Linear Integrated Circuit (4341105) - Summer 2023 Solution

Table of Contents

Question 1(a) [3 marks]
#

Write advantages and disadvantages of negative feedback amplifier

Answer:

Advantages	Disadvantages
Increases bandwidth	Reduces gain
Stabilizes gain	Requires more components
Reduces distortion	Increases cost
Increases input impedance (voltage series)	May cause oscillations if improperly designed
Decreases output impedance (voltage series)	Requires careful phase compensation

Mnemonic: “GRASS Grows Better Despite Dry Soil” (Gain Reduction, Amplifies Stability, Stops distortion, Better impedance)

Question 1(b) [4 marks]
#

Derive the equation of overall gain with negative feedback in amplifier and give application of negative feedback.

Answer:

Derivation of overall gain with negative feedback:

flowchart LR
    I[Input] --> S[Summing\nPoint]
    S --> A[Amplifier\nA]
    A --> O[Output]
    O --> F[Feedback\nNetwork β]
    F --> S

For an amplifier with gain A and feedback factor β:
- Input signal = Vin
- Feedback signal = βVout
- Actual input to amplifier = Vin - βVout
- Output = A(Vin - βVout)
- Therefore, Vout = A(Vin - βVout)
- Vout + AβVout = AVin
- Vout(1 + Aβ) = AVin
- Overall gain = Vout/Vin = A/(1 + Aβ)

Applications of negative feedback:

Operational amplifiers
Voltage regulators
Audio amplifiers
Instrumentation amplifiers

Mnemonic: “AVOI” (Amplifiers, Voltage regulators, Oscillation control, Instrumentation)

Question 1(c) [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 Negative Feedback Amplifier:

flowchart LR
    I[Input] --> S[Current\nSampling]
    S --> A[Amplifier]
    A --> O[Output]
    O --> F[Feedback\nNetwork]
    F -->|Feedback Current| S

In current shunt feedback, the output voltage is sampled and converted to a current that is subtracted from the input current.

Circuit Diagram:

Characteristics:

Feedback type: Current sampling at input, shunt mixing at input
Samples: Output voltage
Feedback to: Input current

Derivation of Input Impedance:

Without feedback: Zin
With current shunt feedback: Zin’ = Zin/(1 + Aβ)
Therefore, input impedance decreases by factor (1 + Aβ)

Derivation of Output Impedance:

Without feedback: Zo
With current shunt feedback: Zo’ = Zo/(1 + Aβ)
Therefore, output impedance decreases by factor (1 + Aβ)

Mnemonic: “DISCO” (Decreased Impedances with Shunt Current Operation)

Question 1(c) OR [7 marks]
#

Draw and Explain voltage series type negative feedback amplifier and Derive the formula of input impedance and output impedance of it.

Answer:

Voltage Series Negative Feedback Amplifier:

flowchart LR
    I[Input] --> S[Voltage\nSampling]
    S --> A[Amplifier]
    A --> O[Output]
    O --> F[Feedback\nNetwork β]
    F -->|Feedback Voltage| S

In voltage series feedback, the output voltage is sampled and fed back in series with the input voltage.

Circuit Diagram:

Characteristics:

Feedback type: Voltage sampling at output, series mixing at input
Samples: Output voltage
Feedback to: Input voltage

Derivation of Input Impedance:

Without feedback: Zin
With voltage series feedback: Zin’ = Zin × (1 + Aβ)
Therefore, input impedance increases by factor (1 + Aβ)

Derivation of Output Impedance:

Without feedback: Zo
With voltage series feedback: Zo’ = Zo/(1 + Aβ)
Therefore, output impedance decreases by factor (1 + Aβ)

Mnemonic: “ISDO” (Increased input impedance, Series feedback, Decreased output impedance, Output voltage sampled)

Question 2(a) [3 marks]
#

Draw and Explain the circuit diagram of UJT as a relaxation oscillator.

