Linear Integrated Circuit (4341105) - Summer 2025 Solution

Table of Contents

Question 1(a) [3 marks]
#

Explain effect of negative feedback on gain and stability.

Answer: Negative feedback significantly improves amplifier performance by reducing gain but enhancing stability and other parameters.

Table:

Parameter	Effect of Negative Feedback
Gain	Reduces overall gain
Stability	Increases stability
Bandwidth	Increases bandwidth

Gain reduction: Makes amplifier more predictable
Stability improvement: Reduces oscillations and distortions
Better control: Provides consistent performance

Mnemonic: “Gain Goes Down, Stability Stays Strong”

Question 1(b) [4 marks]
#

State different types of feedback amplifier and advantages of negative feedback amplifier.

Answer: Four basic feedback topologies exist based on input and output connections.

Table:

Type	Input Connection	Output Connection
Voltage Series	Series	Voltage
Voltage Shunt	Shunt	Voltage
Current Series	Series	Current
Current Shunt	Shunt	Current

Advantages:

Reduced distortion: Minimizes harmonic content
Increased bandwidth: Better frequency response
Improved stability: Consistent operation

Mnemonic: “Very Smart Current Control”

Question 1(c) [7 marks]
#

Derive an equation for overall gain of negative feedback voltage amplifier.

Answer: For negative feedback amplifier, output is fed back to input in opposite phase.

Circuit Analysis: Let A = Open loop gain, β = Feedback factor

Diagram:

graph LR
    A[Input Vi] --> B[Amplifier A]
    B --> C[Output Vo]
    C --> D[Feedback β]
    D --> E[Summing Junction]
    A --> E

Derivation:

Input to amplifier: Vi - βVo
Output: Vo = A(Vi - βVo)
Vo = AVi - AβVo
Vo + AβVo = AVi
Vo(1 + Aβ) = AVi
Overall Gain: Af = A/(1 + Aβ)

Key Points:

Denominator (1 + Aβ): Called loop gain
Stability factor: Determines system response
Gain reduction: Traded for better performance

Mnemonic: “Always Divide by (1 + Loop)”

Question 1(c OR) [7 marks]
#

Draw and explain current shunt type negative feedback amplifier and Derive the formula of input impedance and output impedance of it.

Answer: Current shunt feedback samples output current and feeds back voltage in shunt with input.

Circuit Diagram:

graph TD
    A[Vi] --> B[+]
    B --> C[Amplifier A]
    C --> D[Ro]
    D --> E[RL]
    D --> F[Feedback Network β]
    F --> G[-]
    B --> G

Analysis:

Feedback type: Current sampling, voltage mixing
Input impedance: Decreases due to shunt feedback
Output impedance: Decreases due to current sampling

Formulas:

Input Impedance: Zif = Zi/(1 + Aβ)
Output Impedance: Zof = Zo/(1 + Aβ)

Characteristics:

Low input impedance: Good for current sources
Low output impedance: Good for voltage output
Current-to-voltage converter: Useful in applications

Mnemonic: “Current Shunt Lowers Both Impedances”

Question 2(a) [3 marks]
#

Explain Barkhausen criteria for oscillations.

Answer: For sustained oscillations in feedback circuits, two conditions must be satisfied simultaneously.

Table:

Criteria	Condition	Description
Magnitude	\|Aβ\| = 1	Loop gain unity
Phase	∠Aβ = 0° or 360°	Zero phase shift

Unity loop gain: Ensures signal maintains amplitude
Zero phase shift: Ensures positive feedback
Sustained oscillation: Both conditions create self-sustaining signals

Mnemonic: “One Magnitude, Zero Phase”

Question 2(b) [4 marks]
#

Explain use of tank circuit with neat diagram.

Answer: Tank circuit provides frequency selective positive feedback for oscillator circuits.

Circuit Diagram:

Operation: At resonant frequency, LC tank circuit exhibits:

Table:

Parameter	Value	Effect
Reactance	XL = XC	Resonance
Impedance	Maximum	High selectivity
Phase	0°	Unity feedback

Energy storage: L and C exchange energy
Frequency selection: Sharp resonance characteristic
Oscillation sustenance: Provides positive feedback

Mnemonic: “Tank Stores Energy, Selects Frequency”

Question 2(c) [7 marks]
#

Draw and explain the Hartley Oscillator. Also state equation of oscillation frequency of it.

Answer: Hartley oscillator uses tapped inductor in tank circuit for frequency generation.

Circuit Diagram:

graph TD
    A[Vcc] --> B[RFC]
    B --> C[Collector]
    C --> D[L1]
    D --> E[L2]
    E --> F[Emitter]
    D --> G[C]
    G --> E
    C --> H[Output]

Operation:

Tapped inductor: L1 and L2 provide feedback
Tank circuit: L1+L2 with C determines frequency
Positive feedback: Phase shift through L1-L2 coupling

Frequency Formula: f = 1/[2π√((L1+L2)C)]

Key Features:

Good frequency stability: Inductor-based tuning
Easy tuning: Variable inductor or capacitor
RF applications: Suitable for high frequencies

Mnemonic: “Hartley Has Tapped inductor”

Question 2(a OR) [3 marks]
#

Explain term oscillator as positive feedback amplifier.

Answer: Oscillator generates AC signals using positive feedback without external input signal.

Table:

Parameter	Amplifier	Oscillator
Input	External signal	No external input
Feedback	May use negative	Uses positive
Output	Amplified input	Self-generated AC

Self-sustaining: Positive feedback maintains oscillation
Barkhausen criteria: Loop gain = 1, phase = 0°
Signal generation: Creates AC from DC supply

Mnemonic: “Positive feedback Powers Perpetual signals”

Question 2(b OR) [4 marks]
#

Draw and explain the Crystal Oscillator.

Answer: Crystal oscillator uses piezoelectric effect of quartz crystal for high stability.

Circuit Diagram:

Characteristics:

Table:

Property	Value	Advantage
Stability	±0.01%	Very high
Q factor	>10,000	Sharp resonance
Temperature	Low drift	Stable frequency

Piezoelectric effect: Mechanical vibration creates electrical signal
High Q: Very stable frequency generation
Clock applications: Used in digital systems

Mnemonic: “Crystal Creates Constant frequency”

Question 2(c OR) [7 marks]
#

Draw the Structure, symbol, equivalent circuit of UJT and explain it in brief.

Answer: UJT (Unijunction Transistor) is three-terminal device with unique switching characteristics.

Structure:

Symbol:

Equivalent Circuit:

Operation:

Intrinsic standoff ratio: η = R1/(R1+R2)
Peak point voltage: VP = ηVBB + VD
Negative resistance: After peak point

Applications:

Relaxation oscillator: Sawtooth wave generation
Trigger circuits: SCR firing circuits
Timing applications: RC charging circuits

Mnemonic: “UJT Uses Unique Junction Technology”

Question 3(a) [3 marks]
#

Classify power amplifier based on operating point.

Answer: Power amplifiers are classified based on transistor conduction angle and bias point.

Table:

Class	Conduction Angle	Efficiency	Application
Class A	360°	25-50%	Audio, low power
Class B	180°	78.5%	Push-pull
Class AB	180°-360°	60-70%	Audio power
Class C	<180°	>90%	RF, tuned

Bias point: Determines operating class
Efficiency trade-off: Higher efficiency, more distortion
Application specific: Choose based on requirements

Mnemonic: “All Big Amplifiers Can deliver power”

Question 3(b) [4 marks]
#

Draw and Explain Complementary symmetry push-pull power amplifier.

Answer: Uses NPN and PNP transistors for efficient power amplification without center-tapped transformer.

Circuit Diagram:

graph TD
    A[+Vcc] --> B[NPN Q1]
    B --> C[Output]
    C --> D[RL]
    D --> E[PNP Q2]
    E --> F[-Vcc]
    G[Input] --> B
    G --> E

Operation:

Positive half-cycle: NPN conducts, PNP off
Negative half-cycle: PNP conducts, NPN off
Complementary action: Both transistors handle alternate half-cycles

Advantages:

No transformer: Direct coupling to load
High efficiency: Class B operation
Compact design: Fewer components
Good power transfer: Direct coupling

Mnemonic: “Complementary transistors Complete the cycle”

Question 3(c) [7 marks]
#

Derive an equation for Efficiency of class B push pull amplifier.

Answer: Class B push-pull amplifier has each transistor conducting for 180° of input cycle.

