Linear Integrated Circuit (4341105) - Winter 2024 Solution

Table of Contents

Question 1(a) [3 marks]
#

List advantages and disadvantages of negative feedback

Answer:

Advantages of Negative Feedback	Disadvantages of Negative Feedback
Increases bandwidth	Reduces gain
Improves stability	More components required
Reduces distortion	Complex circuit design
Decreases noise	Possibility of oscillations if improperly designed
Provides better input/output impedance control	Increased power consumption

Mnemonic: “STAND” - Stability, linearity, Amplitude reduction, Noise reduction, Distortion reduction

Question 1(b) [4 marks]
#

Explain effect of negative feedback on gain and stability

Answer:

Effect on Gain	Effect on Stability
Reduces gain by factor (1+Aβ)	Increases stability against temperature variations
Gain equation: A’ = A/(1+Aβ)	Reduces sensitivity to component parameter changes
More predictable gain values	Prevents oscillations in normal operating conditions
Less variation in gain with temperature	Makes circuit performance more consistent over time

Diagram:

graph LR
    A[Input] --> B[Amplifier A]
    B --> C[Output]
    C --> D[Feedback Network β]
    D --> E[Subtractor]
    A --> E
    E --> B

Mnemonic: “GRIP” - Gain Reduction, Improved stability, Predictable performance

Question 1(c) [7 marks]
#

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

Answer:

Step	Equation	Description
1	Vi = Vs - Vf	Input voltage equals source minus feedback
2	Vf = β × Vo	Feedback voltage is β times output voltage
3	Vo = A × Vi	Output voltage is amplifier gain times input voltage
4	Vo = A × (Vs - β × Vo)	Substituting (1) and (2) into (3)
5	Vo + A × β × Vo = A × Vs	Rearranging terms
6	Vo(1 + Aβ) = A × Vs	Factoring Vo
7	Vo/Vs = A/(1+Aβ)	Overall gain equation

Diagram:

graph LR
    Vs[Vs Source] --> Sum((+/-))
    Sum --> A[Amplifier A]
    A --> Vo[Vo Output]
    Vo --> FB[Feedback β]
    FB --> Sum

Mnemonic: “SAFE” - Source, Amplifier, Feedback, Equation A/(1+Aβ)

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

Compare voltage shunt amplifier, voltage series, current shunt and current series amplifier.

Answer:

Parameter	Voltage Series	Voltage Shunt	Current Series	Current Shunt
Input Signal	Voltage	Voltage	Current	Current
Output Signal	Voltage	Current	Voltage	Current
Input Configuration	Series	Parallel	Series	Parallel
Output Configuration	Series	Series	Parallel	Parallel
Input Impedance	Increases	Decreases	Decreases	Increases
Output Impedance	Decreases	Decreases	Increases	Increases
Application	Voltage amplifiers	Transconductance amplifiers	Transresistance amplifiers	Current amplifiers

Diagram:

Mnemonic: “VISC” - Voltage In (Series/shunt), Signal Current (series/shunt)

Question 2(a) [3 marks]
#

Write application of UJT.

Answer:

Applications of UJT
Relaxation oscillators
Timing circuits
Trigger circuits for SCR and TRIAC
Sawtooth wave generators
Pulse generators
Phase control in power electronics

Mnemonic: “ROBOTS” - Relaxation Oscillators, Bistable circuits, Oscillators, Timing, Switching

Question 2(b) [4 marks]
#

Draw circuit diagram of Wein bridge oscillator and Heartly oscillator.

Answer:

Wein Bridge Oscillator:

Hartley Oscillator:

Mnemonic: “WH-RC-LC” - Wein uses RC, Hartley uses LC

Question 2(c) [7 marks]
#

Draw and explain the structure, working and characteristics of UJT.

Answer:

Structure of UJT:

Structure	Working	Characteristics
N-type silicon bar with P-type junction	Acts as voltage divider with intrinsic stand-off ratio η	Negative resistance region in V-I curve
Three terminals: Base1, Base2, Emitter	When VE > ηVBB, it conducts	Peak point and valley point
Single P-N junction	Internal resistance decreases rapidly	Stable switching operation
Single junction but two bases	Generates relaxation oscillations	Temperature sensitivity

V-I Characteristics:

graph LR
    Peak[Peak point] --> Valley[Valley point]
    style Peak fill:#f9f,stroke:#333,stroke-width:2px
    style Valley fill:#bbf,stroke:#333,stroke-width:2px

Mnemonic: “PNVB” - P-N junction, Negative resistance, Valley point, Bases two

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

Classify oscillators based on component used and operating frequency.