Answer:

UJT Relaxation Oscillator:

flowchart TB
    A[UJT Relaxation Oscillator]
    A --- B[Produces pulses with\nRC charging circuit]
    A --- C[UJT triggers when\ncapacitor voltage reaches peak]
    A --- D[Simple timing circuit]

Circuit Diagram:

In this circuit:

C1 charges through R1
When capacitor voltage reaches UJT’s peak point, UJT turns on
Capacitor discharges rapidly through UJT
Process repeats creating oscillations

Mnemonic: “CURD” (Capacitor charges Until Reaching Discharge point)

Question 2(b) [4 marks]
#

Draw circuit diagram of Colpitts oscillator and explain in brief. Give the advantages and disadvantages of it.

Answer:

Colpitts Oscillator:

Circuit Diagram:

Working:

Uses LC tank circuit with capacitive voltage divider (C1 and C2)
Transistor amplifies and provides energy to tank circuit
Oscillation frequency: f = 1/[2π√(L×(C1×C2)/(C1+C2))]

Advantages	Disadvantages
Good frequency stability	Requires two capacitors (C1, C2)
Works well at high frequencies	More difficult to tune than some oscillators
Lower harmonics	Sensitive to transistor parameters
Simple design	Limited frequency range

Mnemonic: “FAST Circuits” (Frequency stable, Appropriate for high frequencies, Simple design, Two capacitors needed)

Question 2(c) [7 marks]
#

Explain the Crystal Oscillator.

Answer:

Crystal Oscillator:

flowchart LR
    A[Crystal Oscillator] --> B[Uses piezoelectric crystal]
    A --> C[Extremely stable frequency]
    A --> D[High Q factor]
    A --> E[Precise timing applications]

Circuit Diagram:

Working Principle:

Based on piezoelectric effect of quartz crystal
Crystal vibrates at natural resonant frequency when voltage applied
Acts as very stable resonator with extremely high Q factor
Provides feedback at precise frequency

Characteristics:

Resonant frequency: Determined by crystal cut and dimensions
Q factor: Typically 10,000-100,000 (much higher than LC circuits)
Frequency stability: Typically 0.001% to 0.01%
Temperature coefficient: Usually low, can be specially cut for zero temp coefficient

Applications:

Clock generation in computers
Frequency standards
Radio transmitters/receivers
Digital watches and clocks
Microcontroller timing

Mnemonic: “STOP Precisely” (Stable, Temperature-resistant, Oscillates, Piezoelectric, Precisely)

Question 2(a) OR [3 marks]
#

Draw and explain the Hartley Oscillator.

Answer:

Hartley Oscillator:

flowchart TB
    A[Hartley Oscillator] --> B[Uses tapped inductor]
    A --> C[LC tank circuit]
    A --> D[RF applications]

Circuit Diagram:

Working:

Uses LC tank circuit with tapped inductor (L1 and L2)
Transistor amplifies and provides energy to tank circuit
Oscillation frequency: f = 1/[2π√(L×C)] where L = L1 + L2
Feedback through inductive coupling

Mnemonic: “TIC” (Tapped inductor Circuit)

Question 2(b) OR [4 marks]
#

Draw and explain Wien Bridge oscillator.

Answer:

Wien Bridge Oscillator:

flowchart LR
    A[Wien Bridge Oscillator] --> B[Uses RC network]
    A --> C[Audio frequency range]
    A --> D[Low distortion]
    A --> E[Stable output]

Circuit Diagram:

Working:

Uses RC Wien bridge network as frequency-selective feedback
R1=R2 and C1=C2 for simplest design
Oscillation frequency: f = 1/(2πRC)
Gain must be ≥ 3 for sustained oscillations
Used for audio frequency generation with low distortion

Mnemonic: “FEAR” (Frequency selective, Equal RC components, Audio range, Reduced distortion)

Question 2(c) OR [7 marks]
#

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

Answer:

Unijunction Transistor (UJT):

Structure:

Symbol:

Equivalent Circuit:

Working Principle:

UJT is a three-terminal device with one emitter and two bases
N-type silicon bar with P-type emitter junction
Forms a voltage divider with internal resistances RB1 and RB2
Emitter current starts flowing when VE > η×VBB + VD
Where η is intrinsic standoff ratio = RB1/(RB1+RB2)

Characteristics:

Intrinsic standoff ratio (η): Typically 0.5 to 0.8
Negative resistance region: Current increases as voltage decreases
Peak point: Beginning of negative resistance region
Valley point: End of negative resistance region

Applications:

Relaxation oscillators
Timing circuits
Trigger generators
SCR triggering circuits
Sawtooth generators

Mnemonic: “NEVER” (Negative resistance, Emitter-triggered, Valley and peak points, Easily timed, Relaxation oscillator)

Question 3(a) [3 marks]
#

Differentiate between voltage and power amplifier.

Answer:

Parameter	Voltage Amplifier	Power Amplifier
Purpose	Amplifies voltage	Delivers power to load
Output impedance	High	Low
Input impedance	High	Relatively low
Efficiency	Not important	Very important
Heat dissipation	Low	High (requires heat sink)
Position in circuit	Early stages	Final stage

Mnemonic: “PEHIP” (Power for Efficiency and Heat, Impedance matters, Position differs)

Question 3(b) [4 marks]
#

Explain class-B push pull power amplifier in detail.

Answer:

Class-B Push-Pull Amplifier:

flowchart TB
    A[Class-B Push-Pull] --> B[Uses two transistors]
    A --> C[Each handles half cycle]
    A --> D[Higher efficiency ~78%]
    A --> E[Crossover distortion]

Circuit Diagram:

Working:

Uses two complementary transistors
Q1 conducts during positive half-cycle
Q2 conducts during negative half-cycle
Each transistor conducts for 180° of input cycle
Theoretical efficiency: 78.5%

Mnemonic: “ECHO” (Efficiency high, Crossover distortion, Half-cycle operation, Output high power)

Question 3(c) [7 marks]
#

Draw and Explain Complementary symmetry push-pull power amplifier in detail also list the disadvantages of it.

Answer:

Complementary Symmetry Push-Pull Amplifier:

flowchart LR
    A[Complementary Push-Pull] --> B[Uses NPN and PNP]
    A --> C[Direct coupling possible]
    A --> D[No center-tapped transformer]
    A --> E[Class B or AB operation]

Circuit Diagram:

Working:

Uses complementary pair (NPN and PNP transistors)
No need for center-tapped transformer
NPN handles positive half-cycle
PNP handles negative half-cycle
Biasing network reduces crossover distortion
Direct coupling to speaker possible

Disadvantages:

Thermal runaway if not properly biased
Requires complementary matched transistors
Crossover distortion in Class-B operation
Needs both positive and negative power supplies
Difficulty finding exact complementary pairs

Mnemonic: “MATCH Precisely” (Matched transistors, Avoids transformers, Thermal issues, Crossover distortion, Heat dissipation needed)

Question 3(a) OR [3 marks]
#

Define the terms related to power amplifier. i)Efficiency ii)Distortion iii)power dissipation capability

Answer:

Term	Definition
Efficiency	Ratio of AC output power delivered to the load to the DC input power drawn from the supply. Mathematically: η = (Pout/Pin) × 100%. Higher efficiency means less power wasted as heat.
Distortion	Unwanted alteration of the output waveform compared to input waveform. Measured as Total Harmonic Distortion (THD). Includes harmonic, intermodulation, crossover, and amplitude distortion.
Power Dissipation Capability	Maximum power that can be dissipated by the amplifier without damage. Depends on heat sink, thermal resistance, and maximum junction temperature of transistors.

Mnemonic: “EDP” (Efficiency converts, Distortion deforms, Power capability protects)

Question 3(b) OR [4 marks]
#

Classify the power amplifier for mode of operation and explain working of different type power amplifier

Answer:

Classification of Power Amplifiers:

flowchart TB
    A[Power Amplifiers] --> B[Class A]
    A --> C[Class B]
    A --> D[Class AB]
    A --> E[Class C]

Class	Conduction Angle	Working
Class A	360°	Amplifier conducts for entire input cycle. Output signal is exact replica of input but amplified. Linear but inefficient (25-30%).
Class B	180°	Two transistors each conduct for half cycle. One handles positive half, other handles negative half. More efficient (70-80%) but has crossover distortion.
Class AB	180°-360°	Compromise between Class A and B. Slight bias to reduce crossover distortion. Good efficiency (50-70%) with acceptable distortion.
Class C	<180°	Conducts for less than half cycle. Very efficient (>80%) but highly distorted. Used mainly in RF tuned amplifiers.

Mnemonic: “ABCE” (A-all cycle, B-both halves separately, C-compromise solution, E-efficiency with distortion)

Question 3(c) OR [7 marks]
#

Derive efficiency of class-B push pull power amplifier.

Answer:

Derivation of Class-B Push-Pull Amplifier Efficiency:

flowchart TB
    A[Class-B Efficiency] --> B[Based on power ratio]
    A --> C[Each transistor conducts half cycle]
    A --> D[Theoretical max efficiency: 78.5%]

Circuit Diagram:

Efficiency Calculation:

DC power input calculation:
- Each transistor conducts for half cycle
- Average DC current: Idc = Imax/π
- DC power input: Pdc = Vcc × Idc = Vcc × Imax/π
AC power output calculation:
- RMS value of current: Irms = Imax/2
- AC power output: Pac = (Irms)² × RL = (Imax/2)² × RL
- For maximum power: Imax × RL = Vcc
- Therefore: Pac = (Vcc)²/(2π × RL)
Efficiency calculation:
- η = (Pac/Pdc) × 100%
- η = [(Vcc)²/(2π × RL)] ÷ [Vcc × Imax/π] × 100%
- η = [(Vcc)²/(2π × RL)] ÷ [Vcc × Vcc/(π × RL)] × 100%
- η = [(Vcc)²/(2π × RL)] × [π × RL/Vcc²] × 100%
- η = π/4 × 100% ≈ 78.5%

Maximum theoretical efficiency of Class-B push-pull amplifier is 78.5%

Mnemonic: “PIPE” (Power ratio, Input DC vs output AC, Pi in formula, Efficiency maximum 78.5%)

Question 4(a) [3 marks]
#

Draw pin diagram and Schematic symbol of IC 741 and explain it in detail.

Answer:

IC 741 Op-Amp Pin Diagram and Symbol:

Pin Diagram:

Schematic Symbol:

Pin Description:

Offset Null (NC1)
Inverting Input (-)
Non-inverting Input (+)
Negative Supply (-Vcc)
Offset Null (NC2)
Output
Positive Supply (+Vcc)
NC (No Connection)

Mnemonic: “ON-INO” (Offset Null, Inverting input, Negative supply, Input non-inverting, Output, No connection)

Question 4(b) [4 marks]
#

Explain differential Amplifier using OPAMP.

Answer:

Differential Amplifier Using Op-Amp:

flowchart LR
    A[Differential Amplifier] --> B[Amplifies difference]
    A --> C[Rejects common mode]
    A --> D[Four equal resistors]
    A --> E[Gain = R2/R1]

Circuit Diagram:

Working:

Output is proportional to difference between inputs
If R1 = R3 and R2 = R4, then: Vout = (R2/R1)(V2-V1)
Rejects signals common to both inputs (common-mode rejection)
Used in instrumentation applications

Mnemonic: “CARE” (Common-mode rejection, Amplifies difference, Resistor matching important, Equal resistors for balance)

Question 4(c) [7 marks]
#

Explain the following parameters of an OP-Amp: 1)Input offset voltage 2) Output Offset Voltage 3) Input Offset Current 4)Input Bias Current 5) CMRR 6) Slew rate 7) Gain.

Answer:

Parameters of an Op-Amp:

Parameter	Description	Typical Value for 741
Input Offset Voltage	Voltage needed at input to zero the output	1-5 mV
Output Offset Voltage	Output voltage when inputs are grounded	Depends on input offset and gain
Input Offset Current	Difference between input bias currents	3-30 nA
Input Bias Current	Average of the two input currents	30-500 nA
CMRR	Ability to reject common-mode signals	70-100 dB
Slew Rate	Maximum rate of output voltage change	0.5 V/μs
Gain (Aol)	Open-loop voltage gain	104-106 (80-120 dB)

Diagram for Input Offset Voltage:

Mnemonic: “VICS BGR” (Voltage offset at Input, Current offset, Slew rate, Bias current, Gain, Rejection ratio)

Question 4(a) OR [3 marks]
#

List characteristics of ideal op-amp.

Answer:

Characteristic	Ideal Value
Open-loop gain	Infinite
Input impedance	Infinite
Output impedance	Zero
Bandwidth	Infinite
CMRR	Infinite
Slew rate	Infinite
Offset voltage	Zero
Noise	Zero

Mnemonic: “ZINC BOSS” (Zero offset, Infinite bandwidth, No noise, CMRR infinite, Bandwidth unlimited, Output impedance zero, Slew rate unlimited, Speed unlimited)

Question 4(b) OR [4 marks]
#

Draw and explain the block diagram of the Operational Amplifier (OPAMP) in detail.

Answer:

Op-Amp Block Diagram:

flowchart LR
    A[Input Stage] --> B[Intermediate Stage]
    B --> C[Output Stage]
    D[Biasing Circuit] --> A
    D --> B
    D --> C
    E[Compensation Network] --> B

Detailed Block Diagram:

Working of Blocks:

Input Stage: Differential amplifier with high input impedance
Intermediate Stage: High-gain voltage amplifier with frequency compensation
Output Stage: Low output impedance buffer, provides current gain
Biasing Circuit: Provides proper DC levels to all stages
Compensation Network: Prevents oscillation, ensures stability

Mnemonic: “DISCO” (Differential stage Input, Second stage amplifies, Compensation network, Output buffer)

Question 4(c) OR [7 marks]
#

Draw & explain Inverting and Non-inverting Op-amp amplifier with the derivation of voltage gain.

Answer:

Inverting Amplifier:

flowchart TB
    A[Inverting Amplifier] --> B[Output 180° out of phase]
    A --> C[Gain = -Rf/Rin]
    A --> D[Virtual ground at inverting input]

Circuit Diagram:

Gain Derivation:

Using virtual ground concept (V- ≈ 0)
Current through Rin: Iin = Vin/Rin
Current through Rf: If = Iin (no current into op-amp input)
Voltage across Rf: Vout = -If × Rf = -Iin × Rf = -Vin × Rf/Rin
Therefore, Gain = Vout/Vin = -Rf/Rin

Non-Inverting Amplifier:

flowchart TB
    A[Non-Inverting Amplifier] --> B[Output in phase with input]
    A --> C[Gain = 1 + Rf/Rin]
    A --> D[Higher input impedance than inverting]

Circuit Diagram:

Gain Derivation:

Due to negative feedback, V- ≈ V+ = Vin
Voltage across Rin: V- = Vin
Current through Rin: IRin = V-/Rin = Vin/Rin
Same current flows through Rf: IRf = IRin
Voltage across Rf: VRf = IRf × Rf = Vin × Rf/Rin
Output voltage: Vout = V- + VRf = Vin + Vin × Rf/Rin = Vin(1 + Rf/Rin)
Therefore, Gain = Vout/Vin = 1 + Rf/Rin

Comparison:

Parameter	Inverting Amplifier	Non-Inverting Amplifier
Gain formula	-Rf/Rin	1 + Rf/Rin
Phase shift	180°	0°
Input impedance	Equal to Rin	Very high (≈ infinite)
Min. possible gain	Can be <1	Always ≥1

Mnemonic: “PING-PONG” (Phase Inverted Negative Gain vs Positive Output Non-inverted Gain)

Question 5(a) [3 marks]
#

Draw and explain integrator using Op-Amp.

Answer:

Op-Amp Integrator:

flowchart LR
    A[Integrator] --> B[RC circuit in feedback]
    A --> C[Output is integral of input]
    A --> D[Acts as low-pass filter]

Circuit Diagram:

Working:

Output voltage is proportional to integral of input
Vout = -1/RC ∫Vin dt
Used in waveform generators, analog computers
Acts as low-pass filter with -20dB/decade slope

Mnemonic: “TIME” (Takes Input and Makes integral over time Exactly)

Question 5(b) [4 marks]
#

Compare different types of power amplifier.

Answer:

Parameter	Class A	Class B	Class AB	Class C
Conduction angle	360°	180°	180°-360°	<180°
Efficiency	25-30%	70-80%	50-70%	>80%
Distortion	Very low	High (crossover)	Low	Very high
Biasing	Above cutoff	At cutoff	Slightly above cutoff	Below cutoff
Applications	High fidelity audio	General purpose	Audio amplifiers	RF amplifiers

Mnemonic: “CABINET” (Conduction angle, Amplification quality, Biasing, Ideal applications, Noise/distortion, Efficiency, Temperature concerns)

Question 5(c) [7 marks]
#

List applications of IC555 and explain any one in detail.

Answer:

Applications of IC 555:

Astable multivibrator
Monostable multivibrator
Bistable multivibrator
Pulse width modulator
Sequential timer
Frequency divider
Tone generator

Astable Multivibrator Using IC 555:

flowchart TB
    A[555 Astable] --> B[Free-running oscillator]
    A --> C[No stable state]
    A --> D[Output continuously switches]
    A --> E[Frequency determined by R1, R2, C]

Circuit Diagram:

Working:

R1, R2, and C determine frequency
Output oscillates between HIGH and LOW
Charging time: t1 = 0.693(R1+R2)C
Discharging time: t2 = 0.693(R2)C
Total period: T = t1 + t2 = 0.693(R1+2R2)C
Frequency: f = 1.44/[(R1+2R2)C]
Duty cycle: D = (R1+R2)/(R1+2R2)

Applications:

LED flashers
Clock generators
Tone generators
Pulse generation

Mnemonic: “FREE” (Frequency determined by Resistors and capacitor, Endless oscillation, Easy to configure)

Question 5(a) OR [3 marks]
#

Draw and explain summing amplifier using Op-Amp.

Answer:

Summing Amplifier Using Op-Amp:

flowchart TB
    A[Summing Amplifier] --> B[Adds multiple inputs]
    A --> C[Weighted sum possible]
    A --> D[Inverting configuration]

Circuit Diagram:

Working:

Uses inverting configuration with multiple inputs
Each input contributes to output based on its resistance
If R1 = R2 = R3 = R, and Rf = R, then Vout = -(V1 + V2 + V3)
If resistors differ, weighted sum is produced: Vout = -Rf(V1/R1 + V2/R2 + V3/R3)
Virtual ground at inverting input simplifies analysis

Mnemonic: “SWIM” (Summing Weighted Inputs with Mixing)

Question 5(b) OR [4 marks]
#

Compare between push-pull amplifier and Complementary push-pull power amplifier.

Answer:

Parameter	Push-Pull Amplifier	Complementary Push-Pull Amplifier
Transistors used	Same type (NPN or PNP)	Complementary pair (NPN and PNP)
Input transformer	Required (center-tapped)	Not required
Output transformer	Required	Not required
Circuit complexity	More complex	Simpler
Cost	Higher due to transformers	Lower
Frequency response	Limited by transformers	Better (wider range)
Phase distortion	Higher	Lower
Power supply	Single polarity	Dual polarity usually required