GATE Analog Circuits by Kanodia

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    Eighth Edition

    GATEELECTRONICS & COMMUNICATION

    Analog CircuitsVol 5 of 10

    RK KanodiaAshish Murolia

    NODIA & COMPANY

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    GATE Electronics & Communication Vol 5, 8eAnalog CircuitsRK Kanodia & Ashish Murolia

    Copyright By NODIA & COMPANY

    Information contained in this book has been obtained by author, from sources believes to be reliable. However,neither NODIA & COMPANY nor its author guarantee the accuracy or completeness of any information herein,and NODIA & COMPANY nor its author shall be responsible for any error, omissions, or damages arising out ofuse of this information. This book is published with the understanding that NODIA & COMPANY and its authorare supplying information but are not attempting to render engineering or other professional services.

    MRP 660.00

    NODIA & COMPANYB 8, Dhanshree Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039Ph : +91 141 2101150,www.nodia.co.inemail : [email protected]

    Printed by Nodia and Company, Jaipur

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    To Our Parents

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    Preface to the Series

    For almost a decade, we have been receiving tremendous responses from GATE aspirants for our earlier books:GATE Multiple Choice Questions, GATE Guide, and the GATE Cloud series. Our first book, GATE MultipleChoice Questions (MCQ), was a compilation of objective questions and solutions for all subjects of GATEElectronics & Communication Engineering in one book. The idea behind the book was that Gate aspirants who

    had just completed or about to finish their last semester to achieve his or her B.E/B.Tech need only to practiceanswering questions to crack GATE. The solutions in the book were presented in such a manner that a studentneeds to know fundamental concepts to understand them. We assumed that students have learned enough ofthe fundamentals by his or her graduation. The book was a great success, but still there were a large ratio ofaspirants who needed more preparatory materials beyond just problems and solutions. This large ratio mainlyincluded average students.

    Later, we perceived that many aspirants couldnt develop a good problem solving approach in their B.E/B.Tech.Some of them lacked the fundamentals of a subject and had difficulty understanding simple solutions. Now,we have an idea to enhance our content and present two separate books for each subject: one for theory, whichcontains brief theory, problem solving methods, fundamental concepts, and points-to-remember. The second bookis about problems, including a vast collection of problems with descriptive and step-by-step solutions that can

    be understood by an average student. This was the origin of GATE Guide(the theory book) and GATE Cloud(the problem bank) series: two books for each subject. GATE Guideand GATE Cloudwere published in threesubjects only.

    Thereafter we received an immense number of emails from our readers looking for a complete study packagefor all subjects and a book that combines both GATE Guideand GATE Cloud. This encouraged us to presentGATE Study Package (a set of 10 books: one for each subject) for GATE Electronic and CommunicationEngineering. Each book in this package is adequate for the purpose of qualifying GATE for an average student.Each book contains brief theory, fundamental concepts, problem solving methodology, summary of formulae,and a solved question bank. The question bank has three exercises for each chapter: 1) Theoretical MCQs, 2)Numerical MCQs, and 3) Numerical Type Questions (based on the new GATE pattern). Solutions are presentedin a descriptive and step-by-step manner, which are easy to understand for all aspirants.

    We believe that each book of GATE Study Package helps a student learn fundamental concepts and developproblem solving skills for a subject, which are key essentials to crack GATE. Although we have put a vigorouseffort in preparing this book, some errors may have crept in. We shall appreciate and greatly acknowledge allconstructive comments, criticisms, and suggestions from the users of this book. You may write to us at [email protected] and [email protected].

    Acknowledgements

    We would like to express our sincere thanks to all the co-authors, editors, and reviewers for their efforts in

    making this project successful. We would also like to thank Team NODIA for providing professional support forthis project through all phases of its development. At last, we express our gratitude to God and our Family forproviding moral support and motivation.

    We wish you good luck !R. K. KanodiaAshish Murolia

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    SYLLABUS

    GATE Electronics & Communications

    Small Signal Equivalent circuits of diodes, BJTs, MOSFETs and analog CMOS. Simple diode circuits, clipping,clamping, rectifier. Biasing and bias stability of transistor and FET amplifiers. Amplifiers: single-and multi-stage,differential and operational, feedback, and power. Frequency response of amplifiers. Simple op-amp circuits. Filters.Sinusoidal oscillators; criterion for oscillation; single-transistor and op-amp configurations. Function generators andwave-shaping circuits, 555 Timers. Power supplies.

    IES Electronics & Telecommunication

    Transistor biasing and stabilization. Small signal analysis. Power amplifiers. Frequency response. Wide bandingtechniques. Feedback amplifiers. Tuned amplifiers. Oscillators. Rectifiers and power supplies. Op Amp, PLL, otherlinear integrated circuits and applications. Pulse shaping circuits and waveform generators.

    **********

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    CONTENTS

    CHAPTER 1 DIODE CIRCUITS

    1.1 INTRODUCTION 1

    1.2 DIODE 1

    1.2.1 Operating Modes of a Diode 1

    1.2.2 Current-Voltage Characteristics of a Diode 2

    1.2.3 Current-Voltage Characteristics of an Ideal Diode 2

    1.3 LOAD LINE ANALYSIS 3

    1.4 PIECEWISE LINEAR MODEL 3

    1.5 SMALL SIGNAL MODEL 4

    1.5.1 Small Signal Resistance 4

    1.5.2 AC and DC Equivalent Model 4

    1.6 CLIPPER AND CLAMPER CIRCUITS 6

    1.6.1 Clippers 6

    1.6.2 Clampers 8

    1.7 VOLTAGE MULTIPLIER CIRCUIT 9

    1.7.1 Voltage Doubler 10

    1.7.2 Voltage Tripler and Quadrupler 11

    1.8 RECTIFIER CIRCUIT 11

    1.8.1 Parameters of Rectifier Circuit 12

    1.8.2 Classification of Rectifiers 12

    1.9 HALF WAVE RECTIFIERS 12

    1.10 FULL WAVE RECTIFIERS 14

    1.10.1 Centre Taped Full wave Rectifier 14

    1.10.2 Bridge Rectifier 15

    1.11 FILTERS 15

    1.12 ZENER DIODE 16

    1.13 VOLTAGE REGULATORS 16

    EXERCISE 1.1 18

    EXERCISE 1.2 36

    EXERCISE 1.3 42

    SOLUTIONS 1.1 47

    SOLUTIONS 1.2 90

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    SOLUTIONS 1.3 111

    CHAPTER 2 BJT BIASING

    2.1 INTRODUCTION 117

    2.2 BASIC BIPOLAR JUNCTION TRANSISTOR 117

    2.2.1 Simplified Structure of BJT 117

    2.2.2 Operating Modes of BJT 118

    2.2.3 Circuit Symbol and Conventions for a BJT 118

    2.3 BJT CONFIGURATION 119

    2.3.1 Common Base Configuration 119

    2.3.2 Common Emitter configuration 120

    2.3.3 Common-Collector Configuration 122

    2.4 CURRENT RELATIONSHIPS IN BJT 122

    2.4.1 Relation between Current Gain 122

    2.4.2 Relation between Leakage Currents 123

    2.5 LOAD LINE ANALYSIS 123

    2.6 BIASING 125

    2.6.1 Fixed Bias Circuit 125

    2.6.2 Emitter Stabilized Bias Circuit 126

    2.6.3 Voltage Divider Bias 128

    2.7 BIAS STABILIZATION 129

    2.7.1 Stability factor 129

    2.7.2 Total Effect on the Collector Current 129

    2.8 EARLY EFFECT 130

    EXERCISE 2.1 132

    EXERCISE 2.2 147

    EXERCISE 2.3 155

    SOLUTIONS 2.1 159

    SOLUTIONS 2.2 201

    SOLUTIONS 2.3 224

    CHAPTER 3 BJT AMPLIFIERS

    3.1 INTRODUCTION 229

    3.2 AC LOAD LINE ANALYSIS 229

    3.3 HYBRID EQUIVALENT MODEL 230

    3.3.1 Current Gain 230

    3.3.2 Voltage Gain 230

    3.3.3 Input Impedance 231

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    3.3.4 Output Impedance 231