Answer:

Based on Components	Based on Operating Frequency
RC Oscillators (Wien bridge, Phase shift)	Audio Frequency (20Hz-20kHz)
LC Oscillators (Hartley, Colpitts, Clapp)	Radio Frequency (20kHz-30MHz)
Crystal Oscillators (Quartz crystal)	Very High Frequency (30MHz-300MHz)
Relaxation Oscillators (UJT based)	Ultra High Frequency (300MHz-3GHz)
Negative Resistance Oscillators (Tunnel diode)	Microwave Frequency (>3GHz)

Mnemonic: “RCLCN” - RC, LC, Crystal, Negative resistance

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

Explain UJT as a relaxation oscillator

Answer:

Operation Stage	Description
Charging Phase	Capacitor charges through resistor R
Threshold Point	When capacitor voltage reaches peak point voltage (ηVBB), UJT turns ON
Discharge Phase	Capacitor discharges rapidly through UJT’s low resistance
Reset	UJT turns OFF after capacitor voltage falls below valley point

Circuit Diagram:

Mnemonic: “CTDR” - Charge, Threshold, Discharge, Repeat

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

Sketch the circuit of Colpitts oscillator and explain working of it in brief

Answer:

Colpitts Oscillator Circuit:

Component	Function
C1 and C2	Voltage divider network that provides feedback
Inductor L	Forms LC tank circuit with C1 and C2
Transistor Q	Provides amplification
RFC (Radio Frequency Choke)	Blocks AC while allowing DC

Working:

Tank circuit (L with C1+C2) determines oscillation frequency
Frequency formula: f = 1/(2π√(L×(C1×C2)/(C1+C2)))
Feedback through capacitive voltage divider
Transistor amplifies and sustains oscillations
Phase shift of 180° through transistor, 180° through feedback network

Mnemonic: “COLTS” - Capacitors form Oscillations with L-Tank circuit Sustainably

Question 3(a) [3 marks]
#

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

Answer:

Term	Definition
Collector Efficiency	Ratio of AC output power to DC power supplied by the collector battery (η = P_out/P_DC × 100%)
Distortion	Unwanted change in waveform shape from input to output (measured as THD - Total Harmonic Distortion)
Power Dissipation Capability	Maximum power that amplifier can safely dissipate as heat without damage (P_D = V_CE × I_C)

Mnemonic: “EDP” - Efficiency measures DC-to-AC conversion, Distortion alters signal, Power dissipation limits operation

Question 3(b) [4 marks]
#

Derive efficiency of class-A power amplifier.

Answer:

Step	Equation	Description
1	P_DC = V_CC × I_C	DC power input
2	P_out = (V_peak × I_peak)/2	AC power output
3	V_peak = V_CC	Maximum voltage swing
4	I_peak = I_C	Maximum current swing
5	P_out = (V_CC × I_C)/2	Substituting max values
6	η = (P_out/P_DC) × 100%	Definition of efficiency
7	η = ((V_CC × I_C)/2)/(V_CC × I_C) × 100%	Substituting power values
8	η = 50%	Maximum theoretical efficiency

Diagram:

graph TD
    A[Class A] --> B["Maximum η = 25-30%"]
    B --> C["Practical η < 50%"]
    style A fill:#f9f,stroke:#333,stroke-width:2px

Mnemonic: “HALF” - Highest Achievable Level Fifty percent

Question 3(c) [7 marks]
#

Explain operation of Complementary symmetry push-pull amplifier

Answer:

Circuit Diagram:

Operation	Description
Positive Half Cycle	NPN transistor Q1 conducts, PNP transistor Q2 is OFF
Negative Half Cycle	PNP transistor Q2 conducts, NPN transistor Q1 is OFF
Crossover Region	Both transistors are almost OFF, causing crossover distortion
Bias Circuit	Reduces crossover distortion by providing slight forward bias
Efficiency	Higher than Class A (theoretically up to 78.5%)
Heat Dissipation	Better than Class A as only one transistor conducts at a time

Mnemonic: “COPS” - Complementary transistors, Opposite conducting cycles, Push-pull operation, Symmetrical output

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

Give classification of Power amplifier

Answer:

Classification Basis	Types
Based on Biasing	Class A, Class B, Class AB, Class C
Based on Configuration	Single-ended, Push-pull, Complementary symmetry
Based on Coupling	RC coupled, Transformer coupled, Direct coupled
Based on Frequency Range	Audio power amplifier, RF power amplifier
Based on Operating Mode	Linear, Switching (Class D, E, F)

Mnemonic: “ABCDE” - A, B, C classes, Direct/transformer coupling, Efficiency increases from A to C

