3(b) [4 marks]
#

Explain class A power amplifier.

Answer:

Parameter	Class A Amplifier
Conduction Angle	360° (full cycle)
Biasing	Q-point at center of load line
Efficiency	Low (25-30% max)
Distortion	Very low

Diagram:

Load Line:

Operating principle: Transistor conducts for entire input cycle
Efficiency calculation: Maximum theoretical efficiency = 50%
Practical efficiency: Typically 25-30% due to losses
Applications: Audio pre-amplifiers, low-power amplifiers where quality matters more than efficiency

Mnemonic: “Class A - Always conducting, All cycle”

Question 3(c) [7 marks]
#

Explain the principle of push pull amplifiers and write short note on class B push pull amplifier.

Answer:

Push-Pull Principle	Class B Push-Pull
Uses two complementary devices	Each transistor conducts for half cycle
Reduces even harmonic distortion	Higher efficiency (78.5% theoretical)
Cancels DC magnetization in transformer	Suffers from crossover distortion
Provides higher output power	Requires proper biasing to minimize distortion

Diagram:

Waveforms:

Working principle: Each transistor conducts for alternate half-cycles
Advantages: Higher efficiency, reduced even harmonics, lower heat generation
Disadvantages: Crossover distortion at transition points
Applications: Audio power amplifiers, output stages of high-power systems

Mnemonic: “Push-Pull: Pair Processes alternate Pulses”

Question 3(a) OR [3 marks]
#

Discuss crossover distortion in push pull amplifier. How it can be removed.

Answer:

Crossover Distortion	Solution Methods
Occurs at signal crossover points	Apply small bias voltage (Class AB)
Due to transistor’s non-linear region	Use diode compensation networks
Creates “dead zone” around zero	Implement feedback correction
Affects small signals more	Use complementary emitter-follower stage

Diagram:

Correction Circuit:

Cause: Transistors require ~0.7V to turn on, creating dead zone
Effect: Distortion particularly noticeable at low volumes
Solution: Class AB biasing with diodes or VBE multiplier
Result: Smoother transition between positive and negative half-cycles

Mnemonic: “Cross to Class AB Smooths the Gap”

Question 3(b) OR [4 marks]
#

Explain complimentary symmetry push-pull amplifier.

Answer:

Component	Purpose
NPN Transistor	Handles positive half-cycle
PNP Transistor	Handles negative half-cycle
Biasing Network	Reduces crossover distortion
Output Coupling	Direct coupling to load

Diagram:

Working Principle:

graph TD
    A[Input Signal] --> B{Voltage Polarity}
    B -->|Positive| C[NPN Conducts]
    B -->|Negative| D[PNP Conducts]
    C --> E[Output]
    D --> E

Key feature: Uses complementary transistors (NPN and PNP) for push-pull operation
Advantage: No output transformer required, direct coupling to load
Efficiency: Typically 78.5% theoretical maximum
Applications: Audio amplifiers, power output stages

Mnemonic: “NPN Pulls-up, PNP Pulls-down”

Question 3(c) OR [7 marks]
#

Derive the equation of efficiency for class B push pull Amplifier.

Answer:

Parameter	Formula	Description
DC Input Power	PDC = 2VCC×IDC	Power drawn from supply
AC Output Power	PAC = Vrms²/RL	Power delivered to load
Maximum Efficiency	η = (π/4)×100% = 78.5%	Theoretical maximum
Practical Efficiency	60-70%	Considering losses

Mathematical Derivation:

For a sinusoidal input: v(t) = Vm sin(ωt)

Step 1: DC Input Power

Input current per transistor: Im/π
Total DC input power: PDC = 2VCC×Im/π

Step 2: AC Output Power

RMS output voltage: Vrms = Vm/√2
Maximum output voltage: Vm = VCC
Output power: PAC = Vrms²/RL = Vm²/2RL

Step 3: Efficiency Calculation

η = (PAC/PDC)×100%
η = ((Vm²/2RL)/(2VCC×Im/π))×100%
Since Vm = VCC and Im = VCC/RL
η = (π/4)×100% = 78.5%

Diagram:

Power dissipation: Most efficient when output voltage swing approaches VCC
Conduction angle: Each transistor conducts for exactly 180°
Practical factors: Biasing current, saturation voltage, and other losses reduce efficiency
Comparison: Much higher than Class A (25-30%), less than Class C (>80%)

Mnemonic: “Pi-over-4 gives 78.5% - Class B’s best”

Question 4(a) [3 marks]
#

Define.(i) CMRR (ii)slew rate.(iii)Input offset Current.

Answer:

Parameter	Definition	Typical Values
CMRR	Ratio of differential gain to common-mode gain	80-120 dB
Slew Rate	Maximum rate of change of output voltage	0.5-20 V/μs
Input Offset Current	Difference between currents into the two inputs	1-100 nA

Diagram:

graph TD
    A[Op-Amp Parameters]
    A --> B[CMRR = Ad/Acm]
    A --> C[Slew Rate = dVo/dt]
    A --> D[IOS = |I+ - I-|]
    
    style B fill:#f9f9f9,stroke:#333
    style C fill:#f9f9f9,stroke:#333
    style D fill:#f9f9f9,stroke:#333

CMRR: Measures op-amp’s ability to reject common-mode signals
Slew Rate: Limits maximum frequency for undistorted output
Input Offset Current: Causes output error even with identical inputs

Mnemonic: “Cancelling Mistakes Requires Ratios”

Question 4(b) [4 marks]
#

Draw and explain the basic block diagram of an operational amplifier.

Answer:

Stage	Function
Differential Input	Accepts and amplifies difference between inputs
High-Gain Intermediate	Provides voltage amplification
Level Shifter	Shifts DC level for output stage
Output Buffer	Provides low output impedance

Diagram:

graph LR
    A[Inverting Input] --> B[Differential Input Stage]
    C[Non-inverting Input] --> B
    B --> D[High-Gain Intermediate Stage]
    D --> E[Level Shifter]
    E --> F[Output Buffer]
    F --> G[Output]
    
    style B fill:#d4f0f0,stroke:#333
    style D fill:#d4f0f0,stroke:#333
    style E fill:#d4f0f0,stroke:#333
    style F fill:#d4f0f0,stroke:#333

Differential input stage: Converts differential input to single-ended output
High-gain stage: Provides most of the open-loop gain
Level shifter: Shifts signal level for proper output operation
Output stage: Provides current gain and low output impedance

Mnemonic: “Diff-Amp Gain Shift Out”

Question 4(c) [7 marks]
#

Explain in detail operational amplifier as integrator.

Answer:

Parameter	Description	Formula
Circuit	Op-amp with capacitor in feedback	-
Transfer Function	Output proportional to integral of input	Vo = -(1/RC)∫Vi dt
Frequency Response	Acts as low-pass filter	Gain = 1/(jωRC)
Phase Shift	-90°	-

Diagram:

Input/Output Waveforms:

Working principle: Capacitor integrates current over time
Mathematical basis: Vo(t) = -(1/RC)∫Vi(t)dt + Vo(0)
Limitations: Capacitor leakage, op-amp input bias current cause drift
Applications: Waveform generators, analog computers, active filters

Mnemonic: “Square-In Triangle-Out, RC sets the Slope”

Question 4(a) OR [3 marks]
#

Explain operational amplifier as summing amplifier.

Answer:

Parameter	Description	Formula
Circuit	Multiple inputs with same feedback	Vo = -(R₁/R₁×V₁ + R₁/R₂×V₂ + …)
Equal Resistors	Simple addition/averaging	Vo = -(V₁ + V₂ + … + Vₙ)
Weighted Sum	Different input resistors	Vo = -(K₁V₁ + K₂V₂ + … + KₙVₙ)
Inverting	Output inverted from inputs	-

Diagram:

Working principle: Each input contributes current to summing junction
Applications: Audio mixers, signal processing, analog computers
Virtual ground: Summing point maintains near-zero voltage
Variations: Inverting, non-inverting, and differential summer

Mnemonic: “Many Inputs, One Output - Sum It All”

Question 4(b) OR [4 marks]
#

State the applications of operational amplifier.

Answer:

Application Category	Examples
Signal Processing	Amplifiers, Filters, Buffers
Mathematical Operations	Adders, Subtractors, Integrators, Differentiators
Waveform Generators	Sine, Square, Triangle, Pulse generators
Instrumentation	Instrumentation amplifiers, Current-to-voltage converters
Comparators	Zero crossing detectors, Window comparators
Precision Rectifiers	Full-wave, Half-wave rectifiers
Voltage Regulators	Series regulators, Shunt regulators

Diagram:

graph TD
    A[Op-Amp Applications]
    A --> B[Signal Processing]
    A --> C[Math Operations]
    A --> D[Waveform Generators]
    A --> E[Instrumentation]
    A --> F[Comparators]
    A --> G[Rectifiers]
    A --> H[Regulators]

Linear applications: Utilize op-amp in linear region for amplification, filtering
Non-linear applications: Use saturation characteristics for comparison, limitation
Analog computation: Perform mathematical operations on analog signals
Signal conditioning: Adapt signals for analog-to-digital conversion

Mnemonic: “SMWIG-CR: Signal, Math, Wave, Instrument, Gate, Convert, Regulate”

Question 4(c) OR [7 marks]
#

Explain op-amp as inverting and non-inverting amplifier.