Analysis: For sinusoidal input: Vi = Vm sin ωt

Output Power:

Peak output voltage: Vom = Vcc
RMS output voltage: Vo(rms) = Vcc/√2
Po = Vo²(rms)/RL = Vcc²/2RL

Input Power:

DC current (average): Idc = 2Im/π
Where Im = Vcc/RL
Pin = Vcc × Idc = 2VccIm/π = 2Vcc²/πRL

Efficiency Calculation: η = Po/Pin = (Vcc²/2RL)/(2Vcc²/πRL) η = π/4 = 0.785 = 78.5%

Key Points:

Maximum theoretical efficiency: 78.5%
Class B advantage: Much higher than Class A (25%)
Practical efficiency: Slightly lower due to losses

Mnemonic: “Push-Pull Provides Pi/4 efficiency”

Question 3(a OR) [3 marks]
#

Differentiate between voltage and power amplifier.

Answer: Voltage and power amplifiers serve different purposes in electronic systems.

Table:

Parameter	Voltage Amplifier	Power Amplifier
Purpose	Increase voltage	Increase power
Load	High impedance	Low impedance
Efficiency	Not critical	Very important
Distortion	Must be low	Moderate acceptable
Coupling	RC/Direct	Transformer

Design priority: Voltage gain vs power delivery
Application: Signal processing vs driving loads
Circuit complexity: Simple vs complex power stages

Mnemonic: “Voltage amplifies signal, Power drives load”

Question 3(b OR) [4 marks]
#

Explain Class AB power amplifier with diagram.

Answer: Class AB operates between Class A and Class B, reducing crossover distortion.

Circuit Diagram:

Operation:

Slight forward bias: Both transistors slightly on
Conduction angle: >180° but <360°
Overlap conduction: Eliminates crossover distortion

Characteristics:

Table:

Parameter	Value	Benefit
Efficiency	60-70%	Better than Class A
Distortion	Low	Better than Class B
Bias	Slight forward	Compromise solution

Mnemonic: “AB Avoids Bad crossover distortion”

Question 3(c OR) [7 marks]
#

Derive an equation for Efficiency of series fed class A power amplifier.

Answer: Series fed Class A amplifier has DC supply connected in series with load.

Circuit Analysis:

DC supply voltage: Vcc
Quiescent current: Icq = Vcc/2RL (for maximum power)
Quiescent voltage: Vceq = Vcc/2

AC Analysis:

Maximum output voltage swing: Vom = Vcc/2
Output power: Po = Vom²/2RL = Vcc²/8RL

DC Power:

DC current: Idc = Icq = Vcc/2RL
Input power: Pin = Vcc × Idc = Vcc²/2RL

Efficiency: η = Po/Pin = (Vcc²/8RL)/(Vcc²/2RL) η = 1/4 = 0.25 = 25%

Key Points:

Maximum theoretical efficiency: 25%
Power wastage: 75% lost as heat
Design limitation: Poor efficiency but good linearity

Mnemonic: “Class A Achieves quarter efficiency”

Question 4(a) [3 marks]
#

Draw pin diagram of IC 741 OP-AMP and explain it.

Answer: IC 741 is 8-pin dual-in-line package operational amplifier with industry standard pinout.

Pin Diagram:

Pin Configuration:

Table:

Pin	Function	Description
1	Offset Null	Offset adjustment
2	Inverting Input	Negative input
3	Non-inverting Input	Positive input
4	-Vcc	Negative supply
5	Offset Null	Offset adjustment
6	Output	Amplifier output
7	+Vcc	Positive supply
8	NC	No connection

Mnemonic: “Null, Negative, Positive, Negative supply, Null, Output, Positive supply, Nothing”

Question 4(b) [4 marks]
#

Define the following OP-AMP parameters. 1. Input offset voltage 2. CMRR

Answer: These parameters define the non-ideal characteristics of practical operational amplifiers.

1. Input Offset Voltage (Vio):

Definition: DC voltage applied between inputs to make output zero
Typical value: 1-5 mV for 741
Cause: Mismatch in input transistors
Effect: Output error in DC applications

2. Common Mode Rejection Ratio (CMRR):

Definition: Ability to reject common signals at both inputs
Formula: CMRR = Ad/Acm
Typical value: 90 dB for 741
Importance: Noise immunity

Table:

Parameter	Symbol	Unit	Ideal	741 Typical
Input Offset Voltage	Vio	mV	0	2
CMRR	-	dB	∞	90

Mnemonic: “Offset creates Output error, CMRR Rejects common signals”

Question 4(c) [7 marks]
#

Explain inverting amplifier using IC 741 in detail.

Answer: Inverting amplifier uses negative feedback with input applied to inverting terminal.

Circuit Diagram:

graph LR
    A[Vin] --> B[R1]
    B --> C["-"]
    D["+"] --> E[Ground]
    C --> F[IC 741]
    F --> G[Vout]
    G --> H[Rf]
    H --> C

Analysis: Using virtual short concept:

V+ = V- = 0V (virtual ground)
Input current: I1 = Vin/R1
Feedback current: If = Vout/Rf
Current balance: I1 = If (no current into op-amp)

Derivation:

Vin/R1 = -Vout/Rf
Voltage Gain: Av = -Rf/R1

Characteristics:

Table:

Parameter	Expression	Notes
Voltage Gain	-Rf/R1	Negative sign
Input Impedance	R1	Low impedance
Output Impedance	~0Ω	Very low
Bandwidth	f = GBW/\|Av\|	Gain-bandwidth product

Applications:

Signal inversion: Phase reversal
Scale factor: Programmable gain
AC amplification: With coupling capacitors

Mnemonic: “Inverting Input gives Inverted output”

Question 4(a OR) [3 marks]
#

List characteristics of ideal OP-AMP.

Answer: Ideal op-amp represents perfect amplifier with theoretical limits for all parameters.

Table:

Parameter	Ideal Value	Practical Impact
Open Loop Gain	∞	Perfect amplification
Input Impedance	∞	No input current
Output Impedance	0Ω	Perfect voltage source
Bandwidth	∞	No frequency limitation
CMRR	∞	Perfect noise rejection
Slew Rate	∞	No slew rate limiting
Input Offset	0V	No DC errors

Perfect performance: All parameters optimized
Design simplification: Analysis becomes easier
Practical approximation: Close to ideal in many applications

Mnemonic: “Infinite Input, Zero Output, Perfect Performance”

Question 4(b OR) [4 marks]
#

Draw and explain summing amplifier using Op-amp.

Answer: Summing amplifier adds multiple input voltages with programmable gain for each input.

Circuit Diagram:

Analysis: Using virtual ground concept (V- = 0V):

Current through R1: I1 = V1/R1
Current through R2: I2 = V2/R2
Current through R3: I3 = V3/R3
Total input current: Iin = I1 + I2 + I3

Output Equation: Vout = -Rf(V1/R1 + V2/R2 + V3/R3)

Special Cases:

Equal resistors: Vout = -(Rf/R)(V1 + V2 + V3)
Unity gain: Rf = R, Vout = -(V1 + V2 + V3)

Applications:

Audio mixing: Multiple signal combination
Digital-to-analog: Weighted resistor DAC
Signal processing: Mathematical operations

Mnemonic: “Sum inputs, Scale by resistor ratios”

Question 4(c OR) [7 marks]
#

Explain differential amplifier using IC 741 in detail.

Answer: Differential amplifier amplifies the difference between two input signals while rejecting common signals.

Circuit Diagram:

Analysis: For the non-inverting input:

V+ = V2 × R3/(R2+R3)

For the inverting input using virtual short:

V- = V+ = V2 × R3/(R2+R3)

Using current balance:

(V1-V-)/R1 = (V–Vout)/Rf

Output Equation: When R1 = R2 and R3 = Rf: Vout = (Rf/R1)(V2 - V1)

Key Features:

Table:

Parameter	Value	Advantage
Differential Gain	Rf/R1	Amplifies difference
Common Mode Gain	~0	Rejects common signals
CMRR	Very high	Excellent noise immunity

Applications:

Instrumentation: Sensor signal processing
Noise rejection: Differential signal transmission
Bridge circuits: Strain gauge measurements

Mnemonic: “Difference amplified, Common rejected”

Question 5(a) [3 marks]
#

Draw the circuit of integrator using Op-amp and its input and output waveforms.

Answer: Op-amp integrator performs mathematical integration of input signal using RC feedback.

Circuit Diagram:

Waveforms:

Operation:

Integration function: Vout = -(1/RC)∫Vin dt
Square wave input: Produces triangular output
Ramp generation: Constant input gives linear ramp

Mnemonic: “Integration creates Triangular from square”

Question 5(b) [4 marks]
#

State advantage and disadvantage of push-pull arrangement of power amplifier

Answer: Push-pull configuration uses two transistors operating in complementary fashion for power amplification.

Advantages:

Table:

Advantage	Benefit	Application
High Efficiency	Up to 78.5%	Battery operated
No Transformer	Compact design	Portable devices
Low Distortion	Better linearity	Audio systems
Heat Distribution	Shared between transistors	Thermal management

Disadvantages:

Disadvantage	Problem	Solution
Crossover Distortion	Dead zone at zero crossing	Class AB bias
Component Matching	Requires matched transistors	Careful selection
Thermal Runaway	Temperature coefficient mismatch	Thermal coupling

Applications:

Audio amplifiers: High fidelity systems
Motor drivers: DC motor control
RF amplifiers: Communication systems

Mnemonic: “Push-Pull Provides Power but Problems exist”

Question 5(c) [7 marks]
#

Draw and explain astable multivibrator using 555 timer IC.

Answer: Astable multivibrator generates continuous square wave output without external trigger using 555 timer.

Circuit Diagram:

Pin Connections:

Pin 1: Ground
Pin 2: Trigger (connected to pin 6)
Pin 3: Output
Pin 4: Reset (+Vcc)
Pin 6: Threshold
Pin 7: Discharge
Pin 8: +Vcc

Operation:

Charging phase: C charges through RA + RB
Threshold reached: At 2/3 Vcc, output goes LOW
Discharging phase: C discharges through RB
Trigger reached: At 1/3 Vcc, output goes HIGH
Cycle repeats: Continuous oscillation

Timing Equations:

HIGH time: t1 = 0.693(RA + RB)C
LOW time: t2 = 0.693(RB)C
Total period: T = t1 + t2 = 0.693(RA + 2RB)C
Frequency: f = 1.44/[(RA + 2RB)C]
Duty cycle: D = (RA + RB)/(RA + 2RB) × 100%

Applications:

Clock generation: Digital systems
LED flasher: Blinking circuits
Tone generation: Audio oscillators
PWM generation: Motor speed control

Mnemonic: “Astable Always oscillates Automatically”

Question 5(a OR) [3 marks]
#

Draw the block diagram of Op-amp and explain it.

Answer: Op-amp internal structure consists of multiple stages for high gain and performance.

Block Diagram:

graph LR
    A[V+] --> B[Differential Amplifier]
    C[V-] --> B
    B --> D[Intermediate Amplifier]
    D --> E[Output Stage]
    E --> F[Output]
    G[Level Shifter] --> E
    D --> G

Stage Functions:

Table:

Stage	Function	Characteristics
Differential Input	High input impedance	Low offset, high CMRR
Intermediate Amplifier	High voltage gain	Most of the gain
Level Shifter	DC level adjustment	Couples AC stages
Output Stage	Low output impedance	Current buffer

High gain: Typically 100,000 or more
Wide bandwidth: MHz range capability
Low output impedance: Drives various loads

Mnemonic: “Differential Input, Intermediate gain, Level shift, Output buffer”

Question 5(b OR) [4 marks]
#

Explain about the terms related to power amplifier. i) Efficiency ii) Distortion.

Answer: These parameters determine power amplifier performance and suitability for applications.

i) Efficiency (η):

Definition: Ratio of AC output power to DC input power
Formula: η = Po(AC)/Pin(DC) × 100%
Importance: Determines heat dissipation and battery life

Efficiency Comparison:

Table:

Class	Efficiency	Application
A	25%	Low power, high fidelity
B	78.5%	Push-pull amplifiers
AB	60-70%	Audio amplifiers
C	>90%	RF applications

ii) Distortion:

Definition: Unwanted changes in output signal shape
Types: Harmonic, intermodulation, crossover
Measurement: Total Harmonic Distortion (THD)

Distortion Sources:

Nonlinearity: Transistor characteristics
Crossover: Dead zone in push-pull
Thermal effects: Temperature variations

Mnemonic: “Efficiency measures Energy use, Distortion shows signal Degradation”

Question 5(c OR) [7 marks]
#

Draw pin diagram of 555 timer IC. Also draw circuit diagram of two stage sequential timer using 555 timer IC.

Answer: 555 timer is versatile IC used for timing applications with standard 8-pin package.

Pin Diagram:

Pin Functions:

Table:

Pin	Name	Function
1	Ground	Common ground
2	Trigger	Starts timing cycle
3	Output	Timer output
4	Reset	Resets timer
5	Control	Voltage reference
6	Threshold	Stops timing cycle
7	Discharge	Discharges timing capacitor
8	Vcc	Supply voltage