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

Derive efficiency of class B push pull amplifier

Answer:

Step	Equation	Description
1	P_DC = (2 × V_CC × I_max)/π	DC power input (each transistor conducts for half cycle)
2	P_out = (V_CC × I_max)/2	AC power output
3	η = (P_out/P_DC) × 100%	Definition of efficiency
4	η = ((V_CC × I_max)/2)/((2 × V_CC × I_max)/π) × 100%	Substituting power values
5	η = (π/4) × 100%	Simplifying
6	η = 78.5%	Maximum theoretical efficiency

Diagram:

graph TD
    A[Class B] --> B["Maximum η = 78.5%"]
    B --> C["π/4 × 100%"]
    style A fill:#bbf,stroke:#333,stroke-width:2px

Mnemonic: “PIPE” - Pi divided by four Equals efficiency

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

Differentiate between class A, B, C and AB power amplifier.

Answer:

Parameter	Class A	Class B	Class AB	Class C
Conduction Angle	360°	180°	180°-360°	<180°
Bias Point	At center of load line	At cutoff	Slightly above cutoff	Below cutoff
Efficiency	25-30%	78.5%	50-78.5%	Up to 90%
Distortion	Lowest	High (crossover)	Low	Very high
Linearity	Best	Poor	Good	Poor
Power Output	Low	Medium	Medium	High
Applications	High-fidelity audio	Audio power amplifiers	Audio power amplifiers	RF power amplifiers

Waveform Comparison:

Mnemonic: “ABCE” - Angle decreases, Bias moves to cutoff, Conduction decreases, Efficiency increases

Question 4(a) [3 marks]
#

Define (i) CMRR (ii) Slew rate

Answer:

Parameter	Definition	Typical Value
CMRR (Common Mode Rejection Ratio)	Ratio of differential mode gain to common mode gain, expressed in dB	90-120 dB
	CMRR = 20 log(Ad/Acm)	Higher is better
Slew Rate	Maximum rate of change of output voltage per unit time	0.5-10 V/μs
	SR = dVo/dt	Higher means faster response

Mnemonic: “CRSR” - Common Rejection Slope Rate

Question 4(b) [4 marks]
#

Explain Op-amp as a Summing amplifier.

Answer:

Circuit Diagram:

Operation	Description
Working Principle	Virtual ground concept - inverting input maintained at ground potential
Output Equation	V_out = -(R_f/R1 × V1 + R_f/R2 × V2 + … + R_f/Rn × Vn)
Special Case	When all input resistors equal (R1=R2=…=Rn=R), V_out = -(R_f/R) × (V1+V2+…+Vn)
Applications	Audio mixers, Analog computers, Signal conditioning circuits

Mnemonic: “SWAP” - Summing With Amplification Property

Question 4(c) [7 marks]
#

Draw noninverting amplifier using op Amp and Derive equation of voltage Gain. Also draw input and output waveform for it

Answer:

Circuit Diagram:

Parameter	Description
Voltage Gain Equation	A_v = 1 + (R_f/R1)
Input Impedance	Very high (typically >10⁶ Ω)
Output Impedance	Very low (typically <100 Ω)
Phase Shift	0° (in phase)

Input and Output Waveforms:

Derivation of Voltage Gain:

Voltage at both input pins is equal (V⁺ = V⁻)
In an ideal op-amp, voltage at the inverting input, V⁻ = V_in
The feedback network forms a voltage divider: V⁻ = V_out × [R1/(R1+R_f)]
Equating the above two equations: V_in = V_out × [R1/(R1+R_f)]
Rearranging: V_out/V_in = (R1+R_f)/R1 = 1 + (R_f/R1)
Therefore, A_v = 1 + (R_f/R1)

Characteristics of Non-inverting Amplifier:

Output is in phase with input (0° phase shift)
High input impedance makes it ideal as voltage amplifier
Gain is always greater than 1
Noise rejection is lower than inverting amplifier

Mnemonic: “UPON” - Unity Plus One plus Noninverting gain

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

Draw symbol of operational amplifier. Draw pin diagram of IC 741.

Answer:

Op-Amp Symbol:

IC 741 Pin Diagram:

Mnemonic: “7-PIN” - 741 Pinout INcludes power, inputs, null, output

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

Draw and explain inverting configuration of op-amp with derivation of voltage gain.