Answer:

Parameter	Inverting Amplifier	Non-Inverting Amplifier
Circuit Configuration	Input to negative terminal	Input to positive terminal
Gain Formula	A = -Rf/Rin	A = 1 + Rf/Rin
Input Impedance	= Rin	Very high (≈ 10⁹ ohms)
Phase Shift	180°	0°
Virtual Ground	At negative input	Not applicable

Inverting Amplifier:

Non-Inverting Amplifier:

Inverting Mode:

Gain equation: Vout = -(Rf/Rin)×Vin
Virtual ground: Negative input maintained at ~0V
Applications: Signal inversion, controlled gain, summing

Non-Inverting Mode:

Gain equation: Vout = (1 + Rf/Rin)×Vin
Minimum gain: Always ≥ 1
Applications: Buffering, voltage amplification with high input impedance

Mnemonic: “Invert: Negative is Input, Non-invert: Positive gets signal”

Question 5(a) [3 marks]
#

Give pin description of IC555.

Answer:

Pin Number	Pin Name	Description
1	Ground	Connected to circuit ground
2	Trigger	Starts timing cycle when < 1/3 VCC
3	Output	Provides output signal
4	Reset	Terminates timing when LOW
5	Control Voltage	Adjusts threshold voltage
6	Threshold	Ends timing cycle when > 2/3 VCC
7	Discharge	Connected to timing capacitor
8	VCC	Positive supply voltage (5-15V)

Diagram:

Input pins: Trigger, Reset, Threshold, Control Voltage
Output pins: Output, Discharge
Power pins: VCC, Ground
Internal structure: Composed of comparators, flip-flop, discharge transistor

Mnemonic: “Ground Triggers Output Reset Control Threshold Discharges Voltage”

Question 5(b) [4 marks]
#

Explain op-amp as differentiator.

Answer:

Parameter	Description	Formula
Circuit	Op-amp with capacitor in input	Vo = -RC(dVi/dt)
Transfer Function	Output proportional to rate of change	H(s) = -sRC
Frequency Response	Acts as high-pass filter	Gain increases with frequency
Phase Shift	+90°	-

Diagram:

Input/Output Waveforms:

Working principle: Output voltage proportional to rate of change of input
Mathematical basis: Vo = -RC(dVin/dt)
Practical limitations: Sensitive to high-frequency noise
Applications: Waveform generation, edge detection, rate-of-change indicator

Mnemonic: “Differentiator Delivers Derivatives - RC determines speed”

Question 5(c) [7 marks]
#

Explain IC 555 as astable and Monostable multivibrator.

Answer:

Parameter	Astable Multivibrator	Monostable Multivibrator
Definition	Free-running oscillator	One-shot pulse generator
Stable States	None (continuously oscillates)	One stable state
Timing	T = 0.693(RA+2RB)C	T = 1.1RC
Trigger	Self-triggering	External trigger required
Output	Continuous square wave	Single pulse of fixed width

Astable Circuit:

Monostable Circuit:

Astable Operation:

Working: Capacitor charges through RA+RB and discharges through RB
Duty cycle: Can be adjusted by proper selection of RA and RB
Frequency: f = 1.44/((RA+2RB)C)
Applications: LED flashers, tone generators, clock pulse generators

Monostable Operation:

Working: Triggered by falling edge on pin 2, outputs HIGH for time T
Time period: T = 1.1RC
Applications: Time delays, pulse width modulation, debouncing

Mnemonic: “Astable Always Alternates, Monostable Makes One pulse”

Question 5(a) OR [3 marks]
#

Explain IC555 as Bistable multivibrator.

Answer:

Parameter	Description
Definition	Flip-flop circuit with two stable states
Triggering	SET by trigger pin (2), RESET by reset pin (4)
Stable States	Two (HIGH or LOW)
Time Period	No timing components needed

Diagram:

Truth Table:

Trigger (Pin 2)	Reset (Pin 4)	Output (Pin 3)
< 1/3 VCC	HIGH	HIGH
> 1/3 VCC	HIGH	No change
Any	LOW	LOW

SET operation: Occurs when trigger pin falls below 1/3 VCC
RESET operation: Occurs when reset pin is pulled LOW
Applications: Latching switches, memory elements, flip-flops
Features: No timing components (R, C) required

Mnemonic: “Bistable Bounces Between two states”

Question 5(b) OR [4 marks]
#

Explain the basic operation of IC555 with internal block diagram.

Answer:

Block	Function
Comparators	Monitor trigger and threshold voltages
Flip-Flop	Controls output state
Discharge Transistor	Discharges timing capacitor
Voltage Divider	Establishes reference voltages

Internal Block Diagram:

Basic Operation:

Voltage divider: Creates 2/3 VCC and 1/3 VCC reference points
Comparator 1: Triggers when pin 2 goes below 1/3 VCC
Comparator 2: Resets when pin 6 goes above 2/3 VCC
Flip-flop: Controls output state based on comparator inputs
Discharge transistor: Connects pin 7 to ground when output is LOW

Versatility: Can be configured in multiple modes (astable, monostable, bistable)
Timing precision: Determined by external RC components
Wide supply range: Functions from 4.5V to 16V

Mnemonic: “Comparators Control Flip-flop For Timing”