    3.4 SMALL SIGNAL PARAMETER 232

    3.4.1 Collector Current and the Transconductance 232

    3.4.2 Base Current and Input Resistance at the Base 233

    3.4.3 Emitter Current and the Input Resistance at the Emitter 233

    3.5 HYBRID-p MODEL 233

    3.5.1 Hybrid p-model Circuit Including the Early Effect 235

    3.6 ANALYSIS OF STANDARD MODELS 235

    3.6.1 Common Emitter Fixed Bias Configuration 235

    3.6.2 Voltage Divider Bias 236

    3.6.3 Common-Emitter Bias Configuration 237

    3.7 FREQUENCY RESPONSE OF COMMON EMITTER AMPLIFIER 238

    3.7.1 Cut-off Frequency 239

    EXERCISE 3.1 241

    EXERCISE 3.2 254

    EXERCISE 3.3 260

    SOLUTIONS 3.1 265

    SOLUTIONS 3.2 295

    SOLUTIONS 3.3 315

    CHAPTER 4 FET BIASING

    4.1 INTRODUCTION 321

    4.2 JUNCTION FIELD EFFECT TRANSISTOR (JFET) 321

    4.2.1 Circuit Symbols of JFET 321

    4.2.2 Characteristics of JFET 322

    4.3 METAL-OXIDE SEMICONDUCTOR FIELD EFFECT TRANSISTOR (MOSFET) 323

    4.3.1 n-channel Enhancement Type MOSFET 323

    4.3.2 p-channel Enhancement Type MOSFET 325

    4.3.3 n-channel Depletion Type MOSFET 326

    4.3.4 p-channel Depletion Type MOSFET 326

    4.4 SOME STANDARD CONFIGURATIONS FOR JFET 328

    4.4.1 Fixed Bias Configuration 328

    4.4.2 Self Bias Configuration 329

    4.4.3 Voltage Divider Biasing 330

    4.5 BIASING CONFIGURATION FOR DEPLETION TYPE MOSFETS 331

    4.6 SOME STANDARD CONFIGURATIONS FOR ENHANCEMENT TYPE MOSFET CIRCUITS 331

    4.6.1 Feedback Biasing Configuration 331

    4.6.2 Voltage Divider Biasing Configuration 332

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    4.6.3 Enhancement Mode NMOS device with the Gate Connected to the Drain 333

    EXERCISE 4.1 334

    EXERCISE 4.2 347

    EXERCISE 4.3 354

    SOLUTIONS 4.1 358

    SOLUTIONS 4.2 388

    SOLUTIONS 4.3 408

    CHAPTER 5 FET AMPLIFIERS

    5.1 INTRODUCTION 413

    5.2 SMALL SIGNAL ANALYSIS OF JFET CIRCUIT 413

    5.2.1 Transconductance 413

    5.2.2 Output Resistance 414

    5.3 SOME STANDARD CONFIGURATIONS 414

    5.3.1 JFET Fixed Bias Configuration 414

    5.3.2 JFET Self Bias Configuration with bypassed Capacitor 415

    5.3.3 JFET Self Bias Configuration with Unbypassed RS 416

    5.3.4 JFET Voltage Divider Configuration 418

    5.3.5 JFET Source Follower (Common Drain) Configuration 418

    5.3.6 JFET Common Gate Configuration 420

    5.4 SMALL SIGNAL ANALYSIS OF DEPLETION TYPE MOSFET 421

    5.5 SMALL SIGNAL ANALYSIS FOR ENHANCEMENT TYPE MOSFET 422

    EXERCISE 5.1 423

    EXERCISE 5.2 432

    EXERCISE 5.3 438

    SOLUTIONS 5.1 442

    SOLUTIONS 5.2 467

    SOLUTIONS 5.3 483

    CHAPTER 6 OUTPUT STAGES AND POWER AMPLIFIERS

    6.1 INTRODUCTION 487

    6.2 GENERAL CONSIDERATION 487

    6.2.1 Power 487

    6.2.2 Power Efficiency 487

    6.3 EMITTER FOLLOWER AS POWER AMPLIFIER 487

    6.3.1 Small Signal Voltage Gain of Emitter Follower 487

    6.3.2 Relation between Input and Output Voltage 488

    6.3.3 Emitter Follower Power Rating 488

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    6.3.4 Power Efficiency 489

    6.4 PUSH-PULL STAGE 489

    6.5 CLASSES OF AMPLIFIERS 490

    6.5.1 Class-AOperation 491

    6.5.2 Class-BOperation : 492

    6.5.3 Class-ABOutput Stage 493

    6.6 AMPLIFIER DISTORTION 494

    6.6.1 Total harmonic Distortion 494

    6.6.2 Relationship Between Total Power and THD 494

    6.7 HEAT SINKS 494

    6.7.1 Junction Temperature 495

    6.7.2 Thermal Resistance 495

    6.7.3 Transistor Case and Heat Sink 495

    EXERCISE 6.1 496

    EXERCISE 6.2 508

    EXERCISE 6.3 513

    SOLUTIONS 6.1 517

    SOLUTIONS 6.2 539

    SOLUTIONS 6.3 551

    CHAPTER 7 OP- AMP CHARACTERISTICS AND BASIC CIRCUITS

    7.1 INTRODUCTION 555

    7.2 OPERATIONAL AMPLIFIER 555

    7.3 IDEAL OP-AMP CIRCUIT 555

    7.3.1 Transfer Characteristic of Ideal Op-amp 556

    7.3.2 Common Mode Signal for Ideal Op-amp 556

    7.4 PRACTICAL OP-AMP CIRCUITS 556

    7.4.1 Inverting Amplifier 556

    7.4.2 Non-inverting Amplifier 558

    7.4.3 Unity Follower 558

    7.4.4 Summing Amplifier 558

    7.4.5 Amplifier with a T-network 559

    7.5 PRACTICAL OP-AMP CIRCUITS WITH FINITE GAIN 559

    7.5.1 Unity Follower 560

    7.5.2 Inverting Amplifier 560

    7.5.3 Non-inverting Amplifier 561

    7.6 SLEW RATE 561

    7.6.1 Maximum Signal Frequency in terms of Slew Rate 562

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    7.7 DIFFERENTIAL AND COMMON-MODE OPERATION 562

    7.7.1 Differential Inputs 562

    7.7.2 Common Inputs 562

    7.7.3 Output voltage 562

    7.7.4 Common Mode Rejection Ratio (CMRR) 562

    7.8 DC OFFSET PARAMETER 563

    7.8.1 Output Offset Voltage due to Input Offset Voltage 563

    7.8.2 Output Offset Voltage due to Input Offset Current 563

    EXERCISE 7.1 565

    EXERCISE 7.2 578

    EXERCISE 7.3 587

    SOLUTIONS 7.1 591

    SOLUTIONS 7.2 619

    SOLUTIONS 7.3 644

    CHAPTER 8 OP - AMP APPLICATION

    8.1 INTRODUCTION 649

    8.2 INVERTING AMPLIFIER 649

    8.3 NON-INVERTING AMPLIFIER 650

    8.4 MULTIPLE-STAGE GAINS 650

    8.5 VOLTAGE SUBTRACTION 650

    8.6 CURRENT TO VOLTAGE CONVERTER 651

    8.7 VOLTAGE TO CURRENT CONVERTER 651

    8.8 DIFFERENCE AMPLIFIER 652

    8.9 INSTRUMENTATION AMPLIFIER 653

    8.10 INTEGRATOR 654

    8.11 DIFFERENTIATOR 655

    8.12 LOGARITHMIC AMPLIFIER 655

    8.13 EXPONENTIAL AMPLIFIER 656

    8.14 SQUARE-ROOT AMPLIFIER 656

    8.15 COMPARATOR 657

    8.16 SCHMITT TRIGGER 657

    8.17 NON INVERTING SCHMITT TRIGGER CIRCUIT 658

    8.18 PRECISION RECTIFIER 659

    8.19 FUNCTION GENERATOR 660

    EXERCISE 8.1 661

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    EXERCISE 8.2 679

    EXERCISE 8.3 684

    SOLUTIONS 8.1 688

    SOLUTIONS 8.2 734

    SOLUTIONS 8.3 751

    CHAPTER 9 ACTIVE FILTERS

    9.1 INTRODUCTION 757

    9.2 ACTIVE FILTER 757

    9.2.1 Low Pass Filter 757

    9.2.2 High Pass Filter 759

    9.2.3 Band pass filter 759

    9.3 THE FILTER TRANSFER FUNCTION 760

    9.3.1 Pole-Zero Pattern of Low Pass Filter 761

    9.3.2 Pole-Zero Pattern of Band Pass Filter 7619.3.3 First-Order Filter Transfer Function 762

    9.3.4 Second-order Filter Transfer Function 763

    9.4 BUTTERWORTH FILTERS 765

    9.5 THE CHEBYSHEV FILTER 765

    9.6 SWITCHED CAPACITOR FILTER 765

    9.7 SENSITIVITY 766

    EXERCISE 9.1 767

    EXERCISE 9.2 778

    EXERCISE 9.3 781

    SOLUTIONS 9.1 786

    SOLUTIONS 9.2 813

    SOLUTIONS 9.3 818

    CHAPTER 10 FEEDBACK AMPLIFIER AND OSCILLATOR

    10.1 INTRODUCTION 821

    10.2 FEEDBACK 821

    10.2.1 Negative Feedback 821

    10.2.2 Positive Feedback 822

    10.3 THE FOUR BASIC FEEDBACK TOPOLOGIES 822

    10.3.1 Voltage Amplifier 822

    10.3.2 Current Amplifier 823

    10.3.3 Transconductance Amplifier 823

    10.3.4 Transresistance Amplifier 824

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    10.4 ANALYSIS OF FEEDBACK AMPLIFIER 824

    10.5 OSCILLATORS 826

    10.6 OP-AMP RC OSCILLATOR CIRCUITS 826

    10.6.1 Wein Bridge Oscillator 826

    10.6.2 Phase Shift Oscillator 827

    10.7 LC OSCILLATOR CIRCUIT 827

    10.7.1 Colpitts Oscillator 827

    10.7.2 Hartley oscillator 828

    10.8 THE 555 CIRCUIT 828

    10.8.1 Monostable Multivibrator 828

    10.8.2 Astable Multivibrator 829

    EXERCISE 10.1 830

    EXERCISE 10.2 840

    EXERCISE 10.3 844

    SOLUTIONS 10.1 849

    SOLUTIONS 10.2 873

    SOLUTIONS 10.3 878

    ***********

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    Page 15

    Chap 1

    Diode Circuit

    Page 15

    Chap 1

    Diode Circuit

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    GATE STUDY PACKAGE Electronics & Communication

    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    1.1 INTRODUCTION

    A general goal of this chapter is to develop the ability to use the piece wiselinear model and approximation techniques in the hand analysis and designof various diode circuits. The chapter includes the following topics: Introduction to diode

    AC and DC analysis of diode.

    Application of diodes to perform signal processing functions: rectification,

    clipping and clamping.

    Zener diode, which operates in the reverse breakdown region

    Application of Zener diode in voltage regulators

    1.2 DIODE

    Diode is a two terminal device with nonlinear i-v (current-voltage)characteristics. Figure 1.1 shows the circuit symbol of a diode. In the diodesymbol, the triangular head denoting the allowable direction of current flowand the vertical bar representing the blocking behaviour for currents in theopposite direction. The corresponding terminals are called the anode (or p-terminal) and the cathode (or n-terminal) respectively.

    Figure 1.1: Diode Circuit Symbol

    1.2.1 Operating Modes of a Diode

    A diode operates in the following two modes:1. Forward bias

    2. Reverse bias

    Forward Bias

    If the p-terminal of a diode is at higher voltage level than the n-terminal(i.e. positive voltage applied across diode), a positive current flows throughthe diode. The diode, operating in this mode, is said to be turned ON orforward biased. Mathematically, we define the condition for a forward biaseddiode as

    CHAPTER 1DIODE CIRCUITS

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    Diode Circuits

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    Diode Circuits

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    General Aptitude Engineering Mathematics

    Forward bias

    V V

    V V

    V V V 0

    >

    >

    >

    anode cathode

    p n

    D p n= -

    _

    `

    a

    bb

    bb

    Reverse Bias

    If the p-terminal of an ideal diode is at lower voltage level than the n

    -terminal (i.e. negative voltage applied across diode), then there is no current

    across the diode. The diode operating in this mode is said to be turned OFFor reverse biased. Mathematically, we define the condition for a reversebiased diode as

    Reverse bias

    V V

    V V

    V V V 0

    > l, so we may write

    Vr V RCT

    m. lb l

    or Vr V RCT

    m. (T T.l )

    or Vr fRCVm= ( /f T1= )

    1.12 ZENER DIODE

    Zener diodes are designed to provide a specified breakdown voltage. Thebreakdown voltage of the zener diode is nearly constant over a wide rangeof reverse bias currents. This makes the zener diode useful in a voltageregulator, or a constant voltage reference circuit. Figure 1.16 shows thezener diode and its equivalent circuit models for ON and OFF states.

    (a) (b)

    Figure 1.16: Equivalent Circuit Model for a Zener Diode for (a) ON State and (b) OFFState

    1.13 VOLTAGE REGULATORS

    A Voltage regulator is a device or combination of devices designed to maintainthe output voltage of a power supply as nearly constant as possible. One of

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    Diode Circuit

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    GATE STUDY PACKAGE Electronics & Communication

    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    the simplest discrete Regulator consists of only a resistor and a zener diode,as shown in Figure 1.17.

    Figure 1.17: Basic Zener Regulator

    METHODOLOGY: TO ANALYSE ZENER REGULATOR CIRCUIT

    The applied dc voltage is fixed, as is the load resistor. The zener regulatorcan be analysed in the following steps.Step 1: Remove the zener diode from the network.

    Step 2: Calculate the voltage across the resulting open circuit.

    V VR R

    RvL

    L

    Li= = +

    Step 3: Determine the state of the zener diode by checking the obtainedvalue of voltage Vfor the following conditions

    If V Vz$ , the zener diode is ON

    If V V< z, the diode is OFFStep 4: Substitute the appropriate equivalent circuit for the resulting

    state of zener diode. For ON state, equivalent mode of Figure1.16(a) can be substituted, while for OFF state the open circuit

    equivalence of Figure 1.16(b) is substituted.

    Step 5: Solve the resulting equivalent circuit for the desired unknowns.For example, assume that the equivalent circuit for the zenerregulator is as shown below.

    So, we get the various parameters for the circuit as

    Load voltage,VL Vz=

    Load current, IL RVL=

    Current through resistance R, IR RV

    RV VR i L= = -

    Zener current, Iz I IR L= -

    Power dissipated by Zener diode, Pz V Iz z=

    ***********

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    Page 32

    Chap 1

    Diode Circuits

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    Diode Circuits

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    General Aptitude Engineering Mathematics

    EXERCISE 1.1

    MCQ 1.1.1 Consider the given circuit and a waveform for the input voltage, shown infigure below. The diode in circuit has cutin voltage 0V =g .

    The waveform of output voltage vois

    MCQ 1.1.2 Consider the given circuit and a waveform for the input voltage, shown infigure below. The diode in circuit has cutin voltage 0V =g .

    The waveform of output voltage vois

    MCQ 1.1.3 Consider the given circuit and a waveform for the input voltage, shown infigure below. The diode in circuit has cutin voltage 0V =g .

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    Page 33

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    Diode Circuit

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    Diode Circuit

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    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    The waveform of output voltage vois

    MCQ 1.1.4 Consider the given circuit and a waveform for the input voltage. The diodein circuit has cutin voltage 0V =g .

    The waveform of output voltage vois

    MCQ 1.1.5 For the circuit shown below, let cut in voltage 0.7V =g V.

    The plot of voverses vifor 10 10vi# #- V is

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    Page 34

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    Diode Circuits

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    Diode Circuits

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    General Aptitude Engineering Mathematics

    MCQ 1.1.6 For the circuit shown below the cutin voltage of diode is 0.7V =g V.

    The plot of voversus viis

    MCQ 1.1.7 For the circuit shown below each diode has 0.7V =g V.

    The vofor 10 10vs# #- V is

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    Page 35

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    Diode Circuit

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    Diode Circuit

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    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    MCQ 1.1.8 A symmetrical 5 kHz square wave whose output varies between 10 V+ and10 V- is impressed upon the clipping shown in figure below

    If diode has 0rf= and 2rr= MWand 0V =g , the output waveform is

    MCQ 1.1.9 In the circuit shown below, the three signals of fig are impressed on theinput terminals.

    If diode are ideal then the voltage vois

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    Diode Circuits

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    General Aptitude Engineering Mathematics

    MCQ 1.1.10 For the circuit shown below the input voltage viis as shown in figure.

    Assume the RCtime constant large and cutin voltage of diode 0V =g . Theoutput voltage vois

    MCQ 1.1.11 For the circuit shown below, the input voltage viis as shown in figure.

    Assume the RCtime constant large and cutin voltage 0V =g . The outputvoltage vois

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    MCQ 1.1.12 In the circuit Iis DC current and capacitors are very large. Using small

    signal model which of following is correct ?

    (A) v vV IR

    Vo s

    T s

    T

    hh

    =+

    (B) v vo s=

    (C) vV IR

    vo

    T S

    s

    h=

    + (D) 0vo=

    Common Data For Q. 13 to 15 :

    Consider the circuit shown below. Assume diodes are ideal.

    MCQ 1.1.13 If 10v1 = V and 5v2 = V, then output voltage vois(A) 9 V (B) 9.474 V

    (C) 0 V (D) 8.943 V

    MCQ 1.1.14 If 10v v1 2= = V, then output voltage vois(A) 9 V (B) 9.474 V

    (C) 4 V (D) 8.943 V

    MCQ 1.1.15 If 5v1 =- V and 5v2 = V then vois(A) 9.474 V (B) 8.943 V

    (C) 4.5 V (D) 9 V

    Common Data For Q.16 and 17 :

    Consider the circuit shown below. Assume diodes are ideal.

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    MCQ 1.1.16 If 10v v1 2= = V, then output voltage vois(A) 0 V (B) 9.737 V

    (B) 9 V (D) 9.5 V

    MCQ 1.1.17 If 5v1 =- V and 10v2 = V, the output voltage vois(A) 9 V (B) 9.737 V

    (C) 9.5 V (D) 4.5 V

    MCQ 1.1.18 In the circuit shown below diodes has cutin voltage of 0.6 V. The diode inON state are

    (A) only D1 (B) only D2

    (C) both D1and D2 (D) None of these

    MCQ 1.1.19 For the circuit shown below cutin voltage of diode is 0.7V =g . What is the

    value of vand i?

    (A) 2.3 V, 2.65 mA (B) 2.65 V, 2.3 mA

    (C) 2 V, 0 mA (D) 0 V, 2.3 mA

    MCQ 1.1.20 For the circuit shown below the value of vand iare (if the diode is ideal)

    (A) 5+ V, 0 mA (B) 1+ V, 0.6 mA

    (C) 5+ V, 0.4 mA (D) 1+ V, 0.4 mA

    MCQ 1.1.21 For the circuit shown below each diode has 0.6V =g V and 0rf= . Bothdiode will be ON if

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    (A) 3.9v >s V (B) 4.9v >s V

    (C) 6.3v >s V (D) 5.3v >s V

    Common Data For Q. 22 and 23 :

    The diode In the circuit shown below has the non linear terminal characteristicas shown in figure. Let the voltage be cosv ti w= V.

    MCQ 1.1.22 The current iDis(A) 2.5(1 )cos tw+ mA (B) 5(0.5 )cos tw+ mA

    (C) 5(1 )cos tw+ mA (D) 5(1 0.5 )cos tw+ mA

    MCQ 1.1.23 The voltage vDis

    (A) 0.25(3 )cos tw+ V(B) 0.25(1 3 )cos tw+ V

    (C) 0.5(3 1 )cos tw+ V

    (D) 0.5(2 3 )cos tw+ V

    MCQ 1.1.24 The circuit inside the box in figure shown below contains only resistor anddiodes. The terminal voltage vois connected to some point in the circuitinside the box.

    The largest and smallest possible value of vomost nearly to is respectively(A) 15 V, 6 V

    (B) 24 V, 0 V

    (C) 24 V, 6 V

    (D) 15 V, 9- V

    MCQ 1.1.25 The Q-point for the Zener diode shown below is

    (A) (0.34 mA, 4 V)

    (B) (0.34 mA, 4.93 V)

    (C) (0.94 mA, 4 V)

    (D) (0.94 mA, 4.93 V)

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    Common Data For Q. 26 to 28 :

    In the voltage regulator circuit shown below the zener diode current is to belimited to the range 5 100iz# # mA.

    MCQ 1.1.26 The range of possible load current is(A) 5 130iL# # mA

    (B) 25 120iL# # mA

    (C) 10 110iL# # mA

    (D) None of the above

    MCQ 1.1.27 The range of possible load resistance is(A) 60 372RL# # W

    (B) 60 200RL# # W

    (C) 40 192RL# # W(D) 40 360RL# # W

    MCQ 1.1.28 The power rating required for the load resistor is(A) 576 mW (B) 360 Wm

    (C) 480 mW (D) 75 Wm

    MCQ 1.1.29 A clipper circuit is shown below.

    Assuming forward voltage drops of the diodes to be 0.7 V, the input-outputtransfer characteristics of the circuit is

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    MCQ 1.1.30 The following circuit has a source voltage Vs as shown in the graph. Thecurrent through the circuit is also shown.

    The element connected between aand bcould be

    MCQ 1.1.31 In the voltage doubler circuit shown in the figure, the switch S is closed att 0= . Assuming diodes D1and D2to be ideal, load resistance to be infiniteand initial capacitor voltages to be zero. The steady state voltage acrosscapacitor C1and C2will be

    (A) 10 , 5V VV Vc c1 2= =

    (B) 10 , 5V VV Vc c1 2= =-

    (C) 5 , 10V VV Vc c1 2= =

    (D) 5 , 10V VV Vc c1 2= =-

    MCQ 1.1.32 What are the states of the three ideal diodes of the circuit shown in figure ?

    (A) , ,D D DON OFF OFF1 2 3

    (B) , ,D D DOFF ON OFF1 2 3

    (C) , ,D D DON OFF ON1 2 3

    (D) , ,D D DOFF ON ON1 2 3

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    MCQ 1.1.33 The equivalent circuit of a diode, during forward biased and reverse biasedconditions, are shown in the figure.

    (I)

    (II)

    If such a diode is used in clipper circuit of figure given above, the outputvoltage Voof the circuit will be

    MCQ 1.1.34 Assuming the diodes D1and D2of the circuit shown in figure to be idealones, the transfer characteristics of the circuit will be

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    MCQ 1.1.35 A voltage signal sin t10 w is applied to the circuit with ideal diodes, as shownin figure, The maximum, and minimum values of the output waveform Voutof the circuit are respectively

    (A) 10 V+ and 10 V-

    (B) 4 V+ and 4 V-

    (C) 7 V+ and 4 V-

    (D) 4 V+ and 7 V-

    MCQ 1.1.36 The forward resistance of the diode shown in Figure is 5 Wand the remainingparameters are same at those of an ideal diode. The dc component of thesource current is

    (A) V50

    m

    p (B) V

    50 2m

    p

    (C) V

    100 2m

    p (D) V

    502 mp

    MCQ 1.1.37 The cut-in voltage of both zener diode DZand diode Dshown in Figureis 0.7 V, while break-down voltage of DZis 3.3 V and reverse break-downvoltage of Dis 50 V. The other parameters can be assumed to be the sameas those of an ideal diode. The values of the peak output voltage ( )Vo are

    (A) 3.3 V in the positive half cycle and 1.4 V in the negative half cycle.

    (B) 4 V in the positive half cycle and 5 V in the negative half cycle.

    (C) 3.3 V in both positive and negative half cycles.

    (D) 4 V in both positive and negative half cycle

    Common Data For Q. 38 and 39

    In the circuit shown in Figure, the source I is a dc current source. Theswitch Sis operated with a time period Tand a duty ratio D. You mayassume that the capacitance Chas a finite value which is large enoughso that the voltage. VChas negligible ripple, calculate the following understeady state conditions, in terms of D, Iand R

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    MCQ 1.1.38 The voltage Vc, with the polarity shown in Figure,

    (A)CI

    (B) ( )CI

    DT1 -

    (C) ( )CI

    D T1 - (D)CI

    T-

    MCQ 1.1.39 The average output voltage Vo, with the polarity shown in figure

    (A)CI

    T- (B)CI

    D T2

    2-

    (C) ( )CI DT

    21 - (D) (1 )

    CI D T

    2 -

    Common Data For Q. 40 and 41:

    Consider the given circuit and the cut in voltage of each diode is 0.6VV =g .

    MCQ 1.1.40 What is the value of ID1, ID2, ID3respectively?(A) 1.94 mA, 0.94 mA- , 1.86mA (B) 1.47 mA, 0.94 mA, 0 mA

    (C) 1.47 mA, 0 mA, 0.94 mA (D) 1.94 mA, 0 mA, 0.94 mA

    MCQ 1.1.41 The value of VD1, VD2, VD3respectively is(A) 0.6, .4 10- , 0.6 (B) 0.6, .8 8- , 0.6

    (C) 0.6, 0.6, 0.6 (D) 0.6, 0.6, .4 10-

    MCQ 1.1.42 The voltage regulator shown below, what are the nominal and worst casevalues of zener diode current if the power supply voltage, zener break downvoltage and registor all have 5 % tolerance?

    Inomz Inomz I min

    worstz h

    (A) 1.2 mA 0 mA 0 mA

    (B) 0.5 mA 0.70 mA 0.103 mA

    (C) 0.5 mA 0.60 mA 0.346 mA

    (D) 0.5 mA 0.796 mA 0.215 mA

    JAEGER/120/3.9

    JAEGER/120/3.9

    JAEGER/168/3.82

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    MCQ 1.1.43 What are the nominal and worst case values of zener power dissipation?(A) 10.8 mW, 0 mW, 0 mW (B) 4.5 mW, 7.52 nW, 1.80 mW

    (C) 4.5 mW, 6.81 mW, 2.03 mW (D) 4.5 mW, 5.13 mW, 5.67 mW

    MCQ 1.1.44 The diode circuit in figure shown below the biasing of the diode D1, D2is

    (A) ON, ON (B) ON, OFF

    (C) OFF, ON (D) OFF, OFF

    MCQ 1.1.45 The diode circuit shown in figure. Assume that diode is ideal, what will bethe biasing modes of diode D1, D2and D3? (FB "forward biased, RB "reverse biased)

    D1 D2 D3(A) FB FB FB(B) FB FB RB(C) FB RB RB(D) FB RB FB

    MCQ 1.1.46 The diode in the circuit shown below have linear parameter of .V 0 7=g (forSi), .V 0 3=g (for Ge) and r 0f= for both the diode. What is the biasingcondition of diode D1and D2?

    (A) OND1 - , OND2 - (B) OND1 - , OFFD2 -

    (C) OFFD1 - , OND2 - (D) OFFD1 - , OFFD2 -

    BOYLSTEAD/68/2.14

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    MCQ 1.1.47 In the voltage regulator circuit below rating of zener diode is given, then therange of values of vithat will maintain the zener diode in the ON state is

    (A) 16.87 36.87 Vvi1 1

    (B) 16.87 23.67 Vvi1 1

    (C) 23.67 36.87 Vvi1 1

    (D) None of the above

    MCQ 1.1.48 Consider the given a circuit and a waveform for the input voltage.

    If the diode has cut in voltage V 0=g , the output waveform of the circuit is

    MCQ 1.1.49 Assume that the diode cut in voltage for the circuit shown below is 0.7VV =g. Which of clamper circuit perform the function shown below ?

    BOYLSTEAD/109/29

    BOYLSTEAD/112/41

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    MCQ 1.1.50 Having lost his 2.4 V cellphone charger, an electrical engineering student

    tries several stores but does not find adaptors with outputs less than 3 V. Hethen decides to put his knowledge of electronics to work and constructs thecircuit shown in figure where three identical diodes in forward bias produce atotal voltage of 3 2.4 VV Vout D -= and resistor R1sustain remaining voltage.Neglect the current drawn by the cellphone and assume 26mVVThermal = .The reverse saturation current ISfor the diode is

    (A) 6 mA (B) 2.602 10 A16# -

    (C) 8.672 10 A14# - (D) 7.598 10 A14#

    -

    Common Data For Q. 25 to 27:

    The circuit diagram of a zener regulator is shown in figure below. The datasheet specification for zener IN4742A provides following values

    12 VVz= at 21mAIzT= , R 9z= (assume constant) 76mAIzM= , 1 mAIzk=.

    MCQ 1.1.51 The values of zener diode voltages V maxz^ h , V minz^ h respectively are(A) 11.82 V, 12.5 V

    (B) 12.5 V, 11.82 V

    (C) 12.18 V, 11.82 V

    (D) 12.5 V, 11.5 V

    MCQ 1.1.52 The maximum value of load current over which the zener diode is in ONstate.(A) 36.88 mA (B) 35.88 mA

    (C) 36.36 mA (D) 35.36 mA

    MCQ 1.1.53 The value of RL(Load resistance) corresponding to maximum load current is(A) 329.43 W (B) 334.44 W

    (C) 325.37 W (D) 320.49 W

    DESHPANDE/169/3.17

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    Common Data For Q. 54 and 55 :

    A breakdown diode has 6.2 VVz= at 25 Cc and 0.02%/ Cz ca = . A silicondiode with 0.7 VVF= and a temperature coefficient of . /mV C1 8 c- isconnected in series with the breakdown diode.

    MCQ 1.1.54 The new value of reference voltage and the temperature coefficient of theseries combination of diode Dand zener diode.

    (A) 6.9 V, 0.008 %/ Cc- (B) 6.9 V, 0.0289 %/ Cc-

    (C) 6.9 V, 0.0 %/ C56 c- (D) 6.2 V, 0.056 %/ Cc-

    MCQ 1.1.55 The new value of Vrefat a temperature of C50c is(A) 6.186 (B) 6.886 V

    (C) 6.914 V (D) 6.700 V

    MCQ 1.1.56 The diode circuit given below. Assume diode is ideal. The operating statesof diodes D1, D2are

    D1 D2(A) ON, ON

    (B) ON, OFF(C) OFF, ON

    (D) OFF, OFF

    MCQ 1.1.57 Consider the given circuit. The diode in circuit has cut in voltage 0.6VV =gand zener diode voltage 9.4 VVz= .

    Plot voversus viis

    REA/29/1.31

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    MCQ 1.1.58 Assume that the diodes in the circuit are ideal. What are the operatingstates of diodes ?

    D1 D2(A) ON ON(B) ON OFF(C) OFF ON(D) OFF OFF

    MCQ 1.1.59 Assuming that the diodes are ideal in the given circuit.

    Status of the diodes D1, D2and D3are respectively(A) ON, OFF, OFF

    (B) ON, OFF, ON

    (C) ON, ON, OFF

    (D) ON, ON, ON

    MCQ 1.1.60 What is the value of voltage Vand current Irespectively ?(A) 0 V, 0 Amp (B) 7.5 V, 7.5 mAmp

    (C) 7.5 V, 0 Amp (D) 0 V, 7.5 mAmp

    **********

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    EXERCISE 1.2

    QUES 1.2.1 In the circuit shown below, D1and D2are ideal diodes. The current i2is_____ mA

    QUES 1.2.2 Let cutin voltage 0.7V =g V for each diode in the circuit shown below.

    The voltage vois _____ volts

    QUES 1.2.3 For the circuit in the figure below. The value of vDis _____volts.

    -

    QUES 1.2.4 The cutin voltage for each diode in circuit shown below is 0.6V =g V. Eachdiode current is 0.5 mA. The value of R3will be _____ kW.

    QUES 1.2.5 The diodes in the circuit shown below has parameters 0.6V =g V and 0rf=. The current iD2is ____ mA.

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    Common Data For Q. 6 to 8 :

    The diodes in the given circuit have linear parameter of 0.6V =g V and 0rf= .

    QUES 1.2.6 If 10v1 = V and 0v2 = V, then vois _____ volts.

    QUES 1.2.7 If 10v1 = V and 5v2 = V, then vois_____ volts.

    QUES 1.2.8 If 0v v1 2= = , then output voltage vois _____ volts.

    Common Data For Q. 9 to 11 :

    The diodes in the circuit shown below have linear parameters of 0.6V =g V and 0rf= .

    QUES 1.2.9 If 0v2 = , then output voltage vois_____ volts.

    QUES 1.2.10 If 5v2 = V, then vois____volts.

    QUES 1.2.11 If 10v2 = V, then vois_____ volts.

    QUES 1.2.12 Ten diodes, each of them provides 0.7 V drop when the current through itis 20 mA, connected in parallel operate at a total current of 0.1 A. What

    current flows in each diode (in Amp) ?

    QUES 1.2.13 In the voltage regulator circuit shown below the maximum load current iLthat can be drawn is _____ mA.

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    QUES 1.2.14 In the voltage regulator shown below the power dissipation in the Zenerdiode is _____ watts.

    QUES 1.2.15 In the voltage regulator circuit shown below the power rating of Zener diodeis 400 mW. The value of RLthat will establish maximum power in Zenerdiode is _____ kW.

    QUES 1.2.16 The secondary transformer voltage of the rectifier circuit shown below is60 2 60sinv ts p= V. Each diode has a cut in voltage of 0.V 7=g V. The

    ripple voltage is to be no more than 2vrip = V. The value of filter capacitorwill be _____ Fm .

    QUES 1.2.17 The input to full-wave rectifier shown below is 120 2 60sinv ti p= V. The

    diode cutin voltage is 0.7 V. If the output voltage cannot drop below 100 V,the required value of the capacitor is _____ Fm .

    QUES 1.2.18 For the circuit shown below diode cutin voltage is 0vin= . The ripple voltageis to be no more than 4vrip = V. The minimum load resistance, that can beconnected to the output is _____ kW.

    QUES 1.2.19 Assuming that the diodes in the given circuit are ideal, the voltage Vois_____ volts.

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    QUES 1.2.20 Assume that D1and D2in given figure are ideal diodes. The value of currentis _____ mA.

    QUES 1.2.21 The current through the Zener diode in figure is _____ mA.

    QUES 1.2.22 Assuming that in given circuit the diodes are ideal. The current in diode D1is ____ mA.

    Common Data For Q. 23 and 24 :

    In the zener diode voltage regulator circuit shown in the figure below, thezener diode has the following parameter.

    Vz 5 V= , 0Rz W=

    QUES 1.2.23 What is the value of R minL (Minimum load resistance) for zener voltageregulator circuit (in kW) ?

    QUES 1.2.24 The value of output voltage vofor 1 kRL W= is ____ volts.

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    General Aptitude Engineering Mathematics

    QUES 1.2.25 If RL 3= , the value of power dissipation in the zener diode is ____ mW.

    Common Data For Q. 26 and 27 :

    Given that a half-wave rectifier circuit driven from a transformer havinga secondary voltage of .V12 6 rms, 60Hzf= with 15R W= and 25 mFC= .Assume the diode on voltage 1 VVon= .

    QUES 1.2.26 The value of the DC output voltage V,dco^ his _____ volts.QUES 1.2.27 The value of ripple voltage is ____ volts.QUES 1.2.28 Assume that in the given circuit diodes are ideal. The value of ID2is __mA.

    QUES 1.2.29 The diode in the circuit shown below have linear parameter of 0.7 VV =gand 0rf W= . The value of the current ID2^ hin diode D2is ____ mA.

    QUES 1.2.30 Given that 26mVVThermal = . A diode is biased with a current of 1 mA. What is

    the value of the current change (in Am

    ), ifV

    D(Diode voltage) changes by 1 mV?QUES 1.2.31 A transformer convert the 110 V, 60 Hz line voltage to a peak to peak swing of 9

    V. A half wave rectifier follows the transformer to supply the power to the laptopcomputer of 0.436RL W= . What is the minimum value of the filter capacitor(in Farad) that maintains the ripple below 0.1 V ? (Assume 0.8VV,D on= )

    QUES 1.2.32 A full wave rectifier is driven by a sinusoidal input cosV V tin o w= , where3 VVo= and 60 Hz2w p= ^ h. Assuming 800 mVV,D on= , what is the value

    of the ripple amplitude (in volt) with a 1000 Fm smoothing capacitor and aload resistance of 30 W.

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    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    Common Data For Q. 33 and 34 :

    A full wave rectifier circuit is powered by ac mains. Power transformerhas a center-tapped secondary. RMS secondary voltage across each half ofsecondary is 20 V. The dc winding resistance of each half of secondary is 5 W. Forward resistance of diode is 2 W. If the equivalent load resistance is 50 W,

    QUES 1.2.33 The value of dc power delivered to load is ____ watt.

    QUES 1.2.34

    The percentage load regulation is _____.QUES 1.2.35 The efficiency of rectification is ____

    QUES 1.2.36 In the circuit shown in figure, both diodes have 1h = , but D1has 10 timesthe junction area of D2. The value of voltage Vis ____ mV.

    QUES 1.2.37 Assume that the voltage drop across each of the diodes in the circuit shownbelow is 0.7 V. The value of current through diode D1, is ID1 =_____ mA

    Common Data For Q. 38 and 39 :

    The circuit shown below has 100R W= , 2 kRC W= , 2 kRL W= . Also, assumea constant diode voltage of 0.6 V and capacitors are very large using thesmall signal model for 1.6VVC= .

    QUES 1.2.38 What is the Q-point value of the diode current (in mA) ?

    QUES 1.2.39 The value ofv t

    v t

    in

    o

    ^^

    hhis _____

    ***********

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    EXERCISE 1.3

    MCQ 1.3.1 Which of the following filter types is suitable only for large values of loadresistance ?(A) Capacitor filter (B) Inductor filter

    (C) Choke-input filter (D) p-type CLC filter

    MCQ 1.3.2 Which of the following filter types is suitable only for small values of loadresistance ?(A) Capacitor filter (B) Inductor filter

    (C) Choke-input filter (D) p-type CLC filter

    MCQ 1.3.3 A rectifier is used to(A) convert a.c. voltage to d.c. voltage

    (B) convert d.c. voltage to a.c. voltage

    (C) both (a) and (b)

    (D) convert voltage to current

    MCQ 1.3.4 The ripple factor of a inductor filter is

    (A)L

    R w

    2 3L (B)

    L

    R

    2 3L

    (C)3 wL

    R

    2L (D)

    wL

    R

    3 2L

    MCQ 1.3.5 A bleeded resistor is used in a d.c. power supply because it(A) keeps the supply OFF (B) keeps the supply ON

    (C) improves filtering action (D) improves voltage regulation

    MCQ 1.3.6

    A half wave diode rectifier and full wave diode bridge rectifier both have aninput frequency of 50 Hz. The frequencies of outputs respectively are(A) 100 Hz and 50 Hz (B) 50 Hz and 100 Hz

    (C) 100 Hz each (D) 50 Hz each

    MCQ 1.3.7 A full wave bridge diode rectifier uses diodes having forward resistance of 50ohms each. The load resistance is also 50 ohms. The voltage regulations is(A) 20% (B) 50%

    (C) 100% (D) 200%

    MCQ 1.3.8 Which of the following components is essential for a voltage multiplier circuit(A) resistor (B) inductor

    (C) capacitor (D) both (A) and (C)

    MCQ 1.3.9 In the circuit of Fig. the approximate value of Voacross the germaniumdiode is

    MAINI/585/2

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    (A) 10 V (B) 5 V

    (C) 2.5 V (D) 0.3 V

    MCQ 1.3.10 When used in a circuit, a zener diode is always(A) forward-biased (B) reverse-biased

    (C) connected in series (D) at a temperature below 0 Cc

    MCQ 1.3.11 For ideal Rectifier and filter circuits, % regulations must be

    (A) 1% (B) 0.1%(C) 5% (D) 0%

    MCQ 1.3.12 A particular green LED emits light of wavelength 5490 Ac . The energybandgap of the semiconductor material used there is (Plancks constant

    6.626 10 J s34# -= - )

    (A) 2.26 eV (B) 1.98 eV

    (C) 1.17 eV (D) 0.74 eV

    MCQ 1.3.13 The function of bleeder resistor in a rectifier with LCfilter is to(A) maintain the minimum current through C

    (B) maintain the minimum current necessary for optimum inductor operation

    (C) maintain maximum current through L(D) Charge capacitor Cto maximum value

    MCQ 1.3.14 The forward resistance of the diode shown in figure is 5 Wand the remainingparameters are same as those of an ideal diode. The dc component of thesource current is

    (A) V50m

    p (B) V

    50 2m

    p

    (C) V100 2

    m

    p (D) V

    502 mp

    MCQ 1.3.15 The relation between diode current, voltage and temperature is given by

    (A) I I e1o VV

    T= - h_ i (B) I I e 1o VVT= -h_ i(C) I I e 1o V

    VT= +h_ i (D) I I eo VVT= h

    MCQ 1.3.16 The Transformer Utilization Factor (TUF) is defined as

    (A)P

    Pratedac

    dc (B)PP

    ac

    dc

    (C)

    I

    I

    dc

    rms (D)

    I

    I

    dc

    m

    MCQ 1.3.17 The Peak inverse voltage (PIV) of an half wave rectifier and full-waverectifier are(A) Vm, V2 mrespectively (B)V2 m, Vmrespectively

    (C) V2m, Vmrespectively (D) Vm,

    V2mrespectively

    MCQ 1.3.18 Inductor filter should be used when(A) load current is high (B) load current is low

    (C) high load resistance RL (D) none of the above

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    General Aptitude Engineering Mathematics

    MCQ 1.3.19 The value of critical inductance in an LCfilter is

    (A) 3L Rc Lw= (B) L R

    3cL

    w=

    (C) LR3

    cL

    w= (D) None of the above

    MCQ 1.3.20 Transformer utilization factor is more for(A) half wave rectifier

    (B) center tapped full wave rectifier

    (C) bridge rectifier

    (D) all of the above

    MCQ 1.3.21 The zener diode in the regulator circuit shown in figure has a Zener voltageof 5.8 Volts and a Zener knee current of 0.5 mA. The maximum load currentdrawn from this circuit ensuring proper functioning over the input voltagerange between 20 and 30 Volts, is

    (A) 23.7 mA (B) 14.2 mA

    (C) 13.7 mA (D) 24.2 mA

    MCQ 1.3.22 The circuit which converts undirectional flow to D.C. is called(A) Rectifier circuit

    (B) Converter circuit

    (C) filter circuit

    (D) Eliminator

    MCQ 1.3.23 The value of current that flows through RLin a p section filter circuit at

    no load is(A) 3 (B) 0.1 mA

    (C) 0 (D) few mA

    MCQ 1.3.24 The voltage atV1andV2of the arrangement shown in Fig. will be respectively.(Assume diode cut in voltage 0.6VV =g )

    (A) 6 V and 5.4 V

    (B) 5.4 V and 6 V

    (C) 3 V and 5.4 V

    (D) 6 V and 5 V

    LAL KISHORE198/4

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    MCQ 1.3.25 For an input 5sinV tS w= (assume ideal diode), the circuit shown in Fig. willbe behave as a

    (A) clipper, sine wave clipped at 2 V-

    (B) clamper, sine wave clamped at 2 V-

    (C) clamper, sine wave clamped at zero volt

    (D) clipped, sine wave clipped at 2 V

    MCQ 1.3.26 A clipper circuit always(A) needs a dc source

    (B) clips both half cycles of input signal

    (C) clips upper portion of the signal

    (D) clips some part of the input signal

    MCQ 1.3.27 The primary function of a clamper circuit is to(A) suppress variations in signal voltage

    (B) raise positive half-cycle of the signal

    (C) lower negative half-cycle of the signal

    (D) introduce a dc level into an ac signal

    MCQ 1.3.28 A zener diode has a dc power dissipation rating of 500 mW and a zenervoltage rating of 6.8 V. The value of IZMis(A) 70 mA

    (B) 72 mA

    (C) 73.5 mA

    (D) 75 mA

    MCQ 1.3.29 When the reverse current in a zener diode increases from 20 mA to 30 mA,the zener voltage changes from 5.6 V to 5.65 V. The zener resistance is(A) 2 W (B) 3 W

    (C) 4 W (D) 5 W

    MCQ 1.3.30 A 4.7 V zener has a resistance of 15 W. When a current 20 mA passesthrough it, then the terminal voltage is(A) 5 V

    (B) 10 V

    (C) 15 V

    (D) 4.7 V

    MCQ 1.3.31 In a full wave rectifier, the current in each of the diodes flows for(A) the complete cycle of the input signal

    (B) half cycle of the input signal

    (C) for zero time

    (D) more than half cycle of input signal

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    MCQ 1.3.32 The ripple factor of power supply is a measure of(A) its filter efficiency

    (B) diode rating

    (C) its voltage regulation

    (D) purity of power output

    MCQ 1.3.33 Larger the value of filter capacitor

    (A) larger the p-pvalue of ripple voltage(B) larger the peak current in the rectifying diode

    (C) longer the time that current pulse flows through the diode

    (D) smaller the d.c. voltage across the load

    MCQ 1.3.34 In a centre-tap full-wave rectifier, if Vm is the peak voltage between thecentre tap and one end of the secondary, the maximum voltage comingacross the reverse-biased diode is(A) Vm (B) V2 m

    (C) /V 2m (D) /V 2m

    MCQ 1.3.35 In a bridge type full wave rectifier, if Vm is the peak voltage across the

    secondary of the transformer, the maximum voltage coming across eachreverse-biased diode is(A) Vm (B) 2Vm

    (C) /V 2m (D) /V 2m

    MCQ 1.3.36 In a half wave rectifier, the peak value of the ac. voltage across the secondary

    of the transformer is 20 V2 . If no filter circuit is used, the maximum d.c.voltage across the load will be(A) 28.28 V (B) 25 V

    (C) 14.14 V (D) 9 V

    ***********

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    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    SOLUTIONS 1.1

    SOL 1.1.1 Option (D) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the diode is operating, and then obtain the output.Step 1: Assume that the diode is OFF, and replace it by open circuit. So,

    equivalent circuit is,

    Step 2: The voltage across the diode terminal is obtained as

    vp vi= at the p-terminal

    vn V5= at the n- terminalStep 3: Now, we have the condition

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we determine the output voltage.

    CASE I:

    If 5 Vv>i , then diode is ON. So the equivalent circuit is

    So, the output voltage is

    vo vi=

    CASE II:

    If 5 Vv

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    SOL 1.1.2 Option (C) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the diode is operating, and then obtain the output.Step 1: Assume that the diode is OFF, and replace it by open circuit. So,

    equivalent circuit is,

    Step 2: The voltage across the diode terminal is obtained as

    vp V2=- at the p-terminal

    vn vi= at the n- terminalStep 3: Now, we have the condition

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we determine the output voltage.

    CASE I:

    If 2 Vvi - , then diode is OFF. So, the equivalent circuit is,

    Since no current flows in the circuit (i.e. i 0= ), so we get

    vo V0=Step 4: From the two results obtained in the above steps, we sketch the

    output waveform as shown below.

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    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    SOL 1.1.3 Option (D) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the diode is operating, and then obtain the output.Step 1: Assume that the diode is OFF, and replace it by open circuit.

    So, equivalent circuit is,

    Step 2: The voltage across the diode terminal is obtained as

    vp V4= at the p-terminal Vn vi= at the n- terminalStep 3: Now, we have the condition

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we determine the output voltage.

    CASE I:

    If Vv 4i , then diode is OFF. So equivalent circuit is

    So, the output voltage is,

    vo vi=Step 4: From the two results obtained in the above steps, we sketch the

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    output waveform as shown below.

    SOL 1.1.4 Option (C) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the two diodes are operating, and then obtain the

    output.Step 1: Assume that both the diodes are OFF, and replace it by open circuit.

    So, equivalent circuit is,

    Step 2: For the assumption, the voltage across diode D1is obtained as

    vp1 vi= at the p-terminal

    vn1 V8= at the n- terminalStep 3: Voltage across the diode D2is obtained as

    vp2 V0= at the p-terminal

    vn1 v 6i= + at the n- terminalStep 4: Now, we have the condition for both the diodes

    vp v> n diode is ON vp v< n diode is OFF Applying these conditions, we determine the output voltage.

    CASE I:

    If v 6 0

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    So, the output voltage is

    vo vi=

    CASE III:

    When 8 Vv>i , then diode D1 is ON and D2 is OFF. So, theequivalent circuit is,

    vo 8 V=Step 5: From the results obtained in the above step, we sketch the output

    waveform as shown below.

    SOL 1.1.5 Option (B) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the diode is operating, and then obtain the output.Step 1: Assume that the diode is OFF, and replace it by open circuit. So,

    the equivalent circuit is

    Step 2: Apply KVL in loop 1,

    10 10 20 10k kI I- - + 0=

    I mA mA3020

    32

    = =

    vo 10 10 3.333 V32

    310

    #= - = =

    Step 3: Now, we have the condition

    vp v> n diode is ON

    vp v< n diode is OFF

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    Using these conditions, we determine the output voltage.

    CASE I

    vi . .3 333 0 7$ + or .v 4 03>i Diode is ON. So, the equivalent circuit is

    So, the output voltage is

    vo .v 0 7i= -

    CASE II

    vi .4 03# Diode is OFF. So, the equivalent circuit is

    So, the output voltage is

    vo .3 33=Step 4: From the results obtained in the above step, we sketch the ransfer

    characteristic is as shown below.

    SOL 1.1.6 Option (C) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the diode is operating, and then obtain the output.Step 1: Assume that the diode is OFF, and replace it by open circuit. So,

    the equivalent circuit is,

    Step 2: Now, we have the condition

    vp v> n diode is ON

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    vp v< n diode is OFF Applying these conditions, we determine the output voltage. By

    using voltage divider rule,

    vn 15 5k kk V

    1 21

    #= + =

    CASE I :

    If 5.7 Vvi# then Diode is OFF, we have the same circuit as shown

    above. So, the output voltage is,

    vo vi=

    CASE II.

    If .v 5 7i$ then Diode is ON and So, the equivalent circuit is,

    By using super node technique

    . .k k k

    v v v v 20 7 15

    10 7

    1o o o i - - +

    -+

    - 0=

    So, the output voltage is

    vo . .v0 4 3 42i= +

    Step 3 From the two results obtained in the above steps, we sketch the Transfercharacteristic is as shown below. So, the transfer characteristic is

    SOL 1.1.7 Option (D) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the diodes are operating, and then obtain the output.Step 1: Assume that the diodes are OFF, and replace it by open circuit

    Step 2: Now, we have the condition

    vp v> n diode is ON

    vp v< n diode is OFF

    Applying these conditions, we determine the output voltage

    CASE I:

    For positive part of vs Vv0 10s# #^ h: For 0 Vv >s , when D1 isOFF current through D2 is (D2is ON). So, the equivalent circuitstage is

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    I . 0.465mA10 1010 0 7

    =+-

    =

    So, the output voltage is

    vo 10 0.465 4.65k mA V#= = If vs .4 65> then diode D4is always OFF. So, the equivalent circuit

    is

    So, the output voltage is

    vo 4.65V= If 0 4.65v<

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    CASE II:

    For negative part of vs v10 0s# #-^ h: For 0 Vv - then D1& D2is ON. So, the equivalent circuit is,

    So, the output voltage is

    vo vs= If 10- .v 4 65<

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    Step 3: From the results obtained in the above steps, we sketch the Transfercharacteristic is as shown below.

    SOL 1.1.8 Option (B) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the two diodes are operating, and then obtain the

    output.Step 1: Assume that the diodes is OFF, and replace it by open circuit. So,

    the equivalent circuit is

    Step 2: Now, we have the condition

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we determine the output voltage.

    CASE I:

    If 2.5 Vvi ,then diode is OFF, 2 Mrr W= . So, the equivalent circuit

    is

    For positive wave waveform,

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    GATE STUDY PACKAGE Electronics & Communication

    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    vi 10 V= Nodal analysis at Node vo,

    .M

    v vM

    v2 1

    2 5o i o- + - 0=

    v3 o v 5i= +

    v3 o 15= So, the output voltage is

    vo 5 V=Step 3: From the two results obtained in the above steps, we sketch the

    output waveform is as shown below.

    SOL 1.1.9 Option (B) is correct.For this circuit KVL gives,

    v v1 2- v vD D1 2= - ,

    v v1 3- v vD D1 3= -

    Suppose that v1is positive and exceeds v2and v3. Then D1must be forward-

    biased with 0vD1 = and consequently 0v

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    vo 0=

    CASE II:

    In positive half cycle diode will be OFF. So, the equivalent circuit is

    By using KVL in loop 1, we have

    vo 10 10 20 V= + =

    From the two results obtained in the above cases, we sketch the output

    voltage as shown below.

    SOL 1.1.11 Option (A) is correct.

    This is a clamper circuit. In this circuit, the diode is in upward direction. So,

    total signal will be clamp above the 5 V+ . From the results obtained in the

    above steps, we sketch the output waveform is as shown below.

    ALTERNATIVE METHOD :

    During 20- V cycle of vi, diode is ON and capacitor will charge up

    instantaneously to 25 V. Output is 5+ V during this cycle.

    During the +10 V of vi, diode is OFF and capacitor will hold on this voltage

    level, giving total output voltage +35 V.

    SOL 1.1.12 Option (A) is correct.In DC equivalent circuit capacitor is open. So, the equivalent circuit is

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    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    Current through diode is I. So, we get the small signal resistance

    rd IV

    IV

    dc

    T Th h= =

    The small signal equivalent circuit is as follows

    For small signal response, open dccurrent source, short capacitor C1and C2,

    and replace diode with it small signal resistance rd. So, the output voltage is

    vo r Rv r

    v

    IV

    R

    IV

    d s

    s ds

    Ts

    T

    h

    h

    =+

    =+

    vV R

    Vv

    V IR

    Vs

    T s

    Ts

    T s

    T

    h

    h

    h

    h=

    + =

    +

    SOL 1.1.13 Option (A) is correct.

    For the given circuit, we first determine the linear region (forward bias orreverse bias) in which the two diodes are operating, and then obtain the

    output.Step 1: Assume that the two diodes are OFF, and replace it by open circuit.

    So, the equivalent circuit is

    Step 2: Now, we have the condition for both the diodes

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we determine the output voltage. So,

    vpn1 10 0 10 V= - =

    vpn2 5 0 5 V= - =Step 3: Assume diode is D1is ON and diode D2is OFF.

    Current through diode D1,

    iD1 1Ampv1 9

    01 910 01=

    +-

    =+-

    =

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    So output voltage,

    vo 1 9 9 V#= = Voltage across diode D2

    vpn2 v vp n2 2= -

    5 9 4= - =- So diode D2is in OFF state.

    Step 4: From the results obtained in the above steps, the output voltage is, Vo 9 V=

    SOL 1.1.14 Option (B) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the two diodes are operating, and then obtain the

    output.Step 1:Assume that the two diodes are OFF, and replace it by open circuit.

    Since, the equivalent circuit is

    Step 2: Now, we have the condition for both the diodes

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we determine the output voltage.

    Step 3: Assume that diode D1and D2 is in ON state. So, the equivalent

    circuit is

    Nodal analysis

    v v v1

    101

    109

    o o0 - + -

    + 0=

    Step 4: From the two results obtained in the above steps, the output voltageis,

    vo 9.474 V1920 9#= =

    In this case iD1and iD2both are positive. So, D1and D2 are ON(assumption is correct)

    SOL 1.1.15 Option (C) is correct.

    For the given circuit, we first determine the linear region (forward bias or

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    reverse bias) in which the two diodes are operating, and then obtain the

    output.Step 1: Assume that the two diodes are OFF, and replace it by open circuit

    So, the equivalent circuit is

    Step 2: Now, we have the condition for both the diodes

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we determine the output voltage.So the

    equivalent circuit is,

    Step 3: From the results obtained in the above steps, the output voltage is,

    vo 9 4.5 V105

    #= =

    SOL 1.1.16 Option (B) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the two diode are operating, and then obtain the

    output.Step 1: Assume that the two diodes are OFF, and replace it by open circuit.

    So, the equivalent circuit is

    Step 2: Now, we have the condition for both the diodes

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we determine the output voltage

    Step 3: Diode D1and D2is ON. So, the equivalent circuit is

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    Using Nodal analysis

    v v v1

    101

    109

    5o o o- + - + - 0=

    So, the output voltage is

    vo 9.737 V= For this voltage, iD1 and iD2 are positive, so D1 and D2 are ON

    (assumption is correct).

    SOL 1.1.17 Option (C) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the two diode are operating, and then obtain the

    output.Step 1: Assume that the two diodes are OFF, and replace it by open circuit

    So, the equivalent circuit is,

    Step 2: Now, we have the condition for both the diodes

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we determine the output voltage.

    Step 3: Diode D1is OFF and D2is ON. So, the equivalent circuit is

    Node analysis at node vo

    1 9

    5 Vv v10o oW W

    -+

    - 0=

    v10 o 95=

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    So, the output voltage is

    vo 9.5V= In this case, iD1is negative, iD2is positive, so D1is OFF and D2is

    ON (i.e. assumption is correct).

    SOL 1.1.18 Option (C) is correct.

    For the given circuit, we determine the linear region (forward bias or reverse

    bias) in which the two diodes are operating.Step 1: Assume that the two diodes are OFF, and replace it by open circuitSo, the equivalent circuit is

    Step 2: Now, we have the condition for both the diodes

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we analyse the circuit.

    Step 3: Both diodes are ON. So, the equivalent circuit is

    Using Nodal analysis at Node 1,

    . .V V V12

    4 818 6

    4 4-+ +

    - 0=

    V 3.71V= In this case, iD1and iD2are positive, so D1and D2are ON (assumption

    is correct).

    SOL 1.1.19 Option (A) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the two diodes are operating, and then obtain the

    output.Step 1: Assume that the two diodes are OFF, and replace it by open circuit

    So, the equivalent circuit is,

    Step 2: Now, we have the condition for both the diodes

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    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we obtain the required unknowns.

    Step 3: When diode D2is ON. So, the equivalent circuit is

    v 3 0.7 2.3 V= - = This diode D1is not conducting because of diode D2is ON. (highly

    forward biased).

    i.

    2.65mA2

    2 3 3=

    - -=

    ^ hSOL 1.1.20 Option (A) is correct.

    For the given circuit, we first determine the linear region (forward bias or

    reverse bias) in which the two diode are operating, and then obtain theoutput.Step 1: Assume that the two diodes are OFF, and replace it by open circuit

    So, the equivalent circuit is,

    Step 2: Now, we have the condition for both the diodes

    vp v> n diode is ON

    vp v< n diode is OFF Applying these conditions, we obtain the required unknowns

    Step 3: Diode D1is OFF & D2is ON. Since the equivalent circuit is

    v 5 V=

    i10 k

    v5W

    = -

    0k

    mA105 5

    = -

    =

    SOL 1.1.21 Option (A) is correct.

    The given circuit is,

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    Sample Chapter of Analog Circuits(Vol-5, GATE Study Package)

    v vs 1- .0 6>

    For both diode is ON when .v 0 6>1 . Using nodal analysis at node v1,

    5.

    5.

    k kv v v v v v 0 6

    500 5000 6s s1 1 1 1+ - + - + + - 0=

    v1. .v

    222 5 4 0 6>s= +

    vs 3.9V>

    SOL 1.1.22 Option (C) is correct.

    The thevenin equivalent circuit for the network to the left of terminal abis

    shown below.

    vTH (2 ) 1 0.5cos cost t200100

    w w= + = + V

    RTH 50100 100100 100# W=

    + =

    The diode can be modeled with 0.5vf= V and

    rf .. . 500 0040 7 0 5 W= - = ,

    iD1 0.5 0.5cos

    R rv v t

    50 50TH fTH f w=

    +-

    =+

    + -

    5(1 )cos tw= + mA

    SOL 1.1.23 Option (A) is correct.

    The voltage, vD r i vf D f= +

    50 5(1 ) 10 0.5cos t 3w# #= + +-

    0.75 0.25 0.25(3 )cos cost tw w= + = + V

    SOL 1.1.24 Option (D) is correct.

    The output voltage cannot exceed the positive po