Answer:

Inverting Amplifier Circuit:

Step	Description
1	Apply virtual ground concept (V⁻ ≈ 0)
2	Current through R_i: I_i = V_in/R_i
3	Current through R_f: I_f = -V_out/R_f
4	By Kirchhoff’s current law: I_i + I_f = 0
5	Therefore, V_in/R_i = V_out/R_f
6	Voltage gain: A_v = V_out/V_in = -R_f/R_i

Mnemonic: “IRON” - Inverting Ratio Of Negative feedback

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

Explain Op-amp as an Integrator.

Answer:

Integrator Circuit:

Parameter	Description
Transfer Function	V_out = -(1/RC) ∫V_in dt
Input Signal	Any waveform (DC, sine, square, etc.)
Output for Constant Input	Ramp (linearly increasing/decreasing)
Output for Square Wave	Triangular wave
Output for Sine Wave	Cosine wave (90° phase shift)

Waveform Transformations:

Practical Considerations:

Need for reset switch across capacitor
Saturation due to input offset voltage
Limited frequency range due to op-amp bandwidth

Mnemonic: “SIRT” - Signal Integration Results in Time-domain transformation

Question 5(a) [3 marks]
#

Draw the diagram of Sequential Timer.

Answer:

Sequential Timer Circuit using IC 555:

Mnemonic: “STTR” - Sequential Timing Through Relay-like operation

Question 5(b) [4 marks]
#

Explain working of timer IC 555 using block diagram

Answer:

Block Diagram of IC 555:

graph TD
    A[Threshold Comparator] --> C[SR Flip-Flop]
    B[Trigger Comparator] --> C
    C --> D[Output Stage]
    C --> E[Discharge Transistor]
    F[Voltage Divider] --> A
    F --> B
    style C fill:#f9f,stroke:#333,stroke-width:2px

Block	Function
Voltage Divider	Creates reference voltages of (2/3)VCC and (1/3)VCC
Threshold Comparator	Compares threshold pin voltage with (2/3)VCC
Trigger Comparator	Compares trigger pin voltage with (1/3)VCC
SR Flip-Flop	Controls output state based on comparator inputs
Output Stage	Provides current to drive external loads
Discharge Transistor	Discharges timing capacitor when output is low

Mnemonic: “VTTDO” - Voltage divider, Two comparators, Toggle flip-flop, Discharge, Output

Question 5(c) [7 marks]
#

Explain astable multivibrator of timer IC 555.

Answer:

Astable Multivibrator Circuit:

Parameter	Formula	Description
Charging Time (HIGH)	t₁ = 0.693 × (Ra + Rb) × C	Output HIGH duration
Discharging Time (LOW)	t₂ = 0.693 × Rb × C	Output LOW duration
Total Period	T = t₁ + t₂ = 0.693 × (Ra + 2Rb) × C	Complete cycle time
Frequency	f = 1.44/((Ra + 2Rb) × C)	Number of cycles per second
Duty Cycle	D = (Ra + Rb)/(Ra + 2Rb)	Ratio of HIGH time to total period

Waveforms:

Mnemonic: “FREE” - Frequency Related to External Elements

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

Draw Pin Diagram of IC 555.

Answer:

IC 555 Pin Configuration:

Pin Name	Pin Number	Function
GND	1	Ground reference
TRIGGER	2	Starts timing cycle when < 1/3 VCC
OUTPUT	3	Output terminal
RESET	4	Resets timing cycle when LOW
CONTROL	5	Controls threshold and trigger levels
THRESHOLD	6	Ends timing cycle when > 2/3 VCC
DISCHARGE	7	Discharges timing capacitor
VCC	8	Positive supply voltage (4.5V-18V)

Mnemonic: “GTORCTDV” - Ground, Trigger, Output, Reset, Control, Threshold, Discharge, Vcc

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

Explain monostable multivibrator of timer IC 555.

Answer:

Monostable Multivibrator Circuit:

Parameter	Description
Trigger	Negative edge triggered at pin 2 (<1/3 VCC)
Pulse Width	T = 1.1 × R × C seconds
Operating States	Stable state (output LOW) and quasi-stable state (output HIGH)
Reset	Can be terminated early by applying LOW to Reset pin

Monostable Operation:

Output normally LOW
Negative trigger pulse initiates timing cycle
Output goes HIGH for duration T
After time T, output returns to LOW
Circuit ignores additional trigger pulses during timing cycle

Mnemonic: “OPTS” - One Pulse Timed by Single trigger

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

Explain bistable multivibrator of timer IC 555.

Answer:

Bistable Multivibrator Circuit:

State	Condition	Output
Set State	Trigger pin (2) momentarily pulled below 1/3 VCC	HIGH
Reset State	Reset pin (4) momentarily pulled LOW	LOW
Memory Function	Maintains state until changed by input	Stable in either state

Bistable Operation: