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Table of contents :
Front Matter ....Pages ixvii
From Physics to Electric Circuits (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 125
Front Matter ....Pages 2727
Major Circuit Elements (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 2987
Circuit Laws and Networking Theorems (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 89141
Circuit Analysis and Power Transfer (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 143200
Operational Amplifier and Amplifier Models (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 201267
Front Matter ....Pages 269269
Dynamic Circuit Elements (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 271317
Transient Circuit Fundamentals (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 319397
Front Matter ....Pages 399399
SteadyState AC Circuit Fundamentals (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 401444
Filter Circuits: Frequency Response, Bode Plots, and Fourier Transform (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 445492
SecondOrder RLC Circuits (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 493534
AC Power and Power Distribution (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 535588
Electric Transformer and Coupled Inductors (Sergey N. Makarov, Reinhold Ludwig, Stephen J. Bitar)....Pages 589650
Back Matter ....Pages 651664
Sergey N. Makarov Reinhold Ludwig Stephen J. Bitar
Practical Electrical Engineering Second Edition
Practical Electrical Engineering
Sergey N. Makarov • Reinhold Ludwig • Stephen J. Bitar
Practical Electrical Engineering Second Edition
Sergey N. Makarov ECE Department Worcester Polytechnic Institute Worcester, WA, USA
Reinhold Ludwig ECE Department Worcester Polytechnic Institute Worcester, MA, USA
Stephen J. Bitar Worcester Polytechnic Institute Worcester, MA, USA
ISBN 9783319966915 ISBN 9783319966922 https://doi.org/10.1007/9783319966922
(eBook)
Library of Congress Control Number: 2018952468 © Springer Nature Switzerland AG 2016, 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional afﬁliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland
To Antonina, Margot, and Juliette
Contents
1
From Physics to Electric Circuits . . . . . . . . . . . . . . . . . . . . . . 1.1 Electrostatics of Conductors . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Charges, Coulomb Force, and Electric Field . . . . 1.1.2 Electric Potential and Electric Voltage . . . . . . . . . 1.1.3 Electric Voltage Versus Ground . . . . . . . . . . . . . 1.1.4 Equipotential Conductors . . . . . . . . . . . . . . . . . . 1.1.5 Use of Coulomb’s Law to Solve Electrostatic Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 SteadyState Current Flow and Magnetostatics . . . . . . . . . 1.2.1 Electric Current . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 Difference Between Current Flow Model and Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3 Physical Model of an Electric Circuit . . . . . . . . . 1.2.4 Magnetostatics and Ampere’s Law . . . . . . . . . . . 1.2.5 Origin of Electric Power Transfer . . . . . . . . . . . . 1.3 Hydraulic and Fluid Mechanics Analogies . . . . . . . . . . . . 1.3.1 Hydraulic Analogies in the DC Steady State . . . . 1.3.2 Analogies for AlternatingCurrent (AC) Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Analogies for Semiconductor Circuit Components . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 3 3 4 5 7 9 11 11 11 13 14 16 18 18 19 20
Part I DC Circuits: General Circuit Theory—Operational Ampliﬁer 2
Major Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Resistance: Linear Passive Circuit Element . . . . . . . . . . 2.1.1 Circuit Elements Versus Circuit Components . . 2.1.2 Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.3 υi Characteristic of the Resistance: Open and Short Circuits . . . . . . . . . . . . . . . . . 2.1.4 Power Delivered to the Resistance . . . . . . . . . . 2.1.5 Finding Resistance of Ohmic Conductors . . . . .
. . . .
29 31 31 31
. . .
34 35 36
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Contents 2.1.6
3
Application Example: Power Loss in Transmission Wires and Cables . . . . . . . . . . 2.1.7 Physical Component: Resistor . . . . . . . . . . . . . 2.1.8 Application Example: Resistive Sensors . . . . . . 2.2 Nonlinear Passive Circuit Elements . . . . . . . . . . . . . . . . 2.2.1 Resistance as a Model for the Load . . . . . . . . . 2.2.2 Nonlinear Passive Circuit Elements . . . . . . . . . 2.2.3 Static Resistance of a Nonlinear Element . . . . . 2.2.4 Dynamic (SmallSignal) Resistance of a Nonlinear Element . . . . . . . . . . . . . . . . . . 2.2.5 Electronic Switch . . . . . . . . . . . . . . . . . . . . . . 2.3 Independent Sources . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Independent Ideal Voltage Source . . . . . . . . . . . 2.3.2 Circuit Model of a Practical Voltage Source . . . 2.3.3 Independent Ideal Current Source . . . . . . . . . . . 2.3.4 Circuit Model of a Practical Current Source . . . 2.3.5 Operation of the Voltage Source . . . . . . . . . . . . 2.3.6 Application Example: DC Voltage Generator with Permanent Magnets . . . . . . . . . . . . . . . . . 2.3.7 Application Example: Chemical Battery . . . . . . 2.4 Dependent Sources and TimeVarying Sources . . . . . . . . 2.4.1 Dependent Versus Independent Sources . . . . . . 2.4.2 Deﬁnition of Dependent Sources . . . . . . . . . . . 2.4.3 Transfer Characteristics . . . . . . . . . . . . . . . . . . 2.4.4 TimeVarying Sources . . . . . . . . . . . . . . . . . . . 2.5 Ideal Voltmeter and Ammeter: Circuit Ground . . . . . . . . 2.5.1 Ideal Voltmeter and Ammeter . . . . . . . . . . . . . . 2.5.2 Circuit Ground: Fluid Mechanics Analogy . . . . 2.5.3 Types of Electric Ground . . . . . . . . . . . . . . . . . 2.5.4 Ground and Return Current . . . . . . . . . . . . . . . 2.5.5 Absolute Voltage and Voltage Drop Across a Circuit Element . . . . . . . . . . . . . . . . . Circuit Laws and Networking Theorems . . . . . . . . . . . . . . . 3.1 Circuit Laws: Networking Theorems . . . . . . . . . . . . . . . 3.1.1 Electric Network and Its Topology . . . . . . . . . . 3.1.2 Kirchhoff’s Current Law (KCL) . . . . . . . . . . . . 3.1.3 Kirchhoff’s Voltage Law (KVL) . . . . . . . . . . . . 3.1.4 PowerRelated Networking Theorems . . . . . . . . 3.1.5 Port of a Network: Network Equivalence . . . . . 3.2 Series and Parallel Network/Circuit Blocks . . . . . . . . . . 3.2.1 Sources in Series and in Parallel . . . . . . . . . . . . 3.2.2 Resistances in Series and in Parallel . . . . . . . . . 3.2.3 Reduction of Resistive Networks . . . . . . . . . . . 3.2.4 Voltage Divider Circuit . . . . . . . . . . . . . . . . . . 3.2.5 Application Example: Voltage Divider as a Sensor Circuit . . . . . . . . . . . . . . . . . . . . .
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39 41 42 46 46 47 48
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49 50 52 52 54 55 57 58
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59 61 64 64 64 66 67 69 69 70 71 71
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72 89 91 91 93 95 98 99 100 100 102 104 105
. 107
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Contents 3.2.6
4
Application Example: Voltage Divider as an Actuator Circuit . . . . . . . . . . . . . . . . . . . . 3.2.7 Current Limiter . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.8 Current Divider Circuit . . . . . . . . . . . . . . . . . . . 3.2.9 Wheatstone Bridge . . . . . . . . . . . . . . . . . . . . . . 3.3 Superposition Theorem and Its Use . . . . . . . . . . . . . . . . . 3.3.1 Linear and Nonlinear Circuits . . . . . . . . . . . . . . 3.3.2 Superposition Theorem or Superposition Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.3 Y (Wye) and Δ (Delta) Networks: Use of Superposition . . . . . . . . . . . . . . . . . . . . . 3.3.4 T and Π Networks: TwoPort Networks . . . . . . . 3.3.5 General Character of Superposition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Circuit Analysis and Power Transfer . . . . . . . . . . . . . . . . . . . 4.1 Nodal/Mesh Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Importance of Circuit Simulators . . . . . . . . . . . . 4.1.2 Nodal Analysis for Linear Circuits . . . . . . . . . . . 4.1.3 Supernode . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.4 Mesh Analysis for Linear Circuits . . . . . . . . . . . 4.1.5 Supermesh . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Generator Theorems and Their Use . . . . . . . . . . . . . . . . . 4.2.1 Equivalence of Active OnePort Networks: Method of Short/Open Circuit . . . . . . . . . . . . . . 4.2.2 Application Example: Reading and Using Data for Solar Panels . . . . . . . . . . . . . . . . . . . . . 4.2.3 Source Transformation Theorem . . . . . . . . . . . . 4.2.4 Thévenin’s and Norton’s Theorems: Proof Without Dependent Sources . . . . . . . . . . . . . . . . . . . . . . 4.2.5 Finding Thévenin and Norton Equivalents and Using Them for Circuit Solution . . . . . . . . . 4.2.6 Application Example: Generating Negative Equivalent Resistance . . . . . . . . . . . . . . . . . . . . 4.2.7 Short Summary of Circuit Analysis Methods . . . . 4.3 Power Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Maximum Power Transfer . . . . . . . . . . . . . . . . . 4.3.2 Maximum Power Efﬁciency . . . . . . . . . . . . . . . . 4.3.3 Application Example: Power Radiated by a Transmitting Antenna . . . . . . . . . . . . . . . . . 4.3.4 Application Example: Maximum Power Extraction from Solar Panel . . . . . . . . . . . . . . . . 4.4 Analysis of Nonlinear Circuits: Load Line Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Iterative Method for Nonlinear Circuits . . . . . . . .
109 111 111 113 115 115 116 120 122 123 143 145 145 145 149 150 152 154 154 155 156 159 161 165 167 168 168 170 171 172 176 178
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Contents 5
Operational Ampliﬁer and Ampliﬁer Models . . . . . . . . . . . . 5.1 Ampliﬁer Operation and Circuit Models . . . . . . . . . . . . 5.1.1 Ampliﬁer Operation . . . . . . . . . . . . . . . . . . . . 5.1.2 Application Example: Operational Ampliﬁer Comparator . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.3 Ampliﬁer Circuit Model . . . . . . . . . . . . . . . . . 5.1.4 IdealAmpliﬁer Model and First SummingPoint Constraint . . . . . . . . . . . . . . . . 5.2 Negative Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Idea of the Negative Feedback . . . . . . . . . . . . . 5.2.2 Ampliﬁer Feedback Loop: Second SummingPoint Constraint . . . . . . . . . . . . . . . . 5.2.3 Ampliﬁer Circuit Analysis Using Two SummingPoint Constraints . . . . . . . . . . . . . . . 5.2.4 Mathematics Behind the Second SummingPoint Constraint . . . . . . . . . . . . . . . . 5.2.5 Current Flow in the Ampliﬁer Circuit . . . . . . . . 5.2.6 MultipleInput Ampliﬁer Circuit: Summing Ampliﬁer . . . . . . . . . . . . . . . . . . . . . 5.3 Ampliﬁer Circuit Design . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Choosing Proper Resistance Values . . . . . . . . . 5.3.2 Model of a Whole Voltage Ampliﬁer Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Voltage Ampliﬁer Versus Matched Ampliﬁer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.4 Cascading Ampliﬁer Stages . . . . . . . . . . . . . . . 5.3.5 Ampliﬁer DC Imperfections and Their Cancellation . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.6 DCCoupled SingleSupply Ampliﬁer: VirtualGround Circuit . . . . . . . . . . . . . . . . . . . 5.4 Difference and Instrumentation Ampliﬁers . . . . . . . . . . . 5.4.1 Differential Input Signal to an Ampliﬁer . . . . . . 5.4.2 Difference Ampliﬁer: Differential Gain and CommonMode Gain . . . . . . . . . . . . . . . . 5.4.3 Application Example: Instrumentation Ampliﬁer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.4 Instrumentation Ampliﬁer in Laboratory . . . . . . 5.5 General Feedback Systems . . . . . . . . . . . . . . . . . . . . . . 5.5.1 SignalFlow Diagram of a Feedback System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 ClosedLoop Gain and Error Signal . . . . . . . . . 5.5.3 Application of General Theory to Voltage Ampliﬁers with Negative Feedback . . . . . . . . . 5.5.4 Voltage, Current, Transresistance, and Transconductance Ampliﬁers with the Negative Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. 201 . 203 . 203 . 206 . 207 . 209 . 211 . 211 . 211 . 213 . 217 . 218 . 219 . 221 . 221 . 223 . 224 . 227 . 229 . 232 . 234 . 234 . 235 . 237 . 240 . 242 . 242 . 242 . 244
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Contents Part II 6
7
Transient Circuits
Dynamic Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Static Capacitance and Inductance . . . . . . . . . . . . . . . . . . 6.1.1 Capacitance, SelfCapacitance, and Capacitance to Ground . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Application Example: ESD . . . . . . . . . . . . . . . . 6.1.3 ParallelPlate Capacitor . . . . . . . . . . . . . . . . . . . 6.1.4 Circuit Symbol: Capacitances in Parallel and in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.5 Application Example: How to Design a 1μF Capacitor? . . . . . . . . . . . . . . . . . . . . . . . 6.1.6 Application Example: Capacitive Touchscreens . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.7 SelfInductance (Inductance) and Mutual Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.8 Inductance of a Solenoid With and Without Magnetic Core . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.9 Circuit Symbol: Inductances in Series and in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.10 Application Example: How to Design a 1mH Inductor? . . . . . . . . . . . . . . . . . . . . . . . 6.2 Dynamic Behavior of Capacitance and Inductance . . . . . . 6.2.1 Set of Passive Linear Circuit Elements . . . . . . . . 6.2.2 Dynamic Behavior of Capacitance . . . . . . . . . . . 6.2.3 Dynamic Behavior of Inductance . . . . . . . . . . . . 6.2.4 Instantaneous Energy and Power of Dynamic Circuit Elements . . . . . . . . . . . . . . . 6.2.5 DC Steady State . . . . . . . . . . . . . . . . . . . . . . . . 6.2.6 Behavior at Very High Frequencies . . . . . . . . . . 6.3 Application Circuits Highlighting Dynamic Behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Bypass Capacitor . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Blocking Capacitor . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Decoupling Inductor . . . . . . . . . . . . . . . . . . . . . 6.3.4 Ampliﬁer Circuits With Dynamic Elements: Miller Integrator . . . . . . . . . . . . . . . . . . . . . . . . 6.3.5 Compensated Miller Integrator . . . . . . . . . . . . . . 6.3.6 Differentiator and Other Circuits . . . . . . . . . . . . Transient Circuit Fundamentals . . . . . . . . . . . . . . . . . . . . . . . 7.1 RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 EnergyRelease Capacitor Circuit . . . . . . . . . . . . 7.1.2 Time Constant of the RC Circuit and Its Meaning . . . . . . . . . . . . . . . . . . . . . . . . 7.1.3 Continuity of the Capacitor Voltage . . . . . . . . . .
271 273 273 275 276 278 279 282 284 285 287 288 290 290 290 293 295 296 297 299 299 301 301 302 303 304 319 322 322 324 325
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Contents 7.1.4
7.2
7.3
7.4
7.5
7.6
Application Example: Electromagnetic Railgun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.5 Application Example: Electromagnetic Material Processing . . . . . . . . . . . . . . . . . . . . . 7.1.6 Application Example: Digital Memory Cell . . . 7.1.7 EnergyAccumulating Capacitor Circuit . . . . . . RL Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 EnergyRelease Inductor Circuit . . . . . . . . . . . . 7.2.2 Continuity of the Inductor Current . . . . . . . . . . 7.2.3 EnergyAccumulating Inductor Circuit . . . . . . . 7.2.4 EnergyRelease RL Circuit with the Voltage Supply . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.5 Application Example: Laboratory Ignition Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Switching RC Oscillator . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 About Electronic Oscillators . . . . . . . . . . . . . . 7.3.2 Bistable Ampliﬁer Circuit with the Positive Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.3 Triggering . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.4 Switching RC Oscillator . . . . . . . . . . . . . . . . . 7.3.5 Oscillation Frequency . . . . . . . . . . . . . . . . . . . 7.3.6 Circuit Implementation: 555 Timer . . . . . . . . . . SingleTimeConstant (STC) Transient Circuits . . . . . . . 7.4.1 Circuits with Resistances and Capacitances . . . . 7.4.2 Circuits with Resistances and Inductances . . . . . 7.4.3 Example of a NonSTC Transient Circuit . . . . . 7.4.4 Example of an STC Transient Circuit . . . . . . . . 7.4.5 Method of Thévenin Equivalent and Application Example: Circuit with a Bypass Capacitor . . . . . . . . . . . . . . . . . Description of the SecondOrder Transient Circuits . . . . 7.5.1 Types of SecondOrder Transient Circuits . . . . . 7.5.2 SeriesConnected SecondOrder RLC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.3 Initial Conditions in Terms of Circuit Current and Capacitor Voltage . . . . . . . . . . . . . 7.5.4 Step Response and Choice of the Independent Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.5 Parallel Connected SecondOrder RLC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Step Response of the Series RLC Circuit . . . . . . . . . . . . 7.6.1 General Solution of the Secondorder ODE . . . . 7.6.2 Derivation of the Complementary Solution: Method of Characteristic Equation . . . . . . . . . . 7.6.3 Finding Integration Constants . . . . . . . . . . . . .
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. 339 . 340 . 342 . 342 . . . . . . . . . .
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. 355 . 358 . 358 . 358 . 361 . 362 . 363 . 366 . 366 . 366 . 368
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Contents 7.6.4 7.6.5 7.6.6
Part III 8
9
Solution Behavior for Different Damping Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 Overshoot and Rise Time . . . . . . . . . . . . . . . . . . 369 Application Example: Nonideal Digital Waveform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
AC Circuits
SteadyState AC Circuit Fundamentals . . . . . . . . . . . . . . . . . 8.1 Harmonic Voltage and Current: Phasor . . . . . . . . . . . . . . 8.1.1 Harmonic Voltages and Currents . . . . . . . . . . . . 8.1.2 Phase: Leading and Lagging . . . . . . . . . . . . . . . 8.1.3 Application Example: Measurements of Amplitude, Frequency, and Phase . . . . . . . . . . 8.1.4 Deﬁnition of a Phasor . . . . . . . . . . . . . . . . . . . . 8.1.5 From Real Signals to Phasors . . . . . . . . . . . . . . . 8.1.6 From Phasors to Real Signals . . . . . . . . . . . . . . . 8.1.7 Polar and Rectangular Forms: Phasor Magnitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.8 Operations with Phasors and Phasor Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.9 Shorthand Notation for the Complex Exponent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 The Concept of Impedance . . . . . . . . . . . . . . . . 8.2.2 Physical Meaning of Impedance . . . . . . . . . . . . . 8.2.3 Magnitude and Phase of Complex Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.4 Application Example: Impedance of a Human Body . . . . . . . . . . . . . . . . . . . . . . . 8.3 Principles of AC Circuit Analysis . . . . . . . . . . . . . . . . . . 8.3.1 AC Circuit Analysis: KVL, KCL, and Equivalent Impedances . . . . . . . . . . . . . . . . 8.3.2 Complete Solution for an AC Circuit: KVL and KCL on Phasor Diagram . . . . . . . . . . . . . . . 8.3.3 Source Transformation . . . . . . . . . . . . . . . . . . . 8.3.4 Thévenin and Norton Equivalent Circuits . . . . . . 8.3.5 Summary of AC Circuit Analysis at a Single Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.6 Multifrequency AC Circuit Analysis: Superposition Theorem . . . . . . . . . . . . . . . . . . . Filter Circuits: Frequency Response, Bode Plots, and Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 FirstOrder Filter Circuits and Their Combinations . . . . . . 9.1.1 RC Voltage Divider as an Analog Filter . . . . . . .
401 403 403 405 408 408 410 411 411 413 416 417 417 419 420 422 423 423 424 425 427 429 429 445 447 447
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Contents 9.1.2
10
HalfPower Frequency and Amplitude Transfer Function . . . . . . . . . . . . . . . . . . . . . . . 9.1.3 Bode Plot, Decibel, and RollOff . . . . . . . . . . . . 9.1.4 Phase Transfer Function and Its Bode Plot . . . . . 9.1.5 Complex Transfer Function: Cascading Filter Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.6 RL Filter Circuits . . . . . . . . . . . . . . . . . . . . . . . 9.2 Bandwidth of an Operational Ampliﬁer . . . . . . . . . . . . . . 9.2.1 Bode Plot of the OpenLoop Ampliﬁer Gain . . . . 9.2.2 UnityGain Bandwidth Versus GainBandwidth Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.3 Model of the OpenLoop AC Gain . . . . . . . . . . . 9.2.4 Model of the ClosedLoop AC Gain . . . . . . . . . . 9.2.5 Application Example: Finding Bandwidth of an Ampliﬁer Circuit . . . . . . . . . . . . . . . . . . . 9.2.6 Application Example: Selection of an Ampliﬁer IC for Proper Frequency Bandwidth . . . . . . . . . . . . 9.3 Introduction to Continuous and Discrete Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Meaning and Deﬁnition of Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 Mathematical Properties of Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3 Discrete Fourier Transform and Its Implementation . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.4 Sampling Theorem . . . . . . . . . . . . . . . . . . . . . . 9.3.5 Applications of Discrete Fourier Transform . . . . . 9.3.6 Application Example: Numerical Differentiation via the FFT . . . . . . . . . . . . . . . . 9.3.7 Application Example: Filter Operation for an Input Pulse Signal . . . . . . . . . . . . . . . . . . 9.3.8 Application Example: Converting Computational Electromagnetic Solution from Frequency Domain to Time Domain . . . . . . . . . . . . . . . . . . . . . . . . SecondOrder RLC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Theory of SecondOrder Resonant RLC Circuits . . . . . . . 10.1.1 SelfOscillating Ideal LC Circuit . . . . . . . . . . . . 10.1.2 Series Resonant Ideal LC Circuit . . . . . . . . . . . . 10.1.3 Series Resonant RLC Circuit: Resonance Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.4 Quality Factor Q of the Series Resonant RLC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.5 Bandwidth of the Series Resonant RLC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.6 Parallel Resonant RLC Circuit: Duality . . . . . . .
452 453 456 457 460 463 463 464 465 466 467 468 470 470 472 473 475 476 476 478
479 493 495 495 497 498 500 502 505
xiv
Contents 10.2
11
Construction of SecondOrder RLC Filters . . . . . . . . . . 10.2.1 SecondOrder BandPass RLC Filter . . . . . . . . 10.2.2 SecondOrder LowPass RLC Filter . . . . . . . . . 10.2.3 SecondOrder HighPass RLC Filter . . . . . . . . . 10.2.4 SecondOrder BandReject RLC Filter . . . . . . . 10.2.5 SecondOrder RLC Filters Derived from the Parallel RLC Circuit . . . . . . . . . . . . . 10.3 RLC Circuits for NearField Communications and Proximity Sensors . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.1 NearField Wireless Link . . . . . . . . . . . . . . . . . 10.3.2 Transmitter Circuit . . . . . . . . . . . . . . . . . . . . . 10.3.3 Receiver Circuit . . . . . . . . . . . . . . . . . . . . . . . 10.3.4 Application Example: NearField Wireless Link in Laboratory . . . . . . . . . . . . . . . . . . . . . 10.3.5 Application Example: Proximity Sensors . . . . . AC Power and Power Distribution . . . . . . . . . . . . . . . . . . . . 11.1 AC Power Types and Their Meaning . . . . . . . . . . . . . . . 11.1.1 Instantaneous AC Power . . . . . . . . . . . . . . . . . 11.1.2 Timeaveraged AC Power . . . . . . . . . . . . . . . . 11.1.3 Application Example: rms Voltages and AC Frequencies Around the World . . . . . . . . . . . . . 11.1.4 rms Voltages for Arbitrary Periodic AC Signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.5 Average AC Power in Terms of Phasors: Power Angle . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.6 Average Power for Resistor, Capacitor, and Inductor . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.7 Average Power, Reactive Power, and Apparent Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.8 Power Triangle . . . . . . . . . . . . . . . . . . . . . . . . 11.1.9 Application Example: Wattmeter . . . . . . . . . . . 11.2 Power Factor Correction: Maximum Power Efﬁciency and Maximum Power Transfer . . . . . . . . . . . . . . . . . . . 11.2.1 Power Factor Correction . . . . . . . . . . . . . . . . . 11.2.2 Application Example: Automatic Power Factor Correction System . . . . . . . . . . . . . . . . . . . . . . 11.2.3 Principle of Maximum Power Efﬁciency for AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . 11.2.4 Principle of Maximum Power Transfer for AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . 11.3 AC Power Distribution: Balanced ThreePhase Power Distribution System . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3.1 AC Power Distribution Systems . . . . . . . . . . . . 11.3.2 Phase Voltages: Phase Sequence . . . . . . . . . . . .
. . . . .
508 508 511 512 514
. 516 . . . .
518 518 519 520
. . . . . .
522 523 535 537 537 538
. 540 . 541 . 543 . 544 . 545 . 547 . 550 . 551 . 551 . 554 . 554 . 556 . 558 . 558 . 559
xv
Contents 11.3.3
Wye (Y) Source and Load Conﬁgurations for ThreePhase Circuits . . . . . . . . . . . . . . . . . . . 11.3.4 Application: Examples of ThreePhase Source and the Load . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3.5 Solution for the Balanced ThreePhase WyeWye Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3.6 Removing the Neutral Wire in LongDistance Power Transmission . . . . . . . . . . . . . . . . . . . . . . 11.4 Power in Balanced ThreePhase Systems: Deltaconnected ThreePhase Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4.1 Instantaneous Power . . . . . . . . . . . . . . . . . . . . . 11.4.2 Average Power, Reactive Power, and Apparent Power . . . . . . . . . . . . . . . . . . . . . 11.4.3 Application Example: Material Consumption in ThreePhase Systems . . . . . . . . . . . . . . . . . . . 11.4.4 Balanced DeltaConnected Load . . . . . . . . . . . . . 11.4.5 Balanced DeltaConnected Source . . . . . . . . . . . 12 Electric Transformer and Coupled Inductors . . . . . . . . . . . . . 12.1 Ideal Transformer as a Linear Passive Circuit Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1.1 Electric Transformer . . . . . . . . . . . . . . . . . . . . . 12.1.2 Ideal OpenCircuited Transformer: Faraday’s Law of Induction . . . . . . . . . . . . . . . . 12.1.3 Appearance of Transformer Currents . . . . . . . . . 12.1.4 Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . 12.1.5 Ideal Loaded Transformer . . . . . . . . . . . . . . . . . 12.1.6 Ideal Transformer Versus Real Transformer: Transformer Terminology . . . . . . . . . . . . . . . . . 12.1.7 Mechanical Analogies of a Transformer . . . . . . . 12.2 Analysis of Ideal Transformer Circuits . . . . . . . . . . . . . . 12.2.1 Circuit with a Transformer in the Phasor Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2.2 Referred (Or Reﬂected) Source Network in the Secondary Side . . . . . . . . . . . . . . . . . . . . 12.2.3 Referred (Or Reﬂected) Load Impedance to the Primary Side . . . . . . . . . . . . . . . . . . . . . . 12.2.4 Transformer as a Matching Circuit . . . . . . . . . . . 12.2.5 Application Example: Electric Power Transfer via Transformers . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Some Useful Transformers . . . . . . . . . . . . . . . . . . . . . . . 12.3.1 Autotransformer . . . . . . . . . . . . . . . . . . . . . . . . 12.3.2 Multiwinding Transformer . . . . . . . . . . . . . . . . . 12.3.3 CenterTapped Transformer: SingleEnded to Differential Transformation . . . . . . . . . . . . . . 12.3.4 Current Transformer . . . . . . . . . . . . . . . . . . . . .
561 562 563 565 568 568 570 570 572 572 589 591 591 591 595 595 596 598 601 602 602 602 604 605 607 610 610 611 612 614
xvi
Contents 12.4
12.5
RealTransformer Model . . . . . . . . . . . . . . . . . . . . . . . . 12.4.1 Model of a Nonideal LowFrequency Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4.2 Model Parameters and Their Extraction . . . . . . . 12.4.3 Analysis of Nonideal Transformer Model . . . . . . 12.4.4 Voltage Regulation and Transformer Efﬁciency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4.5 About HighFrequency Transformer Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Model of Coupled Inductors . . . . . . . . . . . . . . . . . . . . . . 12.5.1 Model of Two Coupled Inductors . . . . . . . . . . . . 12.5.2 Analysis of Circuits with Coupled Inductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5.3 Coupling Coefﬁcient . . . . . . . . . . . . . . . . . . . . . 12.5.4 Application Example: Wireless Inductive Power Transfer . . . . . . . . . . . . . . . . . . . . . . . . . 12.5.5 Application Example: Coupling of Nearby Magnetic Radiators . . . . . . . . . . . . . . . . . . . . . .
616 616 616 618 621 622 624 624 625 628 630 634
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651
xvii
Chapter 1
Chapter 1: From Physics to Electric Circuits
Overview Prerequisites:  Knowledge of university physics: electricity and magnetism Objectives of Section 1.1:  Show that the electric voltage and the electric potential may be treated as two equivalent quantities  Define the electric voltage—work per unit charge—in the form of a line integral and show its independence on the integration path for conservative fields  Relate voltage to the potential energy of the electric field  Introduce threedimensional potential distributions and realize the guiding function of metal wires  Formulate and understand major conditions of electrostatics of conductors  Visualize surface charge distributions in the electrostatic case
Objectives of Section 1.2:  Introduce electric current density as a function of the applied electric field  Visualize steadystate current flow in a single conductor along with the associated electric potential/voltage distribution  Visualize electric and magneticfield distributions for a twowire DC transmission line  Obtain initial exposure to the Poynting vector  Realize that electric power is transferred via Poynting vector even in DC circuits  Indicate a path toward circuit problems where the field effects become important Objectives of Section 1.3:  Review basic hydraulic (fluid mechanics) analogies for DC circuit elements  Present major hydraulic analogies for dynamic circuit elements in AC circuits  Briefly discuss hydraulic analogies for semiconductor components Application Examples:  Human body subject to applied voltage  Human body in an external electric field
© Springer Nature Switzerland AG 2019 S. N. Makarov et al., Practical Electrical Engineering, https://doi.org/10.1007/9783319966922_1
I1
Chapter 1
From Physics to Electric Circuits
Keywords: Electricity, Electric ﬁeld intensity, Electric ﬁeld, Electric ﬁeld magnitude, Lines of force, Electric potential, Electric voltage, Line integral, Contour integral, Conservative ﬁeld, Potential energy of the electric ﬁeld, Voltage drop, Voltage difference, Ground reference, Neutral conductor, Common conductor, Voltage versus ground, Equipotential lines, Volumetric charge density, Surface charge density, Gauss’ theorem, Equipotential surface, Selfcapacitance, Electrostatic discharge, Effect of electrostatic discharge on integrated circuits, Boundary element method, Electric current density, Material conductivity, Transmission line, Direct current (DC), Electric load, Ideal wire, Kirchhoff’s voltage law (KVL), Magnetic ﬁeld, Magneticﬁeld intensity, Ampere’s law, Cross (vector) product, Poynting vector, Poynting theorem, Wireless communications, Wireless power transfer, Fluid mechanics analogy of an electric circuit, Hydraulic analogy of an electric circuit, Voltage source (hydraulic analogy), Resistance (hydraulic analogy), Current source (hydraulic analogy), Capacitance (hydraulic analogy), Inductance (hydraulic analogy), Electric transformer (hydraulic analogy), NMOS transistor (hydraulic analogy), Bipolar junction transistor (hydraulic analogy)
I2
Chapter 1
Section 1.1: Electrostatics of Conductors
Section 1.1 Electrostatics of Conductors This introductory chapter is optional in the sense that the reader does not need its content as a prerequisite for the subsequent chapters. The aim of this chapter is to illustrate why electric circuits trace their origin to electromagnetic ﬁelds. The chapter highlights several ﬁeld concepts which form the theoretical foundation of electric circuits. At the same time it makes clear why, for the majority of electric circuits, the electric and magnetic ﬁelds are often ignored without affecting the ﬁnal results. When this is the case, the electric circuits and components follow useful and simple hydraulic analogies discussed below.
1.1.1 Charges, Coulomb Force, and Electric Field Electric Charges The smallest electric charge is known as the elemental charge of an electron, q ¼1:60218 1019 C (coulombs). In electrical engineering, we deal with much larger charges, which, for this reason, are assumed to be inﬁnitely divisible. There are no positive movable charges in metal conductors. Therefore, when we talk about positive charges, it is implied that we have a lack of electrons at a certain location, e.g., at the surface. Oppositely, the negative charge is the excess of electrons at a certain location. Deﬁnition of the Electric Field Electrostatics plays a key role in explaining the operations of electric capacitors and all semiconductor devices. The word electricity is derived from the Greek word for amber. Probably Thales of Miletus was the ﬁrst who discovered, about 600 B.C., that amber, when rubbed, attracts light objects. An electrostatic force acting on a charge q is known as the Coulomb force. This Coulomb force is a vector; it is measured in newtons (or N) ~ F ¼ q~ E ½N
ð1:1Þ
Equation (1.1) is the deﬁnition of the electric ﬁeld intensity vector, ~ E, often called the electric ﬁeld. This electric ﬁeld is created by other (remote or nearby) charges. In the general case, the electric ﬁeld exists both in free space and within material objects, whether conductors or dielectrics. The electric ﬁeld ~ E is measured in volts (V) per qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 2 2 E, has the same units. From meter (V/m). The ﬁeld magnitude, E ¼ E þ E þ E ¼ ~ x
y
z
Eq. (1.1), 1V ¼
1 N 1m 1 J ¼ 1 C 1 C
ð1:2Þ
I3
Chapter 1
From Physics to Electric Circuits y
lines of force +Q
+
+
+ +
A
+ +
+
+
voltage power 0 supply minus () terminal
VAA’
 
A’
 




plus (+) terminal
E
x
Q
Fig. 1.1. Electric ﬁeld emanating from a voltage supply with opencircuit output terminals.
Electric Field of a Laboratory Power Source As an example, we consider an electric ﬁeld generated by a laboratory voltage source turned on. Figure 1.1 shows the realistic electric ﬁeld distribution modeled numerically. The power supply has two output terminals shaped as two metal cylinders. The plastic cover is ignored. Nothing is connected to the terminals yet. Still, the power supply already performs a “charge separation”: the plus (+) terminal is charged positively (total charge is +Q), whereas the minus () terminal is charged negatively (total charge is Q). Those charges are schematically shown in Fig. 1.1. As a result, an electric ﬁeld is created. The electric ﬁeld in Fig. 1.1 is directed along particular lines, which we call lines of force. This electric ﬁeld surrounds the power supply terminals. Every line of force starts at the positive terminal and ends at the negative terminal. The strength of the electric ﬁeld everywhere in space is linearly proportional to the supply voltage as studied next. However, the ﬁeld shape always remains exactly the same. 1.1.2 Electric Potential and Electric Voltage The electric potential φ measured in volts (V) and electric voltage V measured in volts (V) are two identical quantities once they refer to the same observation point A and to the same reference point A0 . Both terms may be used; the electric potential is frequently denoted by V. The potential is more common in physics. Work is done against the electric forces when a charge is moved in an electric ﬁeld. The electric potential or electric voltage V AA0 between points A (#1) and A0 (#2) is work in joules per coulomb (per unit charge) to bring a positive charge from reference point A0 (#2) to observation point A (#1), i.e., against the electric ﬁeld—see Fig. 1.1. The work per unit charge over a short straight vector distance d~ l is I4
Chapter 1
Section 1.1: Electrostatics of Conductors
1 F d~ l ¼ ~ E d~ l ¼ Edl cos θ ~ q
ð1:3Þ
where θ is the angle between ~ E and d~ l, E ¼ ~ E , and dl ¼ d~ l . The total work or V AA0 is
the sum of all such small contributions conventionally written in the form of an integral ðA
E d~ l¼ V AA0 ¼ ~ A
0
0
A ð
~ E d~ l
ð1:4Þ
A
The integral in Eq. (1.4) is a line integral, also called a contour integral. In the general case, it is evaluated along a curve connecting points A and A0 . In the particular case of Fig. 1.1, this curve is just a straight line. Exercise 1.1: Assume for simplicity that the electric ﬁeld along the line of force from A to A0 in Fig. 1.1 is strictly uniform and has the magnitude of 50 V/m. The line length is 0.02 m. Find voltage (or potential) V AA0 . ð 0:02m ðA V ~ ~ 50 cos π dl ¼ 50 0:02 m ¼ 1 V. Answer: V AA0 ¼ 0 E d l ¼ m A 0
The electrostatic ﬁeld (and any slowly varying electric ﬁeld) is called a conservative ﬁeld. There are two equivalent deﬁnitions of a conservative ﬁeld: 1. Electric voltage or electric potential V AA0 is path independent; it only depends on the position of A and A0 , but not on the shape of the curve between A and A0 . 2. The line integral in Eq. (1.4) over any closed contour is zero. The equivalence of these deﬁnitions is proved by treating two different integration contours between A and A0 as two parts of one closed contour. The independence of the integration path suggests that the voltage is equal to the potential energy of a unit charge in the electric ﬁeld. Strictly speaking, it is the change in the potential energy.
1.1.3 Electric Voltage Versus Ground The voltage between two arbitrary points V AA0 (e.g., between two terminals of a resistor) is a convenient measure when analyzing electric circuits with discrete circuit components. In this case, it is called the voltage drop (or voltage difference) across a circuit element. At the same time, it is equally convenient to deﬁne the “global” or absolute voltage between an arbitrary point in space A and some ﬁxed point A0 , which is assigned the voltage value zero. This ﬁxed point (or the set of points) is called the ground reference. In general, the ground reference may be chosen arbitrarily. In physics of localized conductors, it is I5
Chapter 1
From Physics to Electric Circuits
customary to choose the ground at inﬁnity. In electrical engineering, the ground reference is either the physical (earth) ground or some neutral (common) conductor assigned to zero volts. Thus, the absolute voltage versus ground denoted by V ð~ rÞ is still deﬁned by Eq. (1.4) where point A is now characterized by the position vector ~ r. By deﬁnition, it becomes zero when ~ r approaches ground, i.e., A0 . The equivalent representation of Eq. (1.4) for conservative ﬁelds may be shown to be ~ E ð~ rÞ ¼ gradV ð~ rÞ ¼ ∇V ð~ rÞ
ð1:5Þ
Thus, the electrostatic ﬁeld is expressed as the gradient of the electric potential or of the (absolute) electric voltage everywhere in space. In other words, it means the electric ﬁeld does not have closed loops, but starts and ends at the charges. Equation (1.5) is of signiﬁcant value since it replaces a complicated vector ~ E by the single scalar voltage V. Exercise 1.2: Electric voltage/potential with respect to ground is given in Cartesian coordinates by V ð~ rÞ ¼ y ½V. Determine the electric ﬁeld everywhere in space. Answer: E y ¼ 1V=m, E x ¼ E z ¼ 0.
As an example, we choose the xaxis in Fig. 1.1 as the ground reference. The positive supply terminal is chosen to have a voltage equal to +0.5 V versus ground, and the negative terminal is assigned a voltage equal to 0.5 V versus ground. Those values will uniquely determine charges Q in Fig. 1.1. The function V ð~ rÞ is now plotted using the lines of equal potential, or equipotential lines. The result is shown in Fig. 1.2. It will be proved next that the surface of any metal (or other) conductor in electrostatics is an equipotential surface. All points on this surface have the same value of the electric potential: +0.5 V for the plus terminal and 0.5 V for the minus terminal in Fig. 1.2. Using Eq. (1.5) it can be veriﬁed that the equipotential lines and the lines of force are always perpendicular to each other; you can see this in Fig. 1.2.
I6
Chapter 1
Section 1.1: Electrostatics of Conductors y
0.1 V
+Q
0.2 V 0.3 V 0.4 V
lines of force
plus (+) terminal
+
+
+
+ +
A
+ +
B
voltage power 0 supply minus () terminal Q
E 0V
C x
 
A’
 B B’
equipotential lines
0.4 V 0.3 V 0.2 V
0.1 V
Fig. 1.2. Electric ﬁeld and electric potential (electric voltage) of a 1V voltage supply. Equipotential lines are thin solid curves, while the lines of force are the thicker curves.
Exercise 1.3: In Fig. 1.2, determine voltage differences: V BB0 , V A0 A , V AB , V AC , V B0 C . Answer: V BB0 ¼ 1 V, V A0 A ¼ 1 V, V AB ¼ 0 V, V AC ¼ 0:5 V, V B0 C ¼ 0:5 V.
1.1.4 Equipotential Conductors Consider a conductor, which is characterized by a sufﬁcient number of free charges. The conductor is subject to an applied electric voltage, or to an applied electric ﬁeld, or to those effects combined. The charge density in the conductor is the difference between the concentration of ions (positive charges) and electrons (negative charges) multiplied by the charge of an electron. The charge density can be either volumetric, measured in C/m3, or surface, measured in C/m2. The following is true when dealing with electrostatics: 1. The electric ﬁeld everywhere within the conductor is zero, ~ E¼~ 0. Otherwise, the Coulomb force given by Eq. (1.1) will act on free charges and cause permanent charge motion (current ﬂow), which is impossible. 2. The volumetric charge density everywhere within the conductor is zero. If this were not true, then according to Gauss’ theorem, there would be a nonzero electric ﬁeld within the conductor, which is impossible based on statement #1. 3. The surface charge density is not zero. In fact, the surface charge is distributed so as to assure that statement #1 is satisﬁed. 4. For the electric ﬁeld given by Eq. (1.5), the tangential component of the electric Et ¼ ~ 0 inside the conductor, ~ Et ﬁeld ~ Et is continuous across an interface. Since ~ I7
Chapter 1
From Physics to Electric Circuits
must be zero over the entire surface of the conductor too. This is seen in Figs. 1.1 and 1.2 where the lines of force are perpendicular to the conductor surface. 0 on the conductor surface, any line integral between two points on this 5. Since ~ Et ¼ ~ surface is zero. Consequently the potential or voltage remains the same for any point on this surface. The conductor surface is thus an equipotential surface. These statements have an immediate practical application. Consider two conductors (wires) attached to the power supply terminals as seen in Fig. 1.3. +Q’>Q 0.5 V
+
+
1V
1V

0.5 V minus () terminal
1V

voltage power supply
0.5 V
+
0.5 V
0.5 V

plus (+) terminal 0.5 V
0.5 V 0.5 V
Q’ 0.
2.1.4 Power Delivered to the Resistance Voltage V across the resistance is work in joules necessary to pass 1 C of charge through the resistance. Since there are exactly I coulombs passing through the resistance in one second, the power P delivered to the resistance must be the product of work per unit charge and the number of charges passing through the element in one second: P ¼ V I. The power P has indeed the units of watts (1 V 1 A ¼ 1 J=1 s ¼ 1 W). When the υi characteristic of the resistance is examined, the power is equal to the area of the shaded rectangles in Fig. 2.4. Using Ohm’s law, Eq. (2.3) gives us three equivalent deﬁnitions of the absorbed power by a resistance: P ¼ VI
Basic definition, valid f or any passive circuit element
2
V Power f or resistance in terms of voltage R P ¼ RI 2 Power f or resistance in terms of current
P¼
I
P1
ð2:6Þ ð2:7Þ ð2:8Þ
smaller resistance
larger resistance
P2 0
V
Fig. 2.4. υi Characteristics for the resistances and power rectangles. P1,2 are absorbed powers.
II35
Chapter 2
Major Circuit Elements
Despite their obvious nature, all three equations are useful in practice. In particular, Eq. (2.7) indicates that a small resistance absorbs more power than the large resistance at the same applied voltage; this is seen in Fig. 2.4. Imagine for a moment that we know the voltage across the resistance, but do not know the current. This happens if a number of circuit elements are connected in parallel to a known voltage source. Then Eq. (2.7) is used to ﬁnd the power. However, if the current is known, but the voltage is not (a number of elements connected in series to a current source), then Eq. (2.8) is employed. Example 2.1: A voltage of 10 V is applied to a 2.5Ω resistance. Determine the absorbed electric power in three possible ways as stated by Eqs. (2.6) through (2.8). Solution: To apply two of the three equations, current through the resistance is needed. From Ohm’s law, I ¼ V =R ¼ 4 A. The electric power delivered to the resistance can thus be determined in three ways: P ¼ V I ¼ 40 W Basic power definition, passive reference configuration V2 ¼ 40 W R
Power f or resistance in terms of voltage
P ¼ RI 2 ¼ 40 W
Power f or resistance in terms of current
P¼
2.1.5 Finding Resistance of Ohmic Conductors An ohmic conductor satisﬁes Ohm’s law given by Eq. (2.3). Finding its resistance is equivalent to the derivation of Ohm’s law under certain assumptions. Let us consider a conducting circular cylinder subject to an applied voltage V in Fig. 2.5. The cylinder has length l and a crosssectional area A. equipotential surfaces electric field E
+

+ + + + +
I
V/4
A
V
I
x

+
Fig. 2.5. Finding the resistance of a conducting cylinder.
Finding Total Current The net electric current in a metal or other conductor is deﬁned as the net ﬂux of positive charge carriers directed along the conductor axis x: II36
Chapter 2 I ¼ Aqnυ
Section 2.1: Resistance: Linear Passive Circuit Element ð2:9Þ
Here, qn is the volumetric charge density of free charges q with concentration n in coulombs per cubic meter, C/m3, and υ is the magnitude of the average charge velocity in m/s. In the onedimensional model of the current ﬂow, the average velocity vector is directed along the xaxis seen in Fig. 2.5. Since the electrons have been historically assigned a negative charge, the electric current direction is opposite to the direction of electron motion in a conductor. The electron carries an elemental charge of q ¼ 1.60218 1019 C. Because A and qn are constants for a given conductor, the electric current is simply associated with the charge’s mean velocity υ.
Finding Average Carrier Velocity In order to ﬁnd υ, we use the following method. The total voltage drop V applied to a sufﬁciently long, conducting cylinder is uniformly distributed along its length following the equally spaced equipotential surfaces; this is schematically shown in Fig. 2.5. This fact has been proved in Chapter 1. Such a voltage distribution corresponds to a constant uniform electric ﬁeld within the cylinder, which is also directed along the cylinder axis. The magnitude of the ﬁeld, E, with the units of V/m, is given by E ¼ V =l
ð2:10Þ
The electric ﬁeld creates a Coulomb force acting on an individual positive charge q. The Coulomb force is directed along the ﬁeld; its magnitude F is given by F ¼ qE
ð2:11Þ
The key is a linear relation between the charge velocity υ and force F or, which is the same, a linear relation between the charge velocity υ and the applied electric ﬁeld E, i.e., υ ¼ μE
ð2:12Þ
where μ is the socalled mobility of charge carriers, with the units of m2/(Vs). Carrier mobility plays an important role in semiconductor physics. With the help of Eqs. (2.10) and (2.12), the expression for the total current Eq. (2.9) is transformed to l l l I¼ I ¼ RI, σ ¼ qnμ, R ¼ ð2:13Þ V ¼ Aqnμ Aσ Aσ This is the expression for the resistance of a cylindrical conductor. Material conductivity σ is measured in S/m. Its reciprocal is the material resistivity ρ ¼ 1=σ measured in Ωm.
II37
Chapter 2
Major Circuit Elements
Example 2.2: Estimate resistance R of a small doped Si disk with the length l of 5 μm, cross section of A ¼ 104 cm2, uniform electron, concentration (carrier concentration) of n ¼ 1017 cm3 , and carrier mobility of μn ¼ 1450 cm2/(Vs). Solution: Resistance calculations are usually simple when the onedimensional model of a conducting cylinder or a disk is used. However, one must be careful with the units. Units of cm are customary in semiconductor physics. Therefore, one should ﬁrst convert all different units of length to meters (or to centimeters). After that, we use the deﬁnition of the resistance given by Eq. (2.13) and obtain (units of meters are used):
R¼
l 5 106 ¼ 8 ¼ 0:215 Ω Aqnμn 10 1:602 1019 1023 0:145
ð2:14Þ
Table 2.1 lists conductivities of common materials. What is the major factor that determines the conductivity of a particular conducting material? According to Eq. (2.13), there are two such parameters: charge concentration and charge mobility. Table 2.1. DC conductivities of conductors, semiconductors, and insulators (25 C, multiple sources). Material Silver Copper
Class Conductor Conductor
σ (S/m) 6.1 107 5.8 107
Gold Aluminum Brass Tungsten Zinc
Conductor Conductor Conductor Conductor Conductor
4.1 107 4.0 107 2.6 107 1.8 107 1.7 107
Nickel
Conductor
1.5 107
Iron Tin Lead Graphite Carbon Magnetite
Conductor Conductor Conductor Conductor Conductor Conductor
1.0 107 0.9 107 0.5 107 0.003 107 0.003 107 0.002 107
Material Seawater Human/animal tissues Germanium Fresh water Wet soil Dry soil Intrinsic silicon (Si) Gallium arsenide (GaAs) Glass Porcelain Hard rubber Fused quartz Teﬂon
Class Semiconductor Semiconductor
σ (S/m) 4 0.1–2.0
Semiconductor Semiconductor Semiconductor Semiconductor Semiconductor
2 0.01 0.01–0.001 0.001–0.0001 4.4104
Semiconductor
106
Insulator Insulator Insulator Insulator Insulator
1012 1014 1015 1017 1023
II38
Chapter 2
Section 2.1: Resistance: Linear Passive Circuit Element
It is mostly the different concentration of free charge carriers n that makes the resistance of two materials quite different. For example, n ¼ 8:46 1028 m3 in copper (a good conductor), whereas it may be n ¼ 1016 m3 in a moderately doped silicon crystal (doped semiconductor). However, it is also the difference in mobility μ that represents the “friction” experienced by the “gas” of free charges with density n that is moving through the solid or liquid conductor under the applied voltage (electric ﬁeld).
Exercise 2.5: Using Table 2.1, determine the total resistance of an aluminum wire having a length of 100 m and a crosssectional area of 1 mm2. Answer: 2.5 Ω.
2.1.6 Application Example: Power Loss in Transmission Wires and Cables All metal wires and cables are ohmic conductors. Electric power absorbed by an ohmic conductor is transformed into heat. This is known as electric power loss. We can apply Eqs. (2.6)–(2.8) and Eq. (2.13) in order to determine the loss of electric power in transmission lines and/or cables. This question has signiﬁcant practical importance. Figure 2.6 outlines the corresponding electric circuit. The electric circuit is a closed path for electric current. Resistance RL characterizes the load. Only Ohm’s law is used to analyze this circuit, along with the current continuity. No other circuit laws are necessary. We also consider a voltage source in Fig. 2.6. The voltage sources will be studied next. transmission line (TL)
source
+ 
a load
I
I
I
I
+ I
RL V
b
20 km
Fig. 2.6. A long transmission power line carrying a steadystate current I to the load resistance.
According to Eq. (2.13), the wire resistance is inversely proportional to its diameter. In the USA, the American Wire Gauge (AWG) system was developed to classify the wire diameters of conductors. You probably have heard an electrician refer to a gauge 12 household wiring. This implies a wire diameter of about 2 mm, or 0.0800 . Table 2.2 reports common AWG numbers and maximum current strengths.
II39
Chapter 2
Major Circuit Elements
Table 2.2. American Wire Gauge (AWG) wire parameters. The maximum current is given for solid copper (Source: Handbook of Electronic Tables and Formulas for American Wire Gauge).
AWG # 24 22 20 18 16 14 12 10
Diameter (inches) 0.0201 0.0254 0.0320 0.0403 0.0508 0.0640 0.0808 0.1019
0 (1 aught) 00 (2 aught) 000 (3 aught) 0000 (4 aught)
0.3249 0.3648 0.4096 0.4600
Resistance per 1000 ft Diameter or 304.8 m (Ω) (mm) 0.51054 25.67 0.64516 16.14 0.81280 10.15 1.02362 6.385 1.29032 4.016 1.62814 2.525 2.05232 1.588 2.58826 0.999 Gauges 10 through 1 are not shown 8.252 0.09827 9.266 0.07793 10.404 0.06180 11.684 0.04901
Maximum current in (A) for power transmission 0.577 0.92 1.50 2.30 3.70 5.90 9.30 15.0 150 190 239 302
Example 2.3: An AWG 0 aluminum transmission grid cable schematically shown in Fig. 2.6 has a wire diameter of 8.25 mm and a crosssectional area of 53.5 mm2. The conductivity of aluminum is 4.0 107 S/m. The total cable length (two cables must run to a load) is 40 km. The system delivers 1 MW of DC power to the load. Determine the power loss in the cable when load voltage V and load current I are given by:
1. V ¼ 40 kV and I ¼ 25 A 2. V ¼ 20 kV and I ¼ 50 A 3. V ¼ 10 kV and I ¼ 100 A Why is highvoltage power transmission important in power electronics? Solution: We ﬁnd the total cable resistance from Eq. (2.13):
L L 40 103 ¼ R¼ρ ¼ ¼ 18:7 Ω A σA 4:0 107 53:5 106
ð2:15Þ
The same load current I ﬂows through the load modeled by a resistor RL and through the cables in Fig. 2.6. Therefore, power loss in the cables may be found using Eq. (2.8). Knowing the load voltage (or the voltage across the cable) is not necessary. The power loss in the cables is thus given by P ¼ RI 2. For the three different cases corresponding to the same load power, we obtain
II40
Chapter 2
Section 2.1: Resistance: Linear Passive Circuit Element
Example 2.3 (cont.):
1. P ¼ RI 2 ¼ 11:7 kW or 1.17 % of the load power 2. P ¼ RI 2 ¼ 46:8 kW or 4.68 % of the load power 3. P ¼ RI 2 ¼ 187 kW or 18.7 % of the load power
(2.16)
Clearly, the highvoltage power transmission allows us to reduce the power loss in long cables very signiﬁcantly while transmitting the same power to the load. Therefore, the highvoltage transmission lines passing through the country have typical voltages between 100 kV and 800 kV.
In circuit analysis in the laboratory, we usually consider ideal or perfectly conducting wires whose resistance is zero. This is justiﬁed since the wire lengths for most practical circuit applications are so short that the voltage drop is negligibly small.
2.1.7 Physical Component: Resistor Fixed Resistors Resistance is constructed intentionally, as a separate circuit component. This component is called the resistor. A common axial leaded carbon ﬁlm 0.25W resistor deployed on a solderless protoboard is seen in Fig. 2.7a. Those carbon ﬁlm resistors are typically manufactured by coating a homogeneous layer of pure carbon on highgrade ceramic rods. After a helical groove is cut into the resistive layer, tinned connecting leads of electrolytic copper are welded to the endcaps. The resistors are then coated with layers of tancolored lacquer. The common Surface Mount Device (SMD) thinﬁlm resistor is shown in Fig. 2.7b. Manufacturing process variations result in deviations from the normal resistor values; they are known as tolerances and reported to the end user through an extra color ring (for leaded axial resistors) or an extra digit (for SMD resistors). Typical power ratings for the axial resistors are 1/6 W, ¼ W, ½ W, 1 W, 2 W, and 3 W. When the power delivered to the resistor considerably exceeds the particular rating, the resistor may burn out, releasing a prominent “carbon” smell. The axial resistors have color codes shown in Fig. 2.7c. To ﬁnd the value of the resistor depicted in Fig. 2.7a, we ﬁrst encounter the tolerance code, which will typically be gold, implying a 5 % tolerance value. Starting from the opposite end, we identify the ﬁrst band, and write down the number associated with that color; in Fig. 2.7a it is 9 (white). Then, we read the next band (brown) and record that number; it is 1. After this we read the multiplier black, which is 0. The resistor value in Fig. 2.7a is consequently R ¼ 91 100 ¼ 91 Ω. The surface mount resistors, also known as SMD resistors, do not have color codes. The SMD resistors are labeled numerically as 102 ¼ 10 102 ¼ 1 kΩ, 271 ¼ 27 101 ¼ 270 Ω, etc. Along with this, the SMD resistors, similar to other SMD components, do have codes corresponding to their geometry size. Each size is described as a fourdigit number. The ﬁrst two digits indicate length, and the last two digits indicate II41
Chapter 2
Major Circuit Elements a) axias leaded resistor
b) SMD thinfilm resistor
c) Color codes for axial resistors Black
Brown
Red
Orange
0
1
2
3
Yellow Green 4
5
Blue 6
Violet Gray 7
White
8
9
Fig. 2.7. (a) A leaded axial resistor, (b) a thinﬁlm resistor, and (c) color codes for leaded resistors.
width (in 0.0100 , or 10 mils units). Some popular SMD resistor sizes are 0603 (0.0600 0.0300 , or 60 30 mils, or 1.6 0.8 mm), 0805 (0.0800 0.0500 ), and 1206 (0.1200 0.0600 ).
Variable Resistors (Potentiometers) The simplest variable resistor is a potentiometer. A picture is shown in Fig. 2.8a along with its equivalent electric schematic in Fig. 2.8b. The potentiometer is used either as a voltage divider, discussed later in the text, or as a variable resistor. By rotating the potentiometer shaft, it is possible to obtain any resistance value up to the maximum potentiometer value. The adjustable resistance is obtained between terminals 1 and 2 or 2 and 3 of the potentiometer, respectively. You should remember that the potentiometer is a nonpolar device. This means that it can be placed into the circuit in any orientation. With this knowledge the joking engineer telling you that “all resistors in your circuit are backwards” should not cause any fear. a)
b)
1
25 kΩ
1
2
3
2 3
Fig. 2.8. A rotary 25kΩ potentiometer rated at 0.25 W and its equivalent circuit schematic.
2.1.8 Application Example: Resistive Sensors There are a variety of sensor types—resistive sensors—which use electric resistance variation to measure a mechanical or a thermal quantity. Some of them are shown in Fig. 2.9. As a ﬁrst example, we consider a temperature sensor based on a thermistor (a resistor), with a resistance that varies when ambient temperature changes; see Fig. 2.9a.
II42
Chapter 2
Section 2.1: Resistance: Linear Passive Circuit Element
The word thermistor is a contraction of the words “thermal” and “resistor.” As a second example of a resistance subject to ambient conditions, we will consider a photoresistor (photocell) shown in Fig. 2.9b. The ﬁnal example is a strain gauge shown in Fig. 2.9c.
Fig. 2.9. Sensing elements which change their resistances when ambient conditions change.
Thermistor The thermistor changes its resistance as temperature increases or decreases. Generalpurpose thermistors are made out of metal oxides or other semiconductors. Successful semiconductor thermistors were developed almost simultaneously with the ﬁrst transistors (1950s). For a metaloxide thermistor, its resistance decreases with increasing temperature. Increasing the temperature increases the number of free carriers (electrons) and thus increases the sample conductivity (decreases its resistance). Shown in Fig. 2.9a is a very inexpensive NTC—negative temperature coefﬁcient—leaded thermistor. According to the manufacturer’s datasheet, it reduces its resistance from approximately 50 kΩ at room temperature (about 25 C) by 4.7 % for every degree Celsius (or Kelvin) and reaches about 30 kΩ at body temperature according to the thermistor equation: 1 1 ð2:17Þ R1 ¼ R2 exp B T1 T2 where T1, T2 are two absolute temperatures always given in degrees K. Temperature T2 corresponds to a room temperature of 25 C so that R2¼R25 C, temperature T1 is the observation temperature, and B is the thermistor constant, which is equal to 4200 K in the present case. Equation (2.17) is a nontrivial result of the solidstate physics theory. We emphasize that Eq. (2.17) is more accurate than the temperature coefﬁcient of the thermistor—the above referenced value of 4.7 %. Typical applications include temperature measurement, control, compensation, power supply fan control, and printed circuit board (PCB) temperature monitoring. Inexpensive thermistors operate from 30 C to approximately +130 C. At higher temperatures, thermocouples should be used.
Thermocouple Figure 2.9 does not show one more important temperature sensor—the thermocouple— which is used to measure large temperatures and large temperature differences. It operates
II43
Chapter 2
Major Circuit Elements
based on a completely different principle. The thermocouple does not signiﬁcantly change its resistance when temperature changes. Instead, it generates an electric current and the associated voltage, when the junction of the two metals is heated or cooled, known as the PeltierSeebeck effect; this voltage can be correlated to temperature. Therefore, the thermocouple, strictly speaking, is not a resistive sensor.
Photoresistor (Photocell) An idea similar to the thermistor design applies. Quanta of light incident upon the photocell body create new free charge carriers—new electronhole pairs in a semiconductor. If the concentration of free charges increases, the resistance of the sample decreases according to Eq. (2.13). The resistance is inversely proportional to the concentration. The photocell in Fig. 2.9b is characterized by very large nonlinear variations of the resistance in response to ambient light. Strain Gauge The strain gauge measures mechanical strain. The operation is based on Eq. (2.13), which deﬁnes the resistance through material conductivity σ, the length of the resistor l, and its cross section A. When the resistor, which may be a trace on the base of a metal alloy, is stretched, its length l increases and its cross section A decreases. Hence, the resistance R increases due to both of these effects simultaneously; changes in the resistance may be made visible for small strains. Shown in Fig. 2.9c is an inexpensive uniaxial strain gauge with a nominal resistance of 350 Ω; typical resistances are 120, 350, 600, 700, and 1000 Ω. The gauge changes its resistance R in proportion to the strain sensitivity SG of the wire’s resistance, also called the gauge factor (GF). For a strain gauge, the relative resistance variation, ΔR/R, is estimated based on known values of the strain sensitivity, SG, and strain, ε. The strain gauge equation has the form ΔR=R ¼ S G ε
ð2:18Þ
The dimensionless strain sensitivity SG varies around 2. The strain (a relative elongation) is a dimensionless quantity. It is measured in microstrains, με, where one με is 106. Typical strain values under study are on the order of 1000 με. Using Eq. (2.18) this yields a relative resistance variation as small as 0.2 %. Because of this, the circuits for the strain measurements should be designed and built with great care. Temperature compensation efforts are also required. Since the relative resistance changes are very small, the strain gauge is a linear device: the strain is directly proportional to resistance variations.
Potentiometric Position Sensor Another general resistive sensor is a potentiometric (or potentiometer) position sensor. Its operation becomes apparent when we rotate the potentiometer dial in Fig. 2.8a. A change in the resistance is directly proportional to the rotation angle. The resistance variation can
II44
Chapter 2
Section 2.1: Resistance: Linear Passive Circuit Element
be converted to voltage variation and then measured. Similar potentiometer sensors for measuring linear motion also exist.
Sensitivity of Resistive Sensors One major difference between different resistive sensing elements is a very different degree of the relative resistance variations. For the photocell, the resistance variation is up to 100 times. For the thermistor, the resistance variation can be as much as 50 %. For the strain gauge, the resistance variations do not exceed 0.5 %. Circuit Symbols There are several similar but not identical standards for circuit symbols related to resistance: International standard IEC 60617, American ANSI standard Y32 (IEEE Std 315), etc. Figure 2.10 shows popular circuit symbols for variable resistances. a) generic variable resistor
b) potentiometer
c)
thermistor
d)
photoresistor
e)
strain gauge
Fig. 2.10. Circuit symbols for variable resistances: (a) generic variable resistance, (b) potentiometer, (c) thermistor, (d) photoresistor, and (e) strain gauge.
II45
Chapter 2
Major Circuit Elements
Section 2.2 Nonlinear Passive Circuit Elements 2.2.1 Resistance as a Model for the Load It might appear at ﬁrst glance that the resistance causes mainly power loss and heating. However, the concept of heating elements in household appliances or power losses in long cables covers only a small subset of applications. Important resistance applications are related to the resistive sensors studied previously. Resistances are widely used in the circuits to provide different voltage values, i.e., bias circuit components such as diodes and transistors. Last but not least, resistances model an arbitrary powerabsorbing device, the load, which can be mechanical, acoustical, microwave, optical, etc. From a circuit point of view it does not matter how the electric power supplied to a load is eventually transformed. The circuit delivers certain power to the load, but cares little about whether the power is converted into heat to warm a heating pan, light to illuminate our house, mechanical power to drive a motor, or electromagnetic radiation generated by a cell phone. Circuit analysis is concerned with the power delivered to a powerabsorbing device, the load, leaving its utilization and conversion to other engineering disciplines. Therefore, many practical loads can be replaced by a simple load resistance RL. Such a resistance is often called the equivalent resistance, RL ¼ Req . It is also valid for AC circuits. For AC circuits it is convenient to use rms (root mean square) voltages, which are equivalent to DC voltages and thus provide us with the same power value delivered to the load. Figure 2.11 shows two examples of the load replacement with the equivalent resistance: a light source radiating visible light and an antenna radiating microwaves. a)
monopole antenna
a ground plane
RL a
coaxial cable
b
b b) a
a RL
b
b
Fig. 2.11. (a) Radiating monopole antenna is modeled as a resistance. (b) Light source is approximately modeled as a resistance.
Example 2.4: A small commercial monopole antenna shown in Fig. 2.11a is rated at Req ¼ 50 Ω in the ISM band of 902–928 MHz. When an rms voltage of 10 V is applied to the antenna, what is the total amount of power radiated by the antenna?
II46
Chapter 2
Section 2.2: Nonlinear Passive Circuit Elements
Example 2.4 (cont.):
Solution: We use Eq. (2.7) and obtain P ¼ V 2rms =Req ¼ 2 W. All of this power is radiated in the form of an outgoing electromagnetic wave. There is no heat loss. The resistance Req is called the radiation resistance in such a case. The above analysis is valid only in a certain frequency range.
A load that exactly follows Ohm’s law Eq. (2.3) is called the linear load. While the transmitting antenna in Fig. 2.11a is a linear load, an incandescent light bulb in Fig. 2.11 is not. Most of the loads deviate from the linear Ohm’s law.
2.2.2 Nonlinear Passive Circuit Elements Nonlinear passive circuit elements do not satisfy Ohm’s law with the constant resistance R over a wide range of voltages. Therefore, they are also called nonohmic circuit elements. The nonohmic elements may be described by a similar expression: V ¼ RðV ÞI , RðV Þ
V I ðV Þ
ð2:19Þ
but with a variable resistance R(V). For passive elements, RðV Þ > 0. The resistance R(V) is known as the static or DC resistance. Figure 2.12 depicts the υi characteristics for three distinct passive circuit elements. ohmic conductor
a)
ideal (Shockley) diode
incandescent light bulb
b)
I
c)
I
0
0
0 V
I
V
V
Fig. 2.12. Three υi characteristics: (a) linear—resistance; (b) nonlinear—incandescent light bulb; and (c) nonlinear—ideal or Shockley diode.
The ﬁrst element is an ohmic element (ohmic conductor) with a constant resistance R. The corresponding υi characteristic is a straight line—the circuit element is linear. The second element corresponds to an incandescent light bulb. Its resistance R increases when the applied voltage V increases (the conductivity of the radiating ﬁlament of wire decreases with increasing absorbed power and wire temperature). Hence, the υi characteristic bends down and deviates from the straight line—see Fig. 2.12b. This element only approximately II47
Chapter 2
Major Circuit Elements
follows Ohm’s law. It is therefore the nonlinear circuit element. The third element in Fig. 2.12 corresponds to an ideal (Shockley) diode. The diode does not conduct at negative applied voltages. At positive voltages, its υi characteristic is very sharp (exponential). The diode is also the nonlinear circuit element. Strictly speaking, the υi characteristic of the incandescent light bulb does not belong to the list of circuit elements due to its limited applicability. However, the ideal diode is an important nonlinear circuit element. The nonlinear elements are generally polar (nonsymmetric) as Fig. 2.12c shows.
2.2.3 Static Resistance of a Nonlinear Element Once the υi characteristic is known, we can ﬁnd the static resistance R(V) of the nonlinear circuit element at any given voltage V0. For example, the υi characteristic of the ideal diode shown in Fig. 2.12c is described by the exponential Shockley equation: V 1 ð2:20Þ I ¼ I S exp VT In Eq. (2.20), the constant IS [A] is the diode saturation current. The saturation current is very small. The constant VT [V] is called the thermal voltage. Example 2.5: Give a general expression for the diode resistance R(V) using Eq. (2.20) and ﬁnd its terminal values at V ! 0 and V ! 1, respectively. Then, calculate static diode resistance R0 and diode current I0 when the voltage across the diode is V0 ¼ 0.55 V. Assume that I S ¼ 1 1012 A and V T ¼ 25:7 mV. Solution: Using Eq. (2.20) we obtain
RðV Þ ¼ IS
V V exp 1 VT
ð2:21Þ
When V ! 0, we can use a Taylor series expansion for the exponent. Keeping only the ﬁrst nontrivial term, one has expðV =V T Þ 1 þ V =V T . Therefore,
RðV Þ !
VT when V ! 0 ðor V =V T 0, the polarity of which is shown in Fig. 2.16. The term independent means that voltage VS does not vary because of different parameters of an electric circuit (not shown in the ﬁgure), which is implied to be connected to the voltage source.
Fig. 2.16. Symbol for an ideal voltage source along with the voltage and current behavior.
Current Through the Voltage Source: Active Reference Conﬁguration The current I ﬂowing through the voltage source is shown in Fig. 2.16 by an arrow. The relation between voltage polarity and current direction depicted in Fig. 2.16 is known as the active reference conﬁguration. It is commonly used for all active circuit elements such as voltage and current sources, either dependent or independent. A useful ﬂuid mechanics analogy for the voltage source is water (electric current) that is pushed up the “voltage” hill in Fig. 2.16b by external means. Alternatively, one may think of a water pump that is characterized by a constant pressure drop. The active reference conﬁguration means that the voltage source supplies electric power to the circuit. This conﬁguration is the opposite of the passive reference conﬁguration for the resistance. υi Characteristic of the Voltage Source Figure 2.17a plots the υi characteristic of the ideal voltage source. The term ideal literally means that the υi characteristic is a straight vertical line: the ideal voltage source is capable of supplying any current to any circuit while keeping the same voltage VS across its terminals. In reality, it is not the case since the high currents mean high powers. Therefore, a laboratory power supply—the physical counterpart of the ideal voltage source—will be eventually overloaded as shown in Fig. 2.17b. Figure 2.17c shows a common way of drawing the υi characteristic for the voltage source with the axes interchanged. For the purposes of consistency, the xaxis will always be used as the voltage axis throughout the text. II52
Chapter 2
Section 2.3: Independent Sources
a)
b)
I
I
c)
overload
V
VS 0
0 V
VS
overload
0
I
V
VS
Fig. 2.17. υi Characteristics for (a) ideal voltage source used in the circuit analysis and (b) its physical counterpart—a regulated laboratory power supply. (c) Typical way of drawing the υi characteristic for the voltage source with the axes interchanged.
Symbols for Independent Voltage Source Multiple symbols may be used in a circuit diagram to designate the independent ideal voltage source—see Fig. 2.18. All these symbols are equivalent from the circuit point of view, as long as we imply the ideal source. However, their physical counterparts are quite different. The general symbol in Fig. 2.18a implies either an AC to DC converter (the laboratory power source) or a battery. The symbol in Fig. 2.18b relates to a battery and Fig. 2.18c–d depicts battery banks.
=
5V
d)
=
+
c)
+
+

+

=
5V

b)
5V
5V

a)
Fig. 2.18. (a) Generic DC voltage source, (b) single battery, and (c) and (d) battery banks. All symbols in the circuit diagram are equivalent.
Example 2.7: Solve an electric circuit shown in Fig. 2.19—determine circuit current I and voltage across the resistance V. Solution: We use the graphical solution—plot the υi characteristic of the 2 kΩ resistance and the υi characteristic of the voltage source on the same graph to scale—see Fig. 2.19b. The intersection point is the desired solution: V ¼ 3 V, I ¼ 1:5 mA. Indeed, this simple solution implicitly uses circuit laws (KVL and KCL) studied next.
II53
Chapter 2
Major Circuit Elements b)
a)
I, mA
2
I 1
+
3 V = VS
+ 
R=2 kW
0
V
V, volts

VS 1
I
2 4
2
0
2
4
Fig. 2.19. Electric circuit solution in graphical form.
2.3.2 Circuit Model of a Practical Voltage Source Any practical voltage source is modeled as a combination of an ideal voltage source VS and an ideal resistance R in series—see Fig. 2.20a. The resistance R reﬂects the nonideality of the practical source: it limits the maximum available source current and the maximum available source power by (similar to the currentlimiting resistor): I max ¼ V S =R,
Pmax ¼ V S I max ¼ V 2S =R
ð2:29Þ
Voltage VS is called the opencircuit voltage of the source for an obvious reason. Similarly, current Imax is called the shortcircuit current of the source. Once both quantities are measured in laboratory, resistance R (called the internal source resistance) may be found using Eq. (2.29).
a) practical voltage source
b)
I
R VS
+ 
+
I
V VS
V

Fig. 2.20. Circuit model of a practical voltage source and its υi characteristic.
Exercise 2.8: The opencircuit voltage of a voltage source is 9 V; the shortcircuit current is 2 A. Determine the internal source resistance. Answer: 4.5 Ω.
II54
Chapter 2
Section 2.3: Independent Sources
The υi characteristic of the practical voltage source is the plot of source current I versus voltage V available from the source in Fig. 2.20a. This voltage is generally less than VS since any nonzero current I causes a voltage drop of RI across resistance R. One has V ¼ V S RI ) I ¼
VS V R
ð2:30Þ
This υi characteristic is plotted in Fig. 2.20b by a solid line. The deviation from the straight vertical line characterizes the degree of nonideality. Emphasize that any laboratory power supply is indeed a practical voltage source. However, using a special circuit, its input is regulated so that the output voltage does depend on the output current, at least over a reasonable range of currents. Therefore, instead of Fig. 2.20b we arrive at a more reliable voltage source from Fig. 2.17b. Exercise 2.9: Determine internal source resistance for the source illustrated in Fig. 2.20b given that every horizontal division is 3 V and every vertical division is 1 A. Answer: 0.6 Ω.
b)
1
+ 2
IS

V

a)
+
2.3.3 Independent Ideal Current Source An independent ideal current source is dual of the ideal voltage source. Figure 2.21a shows the corresponding circuit symbol for the DC (constantcurrent) source. As a circuit element, the constantcurrent source is directional: terminals 1 and 2 indicate the direction of the current ﬂow. The current source generates a positive constant current, I S > 0, which ﬂows from terminal 2 to terminal 1 in Fig. 2.21. The term independent means that current IS does not vary because of different parameters of an electric circuit (not shown in the ﬁgure), which is implied to be connected to the current source.
Fig. 2.21. Symbol for the ideal current source along with voltage and current designations.
Voltage Across the Current Source: Active Reference Conﬁguration Once the current source is connected to a circuit, a voltage V will be created across it. The voltage polarity is indicated in Fig. 2.21a. The relation between voltage polarity and current direction shown in Fig. 2.21 is again the active reference conﬁguration, similar to the voltage source. A useful ﬂuid mechanics analogy for the electric current source is a II55
Chapter 2
Major Circuit Elements
water pump that creates a constant water supply (e.g., 0.5 ft3/s). Indeed, this water pump will be characterized by a certain pressure difference across its terminals, which is the analogy of voltage V in Fig. 2.21.
υi Characteristic of the Current Source Figure 2.22a plots the υi characteristic of an ideal current source. Compared to the ideal voltage source, the graph is rotated by 90 degrees. The term ideal again means that the υi characteristic is the straight horizontal line: the ideal current source is capable of creating any voltage across its terminals while keeping the same current IS ﬂowing into the circuit. In reality, it is not the case since high voltages mean high powers. Therefore, a laboratory current power supply—the physical counterpart of the ideal current source—will eventually be overloaded as shown in Fig. 2.17b. The current laboratory supplies are rarely used (one common use relates to transistor testing); they are less common than the voltages supplies. However, the current sources are widely used in transistor circuits, both integrated and discrete. There, the current sources are created using dedicated transistors. Furthermore, photovoltaic sources are essentially current sources. a)
b)
I
I
IS
IS
overload
0
0
V
V
Fig. 2.22. υi Characteristics of (a) an ideal current source and (b) its physical counterpart.
Symbols for Independent Current Source A few equivalent symbols may be used in a circuit diagram to designate the independent ideal current source; see Fig. 2.23. All these symbols are equivalent as long as we imply the ideal source. The symbol in Fig. 2.23a is used in North America, the symbol in Fig. 2.23b is European, and the symbol in Fig. 2.23c may be also found in older texts. a)
b)
c)
= 1 mA
= 1 mA
1 mA
Fig. 2.23. Equivalent symbols of the current source in the circuit diagram.
II56
Chapter 2
Section 2.3: Independent Sources
Example 2.8: Solve an electric circuit shown in Fig. 2.24—determine voltage V across the source and across the resistance. Solution: Similar to the voltage source, we use a graphical solution—plot the υi characteristic of the 2kΩ resistance and the υi characteristic of the current source on the same graph to scale; see Fig. 2.24b. The intersection point gives us the desired solution: V ¼ 3 V. Note that the solutions for this example and the solution for Example 2.7 coincide. This means that, under certain conditions, we can interchange both sources without affecting the circuit performance. Indeed, the graphical solution implicitly uses the circuit laws (KVL and KCL) studied in detail next.
a)
b)
2 R=2 kW
IS
1
+
+
IS
V
V


1.5 mA = IS
I, mA
0 V, volts 1
IS
2 4
2
0
2
4
Fig. 2.24. Electric circuit solution in the graphical form.
2.3.4 Circuit Model of a Practical Current Source Any practical current source is modeled as a combination of the ideal current source IS and the ideal resistance R in parallel—see Fig. 2.25a. The resistance R reﬂects the nonideality of the practical source: it limits the maximum available source voltage and the maximum available source power by V max ¼ RI S ,
Pmax ¼ V max I S ¼ RI 2S
ð2:31Þ
Voltage Vmax is again called the opencircuit voltage of the source. Similarly, current IS is called the shortcircuit current of the source. Once both the quantities are measured, resistance R (called the internal source resistance) may be found using Eq. (2.31). Exercise 2.10: The opencircuit voltage of a current source is 9 V; the shortcircuit current is 2 A. Determine the internal source resistance. Answer: 4.5 Ω.
II57
Chapter 2
Major Circuit Elements a) practical current source
b)
I
I
+
IS
IS
R
V V

Fig. 2.25. Circuit model of a practical current source and its υi characteristic.
The υi characteristic of the practical current source is the plot of current I available from the source versus voltage V across the source in Fig. 2.25b. This current is generally less than IS since a portion of IS ﬂows through the internal resistance R, i.e., I ¼ IS
V R
ð2:32Þ
This υi characteristic is plotted in Fig. 2.25b by a solid line. The deviation from the straight horizontal line characterizes the degree of nonideality. Exercise 2.11: Determine the internal source resistance for the source illustrated in Fig. 2.25b given that every horizontal division is 3 V and every vertical division is 1 A. Answer: 15 Ω.
2.3.5 Operation of the Voltage Source Operation of a voltage power supply of any kind (an electric generator, a chemical battery, a photovoltaic cell, etc.) might be illustrated based on the charge separation principle very schematically depicted in Fig. 2.26. We need to deliver electric power to a load modeled by an equivalent resistance RL. First, we consider in Fig. 2.26 a charged capacitor with a charge Q connected to a load resistor RL at an initial time moment. The capacitor voltage V is related to charge by V ¼ Q=C where C is the (constant) capacitance. The capacitor starts to discharge and generates a certain load current IL. At small observation times, the change in Q is small, so is the change in V. Therefore, the capacitor initially operates as a voltage power supply with voltage V. However, when time progresses, the capacitor discharges and the voltage V eventually decreases.
II58
Chapter 2
Section 2.3: Independent Sources
IL
Q V
IL

+Q
IL

+ +++++++
Charge separation mechanism
RL
Fig. 2.26. Power source schematically represented as a capacitor continuously charged by a charge separation mechanism—the charge pump.
How could we keep V constant, i.e., continuously charge the capacitor? A charge separation mechanism should be introduced between the hypothetic capacitor plates to continuously compensate for the charge leakage. That mechanism may have the forms: 1. For an electromechanical generator, this is the Lorentz force that acts on individual electrons in a conductor and pushes them to one conductor terminal while creating the opposite charge density on the opposite conductor terminal. The macroscopic effect of the Lorentz force is the Faraday’s law of induction. 2. For a battery, these are chemical reactions at the electrodes which cause a charge separation, i.e., positive metal ions dissolve in the electrolyte and leave excess electrons in the metal electrode on the left in Fig. 2.26. 3. For the photovoltaic cell, this is a builtin potential of the semiconductor pnjunction that separates lightgenerated negative carriers (electrons) and positive carriers (holes) as shown in Fig. 2.26. Indeed, the capacitor analogy in Fig. 2.26 is only an illustrative approach, especially for electromechanical power generation. Below, we will consider a few speciﬁc examples.
2.3.6 Application Example: DC Voltage Generator with Permanent Magnets A realistic electromechanical voltage source—a basic DC generator with permanent magnets—is shown in Fig. 2.27. This generator setup makes use of the Lorentz force: ~ f q ~ υ~ B ð2:33Þ The Lorentz force acts on charge q moving with a velocity ~ υ in an external magnetic ﬁeld ~ with the vector ﬂux B measured in tesla (T). The force itself is measured in newtons. The cross symbol in Eq. (2.33) denotes the vector product of two vectors evaluated according to the righthand rule. Shown in Fig. 2.27a are two permanent magnets (stator of the generator) responsible for creating the magnetic ﬂux ~ B emanating from the north pole
II59
Chapter 2
Major Circuit Elements
(N) and terminating at the south pole (S). The armature (rotor) rotates clockwise in Fig. 2.27b with the armature velocity ~ υ. a)
b)
stator
B
v v
B rotor
N
d
+
S
 + brushes   ++
brushes

V
B +
f v
+
V
+

f
A= d
Fig. 2.27. Charge separation in a DC electromechanical generator.
When the ﬂux density ~ B is applied, every positive charge +q in the armature segment l will experience a Lorentz force with the magnitude f ¼ þqυB which will move this charge toward the right terminal of the armature in Fig. 2.27b. Similarly, every negative charge –q in the armature would experience the equal but oppositely directed Lorentz force f ¼ qυB which will move this charge toward the left terminal. Hence, a charge separation occurs along the armature which will give rise to an induced voltage V. Total work W of the Lorentz force on a charge q along the entire armature path in Fig. 2.27b is given by W ¼ 2lf . This work divided by the amount of charge determines the equivalent voltage that will be developed on the generator terminals, i.e., the instantaneous generator voltage V ¼ W =q ¼ 2lυB. If the armature rotates at an angular speed ω (rad/s), the charge velocity perpendicular to the ﬁeld is given by υ ¼ d=2ω cos θ (m/s). Plugging in this expression and averaging over angles θ from 0 to π/2, we obtain the average generator voltage in the form V ¼ ldBhcos θi ¼ ð2=π ÞABω
½V
ð2:34Þ
where A is the armature area. If the rotor has N turns, the result is multiplied by N. The same expression for the voltage is obtained using the Faraday’s law of induction. A regulator circuit is necessary to obtain a ﬂat DC voltage without ripples. Any brushed DC motor operates as a generator when its shaft is rotated with a certain speed. The generated opencircuit voltage may be observed in laboratory with the oscilloscope. Exercise 2.12: Determine average opencircuit generator voltage in Fig. 2.27 given A ¼ 0:1 m2 , B ¼ 0:2 T, ω ¼ 20 rad/s (191 rpm), and the armature with 20 turns. Answer: 5.1 V.
II60
Chapter 2
Section 2.3: Independent Sources
2.3.7 Application Example: Chemical Battery A chemical reaction in a battery induces a continuous charge separation. The “charge pump” so constructed, once connected to a load, is able to create a continuous electric current into a load and a voltage difference across it. You are probably aware of the quest to improve the venerable battery. Extensive coverage in the media and in technical journals frequently reports on new chemical compounds and control circuits. They target smaller, more powerful rechargeable batteries for such diverse devices as portable computers, cell phones, sensors, and automobiles. Speciﬁcally, it is the automotive sector which implements hybrid vehicle technology where powerful electro motors in conjunction with highperformance batteries are supplementing, even completely replacing, conventional combustion engines. For a standard chemical battery, the two important parameters are battery voltage and battery capacity. The capacity, Q, is a new quantity that is needed because of a battery’s inability to provide constant current and power for an inﬁnite time duration. How a battery behaves over time is illustrated in Fig. 2.28 where we monitor the power and current as a function of time. A key time constant is the socalled discharge time, which is critically dependent on the attached load. a)
b)
PB, W #1
IB, A #1
#2
T1
#2
T2
t
T1
T2
t
Fig. 2.28. Generic plots of delivered power, PB, and electric current, IB, for two different loads labeled #1 and #2. The discharge times T1 and T2 correspond to the loads #1 and #2, respectively.
When a load of resistance RL is connected to the battery, and the battery’s internal resistance R is negligibly small compared to that resistance, the circuit current, IB, and the power delivered by the battery, PB, are determined based on Ohm’s law: VB ð2:35Þ , PB ¼ V B I B IB ¼ R The total energy, EB, stored in the battery and then delivered to the circuit is a fixed constant. Its value depends on the battery type and size. The total energy in joules is given by the time integral of delivered power over time, i.e., 1 ð ð2:36Þ EB ¼ PB ðt 0 Þdt 0 0
We can assume that the total energy is a ﬁnite constant; it follows from Eq. (2.36) that the delivered power must drop to zero at a ﬁnite time T. This is schematically shown in
II61
Chapter 2
Major Circuit Elements
Fig. 2.28 for two different load resistances, #1 and #2, which require two different circuit currents. Even though the two power curves in Fig. 2.28a are different, the area under those curves, denoting the total energy stored in the battery, remains the same to a sufﬁcient degree of accuracy. The battery’s terminal voltage VB also remains approximately constant over the entire operation cycle and even afterwards. It is the battery’s current IB that ﬁnally sharply decreases with time and causes a drop in power, as seen in Fig. 2.28b. Let us consider the simplest case where the current is a constant for t < T and at t ¼ T drops to zero and stays zero for t > T. From Eq. (2.36), it follows that 1 ð ðT 0 0 ð2:37Þ EB ¼ PB ðt Þdt ¼ V B I B dt 0 ¼ ½T I B V B 0
0
The expression in the square brackets is the deﬁnition of the battery capacity, Q: EB Q TIB ¼ ð2:38Þ VB Since the battery terminal voltage is always known, its capacity determines the total energy stored in the battery. The capacity is measured in A·h (Ah) or for small batteries in mA·h (mAh). The capacity rating that manufacturers print on a battery is based on the product of 20 h multiplied by the maximum constant current that a fresh battery can supply for 20 h at 20 C while keeping the required terminal voltage. The physical size of batteries in the USA is regulated by the American National Standards Institute (ANSI) and the International Electrotechnical Commission (IEC). Table 2.3 lists the corresponding parameters of some common batteries. Example 2.9: A 12V battery rated at a capacity of Q ¼ 100 A·h may deliver 5 A over a 20h period, 2.5 A over a 40h period, or 10 A over a 10h period. Find the total energy delivered by the battery provided that its internal resistance is negligibly small. Solution: The total energy delivered by the battery is equal to
EB ¼ V B Q ¼ 12 100 V A h ¼ 1200 W h ¼ 4:32 MJ
ð2:39Þ
It remains constant for each case. This example shows that the electric energy can be measured either in joules or in Wh or more often in kWh. Clearly, 1 Wh ¼ 3600 J.
Circuit Model of a Battery As a practical voltage source, a battery always has a small, but ﬁnite internal resistance, R. Battery’s equivalent circuit therefore includes the ideal voltage source and the internal resistance in series—see Fig. 2.29. Even though the values of R are small, the internal resistance has critical implications affecting both the battery’s efﬁciency and its ability to provide a high instantaneous power output. In general, it is difﬁcult to directly measure II62
Chapter 2
Section 2.3: Independent Sources
 + 9V
R
=
9V
+ 
Fig. 2.29. Circuit model of a battery: the ideal voltage source in series with an internal resistor.
the internal resistance of batteries, you would need a calibrated load resistor and sophisticated measurement equipment to precisely measure voltages and currents. Exercise 2.13: A 12V battery has an internal resistance of 10 Ω. What are the maximum current and the maximum power that the battery can output? Answer: I max ¼ 1:2 A, Pmax ¼ 14:4 W
Many battery types have been developed for a wide range of applications. They differ both in battery energy storage per kg of weight, or unit volume, and in power delivery per kg of weight, or per unit volume. In particular, modern heavyduty, deepcycle batteries may sport the following properties: Energy storage : Power density :
150 W h=l;
ð2:40Þ ð2:41Þ
2 kW=l:
You can compare Eqs. (2.40) and (2.41) with the last row of Table 2.3 and establish the approximate density of the battery device. Table 2.3. Characteristics of batteries (from multiple datasheets). Battery size/type AAA AA C D 9V Lithium batteries Lead acid starter battery (automotive, deep cycle) Deepcycle marine, electric vehicles
Rechargeable No No No No No Yes Yes
Yes
Voltage (cell) 1.5 1.5 1.5 1.5 9.0 3.6–3.7 12.6
Capacity (Ah) 1.3 2.9 8.4 20.5 0.6 0.7–1.5 ~600A for 30 s at 32 F before voltage drops to 7.20 V Variable: ~ 30 Wh per kg of weight, or ~ 108 kJ per kg of weight
Resistance (R) 100–300mΩ for alkaline battery per cell ~400 mΩ ~300 mΩ 0 vs. source), gate (with voltage V GS > 0 vs. source), and source itself (grounded). The source is also connected to a metal conductor on the other side of the semiconductor body. Accordingly, there are two types of the electric ﬁeld within the semiconductor body: the horizontal ﬁeld created by VGS, and the vertical ﬁeld created by VDS. The horizontal ﬁeld ﬁlls a conducting channel between the drain and the source with charge carriers, but has no effect on the vertical charge motion. The resulting carrier concentration in the channel is given by n ¼ N ðV GS V Th Þ > 0, N ¼ const, V Th ¼ const: The individual carrier charge is q. Given the channel cross section A, the carrier mobility μ, and the channel length L, determine transistor current ID and transistor resistance (draintosource resistance) RDS. Express both results in terms of quantities listed above including VGS and VTh.
source
transmission line (TL)
load a
Channel with electron carriers
+ 
I
I
I
I
RL b
50 km
II78
Chapter 2
Problems
Problem 2.17. Solve the previous problem when the total cable length (two cables must run to a load) is increased to 200 km. Problem 2.18. An AWG 00 aluminum transmission grid cable has the wire diameter of 9.266 mm. The conductivity of aluminum is 4.0 107 S/m. A power transmission system that uses this cable is shown in the ﬁgure that follows. The load power is 1 MW. Determine the minimum necessary load voltage V that guarantees us a 1 % relative power loss in the cables. source
+ 
transmission line (TL)
load
I
I
I
I
+ V?

100 km
Problem 2.19. An ACdirect microhydropower system is illustrated in the ﬁgure that follows.
length of 1000 m. Each line uses AWG#10 aluminum wire with the diameter of 2.59 mm. The conductivity of aluminum is 4.0 107 S/ m. The house load is an electric range with the resistance of 20 Ω. Determine total power delivered by the generator, Ptotal, total power loss in the transmission lines, Ploss, and total useful power, Puseful (show units).
2.1.7 Physical Component: Resistor Problem 2.20. A leaded resistor has color bands in the following sequence: brown, black, red, gold. What is the resistor value? Problem 2.21. Potentiometer operation may be schematically explained as moving sliding contact #2 in the following ﬁgure along a uniform conducting rod with the total resistance of 20 kΩ. Determine resistance between terminals 1 and 2 as well as between terminals 2 and 3 of the potentiometer, when the sliding contact is at one ﬁfth of the rod length. a)
b)
1
20 kW
1
20 kW 4/5 1/5 2
3
3
2
2.2 Nonlinear Passive Circuit Elements source
+ 
transmission line (TL) to each house
I
I
I
I
load
RL
1 km
Reprinted from MicroHydropower Systems Canada 2004, ISBN 0662358805. The system uses a single phase induction generator with the rms voltage (equivalent DC voltage) of 240 V. The system serves four small houses, each connected to the generator via a separate transmission line with the same
2.2.2NonlinearPassiveCircuitElements 2.2.3 Static Resistance 2.2.4Dynamic(SmallSignal)Resistance 2.2.5 Electronic Switch Problem 2.22. A nonlinear passive circuit element—the ideal diode—is characterized by the hυi characteristic in the form
i I ¼ I S exp
V VT
1 with I S ¼ 1 1013 A
and V T ¼ 25:7 mV. Find the static diode resistance R0 and the diode current I0 when (A) V0 ¼ 0.40 V, (B) V0 ¼ 0.50 V, (C) V0 ¼ 0.55 V, and (D) V0 ¼ 0.60 V.
II79
Chapter 2
Major Circuit Elements Justify your answer.
+
+
I V
V

Problem 2.23. Find the dynamic (smallsignal) resistance r of a nonlinear passive circuit element—the ideal diode—when the operating DC point V0, I0 is given by the solutions to the previous problem. Consider all four cases.
Problem 2.24. A nonlinear passive circuit element is characterized by the υi characteristic in V =V S ﬃ with I S ¼ 1 A and the form I ¼ I S pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2
I
1þðV =V S Þ
V S ¼ 1 V. Plot the υi characteristic to scale. Next, ﬁnd the static element resistance R0, the element current I0, and the corresponding dynamic element resistance r when (A) V0 ¼ 0.1 V, (B) V0 ¼ 1.0 V, and (C) V0 ¼ 5.0 V.
2.3 Independent Sources
I, A
2.3.1 Independent Ideal Voltage Source 2.3.2 Circuit Model of a Practical Voltage Source
0 V, volts
0
Problem 2.27. In the following ﬁgure, determine if the element is a resistance or a voltage source. Find the power delivered to element A or taken from element A in every case.
5
1 mA
A)
+
VA=20 V 1 mA
B)
A

VA=20 V
+
Problem 2.26. A DC circuit shown in the following ﬁgure includes two interconnected passive elements: an ideal diode and a resistance. One possible circuit solution is given by an intersection of two υi characteristics marked by a circle in the same ﬁgure. This solution predicts a nonzero circuit current and a positive voltage across both circuit elements. A. Is this solution an artifact (a mistake has been made somewhere)? B. Is this solution true (the circuit so constructed might function)?
A
1 mA
C)
A
+
Problem 2.25. Repeat the previous problem when I S ¼ 0:5 A and V S ¼ 0:5 V. All other parameters remain the same. Consider the following DC operating points: (A) V0 ¼ 0.05 V, (B) V0 ¼ 0.50 V, and (C) V0 ¼ 2.50 V.
VA=20 V

1 5
V

1
0
Problem 2.28. Based on voltage and current measurements, determine if the circuit element is a resistance or a voltage source. Readings of the ammeter and voltmeter are shown in the following ﬁgure.
II80
Chapter 2 a) 3 V = VS
+1 A +V 
+ 
V I
b)
5 V B)
I
+ +
+A 

A)
Problems
I, A
1
+A 
1 A +V 
0
5 V
C)
V, volts
+A 
1 A +V 
1 5
+5 V
Problem 2.29. The ﬁgure that follows shows a circuit with a passive nonlinear circuit element shown by a rectangle.
V

3V
V

b)
+ 
+
+ +
+ 
1W
I 1
practical voltage source I
I 3 V = VS
5
Problem 2.31. Plot to scale the υi characteristic of the practical voltage source shown in the following ﬁgure. a)
a)
0
I, A
b) 1
I, A
0 V, volts 0 V, volts
1 5
0
5
The polarity (direction of current inﬂow for passive reference conﬁguration) of the element is labeled by a sign plus. The υi characteristic of the element is also shown in the ﬁgure. Determine current I and voltage V. Problem 2.30. Repeat the previous problem for the circuit shown in the ﬁgure that follows.
1 5
0
5
Problem 2.32. The following ﬁgure shows a circuit with a passive nonlinear circuit element labeled by a rectangle. Element’s polarity (direction of current inﬂow for passive reference conﬁguration) of the element is indicated by a sign plus. The υi characteristic of the
II81
Chapter 2
Major Circuit Elements
element is also shown in the ﬁgure. Determine circuit current I. R=2 W
Problem 2.34. The following ﬁgure shows a circuit with a passive nonlinear circuit element shown by a rectangle. Element’s polarity (direction of current inﬂow for passive reference conﬁguration) of the element is labeled by a sign plus. The υi characteristic of the element is also shown in the ﬁgure. Determine current I and voltage V.
+ 
V

4 V = VS
+ +
I
I
b)
I, A
1
a)
0
I
+
V, volts
+
a)
to the circuit element or taken from it in every case.
600 mA = IS
V
I
b) 0
5
2.3.3 Independent Ideal Current Source 2.3.4 Circuit Model of a Practical Current Source Problem 2.33. Readings of the ammeter and voltmeter are shown in the following ﬁgure. A)  A+
+2 mA +V 
5 V +A 
0 V, volts
1 5
0
5
Problem 2.35. Repeat the previous problem for the circuit shown below. a) 600 mA = IS
V

0.5 A  V+
+10 V
C)
I, A
+
B)
1
+
1 5
b) 1
I, A
 A+
+1 A  V+
+10 V
Based on voltage and current measurements, determine if the element is a resistance or a current source. Then, ﬁnd the power delivered
0 V, volts
1 5
0
5
II82
Chapter 2
Problems
Problem 2.36. Plot to scale the υi characteristic of the practical current source shown in the following ﬁgure. practical current source 8.33 W 0.8 A
I
+
a)
800 mA= IS
+ 1
R=2 W
3 V= VS
I, A
V
0
b)
V, volts
I, A
1
1 5
0
5
0 V, volts
2.3.7 Application Example: Chemical Battery
1 5
0
5
Review Problems Problem 2.37. For every circuit element shown in the following ﬁgure, plot its υi characteristic on the same graph. 600 mA= IS
+ 1
R=5 W
4 V= VS
I, A
0 V, volts
1 5
0
5
Problem 2.38. Repeat the previous problem: for every circuit element shown in the ﬁgure below, plot its υi characteristic on the same graph.
Problem 2.39. The electronics aboard a certain sailboat consume 96 W when operated from a 24 V source. A. If a certain fully charged deepcycle marine battery is rated for 24 V and 100 A h, for how many hours can the electronics be operated from the battery without recharging? (The amperehour rating of the battery is the battery capacity—the operating time to discharge the battery multiplied by the current). B. How much energy in kilowatt hours is initially stored in the battery? Problem 2.40. A motor of a small, unmanned electric vehicle consumes 120 W and operates from a 24V battery source. The source is rated for 200 Ah. A. For how many hours can the motor be operated from the source (a battery bank) without recharging? B. How much energy in kilowatt hours is initially stored in the battery source? Problem 2.41. A certain sensing device operates from a 6V source and consumes 0.375 W of power over a 20h time period. The source is a combination of four fully charged AAA batteries, 1.5 V each, assembled
II83
Chapter 2
Major Circuit Elements
+
Problem 2.44. Solve an electric circuit shown in the following ﬁgure—determine current i through the 2kΩ resistance. The independent current source is given by iS ¼ 0:2 1:5 cos 5 t ½mA ; the opencircuit voltage gain of the dependent source is 12 V/V.
iS
vin
3 kW
1 kW
+ 
vout
i

Problem 2.45. Solve an electric circuit shown in the following ﬁgure—determine current i through the 2kΩ resistance. The independent voltage source is given by υS ¼ 0:3 þ 0:7 cos 6t ½V ; the transresistance of the dependent source is 250 V/A.
i
iS
iout
vin
+
30 W
0.1 kW
v
Problem 2.47. Solve an electric circuit shown in the following ﬁgure—determine voltage υ through the 1kΩ resistance. The independent voltage source is given by υS ¼ 0:05 þ 0:1 cos 4t ½V ; the shortcircuit current gain of the dependent source is 10 A/A. iin vS
+ 
50 W
1 kW iout
+
Problem 2.43. Draw circuit diagrams for four major types of dependent sources, label stimulus voltage/current and output voltage/current. Describe operation of each dependent source.
vout
Problem 2.46. Solve an electric circuit shown in the following ﬁgure—determine voltage υ through the 30Ω resistance. The independent current source is given by iS ¼ 0:05 0:2 cos 2t ½mA ; the transconductance of the dependent source is 10 A/V.
+
2.4.1 Dependent Versus Independent Sources 2.4.2 Deﬁnition of Dependent sources 2.4.3 Transfer Characteristics 2.4.4 TimeVarying Sources
2 kW
+ 
v

2.4 Dependent Sources and TimeVarying Sources
40 W
+ 

Problem 2.42. How many Joules are in 1 kWh and how many Nm does this correspond to?
iin vS

in series. The batteries discharge by the end of the 20h period. A. What is the expected capacity of a typical AAA battery used, in mAh? B. How much energy in Joules was stored in each AAA battery?
Problem 2.48. Solve an electric circuit shown in the following ﬁgure—determine current i through a nonlinear passive circuit element shown by a rectangle. Element’s polarity (direction of current inﬂow for passive reference conﬁguration) is labeled by a sign plus. The υi characteristic of the element is also shown in the ﬁgure. The independent current source is given by iS ¼ 0:1 A ; the opencircuit voltage gain of the dependent source is 5 V/V.
II84
Chapter 2
Problems
iS
6 kW
+ 
vin
b)
+
+
a) vout
i
2.5.1 Ideal Voltmeter and Ammeter 2.5.3 Types of Electric Ground 2.5.4 Ground and Return Current
I, mA
1
2.5 Ideal Voltmeter and Ammeter: Circuit Ground
0 V, volts
Problem 2.50. You attempt to measure electric current through a resistance as part of a circuit. Is the following ﬁgure appropriate? What is the current across the 51Ω resistor? I
1 5
0
51 W
5
Problem 2.49. Solve the electric circuit shown in the following ﬁgure—determine voltage υ across the 1kΩ resistance. 1 kW
+
iin
+ 
b) 0.1
v
iout
Problem 2.51. You attempt to measure voltage across a resistance in the circuit. Is the following ﬁgure correct? What is the voltmeter’s reading, assuming an ideal instrument? I=20 mA

vS
+
a)
+A
51 W +V 
I, mA
Problem 2.52. Two circuits with an incandescent light bulb are shown in the following ﬁgure. Will they function? Explain. 0 V, volts
a)
5V
+

0.1 1.5
0
1.5
b)
+
5V

A nonlinear passive circuit element is shown by a rectangle. Element’s polarity (direction of current inﬂow for passive reference conﬁguration) of the element is labeled by a sign plus. The υi characteristic of the element is also shown in the ﬁgure. The independent voltage source is given by υS ¼ 0:6 V; the shortcircuit current gain of the dependent source is 100 A/A.
Problem 2.53. A 9V battery is connected to a 7.5kΩ resistor shown in the following ﬁgure.
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Find current I in every case. (a) The negative terminal is left disconnected; (b) the negative terminal is connected to the positive terminal through the resistor; (c) both terminals are connected to chassis ground. a)
7.5 kW
 +
2.5.5 Absolute Voltage and Voltage Drop Across a Circuit Element Problem 2.55. Determine if the circuit element shown in the following ﬁgure is a resistance, a voltage source, or a wire (short circuit). Absolute voltages at points a and b are measured versus ground. I
I
9V
Va
Vb
a
b
0V 0V
b) 7.5 kW
 +
1. V a ¼ 3 V, V b ¼ 3V, I ¼ 1 A 2. V a ¼ 3 V, V b ¼ 1V, I ¼ 1 A 3. V a ¼ 2 V, V b ¼ 5 V, I ¼ 2 A:
I
9V
c)
Problem 2.56. Determine if the circuit element shown in the following ﬁgure is a resistance, a voltage source, or a wire (short circuit). Absolute voltages at points a and b are measured versus ground.
7.5 kW
 +
I
9V
I
Problem 2.54. What is the voltmeters’ (ammeter’s) reading in the ﬁgure below?
5V
+ 
+V 
#1
Vb
a
b 0V
10 W
a)
Va
V
+ 
#2
+ 
#2
1. V a ¼ 6 V, V b ¼ 3 V, I ¼ 1 A 2. V a ¼ 1 V, V b ¼ 1 V, I ¼ 1 A 3. V a ¼ 7V, V b ¼ 5V, I ¼ 2 A:
0V 10 W
b) 5V
+ 
+V 
A
#1
0V
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0V
+
+
+
+
6
VD 6V
A
VA
E

E VE=7 V
1
D
5V
5
VA=5 V A
B
VB
+
2
+
+
+
D VD=4 V
VE

B
VB=10 V
12 V
C

+
+ 7V
4
VC

C

3
VC=4 V
Problem 2.58. Determine voltages across circuit elements A, B, C, D, and E in the circuit shown in the following ﬁgure.

Problem 2.57. Determine absolute voltages at nodes 1 through 6 in the circuit shown in the following ﬁgure.
0V
0V 0V
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Chapter 3: Circuit Laws and Networking Theorems Overview Prerequisites:  Knowledge of circuit elements, their i characteristics, and Ohm’s law (Chap. 2) Objectives of Sect. 3.1:  Understand the meaning of an electric network and its topology (nodes, branches, loops, meshes)  Review the Kirchhoff’s current law, its use, and value  Review the Kirchhoff’s voltage law, its use, and value  Become familiar with the Tellegen’s theorem and Maxwell’s minimum heat theorem Objectives of Sect. 3.2:  Be able to combine sources and resistances in series and parallel  Practice in the reduction of resistive networks using series/parallel equivalents  Realize the function and applications of the voltage divider circuit  Realize the function of the current divider circuit  Understand the function and applications of the Wheatstone bridge Objectives of Sect. 3.3:  Understand the role and place of linear circuit analysis  Learn the superposition theorem  Understand the decisive value of superposition theorem for linear circuit analysis  Learn about immediate applications of the superposition theorem  Obtain the initial exposure to Y and networks and to T and networks Application examples:  Voltage divider as a sensor circuit  Voltage divider as an actuator circuit  Superposition theorem for a cellphone
© Springer Nature Switzerland AG 2019 S. N. Makarov et al., Practical Electrical Engineering, https://doi.org/10.1007/9783319966922_3
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Keywords: Electric network, Branches of electric network, Nodes of electric network, Loops of electric network, Meshes of electric network, Essential mesh, Branch currents, Branch voltages, Series connection, Parallel connection, Shunt connection, Kirchhoff’s current law, Kirchhoff’s voltage law, Maxwell’s minimum heat theorem, Tellegen’s theorem, Power conservation law for electric networks, Oneport network, Equivalent electric networks, Equivalent electric circuits, Series battery bank, Battery pack, Dualpolarity power supply, Common ground of the dualpolarity power supply, Virtual ground of the dualpolarity power supply, Parallel battery bank, Series and parallel combinations (of resistances, of conductances),, Equivalent resistance, Equivalent circuit element, Reduction of resistive networks, Voltage divider circuit, Voltage division rule, Sensor circuit sensitivity, Maximum sensitivity of the voltage divider circuit, Current limiter, Currentlimiting resistor, Current divider circuit, Current division rule, Wheatstone bridge (deﬁnition of, difference signal, difference voltage, balanced), Linear circuit (deﬁnition of, homogeneity, additivity, superposition), Nonlinear circuit (deﬁnition of, linearization, dynamic or smallsignal resistance), Superposition theorem, Superposition principle, Y network, Δ network, Twoterminal networks, Threeterminal networks, Conversion between Y and Δ networks, Replacing a node by a loop, Δ to Y transformation, Y to Δ transformation, Balanced Y network, Balanced Δ network, Star to delta transformation, T network, T pad, Π network, Π pad, Twoterminal network (deﬁnition of, input port, output port)
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Section 3.1 Circuit Laws: Networking Theorems Electric components have to be interconnected to obtain functional circuits which perform speciﬁc tasks like driving a motor or monitoring a power plant. Interconnected circuit components form an electric network. In turn, any electric network can be evaluated using two simple yet very general laws: Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL). Series, parallel, and other combinations of circuit elements, whether linear or not, can be explained and solved using these laws.1 They were established by Gustav Kirchhoff (1824–1887), a German physicist and mathematician, in 1845, while Kirchhoff was a 21yearold student at the University of Königsberg in East Prussia. The analyses of all electric circuits are based on KCL and KVL.
3.1.1 Electric Network and Its Topology An electric network can be studied from a general mathematical point of view. If the speciﬁc electrical properties are abstracted, there remains a geometrical circuit, characterized by sets of nodes, branches, and loops. These three items form the topology of an electric network, and the interconnection of its elements can be represented as a graph. The study of electric network topology makes it possible to: A. Identify identical circuit blocks in electric circuits which can be drawn in a variety of ways. Examples include series/parallel connections, as well as wye (Y or T) and delta (Δ or Π) blocks as considered in this chapter. B. Analyze general properties of very large electric circuits such as electric power grids. For example, there are important relationships between the power grid topology and risk identiﬁcation and mitigation. C. Relate electric circuits to other disciplines. For example, electric networks have a remarkable ability to model the dynamical behavior of complicated biological systems.
Nodes, Branches, Loops, and Meshes We consider an electric network with four arbitrary circuit elements A to D as shown in Fig. 3.1a. A branch is a twoterminal circuit element. All four twoterminal elements in Fig. 3.1a are therefore branches.
The KCL and KVL concepts are so powerful that they even ﬁnd applications in equivalent form in magnet systems such as transformers and motors. For example, the magnetic ﬂux in a yoke with air gaps can be model according to KCL and KVL.
1
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Fig. 3.1. An electric network and assigning branch voltages/currents.
A node is a point of interconnection of two or more branches. All small six circles in Fig. 3.1a are formally identiﬁed as nodes. Every node i can be assigned a certain voltage Vi with respect to circuit ground. If a short circuit (a connecting wire) connects two or more nodes, these nodes constitute a single node since they have the same voltage. The circuit may be redrawn to reduce the number of nodes and keep only the meaningful nodes (single nodes) with the distinct voltages as shown in Fig. 3.1b. The circuits in Fig. 3.1a and in Fig. 3.1b are identical. A loop is any closed
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path between two or more branches. There are three loops in Fig. 3.1b. A mesh (or essential mesh) is a loop that does not contain any other loops within it. There are two meshes in Fig. 3.1b. A planar (or twodimensional) electric network with b branches, n nodes, and m meshes satisﬁes, after keeping only nodes with distinct voltages (single nodes), the equality b¼nþm1
ð3:1Þ
which is sometimes called the fundamental theorem of network topology. It is proved by considering the electric network as a polygonal graph in two dimensions, where each edge is a branch and each single node is a vertex.
Branch Currents and Voltages The branch currents and their directions may be assigned arbitrarily, see Fig. 3.1c. The physical currents either coincide with them or are directed in opposite directions. This can easily be found by checking the sign of the current value once the analysis is complete. If the branch voltage polarities have to be assigned afterward, they should satisfy the same reference conﬁguration for all branches. Let us say the passive reference conﬁguration with regard to the previously assigned current directions is seen in Fig. 3.1d. The branch voltages (voltage drops) in Fig 3.1d are expressed through the node voltages according to: V A ¼ V 1 V 3,
V B ¼ V 1 V 2,
VC ¼ VD ¼ V2 V3
ð3:2Þ
Conversely, the branch voltages may be assigned arbitrarily at ﬁrst. If the directions of the branch currents have to be assigned afterward, they should again satisfy the same reference conﬁguration. Exercise 3.1: Establish whether or not the networks in Fig. 3.1a and in Fig. 3.1b satisfy Eq. (3.1). Answer: The answer is yes for both ﬁgures if we consider single nodes. However, if we consider every small circle in Fig. 3.1a as a node, Eq. (3.1) will not be satisﬁed.
Series and Parallel Connections Two or more branches (circuit element) are in series if they exclusively share a common node. Elements A and B in Fig. 3.1d are in series. Elements B and C are not since node 2 is also shared by element D. Two or more branches (circuit elements) are in parallel if they are connected to the same two nodes. Elements C and D in Fig. 3.1d are in parallel. The parallel connection is also called the shunt connection; the parallel element may be called shunt (or shunting) elements.
3.1.2 Kirchhoff’s Current Law (KCL) Let us begin our investigation with KCL. It speciﬁcally applies to the nodes in an electric network. Kirchhoff’s current law simply states that the net current entering the node is zero. In other words, the sum of inﬂowing currents is equal to the sum of outﬂowing currents. This statement is also
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known as the current conservation law, which is the electrical counterpart of the mass conservation law in ﬂuid mechanics. For N currents entering a node, KCL can be cast in the form N X
Ii ¼ 0
ð3:3Þ
i¼1
where N denotes the total number of nodal currents. The current directions may be assigned arbitrarily; the same results will eventually be obtained. The nodal current is taken with a plus sign if it is entering the node, i.e., the current arrow is directed toward the node. It carries a minus sign if it is leaving the node, i.e., the current arrow points in the opposite direction. The current value itself (positive or negative) is substituted afterward. Interestingly, if this law did not hold, an uncompensated charge could accumulate in a node over time. This uncompensated charge and its associated Coulomb force would eventually destroy the operation of the underlying electric circuit. To illustrate the use of KCL, we consider Fig. 3.2 with four different node types for a collection of branches A, B, C, D which could represent arbitrary circuit elements.
Fig. 3.2. Different types of the nodes in an electric network subject to KCL.
While the node in Fig. 3.2a is simple, more complicated node conﬁgurations may be observed in a circuit (see the following ﬁgures). You should note that in a node we may move individual joints to one common joint without affecting the circuit’s operation. A node transformation to a single joint is a convenient tool used when working with more complicated nodes.
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Exercise 3.2: Write KLC for the nodes shown in Fig. 3.2. Answer: In the case of Fig. 3.2a, IA + IC ¼ IB for any values of IA, IB, IC. In the case of Fig. 3.2b, IA + IC ¼ IB + ID for any values of IA, IB, IC, ID. In the case of Fig. 3.2c, IA + IB ¼ IC + ID for any values of IA, IB, IC, ID. In the case of Fig. 3.2d, 2A + IC ¼ 5A + IE for any values of IC, IE.
Example 3.1: In some cases, the use of KCL may be sufﬁcient to determine all currents in an electric network. Solve for the unknown currents IA, IC, ID in a network shown in Fig. 3.3. Solution: First, we note that the wire connection on the left states that the current along the wire is preserved. This implies IA ¼ 1 A. KCL for node 1 gives
1A þ I C ¼ 5 A ) I C ¼ 4 A
ð3:4aÞ
Next, KCL applied to node 2 yields
5A ¼ I D þ 2A ) I D ¼ 3 A
ð3:4bÞ
Thus, the circuit is solved. Node 3 has not been used; it can be employed to check the correctness of the solution: IA + ID + 2A ¼ IC or 4 A ¼ 4 A.
Fig. 3.3. Solving in a network using KCL.
3.1.3 Kirchhoff’s Voltage Law (KVL) This law speciﬁcally applies to the loops in an electric network. The Kirchhoff’s voltage law states that the sum of voltages for any closed loop is zero. In other words, the total amount of work needed to move a unit electric charge one loop turn is zero.2 If this law were not applicable, a
The physical counterpart of KVL is Faraday’s law of induction in the static case. When there is a variable magnetic ﬁeld penetrating a wire loop, KVL is no longer valid.
2
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charge moving in a closed loop would constantly accelerate and eventually escape the circuit or constantly decelerate and eventually stop moving. In order to formulate KVL for a closed loop, we need to identify the loop direction. It is usually chosen to be clockwise (see the dotted arrows in Figs. 3.4 and 3.5). KVL in its general form states N X
Vi ¼ 0
ð3:5Þ
i¼1
where N is the total number of circuit elements in a loop. The voltage Vi is taken with a plus sign if the loop arrow is entering the positive voltage polarity and with a minus sign otherwise. The polarities of voltages Vi can be assigned arbitrarily; the same result is obtained. To demonstrate this fact, in Fig. 3.4 we ﬁrst consider a simpler network with different voltage polarities. One network branch is purposely designated as a voltage source, while the rest are arbitrary circuit elements.
Fig. 3.4. KVL applied to a closed loop with one voltage source and three passive circuit elements. The dotted arrow indicates current ﬂow in clockwise direction.
Example 3.2: Write KVL for the circuit shown in Fig. 3.4 that includes an ideal voltage source and three other circuit elements. Solution: In the case of Fig. 3.4a, we start with the source and strictly apply our convention of positive and negative polarity based on the prescribed loop arrow direction:
5V þ V A þ V B V C ¼ 0
ð3:6aÞ
In the case of Fig. 3.4b, we have to change signs of VA, VB, VC and thus obtain
5V V A V B þ V C ¼ 0
ð3:6bÞ
Note that these two cases only differ by voltage polarities.
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Example 3.2 (cont.): Since reversing voltage polarities in Fig. 3.4b was taken into account by changing the sign of the voltage, both ﬁgures yield the identical result after substituting values:
5 V þ 3 V þ 3 V 1 V ¼ 0
ð3:6cÞ
This observation highlights the fact that the voltage polarities for circuit elements may initially be assigned arbitrarily: applying KVL will ultimately lead to the correct signs of the voltage values.
Figure 3.5 shows another electric network with all branches now consistently labeled; the voltages in red denote the actual values with respect to the initially assigned directions.
Fig. 3.5. KVL applied to a network with three loops, two meshes, and ﬁve circuit elements.
Exercise 3.3: Determine the number of branches, nodes, loops, and meshes for the network shown in Fig. 3.5. Answer: There are ﬁve branches, four single nodes, three loops, and two meshes.
Example 3.3: In some cases, the use of KVL may be sufﬁcient to determine all voltages in an electric network. Solve for the unknown voltages in the network shown in Fig. 3.5. Solution: The application of KVL for loop 1 (mesh 1) has the form
5V 7V þ V C ¼ 0 ) V C ¼ 12V
ð3:6dÞ
Moreover, KVL for loop 2 (mesh 2) results in (VC is already known)
12V þ 6V þ V E ¼ 0 ) V E ¼ 6V
ð3:6eÞ
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Example 3.3 (cont.): Finally, applying KVL for loop 3 is the solution check:
5 V 7 V þ 6 V þ 6 V ¼ 0
ð3:6f Þ
If at least one of the voltages VA, VB, VD in Fig. 3.5 is unknown, then a unique solution does not exist. Through the use of KCL more information must be acquired.
3.1.4 PowerRelated Networking Theorems This section would be incomplete without an introductory discussion of two powerrelated networking theorems. The ﬁrst one is Maxwell’s minimum heat theorem formulated by James Clerk Maxwell in 1891. It states that, for a linear electric network of resistive circuit elements and voltage/current sources, the currents distribute themselves in such a way that the total dissipated power (generated heat) in the resistances is a minimum. If there are NR resistive circuit elements, we can state NR X i¼1
V iI i ¼
NR X i¼1
Ri I 2i ¼
NR X V2 i
i¼1
Ri
¼ min
ð3:7aÞ
where voltages Vi and currents Ii must all satisfy the passive reference conﬁguration. Indeed, Eq. (3.7a) is equivalent to the condition where the net power generated by all sources is also minimized. Maxwell’s minimum heat theorem can be proved using KCL, KVL, and Ohm’s law. The second theorem is Tellegen’s theorem formulated by Bernard D.H. Tellegen in 1952. It postulates that, for an arbitrary electric network with a total of N circuit elements of arbitrary (linear or nonlinear, passive or active) nature, the equality N X
V iI i ¼ 0
ð3:7bÞ
i¼1
must hold if all voltages Vi and currents Ii satisfy one (let’s say the passive) reference conﬁguration. Tellegen’s theorem serves as a power (or energy) conservation law for all electric networks. It can be proved based on KCL and KVL only. Tellegen’s theorem has other important implications and generalizations. To illustrate both theorems, we consider two simple examples. Example 3.4: Use Maxwell’s minimum heat theorem and determine the unknown current x through resistance R1 in Fig. 3.6a. Solution: Equation (3.7a) yields
R1 x2 þ R2 ðx I S Þ2 ¼ min
ð3:8aÞ
This function is minimized when its derivative with respect to x is zero. Therefore,
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Example 3.4 (cont.):
x¼
R2 IS R1 þ R2
ð3:8bÞ
which is the current division principle studied in the next section. The voltage division principle may be established similarly.
Fig. 3.6. Examples for Maxwell’s minimum heat theorem and Tellegen’s theorem.
Example 3.5: Prove Tellegen’s theorem for the network shown in Fig. 3.6b. Solution: Both elements follow the passive reference conﬁguration. By KCL we conclude IB ¼ IA. By KVL, VB ¼ VA. Therefore, VAIA + VBIB ¼ 0, which is the simple poof. In practice, element A may be a voltage source, and element B may be a resistance. We can also use the active reference conﬁguration for the source but need to deﬁne its power to be negative in such a case.
3.1.5 Port of a Network: Network Equivalence So far we have studied only closed electric networks. A network can have a port, through which it is interconnected to another network as shown in Fig. 3.7. The network may be active or passive. This speciﬁc network has two (input or output) terminals a and b, which form a single port. Thus, the network in Fig. 3.7 is known as a oneport network. All series/parallel combinations of sources and resistances are typically oneport networks.
Fig. 3.7. Generic system for establishing a network equivalence.
Two arbitrary oneport networks in the form of Fig. 3.7 are said to be equivalent networks (or equivalent electric circuits) when their υi characteristics at terminals a and b coincide. This means that for any given voltage, there is the same current entering the network or leaving it and vice versa. Therefore, these networks are indistinguishable.
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Section 3.2 Series and Parallel Network/Circuit Blocks 3.2.1 Sources in Series and in Parallel SeriesConnected Battery Bank The simultaneous use of KCL and KVL allows us to analyze the behavior of combinations of active circuit elements and establish their equivalence. The physical counterparts are various battery banks, which are interconnections of the identical batteries. Figure 3.8 shows a series battery bank, also called a battery pack, with two or more batteries connected in series. The battery symbol implies an ideal voltage source.
Fig. 3.8. Series combinations of battery cells and their equivalent representations.
We intend to ﬁnd the resulting voltage and current of this combination. To determine the equivalent voltage, we close the circuit loop shown in Fig. 3.8a by introducing a virtual
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circuit element with terminals a and b, with an unknown voltage V between these terminals. This element simulates the rest of the circuit, which closes the current path. Applying KVL for the loop shown in Fig. 3.8a results in V þ 9 V þ 9 V ¼ 0 ) V ¼ 18 V
ð3:9Þ
Moreover, using KCL, we obtain the same current ﬂows throughout the lefthanded circuit of Fig. 3.8a; this current is exactly equal to IB.3 Therefore, the series combination of two batteries in Fig. 3.8a is equivalent to one battery bank that provides double the voltage, or 18 V, compared to the unit cell. However, it delivers a current of a single cell. The same method can now be applied to multiple battery cells connected in series; one such battery bank is shown in Fig. 3.8b.
DualPolarity Voltage Power Supply Two batteries, or other voltage sources, connected in series can be used as a dualpolarity power supply as shown in Fig. 3.9. The middle terminal provides a virtual ground for the circuit or the common ground. Both negative and positive voltages with respect to the common port can now be created in the circuit. Such a source is of particular importance for operational ampliﬁer circuits and for transistor circuits. Every multichannel laboratory power supply may operate as a dual power supply. The common terminal may (but does not have to) additionally be connected to the earth ground.
Fig. 3.9. A dualpolarity power supply constructed with two battery cells.
ParallelConnected Battery Bank As an alternative to the series connected battery bank, we can investigate the parallel battery bank shown in Fig. 3.10. To determine the equivalent voltage of the combination, we again close the circuit loop by introducing a virtual circuit element with terminals a and b and with an unknown voltage V between these terminals. The use of KVL gives V ¼9V
ð3:10Þ
Thus, the voltage of the battery bank in Fig. 3.10 is still equal to the unit cell voltage. However, applying KCL to both rightmost nodes in Fig. 3.10 indicates that the current doubles. Therefore, the
3 If realistic battery cells are capable of delivering different currents when short circuited, then the lowest cell current will ﬂow under short circuit condition.
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parallel combination of two batteries is equivalent to one battery bank that provides the same voltage of 9 V as one unit cell but at twice the current strength.
Fig. 3.10. A parallel combination of battery cells and its equivalent single battery representation.
Series Versus Parallel Connection What is the difference between series and parallel combinations of two 9 V batteries? First, let us ﬁnd the power delivered to the circuit. We assume IB ¼ 1 A in both cases, even though the speciﬁc value of the current is not important. For the series combination in Fig. 3.8a, the delivered power is 18 V 1 A ¼ 18 W. For the parallel combination in Fig. 3.10, the delivered power is again 9 V 2 A ¼ 18 W. Thus, as far as the power rating is concerned, there is no difference. You should, however, remember that we always deliver power to a load. For the series combination, the implied load resistance becomes 18 V/1 A ¼ 18 Ω. For the parallel combination, the anticipated load resistance will be 9 V/2 A ¼ 4.5 Ω. Thus, it is the load resistance that determines which combination should be used. This question is of great practical importance.
Combinations of Current Sources Combinations of current sources are studied similarly. They are important for photovoltaic and thermoelectric semiconductor devices.
3.2.2 Resistances in Series and in Parallel Series Connection After the sources have been analyzed, we turn our attention to series and parallel combinations of resistances (or conductances). Their physical counterparts are various circuit loads, for example, individual households connected to the same power grid or individual motors driven by the same source. Figure 3.11 depicts the series combination of two resistances R1 and R2.
Fig. 3.11. Two resistances in series and their equivalent single resistance representation.
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Note that the current direction for both resistances corresponds to a passive reference conﬁguration: current ﬂows “down the voltage hill.” Again, we close the circuit loop by introducing a virtual circuit element with terminals a and b and with an unknown voltage V between those terminals. This virtual circuit element, which simulates the rest of the circuit, allows us to close the current path. KVL for the loop shown in Fig. 3.11 results in V ¼ V 1 þ V 2 ¼ IR1 þ IR2 ¼ I ðR1 þ R2 Þ ¼ IReq ) Req ¼ R1 þ R2
ð3:11aÞ
Thus, two resistances can be replaced by one equivalent resistance, which is the sum of the individual resistances. It follows from Eq. (3.11a) that the equivalent resistance gives us the same circuit current (and the same power into the load) as the original resistance combination does, for any applied voltage. This is the formal description of the equivalent resistance to be generalized later. Equation (3.11a) can easily be extended to any arbitrary number of resistances connected in series. Equation (3.11a) may be also formulated in terms of conductances, the reciprocals of resistances, 1 1 1 ¼ þ Geq G1 G2
ð3:11bÞ
Parallel Connection Next, we consider the parallel combination of resistances shown in Fig. 3.12. KVL applied to the loop shown in the ﬁgure and for another loop between two resistances indicates that the voltages across every resistance are equal to V. KCL applied to either rightmost node results in
Fig. 3.12. Two resistances connected in parallel and the equivalent resistance.
I ¼ I1 þ I2 ¼
V V V 1 1 1 þ ¼ ) ¼ þ ) Geq ¼ G1 þ G2 R1 R2 Req Req R1 R2
ð3:12Þ
Therefore, the parallel combination of two resistances is equivalent to one resistance, which has a value equal to the reciprocal of the sum of the reciprocal values of both. The equivalent resistance again gives us the same circuit current as the original resistance combination does, for any applied voltage. We emphasize that the equivalent resistance is always smaller in value than each of the resistances to be combined in parallel. Note that the conductances simply add up for the parallel combinations. Equation (3.12) can again easily be extended to any arbitrary number of resistances connected in parallel.
Meaning of Equivalent Circuit Element In summary, the study of series/parallel active and passive circuit elements leads us to the following simple deﬁnition of an equivalent circuit element, either passive or active. The
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equivalent circuit element possesses the same υi characteristic as the υi characteristic of the original circuit, for which voltage and current are acquired at its terminals a and b.
3.2.3 Reduction of Resistive Networks The reduction of a network of many resistances to a single equivalent resistance is a topic of practical importance. There are many examples of distributed resistive networks. A contemporary example is a rear window defroster, which is a distributed resistive heater. The corresponding solution is usually based on: (i) Stepbystep use of series/parallel equivalents (ii) Moving, splitting, or reducing modes (iii) Reliance on ﬂuid mechanics analogies, which may be helpful for resistive networks
Although a unique solution always exists, its practical realization may be quite difﬁcult. The following examples outline the procedure for the reduction of resistive networks. Example 3.6: Find the equivalent resistance between terminals a and b for the resistive network shown in Fig. 3.13a. Solution: We should not start with terminals a and b but with the opposite side of the circuit. First, the three resistances furthest to the right are combined in series in Fig. 3.13a. The next step is the parallel combination of the resulting resistance and the 1.5 kΩ resistance in Fig. 3.13b. The ﬁnal step is another series combination; this results in the ﬁnal equivalent resistance value of 1875 Ω. We need to point out again that it is impossible to: (i) Combine in series two resistances separated by a node. (ii) Move the resistance through a node or move the node through a resistance. These restrictions hold for other circuit elements as well, beside resistors.
Fig. 3.13. Stepbystep circuit reduction via series and parallel resistance combinations.
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Example 3.7: Find the equivalent resistance between terminals a and b for the resistive network shown in Fig. 3.14a. Solution: First, we can split and move a node along the wire, leading to a circuit shown in Fig. 3.14b. Next, we combine two pairs of resistances in parallel. The next step, in Fig. 3.14c, is the series combination of three resistances. The last step is the solution of a parallel circuit, leading to the equivalent resistance of 66.67 Ω.
Fig. 3.14. Stepbystep circuit reduction to a single equivalent resistance.
Exercise 3.4: Using a ﬂuid mechanics analogy of identical water ﬂow in two symmetric channels, ﬁnd the equivalent resistance of the network shown in Fig. 3.15. Answer: 2.5 kΩ
Fig. 3.15. The resistive network discussed in Exercise 3.4.
3.2.4 Voltage Divider Circuit The purpose of the voltage divider circuit is to provide a voltage different from the supply voltage. The voltage divider circuit is associated with resistances in series. It is perhaps the most important
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basic electric circuit or block within another circuit. The voltage division principle is used in sensor circuits, actuator circuits, and bias circuits. Furthermore, any input and output port of a (transistor) ampliﬁer is essentially a voltage divider. We consider a particular voltage divider form, shown in Fig. 3.16, connected to a DC voltage supply. This circuit, as with any other electric circuit, can be analyzed by using KCL and KVL simultaneously. We prefer using this method although the combination of two resistances in series will provide an equivalent solution.
Fig. 3.16. A conventional voltage divider consists of two resistances and an ideal voltage source.
KCL states that the current I through both resistance and the voltage supply is the same. Applying KVL to the circuit loop allows us to ﬁnd the circuit current: V S þ V 1 þ V 2 ¼ 0 ) V S ¼ V 1 þ V 2 ¼ I ðR1 þ R2 Þ ) I ¼
VS R1 þ R2
ð3:13Þ
Once the circuit current I is known, then Ohm’s law can be used. This yields the voltage divider rule V1 ¼
R1 V S, R1 þ R2
V2 ¼
R2 VS R1 þ R2
ð3:14Þ
Equation (3.14) states that the major function of the voltage divider is to divide the voltage of the power source between two resistances in direct proportion so that: (i) The larger resistance always acquires a higher voltage, and the smaller resistance acquires a smaller voltage. (ii) The individual voltages always add up to the supply voltage, i.e., V1 + V2 ¼ VS. Exercise 3.5: A voltage divider circuit uses a 10 V DC source and two resistances: R1 ¼ 5 Ω and R2 ¼ 100 Ω. What are the voltages V1, V2 across the resistances? Answer: V 1 ¼ 0:48 V,
V2 ¼ 9:52 V,
V 1 þ V 2 ¼ 10 V:
The voltage divider with multiple resistances R1, R2, . . ., RN is solved in the form V S ¼ V 1 þ V 2 þ .. . þ V N ) I ¼
VS Ri and V i ¼ VS R1 þ R2 þ . .. þ RN R1 þ R2 þ .. . þ RN
ð3:15Þ
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3.2.5 Application Example: Voltage Divider as a Sensor Circuit Consider a resistive sensing element (thermistor, strain gauge, photoresistor, etc.) as denoted by R2(x) in Fig. 3.17. The element changes its resistance R2(x) when an external parameter x changes. Parameter x could be temperature, pressure, humidity, solar radiation, or any other physical parameter that undergoes an environmental change. A simple sensor conﬁguration is a direct connection to a voltage source and to the DMM for voltage measurements (see Fig. 3.17a). However, no matter how the sensor resistance changes, the sensor will always output the source voltage. A solution to the this problem is a voltage divider circuit shown in Fig. 3.17b. The extra resistance R1 is ﬁxed. According to Eq. (3.14), voltage V2 in Fig. 3.17b varies depending on the inﬂuence of R2(x): V 2 ¼ V 2 ðxÞ ¼
R2 ðxÞ VS R1 þ R2 ðxÞ
ð3:16Þ
Fig. 3.17. (a) Incorrect sensor circuit; (b) a sensor circuit on the basis of a resistive voltage divider where R2(x) changes its resistance depending on the process parameter x.
The variable voltage V2(x) is measured by the voltmeter. The dependence of V2 on R2 is clearly nonlinear in Eq. (3.16). Although Eq. (3.16) can be linearized by choosing a sufﬁciently large R1 to make the denominator nearly constant, we will show later that such an operation greatly decreases device sensitivity. Let us assume that the external parameter x in Eq. (3.16) changes from a lower limit x1 to an upper limit x2, i.e., x1 x x2. As a result, the sensing resistance changes monotonically, but not necessarily linearly, from R0 ¼ R2(x1) to R00 ¼ R2(x2). We also assume that if x1 x x2, then R0 > R00 . The sensor circuit’s sensitivity, S, is given by S¼
V 2 ðx1 Þ V 2 ðx2 Þ x2 x1
V units of x
ð3:17Þ
The sensitivity is expressed in terms of voltage variation per one unit of x. A higher sensitivity implies a larger voltage variation and thus provides a better sensor resolution and improved robustness against noise.
Design of the Sensor Circuit for Maximum Sensitivity Let us pose the following question: what value should the ﬁxed resistance R1 assume in order to achieve the highest sensitivity of the voltage divider sensor? It is clear that R1 cannot be very small
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(otherwise the voltage reading will always be VS and the sensitivity will be zero) and that R1 cannot be very large (otherwise the voltage reading will be always 0 V and the sensitivity will be zero). The sensitivity is thus a positive function that is zero at R1 ¼ 0 and at R1 ¼ 1. According to the extreme value theorem, a global maximum should exist between these two values. We denote the unknown resistance R1 with variable t, substitute V2 from Eq. (3.14), and rewrite Eq. (3.17) in the form 0 VS R R00 ð3:18aÞ S¼ x2 x1 t þ R0 t þ R00 It is convenient to transform this result into a simpler expression S ¼ VSS0f(t), where a constant S0 is called the intrinsic sensitivity of the resistive sensing element and f(t) is the sole function of the ﬁrst resistance, i.e., 0 R R00 t ð3:18bÞ , f ðt Þ ¼ 0 S0 ¼ x2 x1 ðR þ tÞðR00 þ tÞ This function f(t) is to be maximized. At the function’s maximum, the derivative of f(t) versus t should be zero. Using the quotient rule for the differentiation of a fraction, it follows from Eq. (3.18b) that f 0 ðt Þ ¼
R0 R00 t 2 2 0 2 R þ t ðR00 þ tÞ
ð3:18cÞ
The ﬁnal result following from the condition f 0 (t) ¼ 0 is surprisingly simple pﬃﬃﬃﬃﬃﬃﬃﬃﬃ t ¼ R1 ¼ R0 R00
ð3:18dÞ
In other words, the ﬁxed resistance of the voltage divider circuit should be equal to the geometric mean of two extreme resistances of the sensing element itself.
Example 3.8: For the NTC503 thermistor sensing element, x1 ¼ 25 C (room temperature), x2 ¼ 37 C C (approximate temperature of a human body), R0 ¼ 50 kΩ, and R00 ¼ 30 kΩ. What is the sensitivity of the voltage divider sensor if VS ¼ 9 V and (A) R1 ¼ 5 kΩ, pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ (B) R1 ¼ 500 kΩ, and (C) R1 ¼ 30 50 39 kΩ? Solution: We substitute the numbers in Eq. (3.18a) and ﬁnd the sensitivity. The corresponding sensitivity plot as a function of R1 is given in Fig. 3.18a. The particular sensitivity values are (A) S ¼ 39 mV/ C; (B) S ¼ 26 mV/ C; and (C) S ¼ 95 mV/ C.
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Example 3.9: For the SGT1/350TY11 strain gauge, the nominal resistance is 350 Ω. The resistance variation of 0.1% for the tensile strain ε is observed, and the intrinsic device sensitivity is S0 ¼ 700 Ω/ε. What is the sensitivity of the voltage divider sensor if VS ¼ 4.5 V and (A) R1 ¼ 50 Ω, (B) R1 ¼ 5 kΩ, and (C) R1 ¼ 350 Ω? For positive sensitivity numbers, interchange x1,2 in Eq. (3.18a). Solution: In this example, R0 ¼ 350.35 Ω and R00 ¼ 349.65 Ω. We use Eq. (3.18b) and plot the sensitivity as a function of R1; this is seen in Fig. 3.18b. The particular values are (με are microstrain units) (A) S ¼ 0.24 mV/1000με; (B) S ¼ 0.55 mV/1000με; and (C) S ¼ 2.25 mV/1000με.
Fig. 3.18. Sensitivity curves for the divider sensor circuits with a thermistor and a strain gage.
3.2.6 Application Example: Voltage Divider as an Actuator Circuit The circuit shown in Fig. 3.19 contains a voltage divider circuit block with one resistive sensing element, i.e., resistance R1. The block is connected to a threeterminal electronic switch (a transistor). A variable voltage V controlling the switch operation is created by the voltage divider. When the control voltage V reaches a certain threshold voltage VTh or exceeds it, the switch closes. A DC motor is now connected to the source, and the rotor starts to spin. Switches of this type involve ﬁeldeffect transistors. We emphasize that there is no current into the control switch terminal, only the control voltage counts. If R1 is a thermistor and the DC motor is a fan motor, the entire circuit may operate as a basic temperature controller in an enclosure or in a room. The control voltage versus circuit ground, the reference point, is given by (cf. Eq. (3.16)) V ðxÞ V 2 ðxÞ ¼
R2 VS R1 ðxÞ þ R2
ð3:19Þ
where variable x corresponds to the ambient temperature.
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Fig. 3.19. An actuator circuit based on the voltage divider principle.
Example 3.10: The circuit in Fig. 3.19 uses an NTC503 thermistor sensing element with R1 ¼ 50 kΩ at 25 C (room temperature) and R1 ¼ 30 kΩ at 37 C. The ﬁxed resistance (assumed temperature independent) is R2 ¼ 12 kΩ. The threshold voltage of the switch is VTh ¼ 2.3 V. The supply voltage is 9 V. Determine the circuit behavior at 25 C and at 37 C, respectively. Solution: According to Eq. (3.19), the control voltage at 25 C is equal to 1.74 V. This value is below the threshold voltage. Consequently, the switch is open and the motor is not connected to the source. However, the control voltage at 37 C is 2.57 V. This value is above the threshold voltage and the switch is now closed: the motor is connected to the source and is spinning. Figure 3.20 shows the corresponding laboratory setup. Note that in reality the threshold voltage of the transistor switch is not exactly a constant; it depends on the temperature.
Fig. 3.20. Laboratory realization of the circuit shown in Fig. 3.19.
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3.2.7 Current Limiter The current limiter (or simply the currentlimiting resistor) is a particular case of the voltage divider circuit and is of a signiﬁcant practical importance. Its concept is shown in Fig. 3.21. A load resistance RL is connected to the ideal voltage source in series with another (smaller) resistance R, which physically acts as a currentlimiting resistor.
Fig. 3.21. Voltage divider circuit in the currentlimiting conﬁguration. The 10 Ω resistance is used to limit the circuit current.
If the load resistance is ﬁxed at a rather high value, the circuit in Fig. 3.21 does not pose any problem, and the currentlimiting resistance of 10 Ω becomes insigniﬁcant. For example, in Fig. 3.21, the circuit current is 100 mA in the absence of the ﬁrst resistance R. The power delivered to the load resistance is P ¼ RLI2 ¼ 1 W. However, the load resistance may be variable, and it may attain really small values. When this happens, the circuit current increases. In Fig. 3.21, it becomes equal to 10 A when the load resistance decreases to 1 Ω and the currentlimiting resistor is missing. The power delivered to the load resistance also increases to P ¼ 100 W. This large power may overheat and eventually destroy the smallscale load (a thermistor is one example). The role of resistance R is to limit the total current when the load resistance is either variable or constant but small. For example, in the circuit of Fig. 3.21, the maximum possible circuit current is I¼
VS VS ¼ 1A < R þ RL R
ð3:20Þ
irrespective of the value of the load resistance. Therefore, the power delivered to a 1 Ωload becomes always less than 1 W instead of the initial value of 100 W.
3.2.8 Current Divider Circuit The current divider circuit shown in Fig. 3.22 is associated with resistances connected in parallel. It is dual to the voltage divider circuit in the sense that the roles of voltage and current are interchanged. Figure 3.22 shows the concept.
Fig. 3.22. A current divider circuit with an ideal current source.
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To solve the circuit, we again prefer to use KCL and KVL simultaneously. KCL gives IS ¼ I1 þ I2
ð3:21Þ
Based on KVL for both loop 1 and loop 2, the voltage across the current source (voltage between its terminals a and b) is equal to the voltage across either resistor and is equal to V. Application of Ohm’s law gives the expression for this voltage, IS ¼
V V IS R1 R2 þ )V ¼ ¼ IS 1 1 R1 R2 R1 þ R2 þ R1 R2
ð3:22Þ
Therefore, we obtain the current division rule in the form I1 ¼
V R2 ¼ IS, R1 R1 þ R2
I2 ¼
V R1 ¼ IS R2 R1 þ R2
ð3:23Þ
Equation (3.23) teaches us that the major function of the current divider is to divide the current of the power source between two resistances in an inverse proportion so that: (i) The larger resistance always acquires the smaller current, and the smaller resistance acquires the larger current. (ii) The individual currents add up to the source current, i.e., I1 + I2 ¼ IS. In other words, the electric current always chooses a path of least resistance. If one resistance is replaced by a wire the entire source current will ﬂow through the wire; the second resistance will be shorted out by the wire. Exercise 3.6: A current divider uses a 3 mA current source and two resistances: R1 ¼ 200 Ω and R2 ¼ 600 Ω. What are the currents I1, I2 through the resistances? Answer: I1 ¼ 2.25 mA, I2 ¼ 0.75 mA. The smaller resistance acquires the larger current.
Example 3.11: The current divider circuit can be assembled with a voltage source as shown in Fig. 3.23. Find currents I1, I2 and the total circuit current I. Solution: The circuit in Fig. 3.23 can be analyzed in a number of ways. Perhaps the simplest way is to recognize that, according to KVL, the voltages across all three elements in Fig. 3.23 are equal to each other and equal to 10 V. Therefore,
I1 ¼
10V , R1
I2 ¼
10V , R2
I ¼ I1 þ I2 ¼
10V , Req
Req ¼
R1 R2 R1 þ R2
ð3:24Þ
Another way to solve the same circuit is to combine resistances in parallel, ﬁnd the circuit current I, and then apply the current division principle.
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Fig. 3.23. A current divider circuit with a voltage source.
3.2.9 Wheatstone Bridge The Wheatstone bridge was invented by the British scientist and mathematician, Samuel Christie (1784–1865), and ﬁrst used for resistance measurements by Sir Charles Wheatstone in 1843. It is shown in Fig. 3.24. From Fig. 3.24, we can recognize that the Wheatstone bridge is, in fact, a combination of two independent voltage divider blocks: one with two ﬁxed resistances R1, R2 and another with one ﬁxed resistance R3 and some other resistance R4 denoted here by R4 ¼ R(x). When the resistance measurements are implied, R4 is the ﬁxed unknown resistance. However, common modern applications use the Wheatstone bridge as a part of the sensor circuit. In this case, R(x) is a variable resistance (a sensing element), where x is a physical quantity to be measured. The second voltage divider is the voltage divider sensor circuit; the ﬁrst voltage divider is ﬁxed. Circuit ground (absolute voltage reference) may be introduced as shown in Fig. 3.24.
Fig. 3.24. The Wheatstone bridge is a combination of two independent voltage dividers connected to the same voltage source. The second voltage divider is a sensor circuit.
It has been shown that the voltage divider circuit is a basic sensor circuit. Now, why do we need two voltage dividers? The answer to this question will be based on the fact that, with the help of the ﬁxed divider, we can eliminate a DC voltage offset in the sensor voltage reading of the “master” voltage divider and thus enable the use of a difference signal (and a difference ampliﬁer)
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to amplify the likely very weak sensor voltages. The key point is that the output voltage of the Wheatstone bridge is not Va or Vb but the differential voltage between terminals a and b in Fig. 3.24, Vab ¼ Va Vb. Example 3.12: A. A simple voltage divider circuit with R3 ¼ 350 Ω and R(x) ¼ 350 Ω 0.1% (the strain gauge) in Fig. 3.24b is used for mechanical strain measurements with a 4.5 V voltage power supply. What is the output voltage Vb of the sensor circuit? B. The same voltage divider circuit is augmented with another ﬁxed voltage divider having R1 ¼ R2 ¼ 350 Ω to form the Wheatstone bridge shown in Fig. 3.24c. What is the output voltage Vab of the sensor circuit now? Solution: In case A, we use the voltage division rule and obtain
V b ¼ 2:25 V 1:125 mV
ð3:25aÞ
It is difﬁcult to process such voltages since we cannot really amplify them. Ampliﬁcation of 2.251125 V by a factor of 100 gives 225.1125 V; such a large voltage simply cannot be obtained with the common ampliﬁer circuits. In case B, however, the value of Vb should be subtracted from Va ¼ 2.25 V, which yields the sensor voltage in the form
Vab ¼ 1:125 mV
ð3:25bÞ
If we now amplify 1.125 mV by a factor of 100, a conventional value of 0.1125 V would be obtained. Along with this fact, the differential sensor voltage has another signiﬁcant advantage, which is its immunity against circuit noise.
General Model of Wheatstone Bridge Using the voltage division rule twice, the differential voltage Vab of the Wheatstone bridge in Fig. 3.24c becomes (R4 ¼ R(x)) R2 R4 ð3:26Þ VS Vab ¼ R1 þ R2 R3 þ R4 The Wheatstone bridge is balanced when Vab ¼ 0. From Eq. (3.26) one obtains the necessary and sufﬁcient condition for the balanced Wheatstone bridge, R1 R3 ¼ R2 R4
ð3:27Þ
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Section 3.3 Superposition Theorem and Its Use 3.3.1 Linear and Nonlinear Circuits Linear Circuit The superposition theorem studied in this section is only valid for linear circuits. A linear circuit is a circuit that includes only the linear circuit elements (elements with a linear or straight υi characteristic): 1. 2. 3. 4. 5.
Resistance Capacitance (υi relationship is timedependent but still linear) Inductance (υi relationship is timedependent but still linear) Voltage source (independent and linearly dependent) Current source (independent and linearly dependent)
Every linear circuit satisﬁes both the homogeneity and additivity properties. To explain those properties, we consider a linear circuit with an input parameter x (input voltage or current) and an output parameter f(x) (output voltage or current). A function f is the characteristic of the circuit itself; it must be a linear function. Namely, when an input parameter is a linear superposition ax1 + bx2 of two individual stimuli, x1, x2, the output is also a linear superposition of two individual responses, i.e., f ðax1 þ bx2 Þ ¼ af ðx1 Þ þ bf ðx2 Þ
ð3:28Þ
For example, if we double all source strengths in a linear circuit, voltages across every passive circuit element and currents through every circuit element will also double.
Nonlinear Circuit and Circuit Linearization A nonlinear circuit will include nonlinear circuit elements, elements with a nonlinear υi characteristic. Any circuit with semiconductor components (such as diodes, transistors, solar cells) is a nonlinear circuit. Since the vast majority of electronic circuits include semiconductor components, a legitimate question to ask is what value do the linear circuits have in this case? One answer is given by a linearization procedure, which makes it possible to reduce the nonlinear circuit to a linear one, in a certain domain of operating parameters. Mathematically, circuit linearization means that a nonlinear relationship V(I ) is expanded into a Taylor series, dV ðI I 0 Þ þ . . . ð3:29Þ V ðI Þ ¼ V 0 þ dI V ¼V 0 , I¼I 0
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about a certain operating point V0, I0, where only the constant and the linear terms are retained. The derivative in Eq. (3.29), the socalled dynamic or smallsignal resistance r can now be used in place of the familiar resistance R for the linear ohmic circuit elements.
3.3.2 Superposition Theorem or Superposition Principle The superposition theorem, often called the superposition principle, applies to circuits with more than one voltage and/or source. It states that the complete circuit solution is obtained as a linear superposition of particular solutions, for every power source separately. In other words, we are zeroing (or turning off) all the power sources except for one, ﬁnd the solution, and then add up all such solutions. The following rules apply: 1. To turn off a voltage source, we replace it by a short circuit or an ideal wire (see Fig. 3.25a). The voltage across the ideal wire is exactly 0 V. 2. To turn off a current source, we replace it by an open circuit or an air gap (see Fig. 3.25b). The current through the gap is exactly 0 A. 3. The dependent sources do not need to be zeroed. They remain the same for every particular solution and may affect every particular solution.
Fig. 3.25. Superposition theorem for two independent sources (either voltage or current).
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Fig. 3.26. Application of the superposition principle to a circuit with voltage sources.
Example 3.13: Find current I1 in the circuit shown in Fig. 3.26. Solution: The ﬁrst step is shown in the ﬁgure; we apply the superposition theorem and obtain two simpler circuits. Each of those circuits is solved using series equivalents. For the circuit with the 12 V power source,
i1 ¼ 12 V=5 kΩ ¼ 2:4 mA
ð3:30Þ
The minus sign is due to the fact that the actual circuit current ﬂows in the opposite direction. For the circuit with the 18 V source,
i2 ¼ 18 V=5 kΩ ¼ 3:6 mA
ð3:31Þ
The total circuit current I1 is therefore given by I1 ¼ i1 + i2 ¼ 1.2 mA. Circuit voltage across each resistance can be found as well.
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Fig. 3.27. Application of the superposition principle to voltage sources.
Example 3.14: Find current I1 in the circuit shown in Fig. 3.27. Solution: The ﬁrst step is again shown in the ﬁgure; we apply the superposition theorem and obtain two simpler circuits. Each of those circuits is solved using series/parallel equivalents. For the circuit with the 15 V power source,
Req ¼ 1 kΩ þ 3 kΩ þ 0:75 kΩ ¼ 4:75 kΩ
ð3:32Þ
The circuit current is 3.1579 mA; i1 is 75% of this value (from current division). For the circuit with the 10 V source, Req ¼ 3.8 kΩ and i2 is 2.1053 mA. Thus, current I1 is given by
I 1 ¼ i1 þ i2 ¼ 4:4737 mA
ð3:33Þ
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Fig. 3.28. Application of the superposition principle to current sources.
Example 3.15: Find current I1 in the circuit shown in Fig. 3.28. Solution: The ﬁrst step is shown in the ﬁgure; we apply the superposition theorem and obtain two simpler circuits by disconnecting current sources. The ﬁrst circuit predicts i1 ¼ 15 mA, while the second circuit predicts i2 ¼ 10 mA. Therefore,
I 1 ¼ i1 þ i2 ¼ 25 mA
ð3:34Þ
In other words, each of the currents i1, i2 ﬂows in its own loop. The solution does not depend on any particular resistor value.
The superposition theorem is the direct consequence of circuit linearity. Interestingly, the superposition theorem is applicable not only to DC circuits but also to AC and transient circuits. The superposition theorem does not hold for electric power though, since power is the product of voltage and current. The circuit power and the power delivered to individual elements may be correctly obtained only from the ﬁnal solution.
Application Example: Superposition Theorem for a Cellphone Why do we ultimately need to solve circuits with multiple sources? Look at your cellphone. There is a circuit inside, which receives a very weak radiofrequency input voltage signal from the antenna. This is the ﬁrst source. The signal is processed and ampliﬁed by transistors powered by the cellphone battery. This is the second source (or sources). The signal is then demodulated through interaction with an internal highfrequency signal generator. This is the third source.
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Moreover, every component in that circuit is in fact “noisy”; it creates a small thermal noise voltage, which is modeled by its own tiny voltage source. A careful solution of the entire circuit with multiple voltage and current sources including noise sources is possible by superposition. This solution allows electrical engineers to extract the weak input signal from otherwise overwhelming noise and properly design the cellphone circuitry.
3.3.3 Y (Wye) and Δ (Delta) Networks: Use of Superposition The series and parallel resistance conﬁgurations are not the only meaningful network blocks. Situations often arise when the resistances are neither in parallel nor in series. In Fig. 3.29a, b, we introduce two new networking blocks, which are in importance second to the series/parallel equivalents.
Fig. 3.29. (a) Y (wye) and (b) Δ (delta) networks; (c, d) applying the superposition theorem to establish network equivalence.
The ﬁrst block is known as the Y (wye) network. It represents a nodal connection of three arbitrary resistances. The second block is known as the Δ (delta) network. It represents a loop connection of three arbitrary resistances. Both Y and Δ networks have three terminals 1, 2, and 3; they are therefore known as threeterminal networks. This is in contrast to series/parallel resistance circuits, which are usually twoterminal networks. The Y and Δ networks occur either independently or as part of a larger network. Important applications include threephase power electronics circuits, ﬁlter circuits, and impedancematching networks in highfrequency circuits. The theory that follows holds for AC circuits too, when the resistances become general impedances.
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Conversion Between Y and Δ Networks
A problem of signiﬁcant practical importance is the conversion between Y and Δ networks in Fig. 3.29a, b. This conversion is equivalent to replacing a node by a loop (strictly speaking, by a mesh) and vice versa in a more complicated circuit or network. Such a replacement may signiﬁcantly simplify the overall circuit analysis. The conversion is established based on the superposition theorem. Two arbitrary networks are equivalent if their υi characteristics are the same. In other words, by connecting three arbitrary sources to terminals 1, 2, and 3 of the Y network, we must obtain terminal voltages and currents identical to those of the Δ network with the same sources. We select three current sources I1, I2, I3 in Fig. 3.29c, d. The solution with three sources is obtained as a superposition of three partial solutions, with two sources opencircuited at a time. Let us keep the source I1 and replace I2, I3 by open circuits ﬁrst. Voltages V12 for both networks will be the same when the equivalent resistances R12 between terminals 1 and 2 are the same. A similar treatment holds for terminals 1 and 3 (source I3 is kept), and terminals 2 and 3 (source I2 is kept), respectively. Therefore, with reference to Fig. 3.29c, d, we have Rb ðRa þ Rc Þ Ra þ Rb þ Rc Rc ðRa þ Rb Þ ¼ R1 þ R2 ¼ Rc kðRa þ Rb Þ ¼ Ra þ Rb þ Rc Ra ðRb þ Rc Þ ¼ R2 þ R3 ¼ Ra kðRb þ Rc Þ ¼ Ra þ Rb þ Rc
R12 ¼ R1 þ R3 ¼ Rb kðRa þ Rc Þ ¼
ð3:35aÞ
R13
ð3:35bÞ
R23
ð3:35cÞ
Next, we add Eq. (3.35a) and Eq. (3.35b) and subtract from this result Eq. (3.35c). This gives us an expression for R1. To obtain R2, we add Eq. (3.35b) and Eq. (3.35c) and subtract Eq. (3.35a). R3 is obtained by adding Eq. (3.35a) and Eq. (3.35c) and subtracting Eq. (3.35b). The result has the form of a Δ to Y transformation: R1 ¼
Rb Rc , Ra þ Rb þ Rc
R2 ¼
Ra Rc , Ra þ Rb þ Rc
R3 ¼
Ra Rb Ra þ Rb þ Rc
ð3:36Þ
The inverse transformation, a Y to Δ transformation, follows Ra ¼
R1 R2 þ R1 R3 þ R2 R3 R1 R2 þ R1 R3 þ R2 R3 R1 R2 þ R1 R3 þ R2 R3 , Rb ¼ , Rc ¼ R1 R2 R3
ð3:37Þ
Balanced Y and Δ Networks When all resistances of the Y network are equal to RY , the Y network is said to be balanced. When all resistances of the Δ network are equal to RΔ, the Δ network is also balanced. For balanced networks, two previous equations reduce to
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ð3:38Þ
Example 3.16: Find equivalent resistance of a network between terminals a and b in Fig. 3.30a. Solution: To simplify the network, we use the Δ to Y transformation and obtain the network shown in Fig. 3.30b. Since all resistances of the bottom Δ network are equal, we can use the simpliﬁed Eq. (3.38) to ﬁnd the new resistance values. The remaining circuit is solved using series/parallel combinations, which gives us Req ¼ 2.2 kΩ.
Fig. 3.30. Network simpliﬁcation using Δ to Y transformations.
The conversions between Y and Δ networks date back to Arthur E. Kennelly (1861–1939), an Indian American engineer who established them in 1899. Note that the transformation between Y and Δ networks is also called the star to delta transformation.
3.3.4 T and Π Networks: TwoPort Networks
The Y and Δ networks are equivalent to T and Π networks, respectively, which are shown in Fig. 3.31. The T and Π networks are predominantly used as twoport networks or twoport networking blocks. Every port should have two terminals. To create the two ports with two terminals each, we simply split terminal 2 in Fig. 3.31a, c into two terminals: 2 and 4 in Fig. 3.31b, d, respectively. Port #1 is the input port of the network. Port #2 is the output port of the network. Multiple twoport networks may be connected in chains. Equations (3.36), (3.37), and (3.38) allow us to establish the connections between the T and Π networks, which are the same as the connections between the Y and Δ networks. Hence, any twoport T network may be replaced by a twoport Π network and vice versa. The T network is sometimes called the T pad and the Π network the Π pad. Both networks are used as attenuators, ﬁlters, and antenna tuners. In the last two cases, capacitances and inductances occur in place of resistances.
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Section 3.3: Superposition Theorem and Its Use
Exercise 3.7: A twoport T network in Fig. 3.31b is characterized by R1 ¼ 3 kΩ and R2 ¼ R3 ¼ 5 kΩ. Establish resistance values for the equivalent Π network in Fig. 3.31d. Answer: Ra ¼ 18.33 kΩ, and Rb ¼ Rc ¼ 11 kΩ.
Fig. 3.31. Conversion of Y and Δ threeterminal networks to equivalent T and Π twoport, fourterminal networks. (a) Y network. (b) Twoport T network. (c) Δ network. (d) Twoport Π network.
3.3.5 General Character of Superposition Theorem The superposition theorem makes it possible to analyze not only the twoport electric networks but also various networks with multiple ports such as sensor arrays, antenna arrays, multipleinput and multipleoutput (MIMO) communication systems, arrays of magnetic resonance imaging (MRI) coils, etc. It is so widely used that, quite often, we do not even mention its name and consider the corresponding result as “obvious.”
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Circuit Laws and Networking Theorems
Summary
(continued)
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Summary
(continued)
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Circuit Laws and Networking Theorems
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Problems
Problems 3.1 Circuit Laws: Networking Theorems 3.1.1 Electric Network and Its Topology Problem 3.1. In the network graph shown in the ﬁgure below: A. Find the number of branches, single nodes (nodes with distinct voltages), and meshes. B. Prove the equality b ¼ n + m 1 for the number of branches b, meshes m, and single nodes n.
Problem 3.2. Repeat problem 3.1 for the circuits shown in the following ﬁgures.
Problem 3.3. Repeat problem 3.1 for the circuit shown in the ﬁgure that follows. Each straight segment is now a branch.
3.1.2 Kirchhoff’s Current Law (KCL) Problem 3.4. Find current IB for the node shown in the ﬁgure that follows.
Problem 3.5. A. Find current ID for the node shown in the ﬁgure below. B. Redraw this node (and the circuit between terminals a and b) in an equivalent form eliminating the horizontal wire.
Problem 3.6. Determine current iC for the circuit shown in the ﬁgure that follows.
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Problem 3.10. Determine currents iC, iF, iH in the ﬁgure that follows. Problem 3.7. Find a relation between currents iC, iE for the circuit shown in the ﬁgure that follows. Does the problem have a unique solution?
Problem 3.11. Find all unknown currents for the circuit shown in the ﬁgure below.
Problem 3.8. Find currents iA, iB, iD, iH for the circuit shown below.
Problem 3.9. Determine currents iC, iF, iH for the circuit shown in the ﬁgure that follows.
3.1.3 Kirchhoff’s Voltage Law (KVL) Problem 3.12. A. Find voltage VC for the circuit shown in the ﬁgure that follows; B. How would the solution change if VD were equal to 0 V?
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Chapter 3 C. Could the value VE ¼ 0 V be used in this problem?
Problems Problem 3.15. Determine voltage VE for the circuit shown below.
Problem 3.13. Determine voltage VF in the circuit shown in the ﬁgure below.
Problem 3.16. Equipotential lines for a human body subject to a vertical electric ﬁeld with the strength of 1 V/m are shown in the ﬁgure that follows. A. Determine voltages VAB, VBC, VCA. B. Establish the KVL loop and formulate KVL for these three voltages.
Problem 3.14. Find the unknown voltages VA, VB, VG, VH for the circuit shown in the ﬁgure below.
3.1.4 PowerRelated Theorems Problem 3.17. For the circuit shown in the ﬁgure that follows:
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Chapter 3 (i) Determine which circuit elements are resistances and which are sources. (ii) Find the power (delivered to the circuit or taken from the circuit) for every circuit element. (iii) Assuming that the powers of the sources are negative, ﬁnd the sum of all powers in the circuit.
Circuit Laws and Networking Theorems Problem 3.19. For the circuit shown in the ﬁgure: 1. Use KVL to solve for the unknown voltages. 2. Use KCL to solve for the unknown currents. 3. For each of six circuit elements, determine if the element is a resistance or a source. 4. Assuming that the source powers are negative, ﬁnd the algebraic sum of all powers in the circuit.
Problem 3.18. For the circuit shown in the ﬁgure: 1. Use KVL and KCL to solve for unknown currents and voltages. 2. Find the power (delivered to the circuit or taken from the circuit) for every circuit element. 3. Assuming that the powers of the sources are negative, ﬁnd the algebraic sum of all powers in the circuit.
3.2 Series and Parallel Network/Circuit Blocks 3.2.1 Sources in Series and in Parallel Problem 3.20. The electronic circuits onboard an 18foot long Parker motor boat consume 96 W when operated from a 24V source. The source is a combination of two fully charged deepcycle batteries, each of which is rated for 12 V and 100 ampere hours. 1. Should the batteries be connected in series or in parallel? 2. For how many hours can the electronics be operated from the battery bank without recharging? 3. How much energy in kilowatt hours is initially stored in each battery?
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Chapter 3 Problem 3.21. A certain sensing device consumes 0.375 W of power over a 20 h period and operates from a 6 V source. The source is a combination of four fully charged AAA batteries, 1.5 V each. The batteries discharge by the end of the 20 h period. 1. Should the batteries be connected in series or in parallel? 2. What is a typical capacity of the AAA battery? 3. How much energy in watt hours is initially stored in each battery?
Problems below.All resistancesare equal: R1 ¼ . . .¼ R15 ¼10 Ω. Determine the heat power (power delivered to the defroster).
Problem 3.22. A load has the resistance of 1.5 Ω and requires the applied voltage of 3 V. A number of battery cells are given, each of which is rated for 1.5 V. Each cell may deliver no more than 1 A of current. 1. Construct and draw a battery bank that could be used to drive the load. 2. Is the solution to the problem unique? Problem 3.23. Repeat the previous problem when the required load voltage is changed to 6 V.
Problem 3.26. Find the equivalent resistance between terminals a and b.
3.2.2 Resistances in Series and in Parallel 3.2.3 Reduction of Resistive Networks Problem 3.24. Determine the equivalent resistance between terminals a and b.
Problem 3.27. Find the equivalent resistance between terminals a and b.
Problem 3.25. The equivalent electric circuit for a car rear window defroster is shown in the ﬁgure
Problem 3.28. Find the equivalent resistance between terminals a and b.
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Circuit Laws and Networking Theorems Problem 3.33. Find the equivalent resistance between terminals a and b.
Problem 3.29. Determine the equivalent resistance between terminals a and b.
Problem 3.30. Find the equivalent resistance between terminals a and b.
Problem 3.31. Determine the equivalent resistance (resistance between ports a and b) of the network shown in the ﬁgure that follows.
Problem 3.34. Determine the equivalent resistance of the network (resistance between terminals a and b) shown in the ﬁgure below.
Problem 3.35. Determine the equivalent resistance between terminals a and b (show units).
Problem 3.32. Determine the equivalent resistance between terminals a and b. Problem 3.36. A network shown in the ﬁgure below is known as a ladder. The ladder network includes one particular section with a series (R1 ¼ 1 Ω) and a shunt (R2 ¼ 1 Ω) resistance, known as the Lsection. This section is then repeated an inﬁnite number of times to the right. Determine the equivalent resistance between terminals a and b of the ladder network.
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Problems A. Determine the equivalent resistance between terminals a and b of the ladder network when R1 ¼ 2R3 ¼ 10 Ω and R2 ¼ 10 Ω. B. How different is your result from the equivalent resistance of the ﬁnite ladder network with only four sections?
Hint: For the semiinﬁnite ladder network in the previous ﬁgure, the equivalent resistance, Req, will not change after adding a new section up front as shown in the ﬁgure that follows.
3.2.4 Voltage Divider Circuit Problem 3.39. Redraw the circuit shown in the ﬁgure that follows and plot the distribution of the circuit voltage versus ground reference point to scale between two circuit points a and b as a function of distance x from point a.
Problem 3.37. A. Repeat the previous problem for the semiinﬁnite ladder network circuit shown in the same ﬁgure when R1 ¼ R3 ¼ 10 Ω and R2 ¼ 25 Ω. B. How different is your result from the equivalent resistance of the ﬁnite ladder network with only four sections? Problem 3.38. Another important ladder network type (with the Tsection) is shown in the ﬁgure.
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Problem 3.40. Repeat the task of the previous problem for the disconnected circuit shown in the ﬁgure below.
Problem 3.41. For the circuit shown in the ﬁgure that follows, use the voltage division principle to calculate V1, V2, V3.
Problem 3.42. For the circuit shown in the following ﬁgure: A. Calculate V1, V2, V3. B. Find the voltage across the current source.
Problem 3.43. For the circuits shown in the ﬁgure that follows, determine voltages V1, V2, absolute voltage Va at node a versus ground (show units), and circuit current I.
3.2.5 Application Example: Voltage Divider as a Sensor Circuit 3.2.6 Application Example: Voltage Divider as an Actuator Circuit Problem 3.44. An NTC thermistorbased temperature sensor should operate between 25 C and 65 C from a 6 V DC power supply. The thermistor’s resistance changes from R0 ¼ 50 kΩ to R00 ¼ 20 kΩ in this temperature range. A. Present a circuit diagram for the simple temperature sensor. B. Determine the value of the unknown resistance for the maximum circuit sensitivity.
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Chapter 3 C. Determine the maximum circuit sensitivity. D. What is the circuit sensitivity when the unknown resistance is set to 1 kΩ?
Problem 3.45. In the previous problem, using software of your choice (MATLAB is recommended), plot the circuit sensitivity to scale as a function of the value of the unknown resistance in the range from 1 kΩ to 100 kΩ. Problem 3.46. A strain gauge with nominal resistance of 120 Ω is used in conjunction with a 2.5 V DC voltage source. Its nominal resistance changes by 0.2% when the gauge operates in the permissible strain range, which is 1000με. A. Present a circuit diagram for the simple strain gauge sensor. B. Determine the value of the unknown resistance for the maximum circuit sensitivity. C. Determine the maximum circuit sensitivity. D. What is the circuit sensitivity when the unknown resistance is set to1 kΩ? Problem 3.47. In the previous problem, using software of your choice (MATLAB is recommended), plot the circuit sensitivity to scale as a function of the value of the unknown resistance in the range from 10 Ω to 1000 Ω.
Problems B. Can you derive an analytical formula that gives the voltage variation across R2 ¼ R1 Δ as a linear function of an arbitrary (but very small) resistance variation Δ? [Hint: use your calculus background—the Maclaurin series versus a small parameter.]
3.2.7 Current Limiter Problem 3.49. A thermistor is connected to an ideal voltage power source of 9 V. Determine the value of the currentlimiting resistor R based on the requirement that the power delivered to the thermistor should be always less than 0.1 W. The lowest possible value of R should be chosen. Consider two cases: 1. Thermistor resistance is exactly 100 Ω. 2. Thermistor resistance changes from 200 Ω to 100 Ω.
3.2.8 Current Divider Circuit Problem 3.50. For the circuit shown in the ﬁgure that follows: A. Calculate voltage between terminals a and b. Show its polarity on the ﬁgure. B. Use the current division principle to calculate branch currents i1, i2.
Problem 3.48. A voltage divider circuit with R1 ¼ 700 Ω (the ﬁxed resistance) and R2 ¼ 700 Ω 0.1% (the strain gauge) is used. The voltage power supply is rated at 4.5 V. A. Show that voltage across the strain gauge varies in the range 2.25 V 1.125 mV.
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Chapter 3 Problem 3.51. Find branch currents i1, i2 for the circuit depicted in the ﬁgure below.
Circuit Laws and Networking Theorems Problem 3.56. Find the voltage between terminals a and b for the circuit shown in the ﬁgure that follows.
Problem 3.52. Find branch currents i1, i2 for the circuit shown in the ﬁgure that follows. Problem 3.57. Find the voltage between terminals a and b for the circuits shown in the ﬁgure that follows.
Problem 3.53. Find the voltage between terminals a and b (voltage across the current power source) for the circuit shown in the ﬁgure that follows.
Problem 3.54. The voltage source in the circuit is delivering 0.2 W of electric power. Find R. Problem 3.58. You are given: (i) A photoresistor that changes its resistance from 20 kΩ for brightness to 500 kΩ for darkness (ii) A 9 V battery (iii) A voltmeter (DMM) (iv) Any other necessary precise resistors
3.2.9 Wheatstone Bridge Problem 3.55. Describe in your own words the function and the scope of the Wheatstone bridge.
Construct (present the circuit diagram) a Wheatstone bridge sensor circuit that: (i) Has zero voltage reading for brightness
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Chapter 3 (ii) Has maximum possible voltage reading for darkness (has maximum sensitivity)
Problem 3.59. You are given: 1. A SGT1/350TY11 uniaxial strain gauge with the nominal resistance of 350 Ω (no strain) 2. A 4.5 V voltage source 3. Any number of precise resistors, of any value When tensile strain is applied, a resistance variation up to +0.1% is observed. Present the circuit diagram for the Wheatstone bridge sensor circuit that: (i) Has zero voltage reading at no strain (ii) Has maximum possible voltage response when strain is present (has maximum sensitivity) (iii) Outputs positive voltages when the resistance of the strain gauge increases
Problem 3.60. Resistance of a strain gauge increases when its length increases (a bended surface under test becomes convex) and decreases when its length decreases (a bended surface under test becomes concave). The corresponding strains are the tensile strain and the compressive strain. You are given two strain gauges (#1 and #2), which are attached to opposite sides of a thin bended surface under test, at the same position. The gauge resistance at normal conditions (no bending) is R ¼ 100 Ω. You are also given any number of ﬁxed resistors, of any value. 1. Suggest and sketch a sensor circuit which will convert changes in the resistance into measurable voltage changes. This circuit should possibly have: (a) Zero sensor output voltage at normal sensor conditions (no bending) (b) Maximum voltage sensitivity to changes in resistance (sensitivity to the strain)
Problems 2. Label one (or two) strain gauges used, and specify the values of all deployed resistances. 3. Show power supply and DMM connections.
Combined Voltage and Current Dividers Problem 3.61. For the circuit shown in the ﬁgure: A. Find currents i, i1, i2 (show units). B. Find power P delivered by the voltage source to the circuit. C. Find voltages V1, V2.
Problem 3.62. Find voltage V across the 20 kΩ resistance and current i for the circuit shown in the ﬁgure that follows.
Problem 3.63. For the circuit that follows determine: A. Current i2 through the 0.25 kΩ resistance B. Power P absorbed by the 600 Ω resistance
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3.3 Superposition Theorem and Its Use 3.3.2 Superposition Theorem or Superposition Principle Problem 3.67. Solve the circuits shown in the ﬁgure that follows (ﬁnd current I and show units) using superposition theorem. Problem 3.64. For the circuit shown in the ﬁgure determine: A. Current i2 through the 0.25 kΩ resistance, B. Power P absorbed by the 0.3 kΩ resistance.
Problem 3.68. Find current I (show units) using the superposition theorem. Problem 3.65. Determine current i (show units) in the circuit that follows.
Problem 3.69. Find current I (show units) using the superposition theorem. Problem 3.66. Determine current i and voltage υ for the circuit shown in the ﬁgure (show units).
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Problems Problem 3.71. For the circuit shown in the ﬁgure, determine the absolute voltage (versus chassis ground) and the electric current at the circuit point a.
Problem 3.72. Determine the absolute voltage (versus chassis ground) and the current i at the circuit point a.
Problem 3.70. (a) Find current I (make sure to show units) using the superposition theorem. (b) Find voltage V (make sure to show units) using the superposition theorem.
Problem 3.73. For the circuit shown in the ﬁgure that follows, ﬁnd the current across the 10 Ω resistance. Show its direction in the ﬁgure.
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Problem 3.74. Determine the unknown current I in the circuit (solve the problem by superposition).
Circuit Laws and Networking Theorems
Problem 3.77. Convert the network shown in the ﬁgure that follows from Δ to Y: A. Draw the corresponding Y network. B. Label its terminals. C. Determine and label the corresponding resistance values.
Problem 3.75. Determine the unknown current I in the circuit (solve the problem by superposition). Problem 3.78. Convert the twoport network shown in the ﬁgure that follows from T to Π: 1. Draw the corresponding Π network. 2. Label its terminals and ports. 3. Determine and label the corresponding resistance values.
3.3.3 Y (Wye) and Δ (Delta) Networks: Use of Superposition 3.3.4 T and Π Networks: TwoPort Passive Networks Problem 3.76. Convert the network shown in the ﬁgure that follows from Y to Δ: A. Draw the corresponding Δ network. B. Label its terminals. C. Determine and label the corresponding resistance values in the ﬁgure.
Problem 3.79. Convert the twoport network shown in the ﬁgure that follows from Π to T:
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Problems
A. Draw the corresponding T network. B. Label its terminals and ports. C. Determine and label the corresponding resistance values.
Problem 3.80. For the bridge network, determine the equivalent resistance between terminals a and b.
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Chapter 4
Chapter 4: Circuit Analysis and Power Transfer Overview Prerequisites: • Knowledge of major circuit elements, their i characteristics, and Ohm’s law (Chap. 2) • Knowledge of basic networking theorems (Chap. 3) Objectives of Sect. 4.1: • Become familiar with the nodal analysis and be able to apply it to solve in arbitrary linear circuits • Become familiar with the mesh analysis and be able to apply it to solve in arbitrary linear circuits Objectives of Sect. 4.2: • Become familiar with the method of short/open circuit • Establish and prove the source transformation theorem • Establish and prove Thévenin and Norton’s theorems • Be able to apply Thévenin and Norton’s equivalents for circuit solution Objectives of Sect. 4.3: • Establish the maximum power theorem and become familiar with the power efficiency concept • Be able to apply the concepts of Thévenin and Norton’s equivalents and maximum power theorem in practice Objectives of Sect. 4.4: • Obtain an initial exposure to nonlinear circuit analysis • Be able to solve in a simple nonlinear circuit
© Springer Nature Switzerland AG 2019 S. N. Makarov et al., Practical Electrical Engineering, https://doi.org/10.1007/9783319966922_4
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Application Examples: Reading and using data for solar panels Power radiated by a transmitting antenna Maximum power extraction from solar panel Solving the circuit for a generic solar cell
Keywords: Nodal analysis, Mesh analysis (mesh current analysis), Supernode, Supermesh, Method of short/open circuit (deﬁnition of opencircuit network voltage, shortcircuit network current), Source transformation theorem, Circuit equivalent (see equivalent circuit), Thévenin’s theorem (formulation, proof, special cases), Thévenin equivalent, Norton’s theorem, Norton equivalent, R2R ladder network, negative equivalent (Thévenin) resistance, Maximum power theorem (principle of maximum power transfer), Power efﬁciency, Analysis of nonlinear circuits, Load line (deﬁnition, method of), Iterative method for nonlinear circuits (deﬁnition of explicit iterative scheme, implicit iterative scheme), Solar cell (cSi, opencircuit voltage, shortcircuit photocurrent density, ﬁll factor, characteristic equation of), Solar panel (series cell connection, opencircuit voltage, shortcircuit photocurrent, ﬁll factor, maximum power load voltage, maximum power load current)
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Chapter 4
Section 4.1: Nodal/Mesh Analysis
Section 4.1 Nodal/Mesh Analysis 4.1.1 Importance of Circuit Simulators The series and parallel equivalent circuits along with Y and Δ transformations provide a practical tool for solving simple circuits involving typically only a few elements. However, for more elaborate circuits, circuit simulators such as SPICE (Simulation Program with Integrated Circuit Emphasis) and their various modiﬁcations become indispensable tools for the professional engineer. SPICE was developed by the Electronics Research Laboratory at the University of California, Berkeley, and ﬁrst presented in 1973. These circuit simulators are quite general and allow us to model circuits with passive and active elements including semiconductor devices such as diodes, transistors, and solar cells. Since those elements typically exhibit nonlinear currentvoltage behaviors, elaborate solution strategies are needed. The circuit simulators use quite interesting algorithms: they often operate in the time domain, even for DC circuits. For example, a solution for a DC circuit is obtained as the steadystate limit of a transient solution, for voltage and/or current sources turned on at a certain time instance. The key of the timedomain approach is its inherent ability to solve nonlinear problems, with passive and active circuit elements. In this section, we will not discuss the speciﬁcs of the numerical circuit simulation. Instead, we will provide the foundation of the nodal analysis (or node analysis) and the mesh analysis (or the mesh current analysis), which are two important features of any professional circuit simulator. The nodal and mesh analyses in their pure form do not involve timedomain methods. They are primarily applicable only to linear circuits, also referred to as linear networks.
4.1.2 Nodal Analysis for Linear Circuits The nodal analysis is a general method of solving linear networks of arbitrary complexity, which is based on KCL and Ohm’s law. Let us consider a circuit shown in Fig. 4.1a, which is a resistive bridge circuit with a bridging resistance. This circuit may be solved using Δ to Y conversion; see, for instance, example 3.16 of Chap. 3. Here, we prefer to use the nodal analysis directly. The nodal analysis operates with the absolute values of the node voltages in the circuit with respect to ground reference. It may be divided into a number of distinct steps: 1. A ground reference needs to be assigned ﬁrst; this is a node where the voltage is set to 0 V. To this end, we ground the negative terminal of the voltage power supply. 2. Next, we select nontrivial (also called nonreference) nodes for which we do not know the voltages. These are nodes 1 and 2 in Fig. 4.1b. The two additional nodes are eliminated from the analysis since the voltages there are already known.
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3. We label absolute node voltages versus ground reference as V1, V2—see Fig. 4.1c. 4. Then currents for every nontrivial node are labeled, assuming that all currents are outﬂowing— see Fig. 4.1c. This condition can be replaced by all inﬂowing currents. 5. Next, KCL is applied to every nontrivial node. We express the currents as the difference of two absolute voltages: the voltage at the beginning of the current arrow (voltage at the master node) minus the voltage at the end of the current arrow (voltage at any other node) and then divide this difference by the appropriate resistance. Hence, we arrive at a system of linear equations for the nodal voltages. Currents are no longer involved. 6. After the resulting system of linear equations is solved, all circuit parameters can be determined as necessary. a) 1 kW
1 kW
5V
1 kW
+ 3 kW
1 kW
1 kW
1 kW
5V
1. Select ground reference 2. Select and label nontrivial nodes
5V
b)
1 kW
+ 
1
2
3 kW
1 kW
3. Label unknown node voltages 4. Label outflowing currents for each node
0V 0V 5V
c)
1 kW
1 kW
5V
1 kW
+ 
V2
V1 3 kW
1 kW
0V 0V
5. Write KCL for each node in terms of node voltages 6. Solve the resulting system of equations
Fig. 4.1. Major steps of the nodal analysis applied to a bridge circuit.
Note that the resulting system of linear (or nodal) equations may have one, two, three, or more equations depending on the complexity of the circuit as well as on your own “smart” choice of the nontrivial nodes. The following two examples will apply the nodal analysis to a circuit with a voltage source.
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Section 4.1: Nodal/Mesh Analysis
Example 4.1: Solve the circuit shown in Fig. 4.1a using nodal analysis and ﬁnd the supply current. Solution: Steps 1 to 4 are indicated in Fig. 4.1b, c. Applying KCL to node 1 and then to node 2 (the order is not important), one has.
V1 5 V V1 0 V V1 V2 þ þ ¼0 1 kΩ 3 kΩ 1 kΩ
ð4:1aÞ
V2 5 V V2 0 V V2 V1 þ þ ¼0 1 kΩ 1 kΩ 1 kΩ
ð4:1bÞ
This is a system of linear equations for two unknown voltages. It simpliﬁes to
7=3V 1 V 2 ¼ 5 V
ð4:2aÞ
3V 2 V 1 ¼ 5 V
ð4:2bÞ
Equations (4.2) are solved, for example, via Gaussian elimination. This yields
V 1 ¼ 3:33 V, V 2 ¼ 2:78 V
ð4:3Þ
The circuit current (current of the voltage source) is therefore
ð5 V V 1 Þ=1 kΩ þ ð5 V V 2 Þ=1 kΩ ¼ 3:89 mA:
a)
b)
1 kW 1 kW
1 kW 1 kW
1 kW
V1
10 V
+ 
10 V
1 kW
1 kW
+ 
10 V
0V
1 kW V2
1
2
1 kW
1 kW
0V
0V
Fig. 4.2. Major steps of the nodal analysis applied to a circuit with a voltage source.
Example 4.2: Solve the circuit shown in Fig. 4.2a employing the nodal analysis. Solution: Steps 1 to 4 are indicated in Fig. 4.2b. Applying KCL to node 1 and then to node 2, one obtains a system of equations with two unknown voltages:
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Example 4.2 (cont.):
V 1 10 V V 1 0 V V 1 V 2 þ þ ¼0 1 kΩ 1 kΩ 1 kΩ V 2 10 V V 2 0 V V 2 V 1 þ þ ¼0 1 kΩ 1 kΩ 1 kΩ
ð4:4aÞ ð4:4bÞ
In setting up these equations, it does not matter which sequence of nodes are selected. Simplifying Eqs. (4.4) gives
3V 1 V 2 ¼ 10 V
ð4:5aÞ
3V 2 V 1 ¼ 10 V
ð4:5bÞ
The solution is obtained by symmetry, i.e., V1 ¼ V2 ¼ 5 V. The circuit current provided by the power supply is 10 mA. All other branch currents can then be found using Ohm’s law. An interesting feature of the circuit shown in Fig. 4.2a is that the marked 1 kΩ resistor can be considered as “dead,” since there is no current ﬂowing through it (the voltage difference across this resistor is exactly zero). This resistor can be removed from the circuit without affecting the behavior of the circuit in terms of voltages and currents. It might appear at ﬁrst sight that the circuits shown in Figs. 4.1 and 4.2 have a different network topology. In fact, they do not. To prove this, you should attempt to redraw the circuit in Fig. 4.2a; your redrawn topology should be identical with the circuit in Fig. 4.1a.
Circuits with a Current Source When a current source is present in a circuit, the solution becomes even simpler: one makes use of the existing current and substitutes its value into the KCL equation written for a particular node. For example, KLC applied to node 1 in Fig. 4.3 includes the outﬂowing current of 1 mA; the current sign must be taken into account. The same idea is applied to circuits with multiple current power supplies. When only the current sources are present, the ground may be connected to the incoming terminal of a current source.
1 kW
1
1 mA
1 kW
1 kW
10 V
+ 
2
1 kW
Fig. 4.3. A circuit with a current source solved via nodal analysis.
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Section 4.1: Nodal/Mesh Analysis
Exercise 4.1: Determine the voltage across the current source in Fig. 4.3 using the nodal analysis. Answer: 8.6 V.
4.1.3 Supernode The nodal analysis requires a “good” eye to see possible simpliﬁcations when labeling the nodes. Let us examine a particular case and point out a few useful subtleties.1 Figure 4.4a depicts a network with two voltage sources. The property of the 5 V source is such that it is not ﬁxed to a particular ground connection—we therefore call it a ﬂoating source. Setting up the nodal analysis becomes tricky, since we do not know the current through this source. However, a supernode may be formed as shown in Fig. 4.4b. 1 kW
a)
5V
1 kW
+

10 V
+ 
1 kW
1 kW
supernode 1 kW
b)
5V
1 kW
10 V
+ 
1 kW
0V

V2
+
V1
10 V
1 kW
0V
0V
Fig. 4.4. A network with a ﬂoating voltage source between nodes 1 and 2.
When KCL is applied to any closed contour around the supernode, the net current must add to zero in such a case. With reference to Fig. 4.4b this yields V 1 10 V V 1 0 V V 2 0 V V 2 10 V þ þ þ ¼0 1 kΩ 1 kΩ 1 kΩ 1 kΩ
ð4:6aÞ
Subtleties are often euphemism for “playing” around with the circuit, like redrawing the wire connections and rearranging the circuit elements. This is done to ﬁnd simpler solution approaches.
1
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Chapter 4
Circuit Analysis and Power Transfer
What is the second equation for the two unknowns V1 and V2? It simply turns out to be the relation between the supernode voltages themselves. Since V2 V1 is the voltage of the power source, we obtain V2 ¼ V1 þ 5 V
ð4:6bÞ
Equations (4.6) can now be solved even without the calculator, eliminating one of the unknowns, which yields V 1 ¼ 2:5 V,
V 2 ¼ 7:5 V
ð4:7Þ
The circuit is solved. All currents can be found using the nodal voltages and Ohm’s law. Example 4.3: Solve the circuit shown in Fig. 4.4a by using the standard nodal analysis, without the supernode concept. Solution: We have to specify an unknown current Ix through the 5 V source, which ﬂows, say, from left to right in Fig. 4.4a. This convention results in the following two equations for the two nodes.
V 1 10 V V 1 0 V þ þ Ix ¼ 0 1 kΩ 1 kΩ V 2 0 V V 2 10 V þ Ix ¼ 0 1 kΩ 1 kΩ
ð4:8aÞ ð4:8bÞ
Next, we can add both equations and thereby eliminate Ix. The result is exactly Eq. (4.6a) for the supernode. We must add one more condition to solve this equation. Eq. (4.6b) is the only choice, i.e.,
V2 ¼ V1 þ 5 V
ð4:8cÞ
Interestingly, we arrive at the supernode concept again but in a more complicated way. This is the reason why the supernode approach is such useful tool.
4.1.4 Mesh Analysis for Linear Circuits The mesh analysis (or the mesh current analysis) is using loops instead of nodes. Only loops that do not contain any other loops—the meshes—are employed. The meshes as elements of the networking topology were deﬁned in Sect. 3.1.1 of Chap. 3. Accordingly, instead of KCL the mesh analysis makes use of KVL. Hence, we need to choose mesh currents for every mesh. Figure 4.5 depicts the concept for a circuit with three meshes. Note that this circuit is identical to the circuit from Fig. 4.1. A ground connection does not have to be introduced for the mesh
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Section 4.1: Nodal/Mesh Analysis
method. Let us denote the mesh current for mesh 1 in Fig. 4.5 by I1, the mesh current for mesh 2 by I2, and the mesh current for mesh 3 by I3. mesh #3
mesh #1
1 kW
1 kW
I1 5V
+ 
1 kW
I3 3 kW
I2
1 kW
mesh #2
Fig. 4.5. Circuit analysis based on the mesh analysis; circuits in Fig. 4.5 and in Fig. 4.1 coincide.
The KVL equations for the three meshes are based on Ohm’s law for the proper reference conﬁguration. We do not need the fourth (large) loop encompassing the entire circuit. For resistors that are shared by two adjacent meshes, we combine either the difference or the sum of the two adjacent mesh currents. The mesh equations become Mesh1 :
1 kΩ ðI 1 I 3 Þ þ 1 kΩ I 1 þ 1 kΩ ðI 1 I 2 Þ ¼ 0
ð4:9aÞ
Mesh2 :
3 kΩ ðI 2 I 3 Þ þ 1 kΩ ðI 2 I 1 Þ þ 1 kΩ I 2 ¼ 0
ð4:9bÞ
Mesh3 :
5V þ 1 kΩ ðI 3 I 1 Þ þ 3 kΩ ðI 3 I 2 Þ ¼ 0
ð4:9cÞ
We have arrived at a system of three equations for the three unknown mesh currents I1, I2, and I3. After division by 1 kΩ and combining similar terms it simpliﬁes to Mesh1 :
þ 3I 1 I 2 I 3 ¼ 0
ð4:10aÞ
Mesh2 :
I 1 þ 5I 2 3I 3 ¼ 0
ð4:10bÞ
Mesh3 :
I 1 3I 2 þ 4I 3 ¼ 5 mA
ð4:10cÞ
In contrast, the nodal analysis applied to the same circuit requires only two equations for two unknown node voltages, see Example 4.1. The ﬁnal solution is indeed the same. Thus, the nodal analysis is more beneﬁcial for small networks when a voltage source or sources are present. If, however, instead of a voltage source, a current source were present in Fig. 4.5, the nodal analysis would require three equations. By contrast, the mesh analysis would require only two equations, because I3 is deﬁned by the current source. Reasoning like this gives us clues which method is most suitable. When mixed power supplies like voltage and current sources are involved, there is usually no real difference between the two methods. The choice often becomes a matter of taste.
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Circuit Analysis and Power Transfer
Exercise 4.2: Determine the mesh currents for the circuit in Fig. 4.5. Answer: I1 ¼ +2.22 mA, I2 ¼ +2.78 mA, I3 ¼ +3.89 mA.
4.1.5 Supermesh Consider the circuit shown in Fig. 4.6. The straightforward mesh analysis should use KVL written for meshes 1 and 2. However, KVL cannot be formulated directly since we do not know the voltage across the current source. A solution is to combine meshes 1 and 2 into a supermesh and write KVL around its periphery. The mesh equations become Supermesh : Mesh3 :
1 kΩ ðI 1 I 3 Þ þ 1 kΩ I 1 þ 1 kΩ I 2 þ 3 kΩ ðI 2 I 3 Þ ¼ 0
5 V þ 1 kΩ ðI 3 I 1 Þ þ 3 kΩ ðI 3 I 2 Þ ¼ 0
KCL for the central branch of the source :
I 1 I 2 ¼ 1 mA
ð4:11aÞ ð4:11bÞ ð4:11cÞ
mesh #1
mesh #3
1 kW
1 kW
I1 5V
+ 
1 mA
I3 3 kW
I2
1 kW
mesh #2
Fig. 4.6. Circuit solved with the supermesh method.
After division by 1 kΩ and combining similar terms, the system of Eq. (4.11) is simpliﬁed to þ2I 1 þ 4I 2 4I 3 ¼ 0 I 1 3I 2 þ 4I 3 ¼ 5 mA þI 1 I 2 ¼ 1 mA
ð4:12Þ
Exercise 4.3: Determine the mesh currents for the circuit in Fig. 4.6. Answer: I1 ¼ +3 mA, I2 ¼ +2 mA, I3 ¼ +3.5 mA.
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Section 4.1: Nodal/Mesh Analysis
Example 4.4: Outline the solution approach for the circuit shown in Fig. 4.6 using the standard mesh analysis, without the supermesh concept. Solution: The voltage across the current source is introduced as an extra unknown, Vx. Then, we write three KVL equations for three meshes in Fig. 4.6, which will contain four unknowns: I1, I2, I3, and Vx. An extra equation is needed, which is KCL applied to the central branch: I1 I2 ¼ 1 mA. As a result, we need to solve a system of four simultaneous equations. This is considerably more work than in the previous case. This is why the supermesh approach is such a useful tool for the mesh analysis.
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Section 4.2 Generator Theorems and Their Use 4.2.1 Equivalence of Active OnePort Networks: Method of Short/Open Circuit In Chap. 3, we considered passive linear networks involving only resistors, which we have transformed into equivalent circuits. Active linear networks, which include sources and resistors simultaneously, can be subjected to similar transformations. We know that two electric singleport networks are equivalent when their terminal υi characteristics are identical. For passive resistive networks studied in Chap. 3, we connected arbitrary source(s) across the network terminals and checked the resulting υi characteristics. For active networks with sources and resistances, we can use the same method. Alternatively, we can connect arbitrary resistance(s) across the network terminals and check either the resulting voltage or current. A test resistance to be connected will be denoted here by R. If for two networks the voltages across the resistance R (or currents through it) are identical for all values of R, the networks are equivalents.
Method of Short/Open Circuit However, testing all possible values of resistance R connected to terminals a and b of a network in Fig. 4.7 is not necessary. Note that an active linear network may ultimately have only two elements: a source and a resistance. To uniquely determine the two elements (their values), only two equations are necessary. It is therefore customary to check only two (limiting) values of the test resistance: R ! 1 and R ¼ 0
ð4:13Þ
test of shortcircuit current a
arbitrary electric network
VOC

b
a
+
test of opencircuit voltage
arbitrary electric network
ISC b
Fig. 4.7. Method of short/open circuit for an active, oneport network.
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Conveniently, this corresponds to open and shortcircuit conditions. In the ﬁrst case, the voltage between terminals a and b is the opencircuit network voltage VOC. In the second case, the current ﬂowing from terminal a to terminal b is the shortcircuit network current ISC. The pair VOC, ISC is the key for the method of short/open circuit. This method states that two active linear circuits are equivalent when their VOC and ISC coincide. Network equivalency relates not only to the linear active networks with two components, but, as will be shown soon, it is valid for all active linear networks.
4.2.2 Application Example: Reading and Using Data for Solar Panels The method of short/open circuit is also very useful for active nonlinear networks, including nonlinear sources. An example is a solar cell or a combination thereof, a solar panel. Every solar panel has the measured data for VOC and ISC listed on its backside. The shortcircuit current is simultaneously the photocurrent of the solar cell. Table 4.1 collects this data for common crystalline silicon (or cSi) solar panels. It is organized in such way that VOC is given per one cell in the panel and ISC is given in terms of photocurrent density, JP, per unit cell area. The cells in the panel are connected in series. Table 4.1. Manufacturers’ speciﬁed parameters for different cSi solar panels from ﬁve different manufacturers (1–230 W output power range, circa 2012–2015). The cell area is either measured directly or extracted from the datasheet.
Solar panel 1 W BSPI12 power up cSi panel 10 W BSP1012 power up cSi panel 65 W BSP1012 power up cSi panel 230 W sharp NDU230C1 cSi panel 175 W BP solar SX3175 cSi panel 6 W global solar GSE6 cSi panel 200 W GE Energy GEPVp200 cSi panel Average
Cells, N 36 36 36 60 72 44 54
VOC/N, V 0.59 0.59 0.61 0.62 0.61 0.52 0.61 0.593
Cell area A, cm2 2.36 ~22.0 121.7 241.0 156.25 16.6 249.3
JP ¼ ISC/A A/cm2 0.030 0.030 0.032 0.034 0.033 0.027 0.032 0.0311
Table 4.1 demonstrates that cSi solar cells have approximately the same opencircuit voltage of 0.6 V per cell. The opencircuit voltage does not depend on the area of the cell. The shortcircuit photocurrent density is also approximately the same for cSi solar cells from different manufacturers. On average, it is given by JP ¼ 0.03 A/cm2. These values corresponds to an incident light intensity of 1000 W/m2 at T ¼ 25 C. The photocurrent density does not depend on the area of the cell. Obviously, the total photocurrent does. Exercise 4.4: A cSi solar panel (or solar module) has the opencircuit voltage of 23.4 V? How many individual solar cells does it have? Answer: Approximately 39.
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Exercise 4.5: A cSi solar panel is needed with the opencircuit voltage of 12 V and a shortcircuit current of 3 A. Design the panel: give the number of cells to be connected in series and the required unit cell area. Answer: Twenty cells with the area of 100 cm2 (10 10 cm) each.
4.2.3 Source Transformation Theorem The most fundamental transformation of active linear networks is the source transformation theorem. The source transformation theorem involves the substitution of an independent voltage source VT in series with resistance RT for an independent current source IN with resistance RN and vice versa, as seen in Fig. 4.8a, b. The meaning of indexes N and T will become apparent soon. The identical theorem applies to the dependent sources shown in Fig. 4.8c, d. a)
b) RT a
a
+ 
VT
IN
RN
b c)
b d)
RT a
a
+ 
vT
iN b
RN b
Fig. 4.8. Transformation of dependent and independent sources.
Let us prove this theorem by establishing the circuit equivalence. The pair VOC, ISC is to be found for every network. For the two networks in Fig. 4.8a, b, we have V OC ¼ V T , I SC ¼ V OC ¼ RN I N , I SC
VT RT ¼ IN
ð4:14aÞ ð4:14bÞ
Equations (4.14) have a unique solution in the form of the source transformation theorem RN ¼ RT , I N ¼
VT RT
ð4:15Þ
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Section 4.2: Generator Theorems and Their Use
If Eqs. (4.15) are satisﬁed, both networks in Fig. 4.8a, b have equal VOC and ISC. This ensures that their entire υi characteristics are also the same. To conﬁrm this fact, an arbitrary resistance R could be connected across the port. The resulting voltages is found directly by solving the voltage divider and the current divider circuits, respectively. Both voltages are equal to VTR/(R+RT). Thus, the source transformation theorem is proved. Exercise 4.6: A network has a 10 V voltage source in series with a 20 Ω resistance. It is replaced by a current source IN in shunt with resistance RN. Find IN and RN. Answer: IN ¼ 0.5 A, RN ¼ 20 Ω.
Exercise 4.7: A linear active circuit records an opencircuit voltage of 5 V and a shortcircuit current of 1 mA. Determine its equivalents in the form of a voltage source in series with a resistance and in the form of a current source in shunt with a resistance. Answer: VT ¼ 5 V, RT ¼ 5 kΩ and IN ¼ 1 mA, RN ¼ 5 kΩ.
Often, the source transformation theorem allows us to simplify the circuit analysis through simple network manipulations. Example 4.5: Find current I1 in the circuit shown in Fig. 4.9a. Solution: The circuit can be solved using the superposition theorem. Another way is to employ the source transformation theorem. The corresponding steps are outlined in Fig. 4.9b, c. We use the source transformation three times and end up with the parallel combination of two current sources and three resistances. The three resistances in parallel are equivalent to the 0.75 kΩ resistance; the voltage across every element in parallel is then 0.75 kΩ 2 mA ¼ 1.5 V. Therefore, I1 ¼ 0.75 mA.
Example 4.6: The circuit in Fig. 4.10a includes a currentcontrolled voltage source with the strength of 4000ix [V]. Find current ix by using source transformation. Solution: The corresponding circuit transformation is shown in Fig. 4.10b. The circuit with the currentcontrolled current source in Fig. 4.10b is solved using KCL and KVL. KCL applied to the bottom node states that the current of 3 mA+3ix ﬂows through the rightmost 1 kΩ resistance (directed down). Since, by KVL, the voltages across both resistances must be equal, one has
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Circuit Analysis and Power Transfer
Example 4.6 (cont.):
3 mA þ 3ix ¼ ix ) ix ¼ 1:5 mA
ð4:16Þ
Alternatively, one could convert the independent current source to an independent voltage source. However, this method would hide ix.
a)
2.5 kW
0.5 kW
6 mA
2 kW
I1
+ 
2 kW
2V
b) 2.5 kW
0.5 kW
3V
I1
+ 
2 kW
2 kW
1 mA
2 kW
1 mA
c) I1 3 kW
1 mA
2 kW
Fig. 4.9. Circuit modiﬁcations using the source transformation theorem.
a) ix
1 kW
3 mA
+ 
1 kW
4000ix
b) ix 3 mA
1 kW
1 kW
4ix
Fig. 4.10. Using source transformation for a circuit with dependent sources.
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Chapter 4
Section 4.2: Generator Theorems and Their Use
4.2.4 Thévenin’s and Norton’s Theorems: Proof Without Dependent Sources The origin of Thévenin’s theorem is due to Léon Charles Thévenin, a French engineer (1857–1926). The theorem is illustrated in Fig. 4.11a, b and can be expressed in the following form: 1. Any linear network with independent voltage and current sources, dependent linear sources, and resistances as shown in Fig. 4.11a, can be replaced by a simple equivalent network: a voltage source VT in series with a resistance RT. 2. The equivalent network in Fig. 4.11b is called the Thévenin equivalent. 3. Voltage VT is the opencircuit voltage VOC of the original network (with no current ﬂowing into either terminal). 4. When dependent sources are absent, Thévenin resistance RT is the equivalent resistance Req of the original network with all independent sources turned off (voltage sources are replaced by short circuits, current sources by open circuits). 5. When both dependent and independent sources are present, the independent sources are not turned off. Resistance RT is given by RT ¼ VOC/ISC, where ISC is the shortcircuit current of the original network. 6. When only the dependent sources are present, a current (or voltage) source is connected to the network terminals. Thévenin or equivalent resistance is given by RT ¼ V/I where I is the source current and V is the voltage across the source. Thévenin voltage is strictly speaking not deﬁned in this case.
The Norton’s theorem is dual to the Thévenin’s theorem. It was named in honor of Edward L. Norton (1898–1983), an engineer at Bell Labs in New Jersey.2 Norton’s theorem is illustrated in Fig. 4.11c, d. The equivalent circuit (Norton’s equivalent) is now the current source in shunt with the resistance as shown in Fig. 4.11d. Since the equivalence of both networks in Fig. 4.11b, d has already been established, the Norton’s theorem will follow from the Thévenin’s theorem and vice versa. b)
a)
a
a any linear network of sources (independent and linear dependent) and resistances
RT
Thevenin VT
+ 
b
b
c)
d) any linear network of sources (independent and linear dependent) and resistances
a
a Norton IN b
RN b
Fig. 4.11. Thévenin’s and Norton’s theorems: replacing linear active circuits by its Thévenin and Norton equivalents.
The ﬁrst publication that discusses this equivalent circuit concept is actually due to Hans F. Mayer (1895–1980) who made the discovery in 1926 while a researcher at Siemens Company.
2
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Chapter 4
Circuit Analysis and Power Transfer
Proof of Thévenin’s Theorem for Active Networks Without Dependent Sources The proof is based on the assumption of circuit linearity. The υi characteristics of both networks is established using a current source of strength I connected as seen in Fig. 4.12. a)
b) a
a
I
V
b
VT
+ 
+
+
RT
I
V

any linear network of sources (independent and linear dependent) and resistances
b
Fig. 4.12. Derivation of Thévenin’s theorem by establishing the υi characteristics.
Since the entire circuit is assumed linear, the υi characteristic of the current source in Fig. 4.12a must have the form of linear function, V ¼ AI þ B
ð4:17Þ
where V is the voltage across the current source. A and B are some “constant” coefﬁcients, which do not depend on I, but do depend on the network parameters. Our goal is to ﬁnd A and B, respectively. First, we check the value I ¼ 0 when the external current source is turned off, i.e., replaced by an open circuit. From Eq. (4.17), voltage V equals B. Therefore, it is equals VOC of the original network. Therefore, B ¼ V OC
ð4:18Þ
Next, let us turn off all internal sources. The network becomes an equivalent resistance Req. The constant B (its opencircuit voltage) is zero. Equation (4.17) therefore yields V ¼ AI for any value of I. On the other hand, for the current source I connected to the resistance Req, it must be V ¼ ReqI. Comparing the two expressions we obtain, A ¼ Req
ð4:19Þ
The simpler network in Fig. 4.12b is also described by the υi characteristic in the form of Eq. (4.17). In this case, B ¼ VOC ¼ VT, A ¼ Req ¼ RT. We ﬁnally compare the two υi characteristics, V ¼ Req I þ V OC for linear active network;
V ¼ RT I þ V T for The venin equivalent ð4:20Þ
and establish the Thévenin’s theorem. A test voltage source could be used in place of the current source in Fig. 4.12, with the same result obtained. The physical background of Thévenin’s theorem is thus the fact that the terminal response of any linear network has a linear υi characteristic—a linear function with only two independent coefﬁcients: A and B. A simpler network with exactly two independent parameters, VT and RT, allows us to model this response.
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Chapter 4
Section 4.2: Generator Theorems and Their Use
Equivalence of Arbitrary Linear Networks with Identical VOC, ISC On the one hand, linear active networks with only two elements (a source and a resistance) are equivalent when their VOC, ISC are identical. On the other hand, any active linear network is equivalent to a linear network with only two elements. Therefore, we conclude that two arbitrary linear networks are equivalent when their VOC and ISC coincide.
4.2.5 Finding Thévenin and Norton Equivalents and Using Them for Circuit Solution Along with the superposition and source transformation theorems, Thévenin’s and Norton’s theorems (or their outcome, Thévenin and Norton equivalent circuits or simply equivalents) are powerful tools to solve more complicated electric circuits. The idea is to replace a larger part of the circuit by its equivalent and, subsequently, ﬁnd the required voltage/current/power for a remaining small circuit part. However, a bit of practice is required to implement such a solution. We will ﬁrst establish Thévenin (and implicitly Norton) equivalents for some “large” circuits—oneport networks—shown in Figs. 4.13 and 4.14, and then implement this idea using a few examples. Example 4.7: Establish Thévenin equivalent circuits in the form of Fig. 4.11b for the oneport networks shown in Fig. 4.13. This ﬁrst network is a battery bank or a network consisting of seriesconnected practical voltage sources, with RB1 ¼ RB2 ¼ RB3 ¼ 1 Ω, VB1 ¼ VB2 ¼ VB3 ¼ 9 V. Solution: Case (a): The Thévenin voltage is the opencircuit voltage between terminals a and b. Since no current ﬂows through any of the resistors, KVL gives VT ¼ Vab ¼ 27 V. The Thévenin resistance is found by turning off the voltage sources, i.e., replacing them by short circuits. Therefore, RT ¼ Rab ¼ 3 Ω. Case (b): The Thévenin voltage is the opencircuit voltage between terminals a and b. Current cannot ﬂow through open terminals a and b, but it still can ﬂow within the loop containing the source and the two resistors. Therefore, VT ¼ Vab ¼ V3kΩ ¼ 9 V. The Thévenin resistance is found by turning off the voltage source (replacing it by a wire). Since the two resistors are combined in parallel, RT ¼ Rab ¼ 1.2 kΩ. Case (c): Although the source transformation theorem immediately gives us VT ¼ 3 V, RT ¼ 3 kΩ, we would like to solve the problem directly. The Thévenin voltage is the opencircuit voltage between terminals a and b. Current cannot ﬂow through open terminals a and b, but it still can ﬂow within the loop containing the current source and the resistor. Therefore, VT ¼ Vab ¼ V3kΩ ¼ 3 V. The Thévenin resistance is found by turning off the current source (replacing it by an open circuit). This gives RT ¼ Rab ¼ 3 kΩ.
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Circuit Analysis and Power Transfer
Example 4.7 (cont.): Case (d): The Thévenin voltage is the opencircuit voltage between terminals a and b. Current cannot ﬂow through open terminals a and b and through a 0.2 kΩ resistor, but it still can ﬂow within the loop containing the voltage source and both 2 kΩ and 3 kΩ resistors. Therefore, VT ¼ Vab ¼ V3kΩ ¼ 9 V. The Thévenin resistance is found by turning off the voltage source (replacing it by a short circuit). This gives us RT ¼ Rab ¼ 1.2 kΩ + 0.2 kΩ ¼ 1.4 kΩ. Case (e): First, note that the 1 kΩ resistor placed in shunt with the voltage source has no effect on the solution whatsoever; it is either shorted out by a wire (when ﬁnding Thévenin resistance) or simply possesses the supply voltage (when ﬁnding Thévenin voltage). In practice, such a combination only dissipates power and should be avoided when possible. The rest of the solution follows case (b) with interchanged resistors and yields VT ¼ Vab ¼ V2kΩ ¼ 6 V, RT ¼ Rab ¼ 1.2 kΩ.
Fig. 4.13. Active linear networks to be converted to their Thévenin equivalents as given in the form of Fig. 4.11b.
Exercise 4.8: Along with the combination of a voltage source in parallel with a resistance as in Fig. 4.13e, give another example of a “meaningless” circuit conﬁguration. Answer: A current source in series with a resistance.
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Section 4.2: Generator Theorems and Their Use
Example 4.8: A twoterminal network shown in Fig. 4.14a is a twobit R2R ladder network used for digitaltoanalog conversion. Express (i) Thévenin (or equivalent) voltage VT (ii) Thévenin (or equivalent) resistance RT
in terms of (digital) voltages D0,D1, and resistance R. Solution: One way to solve the problem is to ﬁnd VT and RT directly from the circuit in Fig. 4.14a. While the solution for RT is straightforward, ﬁnding VT requires more work. Yet another method is to apply Thévenin equivalent to the leftmost section of the ladder network ﬁrst. The result is the circuit shown in Fig. 4.14b. The ﬁnal Thévenin equivalent has the form RT ¼ R, V T ¼ D21 þ D40 . This method may be applied to ladder networks with multiple sections.
Exercise 4.9: Repeat the previous example for the ladder shown in Fig. 4.14c. Answer: RT ¼ R, V T ¼ D22 þ D41 þ D80 .
R
a)
a 2R 2R D0
+ 
2R D1
c)
+ 
R
R
a b
2R 2R
2R
2R
R
b)
a R D0/2
D0
+ 
D1
+ 
D2
+ 
2R
+ 
D1
b
+ b
Fig. 4.14. Twobit and threebit R2R ladder networks.
Using Thévenin and Norton Equivalents for Circuit Solution In this section, we will use the method of Thévenin equivalent and conduct a circuit solution. Note that, in contrast to a nodal analysis, this is a semianalytical method which may not always work out for every network. Consider two circuits shown in Fig. 4.15a, b. In either case, the goal is to ﬁnd the current I. We will implement the idea discussed above, i.e., replace a larger part (or parts) of the circuit by its equivalent and then solve for the current in the remaining part.
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Circuit Analysis and Power Transfer
Example 4.9: For the two circuits shown in Fig. 4.15a, b, ﬁnd the current I. Solution: Case (a): One way to solve the problem is to apply the current division principle multiple times. Another method is to replace the circuit to the left of the dashed line (also called “reference plane” in more advanced applications) by its Thévenin equivalent as shown in the ﬁgure. The remaining solution becomes trivial and yields a current value of 0.25 mA. Case (b): One way to solve the problem is to apply the superposition theorem with or without source transformation and then use the current division again. Yet another method is to replace the circuits to the left/right of two dashed lines (two “reference planes”) by its Thévenin equivalents as shown in the same ﬁgure. The remaining solution is straightforward; it yields the current value of 3.214 mA.
Fig. 4.15. Two circuits solved using the method of Thévenin equivalent.
Thévenin and Norton Equivalents with Dependent Sources Thévenin and Norton equivalents with dependent sources (ampliﬁers) are a fascinating subject that can lead to unexpected concepts such as the negative equivalent resistance.
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Section 4.2: Generator Theorems and Their Use
Example 4.10: Find Thévenin and Norton equivalent circuits for the network in Fig. 4.16a. Solution: The network includes a voltagecontrolled voltage source. Therefore, its analysis should be performed in a general form, by ﬁnding the pair VOC, ISC. The shortcircuit current ISC is found straightforwardly. Since the rightmost resistance is shorted out, υx ¼ 0, and ISC ¼ 10 mA. To ﬁnd the opencircuit voltage, which is equal to υx, we use the source transformation theorem and arrive at the equivalent circuit in Fig. 4.16b. Next, we solve this circuit. Employing KVL, the voltage across the leftmost resistance is equal to 10 V 4υx. By KCL, the currents through both resistances must be the same. Since the resistors are equal, we obtain the equality 10 V 4υx ¼ υx so that υx ¼ 2 V. The opencircuit voltage has the same value. The Thévenin and Norton equivalents are
V OC ¼ 200 Ω I SC V OC ¼ 10 mA, RN ¼ ¼ 200 Ω I SC
V T ¼ V OC ¼ 2 V, RT ¼
ð4:21aÞ
I N ¼ I SC
ð4:21bÞ
a) 3vx a
+

+ 1 kW
10 mA
1 kW
vx
b b) 1 kW
3vx a
+

+ 
+ 10 V
1 kW
vx
b
Fig. 4.16. A network with dependent sources to be converted to its equivalent forms.
4.2.6 Application Example: Generating Negative Equivalent Resistance Consider a circuit shown in Fig. 4.17a. It includes only a dependent source: the voltagecontrolled voltage source; it does not have independent sources. Therefore, the circuit analysis has to be done
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Circuit Analysis and Power Transfer
by connecting a current (or voltage) source between terminals a and b as shown in Fig. 4.16b. The quantity of interest is the Thévenin (or equivalent) resistance. The KVL for the circuit in Fig. 4.17b gives Aυx RI+υx ¼ 0. Therefore, by deﬁnition, RT
υx R ¼ 1A I
ð4:22Þ
As long as the opencircuit voltage gain of the dependent source, A, is greater than one, Eq. (4.22) states the negative equivalent or Thévenin resistance. Physically, this means that the Thévenin equivalent circuit is delivering power instead of absorbing it.
Construction and the Use of Negative Equivalent Resistance A circuit block, which is equivalent to the negative resistance, may be constructed using the operational ampliﬁer studied in the next chapter. This block may be used for different purposes including signal generation. The difference between the negative resistance and the power source is that the negative resistance may supply power of any type (DC, AC, or an arbitrary waveform), i.e., support the selfoscillating circuits. Figure 4.17c summarizes Thévenin resistances generated by their basic networks with only dependent sources. The same method of analysis (simultaneous use of KCL and KVL) has been applied to every network. Although all of the networks in principle generate the negative equivalent resistance values, the realization of some particular circuits may be difﬁcult.
b)
a)
a
a
+ Avx

R
+ 
+
+
vx
Avx


R
+
+ 
vx
I
b
b c)
RT=R/(1A)
+ Avx

R
+ RT=RZ
+ Zix

RT=R/(1GR)
+
vx
Gvx


+ vx
R
RT=R(1A)
ix R
+ 
+
ix
+ Aix
R

Fig. 4.17. Thévenin equivalent resistances for basic networks with dependent sources.
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Section 4.2: Generator Theorems and Their Use
4.2.7 Short Summary of Circuit Analysis Methods In summary, a linear circuit can be solved using any of the methods studied in this and in the previous chapters: 
Superposition theorem Nodal/mesh analysis Source transformation theorem Method of Thévenin and/or Norton equivalent circuit(s)
or a combination of those. While the nodal/mesh analysis is always applicable, other methods may turn out to be even more useful since they often provide a physical insight into the circuit behavior. Exercise 4.10: How could you ﬁnd the opencircuit voltage Vab in Fig. 4.14a? Answer: A. When the superposition theorem is applied, shorting out D0 gives Vab ¼ D1/2. Shorting out D1 gives Vab ¼ D0/4. B. When the nodal analysis is applied, we ground negative terminals of both sources and ﬁnd the unknown voltage of the upper left node via KCL. Only one equation needs to be solved. This is perhaps the simplest solution method. C. The source transformation theorem cannot easily be applied. D. The method of Thévenin equivalent circuits is described in example 4.8.
Exercise 4.11: How could you ﬁnd current I for the circuit in Fig. 4.15b? Answer: A. When the superposition theorem is applied, shorting out the 30 V source gives I1 ¼ 3.214 mA. Shorting out the 15 V source gives I2 ¼ 2I1(due to symmetry!). The resulting current is I ¼ I1+I2 ¼ 3.214 mA. B. When the nodal analysis is applied, we ground the entire bottom node and combine both 0.2 kΩ resistors in series. There are two nontrivial nodes on top and two nodal equations. After the solution is obtained (V1 ¼ 12.86 V for the left node 1 and V2 ¼ 14.14 V for the right node 2), current I is found as the difference of two nodal voltages divided by 0.4 kΩ. C. The source transformation theorem followed by superposition could be applied too. D. The method of Thévenin equivalent circuits is described in example 4.9.
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Chapter 4
Circuit Analysis and Power Transfer
Section 4.3 Power Transfer 4.3.1 Maximum Power Transfer The principle of maximum power transfer from a source to a load will now be quantiﬁed. This principle is also known as a maximum power theorem. The circuit under study is shown in Fig. 4.18. It involves an arbitrary linear source (a battery, a generator, etc.), which is represented by its Thévenin equivalent, and a load, which is characterized by its equivalent resistance RL. All other load parameters (dynamic, mechanical, and thermal) are implicitly included into the load’s resistance. a
 +
RT VT
+ 
RL
b
Fig. 4.18. A battery (or another practical voltage source) connected to a load.
The key question you have to ask yourself is this: For a given ideal voltage source VT and a given internal resistance RT, can the electric power delivered to the load be maximized, and at which value of RL does the maximum occur? The answer is found by solving the circuit in Fig. 4.18. First, the current is determined from the given voltage source VT and the total resistance using the series equivalent, I¼
VT RT þ RL
ð4:23Þ
This allows us to compute the power at the load based on PL ¼ RL I 2 ¼
RL V 2T ðRT þ RL Þ2
ð4:24Þ
When VT and RT are ﬁxed, the magnitude of the load resistance determines the delivered power PL. This power tends to zero when either RL ! 0 or RL ! 1; moreover, it is always
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Chapter 4
Section 4.3: Power Transfer
positive. Therefore, according to Rolle’s theorem of calculus, the power must have a maximum at a certain value of RL. For example, Fig. 4.19 shows a plot of the load power as a function of RL for VT ¼ 9 V and RT ¼ 5 Ω.
Load power, W
4
3
2
1
0
0
10
20 30 40 Load resistance, W
50
Fig. 4.19. Load power as a function of the load resistance for ﬁxed VT ¼ 9 V, RT ¼ 5 Ω.
We will ﬁnd the maximum of the load power analytically. We treat PL in Eq. (4.24) as a function of RL, i.e., PL ¼ PL(RL). It is known that a function has a maximum when its ﬁrst derivative is set to zero. Consequently, differentiating PLwith respect to RL gives " # " # dPL 1 R R R L T L ¼ V 2T 2 ¼ V 2T ¼0 ð4:25Þ dRL ðRT þ RL Þ2 ðRT þ RL Þ3 ðRT þ RL Þ3 The necessary and sufﬁcient condition for Eq. (4.25) to hold is RL ¼ RT ) PL ¼ 0:25V 2T =RT
ð4:26Þ
This result is of great practical value despite, or maybe because of, its simplicity. The maximum output power is achieved when the load resistance is equal to the internal resistance of the power source. In other words, the load is matched to the source; it is called the matched load. In power engineering and in RF and microwave engineering, the problem of load matching is very important. However, it must be clearly stated that no more than 50% of the total circuit power can be extracted even in the best case. This statement makes sense if we again examine the circuit in Fig. 4.18 and assume two equal resistances. The power is divided equally; half of the total power is spent to heat up the power source. The power maximum in Fig. 4.19 is relatively ﬂat over the domain RL > RT; however, the power drops sharply when RL < RT. This last condition should be avoided if at all possible.
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Chapter 4
Circuit Analysis and Power Transfer
Example 4.11: An audio ampliﬁer produces an rms output of 20 V. The ampliﬁer’s output resistance is rated at 4 Ω. You are given four 4 Ω speakers. How should you connect the speakers for the maximum acoustic power—in series, parallel, or a single speaker only? Solution: The rms voltage simply means the equivalent DC voltage that provides the same power to the load as the average power of the primary AC voltage. Hence, the sophisticated AC audio ampliﬁer circuit is essentially replaced by its DC Thévenin equivalent with VT ¼ 20 V and RT ¼ 4 Ω. Similarly, the dynamic speakers are replaced by a DC load with RL ¼ 16 Ω if connected in series combination, RL ¼ 1 Ω if connected in parallel, or with RL ¼ 4 Ω if only a single speaker is employed. The output (audio) powers are as follows:
PL ¼
1 400
¼ 16 W four speakers in parallel
ð4:27aÞ
¼ 25 W single speaker ð4 þ 4Þ2 16 400 ¼ 16 W four speakers in series PL ¼ ð4 þ 16Þ2
ð4:27bÞ
PL ¼
ð4 þ 1Þ2 4 400
ð4:27cÞ
The best (loudest) choice would be surprisingly one single speaker.
4.3.2 Maximum Power Efﬁciency A power analysis would be incomplete without discussing the efﬁciency of the power transfer. Consider an electric boat driven by a marine battery. The optimization of the batterymotor system for maximum power transfer implies that we will move fast, but perhaps not very far. Another optimization is possible for maximum power efﬁciency. In this case, we could tolerate a smaller speed in order to travel a longer distance. The circuit to be analyzed is again the circuit in Fig. 4.18. The useful power delivered to the load is given by Eq. (4.24). The total power delivered by the source is P ¼ ðRT þ RL ÞI 2 ¼
V 2T RT þ RL
ð4:28Þ
The power efﬁciency E is deﬁned as the ratio of the useful power to the total power, E¼
PL RL ¼ ðRT þ RL ÞI 2 ¼ P RT þ RL
ð4:29Þ
Thus, the power efﬁciency is a simple function of the load resistance and the source resistance. It does not depend on the source voltage. The efﬁciency is zero when the load resistance is zero. It monotonically increases and approaches maximum (the maximum value is unity, which
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Chapter 4
Section 4.3: Power Transfer
corresponds to an efﬁciency of 100%) when the load resistance becomes large enough when compared to the source resistance. The maximum power efﬁciency is thus achieved when RL > > RT. For example, Fig. 4.20 augments the load power graph from Fig. 4.19 with the corresponding
Load power, W
3
80
60 load power
2
40
20
1
0 0
10
20 30 40 Load resistance, W
Power efficiency, %
power efficiency
4
0 50
Fig. 4.20. Load power and power efﬁciency for ﬁxed VT ¼ 9V, RT ¼ 5 Ω.
efﬁciency curve. Example 4.12: A battery with the stored energy of EB ¼ 0.1 MJ, and VT ¼ 12 V, RT ¼ 5 Ω, delivers its entire energy during the time period 0 t T and discharges with a constant output voltage/current. Two loads are used: RL ¼ 5 Ω and RL ¼ 50 Ω. Determine discharge time T and total energy delivered to the load in each case. Solution: The discharge time, T ¼ EB/P, is determined ﬁrst when the total power P follows based on Eq. (4.28). Assuming constant battery discharge rate, we obtain
T 1:9 hours for the 5 Ω load;
T 10:6 hours for the 50 Ω load
ð4:30Þ
The total energy delivered to the load, E ¼ TPL, in each case is given by
E 5Ω ¼ 50 kJ for the 5 Ω load;
E 50Ω ¼ 91 kJ for the 50 Ω load
ð4:31Þ
Thus, the total energy extracted from the battery is nearly twice as high in the second case. However, it takes about ﬁve times longer to extract this energy.
4.3.3 Application Example: Power Radiated by a Transmitting Antenna A transmitting antenna in a radio handset features a monopole antenna. It is connected to a source that has the same basic form as in Fig. 4.18 but with an AC generator instead of the DC source and with an internal (generator) resistance of 50 Ω. The antenna represents a load with a “radiation” resistance of 50 Ω. This resistance describes the power loss in terms of the electromagnetic
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Chapter 4
Circuit Analysis and Power Transfer
radiation from the antenna. Thus, the antenna, if properly matched to the power source, will radiate 50% of the total power as electromagnetic waves into space. Now, an inexperienced electrical engineer decides to “modify” the handset by cutting the monopole antenna to only one third of its length. In this case, the antenna’s radiation resistance is reduced to one ninth of its original value. How does this affect the radiated signal power? To answer this question, we ﬁnd the instantaneous load power, which according to Eq. (4.24), is PL ðt Þ ¼
RL V 2T ðt Þ ðRT þ RL Þ2
ð4:32Þ
The ratio of the power levels for the two antenna conﬁgurations does not depend on time and we obtain PLshort 50=9 50 0:0018 ¼ 0:36 ¼ = ¼ 2 2 0:0050 PLoriginal ð50 þ 50=9Þ ð50 þ 50Þ
ð4:33Þ
Thus, for the shorter antenna, we will only achieve about 36% of the radiated power compared to the original handset. In practice, this estimate becomes even lower due to the appearance of a very signiﬁcant antenna capacitance.
4.3.4 Application Example: Maximum Power Extraction from Solar Panel Every solar panel reports the measured data for VOC and ISC listed on its backside. For linear circuits, VT ¼ VOC, RT ¼ VOC/ISC. If the solar panel were a linear circuit, the maximum extracted power would be exactly equal to 0.25VOCICS according to Eq. (4.26). Fortunately, this is not the case. The maximum extracted power is signiﬁcantly greater than this value. However, it is still less than the “best” possible value of VOCICS. To quantify the maximum power output, every solar panel has another set of measured data, VMP and IMP, also listed on its backside. VMP stands for maximum power load voltage, and IMP stands for the maximum power load current. The maximum extracted power is the product VMPIMP, which is always less than VOCICS. The ratio of two powers F¼
V MP I MP > 1. The area of the shaded rectangle is the load power; it is clearly maximized at a certain operating point Q, where V ¼ VMP, I ¼ IMP. I, A IP Q
IMP
Characteristic equation of the cell I = I(V)
PL
0
VMP
VOC
V, Volts
Fig. 4.25. Finding operating point Q of the solar cell for the maximum power transfer.
Finding maximum power parameters is a straightforward but lengthy procedure. The load power is found as a function of V; its derivative must be set equal to zero to maximize the load power. This is essentially the maximum power theorem for nonlinear circuits. The ﬁnal expressions are V OC nV T , I MP ¼ I SC 1 90%of I SC ð4:45Þ V MP ¼ V OC nV T ln 1 þ nV T V MP For n ¼ 1.75 and VOC ¼ 0.6V, VMP 80 % of VOC, IMP 90 % of ISC which is close to the data in Table 4.2 given that the ﬁll factors for the cell and the panel are approximately the same. Note that the load resistance is found as RL ¼ VMP/IMP.
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Chapter 4
Summary
Summary Nodal analysis
Supernode
Circuit analysis techniques: nodal/mesh analysis Based on the KCL and Ohm’s law: V 1 V S R1
þ V 1 0 R3
V
2 þ V 1RV ¼0 2
V 2 V S R5
þ V 2 0 R4
V
1 þ V 2RV ¼0 2
The KCL for the supernode: V1 VS V1 þ R1 R3 V2 V2 VS þ þ ¼0 R4 R5
Mesh analysis
Plus the KVL:V2 ¼ V1+V0 Based on the KVL and Ohm’s law:
Supermesh
R1(I1 I3)+R3I1+R5(I1 I2) ¼ 0 R2(I2 I3)+R5(I2 I1)+R4I2 ¼ 0 VS+R1(I3 I1)+R2(I3 I2) ¼ 0 For meshes 1, 2, and 3. The KVL for the supermesh: R1 ðI 1 I 3 Þ þ R3 I 1 þ R4 I 2 þ R 2 ðI 2 I 3 Þ ¼ 0
Source transformation theorem
Plus Eq. for mesh 3 And the KCL: I1 I2 ¼ IS Circuit analysis techniques: source transformation theorem Substitution of voltage source VT in series with resistance RT for current source IN with resistance RN: RN ¼ RT, I N ¼ VRTT VOC ¼ VT, I SC ¼ VRTT (continued)
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Chapter 4
Circuit Analysis and Power Transfer
Circuit analysis techniques: Thévenin/Norton theorems/equivalents Thévenin Any linear network with indepenand Norton dent sources, dependent linear theorems sources, and resistances can be replaced by a simple equivalent network in the form: i. A voltage source VT in series with resistance RT ii. A current source IN in parallel with resistance RN Summary of major circuit analysis methods (linear circuits)  Superposition theorem (pervious chapter)  Nodal/mesh analysis (this chapter)  Source transformation theorem (this chapter)  Thévenin and Norton equivalent circuits (this chapter) Linear networks: Measurements/equivalence – Two arbitrary linear networks are Method equivalent when their VOC and of short/ ISC coincide open – This method is also used for circuit nonlinear circuits Linear networks: maximum power theorem Maximum – Power delivered to the load is power theorem maximized when RL ¼ RT (load – For highfrequency circuits it matching) also means no “voltage/current wave reﬂection” from the load
Power efﬁciency is maximized when the load resistance is very high (load bridging)
Linear networks: maximum power efﬁciency Power transfer efﬁciency: L E ¼ PPL ¼ RTRþR L
(continued)
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Chapter 4
Summary
Linear networks: dependent sources and negative equivalent resistance R Equivalent Case (a): RT ¼ 1A resistances Case (b): RT ¼ R Z R of basic Case (c): RT ¼ 1GR linear networks Case (d): RT ¼ R(1 A) with dependent Equivalent resistance may become sources negative
Basic nonlinear circuits
Load line method
Nonlinear networks: four basic topologies – Linear voltage source connected to a nonlinear load (diode circuit) – Linear current source connected to a nonlinear load – Nonlinear voltage source connected to a linear load; – Nonlinear current source connected to a linear load (photovoltaic circuit) Nonlinear networks: circuit analysis via load line method Solution: The load line (υi characteristic of the linear source, which is I ¼ V TRV ) intersects the T υi characteristic of the nonlinear load, I(V )
Nonlinear networks: iterative solution Finding the For the circuit in the previous row: Implicit scheme: intersection Initial guess V0may be 0 V n point VT V VT V ) V nþ1 ¼ I 1 I ðV Þ ¼ , iteratively RT RT n ¼ 0, 1, .. . Nonlinear networks: ﬁnding resistive load for maximum power extraction Finding Load power is computed as maximum PL ¼ V I(V ). Then, it is maxiload power mized, which is equivalent to for the solving equation: dPL equivalent dV ðV Þ ¼ 0 model of a For unknown voltage V solar cell (continued)
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Chapter 4
Circuit Analysis and Power Transfer
Some useful facts about power extraction from solar cells/modules Typical values of opencircuit voltage VOC and Crystalline silicon or cSi cell: photocurrent density JP of a cSi cell VOC 0.6 V; JP 0.03 A/cm2 – Opencircuit module voltage is N times the cell voltage, N VOC – Shortcircuit module current ISC is the cell shortcircuit current ISC ¼ AJP where A is cell area
Typical values of maximum power parameters and ﬁll factor for cSi cells/modules; the load resistance must be RL ¼ VMP/IMP Lossless singlediode model of a solar cell: IP ¼ AJP—photocurrent (A) A—cell area (cm2) VT—thermal volt. (0.0257 V) n—ideality factor (1 < n < 2) I S I P exp VnVOCT (A)
Maximum power analytical solution for lossless singlediode model of a solar cell
I MP Fjmodule ¼ VVMP Fjcell (for lowloss OC I SC modules) VMP 0.8VOC, IMP 0.9ISC, F 0.72
h i I ¼ I P I S exp nVV T 1
V MP V OC nV T ln 1 þ VnVOCT ,
I MP ¼ I SC 1 VnVMPT
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Chapter 4
Problems
Problems B. Show the current directions for every resistance in the circuit.
4.1 Nodal/Mesh Analysis 4.1.2 Nodal Analysis Problem 4.1. Using the nodal analysis, determine the voltage V1 in the circuit shown in the following ﬁgure.
I
1 kW
+ 
5V
1 kW
1 kW
3 kW
2 kW
Problem 4.4. A. Using the nodal analysis, determine the total circuit current I for the circuit shown in the ﬁgure below. B. Show the current directions for every resistance in the circuit. Problem 4.2. Using the nodal analysis, determine the voltage V1 or current I in the circuits shown in the following ﬁgure.
1 kW 1 kW
1 kW
I
+ 
10 V
5 kW
1 kW
Problem 4.5. Introduce the ground termination, write the nodal equations, and solve for the node voltages for the circuit shown in the ﬁgure below. Calculate the current I shown in the ﬁgure. 1 kW 1 kW
1 kW I
1 mA 10 V
Problem 4.3. A. Using the nodal analysis, determine the supply current I for the circuit shown in the ﬁgure.
+ 
1 kW
Problem 4.6. Introduce the ground termination, write the nodal equations, and solve for the node voltages for the circuit shown in the ﬁgure. Calculate the current through the resistance Rx and show its direction in the ﬁgure.
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Chapter 4
Circuit Analysis and Power Transfer
1W
1 kW
5V
1 kW
1W
+

I
1W
Rx=1 W
+ 
1A
Problem 4.7. For the circuit shown in the ﬁgure below (i) Determine the current ix through resistance Rx. (ii) Show its direction on the ﬁgure. 1 kW
0V 10 V
1 kW Rx=1 kW
+ 3 kW
4 kW
Problem 4.8. For the circuit shown in the ﬁgure below A. Write nodal equations and solve for the node voltages. Then, ﬁnd the value of i1. B. Could this problem be solved in another (simpler) way? 10 W
1A
i1 20 W 5 W
1 kW
10 V
1 kW
5 kW 1 kW
1 kW
Problem 4.10. The ﬁgure below shows the DC equivalent of a residential threewire system, which operates at 240 V rms (do not confuse it with the threephase system, which carries a higher current). Two 120 V rms power supplies are connected as one dualpolarity power supply, i.e., in series. The 10 Ω load and the 20 Ω are those driven by the twowire (and one ground) standard wall plug with 120 V rms—the lights, a TV, etc. The 6 Ω load consumes more power, and it is driven with 240 V rms using a separate bigger wall plug (+/ and neutralnot shown)—the stove, washer, dryer, etc. Determine the power delivered to each load. Hint: Use a calculator or software of your choice for the solution of the system of linear equations (MATLAB is recommended). 0.1 W
+ 
hot
10 W
120 V 0.1 W neutral
2A
+ 
6W
20 W
120 V 0.1 W hot
Problem 4.9. A. Write the nodal equations and solve for the node voltages for the circuit shown in the ﬁgure below. Then, ﬁnd the value of i1. B. Use MATLAB or other software of your choice for the solution of the system of linear equations; attach the code to the solution.
4.1.3 Supernode Problem 4.11. Introduce the ground terminal, write the nodal equations, and solve for the node
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Chapter 4
Problems
voltages for the circuit shown in the ﬁgure. Calculate the current I shown in the ﬁgure.
1 kW
+
Vx
3 kW
5V

1mA
1 kW 1 kW
2 kW
1 kW
2 kW
+

15 V
+ 
1 kW
I
1 kW
Problem 4.12. Introduce the ground terminal, write the nodal equations, and solve for the node voltages for the circuit shown in the ﬁgure. Calculate the current I shown in the ﬁgure.
Problem 4.15. For the circuit shown in the ﬁgure A. Determine the circuit current I using either the nodal analysis or the mesh analysis. B. Explain your choice for the selected method. 1 kW 1 kW
1 kW
1 kW I
+ 
5 kW
5V
1 kW
1 kW

+
10 V
10 V
+ 
1 kW
I
1 kW
5 kW
4.1.4 Mesh Analysis 4.1.5 Supermesh Problem 4.13. For the circuit shown in the ﬁgure, determine the current i1 A. Using the mesh analysis B. Using the nodal analysis
Problem 4.16. For the circuit shown in the ﬁgure, determine its equivalent resistance between terminals a and b. Hint: connect a power source and use a mesh current analysis or the nodal analysis. 1 kW a
1 kW
1 kW
1 kW
1 kW
Which method is simpler? b
5W
2 mA
i1 20 W 5 W
3 mA
1 kW
1 kW
Problem 4.17. For the circuit shown in the ﬁgure, determine the current i1 of the 20 V voltage source.
Problem 4.14. For the circuit shown in the ﬁgure, determine the voltage Vx A. Using the mesh analysis B. Using the nodal analysis Which method appears to be simpler?
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Chapter 4
Circuit Analysis and Power Transfer 9W
2W
8W
Problem 4.20. Given two networks shown in the ﬁgure below: A. Determine their opencircuit voltage, VOC, and shortcircuit current, ISC, for each of them. B. Are the networks equivalent?
i1
+ 
20 V
10 W
4W
0.5 A
Problem 4.18. Determine the voltage across the current source for the circuit shown in the ﬁgure below using the mesh analysis and the supermesh concept.
a) a 1 kW
3V
+ 
1 kW
b
b) 2 kW
10 V
1 kW 2 mA
+ 
0.5 kW
3 kW
1 kW
4.2 Generator Theorems 4.2.1 Equivalence of Active OnePort Networks: Method of Short/Open Circuit Problem 4.19. A linear active network with two unknown circuit elements measures VOC ¼ 5 V, ISC ¼ 10 mA. A. Determine the parameters VT, RT and IN, RN of two equivalent networks shown in the ﬁgure below. B. Could you identify which exactly network is it? a) a RT VT
a 6 mA
+ 
b
4.2.2 Application Example: Reading and Using Data for Solar Panels Problem 4.21. The area of a single cell in the 10 W BSP1012 Power Up cSi panel is ~22 cm2. Predict: A. Opencircuit voltage of the solar cell, VOC B. Photocurrent density of the cell, JP C. Shortcircuit current of the cell, ISC Compare the above value with the value ISC ¼ 0.66 A reported by the manufacturer.
Problem 4.22. The area of a single cell in the 175 W BP Solar SX3175 cSi panel is 156.25 cm2. Predict: A. Opencircuit voltage of the solar cell, VOC B. Photocurrent density of the cell, JP C. Shortcircuit current of the cell, ISC
b
b) a IN
RN b
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Chapter 4
Problems
Compare the above value with the value ISC ¼ 5.1A reported by the manufacturer. What value should the photocurrent density have in order to exactly match the shortcircuit current reported by the manufacturer?
Problem 4.23. Are the solar cells in the solar module connected in parallel or in series? Why is one particular connection preferred? Problem 4.24. You are given three cSi solar cells shown in the ﬁgure below, each of area A/ 3. Draw wire connections for a cell bank, which has the performance equivalent to that of a large solar cell with the area A.
A/3
A. Estimate the area of the single solar cell using the common photocurrent density value for cSi solar cells. B. Estimate the opencircuit voltage for the single cell.
a n p
b
Problem 4.27. A 20 W BSP2012 PV cSi module shown in the ﬁgure has 36 unit cells connected in series, the shortcircuit current of 1.30 A and the opencircuit voltage of 21.7 V.
A/3 c n p
d
A/3 e n p
f
Problem 4.25. Individual solar cells in the ﬁgure to the previous problem are to be connected into a standard solar module. Draw the corresponding wire connections. Problem 4.26. A 10 W BSP1012 PV cSi module shown in the ﬁgure has 36 unit cells connected in series, the shortcircuit current of 0.66 A and the opencircuit voltage of 21.3 V.
A. Estimate the area of the single solar cell using the common photocurrent density value for cSi solar cells. B. Estimate the opencircuit voltage for the single cell.
Problem 4.28. A cSi solar module is needed with the opencircuit voltage of 10 V and short
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Chapter 4
Circuit Analysis and Power Transfer
circuit current of 1.0 A. A number of individual solar cells are available, and each has the area of 34 cm2, opencircuit voltage of 0.5 V, and shortcircuit current of 1.0 A. Identify the proper module conﬁguration (number of cells) and estimate module’s approximate size.
Problem 4.29. A cSi solar module is needed with the opencircuit voltage of 12 V and shortcircuit current of 3.0 A. A number of individual solar cells are available, and each has the area of 34 cm2, the opencircuit voltage of 0.5 V, and shortcircuit current of 1.0 A. Identify the proper module conﬁguration (number of cells) and estimate module’s approximate size. 4.2.3 Source Transformation Theorem Problem 4.30. Find voltage V in the circuit shown in the ﬁgure below using source transformation. 1 kW
2 kW
+ 3 mA
1 kW
V
2 kW

+ 
2V
Problem 4.31. The circuit shown in the ﬁgure below includes a currentcontrolled voltage source. Find current ix using source transformation.
2 kW
+ vx
3 mA
1 kW

+ 
2vx
Problem 4.33. Repeat the previous problem for the circuit shown in the ﬁgure below. 2 kW
+ vx
2 mA
1 kW

+
2vx
4.2.4 Thévenin’s and Norton’s Theorems 4.2.5 Finding Thévenin and Norton Equivalents and Using them for Circuit Solution Problem 4.34. Find (i) Thévenin (or equivalent) voltage (ii) Thévenin (or equivalent) resistance for the twoterminal network shown in the ﬁgure below (three practical voltage sources or batteries in series) when RB1 ¼ 0.5 Ω, RB2 ¼ 0.5 Ω, RB3 ¼ 0.5 Ω, VB1 ¼ 6 V, VB2 ¼ 6 V,VB3 ¼ 6 V. a VB3
+ 
RB3 a
5 kW
ix 3 mA
1 kW
+ 
4000ix
Problem 4.32. The circuit shown in the ﬁgure below includes a voltagecontrolled voltage source. Find voltage υx using source transformation.
VB2
+ 
VB1
+ 
RB2
RT VT
+ b
RB1 b
Problem 4.35. Find (i) Thévenin (or equivalent) voltage
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Chapter 4
Problems
(ii) Thévenin (or equivalent) resistance
for the twoterminal network shown in the ﬁgure below (three practical voltage sources or batteries in parallel) when RB1 ¼ 1 Ω, RB2 ¼ 1 Ω, RB3 ¼ 1 Ω, VB1 ¼ 6 V, VB2 ¼ 6 V, VB3 ¼ 6 V.
Problem 4.36. Find (i) Thévenin (or equivalent) voltage (ii) Thévenin (or equivalent) resistance
Problem 4.38. For the circuits shown in the ﬁgure below, determine their equivalent (Thévenin) circuit by ﬁnding VT and RT (show units).
for the twoterminal network shown in the ﬁgure below. Assume three 9 V sources and resistances of 1 Ω each. a RB3
RB2
RB1
+ 
+ 
VB1
VB2
+ 
VB3
b a RT VT
+ b
Problem 4.37. For the circuits shown in the ﬁgure below, determine their equivalent (Thévenin) circuit by ﬁnding VT and RT (show units).
Problem 4.39. Find (i) Thévenin (or equivalent) voltage
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Chapter 4
Circuit Analysis and Power Transfer
(ii) Thévenin (or equivalent) resistance
for the twoterminal networks shown in the ﬁgure below. a)
1 kW a
+ 
Problem 4.41. A twoterminal network shown in the ﬁgure is a starting section of a ladder network used for digitaltoanalog conversion. Express (i) Thévenin (or equivalent) voltage (ii) Thévenin (or equivalent) resistance in terms of (digital) voltage D0and resistance R.
6 kW
10 V
a 2R 2R
b 3 kW
b)
+ 
a
b
6 kW
1 kW
10 V
Problem 4.42. A twoterminal circuit network in the ﬁgure is a twobit ladder network used for digitaltoanalog conversion. Express parameters of the corresponding Norton equivalent circuit in terms of (digital) voltages D0,D1 and resistance R.
b 2 kW 1 kW
c)
+ 
D0
2 kW
a 6 kW
1 mA
R a
b 3 kW 5W
d)
+ 
48 V
2R
a
D0
+ 
D1
+ b
16 W
20 W
2R
2R
4W
b
Problem 4.40. Find (i) Thévenin voltage (ii) Thévenin resistance for the twoterminal network shown in the ﬁgure below when R1 ¼ R2 ¼ R3 ¼ 1 kΩ, VS1 ¼ VS2 ¼ 10 V.
Problem 4.43. A twoterminal network shown in the ﬁgure is a fourbit ladder network used for digitaltoanalog conversion. Express (i) Thévenin (or equivalent) voltage (ii) Thévenin (or equivalent) resistance in terms of four (digital) voltages D0,D1,D2,D3 and resistance R.
R2 a R1
R3 VS2
+
+ 

VS1
b
IV192
Chapter 4
Problems R
R
R
1
a)
+
2R
D0
+ 
D1
+ 
+ 
D2
Avx
2R
2R
2R
2R
D3
+ 
2
vx

b)
ix
+
Problem 4.44. Determine the Norton equivalent for the circuit shown in the ﬁgure. Express your result in terms of I and R.
+
R
+ 
Zix

R
+ 
c)
+
+ a
Gvx
vx
R

I
I
R
d)
R
ix
+ Aix
b
R

Problem 4.45. Establish Thévenin and Norton equivalent circuits for the network shown in the ﬁgure below. a
+

2vx
+ 2 kW
0.1 mA
1 kW
vx
Problem 4.48. Each of three identical batteries is characterized by its Thévenin equivalent circuit with VT ¼ 9 V and RT ¼ 1 Ω. The batteries are connected in series. The entire battery bank is connected to a 5 Ω load. Find the power delivered to the load.
b
4.2.5 Application Example: Generating Negative Equivalent Resistance Problem 4.46. Establish the equivalent (Thévenin) resistance for the network shown in the ﬁgure below. Carefully examine the sign of the equivalent resistance.
Problem 4.49. Each of three identical batteries is characterized by its Thévenin equivalent circuit with VT ¼ 9 V and RT ¼ 1 Ω. The batteries are connected in parallel. The entire battery bank is connected to a 5 Ω load. Find the power delivered to the load. Problem 4.50. For two circuits shown in the ﬁgure below, ﬁnd the current I using the method of Thévenin equivalent.
a
+ Avx

+ 
R1 R2
+ vx
b
Problem 4.47. Derive the equivalent (Thévenin) resistance for the networks shown in the ﬁgure below (conﬁrm Fig. 4.16 of the main text).
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Chapter 4
Circuit Analysis and Power Transfer
Problem 4.54. Solve the circuits shown in the ﬁgure below and determine the current I (show units). You can use any of the methods studied in class:
Problem 4.51. For the circuit shown in the ﬁgure below, ﬁnd the current I using the method of Thévenin equivalent.

Superposition theorem Nodal/mesh analysis Source transformation theorem Thévenin and Norton equivalents
or their combinations. a)
I
1 mA
2 kW
1 kW
5 mA
1 kW
4 mA
3V
+

b) 1 mA
2 kW
3V
+

Problem 4.52. For the circuit shown in the ﬁgure below, ﬁnd the current I using the method of Thévenin equivalent.
I
6 kW
c)
2 kW I
6V
+ 
3 kW
+ 
3V
Problem 4.55. Solve the circuits shown in the ﬁgure below and determine unknown voltage Vor current I (show units). You can use any of the methods studied in class:
4.2.6 Summary of Circuit Analysis Methods Problem 4.53. Solve the circuit shown in the ﬁgure below and determine the current I (show units). You can use any of the methods studied in class:  Superposition theorem  Nodal/mesh analysis  Superposition theorem  Source transformation theorem  Nodal/mesh analysis  Thévenin and Norton equivalents  Source transformation theorem  Thévenin and Norton equivalents or their combinations. or their combinations.
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Chapter 4 a)
+ 
1 kW
Power delivered to the load has a maximum at exactly one value of the load resistance. Find that value and prove your answer graphically using software of your choice (MATLAB is recommended).
2 kW
+ 4V
2 kW
6 mA
V

b)
+ 
Problems
0.5kW
1.5 kW
I
+ 
1 kW
70V 0.5kW
1.5 kW
24V
+ 
RL?
1W
c)
10 V
1W
+
1W
I 1W
2A
Problem 4.56. Solve the circuit shown in the ﬁgure below and determine the current I (show units). You can use any of the methods studied in class: 
1W
14V
Superposition theorem Nodal/mesh analysis Source transformation theorem Thévenin and Norton equivalents
or their combinations.
Problem 4.58. A power supply for an electric heater can be modeled by an ideal voltage source of unknown voltage in series with the internal resistance RT ¼ 4 Ω. A. Can you still determine when the power delivered to a load (a heating spiral with resistance RL) is maximized? B. Does the answer depend on the source voltage? Problem 4.59. For the circuit shown in the ﬁgure, when is the power delivered to the load maximized?
1.5 W 12 V
+ 
RL ? 1.5 W
Problem 4.60. A battery can be modeled by an ideal voltage source VT in series with a resistance RT.
4.3 Power Transfer 4.3.1 Maximum Power Transfer 4.3.2 Maximum Power Efﬁciency Problem 4.57. A deepcycle marine battery is modeled by an ideal voltage source of 24 V in series with a 1 Ω resistance shown in the ﬁgure below. The battery is connected to a load, and the load’s resistance, RL, needs to be optimized.
RT VT
+ 
RL?
IV195
Chapter 4
Circuit Analysis and Power Transfer
(i) For what value of the load resistance RL is the power delivered to the load maximized? (ii) What percentage of the power taken from the voltage source VT is actually delivered to a load (assuming RL is chosen to maximize the power delivered)? (iii) What percentage of the power taken from the voltage source VT is delivered to a load when RL ¼ 0.1RT?
Problem 4.61. A certain micropower photovoltaic device can be modeled under certain conditions as an ideal current power source and a resistance in parallel—see the ﬁgure below. At which value of the load resistance, RL, is the power delivered to the load maximized?
5 mA
50 kW
RL
Problem 4.62. A lowcost polycrystalline Power Up BSP1–12 1 W Solar panel lists ratings for the output voltage and current, which give maximum load power: VLmax¼17.28 V, ILmax¼0.06A. Based on these cell speciﬁcations, which value of the equivalent resistance should the load to be connected to the solar cell have for maximum power output? Problem 4.63. The solar panel from the previous problem generates a signiﬁcant voltage of ~13 V in a classroom without direct sun light, but the resulting current is small, ~1 mA. A. Now, which value of the equivalent resistance should the load have for maximum power output?
B. How is the maximum load power different, compared to the previous problem?
Problem 4.64. The heating element of an electric cooktop has two resistive elements, R1 ¼ 50 Ω and R2 ¼ 100 Ω, that can be operated separately, in series, or in parallel from a certain voltage source that has a Thévenin (rms) voltage of 120 V and internal (Thévenin) resistance of 30 Ω. For the highest power output, how should the elements be operated? Select and explain one of the following: 50 Ω only, 100 Ω only, series, and parallel. Problem 4.65. You are given two speakers (rated at 4 Ω and 16 Ω, respectively) and an audio ampliﬁer with the output resistance(impedance) equal to 8 Ω.
8W
4W
16 W
A. Sketch the circuit diagram that gives the maximum acoustic output with the available components. Explain your choice. B. Sketch the circuit diagram for the maximum power efﬁciency. Explain your choice.
4.3.4 Application Example: Maximum Power Extraction from Solar Panel Problem 4.66. A. Describe in your own words the meaning of the ﬁll factor of a solar cell (and solar module). B. A 200 W GE Energy GEPVp200 cSi panel has the following reading on its back: VOC ¼ 32.9 V, ISC ¼ 8.1 A,
IV196
Chapter 4
Problems
VMP ¼ 26.3 V, and IMP ¼ 7.6 A. What is the module ﬁll factor? What is approximately the ﬁll factor of the individual cell?
Problem 4.67. Using two Web links, http://powerupco.com/site/ http://www.affordablesolar.com/, identify the solar panel that has the greatest ﬁll factor to date. Problem 4.68. A 10 W BSP1012 PV cSi module shown in the ﬁgure has 36 unit cells connected in series, the shortcircuit current of 0.66 A and the opencircuit voltage of 21.3 V. The maximum power parameters are VMP ¼ 17.3 V and IMP ¼ 0.58 A.
A. Estimate the area of the single solar cell using the common photocurrent density value for cSi solar cells (show units). B. Estimate the opencircuit voltage for the single cell. C. Estimate the ﬁll factor of the module and of the cell. D. Estimate the value of the equivalent load resistance R required for the maximum power transfer from the module to the load.
Problem 4.70. A REC SCM220 220 Watt 20 V cSi solar panel shown in the ﬁgure has the following readings on the back: the short circuit current ISC of ~8.20 A, the open circuit voltage VOC of ~36.0 V, the maximum power voltage VMP of ~28.7 V, and the maximum power current IMP of ~7.70 A. Repeat four tasks of problem 4.57.
Problem 4.71. A 14.4 W load (a DC motor) rated at 12 V is to be driven by a solar panel. A cSi photovoltaic sheet material is given, which has the opencircuit voltage of 0.6 V and the photocurrent density of JP ¼ 0.03 A/cm2.
Problem 4.69. A 20 W BSP2012 PV cSi module shown in the ﬁgure has 36 unit cells connected in series: the shortcircuit current of 1.30 A, the opencircuit voltage of 21.7 V, the maximum power voltage VMP of 17.3 V, and the maximum power current IMP of 1.20 A. Repeat four tasks of the previous problem.
IV197
Problem 4.73. You are given the generic ﬁll factor F ¼ 0.72 for the cSi solar panels, the generic opencircuit voltage VOC ¼ 0.6 V of the cSi cell, and the generic photocurrent density JP ¼ 0.03 W/cm2. A. Derive an analytical formula that expresses the total area Amodule in cm2 of a solar panel, which is needed to power a load, in terms of the required load power PL. B. Test your result by applying it to the previous problem. Problem 4.74. You are given a lowcost lowpower ﬂexible (with the thickness of 0.2 mm) aSi laminate from PowerFilm, Inc., IA with the following parameters: the ﬁll factor of F ¼ 0.61, the singlecell opencircuit voltage, VOC ¼ 0.82 V, and the photocurrent density, JP ¼ 0.0081 W/cm2. A. Derive an analytical formula that expresses the total module area Amodule in cm2, which is needed to power a load, in terms of the required load power PL.
4.4 Analysis of Nonlinear Circuits: Generic Solar Cell 4.4.1 Analysis of Nonlinear Circuits: Load Line Method 4.4.2 Iterative Solution for Nonlinear Circuits Problem 4.75. A circuit shown in the ﬁgure below contains a nonlinear passive element. Using the load line method approximately, determine the voltage across the element and the current through it for the two types of the υi characteristic, respectively. 10.55 kW
I
a
9.5 V
+ 
+
Problem 4.72. A custom 100 W load (a DC motor) rated at 24 V is to be driven by a solar panel. A cSi photovoltaic sheet material is given, which has the opencircuit voltage of 0.6 V and the photocurrent density of JP ¼ 0.03 A/cm2. Outline parameters of a solar module (number of cells, cell area, and overall area) which is capable of driving the motor at the above conditions and estimate the overall panel size.
B. Compare your solution with the solution to the previous problem.
+
Outline parameters of a solar module (number of cells, cell area, and overall area) which is capable of driving the motor at the above conditions and estimate the overall panel size.
Circuit Analysis and Power Transfer
vi
V

Chapter 4
b
a) 1
I, mA
0 V, volts
1 0
b) 1
5
10
I, mA
0 V, volts
1 0
5
10
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Chapter 4
Problems
I ¼ 5 10
vi
V 1 ½A : exp 0:025 V
Using the iterative solution, determine the voltage across the element and the current through it. I
100 W
667 W
V

1V
b
a
+
+
9 mA
10
+
a
υi characteristic of the nonlinear element (the ideal Shockley diode) is
+ 
V
b
0 V, volts
10 0
vi

I, mA
Problem 4.79. Repeat the previous problem for the circuit shown in the ﬁgure below. The υi characteristic of the nonlinear element is the same. a
10
+
5
+
10
+
Problem 4.76. A circuit shown in the ﬁgure below contains a nonlinear passive element as a part of a current source. Using the load line method approximately, determine the voltage across the element and the current through it for the υi characteristic of the nonlinear element shown in the same ﬁgure.
8 mA
vi
Problem 4.77. Repeat the previous problem for the circuit shown in the ﬁgure that follows.
V
1 kW
b
+
8 mA
+
a
vi
1 kW
V
b
4.4.3 Application Example: Solving the Circuit for a Generic Solar Cell Problem 4.80. The I(V )dependence for a resistive load in a circuit is shown in the ﬁgure below.
I, mA
10
I, A 16A
0 V, volts
PL 10 0
5
10
Problem 4.78. A circuit shown in the ﬁgure below contains a nonlinear passive element. The
0
10V
V, Volts
A. At which value of the load voltage is the power delivered to the load maximized?
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Chapter 4
Circuit Analysis and Power Transfer
B. What is the related value of load resistance?
Problem 4.81. A hypothetic thermoelectric engine developed by US Navy has the I(V ) dependence shown in the ﬁgure below. a)
I, A 14.1A 10(2V)
1/2
PL
2V
0
b)
V, Volts
I, A 17.3A 10(3V)
1/2
PL V, Volts 0
3V
1. At which value of the load voltage is the power, PL, delivered to the load maximized? 2. What is the related value of load resistance for maximum power transfer?
Problem 4.82. Estimate the values of VMP and IMP versus VOC and ISC for a set of generic cSi solar cells. Every cell has VT ¼ 0.026V (room temperature of 25 C) and VOC ¼ 0.6 V. The ideality factor n in Eqs. (4.43)–(4.45) is allowed to vary over its entire range as shown in the Table below. n 1.00 1.25 1.50 1.75 2.00
VMP/VOC,%
IMP/IOC,%
F
IV200
Chapter 5
Chapter 5: Operational Ampliﬁer and Ampliﬁer Models Overview Prerequisites:  Knowledge of major circuit elements (dependent sources) and their i characteristics (Chapter 2)  Knowledge of basic circuit laws (Chapter 3) and Thévenin equivalent (Chapter 4) Objectives of Section 5.1:  Learn and apply the model of an operational amplifier including principle of operation, opencircuit gain, power rails, and input and output resistances  Correlate the physical operational amplifier with the amplifier circuit model  Establish the idealamplifier model  Learn the first practical amplifier circuit—the comparator Objectives of Section 5.2:  Understand and apply the concept of negative feedback to an operational amplifier circuit  Construct three canonic amplifier circuit configurations with negative feedback: the noninverting amplifier, the inverting amplifier, and the voltage follower  Understand the current flow in the amplifier circuit including the power transfer from the power supply to the load Objectives of Section 5.3:  Choose the proper resistance values for the feedback loop and learn how to cascade multiple amplifier stages  Learn about input/output resistances of the amplifier circuit and establish load bridging and load matching conditions important in practice  Find ways to eliminate the DC imperfections of the amplifier that become very apparent at high amplifier gains  Use an amplifier IC with a single voltage supply (a battery) Objectives of Section 5.4:  Obtain the initial exposure to differential signals and difference amplifiers  Build an instrumentation amplifier  Connect an instrumentation amplifier to a resistive sensor
© Springer Nature Switzerland AG 2019 S. N. Makarov et al., Practical Electrical Engineering, https://doi.org/10.1007/9783319966922_5
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Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
Objectives of Section 5.5:  Learn a general feedback system including closedloop gain and error signal  Apply the general feedback theory to voltage amplifier circuits  Construct current, transresistance, and transconductance amplifiers with the negative feedback Application Examples: Operational amplifier comparator Instrumentation amplifier in laboratory
Keywords: Operational ampliﬁer: (abbreviation opamp, integrated circuit, dual inline package, noninverting input, inverting input, output terminal, power terminals, offsetnull terminals, differential input voltage, opencircuit voltage gain, openloop voltage gain, openloop conﬁguration, closedloop conﬁguration, power rails, voltage transfer characteristic, railtorail, comparator, digital repeater, zerolevel detector, circuit model, input resistance, output resistance, ideal ampliﬁer, idealampliﬁer model, marking, summing point, commonmode input signal, differential input signal, summingpoint constraints, ﬁrst summingpoint constraint, second summingpoint constraint, sourcing current, sinking current, DC imperfections, input offset voltage, input bias current, input offset currents), Negative feedback, Feedback loop, Feedback as a dynamic process, Noninverting ampliﬁer, Inverting ampliﬁer, Voltage follower (buffer) ampliﬁer, Summing ampliﬁer, Digitaltoanalog converter, Binary counter, DCcoupled ampliﬁer, ACcoupled ampliﬁer, Capacitive coupling of an ampliﬁer, Gain tolerance of an ampliﬁer, Circuit model of a voltage ampliﬁer, Input resistance of ampliﬁer circuit, Output resistance of ampliﬁer circuit, Load bridging (impedance bridging), Load matching (impedance matching), Cascading ampliﬁer stages, Virtualground (integrated) circuit, Differential voltage of a sensor, Commonmode voltage of a sensor, Differential sensor, Singleended sensor, Difference ampliﬁer, Differential ampliﬁer circuit gain, Commonmode ampliﬁer circuit gain, Commonmode rejection ratio (CMRR), Unity commonmode gain stage, Instrumentation ampliﬁer, Load cell, Current ampliﬁer using opamp, Transconductance ampliﬁer using opamp, Transresistance ampliﬁer using opamp, Howland current source (Howland current pump) Linear feedback system: (forward gain? openloop gain, feedback gain, feedback factor, summing node, difference node, closedloop gain, error signal)
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Chapter 5
Section 5.1: Ampliﬁer Operation and Circuit Models
Section 5.1 Ampliﬁer Operation and Circuit Models The lowpower ampliﬁer integrated circuit (IC) is arguably the most widely employed discrete circuit component encountered in common electronic audio, control, and communication systems. Among ampliﬁers, the differential input, highgain ampliﬁer called the operational ampliﬁer (or simply opamp) has become a popular choice in many circuit applications. At this point, it is impossible for us to understand the internal operation of the ampliﬁer IC without basic knowledge of semiconductor electronics, especially the junction transistor studied in the following courses. Fortunately, the circuit model of an operational ampliﬁer does not require knowledge of the IC fabrication steps, nor does it require an understanding of the internal transistor architecture. Conceptually, operational ampliﬁers can be introduced early in the book, which enables us to immediately proceed toward our goal of designing and building practical circuits.
5.1.1 Ampliﬁer Operation Symbol and Terminals After the ampliﬁer chip is fabricated as an integrated circuit and the bond wires are attached, it is permanently sealed in a plastic package. Often the encasing is done in a dual inline (DIPN) package with N denoting the number of IC pins. Figure 5.1 on the right shows an example of a DIP package. One IC chip may contain several independent individual ampliﬁers. We start analyzing the ampliﬁer model by ﬁrst labeling the terminals and introducing the ampliﬁer circuit symbol (a triangle) as shown in Fig. 5.1 on the left. The ampliﬁer is typically powered by a dualpolarity voltage power supply with three terminals: V CC and common (ground) port of 0 V, see Section 3.2. The index C refers to the collector voltage of the internal transistors. +VCC v+ v
+

+
=
vout
VCC common
v v+
VCC
offset null
common
NC +VCC vout
Fig. 5.1. Terminals of the operation ampliﬁer (left); they also denote pins of the ampliﬁer IC package (see a common LM 741 chip on the right). All voltages are referenced with respect to a common port of the dualpolarity voltage supply.
The ampliﬁer has a total of ﬁve terminals, notably: 1. A noninverting input with the input voltage υþ with respect to common 2. An inverting input with the input voltage υ with respect to common 3. An output terminal with the output voltage υout with respect to common V203
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
4. Power terminal þV CC with a positive voltage VCC (+9 V) with respect to common 5. Power terminal V CC with the negative voltage V CC (9 V) with respect to common Each of the ﬁve terminals corresponds to a particular metallic pin of the IC package. All of the ampliﬁer’s terminals are used in an ampliﬁer circuit and none of them should be left disconnected. However, the chip itself could have some not connected (NC) terminals that maintain symmetry and which are used as heat sinks, see Fig. 5.1 on the right. Also note that a number of ampliﬁer ICs, including the LM74, may have extra terminals or pins, the socalled offset or offsetnull terminals. These terminals are used to control the input offset voltage (an imperfection) of the ampliﬁer. Historical: The abbreviation for the operational ampliﬁer is opamp; this abbreviation is not quite ofﬁcial but is used by most practitioners. The term operational ampliﬁer ﬁrst appeared in a 1943 paper by John R. Ragazzini, an American electrical engineer and ECE professor. One of his students introduced the terms inverting and noninverting inputs. One of his most notable students was Rudolf Kalman who became famous for the invention of the Kalman ﬁlters.
OpenCircuit or OpenLoop Voltage Gain Once the ampliﬁer chip is properly powered, its operation is quite simple: the output voltage is expressed through the two input voltages in the form υout ¼ Aðυþ υ Þ
ð5:1Þ
which is identical to the operation of the voltagecontrolled voltage source introduced in Section 2.4. Here,υþ υ is the differential input voltage to the ampliﬁer. The dimensionless constant A is called the opencircuit voltage gain of the ampliﬁer. Quite frequently, the term openloop gain is used and A is replaced by AOL. Equation (5.1), which will be called the ampliﬁer equation, is always valid. It does not matter if the ampliﬁer is in the openloop conﬁguration, (i.e., no feedback loop is present) or in a closedloop conﬁguration (a feedback loop is present; see the next section). The ampliﬁer IC is intentionally built in such a way as to provide the highest possible opencircuit gain; it is achieved using transistors connected in series such as the Darlington pair. Typically, A 105 108
ð5:2Þ
The exact gain value cannot be controlled precisely due to manufacturing tolerances. The opencircuit gain is often measured in V/mV. For example, the value of 160 V/mV corresponds to the gain value of 160,000. The opencircuit gain is difﬁcult to measure.
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Chapter 5
Section 5.1: Ampliﬁer Operation and Circuit Models
Power Rails and Voltage Transfer Characteristic in the OpenLoop Conﬁguration Two power interconnects of the ampliﬁer are often called rails. The term “rail” appears simply because the power interconnections are represented by two long horizontal wires in the circuit diagram connected to þV CC and V CC , respectively, which resemble long metal rails. The positive rail is þV CC , and the negative rail is V CC . The power rails are interfaced to a laboratory dualpolarity voltage power supply that also provides a common (ground) port to be used later. The output ampliﬁer voltage can never exceed the positive rail voltage or be less than the negative rail voltage. In other words, V CC υout V CC
ð5:3Þ
Should the output voltage found in Eq. (5.1) exceed VCC, it will be forced to VCC. Likewise, should the output voltage drop to less than V CC , it will be forced to V CC . In view of these physical constraints, Eq. (5.1) may be rewritten in the form υout ¼ Aðυþ υ Þ, υout ¼ þV CC , υout ¼ V CC ,
jυout j < V CC Aðυþ υ Þ > þV CC Aðυþ υ Þ < V CC
ð5:4Þ
Example 5.1: Plot to scale the output voltage of an operational ampliﬁer with an opencircuit gain of A ¼ 105 when the noninverting input voltage υþ changes from 1 mV to +1 mV and the inverting input voltage υ is set to zero. The ampliﬁer is powered by a 16V dual voltage supply. This plot will give us the voltage transfer characteristic of the openloop ampliﬁer. Solution: Ampliﬁer Eq. (5.4) gives the result shown in Fig. 5.2 by a thick piecewiselinear curve. Due to the extremely high openloop gain, the ampliﬁer output is almost always saturated. This means that, except for a very narrow domain of input voltages on the order of 0.2 mV, the output simply follows the power rail voltage, either positive or negative. This is a very remarkable feature of the openloop ampliﬁer.
Power Rails in Practice The power rail(s) of the ampliﬁer or the supply voltage is speciﬁed in the datasheet. For example, the LM358 ampliﬁer IC operates using a single supply 3 V to 32 V or dual supplies 1.5 V to 16 V. As we can see from this data, the ampliﬁer does not necessarily operate using a dual voltage supply; a single supply (“single rail”) can be used as well. The same ampliﬁer chip (e.g., LM358) can be used either with the single voltage supply or with a dual supply. This question, although less important in theory, is very important in practice. Also note that, in practice, the output never exactly reaches the positive or negative rail
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Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
voltages; there is always a voltage offset; it can vary from a minimum value of between 0.01 Vand 0.05 V for certain special ICs (called the railtorail ampliﬁers) all the way up to 1.8 V for other ampliﬁers (e.g., LM741). vout, V 20
+VCC
positive power rail
10
0
10 negative power rail
VCC
20 0.8
0.4
0
0.4
0.8
v+, mV
Fig. 5.2. Ampliﬁer output voltage in the openloop conﬁguration. The openloop gain is AOL ¼ 105 and the supply voltage is 16 V. Note that the scale for the input voltage is in mV.
5.1.2 Application Example: Operational Ampliﬁer Comparator A comparator is a circuit or a device that compares two input voltages and outputs a digital voltage (e.g., 10 V) as an indication of which input voltage is larger. Due to a very high gain, the operational ampliﬁer in the openloop conﬁguration shown in Fig. 5.3a may operate as a basic comparator. Figure 5.3b shows one possible application of the comparator: a digital repeater. We assume that υ ¼ V threshold ¼ 0. The input voltage to the comparator υþ ðt Þ is a weak noisy digital signal shown in Fig. 5.3b. This signal is compared to a threshold level of zero volts (the threshold voltage in Fig. 5.3). a)
b)
Vthreshold=0 V
Output voltage vout(t), V
15
v+(t)
10
A =1,000,000
vout
+

5 0
Vthreshold
5
0V
10 15 0
10
20
30
Input voltage v+(t), V
40 50 time, µs
Fig. 5.3. A simple operational ampliﬁer as a voltage comparator.
When the ampliﬁer opencircuit gain tends to inﬁnity (the transfer characteristic in Fig. 5.2 becomes a straight vertical line), Eq. (5.4) applied to the present case is reduced to
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Chapter 5
Section 5.1: Ampliﬁer Operation and Circuit Models υ þ ðt Þ > 0 υ þ ðt Þ < 0
υout ¼ þV CC , υout ¼ V CC ,
ð5:5Þ
Figure 5.3b shows the resulting output voltage for V CC ¼ 10 V. The weak input digital signal will thus be ampliﬁed and cleaned from noise, which is one major function of a digital repeater. In practice, dedicated comparators are used instead of this simple setup, which are much faster and have useful additional features. The comparator ampliﬁer may also be employed for other purposes such as a zerolevel detector. Exercise 5.1: In Fig. 5.3, the threshold voltage of the comparator ampliﬁer is changed to +5 V. What will be the output of the comparator circuit? Answer: 10 V at any time instant.
5.1.3 Ampliﬁer Circuit Model Circuit Model An equivalent circuit model of an ampliﬁer is shown in Fig. 5.4. This circuit model is a twoport electric network. It includes three single circuit elements: an ideal voltagecontrolled voltage source Aðυþ υ Þ (Section 2.4), input resistance Rin of the ampliﬁer, and output resistance Rout of the ampliﬁer. Amplifier
v+
+ v+  v
v
+
Rout Rin
+ 
A(v+
v)
vout= A(v+ v)

Fig. 5.4. Equivalent circuit model of an ampliﬁer is in the shadow box as a twoport network. No load is connected. The ground of the output terminal is the common port.
Analysis of the Ampliﬁer Circuit Model: Effect of Input/Output Resistances The input/output resistances of an ampliﬁer in Fig. 5.4 impose rather severe limitations on its desired operation. First, a large but ﬁnite input resistance always implies that some input current,iin ¼ ðυþ υ Þ=Rin , will ﬂow into the ampliﬁer as long as the input voltage signal is different from zero. Consequently, the ampliﬁer would require not only the input voltage but also a certain amount of input power. As a result, the ampliﬁer may appreciably load a sensor connected to its input, i.e., require more power than (a tiny) sensor can actually provide. Second, a ﬁnite output resistance limits the output current iout to the ampliﬁer; this resistance operates as a current limiting resistor which is studied in Chapter 3. Along with V207
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
this, it also leads to the fact that the voltage across any load connected to the ampliﬁer’s output will not be equal to the desired output voltage given by Eq. (5.1), except for an open circuit. These limitations are quantiﬁed when we consider a circuit shown in Fig. 5.5. The circuit includes the ampliﬁer model, an arbitrary source represented by its Thévenin equivalent υS, RS, and a load represented by its equivalent resistance RL. Amplifier
iin
RS vS
+ 
Rout
+ vx

iout
Rin
+ 
Avx
+ RL
vout

Fig. 5.5. Ampliﬁer circuit model with connected source and load resistances.
Using the voltage division principle twice, the output voltage in Fig. 5.5 is expressed as Rin RL A ð5:6Þ υout ¼ υS Rin þ RS RL þ Rout ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} υX
This result is quite different from the ideal behavior of the ampliﬁer described by the perfect ampliﬁcation of the source signal υout ¼ AυS
ð5:7Þ
using the available opencircuit gain A of the ampliﬁer. Exercise 5.2: For the ampliﬁer circuit in Fig. 5.5 with A ¼ 1000, determine the output voltage given that υS ¼ 1 mV, RS ¼ 50 Ω, and RL ¼ 50 Ω for two cases: A. Rin ¼ 1 MΩ and Rout ¼ 1 Ω. B. Rin ¼ 50 Ω and Rout ¼ 50 Ω. Answer: Case A: υout ¼ 0:98 V (which is close to the ideal behavior, υout ¼ 1:00 V). Case A: υout ¼ 0:25 V (three quarters of the voltage gain are lost).
According to Eq. (5.6), υout < AυS for any positive ﬁnite values of Rin and Rout. In order to make use of the full available opencircuit gain A of the ampliﬁer, we should:
V208
Chapter 5
Section 5.1: Ampliﬁer Operation and Circuit Models
1. Design Rin as large as possible, ideally an open circuit, that is, Rin ¼ 1
ð5:8Þ
2. Design Rout as small as possible, ideally a short circuit, that is, Rout ¼ 0
ð5:9Þ
In this and only this case, the equality υout ¼ AυS will be satisﬁed exactly.
5.1.4 IdealAmpliﬁer Model and First SummingPoint Constraint The ampliﬁer IC can then be described with a high degree of accuracy by using the socalled idealampliﬁer model. It is based on the best possible choices for input/output resistances as described by Eqs. (5.8) and (5.9), respectively. It is also based on the assumption that the openloop gain in Eq. (5.1) is made as high as possible, i.e., equal to inﬁnity. The idealampliﬁer model is an important theoretical and practical tool for the analysis of microelectronic ampliﬁer circuits. This model will be used in the following sections of this chapter and in subsequent chapters. We can summarize the model of an ideal operational ampliﬁer in concise form: 1. 2. 3. 4.
No current can ﬂow into the ampliﬁer (into either input terminal). The openloop gain A is inﬁnitely high. The input resistance Rin is inﬁnitely high. The output resistance Rout is zero.
Property 1 follows from property 3 and vice versa. One more condition of the idealampliﬁer model could be added, namely, that the power rails V CC are exactly reached when operated in saturation. The idealampliﬁer model does not use the accurate internal ampliﬁer circuit shown in Fig. 5.4 or in Fig. 5.5, respectively. Instead, a simple triangle symbol may be used for the ideal ampliﬁer, which is shown in Figs. 5.1 and 5.3. Exercise 5.3: Solve the previous exercise for the idealampliﬁer model. Answer: Case A, B: υout ¼ 1:00 V.
First SummingPoint Constraint The summing point of an ampliﬁer is the connection of the two inputs to the ampliﬁer. The commonmode input signal is the half sum of the two input voltages, ðυþ þ υ Þ=2. The differential input signal is the input voltage difference, υx ¼ υþ υ . Conditions applied to the ampliﬁer’s input are called summingpoint constraints. The ﬁrst summingpoint constraint is applied to the idealampliﬁer model. It states that no current can ﬂow into either of the ampliﬁer terminals as shown in Fig. 5.6. This is consistent with an inﬁnitely high input resistance. The condition of no input current into the ampliﬁer means that virtually no input power is necessary. For example, V209
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models summing point i+ v+
+
vout

vi
Fig. 5.6. The ﬁrst summingpoint constraint stipulates that no current ﬂows into the ideal ampliﬁer.
a tiny sensor, which does not deliver any appreciable power, could directly be connected to the input. The voltage from the sensor will still be accepted as the input of the ampliﬁer. This condition is a convenient abstraction of the idealampliﬁer model. In reality, a very small input current does exist, typically on the order of nanoamperes (nA) for common ampliﬁer ICs or picoamperes (pA) for ICs with an input JFET stage.
Realistic Values of Input/Output Resistances and Output Current How far off from reality is the assumption of inﬁnite input resistance? A review of the datasheets reveals that the input resistance of the older ampliﬁer IC (e.g., LM741, LM1458) varies from 0.3 to 6 MΩ. The input resistance of JFETinput stage ampliﬁers (TL082) is on the order of 1 TΩ (1012 Ω). Now, how realistic is the assumption of zero output resistance? Note that if the output resistance were exactly zero, the ampliﬁer would be able to source an inﬁnite current (power) into a lowresistance load. Clearly, we cannot expect a large output power from a physically small ampliﬁer IC. Therefore, we have to introduce a small internal output resistance, which appears to be on the order of 1–100 Ω. The corresponding output shortcircuit current of the common ampliﬁer ICs (LM741, LM1458, LM358) cannot exceed 40–60 mA; the current into a load is smaller. The output current of faster ampliﬁer ICs (TL082) is even smaller. If the load requires more current than the chip can provide, then the output voltage will notably be clipped.
V210
Chapter 5
Section 5.2: Negative Feedback
Section 5.2 Negative Feedback In most practical circuits, the ampliﬁer IC is not used in openloop conﬁguration. Engineers have modiﬁed the openloop condition into a negative feedback loop in order to set the gain to a desired value and ensure the ampliﬁer’s stability. This section provides you with all the essential knowledge needed to design an ampliﬁer with negative feedback. The mathematical model introduced in this section is based on two conditions imposed at the ampliﬁer’s input; we term them the summingpoint constraints: a) No electric current ﬂows into or out of the ampliﬁer inputs. b) The differential voltage at the ampliﬁer’s input is zero. The ﬁrst summingpoint constraint has already been introduced in the previous section. The second summingpoint constraint has yet to be derived. We will show that the two summingpoint constraints, along with KCL and KVL, will enable us to solve any ampliﬁer circuit that involves a negative feedback, no matter what the speciﬁc nature of the feedback loop is and regardless of whether it is DC, AC, or a transient circuit.
5.2.1 Idea of the Negative Feedback The idea of the negative feedback goes way back—we may say almost to the Stone Age. Take a wooden rod of 1–2 feet in length. Hold the rod in the vertical position at the tip of your ﬁnger. You will probably succeed. Now, close your eyes and try to do the same. You will most likely fail. The reason for the failure is a breakdown of the feedback loop. This loop is created by visual control of the rod’s position; you automatically apply a compensating acceleration to the bottom tip of the rod when it begins to fall. Another good example is driving a car and trying to stay in the center of the lane. The negative feedback for electronic ampliﬁers was ﬁrst invented and realized by Harold S. Black (1898–1983), a 29yearold American electrical engineer at Bell Labs. To many electrical engineers, this invention is considered perhaps the most important breakthrough of the twentieth century in the ﬁeld of electronics because of its wide applicability. We will construct simple ampliﬁer circuits of a given gain, using a resistive feedback loop. Being able to perform this task is already critical from the practical point of view. 5.2.2 Ampliﬁer Feedback Loop: Second SummingPoint Constraint We construct the feedback loop, as shown in Fig. 5.7, by connecting the output to the inverting input terminal. This was exactly the idea of Harold Black. The shadowed box in the feedback loop may represent one or more circuit elements. The feedback loop may be a simple wire, a resistance, a network of circuit elements (resistances, inductances, capacitances), etc. The negative feedback simply means that the output voltage, or rather a portion of it, is returned back to the inverting input.
V211
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models v+
+
vout

+
vx

v
Fig. 5.7. A feedback loop around an ampliﬁer.
Feedback as a Dynamic Process According to Eq. (5.1) of the previous section, the output voltage is proportional to υx ¼ υþ υ
ð5:10Þ
where υx is the differential input voltage. Hence, υx or υx multiplied by a constant is returned to the input during a very short period of time. The feedback effect is inherently a very fast dynamic process, which leads to a static solution with quite remarkable properties. In the example that follows, we will attempt to model the effect of the feedback loop using several very simplifying assumptions. Example 5.2: An ampliﬁer with a feedback loop in Fig. 5.7 has υþ ﬁxed at +10 V. υ is equal to 0 V at t ¼ 0. We shall assume that 50 % of υþ is returned back to the input in 1 μs. How does the differential voltage υx change with time? Solution: 1. At t ¼ 0, υx ¼ 10 V 0 V ¼ 10 V. Next, 50 % of 10 V is returned in 1 μs. The voltage υ becomes equal to 0 V þ 5 V ¼ 5 V after 1 μs. 2. At t ¼ 1 μs, υx ¼ 10 V 5 V ¼ 5 V. Next, 50 % of 5 V is returned in 1 μs. The voltage υ becomes equal to 5 V þ 2:5 V ¼ 7:5 V after 2 μs. 3. At t ¼ 2 μs, υx ¼ 10 V 7:5V ¼ 2:5 V. Next, 50 % of 2.5 V is returned in 1 μs. The voltage υ becomes equal to 7:5 V þ 1:25 V ¼ 8:75 V after 3 μs. The process further continues so that voltage υx halves every microsecond. The process dynamic is shown in Table 5.1 and visualized in Fig. 5.8.
Table 5.1. Dynamics of the differential input voltage as a function of time for Example 5.2. Time, μs 0 1 2 3 4
υþ 10 V 10 V 10 V 10 V 10 V
υ 0V 5V 7.5 V 8.75 V 9.375 V
υx ¼ υþ υ 10 V 5V 2.5 V 1.25 V 0.625 V
V212
Chapter 5
Section 5.2: Negative Feedback vx 10V
5V 2.5V 0V 0
1
2
3
time, µs
Fig. 5.8. Dynamics of the differential input voltage as a function of time.
Both Table 5.1 and Fig. 5.8 make clear that the differential voltage υx decays to zero very rapidly, once the feedback loop is introduced. Hence we arrive at the second summingpoint constraint, which is valid only for the ampliﬁers with the negative feedback loop: the differential input voltage to the ampliﬁer is exactly equal to zero. The second summingpoint constraint is a close approximation to reality. Its accuracy depends on the value of the openloop gain of the ampliﬁer. If the openloop gain were inﬁnite, the second summingpoint constraint would be exact.
5.2.3 Ampliﬁer Circuit Analysis Using Two SummingPoint Constraints Next, we will solve an ampliﬁer circuit with negative feedback using the two summingpoint constraints (SPC): (i) no current into or out of the input ampliﬁer terminals and (ii) the differential input voltage is zero. The method of two summingpoint constraints is an accurate solution method for a wide variety of ampliﬁer circuits with the negative feedback. For ampliﬁer circuits with a single input, we will denote the input voltage to the ampliﬁer circuit by υin. Voltage υin may be equal to υþ or to υ , depending on ampliﬁer type to be used. Noninverting Ampliﬁer The ﬁrst ampliﬁer conﬁguration is the socalled noninverting ampliﬁer shown in Fig. 5.9. The feedback loop contains one resistance R2. Another resistance R1 shunts the inverting input to ground. The input voltage to the ampliﬁer circuit is the voltage υin with respect to ground, or common in this case, which implies the use of the dualpolarity voltage power supply. The output voltage with respect to common is υout. We apply the ﬁrst summingpoint constraint and KCL to the node “*” in Fig. 5.9 and obtain i1 ¼ i2
ð5:11Þ
Equation (5.11) is further transformed using Ohm’s law in the form υ* 0 υout υ* ¼ R1 R2
ð5:12Þ
V213
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models vin
+ vx
v* common
+
+

vout
i2

R2
i1 R1
common
common
Fig. 5.9. Circuit diagram of the noninverting ampliﬁer. A dual power supply is not shown.
The second summingpoint constraint yields υ* ¼ υin
ð5:13Þ
since υx ¼ 0. Equation (5.12) thus reads υin υout υin υout υin υin ¼ ) ¼ þ R1 R2 R2 R1 R2 As a result, we ﬁnd that the voltage inputtooutput relation becomes R2 υout ¼ 1 þ υin R1
ð5:14Þ
ð5:15Þ
The ampliﬁer circuit is solved: we have expressed the output voltage in terms of the input voltage and a resistor ratio. Equation (5.15) is the basic result in ampliﬁer theory. It shows that the feedback loop allows us to precisely control the gain with two arbitrary resistances. One chooses the proper resistance combination to achieve any ﬁnite gain between one (setting R2 ¼ 0) and the openloop (inﬁnite) gain (setting R1 ¼ 0). In the last case, the negative input terminal becomes grounded; the feedback loop is irrelevant and can be replaced by an open circuit so that the ampliﬁer again becomes the comparator. The gain expression R2 1 ð5:16Þ ACL ¼ 1 þ R1 is called the closedloop gain of the ampliﬁer; it clearly relates the output voltage to the input voltage. Equation (5.16) is a dramatic illustration of the negative feedback. We started with an ampliﬁer having a very large yet loosely predictable openloop gain. Through applying the negative feedback, we arrived at a gain that is much smaller than the openloop gain; however, it is controllable and stable. Equation (5.16) can be derived more simply using the voltage divider concept. Namely, resistors R1, R2 form a voltage divider between 0 V and the output voltage. Hence, the voltage at node (*) may be found. Equating this voltage to the input voltage gives us Eq. (5.16). V214
Chapter 5
Section 5.2: Negative Feedback
Exercise 5.4: Solve the circuit shown in Fig. 5.10, i.e., ﬁnd the output voltage υout with respect to common. R2 1 106 υin ¼ 1 þ 1 mV ¼ 197 mV. Answer: υout ¼ 1 þ R1 5:1 103
+
+
+

1 mV
vout

R2=1 MW R1=5.1 kW
Fig. 5.10. A noninverting ampliﬁer circuit with an input voltage of 1 mV.
Inverting Ampliﬁer The next ampliﬁer circuit conﬁguration is the inverting ampliﬁer shown in Fig. 5.11. Note that the input terminals are now ﬂipped. The negative feedback loop is still present; it involves resistance R2. Another resistance, R1, forms a voltage divider. R2

v*

i1
common
+ vx
i2
+
R1
vin
+ vout

common
Fig. 5.11. Circuit diagram of the inverting ampliﬁer; a dual power supply is used (not shown).
The input voltage to the ampliﬁer circuit is the voltage υin with respect to ground or common. The output voltage with respect to common is υout. To solve the ampliﬁer circuit, we use the same solution procedure as for the noninverting ampliﬁer. However, the ﬁnal result will be quite different. We apply the ﬁrst summingpoint constraint and KCL to the node labeled “*” in Fig. 5.11 and again obtain i1 ¼ i2
ð5:17Þ
Equation (5.17) is transformed using Ohm’s law, υin υ* υ* υout ¼ R1 R2
ð5:18Þ
V215
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
The second summingpoint constraint yields υ* ¼ 0 since υx ¼ 0. Equation (5.18) then gives R2 υout ¼ υin R1
ð5:19Þ
The ampliﬁer circuit is solved: we have expressed the output voltage in terms of the input voltage. Equation (5.19) is another key result in ampliﬁer theory. The expression ACL ¼
R2 R1
ð5:20Þ
is also called the closedloop gain of the inverting ampliﬁer; the gain again relates the output voltage to the input voltage. It is now negative, which means that the output voltage is inverted. This circumstance is hardly important for the AC signals where the voltage inversion is equivalent to a phase shift of π radians or 180 degrees. The feedback loop of the inverting ampliﬁer also enables us to control the gain of the ampliﬁer with two standard resistors. We can choose the proper resistance combination to achieve any ﬁnite gain between zero (R2 ¼ 0) and negative inﬁnity (R1 ¼ 0). In Fig. 5.11 we clearly see how the ampliﬁer gain is controlled by the voltage divider with resistors R1 and R2. Exercise 5.5: Solve the invertingampliﬁer circuit shown in Fig. 5.12, i.e., ﬁnd the output voltage υout with respect to common. R2 1 104 1 mV ¼ 196 mV. Answer: υout ¼ υin ¼ R1 51
R2=10


R1=51
+
+ 1 mV
+v

out
Fig. 5.12. An inverting ampliﬁer circuit with an applied input voltage of 1 mV.
Voltage Follower or Buffer Ampliﬁer The third important member of the ampliﬁer family is the voltage follower or buffer ampliﬁer whose circuit is shown in Fig. 5.13. The negative feedback loop is just a wire.
V216
Chapter 5
Section 5.2: Negative Feedback vin
+ vx

+
+

vout

common
common
Fig. 5.13. Circuit diagram of the buffer ampliﬁer; a dual power supply is used.
The use of the second summingpoint constraint immediately leads to υout ¼ υin ,
ACL ¼ 1
ð5:21Þ
so that the gain of this ampliﬁer type is simply unity. Why do we need a unitygain ampliﬁer? The reason is that, while the buffer ampliﬁer in Fig. 5.13 passes the voltage without change, it requires virtually no current at the input (virtually no input power) but, at the same time, could source a signiﬁcant current (on the order of 20–40 mA) at the output, i.e., provide signiﬁcant output power. In order words, it becomes in a certain sense a power ampliﬁer. A simple example would be a capacitive sensor that cannot deliver currents on the order of 10 mA or even smaller currents; otherwise, the corresponding capacitor would immediately discharge. Such a sensor cannot directly be connected to an LED indicator that requires at least 10 mA. However, this sensor may deliver signiﬁcant voltages, on the order of 1–5 V, which do not need to be ampliﬁed. The use of a buffer ampliﬁer can nicely solve this connection problem. The above discussion directly leads us to the concept of input resistance of the ampliﬁer circuit studied in the next section. Exercise 5.6: Solve the voltagefollower circuit shown in Fig. 5.13, i.e., ﬁnd the output voltage υout with respect to common when the input voltage with respect to common is a) 1 V and b) 10 V. The ampliﬁer is powered by a 6V dual supply. Answer: a) υout ¼ 1 V; b) υout ¼ 6 V.
5.2.4 Mathematics Behind the Second SummingPoint Constraint The second summingpoint constraint might appear to be mysterious, at least at ﬁrst sight. How does the ampliﬁer accept the input signal if there is no current at the input and the differential input voltage is zero? Can we avoid using the second SPC, and at what cost? We will show that the second SPC is nothing but a handy tool to solve the ampliﬁer circuit with the negative feedback, with a high degree of accuracy. Mathematically, the second SPC gives a leading (and usually very accurate) term of what is known as an asymptotic expansion with regard to a small parameter, here the inverse openloop gain A1 . Let us now ignore the second SPC and derive the gain equation for the buffer ampliﬁer exactly. A similar derivation for the noninverting ampliﬁer is given as a homework problem. Looking at Fig. 5.13, we conclude that υ ¼ υout since the V217
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
negative ampliﬁer terminal is directly connected to the output. According to the ampliﬁer equation, Eq. (5.1) of the previous section, we have υout ¼ Aðυþ υ Þ ¼ Aðυin υout Þ
ð5:22Þ
Solving Eq. (5.22) for the output voltage yields υout ¼
A υin 1þA
ð5:23Þ
Using a Maclaurin series expansion, we obtain with A >> 1 the result A 1 ¼ 1 1=A 1 1 þ A 1 þ 1=A
ð5:24Þ
which is consistent with Eq. (5.21) and is very accurate since typically A > 105. A similar derivation holds for the noninverting (or the inverting) ampliﬁer conﬁguration. With this in mind, the second SPC is clearly optional. Instead, the ampliﬁer deﬁnition Eq. (5.1) may be used, along with the condition of the high openloop gain. However, it is rather tedious to repeat the asymptotic analysis every time; so we prefer to use the accurate and simple summingpoint constraint. The ﬁnite value of the opencircuit gain A becomes important for highspeed ampliﬁers with the feedback loop; see Chapter 10.
5.2.5 Current Flow in the Ampliﬁer Circuit The current ﬂow in the complete ampliﬁer circuit is illustrated in Fig. 5.14. The output current through the load resistance RL of the ampliﬁer circuit in Fig. 5.14 is provided by the dualpolarity power supply. In this sense, the ampliﬁer is also a “valve” (similar to its building block, the transistor), which “opens” the power supply in response to the lowpower (or virtually nopower) input voltage signal. In Fig. 5.14, you should note that standard resistor values (5 % or 1 % tolerance) may be slightly different from the values used in this ﬁgure for convenience. We consider the positive input voltage of 100 mV in Fig. 5.14a ﬁrst. The noninverting ampliﬁer has a closedloop gain ACL equal to 50. The output voltage is thus +5 V, which is the push mode. The load current of 10 mA is found from Ohm’s law. The feedback current of 0.1 mA is found using the second SPC and Ohm’s law. The power supply current is the sum of both. The feedback current controls the gain of the ampliﬁer, and the load current drives the load. The overall ampliﬁer circuit efﬁciency (neglecting the loss in the IC itself) depends on the ratio of these two currents. Therefore, we should keep the feedback current small. The power current path is shown in Fig. 5.14a by a thick trace. The current at node A can only enter the upper power supply. Thus, it is the upper power supply being used. The ampliﬁer is operating in the “push” mode, i.e., the ampliﬁer sources the current. When the input voltage is negative as in Fig. 5.14b, the lower power supply is delivering power. Now, the ampliﬁer sinks the current; it is operating in the “pull” mode. V218
Chapter 5
Section 5.2: Negative Feedback
However, the ratio of the load current and feedback loop current remains the same, at least for the ideal ampliﬁer. Similar results are obtained for the inverting ampliﬁer and the voltage follower, respectively. a) positive input voltage  “push” vin
+

+ 100 mV

10.1mA 10.1mA
vout=5 V 10.1mA
0.1mA
0.1mA
100 mV
R2=49 k
10mA
RL=500
common
+ 9V

R1=1 k
A 0V common
0.1mA
10.1mA
common
+ 9V

not used
b) negative input voltage  “pull” vin
+

10.1mA
vout=5 V

100 mV
0.1mA
+
100 mV 0.1mA
R2=49 k
10mA
RL=500
R1=1 k
+ 9V
not used

common
10.1mA
+
A 0V common
0.1mA
common
9V
10.1mA
Fig. 5.14. Current ﬂow in the noninverting ampliﬁer circuit operating in the (a) push mode and (b) pull mode. The path of the (relatively high) load current is marked in bold.
5.2.6 MultipleInput Ampliﬁer Circuit: Summing Ampliﬁer Figure 5.15 shows an important ampliﬁer type on the basis of the inverting ampliﬁer—the summing ampliﬁer. A summing ampliﬁer performs a simple mathematical operation: it sums several weighted input voltages. The summing ampliﬁer is a prototype of the binaryweightedinput digitaltoanalog converter. According to KCL and to the ﬁrst summingpoint constraint, one has with reference to Fig. 5.15 iF ¼ i1 þ i2 þ i3
ð5:25Þ
On the other hand, the second summingpoint constraint (the differential voltage to the ampliﬁer is zero and the inverting input is the common or virtual ground) yields
V219
Chapter 5 i1 ¼
Operational Ampliﬁer and Ampliﬁer Models
υ1 υ2 , i2 ¼ , R1 R2
i3 ¼
υ3 R3
ð5:26Þ
in terms of input voltages υ1, υ2, υ3. Therefore, voltage υout in Fig. 5.15 found from Eq. (5.25) is now written in the form 0 υout iF ¼ ¼ i1 þ i2 þ i3 ¼ RF
υ1 υ2 υ3 þ þ R1 R2 R3
RF RF RF ) υout ¼ υ1 υ2 υ3 R1 R2 R3
ð5:27Þ
Example 5.3: An input to the ampliﬁer circuit in Fig. 5.15a is a timing sequence shown in Fig. 5.15b. Such a sequence is known as a binary counter; it represents all threebit binary numbers in an ascending order, with the time interval of 1 μs. The ampliﬁer circuit is characterized by RF ¼ 2 kΩ, R1 ¼ 40 kΩ, R2 ¼ 20 kΩ, and R3 ¼ 10 kΩ. Plot the absolute output voltage to scale. Solution: After plugging in the numbers, Eq. (5.27) is transformed to jυout j ¼ 0:05υ1 þ 0:1υ2 þ 0:2υ3
ð5:28Þ
Figure 5.15c shows the result. This is a staircase approximation of the straight line.
v1 v2 v3
R1
i1
v2
R2
i2
v3
R3
i3
iF
RF 0V

b)
v1
+v
+
a)

c)
Input
out
Output
2V
Vout=2.5e5*t
5V 0V 5V
1V
0V 5V 0V 0
1
2
3 4 5 time, s
6
7
8
0V 0
1
2
3 4 5 time, s
6
7
8
Fig. 5.15. (a) Circuit diagram of a summing ampliﬁer. (b) and (c) Typical input and output voltages.
A large collection of practical ampliﬁer circuits with the negative feedback exists. Some of them are DCcoupled ampliﬁers (considered here), some are intended for ampliﬁcation of AC voltage signals with zero mean (the socalled ACcoupled ampliﬁers).
V220
Chapter 5
Section 5.3: Ampliﬁer Circuit Design
Section 5.3 Ampliﬁer Circuit Design Now that the theory of the negative feedback loop has been established, we can turn our attention to the laboratory. Our hope is to be immediately successful with our designs. However, a number of questions will arise almost instantly. They raise issues such as how to choose the resistor values, how to connect the sensor as part of the input load, and how to use an ampliﬁer chip with a single power supply (a battery).
5.3.1 Choosing Proper Resistance Values There are several rules regarding how to choose resistances R1, R2 controlling the feedback loop in both noninverting and inverting conﬁgurations. They are: 1. Resistances R1, R2 cannot be too small. Imagine that in Fig. 5.14 of Section 5.2, the resistor values are changed to R1 ¼ 1 Ω, R2 ¼ 49 Ω. The same noninverting gain will be achieved and the same output voltage will be obtained. However, the feedback loop current now becomes 100 mA instead of 0.1 mA. The generalpurpose opamp chips are not capable of delivering such large currents. Furthermore, the ohmic losses in the feedback loop become high. Therefore, one should generally use R1 , R2 50 100 Ω
ð5:29aÞ
2. Resistances R1, R2 cannot be too large. Let us assume that resistance R2 equals 100 MΩ. This means that this physical resistor and the feedback loop represent almost an “open circuit.” Unwanted electromagnetic signals may couple into such a circuit through the related electric ﬁeld difference across its terminals. This effect is known as capacitive coupling. Furthermore, the very large resistances increase the parasitic effect of the input offset current. Plus, very large resistances are unstable—their values depend on moisture, temperature, etc. Therefore, one should generally use R1 , R2 1 MΩ
ð5:29bÞ
3. When a precision design is not warranted, inexpensive 5 % tolerance resistors may be used. Otherwise, 1 % or even 0.1 % tolerance resistors are employed. Moreover, in lieu of ﬁxed resistors, we may use one or two potentiometers to make the gain adjustable. 4. The load resistance should be sufﬁciently large in order not to overdrive the ampliﬁer. A good choice is RL 100 Ω
ð5:29cÞ
This requires an output current of 20 mA at υout ¼ 2 V when RL is exactly 100 Ω. If RL < 100 Ω, the ampliﬁer output voltage may decrease compared to the expected value due to ampliﬁer’s inability to source/sink sufﬁcient current.
V221
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
Example 5.4: A noninverting ampliﬁer circuit with a gain of 11 is needed in the conﬁguration depicted in Fig. 5.16. Identify one set of proper resistance values. Solution: To satisfy Eqs. (5.29), we simply choose the round numbers R1 ¼ 1 kΩ, R2 ¼ 10 kΩ, RL ¼ 100 Ω
ð5:30Þ
However, other choices are indeed possible. For example, the set R1 ¼ 100 kΩ, R2 ¼ 1 MΩ, RL ¼ 100 Ω
ð5:31Þ
will solve the problem too.
+ 100 mV> 1 ) ACL ¼
1 > 1, one obtains xe 0
ð5:66Þ
V243
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
As applied to the ampliﬁer circuits, Eq. (5.66) is exactly the second summingpoint constraint or the condition of the zero differential input ampliﬁer voltage under presence of the negative feedback. Exercise 5.10: The openloop gain A in Fig. 5.36 is 10,000. The forward gain block is used in the closedloop conﬁguration with the feedback factor β of 0.1. Determine the error signal xe if the input voltage signal is 1 mV. Answer: xe ¼ 0:999 μV 1 μV.
5.5.3 Application of General Theory to Voltage Ampliﬁers with Negative Feedback Two circuits shown in Fig. 5.37 are the buffer ampliﬁer circuit and the noninverting ampliﬁer circuit, respectively. The goal is to ﬁnd the closedloop gain ACL of the ampliﬁer circuit when the given opencircuit voltage gain A is large but ﬁnite. In this case, the second summingpoint constraint cannot be applied. Therefore, simple Eqs. (5.16) and (5.21) obtained previously need to be modiﬁed. a)
vin
b)
+ vout

vout
+
vin
vout
R1

vout
R2
Fig. 5.37. Two ampliﬁer circuits with negative feedback networks indicated by a shaded rectangle.
Both circuits from Fig. 5.37 have the form of the feedback system as in Fig. 5.36. The feedback network is indicated by a shaded rectangle. The signal x is now the voltage. Since A is given, the only problem is to ﬁnd the feedback factor, β. For the buffer ampliﬁer, the feedback factor is clearly one. For the noninverting ampliﬁer circuit, the feedback loop is the voltage divider, where β is determined by the resistance ratio. Note that the voltage divider model implies no current into ampliﬁer’s input terminals. Speciﬁcally, βυout is equal to R1 =ðR1 þ R2 Þυout . Therefore, β ¼ 1 buffer ampl: circuit;
β¼
R1 noninv: ampl: circuit R1 þ R2
ð5:67Þ
Substitution into Eq. (5.62) gives us two expressions for the closedloop gain: ACL ¼
A A noninv: ampl: circuit buffer ampl: circuit; ACL ¼ 1 1þA 1 þ AR RþR 1
ð5:68Þ
2
V244
Chapter 5
Section 5.5: General Feedback Systems
The ﬁrst equation (5.68) coincides with Eq. (5.23) obtained in Section 5.2 using the accurate circuit analysis. So does second equation (5.68) when we repeat the same analysis for the noninverting ampliﬁer conﬁguration. If A ! 1, then the simple gain expressions—Eqs. (5.16) and (5.21)—derived with the help of the second summingpoint constraint are obtained from Eqs. (5.68). The analysis of the inverting ampliﬁer requires more efforts since this ampliﬁer type is not exactly the voltage ampliﬁer but rather a transresistance ampliﬁer considered next. Exercise 5.11: The openloop (opencircuit) gain A of a noninverting ampliﬁer circuit with R1 ¼ 1 kΩ, R2 ¼ 9 kΩ is 10,000. Determine the closedloop gain. Answer: ACL ¼ 9:99, which is by 0.1 % different from ACL ¼ 1 þ RR21 ¼ 10.
Last but not least, we emphasize another signiﬁcant advantage of the negative feedback. When the input and output resistances of the ampliﬁer model in Fig. 5.5 have ﬁnite values (which occurs in practice), the negative feedback loop effectively increases the input resistance and decreases the output resistance, i.e., makes the entire ampliﬁer circuit look closer to the idealampliﬁer model.
5.5.4 Voltage, Current, Transresistance, and Transconductance Ampliﬁers with the Negative Feedback At the end of this short section, we consider four basic ampliﬁer circuits with negative feedback, which correspond to the four basic dependent sources studied in Chapter 2: voltage ampliﬁer, transconductance ampliﬁer, transresistance ampliﬁer, and current ampliﬁer. Figure 5.38 shows the corresponding circuit diagrams. Load resistance RL is introduced for the transconductance ampliﬁer and the current ampliﬁer, respectively, where the output is the load current. Although every ampliﬁer circuit may be represented in the form similar to the feedback diagram in Fig. 5.36 and analyzed accordingly, only a simpliﬁed treatment will be given here. It utilizes the condition A ! 1 and the resulting second summingpoint constraint. Using the both summingpoint constraints, we may obtain, with reference to Fig. 5.38, R2 voltage amplifier ð5:69aÞ υin υout ¼ 1 þ R1 iout ¼ GF υin , GF ¼ 1=RF
transconductance amplifier
ð5:69bÞ
υout ¼ RF iin
transresistance amplifier
ð5:69cÞ
V245
Chapter 5
iout
Operational Ampliﬁer and Ampliﬁer Models
R2 iin ¼ 1þ R1
ð5:69dÞ
current amplifier
Note that other more elaborate circuits may be considered; some of them are analyzed in the corresponding homework problems. voltage amplifier
a)
transconductance amplifier
b)
+
+


vout
RL
vin
+ 
vin
R2 R1
c)
RF
transresistance amplifier RF
d)
current amplifier R1 R2 RL
iout


+
+ iin
iout
+ 
vout iin
Fig. 5.38. Four basic ampliﬁer types with negative feedback.
V246
Chapter 5
Problems
Summary Ampliﬁer circuit Openloop operational ampliﬁer (comparator)
Operation
Formulas Operation with V CC power rails: υout ¼ Aðυþ υ Þ,
jυout j < V CC
Opencircuit (openloop) gain A is very high
Ampliﬁer circuit model
– Valid for any voltage ampliﬁer or voltage ampliﬁer circuit, but Rin, Rout, A are different in every case; – Ideal ampliﬁer model (useful simpliﬁcation): Rin ¼ 1, Rout ¼ 0, A ¼ 1
Negative feedback for the idealampliﬁer model: differential input voltage is zero (2nd SPC) For idealampliﬁer model: Noninverting ampliﬁer υout ¼ ACL υin , ACL ¼ 1 þ
R2 R1
Rin ¼ 1, Rout ¼ 0 1 R1 Exact: ACL ¼ A 1 þ A R1 þ R2 Voltage follower (buffer) ampliﬁer
For idealampliﬁer model: υout ¼ ACL υin , ACL ¼ 1 Rin ¼ 1, Rout ¼ 0 Exact: ACL ¼
Inverting ampliﬁer
A 1þA
For idealampliﬁer model: υout ¼ ACL υin , ACL ¼
R2 R1
Rin ¼ R1 , Rout ¼ 0
ACL
Exact: R2 R2 1 ¼ A Aþ1þ R1 R1 (continued)
V247
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
Cascaded ampliﬁer
Load bridging
– Gains of individual stages multiply; – Input resistance of the ampliﬁer circuit is the input resistance of stage 1; – Individual stage gain should not exceed 100
Input load bridging versus input load matching – Source (sensor) sees the ampliﬁer as an open circuit: υout ¼ υS ACL – No current from the source can ﬂow into ampliﬁer circuit
Load matching
– Source (sensor) sees the ampliﬁer as resistance Rin ¼ R1 : υout ¼
Rin υS ACL RS þ Rin
– Matching condition RS ¼ Rin is important for highfreq. circuits
Noninverting ampliﬁer
Inverting ampliﬁer
DC imperfections and their cancellation – Shortcircuited output voltage may be trimmed to zero using the offset null terminal – Shortcircuited output voltage may be trimmed to zero by adjusting commonterminal voltage; – Extra resistance R eliminates the effect of the input bias current The same as above
(continued)
V248
Chapter 5
Summing ampliﬁer
Problems
Multipleinput ampliﬁer circuits – The summing ampliﬁer sums several weighted input voltages: υout ¼
RF RF RF υ1 υ2 υ3 R1 R2 R3
– Used as a prototype of the digital to analog converter Difference ampliﬁer
– True difference ampliﬁer: R2 R4 R2 ¼ , υout ¼ ðυa υb Þ R1 R3 R1 – Rejects commonmode voltage – For the general difference ampliﬁer circuit see Eq. (5.52c)
Unity commonmode gain stage
– Differential gain: υa * υb * ¼ AD ðυa υb Þ AD ¼ 1 þ
R4 R3
– Commonmode gain: υa * þ υb * ¼ ACM ðυa þ υb Þ ACM ¼ 1 Instrumentation ampliﬁer υout
Output voltage: R2 R4 ¼ 1þ ðυa υb Þ R1 R3
Closedloop differential gain: R2 R4 1þ ACL ¼ R1 R3 Simpler instrumentation ampliﬁer
– Rejects commonmode voltage Output voltage: R2 R2 ðυa υb Þ υout ¼ 2 þ 1 þ RG R1 – Closedloop differential gain: R2 R2 ACL ¼ 2 þ 1 þ RG R1 – Rejects commonmode voltage (continued)
V249
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
General feedback systems and ampliﬁers with negative feedback Closedloop gain: Signal—ﬂow diagram xout ¼ ACL xin , ACL ¼
A 1 1 þ Aβ β
Error signal: xe ¼ Transconductance ampliﬁer
1 xin 0 1 þ Aβ
Closedloop operation: iout ¼ GF υin , GF ¼ 1=RF Variations of this simple circuit are possible
Transresistance ampliﬁer
Closedloop operation: υout ¼ RF iin Variations of this simple circuit are possible
Current ampliﬁer
Closedloop operation: R2 iout ¼ 1 þ iin R1 Variations of this simple circuit are possible
Howland ampliﬁer (current pump)
Closedloop operation: (R1 =R3 ¼ R2 =R4 ) iout ¼ GF ðυa υb Þ, GF ¼ 1=R2 Variations of this clever circuit are possible
V250
Chapter 5
Problems 5.1 Amplifier operation and circuit models 5.1.1 Ampliﬁer Operation
Problem 5.1. An operational ampliﬁer has ﬁve terminals. A. Sketch the ampliﬁer symbol. B. Name each of the opamp terminals and describe its function in one sentence per terminal. C. Can the ampliﬁer IC have more than ﬁve terminals? Explain. Problem 5.2. You may wonder about the meaning of the two letters preceding ampliﬁer marking, e.g., LM741. Each of the semiconductor companies has its own abbreviation, e.g., LM for an ampliﬁer designed and manufactured by the National Semiconductor Corporation (acquired by Texas Instruments in 2011), AD for an ampliﬁer manufactured by Analog Devices, MC for STMicroelectronics, TL for Texas Instruments, etc. The same chip, e.g., LM741, may be manufactured by several semiconductor chip makers. The part number is given by a numerical code that is imprinted on the top of the package. An MC1458 ampliﬁer IC chip is shown in the ﬁgure below. This IC is a dual operational ampliﬁer. In other words, one such IC package contains two separate operational ampliﬁers.
Problems Problem 5.3. What is the minimum number of pins required for: A. The dual operational ampliﬁer (the corresponding IC package contains two separate operational ampliﬁers)? B. The quad operational ampliﬁer (the corresponding IC package contains four separate operational ampliﬁers)? Problem 5.4. An operational ampliﬁer has an opencircuit gain of A ¼ 2 105 and is powered by a dual source of 10 V. It is operated in the opencircuit conﬁguration. What is the ampliﬁer’s opencircuit output voltage υout if A. υþ ¼ 0V, υ ¼ 0V B. υþ ¼ þ1V, υ ¼ þ1V C. υþ ¼ þ1V, υ ¼ 0V D. υþ ¼ 0V, υ ¼ 1V E. υþ ¼ þ1mV, υ ¼ 0V F. υþ ¼ 1mV, υ ¼ 0V G. υþ ¼ 10μV, υ ¼ 0V H. υþ ¼ 0V, υ ¼ 10μV Problem 5.5. Based on the solution to Problem 5.4, why do you think the operational ampliﬁer is seldom used in the openloop conﬁguration, at least in analog electronics? Problem 5.6. Using the website of the National Semiconductor Corporation, determine the maximum and minimum supply voltages (operating with the dualpolarity power supply) for the following ampliﬁer’s ICs: A. LM358 B. LM1458 C. LM741 Which ampliﬁer IC from the list may be powered by two AAA batteries?
A. Download the ampliﬁer’s datasheet from http://www.datasheetcatalog.com B. Redraw the ﬁgure to this problem in your notes and label the pins for the noninverting input, the inverting input, and the output of the operational ampliﬁer #1. C. Label pins for þV CC and V CC .
Problem 5.7. Plot to scale the output voltage of the operation ampliﬁer with an opencircuit gain A ¼ 5 104 when the noninverting input voltage υþ changes from 2 mV to +2 mV and the inverting input voltageυ is equal to 1 mV. The ampliﬁer is powered by a 12V dual voltage supply. Label the axes.
V251
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
Problem 5.8. Repeat the previous problem for A ¼ 5 105 .
5.1.2 Operational Ampliﬁer Comparator
Problem 5.9. In a circuit shown in the ﬁgure below, an operational ampliﬁer is driven by a 10V dual power supply (not shown). The opencircuit DC gain of the ampliﬁer is A ¼ 1, 000, 000. Sketch to scale the output voltage to the ampliﬁer when a) V threshold ¼ 30 mV b) V threshold ¼ þ30 mV Assume that the ampliﬁer hits the power rails in saturation. A=1,000,000
v+
Problem 5.10. Based on the solution to the previous problem, why do you think the operational ampliﬁer in the openloop conﬁguration may be useful for digital circuits? Problem 5.11. Solve Problem 5.9 when the input voltage is applied to the inverting input and the threshold voltage is applied to the noninverting input. Problem 5.12. The circuit shown in the ﬁgure is a zerolevel detector. An operational ampliﬁer in the openloop conﬁguration is driven by a 10V dual power supply (not shown). The opencircuit ampliﬁer gain is 100,000. Sketch the output voltage to scale. Assume that the ampliﬁer hits the power rails in saturation.
vout A =100,000
+

Vthreshold Input voltage v+, mV
v out

0V
80
Input and output voltages, V
60
15
40
10
20
5
0
0
20
5
40
+
v in
v in
10
0
10
20 30 time, ms
a) Output voltage vout, V
40
50
10 5 0 5 10 15 10
20 30 time, ms
b)
0
10
20 30 time, ms
40
50
Vthreshold=30mV
15
0
15
40
50
Output voltage vout, V
Vthreshold=+30mV
10
40
Problem 5.13. In a circuit shown in the ﬁgure below, an operational ampliﬁer is driven by a 15V dual power supply (not shown). The opencircuit gain of the ampliﬁer is A ¼ 100, 000. Sketch to scale the output voltage to the ampliﬁer when a) V threshold ¼ 0 mV b) V threshold ¼ þ4 mV
15 10 5 0 5 10 15 0
20 30 time, ms
50
V252
Chapter 5
Problems
Assume that the ampliﬁer hits the power rails in saturation. A =100,000
v+
A. Rin ¼ 100 kΩ and Rout ¼ 2 Ω. B. Rin ¼ 50 Ω and Rout ¼ 25 Ω. C. Rin ¼ 1 and Rout ¼ 0.
vout
+

Vthreshold
iout
iin
RS
0V
Input voltage v+, mV
+
8

6
vS
Rout
+ vx
+ 
Rin

Avx
RL
+ vout

4 2 0 2 4 20
30 time, ms
a) Output voltage vout, V
40
50
Vthreshold=0 mV
15 10 5 0 5 10 15 0
10
b)
20 30 time, ms
Output voltage vout, V
40
50
10 5 0 5 10 0
10
Problem 5.17 A. List all conditions of the idealampliﬁer model. B. What is the shortcircuit output current of the ideal ampliﬁer?
Vthreshold=+4 mV
15
15
Problem 5.16. Name one reason why we should attempt to: A. Make the input resistance (impedance) to the ampliﬁer as high as possible. B. Make the output resistance (impedance) to the ampliﬁer as low as possible.
20 30 time, ms
40
50
5.1.3 Ampliﬁer Circuit Model 5.1.4 IdealAmpliﬁer Model Problem 5.14 A. Draw the circuit model of an operational ampliﬁer. B. Describe the meaning of the ampliﬁer as the voltagecontrolled voltage source in your own words.
Problem 5.18 A. What is the ﬁrst summingpoint constraint? B. What is the equivalent formulation of the ﬁrst summingpoint constraint in terms of the input resistance (impedance) to the ampliﬁer? Problem 5.19. An ampliﬁer circuit is shown in the ﬁgure below. The ﬁrst summingpoint constraint applies. Determine current i2. R2=500 W R1=100 W 1V
0V

10
i2
v out
+
0
common
Problem 5.15. For an equivalent ampliﬁer circuit with A ¼ 1500 shown in the ﬁgure below, determine the output voltage given that υS ðt Þ ¼ 1 cos ω t ½mV , RS ¼ 50 Ω, RL ¼ 50 Ω for three cases:
Problem 5.20. An ampliﬁer circuit is shown in the ﬁgure below. The ﬁrst summingpoint constraint applies. An ideal operational ampliﬁer has an opencircuit gain of A ¼ 2 105 .
V253
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
Determine the output voltage, Vout. You are not allowed to use any of the materials of the next section! Hint: Denote the unknown voltage at node * by υ*, express υ* in terms of υout, and then solve for υout. R2=500 R1=100
i2
v out

+
1V i1
common
Problem 5.21. An ECE laboratory project uses the LM358 ampliﬁer IC. A. What semiconductor company has developed this chip? B. Is the chip from the lab project necessarily manufactured by this company? (See http://www.datasheetcatalog.com/ for manufacturers’ datasheets related to this product.) C. Use the DigiKey distributor’s website and estimate average cost for this ampliﬁer chip (DIP8 package) in today’s market. Problem 5.22. An ECE laboratory project uses the TL082 ampliﬁer IC. A. What semiconductor company has developed this chip? B. Is the chip from the lab project necessarily manufactured by this company? (See http://www.datasheetcatalog.com/ for manufacturers' datasheets related to this product.) C. Use the DigiKey distributor’s website to estimate the average cost for this ampliﬁer chip (DIP8 package) in today’s market.
5.2 Negative Feedback 5.2.2 Ampliﬁer Feedback Loop. Second SummingPoint Constraint 5.2.3 Ampliﬁer Circuit Analysis Using Two Summingpoint Constraints Problem 5.23 A. Name the two summingpoint constraints used to solve an ampliﬁer circuit. B. Which summingpoint constraint remains valid without the negative feedback?
Noninverting Ampliﬁer Problem 5.24 A. Draw the circuit diagram of the basic noninverting ampliﬁer conﬁguration. B. Accurately derive the expression for the ampliﬁer gain in terms of the resistances, assuming an ideal operational ampliﬁer. Problem 5.25. Using the two summingpoint constraints, solve the idealampliﬁer circuit shown in the ﬁgure if the input voltage has the value of 2 mV. A. Label and determine the currents in the feedback loop. B. Determine the output voltage of the ampliﬁer versus the common port. vin
+ 2 mV

pin 5
+

pin 7
+
vx
pin layout  LM1458(#2)

vout
+
pin 6

R2=51 kW common
R1=100 W
common
V254
Chapter 5
Problems R2=33 kW pin layout  LM1458 (#1)
R1=1 kW
pin 2

vin
+
pin 1
vout
+
Problem 5.26. Determine the output voltage of the ideal operational ampliﬁer shown in the ﬁgure. The ampliﬁer is driven by a 10V dual power supply.
0.5V

+
pin 3

pin layout  LM1458 (#1)
vin
+ 0.5 V

pin 3
+
pin 1
+
vx


vout
common common
+
pin 2

Voltage Follower
R2=100 kW common
R1=1 kW
common
Inverting Ampliﬁer Problem 5.27 A. Draw the circuit diagram of the basic inverting ampliﬁer conﬁguration. B. Give the expression for the ampliﬁer gain in terms of the resistances, assuming an ideal operational ampliﬁer.
Problem 5.30 A. Using only the ﬁrst summingpoint constraint (SPC), solve the circuit shown in the ﬁgure, i.e., determine the output voltage of the ampliﬁer versus the common port. B. What function does this ampliﬁer have? Why is it important? pin layout  LM1458 (#1)
vin
+
+
Problem 5.28. Using the two summingpoint constraints, solve the idealampliﬁer circuit shown in the ﬁgure that follows if the input voltage is 1 mV. A. Label and determine the currents in the feedback loop. B. Determine the output voltage of the ampliﬁer versus the common port. R2=10 kW R1=100 W
+
+
1 mV

pin 5
pin 7
1 mV

vout pin 1

+
pin 2

common
Exercises on the Use of the Negative Feedback Problem 5.31. (A review problem) For three basic idealampliﬁer circuits:
pin layout  LM1458 (#2)
pin 6

vin
pin 3
vout
+

common common
Problem 5.29. Determine the output voltage of the ideal operational ampliﬁer shown in the ﬁgure. The ampliﬁer is driven by a 10V dual power supply.
Inverting ampliﬁer Noninverting ampliﬁer Voltage follower
(each includes negative feedback) present 1. A circuit diagram 2. Expression for the ampliﬁer gain Problem 5.32. Determine the output voltage of ampliﬁer conﬁgurations shown in the ﬁgure that follows. The ampliﬁer is powered by a 9V dualpolarity voltage power supply. Assume an ideal operational ampliﬁer.
V255
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
a) vin
vout
+
+
+

1 mV

common
common
b) v in
Problem 5.34. Each of the circuits shown in the ﬁgures below employs negative feedback. Find the output voltage Vout vs. ground (or common). Hint: The ground symbol in the ampliﬁer circuit usually has the same meaning as the common port.
vout
+
+
+

1 mV
a)
4 kW 1 kW



+
+ common
1V
a)
+
vout
+  7V

+


+
+
vout

50 W
1 kW
+


+
+
vout
+  3V
+ 5 kW
1 kW
d)
+

3 kW
+
vout

3.5 mA
2 kW
+
vout


Problem 5.35. Each of the circuits shown in the ﬁgures below employs an inverting ampliﬁer. 1. Solve each circuit (ﬁnd υout) with an input voltage of 1 mV. 2. Based on this solution, ﬁnd the closedloop voltage gain ACL of the corresponding ampliﬁer circuit.
1V

5 kW
d)
1 kW
c)
2 mA
2 mA
1V
b)
2 kW
1 kW
b)
c)
3 kW
2 mA
3 mA


Problem 5.33. Each of the circuits shown in the ﬁgures below employs negative feedback. Find the output voltage υout vs. ground (or common). Hint: The ground symbol in an ampliﬁer circuit usually has the same meaning as the common port.
vout
+ 
+
common
+
vout

+
vout

V256
Chapter 5
Problems R2=5.1 kW
a) R1=100 W
vout

vin
+ b) R1=100 W
vout

vin
R2=5.1 kW
A. Find the value of the output current iout if the input current is 1 mA, R1 ¼ 9 kΩ, R2 ¼ 1 kΩ. B. Why do you think this ampliﬁer type is known as the current ampliﬁer? To answer this question quantitatively, analytically express the output current iout (current through the load) in terms of the unknown input current iout and two arbitrary resistor values, R1,2.
+ iin
RL=51 W

+
c)
R2=5.1 kW
RL
iout
R1
R1=100 W
vout

vin
R2
+ RL=51 W
R1=100 W
R2=5.1 kW
d) R1=100 W
vout

vin
5.2.4 Mathematics Behind the Second SummingPoint Constraint
+ R1=100 W
RL=51 W
Problem 5.36. An inverting ampliﬁer that achieves highgain magnitude with a smaller range of resistance values is shown in the ﬁgure below. Find its output voltage υout vs. ground (or common port) and the resulting ampliﬁer gain.
Problem 5.38 A. Derive an expression for the closedloop gain of the noninverting ampliﬁer based only on the þdeﬁnition of the output voltage υout ¼ A υin υin , without using the second summingpoint constraint. B. Determine the exact gain value when A ¼ 2 105 R1 ¼ 1 kΩ,R2 ¼ 9 kΩ +
vin
1 kW vin=1 mV
v* vout
+
vout
*
R2
R1

+
1 kW

10 kW
10 kW
Problem 5.37. The ampliﬁer circuit shown in the ﬁgure employs negative feedback.
Problem 5.39 A. Derive an expression for the closedloop gain of the inverting ampliﬁer based only on the deﬁnition of the output voltage υout
V257
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models
¼ A υþ in υin , without using the second summingpoint constraint. B. Determine the exact gain value when
vin=1 V
vout
+

RL=100 W
A ¼ 2 105
R2=4 kW
R1 ¼ 1 kΩ,R2 ¼ 10 kΩ
+
R1=1 kW
common 0 V
5.2.5 Current Flow in the Ampliﬁer Circuit
Hint: Change the polarity of the input voltage and the voltage sign if you have trouble operating with negative values.
vout
+
Problem 5.42. The ampliﬁer shown in the ﬁgure below is powered by a 9V dualpolarity voltage power supply. A. Redraw the ampliﬁer schematic in your notes. B. Show the current direction in every wire of the circuit by an arrow and write the corresponding current value close to each arrow. R2=10 kW
vin R1=1 kW
vout


100 mV
+100 mV
R2=49 kW
+
R1=1 kW
common 0 V
RL=100 W
+ 
RL=500 W

+
+

Problem 5.40. The ampliﬁer circuit shown in the ﬁgure below is powered by a 9V dualpolarity voltage power supply. A. Redraw the ampliﬁer schematic in your notes. B. Show the current direction in every wire of the circuit by an arrow and write the corresponding current value close to each arrow.
+
+ + 

Problem 5.41. Repeat the previous problem for the circuit shown in the ﬁgure below.
Problem 5.43. Repeat the previous problem when the input voltage to the ampliﬁer is 100 mV.
V258
Chapter 5
Problems
5.2.6 MultipleInput Ampliﬁer Circuit: Problem 5.46. By solving the ampliﬁer circuit shown in the following ﬁgure, Summing Ampliﬁer Problem 5.44. By solving the ampliﬁer circuit shown in the ﬁgure, ﬁll out the table that follows. Assume that R0 ¼ RF =4, R1 ¼ RF =2, and R2 ¼ RF .
D2
R2
iF
0V
2R
vout
0V
common
υout, V
D 0, V 0 5 0 5 0 5 0 5
R
0V
Problem 5.48. The noninverting ampliﬁer shown in the ﬁgure below has been wired in laboratory. A. Do you have any concerns with regard to this circuit? B. If you do, draw the corrected circuit diagram. +

v in=100mV
R
c
+ 10 10

2R
0V
Problem 5.47. State the limitations on the feedback resistances and the output load resistance of an ampliﬁer circuit.
2R
2R
b
c
R
D2=5 V
0V a
b
5.3.1 Choosing Proper Resistance Values
+
2R
R
5.3 Amplifier Circuit Design
Problem 5.45. The ampliﬁer circuit shown in the ﬁgure below employs negative feedback. This conﬁguration is known as a threebit digitaltoanalog converter (DAC) on the base of an R/2R ladder. By solving the ampliﬁer circuit, determine its output voltage in terms of resistances R, RF, given the input voltages D0 ¼ 0 V, D1 ¼ 0 V, D2 ¼ 5 V. D1=0 V
a
vout
common
D1, V 0 0 5 5 0 0 5 5
D0=0 V
RF
2R
2R
+ 2R
D2, V 0 0 0 0 5 5 5 5
D2

R1
D1
+
D1
D0
RF

R0
+
D0
determine its output voltage in terms of resistances R, RF, given the input voltages D0 ¼ 0 V, D1 ¼ 5 V, D2 ¼ 0 V.
iF
1
RF
common
vout
v out

5.3.2 Model of a Whole Ampliﬁer Circuit 5.3.3 Input Load Bridging or Matching Problem 5.49. For three basic ampliﬁer circuits
0V
0V
Inverting ampliﬁer Noninverting ampliﬁer Voltage follower
V259
Operational Ampliﬁer and Ampliﬁer Models
Problem 5.50 A. Explain in your own words the concept of load bridging (impedance bridging). B. Which ampliﬁer, the noninverting or inverting, should be subject to load bridging?
R2=5.1 kW v in
R1=100 W
v out

(each includes negative feedback), present 1. A circuit diagram 2. An expression for the closedloop ampliﬁer circuit gain 3. An expression for the input resistance (impedance) 4. An expression for output resistance (impedance)
+
Chapter 5
R*=100 W
RL=51 W
5.3.4 Cascading Ampliﬁer Stages Problem 5.55. For the ampliﬁer circuit shown in the ﬁgure, ﬁnd the output voltage and the input resistance (show units). R2=10 k R1=100

+
+
1 mV=vin

R4=10 k R3=1 k

+
Problem 5.51. An electromechanical sensor is given by its Thévenin equivalent wherein the sensor voltage υS is small. The sensor’s equivalent resistance RS may vary in time; but it is always less than 1 kΩ. An inverting ampliﬁer is needed that generates an ampliﬁed version of the sensor’s voltage. The output voltage should be 100υS . Draw the corresponding circuit diagram and specify one possible set of resistor values.
+

Problem 5.52. Construct an ampliﬁer circuit matched to a 100Ω load. The ampliﬁer’s gain is jACL j ¼ 100. The sign of the gain (positive or negative) is not important and the input AC signal. Hint: Multiple solutions many exist. Present at least two solutions. Problem 5.53. Find the input resistance (impedance) to the ampliﬁer circuit shown in the ﬁgure below. vin
200
+
vout
Problem 5.56. A sensor with Thévenin voltage (source voltage) υS ¼ 2:5 mV shown in the ﬁgure is to be connected to an ampliﬁer circuit. An ampliﬁed replica of the sensor’s voltage, υout 1000 υS , is needed at the circuit’s output. A. Present one possible circuit diagram and specify all necessary resistor values. B. Present another (distinct) circuit diagram and specify all necessary resistor values.

R2=51 k 200
RS=25 kW R1=1 k
vS
+ 0V
Problem 5.54. Find the input resistance (impedance) to the ampliﬁer circuit shown in the ﬁgure below.
Problem 5.57. An ampliﬁer in the conﬁguration shown in the ﬁgure below is connected to a
V260
Chapter 5
Problems
sensor with Thévenin (source) voltage υS ¼ 25 mV. An ampliﬁed replica of the sensor’s voltage, υout 100 υS , is needed at the output. A. Do you have any concerns with regard to this circuit? B. If you do, draw an appropriate circuit diagram. vin RS=50 k vS
+ 
Problem 5.61. An ampliﬁer circuit is needed with a positive gain of 5000 5 %. The input resistance (impedance) should be as high as possible. Present one possible circuit diagram and specify the necessary resistor values including tolerance.
5.3.5 Ampliﬁer DC Imperfections and Their Cancellation Problem 5.62. Determine the output voltage to nonideal operational ampliﬁer circuits (with the nonzero input offset voltage) shown in the ﬁgures below. actual opamp
a)
0V
ideal opamp
10 M
10mV
vout
+
1mV
vout
+

+
100 k

+

+


1
5 kW=R2 1 kW=R1
Problem 5.59. An ampliﬁer circuit is needed with the closedloop gain ACL ¼ þ10, 000. The input resistance (impedance) to the circuit should be as high as possible. Present the corresponding circuit diagram and specify the necessary resistor values. Problem 5.60. An ampliﬁer circuit is needed with a positive gain of 1000 20 %. The input resistance (impedance) to the circuit should be 1 kΩ. Present the circuit diagram and specify the necessary resistor values including tolerance.
b)
actual opamp ideal opamp
10mV
+

+ 
+
Problem 5.58. An ampliﬁer circuit is needed with the closedloop gain ACL ¼ þ1000. The input resistance (impedance) to the circuit should be 5 kΩ. Present two alternative circuit diagrams and specify the necessary resistor values. The ﬁrst circuit must use inverting ampliﬁers and the second circuitnoninverting ampliﬁers.

vout
+
5mV
5 kW=R2 1 kW=R1
Problem 5.63. Determine the output voltage to nonideal operational ampliﬁer circuits (with nonzero input currents) shown in the ﬁgures below. The input terminal is connected directly to the common terminal (grounded). The strength of every bias current source is 100 nA. Hint: The upper bias current source does not contribute to the solution.
V261
Chapter 5 a)
Operational Ampliﬁer and Ampliﬁer Models R2
actual opamp ideal opamp
5 mV=vin
vout

+
R1
vout

+
0V
+
+

10 k
0V
2 kW
b)
Problem 5.67. In the previous problem, denote the terminal voltage of 4 mV by Voff, the input voltage of 5 mV by υin, the output voltage by υout, and the ampliﬁer gain by ACL. Derive an analytical formula that determines Voff in terms of υin given that the output voltage to the ampliﬁer is exactly zero.
actual opamp ideal opamp
0 V 5/3 kW
+

0V
4 mV
vout
+
10 kW 2 kW
Problem 5.64. For the ampliﬁer circuit shown in the ﬁgure below, determine the output voltage. Use R1 ¼ R2 ¼ 1 kΩ. 1 mV=v in
v out
+
+

+

Problem 5.68. For the circuit shown in the ﬁgure below: A. Determine the output voltage of the nonideal operational ampliﬁer circuit (with the nonzero input offset voltage). B. Does the ampliﬁer circuit really follow the idealampliﬁer circuit law: υout ¼ ACL υin ? C. What happens if the input voltage changes from 10 mV to 20 mV?

actual opamp
R2 0V
R1
ideal opamp
0V
Problem 5.65. In the previous problem, denote the terminal voltage of 2 mV by Voff, the input voltage of 1 mV by υin, the output voltage by υout, and the ampliﬁer gain by ACL. Derive an analytical formula that determines Voff in terms of Vin given that the output voltage υout to the ampliﬁer is exactly zero. Problem 5.66. For the ampliﬁer circuit shown in the ﬁgure below, determine the output voltage. Use R1 ¼ 1 kΩ and R2 ¼ 4 kΩ.
10 mV
+
v out
5mV

+
2mV
+


+
5kW=R2
0V 1 kW=R1
0V
6 mV
Problem 5.69. Solve the previous problem with the offset voltage in the feedback loop changed from 6 mV to 5 mV. Problem 5.70. In problem 5.68, denote the terminal voltage of 6 mV by Voff, the input
V262
Chapter 5
Problems
voltage of 10 mV by υin, the input offset voltage by VOS, the output voltage by υout, and the ampliﬁer gain by ACL. Derive an analytical formula that determines Voff in terms of VOS given that the output voltage υout to the ampliﬁer must exactly follow the idealampliﬁer gain law: υout ¼ ACL υin . R2
vin
VOS 

+
+
+
actual opamp ideal opamp
+
R1 0V
a)
R2
VOS 
actual opamp ideal opamp
+

R1
+
+
Voff R2
actual opamp ideal opamp
R1

Voff
+
+
VOS
2k
100 k
+ 
1k 0V
Problem 5.73. For the singlesupply ampliﬁer circuit shown in the ﬁgure: A. Determine the output voltage versus circuit ground (the negative terminal of the voltage power supply). B. Do you see any problem with this circuit? 12 V=VCC
+
0V
0V
0V 0.9 M
100 mV=vS


virtual ground circuit
+
+

b)

=RS
vout
vout
0V
+
9 V=VCC
Problem 5.71. For two nonideal operational ampliﬁer circuits (with the nonzero input offset voltage) shown in the ﬁgures below, determine the necessary offset voltage, Voff, which ensures that the output voltage, υout, to the ampliﬁer exactly follows the idealampliﬁer gain law: υout ¼ ACL υin . You need to express this voltage in terms of other circuit parameters that are given in ﬁgures a) and b).
+
100
+ 
Voff
vin
9 V=VCC
+

0V
SingleSupply
Problem 5.72. For the singlesupply ampliﬁer circuit shown in the ﬁgure: A. Determine the output voltage versus circuit ground (the negative terminal of the voltage power supply). B. What potential problem do you see with this circuit? How could you ﬁx it?
vout

vin
5.3.6 DCCoupled Ampliﬁer
100
+ 
=RS


0V 0.9 M
1 mV=vS
vout
100 k
1k
vx
12 V=VCC
+ 
vout
+

virtual ground circuit
100 k 0V
0V
V263
Chapter 5
Operational Ampliﬁer and Ampliﬁer Models Problem 5.79. Find the output voltage to the differenceampliﬁer circuits shown in the ﬁgures below. Assume the idealampliﬁer model and exact resistance values. a)
Problem 5.74. The model of an input signal from a threeterminal sensing device is shown in the ﬁgure below. What are the differential and commonmode voltages at terminals a and b?
1V
a
+ 
0.1 V
+ 
0.1 V
5 kW 1 kW
0.5 V
50 kW

0V
b)
5 kW 1 kW 1V

+
1V
+
10 kW
1V
+ 
v out

5.4.1 Differential Input Signal to an Ampliﬁer
+
5.4 Difference and Instrumentation Amplifiers
10 kW 25 kW
v out
+

b 0V 0V
Problem 5.75. The Wheatstone bridge in Fig. 5.27 is connected to V CC rails instead of þV CC and ground. Furthermore, R2 ¼ 1:1R1 , RðxÞ ¼ 1:1R3 , and V CC ¼ 6 V. What are the differential and commonmode voltages? Problem 5.76. The Wheatstone bridge in Fig. 5.27 is connected to ground and V CC rails instead of þV CC and ground. Given that R2 ¼ 1:05R1 , RðxÞ ¼ 1:05R3 , and V CC ¼ 6 V, determine the differential and commonmode voltages.
5.4.2 Difference Ampliﬁer Problem 5.77. Design a difference ampliﬁer with a differential gain of 20. Present the circuit diagram and specify one possible set of resistor values. In the circuit diagram, label the input voltages as υa, υb and express the output voltage in terms of υa, υb. Problem 5.78. Repeat the previous problem for a differential gain of 100.
Problem 5.80. Your technician needs to control a process using two sensors with output voltages υ1 and υ2, respectively. The weighted difference in sensor reading, υ ¼ 1υ1 0:5υ2 , is critical for the product quality. The technician reads voltage υ1 and then voltage υ2 and then uses a calculator to ﬁnd υ. Help the technician, i.e., sketch for him a differenceampliﬁer circuit that will directly output υ to the DMM. The negative terminal of the DMM is always grounded. Specify one possible set of resistor values. Problem 5.81. Repeat the previous problem when the weighted difference in sensor reading, υ ¼ 10υ1 5υ2 , needs to be processed. Problem 5.82. For the circuit shown in the ﬁgure, ﬁnd the output voltage if the input voltages are υb ¼ 1 V and υa ¼ 1:01 V, respectively. Assume the idealampliﬁer model and exact resistance values.
V264
Chapter 5
Problems 5.4.3 Instrumentation Ampliﬁer
10R
R vb
v out

+
+

va
10R
R
0V
Problem 5.83 A. For the circuits shown in the ﬁgures below, ﬁnd the output voltage if the input voltages are υa ¼ 1 V and υb ¼ 1 V, respectively Assume the ideal ampliﬁer and exact resistance values. B. What is the value of the commonmode gain in every case? a)
2R
R
vb v out

+
+

va
2R
1.05R
0V
b) 2R
1.05R
Problem 5.85 A. Why is the original difference ampliﬁer not used as an instrumentation ampliﬁer? B. Why is the circuit in Fig. 5.31 not used as the instrumentation ampliﬁer? Problem 5.86 A. Find the differential gain and the commonmode gain for the ampliﬁer circuit shown in Fig. 5.32. The differential output voltage is υa * υb *, and the commonmode output voltage is 0:5ðυa * þ υb *Þ. B. Find the differential gain and the commonmode gain for the ampliﬁer circuit shown in Fig. 5.31. Problem 5.87. Design an instrumentation ampliﬁer with a differential gain of 210. Present the corresponding circuit diagram and specify one possible set of resistance values. In the circuit diagram, label the input voltages as υa, υb and express the output voltage in terms of υa, υb.
vb v out

+
+

va
2R
R
0V
Problem 5.84. For the differenceampliﬁer circuit shown in the ﬁgure below, ﬁnd the differentialmode resistance (impedance) to the ampliﬁer. The differentialmode resistance is deﬁned as the ratio of a voltage of a power supply placed between terminals a and b to the current that ﬂows through this power supply.
Problem 5.88. Design an instrumentation ampliﬁer with a differential gain of 1010. Present the corresponding circuit diagram and specify one possible set of resistor values. In the circuit diagram, label the input voltages as υa, υb and express the output voltage in terms of υa, υb. Problem 5.89. The following voltages are measured: υa ¼ 3:750 V and υb ¼ 3:748 V. Find voltages versus circuit ground (common port of the dual supply) for every labeled node in the circuit shown in the ﬁgure below. The ampliﬁer circuit is powered by a 10 V dual supply. Assume exact resistance values and the idealampliﬁer model.
XR
R
v out

+
+
R
XR
0V
V265
Chapter 5
vb
2

100 kW

100 kW
100 kW
6
1
+ 10 kW
1.05 kW
5
3
100 kW
100 kW
va
3
Problem 5.90. Repeat the previous problem when node 1 is grounded. Problem 5.91. The following voltages are measured: υa ¼ 5:000 V and υb ¼ 5:001 V. Find voltages versus circuit ground (common port of the dual supply) for every labeled node in the circuit shown in the ﬁgure below. The ampliﬁer circuit is powered by a 10 V dual supply. Assume exact resistance values and the idealampliﬁer model.
6
5
100kW
+
25 kW
+
Problem 5.94 A. Find the output voltage for the ampliﬁer circuit shown in the ﬁgure below. B. Denote the input voltage of 0.1 V by υa, the input voltage of 0.08 V by υb, the 10kΩ resistor by R1, the 40kΩ resistor by R2, and the 10kΩ resistor by RG. Express the output voltage in the general form, in terms of two input voltages and the resistances. +
0.1 V

+

v out
40 kW
100 kW
25 kW 2
1 kW
4
1 kW

1
4 kW
6
10 kW
+ 1 kW 3
100kW

va
1 kW


vb
4
1 kW
0.95 kW
4
1
1 kW
25 kW 2

10 kW
1 kW
va
+

+
+
vb
Operational Ampliﬁer and Ampliﬁer Models
1 kW
0.08 V
5
+

25 kW
+
Problem 5.92. Repeat the previous problem when node 1 is grounded. Problem 5.93. The following voltages are measured: υa ¼ 5:000 V and υb ¼ 5:001 V. Find voltages versus circuit ground (common port of the dual supply) for every labeled node in the circuit shown in the ﬁgure below. The ampliﬁer circuit is powered by a 10 V dual supply. Assume exact resistance values and the idealampliﬁer model.
40 kW
10 kW
5.5 General Feedback Systems 5.5.1 Signalﬂow Diagram of a Feedback System 5.5.2 ClosedLoop Gain and Error Signal Problem 5.95. The block diagram of Fig. 5.36 is applied to a voltage ampliﬁer.
V266
Chapter 5 A. Given the input signal xin ¼ 10 mV, the error signal xe ¼ 1 μV, and the output signal xout ¼ 1 V, determine the openloop gain and the feedback factor. B. Given the ratio of input to error signal xin =xe ¼ 100 and the feedback factor of 0.1, determine the openloop gain. Problem 5.96. The openloop gain A in Fig. 5.36 varies between two extreme values of A ¼ 10, 000 2, 000 (20 % gain variation) depending on the system parameters. The forward gain block is used in the closedloop conﬁguration with the feedback factor β of 0.1. Determine the two extreme values of the closedloop gain, ACL. Problem 5.97. The openloop gain A in Fig. 5.36 is 100,000. The forward gain block is used in the closedloop conﬁguration with the feedback factor β of 1. Determine the error signal, xe, if the input voltage signal is 1 mV. Problem 5.98. The closedloop gain of a noninverting ampliﬁer circuit with R1 ¼ 1 kΩ, R2 ¼ 100 kΩ is 99. Determine the opencircuit gain A of the ampliﬁer chip.
Problems Problem 5.100. The circuit shown in the ﬁgure that follows is a feedback transconductance ampliﬁer. Express iout in terms of υin. transconductance amplifier
+

v in
RL
R2
+ 
iout
RF
R1
Problem 5.101. The ampliﬁer circuit shown in the ﬁgure that follows is the Howland current source widely used in biomedical instrumentation; its output is the current through the load resistance. A. Classify the ampliﬁer circuit in terms of four basic ampliﬁer topologies and mention the most important circuit features. B. Derive its gain equation iout ¼ ðυa υb Þ =R2 given that R1 =R3 ¼ R2 =R4 . R3 R1 vb
+

va
R2
5.5.3 Application of General Theory to Voltage Ampliﬁers with Negative Feedback 5.5.4 Voltage, Current, Transresistance, and Transconductance Ampliﬁers with the Negative Feedback
iout =iL
R L R4
Problem 5.99. Derive the gain Eqs. (5.69) for the ampliﬁer circuits shown in Fig. 5.38.
V267
Part II Transient Circuits
Chapter 6
Chapter 6: Dynamic Circuit Elements
Overview Prerequisites:  Knowledge of basic circuit theory (Chapters 2 and 3)  Knowledge of operational amplifiers with negative feedback (Chapter 5) Objectives of Section 6.1:  Define types of capacitance encountered in electric circuits  Define selfinductance and mutual inductance from the first principles  Define field energy stored in a capacitor/inductor  Be able to combine capacitances/inductances in series and in parallel  Understand construction of practical capacitors/inductors  Understand fringing effect and its use in sensor circuits Objectives of Section 6.2:  Derive dynamic equations for capacitance/inductance from the first principles  Establish how the capacitance may create large transient currents  Establish how the inductance may create large transient voltages  Define instantaneous energy and power of dynamic circuit elements  Establish the behavior of dynamic circuit elements in the DC steady state and at a very high frequency Objectives of Section 6.3:  Obtain initial exposure to bypass/blocking capacitor and decoupling inductor  Obtain initial exposure to amplifier circuits with dynamic circuit elements Application Examples: Electrostatic discharge and its effect on integrated circuits How to design a 1F capacitor? How to design a 1mH inductor? Capacitive touchscreens Bypassing a DC motor
© Springer Nature Switzerland AG 2019 S. N. Makarov et al., Practical Electrical Engineering, https://doi.org/10.1007/9783319966922_6
VI271
Chapter 6
Dynamic Circuit Elements
Keywords: Capacitance, Capacitance of two conductors, Selfcapacitance, Capacitance to ground, Capacitance of two equal conductors separated by large distances, Energy stored in a capacitance, Electrostatic discharge (ESD), ESD effect on integrated circuits, Device under test (DUT), Parallelplate capacitor (base formulas, fringing effect, fringing ﬁelds), Capacitor (absolute dielectric permittivity, relative dielectric permittivity, dielectric strength, normalized breakdown voltage, electrolytic, tantalum, ceramic, marking, set of base values), Capacitive touch screens (selfcapacitance method, mutualcapacitance method), Magnetic ﬂux density, Magnetic ﬁeld, Absolute magnetic permeability, Relative magnetic permeability, Magnetic induction, Magnetic ﬂux, Selfinductance, Inductance, Mutual inductance, Energy stored in an inductance, Solenoid (air core, toroidal magnetic core, straight magnetic core, short, fringing ﬁelds), Inductor (marking, set of base values, also see solenoid), Dynamic equation for capacitance (deﬁnition, derivation, ﬂuid mechanics analogy), Capacitance (instantaneous energy, instantaneous power, behavior in the DC steady state, behavior at very high frequencies), Dynamic equation for inductance (deﬁnition, derivation, ﬂuid mechanics analogy), Inductance (instantaneous energy, instantaneous power, behavior in the DC steady state, behavior at very high frequencies), Bypass capacitor, Decoupling capacitor, Shunt capacitor, Snubber RC circuit, Decoupling inductor, Inductor choke, Transient circuit, Ampliﬁer circuits with dynamic circuit elements, Active ﬁlters, Miller integrator (circuit, DC gain, compensation, time constant), Analog pulse counter, Analog computer, Differentiator ampliﬁer (circuit, gain at very high frequencies), Active differentiator
VI272
Chapter 6
Section 6.1: Static Capacitance and Inductance
Section 6.1 Static Capacitance and Inductance 6.1.1 Capacitance, SelfCapacitance, and Capacitance to Ground Capacitance reﬂects the ability of arbitrary conductors to store electric charge and, simultaneously, the store energy of the electric ﬁeld in the surrounding space. When no dielectric is present, capacitance is determined entirely by the geometry of conductors. When a dielectric material is present, its permittivity becomes important. Capacitance deﬁnitions will be given with reference to Fig. 6.1. a)
V=0
b)
c)
+Q +Q
1 +V

+Q
1
d
          
V2
d
V
V Q
+V
1
1
V1
+Q
d)
V=0
2
2 Q
V Q
Fig. 6.1. Conductor geometry for capacitance deﬁnitions.
1. Capacitance, C, of Two Conductors. Two arbitrary insulated conductors near together in Fig. 6.1a constitute a simple capacitor. Its capacitance, C, is found with the help of electrostatic theory. Further, it is used in various dynamic models. Capacitance C of two insulated conductors 1 and 2 is deﬁned by the ratio C
Q >0 V
ð6:1aÞ
where Q > 0 is the (absolute) net charge of either conductor given that the net charge of the system with both conductors is zero and V is the potential difference or voltage between two conductors 1 and 2, i.e., V ¼ V 1 V 2 . This ratio does not depend on V; it is always taken so as to make the capacitance positive. 2. SelfCapacitance, Cself, of a Conductor. When an electric charge Q is added to a single isolated conductor in Fig. 6.1b, its surface will possess a certain absolute voltage V versus 0 V at inﬁnity. The ratio C self
Q >0 V
ð6:1bÞ
is the selfcapacitance of the conductor. The selfcapacitance is the capacitance when the second conductor is a hollow conducting sphere of inﬁnite radius subject to 0 V. 3. Capacitance to Ground, C, of a Conductor. For conductor 1 in Fig. 6.1c with charge þQ, its capacitance to ground is the capacitance when the second conductor is an inﬁnite conducting ground plane in Fig. 6.1c subject to 0 V (and charged to –Q). VI273
Chapter 6
Dynamic Circuit Elements
Capacitance to ground, C, is always greater than the selfcapacitance, Cself; their ratio becomes quite large when the separation distance d from the plane is small. On the other hand, C ! C self when d ! 1 in Fig: 6:1c:
ð6:1cÞ
4. Capacitance, C, of Two Equal Conductors Separated by Large Distances. For conductors 1 and 2 in Fig. 6.1d, the capacitance approaches 1 C ! C self when d ! 1 in Fig: 6:1d: 2
ð6:1dÞ
The separation distance d must be large compared to the conductor’s size. Equation (6.1d) will be proved shortly. The capacitance is recorded in units of farads or F. This unit is named in honor of Michael Faraday (1791–1867), a British physicist and chemist, who was known to many as “the best experimentalist in the history of science.” Typical capacitance values in electronics are pF (picofarad or simply paf), nF (nanofarad), and μF (microfarad). In power electronic circuits, larger capacitances might be used. The capacitance unit is linked to other MKS units as follows: 1 F¼1
As J C ¼1 2¼1 V V V
ð6:2Þ
where C is the unit of coulomb. The total electric ﬁeld energy stored between two conductors and in the surrounding space is given by 1 E ¼ CV 2 2
ð6:3Þ
This result can be derived from the deﬁnition of the electric potential (voltage). The energy is equal to work, which is necessary to put all charges of the capacitor in place. Equation (6.3) is valid for any conﬁguration shown in Fig. 6.1. Example 6.1: Prove Eq. (6.1d). Solution: For a conductor in Fig. 6.1b, the stored electric energy is given by 1 E self ¼ C self V 2 2
ð6:4aÞ
For two conductors in Fig. 6.1d separated by a very large distance d, the stored electric energy is approximately given by
VI274
Chapter 6
Section 6.1: Static Capacitance and Inductance
Example 6.1 (cont.): 1 E ¼ 2E self ¼ C ð2V Þ2 2
ð6:4bÞ
Comparing Eqs. (6.4a) and (6.4b) we obtain the necessary result.
Exercise 6.1: A metal circle or radius r ¼ 0:1 m has the selfcapacitance C self ¼ 8ε0 r where ε0 ¼ 8:85419 1012 F=m is the permittivity of vacuum. Estimate capacitance of a capacitor formed by two coaxial circles separated by 1 m. Answer: ~3.54 pF from Eq. (6.1d). A precise numerical solution predicts 3.77 pF.
Exercise 6.2: How large is the stored energy in a 100μF laboratory capacitor at 10 V? Answer: 0.005 J or, which is the same on the power basis, 5 mW of power delivered during one second. However, this power will not be delivered uniformly.
Exercise 6.3: How large is the stored energy in a 20F ultracapacitor charged to 25 V? Answer: 6250 J or 6.25 kJ. This is certainly a signiﬁcant value. At the same time, the discharge rate (available current or power) is much less in this case than the current or power delivered by laboratory electrolytic capacitors.
6.1.2 Application Example: ESD Selfcapacitance results may be applied for the prediction of ESD (electrostatic discharge) effects on integrated circuits (ICs). One of the most common causes of electrostatic damage is the direct transfer of electrostatic charge through a signiﬁcant series resistor from the human body or from a charged material to the electrostatic dischargesensitive (ESDS) device. The concept is shown in Fig. 6.2a. A metal ground plane and the highly conducting human body naturally form a capacitor. The body capacitance C in Fig. 6.2b is deﬁned as the capacitance between the body, assumed to be a conductor, and the large (ideally inﬁnite) ground plane. Its value depends signiﬁcantly on the posture of the body with respect to the ground surface. The typical separation distance is 2 cm. It may be shown that, at such distances, C 2C self . Therefore, instead of calculating C directly, we can ﬁnd the selfcapacitance of the human body, Cself, and then multiply it by 2.
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Typical selfcapacitances of a 177cmtall male student are indicated in Fig. 6.3. These values were obtained by the numerical method described in Chapter 1. a)
b) DUT
t=0
t=0
=
series body resistance charged body to ground capacitance
R C
Fig 6.2. Equivalent circuit for understanding ESD and its effect on a device under test (DUT).
The simplifying assumption C 2C self and Fig. 6.3 predict body capacitances in the range 86–95 pF. These values are in a good agreement with the generally accepted human body model (HBM), which, with reference to Fig. 6.2b, uses R ¼ 1:5 kΩ, C ¼ 100 pF Cself=47.2 pF
ð6:5Þ Cself=44.3 pF
Cself=43.4 pF
Cself=43.2 pF
Fig 6.3. Typical selfcapacitance values for a 177cmtall male person. Note how the selfcapacitance changes when the body poses change.
6.1.3 ParallelPlate Capacitor Consider a parallelplate capacitor shown in Fig. 6.4a, b. Both inﬁnitely thin conducting square plates with the side a and area A ¼ a2 are separated by distance d. The upper plate has a total charge +Q; the lower plate has the opposite charge –Q; the net charge of the capacitor is zero. Feeding conductors are implied to be disconnected; they are excluded from consideration. Assuming that the entire electric ﬁeld is concentrated only within the capacitor and that it is uniform in space (equal to V/d), the approximate capacitance is established as
VI276
Chapter 6
C¼
Section 6.1: Static Capacitance and Inductance
ε0 A d
ð6:6Þ
where ε0 is the dielectric permittivity of vacuum if the capacitor is situated in vacuum. For Eq. (6.6) to hold, the plates do not have to be square. If the capacitor does not have a highε dielectric inside, Eq. (6.6) is a good approximation only if d is very small compared to the dimensions of the plates. Otherwise, the fringing effect must be taken into account. The fringing effect is illustrated in Fig. 6.4a, b. Fringing means that the electric ﬁeld extends outside the physical capacitor. The electric ﬁeld outside the capacitor possesses certain extra energy. Therefore, according to Eq. (6.3) where voltage V is ﬁxed, the capacitance must increase compared to the nonfringing case. Figure 6.4c given below illustrates numerically found capacitance values Cexact for the parallelplate capacitor with fringing. These values have been accurately computed using a rigorous numerical adaptive procedure. Figure 6.4c predicts a nearly linear increase of the ratio Cexact/C as a function of the separation distance. Therefore, the wrong result, C ! 0 when d ! 1, which is predicted by Eq. (6.6), is corrected. Instead, one will have C ! 0:5 C self when d ! 1. a)
b)
Potential (voltage) distribution, V 0.3
+1 V
0.4
0.5 0.6 0.7 0.8 0.9
lines of force (Efield)
1 V
c) 3.5
equipotential lines
Cexact/C
3
0.9 0.8 0.7
2.5
0.6
2
0.5 0.4 0.3
1.5 1 0
0.2
0.4
0.6
0.8 d/a 1
Fig. 6.4. (a) Equipotential lines and lines of force for a capacitor with d=a ¼ 0:2 in the central crosssectional plane (the plates are at 1 V). (b) Fringing electric ﬁeld for the same capacitor observed in the central crosssectional plane (the plates are at 1 V). (c) Ratio of the accurate capacitance values (found numerically) to the values predicted by Eq. (6.6).
The fringing ﬁeld of capacitors is utilized in capacitive touch screens. In this case, the signiﬁcant fringing ﬁeld is a desired effect. Therefore, conﬁgurations other than the parallelplate capacitor are used. These conﬁgurations will be studied later in this section.
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Dynamic Circuit Elements
When a dielectric material of relative permittivity εr > 1 is inserted between the capacitor plates, a substitution ε0 ! εr ε0 has to be made in Eq. (6.6). The fringing effect is less apparent for higher values of εr. Exercise 6.4: Estimate the static capacitance of a parallelplate capacitor with a ¼ 1 cm2 and d ¼ 1:57 mm using the basic formula. The substrate material is Rogers 4003 laminate with εr ¼ 3:55. Answer: 2.00 pF.
6.1.4 Circuit Symbol: Capacitances in Parallel and in Series Figure 6.5 shows the capacitances in parallel and in series, along with the capacitance circuit symbol. This symbol is reserved for the capacitance as a circuit element. Such an element is an ideal capacitor excluding manufacturing imperfections (parasitic resistance and inductance). In the following text, we will frequently employ both words—capacitance and capacitor—to denote the same ideal circuit element. The parallel and series connections of capacitances are opposite when compared to the resistances. To establish this fact we consider two combinations in Fig. 6.5. a)
b)
C1
+Q1 Q1 +Q2 Q2 +Q3 Q3
C2 A
B
C3
+
VC
+ vC1  + vC2  + vC3 

A
C1
C2
+
VC
C3
B

Fig. 6.5. Capacitances in parallel and in series: (a) capacitances in parallel are added; they behave similarly to resistances in series and (b) capacitances in series are combined in the same way as resistances in parallel.
For the parallel conﬁguration in Fig. 6.5a, the same voltage VC is applied to every capacitance. One has for the charges on the capacitor plates, Q1 ¼ C 1 V C , Q2 ¼ C 2 V C , Q3 ¼ C 3 V C ) Q1 þ Q2 þ Q3 ¼ Qtotal ¼ ðC 1 þ C 2 þ C 3 Þ V C
ð6:7Þ
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Section 6.1: Static Capacitance and Inductance
Thus, the capacitances connected in parallel behave like a single capacitance Ceq, C eq ¼ C 1 þ C 2 þ C 3
ð6:8Þ
Equation (6.8) also makes intuitive sense if we take a closer look at Fig. 6.5a where the three individual capacitors visually form a bigger capacitor comprised of larger plate areas. This clearly increases the capacitance accordingly. In Fig. 6.5b, however, the situation is different. The thicknesses of each capacitor add together, which decreases the overall capacitance since thickness varies inversely with capacitance. Assume that every capacitor was initially uncharged and apply voltage VC between terminals A and B. Since each pair of inner conductors in Fig. 6.5b has remained insulated, Q1 ¼ Q2 ¼ Q3 ¼ Q. Next, by KVL, Q1 Q2 Q3 1 1 1 Q þ þ ¼ þ þ VC ¼ V1 þ V2 þ V3 ¼ Q¼ ð6:9Þ C1 C2 C3 C eq C1 C2 C3 and for the series combination of the capacitances, one has 1 1 1 1 ¼ þ þ C eq C 1 C 2 C 3
ð6:10Þ
Exercise 6.5: Find the equivalent capacitance of the circuit shown in Fig. 6.6. Answer: C eq ¼ 44 μF
33 F a 33 F
33 F
b 33 F
Fig. 6.6. A capacitive network which includes series and parallel combinations of capacitances.
6.1.5 Application Example: How to Design a 1μF Capacitor? Let us to design a capacitor. Our modest goal is a 1μF capacitor. We consider two aluminum plates separated by 1 mm. Equation (6.6) then allows us to predict the required plate area: A¼
dC 103 106 ¼ ¼ 113 m2 ! ε0 8:854 1012
ð6:11aÞ
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Chapter 6
Dynamic Circuit Elements
Well, such a capacitor will certainly occupy a signiﬁcant fraction of a lecture hall and is hardly practical. How, then, do manufactures design a capacitor of 1 μF? The ﬁrst step is to use a dielectric material sandwiched within the capacitor. A dielectric medium increases charges stored on the two metal capacitor plates depending on εr 1, the relative dielectric permittivity of the dielectric medium. Table 6.1 gives us a list of permittivities for a number of dielectric materials. For each material, a dielectric strength, or normalized breakdown voltage, is also given. This is actually the maximum electric ﬁeld (notice the unit of V/m) that the capacitor can handle. It is for this reason that capacitors carry a voltage rating that you should not exceed in your circuit. From a practical point of view, the higher the capacitance, the lower the voltage rating. The wellknown dilemma with the capacitor is that a decrease in the separation distance increases the capacitance and the stored energy. However, as already mentioned, it simultaneously decreases the maximum applied voltage due to the dielectric breakdown effect. For our capacitor, we will again use the mica dielectric material listed in Table 6.1. Equation (6.6) now transforms to A¼
dC 103 106 ¼ ¼ 16 m2 ε0 εr 8:854 1012 7
ð6:11bÞ
Table 6.1. Relative dielectric permittivity and dielectric strength of some common materials. Material Air Aluminum oxide Fused silica (glass) Gallium arsenide (GaAs) Germanium (Ge) crystal Mica Nylon Plexiglas Polyester Quartz Rutile (titanium dioxide) Silicon (Si) crystal Styrofoam Teﬂon Water (distilled, deionized)
Relative permittivity 1.0 8.5 3.8 13 16 7.0 3.8 3.4 3.4 4.3 100–200 12 1.03–1.05 2.2 ~80
Dielectric strength in V/m 0.4–3.0 106 Up to 1000 106 470–670 106 (or lower) ~10 106 Up to 400 106 ~20 106 ~30 106 8 106 (fused quartz) 10–25 106 ~30 106 87–173 106 65–70 106
Even though the result looks a bit better, it is still far from practical. However, what if we try to make the dielectric layer very thin? An oxide is a dielectric, so could we just oxidize one top aluminum plate with a very thin (i.e., d ¼ 10 μm) oxide layer and pressﬁt it to
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Chapter 6
Section 6.1: Static Capacitance and Inductance
the second plate? The result becomes (the relative dielectric constant of 8.5 is now that for aluminum oxide from Table 6.1) A¼
hC 105 106 ¼ ¼ 13 cm2 ε0 εr 8:854 1012 8:5
ð6:11cÞ
Electrolytic Capacitors Once such a thin ﬁlm is rolled into a cylinder, it will clearly become a compact design, similar in size to a 1μF electrolytic capacitor routinely used in the laboratory. Unfortunately, one problem still remains: the permanent oxide layer is fragile and rough in shape. A better idea is to chemically grow such a layer using a socalled anodization process. This process occurs when the aluminum foil is in contact with an electrolyte as a second conductor and an appropriate voltage is applied between them. This is the smart idea behind an electrolytic capacitor. And this is also the reason why an electrolytic capacitor is polarized. The term electrolytic capacitor is applied to any capacitor in which the dielectric material is formed by an electrolytic method; the capacitor itself does not necessarily contain an electrolyte. Along with aluminum capacitors, tantalum capacitors (both wet and dry) are also electrolytic capacitors. Ceramic Capacitors A competitor to the electrolytic capacitor is a nonpolarized ceramic capacitor. Ceramic capacitors consist of a sandwich of conductor sheets alternated with ceramic material. In these capacitors the dielectric material is a ceramic agglomerate whose relative static dielectric permittivity, εr, can be changed over a very wide range from 10 to 10,000 by dedicated compositions. The ceramic capacitors with lower εr values have a stable capacitance and very low losses, so they are preferred in highprecision circuits and in highfrequency and RF electronic circuits. Typically, these “fast” ceramic capacitors have very small capacitances, on the order of pF and nF, and they can hold a high voltage. At the same time, the “slow” ceramic capacitors may have values as high as 1 μF. Therefore, the task of the above example can be solved with the ceramic capacitor as well. Capacitor Marking Figure 6.7 shows two examples of ceramic capacitors, with 100pF and 1.0μF capacitance from two different companies. To read the capacitance in the ﬁgure, we use the following rule: 101 ¼ 10 101 pF ¼ 100 pF, and 105 ¼ 10 105 pF ¼ 1 μF. Indeed, 473 ¼ 47 103 pF ¼ 47 nF, and so forth. The tolerance letters may be present: F ¼ 1 %, G ¼ 2 %, J ¼ 5%, K ¼ 10 %, and M ¼ 20 %. Also, the voltage rating should be given.
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Dynamic Circuit Elements
Fig. 6.7. Left, ceramic capacitors of 100 pF. Right, radial leaded ceramic capacitors of 1.0 μF.
A standardized set of capacitance base values is deﬁned in the industry. The capacitance of any (electrolytic or not) capacitor can then be derived by multiplying one of the base numbers 1.0, 1.5, 2.2, 3.3, 4.7, or 6.8 by powers of ten. Therefore, it is common to ﬁnd capacitors with capacitances of 10, 15, 22, 33, 47, 68, 100, 220 μF, and so on. Using this method, values ranging from 0.01 to 4700 μF are customary in most applications. The value of the capacitance and the allowed maximum voltage are prominently written on the case of the electrolytic capacitor so reading those does not constitute any difﬁculties.
6.1.6 Application Example: Capacitive Touchscreens Capacitive touchscreens use the fringing ﬁeld of a capacitor studied previously. Many small capacitors with a signiﬁcant fringing ﬁeld are involved. If a conducting ﬁnger (an extra conductor) is placed in the fringing ﬁeld, the corresponding capacitance changes. There are two possible solutions called the selfcapacitance method and the mutualcapacitance method, respectively. The difference is in the measurement nodes for the capacitance. In the ﬁrst case, the capacitance is measured between the touch pad electrode and a ground. In the second case, the capacitance is measured between two pad electrodes, neither of which is grounded. Both methods may be combined. SelfCapacitance Method Consider a human ﬁnger in the proximity of a touchscreen as shown in Fig. 6.8a. The touchscreen itself may be a lattice of circular touch pads surrounded by a ground plane and separated from it by an airgap ring—see Fig. 6.8b. When the ﬁnger is not present, each pad has capacitance CP to ground, which is called a parasitic capacitance. When the (grounded) ﬁnger appears in the vicinity of the touchpad, there appears another capacitance, CF, which is called the ﬁnger capacitance. Figure 6.8a indicates that both capacitances are in parallel so that the resulting ground capacitance increases as CP ! CP þ CF > CP
ð6:12Þ
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Chapter 6
Section 6.1: Static Capacitance and Inductance
This change in capacitance is recorded. Physically, the presence of the ﬁnger (or hand) increases the size of the ground conductor and thus increases the resulting capacitance. Typical values of CP are on the order of 100 pF; CF is on the order of 1–0.1 pF. Now assume that the desired resolution along one dimension of the screen is N. Then, N2 individual touch pads are needed including the corresponding sensing circuitry. This may be a signiﬁcant disadvantage of the selfcapacitance method. a)
b) CP
CF
CP
CP
CP
CP 0V pad
0V
o v e r l a y
ground plane
Fig. 6.8. Selfcapacitance method for a capacitive touchscreen. The touchscreen is enlarged.
MutualCapacitance Method The electrodes are typically interleaving rows and columns of interconnected square patches, which are shown in Fig. 6.9. Neither of them is connected to circuit ground (the third conductor) or to each other. When a ﬁnger touches the panel, the mutual capacitance CM between the row and column, which mostly concentrates at the intersection, decreases, in contrast to the previous case. This change in capacitance is recorded. Assume again that the desired resolution along one dimension of the screen is N. Then, only 2N individual touch pads (electrodes) are needed including the corresponding sensing circuitry. This is a signiﬁcant advantage of the mutualcapacitance method. sensing column
finger position
CM
driven row
Fig. 6.9. Mutualcapacitance method for a capacitive touchscreen. Surface charge distribution is illustrated when the driven row is subject to an applied voltage. Finger projection is a circle.
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Chapter 6
Dynamic Circuit Elements
6.1.7 SelfInductance (Inductance) and Mutual Inductance As long as the physical capacitor stores energy of the electric ﬁeld, a physical inductor stores energy of the magnetic ﬁeld. The inductor stores magneticﬁeld energy only when an electric current I ﬂows through it. This is in contrast to a capacitor, which, once charged, stores the electric ﬁeld energy in vacuum indeﬁnitely, even when disconnected from the charging circuit. To use the inductor as an energystorage element, one therefore needs to maintain a current in the circuit. The primary quantity is the magnetic ﬂux density, ~ B, which is measured in webers per m2, or tesla (1 T ¼ 1 Wb=m2 ¼ 1 V s=m2). ~ by ~ ~ where μ is The magnetic ﬂux density is related to the magnetic ﬁeld, H, B ¼ μH magnetic permeability. In older power electronics texts, ~ B, may be called magnetic ~ induction. In Fig. 6.10, the magnetic ﬂux density B is created by circuit #1 (a closed loop of current I). Instead of the vector ﬁeld ~ B, it is convenient to use a simpler scalar quantity known as magnetic ﬂux or simply ﬂux, Φ. For a constant ~ B, which is strictly perpendicular to the plane of circuit #1 with area A, the magnetic ﬂux would be equal to Φ ¼ AB
ð6:13Þ
where B is the magnitude (length) of vector ~ B. The ﬂux is measured in webers or in V s (1 Wb ¼ 1 V s).
circuit #1 circuit #2
A I
lines of magnetic flux (B= H )
Fig. 6.10. Magnetic ﬂux density generated by circuit #1.
Generally, Eq. (6.13) is only approximately valid for circuit #1 in Fig. 6.10. The exact ﬂux is given by a surface integral over the area of the circuit, ðð ~ B~ nda ð6:14Þ Φ¼ A
where ~ n is a unit normal to the surface A. Flux is an algebraic quantity and hence could be positive or negative depending on the chosen direction of ~ n. We assume that positive current in the circuit produces positive ﬂux—see Fig. 6.10. The selfinductance of circuit VI284
Chapter 6
Section 6.1: Static Capacitance and Inductance
#1 in Fig. 6.10 is its inductance; both terms have the same meaning. The inductance, L, of circuit #1 is given by L
Φ >0 I
ð6:15Þ
Thus, the inductance is the magnetic ﬂux through circuit #1 produced by a unit current in the same circuit. The mutual inductance, M, between circuits #2 and #1 in Fig. 6.10 is the magnetic ﬂux, Φ0 , through circuit #2 produced by unit current in circuit #1, i.e., 0
Φ M I
ð6:16Þ
Both L and M have the units of henry, or H. This unit is named in honor of Joseph Henry (1797–1878), an American scientist. Typical inductance values in electronics are nH (nanohenries) and μH (microhenries). In power electronics, larger inductances may be used. Henry is converted to V, A, and energy, J, as follows: 1 H¼1
Vs J ¼1 2 A A
ð6:17Þ
One may observe a close similarity between Eqs. (6.17) and (6.2). Both equations become identical if we interchange V and A. Equation (6.17) also has a number of simple and important implications related to energy and power. Total magneticﬁeld energy stored in space surrounding circuit #1 in Fig. 6.10 is given by 1 E ¼ LI 2 2
ð6:18Þ
Equation (6.18) may be considered as another deﬁnition of selfinductance (or inductance). As such, it is frequently used in practice. Exercise 6.6: A ﬂux linking the circuit is 0.1 Wb. Find the circuit’s inductance and magneticﬁeld energy stored if the circuit current is 1 A. Answer: L ¼ 100 mH, E ¼ 0:05 J.
6.1.8 Inductance of a Solenoid With and Without Magnetic Core Consider a solenoid (a long helical coil of length l) with applied current I shown in Fig. 6.11. The case of airﬁlled coil in Fig. 6.11a is studied ﬁrst. Magnetic ﬂux density ~ B within the solenoid is nearly uniform and is directed along its axis. Therefore, the ﬂux through one turn of the coil (one loop) is given by Eq. (6.13). It is equal to AB, where A is the loop area. The net ﬂux Φ through the entire solenoid is AB times the number of turns comprising the coil, N. The inductance is therefore obtained from Eq. (6.15) as VI285
Chapter 6
L¼N
Dynamic Circuit Elements
AB I
ð6:19Þ
The magnetic ﬂux, B, within the solenoid is found in physics courses: B¼
μ0 N I l
ð6:20Þ
where the natural constant μ0 ¼ 4π 107 H=m is the magnetic permeability of vacuum (or air, which is very close to a vacuum with regard to magnetic properties). Substitution of Eq. (6.20) into Eq. (6.19) yields a simple equation for the inductance L¼
μ0 AN 2 l
½H
ð6:21Þ
Thus, strong inductances can be created by a large number of turns (a quadratic dependence), a large coil cross section, and a smaller coil length. Equation (6.21) also holds for various bent solenoids (such as toroidal coils). Equation (6.21) makes clear that the inductance, like capacitance and resistance, is independent of externally applied circuit conditions.
a)
b) I
B= 0H
I
magnetic core ( r) B= 0 rH
magnetic core ( r)
c)
I
N
N
A
A
I
I
Fig. 6.11. Three types of a solenoid.
The above derivation is only valid for a solenoid that is long compared to its diameter. When this is not the case, a modiﬁcation to Eq. (6.22) is made, namely, μ0 AN 2 8w w2 w4 r ð6:22Þ þ 1 , w¼ 2 μs, the voltage is 100 mV. Therefore, the current is found in the form: iC ðtÞ ¼ 104 105 ¼ 10 A when t changes from 1 to 2 μs and iC ðtÞ ¼ 0 otherwise. The result is shown in Fig. 6.14 by a dashed curve. We observe a strong current spike when the voltage across the capacitor changes rapidly, and we observe no current ﬂow when the voltage across the capacitor remains constant.
Note that the relatively small voltage on the order of 100 mV in Fig. 6.14 leads to a very large current spike of 10 A(!) through the capacitance. The key point here is that the current VI291
Chapter 6
Dynamic Circuit Elements
increase is due to the rapid change in voltage. Such a change can be created when the capacitor discharges through a small resistance. This is the reason why capacitors are routinely employed to deliver large currents, or high power levels, for a very short period of time. The high currents are common in motor starting circuits, in electronic ﬂashes, in solenoids, and in various electromagnetic propulsion systems. 100 F capacitance
vC
+ vC , V
iC, A
iC 0.2
10
0.1
5
0
1
2
3
t, s
Fig. 6.14. Applied voltage across the capacitor (solid curve) and resulting current (dashed curve).
The capacitor is charged with an electric current. The voltage across the capacitor, from Eqs. (6.28) and (6.29), is given by ðt 1 q ðt Þ iC ðt 0 Þdt 0 ¼ C υC ¼ C C
ð6:30Þ
0
Equation (6.30) tells us that the capacitor voltage is equal to zero at the initial time, i.e., at t ¼ 0. Once the current iC(t) is known, the voltage at any point in time is obtained by carrying out the integration in Eq. (6.30). At any time instant, the voltage is equal to the instantaneous stored charge qC(t) divided by capacitance. The current in Eq. (6.30) is either predeﬁned or found from circuit considerations. The example that follows illustrates voltage calculations. Example 6.3: A 1μF capacitor is charged with an electric current, iC ðt Þ ¼ 2 t ½mA . The capacitance voltage is equal to zero at the initial time instance t ¼ 0. When will the capacitor be charged to 10 V? Solution: The integration yields ðt 1 0:001 2 υC ¼ t ¼ 10 ) t ¼ 0:1 s iC ðt0 Þdt 0 ¼ C C
ð6:31Þ
0
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Section 6.2: Dynamic Behavior of Capacitance and Inductance
6.2.3 Dynamic Behavior of Inductance We use the passive reference conﬁguration for the inductance in Fig. 6.15a. Lowercase letters denote timevarying voltage and current. The dynamic behavior of the inductance is described by the wellknown voltagetocurrent relation (dynamic equation), which plays the role of “Ohm’s law” for the inductance: υL ¼ L
d iL dt
ð6:32Þ
Equation (6.32) follows from the inductance deﬁnition, Φ ¼ LiL , given in the previous section. We obtain Eq. (6.32) after differentiation and using Faraday’s law of induction for the time derivative of the magnetic ﬂux (the plus sign is used in Faraday’s law to be consistent with the passive reference conﬁguration): dΦðt Þ ¼ υL dt
ð6:33Þ
a)
b)
massive wheel of mass m
inductance L + vL 
iL
Fig. 6.15. Passive reference conﬁguration for inductance and its ﬂuid mechanics analogy.
A ﬂuid mechanics analogy of the dynamic inductance effect is given here in terms of alternating current, which corresponds to alternating ﬂuid motion in Fig. 6.15b. A massive wheel with rotational inertia in Fig. 6.15b represents inductance. The inductance value, L, corresponds to the mechanical mass m of the wheel. When m ! 1 or L ! 1, the wheel does not responds to ﬂuid oscillations and blocks the alternating ﬂuid ﬂow entirely. In the opposite case (m ! 0 or L ! 0), the wheel has no effect on the ﬂuid ﬂow. Intermediate cases correspond to a partial blocking. Example 6.4: The current through a 2mH inductor is shown in Fig. 6.16 by a solid curve. At t ¼ 0, the current is zero. Sketch the voltage across the inductance to scale versus time. Solution: We use Eq. (6.32) to ﬁnd the voltage across the inductance. In Fig. 6.16, iL ðt Þ ¼ 103 t 106 A when t changes from 1 to 2 μs and iL ðt Þ ¼ 0 at t < 1 μs. At t > 2 μs, the current is 1 mA. Therefore, the voltage is found in the form: υL ðt Þ ¼ 2 103 103 ¼ 2 V when t changes from 1 to 2 μs and υL ðt Þ ¼ 0 otherwise. The result is shown in Fig. 6.16 by a dashed curve.
VI293
Chapter 6
Dynamic Circuit Elements 2 mH inductance
+
vL

iL, mA
v L, V
iL 2
2
1
1
0
1
t, µs
2
3
Fig. 6.16. Impressed current through an inductance (solid curve) and resulting voltage across the inductance (dashed curve).
Note that relatively small current, on the order of 1 mA, leads to a large voltage spike of 2 V across the inductance in Fig. 6.16. The key point here is again the rapid change in the current. If the current in the present example were on the order of 1 A, a voltage spike of 2000 V would be observed. This is the reason why inductors are routinely used to boost the voltage to a higher level. These high voltages are common in electric and electronic ignition systems including the most common car ignition plug. From Eqs. (6.32) and (6.33) the current through the inductance is given by ðt 1 Φðt Þ iL ¼ υL ðt 0 Þdt 0 ¼ L L
ð6:34Þ
0
Equation (6.34) implies that the current is equal to zero at the initial time, i.e., at t ¼ 0. Once the voltage is known as a function of time, the current through the inductance at any time moment is obtained by the calculation of the integral in Eq. (6.34). At any time instant, the current is equal to the instantaneous magnetic ﬂux Φ(t) divided by inductance. Example 6.5: A 1mH inductor is subject to applied voltage, υL ðt Þ ¼ 2 t ½mV . The inductance current is equal to zero at the initial time instance t ¼ 0. When will the magneticﬁeld energy stored in the inductance reach 1 J? Solution: The integration in Eq. (6.34) and using Eq. (6.18) for the energy stored in the inductor yield ðt 1 0:001 2 1 2 iL ¼ t ¼ t 2 ½A ) L t2 ¼ 1 J ) t ¼ 6:7 s υL ðt 0 Þdt 0 ¼ L L 2
ð6:35Þ
0
VI294
Chapter 6
Section 6.2: Dynamic Behavior of Capacitance and Inductance
So, if the capacitance is associated with a spring, the inductance is associated with the mass, and the resistance is associated with a dash pot (damping element), then the entire electric circuit containing dynamic elements is nothing else but a mechanical system. Is this correct? Clearly the same analysis methods are applicable to both systems, electrical and mechanical! The model of an entire building in terms of lumped mechanical elements is in theory the same as the model of a complicated electric circuit. Both models can be analyzed by using the theory of linear systems, and both models follow the same control theory. A more difﬁcult issue is related to nonlinear circuit elements.
6.2.4 Instantaneous Energy and Power of Dynamic Circuit Elements An elegant derivation of the energy stored in a capacitance can also be obtained by integrating the power delivered (or taken) by the capacitance. The instantaneous electric power pC(t) can be written in the form pC ðt Þ ¼ υC iC ¼ υC C
dυC 1 dυ2C ¼ C 2 dt dt
ð6:36Þ
The stored energy is then the time integration of the power, i.e., ðt
EC ðt Þ ¼ pC ðt 0 Þdt 0 ¼ 0
ðt
1 dυ2C 0 1 2 C 0 dt ¼ C υC ðt Þ υ2C ð0Þ 2 dt 2
ð6:37Þ
0
where the lower limit, υC(0), is the initial state of the capacitance. Suppose that υC ð0Þ ¼ 0, i.e., the capacitance is initially uncharged and has zero stored energy. Then, 1 EC ðt Þ ¼ Cυ2C ðt Þ 2
ð6:38Þ
Equation (6.38) is the formal proof of the corresponding static result, Eq. (6.3), postulated in the previous section. We can derive the energy stored in the inductance using the same method—by integrating the power. The instantaneous power supplied to or obtained from the inductance has the form pL ðt Þ ¼ υL iL ¼ iL L
diL 1 di2L ¼ L 2 dt dt
ð6:39Þ
The energy stored in the inductance is the integral of Eq. (6.39), i.e., ðt
0
0
ðt
EL ðt Þ ¼ pL ðt Þdt ¼ 0
1 di2L 0 1 2 L 0 dt ¼ L iL ðt Þ i2L ð0Þ 2 dt 2
ð6:40Þ
0
Suppose that iL ð0Þ ¼ 0, i.e., the inductance is initially “uncharged” or has no stored energy. Then, VI295
Chapter 6
Dynamic Circuit Elements
1 EL ðt Þ ¼ Li2L ðt Þ 2
ð6:41Þ
Equation (6.41) is the formal proof of the corresponding static result, Eq. (6.18), postulated in the previous section. The series and parallel combinations of inductances and capacitances may also be analyzed using the dynamic element equations; the laws obtained in the previous section will be conﬁrmed. Exercise 6.8: Determine instantaneous power supplied to the capacitance in Fig. 6.14 at A. t ¼ 2 μs and B. t ¼ 1 μs. Answer: 1 W and 0 W, respectively.
Exercise 6.9: Repeat the previous exercise for the inductance shown in Fig. 6.16. Answer: 0.002 W and 0 W, respectively.
6.2.5 DC Steady State According to Eq. (6.28) when voltage across the capacitance does not change with time, the capacitance becomes an open circuit (no current) under DC steadystate condition, i.e., dυC dυC ¼ 0 ) iC ¼ C ¼0 dt dt
ð6:42Þ
This is to be expected since a DC current cannot ﬂow through empty space between two capacitor plates. Similarly, according to Eq. (6.32) the inductance becomes a short circuit for the DC steady state when current across the inductance does not change in time, i.e., diL diL ¼ 0 ) υL ¼ L ¼0 dt dt
ð6:43Þ
In other words, there is no voltage drop across a (long) bent piece of wire, which is the inductor, for DC currents. Equations (6.42) and (6.43) allow us to establish the behavior of any transient electric circuit in the long run, after the circuit behavior has been stabilized. The transient circuit is a circuit with dynamic elements and a switch. Figure 6.17 shows one such circuit that consists of a number of dynamic (and static) elements and a switch. The switch connects the voltage source to the rest of the circuit as the switch closes at t ! 0.
VI296
Chapter 6
Section 6.2: Dynamic Behavior of Capacitance and Inductance 10 H
t=0
40 5V
+ 
1 F
I 1 H
Fig. 6.17. A circuit is used to study the DC steady state. The switch closes at t ¼ 0. The behavior of the circuit at t ! 1 is sought after the circuit has stabilized.
Example 6.6: Find current I in Fig. 6.17 at t ! 1, i.e., under DC steady state. Solution: Immediately after the switch in Fig. 6.17 closes, the voltages and currents in the circuit may be subject to a complicated response. In particular the voltage across certain dynamic elements may be higher than the voltage of the power supply of 5 V. However, in the long run as t ! 1, the circuit behavior stabilizes and we reach the DC steady state. The capacitance in Fig. 6.17 becomes an open circuit and may be ignored. Both inductances can be replaced by a wire (short circuit). The resulting DC circuit is shown in Fig. 6.18. Thus, we obtain I¼
5 V ¼ 125 mA 40 Ω
ð6:44Þ
Other more complicated circuits can be analyzed in exactly the same way.
5V
+ 
I
40
Fig. 6.18. A DC equivalent of the circuit in Fig. 6.17 under steadystate conditions.
6.2.6 Behavior at Very High Frequencies At very high frequencies, the behavior of the two dynamic circuit elements is exactly the opposite: the capacitance becomes a short circuit, whereas the inductance becomes an open circuit. To establish this fact, we can either use the ﬂuid mechanics analogies or the dynamic equations themselves. For example, the inertia of a massive wheel (inductance) will prevent any very fast movements of it so that the oscillating ﬂuid ﬂow will be entirely blocked when the oscillation frequency tends to inﬁnity. On the other hand, the forces on a ﬂexible membrane of zero mass will be so high for a rapidly oscillating ﬂuid ﬂow that its ﬁnite stiffness no longer matters. The membrane will simply be moving along with the ﬂuid, which means the full transmission through its counterpart—the capacitance. VI297
Chapter 6
Dynamic Circuit Elements
The behavior of dynamic elements at very high frequencies is exactly as important as the behavior at DC; it will be studied quantitatively in Chapter 9. Example 6.7: Illustrate how is the capacitance becoming a short circuit at very high frequencies using the capacitor’s dynamic equation as a starting point. Solution: Assume that there is a periodic current with the amplitude of 1 A, iC ðtÞ ¼ 1 A cos ωt through a 10μF capacitance. The resulting capacitor voltage is given by Eq. (6.30), υC ðt Þ ¼ 1=ðωCÞ sin ω t. When ω ¼ 108 rad=s, the capacitor voltage has the amplitude of 1 mV. This small voltage approximately corresponds to a short circuit. When ω increases, the voltage amplitude is reduced even further.
VI298
Chapter 6
Section 6.3: Application Circuits Highlighting Dynamic Behavior
Section 6.3 Application Circuits Highlighting Dynamic Behavior 6.3.1 Bypass Capacitor Let us consider the circuit shown in Fig. 6.19a. It includes a voltage source represented by its Thévenin equivalent and a load represented by its equivalent resistance RL. The source generates a voltage in the form of a (large) DC component VS and superimposed (small) AC signal υS(t). This setup could model a nonideal DC voltage power supply, which does not create the exactly DC voltage. In fact, a weak AC component may be present. This AC component (also called the noise component) has a frequency of either 60 n Hz, where n is an integer (USA, Canada, parts of South America, Saudi Arabia, etc.), or 50 n Hz (the rest of the world) and appears due to a not quite perfect rectiﬁcation of the primary AC power. As an aside, switching power supplies create noise spikes at much higher frequencies. The weak AC component may lead to circuit oscillations, especially when dealing with highgain ampliﬁers. It therefore should be removed from the load, or “ﬁltered out” as engineers often say. The idea is to use a capacitor C in parallel with the (imperfect) power supply and in parallel with the load, the socalled bypass capacitor. This capacitor ideally becomes a short circuit for the highfrequency noise component of the source and shorts it out (or bypasses). The corresponding circuit diagram is shown in Fig. 6.19b. An electrolytic capacitor is typically used as the bypass capacitor. a)
b)
source RS
vS(t) VS
+

+ 
source RS
load
DC AC
vS(t)
+
VS
+ 
RL

load
DC AC
C
RL
Fig. 6.19. Model of a voltage source connected to a load with a bypass capacitor.
In many cases, the undesired noise source in a circuit is not a lowfrequency noise source of a nonideal power supply, but rather a highfrequency noise generator. Examples include highspeed DC motors, analogtodigital converters, and other digital circuits. Radiofrequency (RF) highspeed ampliﬁers are also very sensitive to RF noise that is created by connectors and wires which can act like antennas. By placing a bypass capacitor as closely as possible to the power supply pins of every chip, such RF noise sources may be eliminated. Bypass capacitors are so prevalent that they are encountered in virtually every working piece of electronic equipment.
VI299
Chapter 6
Dynamic Circuit Elements
Example 6.8: Explain the operation of bypassing a DC motor. Solution: A DC motor with a bypass capacitor is also described by the model shown in Fig. 6.19b. In this case, the source voltage V S þ υS ðt Þ becomes the induced electromotive force (emf), E, of the motor, The induced emf is still a DC voltage but with quite a signiﬁcant highfrequency noise component created by the spinning rotor comprised of a ﬁnite number of individual switched coils. The source resistance becomes the armature and brush resistance RM, that is, RS ¼ RM . The load resistance RL may, for example, be the oscilloscope resistance. We consider a small DC fan motor directly connected to a 5V power supply shown in Fig. 6.20. The motor creates a substantial highfrequency noise seen on the oscilloscope in Fig. 6.20, left (with 100 mV per division resolution). The oscilloscope measures the voltage across the motor, which is the 5V DC component plus the noise component. Once a 1000μF capacitor is connected in parallel with the motor (one may call it a shunt capacitor), the resulting voltage becomes a highly stable 5V DC (see Fig. 6.20, right).
Fig. 6.20. Effect of bypass capacitor on the highfrequency noise created by a DC motor.
The bypass capacitor in Fig. 6.20 may be considered as a part of the snubber RC circuit, which includes a capacitance and a small series resistance. The snubber circuits are used to suppress highvoltage spikes in inductive switching systems like electric motors.
VI300
Chapter 6
Section 6.3: Application Circuits Highlighting Dynamic Behavior
6.3.2 Blocking Capacitor Quite often, an opposite scenario is desired—we want to block a DC component at the load—see Fig. 6.21. An example is an audio ampliﬁer (the source) connected to a speaker (the load). The audio ampliﬁer may generate an unwanted DC component, which may overheat the speaker coil made of a very thin wire. The idea is to use a capacitor C in series with the (imperfect) ampliﬁer and in series with the load. This blocking or decoupling capacitor will block the DC current at the load as shown in Fig. 6.21b. a)
b)
source RS
vS(t) VS
+
vS(t)
DC AC
+ 
source RS
load
RL VS
load
C
+
+ 
RL
AC
Fig. 6.21. Model of a voltage source connected to a load with a blocking capacitor.
6.3.3 Decoupling Inductor A decoupling inductor is the complement of the bypass capacitor. Consider the circuit shown in Fig. 6.22a. It includes a source represented by its Norton equivalent I S þ iS ðt Þ, RS and a load with an equivalent resistance RL. The source generates an electric current in the form of a DC component and a superimposed AC signal. Assume that we would like to have only the direct current at the load. The idea is to use an inductor L in series with the load, the socalled decoupling inductor as shown in Fig. 6.22b. At a sufﬁciently high frequency, this inductor will block the AC component at the load. source
a)
load
source
b)
load L
iS(t)
AC
iS(t) RS
IS
AC RS
RL
RL
IS
Fig. 6.22. Model of a current source connected to a load with a decoupling inductor.
A common application of the decoupling inductor is the socalled radiofrequency (RF) inductor choke. Here, we’d rather intend to redirect the alternating current. The inductor choke prevents the very weak alternating current received by an antenna to be lost in the lowresistance DC power supply, which powers an ampliﬁer. Instead, it forces the current to ﬂow directly into the input port of the ampliﬁer. In order to model the choke effect in Fig. 6.22, we should in fact interchange the role of two resistances: we consider the load resistance as the DC supply resistance and the source resistance as the desired input resistance of the ampliﬁer. VI301
Chapter 6
Dynamic Circuit Elements
6.3.4 Ampliﬁer Circuits With Dynamic Elements: Miller Integrator Ampliﬁer circuits with the dynamic elements in the negative feedback loop can serve different purposes. In particular, they operate as active ﬁlters. Here, we will introduce the operation concept and present simple examples. The Miller integrator circuit is an invertingampliﬁer circuit considered in Chapter 5, but with the feedback resistance R2 replaced by a capacitance C—see Fig. 6.23. Given a timevarying input voltage signal, the capacitor will conduct a current. Therefore, the negative feedback is still present, even though we now have a capacitance instead of the resistance in the feedback loop. + vC 
+ i1

vx

v*
i2

R
+
vin
+
vout
+

Fig. 6.23. Miller integrator circuit.
The circuit analysis uses two summingpoint constraints: no current into the ampliﬁer and zero differential input voltage. Therefore, the node voltage υ * in Fig. 6.23 is also zero. The currents i1 ¼ iR and i2 ¼ iC in Fig. 6.23 are equal to each other. This yields iC
iR zﬄﬄ}ﬄﬄ{ z}{ dυC d ðυ* υout Þ dυout υR υin υ* υin ¼ C ¼ ¼ ) C ¼C ¼ dt R dt dt R R ðt dυout υin 1 υin ðt 0 Þdt 0 V C ¼ ) υout ¼ C RC dt R
ð6:45Þ
0
where VC is a constant (the initial voltage across the capacitor at t ¼ 0). Thus, an integral of the input voltage (weighted by 1=ðRC Þ) is provided at the output. Interestingly, the time constant τ of the integrator, τ ¼ RC, has the unit of seconds. Example 6.9: The analog pulse counter is an integrator circuit shown in Fig. 6.23 that counts monopolar voltage pulses simply by integrating the input voltage as time progresses. Assume that the input to the ampliﬁer is the voltage shown in Fig. 6.24, where every rectangular voltage pulse of 10ms duration corresponds to a car passing through a gate. Given that R ¼ 10 kΩ, C ¼ 0:1 μF, and that the initial value of the output voltage is reset to zero, how many cars should pass the gate in order to reach the output voltage threshold of 6 V? Solution: The time constant τ of the integrator, τ ¼ RC, is equal to 1 ms.
VI302
Chapter 6
Section 6.3: Application Circuits Highlighting Dynamic Behavior
Example 6.9 (cont.): According to Eq. (6.45), υout ¼ 1000 s1
N X
0:06 V 0:01 s ¼ N 0:6 V
ð6:46Þ
n¼1
where N is the number of pulses (cars). Equating the above expression to 6 V gives N ¼ 10. The time interval between passing cars is not important.
Input voltage, mV 80 60 40 20 0 0
10
30 20 time, ms
40
50
Fig. 6.24. Input signal to the ampliﬁer in Example 6.9.
Along with Example 6.9, other applications of the Miller integrator include various waveshaping circuits.
6.3.5 Compensated Miller Integrator The circuit in Fig. 6.23 will not function in the laboratory, when a realistic ampliﬁer chip is used that is different from the idealampliﬁer model. The reason is that the capacitance is equivalent to an open circuit at DC. Therefore, the feedback loop is simply missing in the Miller integrator at DC, and the entire ampliﬁer circuit becomes a comparator with a very high openloop gain. A small random input offset voltage, VOS, which is present for open circuit
actual opamp ideal opamp

VOS
+
+
R

vout
+

Fig. 6.25. Effect of the input offset voltage on the integrator circuit at DC.
VI303
Chapter 6
Dynamic Circuit Elements
any realistic ampliﬁer IC, will saturate the ampliﬁer toward one of the power rails (depending on the sign of VOS) even if its input is at zero volts (grounded)—see Fig. 6.25. Note that the input offset voltage source, VOS, may be added to either ampliﬁer terminal. Hence, the voltage across the capacitance will approach the rail voltage and the capacitance itself will become permanently charged. The dynamics of this process can be analyzed explicitly, starting with some initial voltage value, say υC ¼ 0 V. In this case, we are allowed to use the negative feedback. For example, given that R ¼ 10 kΩ, C ¼ 0:1 μF, V OS ¼ 5 mV; it takes exactly 1 s to reach the output voltage of 5 V! A similar effect is created by input bias currents to the ampliﬁer. To overcome this issue, a large resistance, RF, is introduced in parallel with C in order to maintain the negative feedback at DC and discharge the capacitance as needed—see Fig. 6.26. If, for example, RF ¼ 10 MΩ, then the capacitance will discharge over time on the order of RF C ¼ 1 s. This estimate is comparable with the estimate for the charging time. As a result, a balance will be established that results in a certain nonzero υout with the output of the ampliﬁer grounded. A further quantitative discussion may be carried out. RF

v*
C

+
R
+
vin
vout
+

Fig. 6.26. Miller integrator improved with a large resistance, RF, in the feedback loop.
6.3.6 Differentiator and Other Circuits When the resistance and the capacitance in Fig. 6.23 are interchanged, a differentiator ampliﬁer circuit (or active differentiator) is obtained; its output signal is a derivative of the input signal. The corresponding solution is studied in one of the homework problems. The differentiators are rarely used in practice since they attempt to amplify any input noise (they become “noise magniﬁers”). The reason for this is an inﬁnitely high gain of the ampliﬁer circuit at high frequencies, when the capacitance becomes a short circuit. A small resistance added in series with the capacitance reduces this effect and assures the ﬁnite gain similar to the standard inverting ampliﬁer. Exercise 6.10: Draw an integrator circuit with an inductance instead of the capacitance. Answer: The circuit in Fig. 6.23 but with the resistance replaced by an inductance and with the capacitance replaced by the resistance.
VI304
Chapter 6
Section 6.3: Application Circuits Highlighting Dynamic Behavior
Inductances might be used instead of capacitances; the ampliﬁer circuits so constructed will be either a differentiator or an integrator. We again pass the corresponding analysis to the homework exercises. However, the physical inductors tend to have a signiﬁcant series resistance and are more bulky. Last but not least, we may ask ourselves a question: as long as the ampliﬁer circuits can perform multiplication, addition (or subtraction), and integration (or differentiation), can we now build an analog computer, which operates with analog voltages and replaces its digital counterpart at least for simple computational tasks? The answer is yes, we can. In fact, this was done a long time ago, in the mid1960s.
VI305
Chapter 6
Dynamic Circuit Elements
Summary Static capacitance and inductance Capacitance
Property
Inductance
Deﬁnition
Q >0 V Q—charge of either conductor; V—volt. between two conductors Units: F ¼ C=V
Φ >0 I Φ—magnetic ﬂux through the circuit; I—circuit current Units: H ¼ ðV sÞ=A
Physical meaning
Charge on either conductor produced by 1 V voltage difference between the two conductors 1 1 E ¼ CV 2 or E ðt Þ ¼ Cυðt Þ2 2 2 1 1 1 1 ¼ þ þ in series C eq C 1 C 2 C 3 C eq ¼ C 1 þ C 2 þ C 3 in parallel
Magnetic ﬂux through the circuit produced by 1 A of current in the same circuit 1 1 E ¼ LI 2 or E ðt Þ ¼ Liðt Þ2 2 2
Stored energy, J (static or dyn.) Series/parallel combinations
L¼
C¼
Leq ¼ L1 þ L2 þ L3 in series 1 1 1 1 ¼ þ þ in parallel Leq L1 L2 L3
Basic models (no fringing ﬁelds) ε0 A εr ε0 A or C ¼ (diel. d d material inside) A ¼ ab, ε0 ¼ 8:854187 1012 F=m C¼
μ0 AN 2 μ μ AN 2 or L ¼ 0 r l l (closed magnetic core ) μ0 ¼ 4π 107 H=m
L¼
Models with fringing Capacitance of a parallelplate square capacitor (Table 6.1) Inductance of a ﬁniteradius solenoid Inductance of a solenoid with a ﬁnite core
8w w2 w4 þ Lexact ¼ L 1 , 3π 2 4
r w ¼ < 1; r is the radius of the coil l
0:5πμ0 l * N 2 l i 1 * , μr 100, see Fig. 6.12c L h* 2l ln lr 1 (continued)
VI306
Chapter 6
Property Dynamic modelPassive ref. conf. Dynamic modelυ–i characteristic Dynamic modelcharge and ﬂux
Summary Dynamic behavior Capacitance
iC ¼ C
Inductance
dυC dt
dqC ¼ iC dt
υL ¼ L
diL dt
dΦðt Þ ¼ υL (passive ref. conf.) dt
Behavior at DC Behavior at very high frequencies Bypassing/Decoupling
Bypass capacitor
Blocking capacitor
Decoupling inductor
Property Miller Integrator (openloop ampliﬁer at DC)
Timedomain operation
Ampliﬁer circuits with capacitor/inductor Capacitance
dυout υin ¼ , τ ¼ RC dt τ
Inductance
dυout υin L ¼ , τ¼ R dt τ (continued)
VI307
Chapter 6
Dynamic Circuit Elements
Compensated integrator with a ﬁnite gain at DC Timedomain operation
dυout υout υin þ ¼ , τ ¼ RC dt Rf C τ
dυout Rin υin L þ υout ¼ , τ ¼ R dt L τ
Differentiator circuit (inﬁnite gain at very high frequencies) Timedomain operation
υout ¼ τ
dυin , τ ¼ RC dt
υout ¼ τ
dυin L , τ¼ R dt
Differentiator with a ﬁnite gain at very high frequencies Timedomain operation
Rin C
dυout dυin þ υout ¼ τ , τ ¼ RC dt dt
L dυout dυin þ υout ¼ τ , τ ¼ L=R Rf dt dt
VI308
Chapter 6
Problems
Problems 6.1 Static Capacitance and Inductance
+ a
0 r
h E
6.1.1 Capacitance, Selfcapacitance, and Capacitance to Ground 6.1.2 Application Example: ESD Problem 6.1. A. Describe in your own words the physical meaning of capacitance. B. Suggest a way to memorize the expression for the capacitance of two conductors. C. What is approximately the selfcapacitance of a human body? D. What is approximately the capacitance of a human body (to ground)? E. How does the humanbody selfcapacitance change in embryo pose (yoga)? Problem 6.2. A metal square plate with the side of 10 mm has the selfcapacitance C self ¼ 0:41 pF. A. Estimate the capacitance of a capacitor formed by two such parallel plates separated by 30 mm. Compare this value to the exact result of 0.23 pF. B. Estimate the capacitance of the plate to ground when the separation distance is 30 mm. Compare this value to the exact result of 0.43 pF. Problem 6.3. Draw the basic electriccircuit model of a human body and specify the generic element values.
b
vC
A. Determine capacitance of the capacitor, C. B. Determine the electric ﬁeld strength, E (in V/m), within the capacitor volume and total charge, Q, on either capacitor plate if the applied voltage is 25 V. C. Determine the electric ﬁeld energy stored in the capacitor if the applied voltage is 25 V. Problem 6.5. Solve the previous problem when the separation distance, h, between the plates is reduced to 100 μm. Problem 6.6. For the enclosedcylinder capacitor shown in the ﬁgure, a ¼ 10 cm, b ¼ 9:99 cm, H ¼ 5 cm, εr ¼ 16, the electrodes are the inner and outer cylinder surfaces, respectively. A. Determine the capacitance of the capacitor, C. B. Determine the electric ﬁeld strength, E (in V/m), within the capacitor volume and total charge, Q, on either capacitor plate if the applied voltage is 50 V. C. Determine the electric ﬁeld energy stored in the capacitor if the applied voltage is 25 V. Hint: The capacitance per unit area of the
device is that of the parallelplate capacitor. 6.1.3 ParallelPlate Capacitor 6.1.4 Capacitances in Parallel and V + in Series C
Problem 6.4. For the parallelplate capacitor schematically shown in the ﬁgure,
a
0 r
b
H
a ¼ 10 cm, b ¼ 20 cm, h ¼ 1 mm, εr ¼ 12
VI309
Chapter 6
Dynamic Circuit Elements
Problem 6.7. Solve the previous problem when the separation distance between the two electrodes is reduced to 25 μm. Problem 6.8. A cross section of the rolled capacitor is approximated by a rational spiral shown in the ﬁgure. Here, a ¼ 0:25 cm, εr ¼ 10, the separation distance between the two conductors is 20 μm, and the height of the entire roll is 1.5 cm. A. Determine capacitance of the capacitor, C. B. Determine the electric ﬁeld strength, E (in V/m), within the capacitor and total charge, Q, on either capacitor plate if the applied voltage is 12 V. C. Determine the electric ﬁeld energy stored in the capacitor if the applied voltage is 12 V.
47 F a 47 F b 47 F
Problem 6.11. Find the equivalent capacitance for each circuit shown in the ﬁgure below. a)
0 r
VC
+ 
10 F
10 F
a
15 F 15 F
47 F
b
b)
10 F
a
10 F
47 F 15 F
b
Problem 6.12. Find the equivalent capacitance for each circuit shown in the following ﬁgure. a)
33 nF
33 nF
a
Problem 6.9. Determine the capacitance of the three leaded capacitors shown in the ﬁgure (from left to right).
15 F
15 F
Hint: The capacitance per unit area of the device is that of the parallelplate capacitor. a
47 F
b
33 nF
33 nF
2.2 pF
2.2 pF
2.2 pF
2.2 pF
b) a
b
Problem 6.10. Find the equivalent capacitance for the circuit shown in the following ﬁgure.
Problem 6.13. Two 1μF capacitances have an initial voltage of 10 V and 0 V, respectively (before the switch is closed), as shown in the ﬁgure. Find the total electric energy stored in the system before the switch is closed. Find the
VI310
Chapter 6
Problems
voltage across each capacitance and the total stored energy after the switch is closed. What could have happened to the missing energy?
surrounding space if the applied current, IL, is 0.5 A.
t=0
+ 10 V
1 F
+
1 F
0V


6.1.7 Selfinductance (Inductance) and Mutual Inductance 6.1.8 Inductance of a Solenoid With and Without Magnetic Core 6.1.9 Circuit Symbol. Inductances in Series and in Parallel Problem 6.14. A. Describe in your own words the physical meaning of inductance. B. Do you think a straight wire has a certain inductance per unit length? You might want to ask the TA and/or browse the Web and present the corresponding expression (if any). Problem 6.15. Three aircore inductors of the same cross section are shown in the following ﬁgure. The inductor length is proportional to the number of turns. Find the ratios of inductances: L2/L1, L3/L2, L3/L1. + +
+ L1
L3
L2


Problem 6.16. The solenoid shown in the ﬁgure has a diameter d ¼ 1 cm and a length l ¼ 10 cm. A. Find the solenoid’s inductance, L, using the common assumption d=l > > : α 1 þ ς2 where the new constant ζ ¼ α=ω0 is the damping ratio of the RLC circuit. Formally, this constant has units of 1/rad; it is often considered dimensionless. We must distinguish between three separate cases depending on the value of the damping ratio: Case A This situation (overdamping) corresponds to ς > 1. In this case, s1,2 are both real and negative. Since the original ODE is linear, the general solution is simply the combination of two independent decaying exponential functions:
xc ðt Þ ¼ K 1 expðs1 t Þ þ K 2 expðs2 t Þ
ð7:84aÞ
Case B This case (critical damping) corresponds to ς ¼ 1. Both roots s1,2 become identical. Therefore, a solution in the form of Eq. (7.84a) with two independent constants can no longer be formed. Only one independent constant may be available. Fortunately, another solution in the form t exp(s1t) exists in this special case. This fact is proved by direct substitution. Thus, the general solution becomes
xc ðt Þ ¼ K 1 expðs1 t Þ þ K 2 texpðs1 t Þ
ð7:84bÞ
Case C This case (underdamping) corresponds to ς < 1. Both roots s1,2 become complex. This means that our initial simple guess Eq. (7.81) is no longer correct. One can prove by direct substitution that the general solution now has the oscillating form
xc ðt Þ ¼ K 1 expðα t Þ cos ωn t þ K 2 expðα t Þ sin ωn t
ð7:84cÞ
pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ where ωn ¼ ω20 α2 is the (radian) natural frequency of the circuit. 1/α is also called the time constant or the time constant of the decay envelope. The complementary solution in the form of Eqs. (7.84) always contains two independent integration constants. They should be used to satisfy the initial conditions, which complete the solution.
VII367
Chapter 7
Transient Circuit Fundamentals
7.6.3 Finding Integration Constants According to Eqs. (7.79) and (7.80), the capacitor voltage is υC ðt Þ ¼ xc ðt Þ þ V S where xc(t) is given by Eqs. (7.84). The integration constants may be found using Eq. (7.78b), which dictates that both the capacitor voltage and its derivative must vanish at the initial time t ¼ 0. We then have from Eqs. (7.84) Case A: K 1 þ K 2 þ V S ¼ 0, s1 K 1 þ s2 K 2 ¼ 0 Case B: K 1 þ V S ¼ 0, s1 K 1 þ K 2 ¼ 0 Case C: K 1 þ V S ¼ 0, αK 1 þ ωn K 2 ¼ 0
ð7:85aÞ ð7:85bÞ ð7:85cÞ
The solution of Eqs. (7.85) has the form s2 V S s1 V S , K2 ¼ s1 s2 s2 s1 Case B: K 1 ¼ V S , K 2 ¼ s1 V S α Case C: K 1 ¼ V S , K 2 ¼ V S ωn Case A:
K1 ¼
ð7:86aÞ ð7:86bÞ ð7:86cÞ
Equations (7.83) through (7.86) complete the step response solution for the series RLC circuit. The circuit may behave quite differently depending on the value of the damping ratio ς.
7.6.4 Solution Behavior for Different Damping Ratios We consider the series RLC circuit shown in Fig. 7.35. We will choose round numbers L ¼ 1 mH, C ¼ 1 nF. These values approximately correspond to an RLC transient circuit operating in the 100 kHz–1 MHz frequency band.
+ 10 V=VS
R vR

t=0
+ 
i(t) 1 mH
1 nF

+ vL 
vC +
Fig. 7.35. RLC series circuit; the resistance value R may be varied.
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Chapter 7
Section 7.6: Step Response of the Series RLC Circuit
Example 7.9: Determine the solution for the capacitor voltage for the circuit shown in Fig. 7.35 over a time interval from 0 to 25 μs for R ¼ 200 Ω, 1 kΩ, 2 kΩ, and 20 kΩ. Solution: Equations (7.83) through (7.86) give R ¼ 200 Ω ) α ¼ 105 , ω0 ¼ 106 ) ζ ¼ 0:1 (Case C—underdamped circuit) ) υC ðt Þ ¼ 10 10exp 105 t cos 9:95 105 t 1:005exp 105 t sin 9:95 105 t
ð7:87aÞ
R ¼ 1 kΩ ) α ¼ 5 105 , ω0 ¼ 106 ) ζ ¼ 0:5 (Case C—underdamped circuit) ) υC ðt Þ ¼ 10 10exp 5 105 t cos 8:66 105 t ð7:87bÞ 5:77exp 5 105 t sin 8:66 105 t R ¼ 2 kΩ ) α ¼ 106 , ω0 ¼ 106 ) ζ ¼ 1 (Case B—critically damped circuit) ) υC ðt Þ ¼ 10 10exp 106 t 107 texp 106 t
ð7:87cÞ
R ¼ 20 kΩ ) α ¼ 107 , ω0 ¼ 106 ) ζ ¼ 10 (Case A—overdamped circuit) ) υC ðt Þ ¼ 10 10:0252exp 5:013 104 t þ 0:0252exp 1:995 107 t
ð7:87dÞ
Equations (7.87) satisfy both the initial conditions to within numerical rounding error. Figure 7.36 shows the solution behavior for four distinct cases.
The ﬁrst person who discovered and documented the oscillatory transient response of an electric circuit similar to that depicted in Fig. 7.36 was probably Félix Savary (1797– 1841). A renowned astronomer and French academician, he worked with Ampère and discovered an oscillatory discharge of a Leyden jar (an early prototype of the battery) in 1823–1826. Some ﬁfteen years later, the similar observation has been made by Joseph Henry.
7.6.5 Overshoot and Rise Time One important result seen in Fig. 7.36a, b is the socalled dynamic overshoot caused by a sudden application of a voltage pulse and the associated voltage ringing. The dimensionless overshoot (overshoot percentage after multiplying by 100) Mp is the maximum voltage value minus the supply voltage divided by the supply voltage. For a slightly
VII369
Chapter 7
Transient Circuit Fundamentals
damped circuit, the overshoot may be quite large—see Fig. 7.36a, b. The rise time tr (which is sometimes called the “transition time”) in digital circuits is the time taken for the voltage to rise from 0.1VS to 0.9VS (T. L. Floyd, Digital Fundamentals, 9th, p. 8); see Figs. 8.14a–c. The circuit designer typically attempts to minimize both the rise time and the overshoot. An important example considered later in our text is a pulse train to be transmitted at a maximum speed (which requires minimum rise time) and with minimum distortion (which requires minimum overshoot). Figure 7.36 indicates that those goals are in fact conﬂicting. Decreasing the rise time increases the overshoot and vice versa. Designing the damping ratio close to unity, or slightly below it, is a reasonable compromise to quickly achieve the desired voltage level without a signiﬁcant overshoot and ringing. The overshoot and rise time may be estimated analytically. We present here the estimates found in common control theory textbooks: expðπζ Þ M p ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ f or ζ < 1, M p ¼ 0 f or ζ 1 1 ζ2 t r ¼ 1 0:4167ζ þ 2:917ζ 2 =ωn f or ζ < 1
ð7:88aÞ ð7:88bÞ
Exercise 7.15: The damping coefﬁcient of 15,000 neper/s and the natural frequency of 10 kHz are measured for an unknown series RLC circuit in laboratory via its step response. Given R ¼ 10 Ω, determine L and C. Answer: L ¼ 0:333 mH, C ¼ 0:719 μF.
As to mechanical engineering, it is interesting to note that an automotive suspension system is described by the same step response model and behaves quite similarly to the RLC circuit in Fig. 7.36. Well, driving a car everyday should certainly be a motivation for studying this topic.
7.6.6 Application Example: Nonideal Digital Waveform Modeling Circuit The series RLC block is an appropriate model to study the voltage pulse as it realistically occurs in digital circuits and in power electronic circuits involving pulse width modulation (PWM). The ideal square voltage pulse, shown in Fig. 7.33 of the previous section, is a crude approximation of reality. Parasitic capacitance, resistance, and inductance are always present in the circuit. As a result, the pulse form is distorted. To model the pulse form distortion, we consider the circuit shown in Fig. 7.37.
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Chapter 7
Section 7.6: Step Response of the Series RLC Circuit
R=200 , =0.1 Case C underdamped circuit
Capacitor voltage, V
20 overshoot
16 12
VS
8 rise time
4
R= k , =0.5 Case C underdamped circuit
Capacitor voltage, V
0 20 16 overshoot
12
VS
8 rise time
4 0
R=2k , =1.0 Case B critically damped circuit
Capacitor voltage, V
20 16 no overshoot
12
VS
8 rise time
4 0
R=20k , =10 Case A overdamped circuit
Capacitor voltage, V
20 16 no overshoot
12
VS
8 rise time 4 0 0
5.0
10
15
20
t, s
25
Fig. 7.36. Circuit responses in terms of capacitor voltages for different damping factors.
VII371
Chapter 7
Transient Circuit Fundamentals a)
b)
R
vS(t)
VS vS(t)
+ 
L C

vC +
Falling (trailing) edge
Rising (leading) edge
0
T
t
Fig. 7.37. The RLC circuit for studying the nonideal digital (pulse) waveform.
The switch is now removed and a timevarying voltage source υS(t) is introduced; it generates the voltage pulse (one bit) of duration T with amplitude VS as seen in Fig. 7.37b. One may think of the source voltage as an “ideal” digital waveform and, for example, of the capacitor voltage as a nonideal (realistic) waveform inﬂuenced by parasitic capacitance, resistance, and inductance.
Solution The solution to the pulse problem is derived as described at the end of the previous section. We know that Eqs. (7.83) through (7.86) determine the step response—the capacitor voltage υC(t) for the circuit with the DC voltage source shown in Fig. 7.35 after closing the switch. To obtain the solution υpulse C (t) for the voltage pulse shown in Fig. 7.37, we simply combine two such step responses, i.e., υCpulse ðt Þ ¼ υC ðt Þ υC ðt T Þ
ð7:89Þ
This operation again underscores the importance of the fundamental step response solution. Close inspection of Eq. (7.89) shows that the pulse will possess the dynamic overshoot and the nonzero rise time similar to the step response solution. This happens at the rising (or leading) edge of the pulse. At the same time, a dynamic undershoot and a nonzero settling time will happen at the falling or trailing edge as depicted in Fig. 7.38. Example 7.10: Determine the solution for the capacitor voltage, υpulse C (t), for the circuit shown in Fig. 7.37 with L ¼ 1 μH, C ¼ 1 nF, V S ¼ 10 V, T ¼ 0:5 μs over the time interval from 0 to 1 μs for R ¼ 15 Ω, 30 Ω, and 60 Ω. Solution: We ﬁnd the step response υC(t) following Eqs. (7.83) through (7.86) ﬁrst and then obtain the ﬁnal solution using Eq. (7.89). For the step response, we obtain R ¼ 15 Ω ) ζ ¼ 0:24 (Case C—underdamped circuit) ) υC ðt Þ ¼ 10 10exp 7:5 106 t cos 3:07 107 t 2:44exp 7:5 106 t sin 3:07 107 t ð7:90aÞ
VII372
Chapter 7
Section 7.6: Step Response of the Series RLC Circuit
Example 7.10 (cont.): R ¼ 30 Ω ) ζ ¼ 0:47 (Case C—underdamped circuit) ) υC ðt Þ ¼ 10 10exp 1:5 107 t cos 2:78 107 t 5:39exp 1:5 107 t sin 2:78 107 t
ð7:90bÞ
R ¼ 60 Ω ) ζ ¼ 0:95 (Case C—underdamped circuit) ) υC ðt Þ ¼ 10 10exp 3:0 107 t cos 1:0 107 t 30exp 3:0 107 t sin 1:0 107 t
ð7:90cÞ
Figure 7.38a–c shows the distorted pulse forms for three particular cases. Figure 7.38a outlines the major pulse parameters: rise time, fall time, overshoot, undershoot, and pulse width. One can see that there is again a conﬂict between the desire to simultaneously decrease the rise time and the overshoot.
The overshoot and undershoot in Fig. 7.38 approximately coincide, and so do the rise time and the fall or settling time. Note that this is not always the case. The voltage pulse may be very signiﬁcantly and unsymmetrically distorted when the initial pulse width, T, is comparable with the rise time. A good illustration is the previous example solved for R ¼ 15 Ω when T ¼ 0:25 μs or less. Exercise 7.16: Using a theoretical approximation, ﬁnd the overshoot for the case of Fig. 7.38a and compare this value with value observed on the ﬁgure. Answer: 48 % (theory) versus 50 % (observation).
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Chapter 7
Transient Circuit Fundamentals
overshoot ringing
R=15W, =0.24 Case C underdamped circuit
Capacitor voltage, V
a)
10
VS 90%
90%
5
50%
50%
pulse width
ringing
10%
10%
0
undershoot fall time
5 rise time 10
b)
R=30W, =0.47 Case C underdamped circuit
Capacitor voltage, V
overshoot 10
VS
90%
5
50%
0
10% undershoot
5 rise time 10
R=60W, =0.95 Case C underdamped circuit
Capacitor voltage, V
c)
VS
10
90%
5
50%
0
10%
5 rise time 10 0
0.2
0.4
0.6
0.8
t, ms
Fig. 7.38. Distorted pulse forms for three different values of the damping ratio. When the damping ratio increases, the overshoot decreases but the rise time increases.
The solution for the secondorder circuits with arbitrary sources and arbitrary initial conditions can quite simply be obtained numerically. A straightforward ﬁnitedifference secondorder method may be implemented in MATLAB or in other software packages with a few lines of the code. This method is the extension of the Euler method used for ﬁrstorder transient circuits. Interestingly, the same method may be applied to radiofrequency pulse propagation in transmission lines and in free space, including problems such as signal penetration through walls.
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Chapter 7
Summary
Summary General facts about transient circuits Voltage across the capacitor(s) remains continuous for all times—use this voltage as an unknown function Current through the inductor(s) remains continuous for all times—use this current as an unknown function When discharged through a small resistance, the (ideal) capacitor is able to deliver an extremely high power (and current) during a very short period of time When disconnected from the source, the (ideal) inductor in series with a large resistance is able to deliver an extremely high power (and voltage) during a very short period of time Transient circuit Generic circuit diagram Solution plot Energyrelease RC circuit t υC ðt Þ ¼ V 0 exp τ τ ¼ RC ODE: dυC υC þ ¼0 dt τ Energyaccumulating RC circuit h t i υC ðt Þ ¼ VS 1 exp τ τ ¼ RC ODE: dυC υC VS þ ¼ dt τ τ Energyrelease RL circuit t iL ðt Þ ¼ I S exp τ τ ¼ L=R ODE: diL iL þ ¼0 dt τ Energyaccumulating RL circuit h t i iL ðt Þ ¼ I S 1 exp τ τ ¼ L=R ODE: diL iL I S þ ¼ dt τ τ Energyrelease RL circuit t iL ðt Þ ¼ I 0 exp τ τ ¼ L=R, I 0 ¼ VS =R0 ODE: diL iL VS þ ¼ dt τ R0 τ (continued)
VII375
Chapter 7
Transient Circuit Fundamentals Voltage or current
Power
Generic energyrelease curves for either dynamic element
Switching RC oscillator (relaxation oscillator)—RC timer Bistable ampliﬁer circuit with positive feedback. Two stable states: υout ¼ þVCC υout ¼ VCC Threshold voltage(s): R1 υ* ¼ υout R1 þ R2 Relaxation oscillator with positive feedback: υout ¼ þVCC if υ* > υC υout ¼ VCC if υ* < υC Threshold voltage(s): R1 υ* ¼ υout R1 þ R2 Period and frequency of the relaxation oscillator ðβ ¼ R1 =ðR1 þ R2 ÞÞ
T ¼ 2τln
1þβ 1β
1 1 1 þ β 1 ln f ¼ ¼ T 2τ 1β
Singletimeconstant (STC) transient circuits τ ¼ RC or L τ¼ R τ ¼ R1 R2 C or τ¼
L R1 þ R2
τ ¼ RðC 1 þ C 2 Þ or τ¼
L1 þ L2 R (continued)
VII376
Chapter 7
Summary
L1 þ L2 R1 þ R2 τ ¼ R1 R2 ðC 1 þ C 2 Þ τ¼
Arbtr. initial conditions are not allowed Arbtr. initial conditions are not allowed STC circuits with general sources
Bypass capacitor and decoupling inductor
Solution for load voltage or load current
υS ðt Þ ¼ VS þVm cos ω t, τ ¼ RS RL C t RL VS υL ¼ 1 exp τ RS þ RL þ h
RL V m 1 RS þ RL 1 þ ðωτÞ2
iS ðt Þ ¼ I S þ I m cos ω t, τ ¼ L=ðRS þ RL Þ t RS I S 1exp iL ¼ τ RS þRL RS I m 1 þ RS þRL 1þ ðωτÞ2 h t i cosωtþωτsinωtexp τ
t i cosωt þ ωτ sinωt exp τ Secondorder transient circuits
With two identical dynamic elements • Damping coefﬁcient: α ¼ R=ð2LÞ pﬃﬃﬃﬃﬃﬃﬃ • Undamped res. freq.: ω0 ¼ 1= LC With series LC network • Step response with zero initial conditions: use capacitor voltage υC(t)
• Damping ratio: ζ ¼ α=ω0 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ • Natural freq.: ωn ¼ ω20 α2 Characteristic Eq.: s2 þ 2αs þ ω20 ¼ 0 • Damping coefﬁcient: α ¼ 1=ð2RC Þ pﬃﬃﬃﬃﬃﬃﬃ • Undamped res. freq.: ω0 ¼ 1= LC
With parallel LC network
• Step response with zero initial conditions: use inductor current iL(t)
• Damping ratio: ζ ¼ α=ω0 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ • Natural freq.: ωn ¼ ω20 α2 Characteristic Eq.: s2 þ 2αs þ ω20 ¼ 0 (continued)
VII377
Chapter 7
Transient Circuit Fundamentals • Overdamped (ζ > 1): xc ðt Þ ¼ K 1 expðs1 t Þ þ K 2 expðs2 t Þ K1 ¼
Overdamped, criticallydamped, and underdamped RLC circuits
s2 V S s1 V S , K2 ¼ s1 s2 s2 s1
• Critically damped (ζ ¼ 1): xc ðt Þ ¼ K 1 expðs1 t Þ þ K 2 t expðs1 t Þ K 1 ¼ V S , K 2 ¼ s1 V S • Underdamped (ζ < 1): xc ðt Þ ¼ K 1 expðα t Þ cos ωn tþ K 2 expðα t Þ sin ωn t K 1 ¼ VS , K 2 ¼ ωαn VS
Nonideal digital waveform: secondorder circuit
• Overshoot expðπζ Þ M p ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ f or ζ < 1 1 ζ2 • Undershoot is approximately overshoot for rise times small compared to pulse width • Rise time t r ¼ 10:4167ζþ2:917ζ 2 =ωn f or ζ