Schaum's Outline of Electronic Devices and Circuits, Second Edition [by Jimmie J Cathey] pdf

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Theory and Problems of

ELECTRONIC

DEVICES

AND CIRCUITS

Second Edition

JIMMIE J. CATHEY, Ph.D.

Professor of Electrical Engineering

University of Kentucky

Schaum’s Outline Series

McGRAW-HILL

New York Chicago San Francisco Lisbon London Madrid Mexico City

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may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher.

0-07-139830-9

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control devices in circuit applications. The emphasis in this book is on the latter category, beginning with the terminal characteristics of electronic control devices. Other topics are dealt with only as necessary to an understanding of these terminal characteristics.

This book is designed to supplement the text for a first course in electronic circuits for engineers. It will also serve as a refresher for those who have previously taken a course in electronic circuits. Engineering students enrolled in a nonmajors’ survey course on electronic circuits will find that portions of Chapters 1 to 7 offer a valuable supplement to their study. Each chapter contains a brief review of pertinent topics along with governing equations and laws, with examples inserted to immediately clarify and emphasize principles as introduced. As in other Schaum’s Outlines, primary emphasis is on the solution of problems; to this end, over 350 solved problems are presented.

Three principal changes are introduced in the second edition. SPICE method solutions are presented for numerous problems to better correlate the material with current college class methods. The first-edition Chapter 13 entitled ‘‘Vacuum Tubes’’ has been eliminated. However, the material from that chapter relating to triode vacuum tubes has been dispersed into Chapters 4 and 7. A new Chapter 10 entitled ‘‘Switched Mode Power Supplies’’ has been added to give the reader exposure to this important technology.

SPICE is an acronym for Simulation Program with Integrated Circuit Emphasis. It is commonly used as a generic reference to a host of circuit simulators that use the SPICE2 solution engine developed by U.S. government funding and, as a consequence, is public domain software. PSpice is the first personal computer version of SPICE that was developed by MicroSim Corporation (purchased by OrCAD, which has since merged with Cadence Design Systems, Inc.). As a promotional tool, Micro-Sim made available several evaluation versions of PSpice for free distribution without restriction on usage. These evaluation versions can still be downloaded from many websites. Presently, Cadence Design Systems, Inc. makes available an evaluation version of PSpice for download by students and professors atwww.orcad.com/Products/Simulation/PSpice/eval.asp.

The presentation of SPICE in this book is at the netlist code level that consists of a collection of element-specification statements and control statements that can be compiled and executed by most SPICE solution engines. However, the programs are set up for execution by PSpice and, as a result, contain certain control statements that are particular to PSpice. One such example is the .PROBE statement. Probe is the proprietary PSpice plot manager which, when invoked, saves all node voltages and branch currents of a circuit for plotting at the user’s discretion. Netlist code for problems solved by SPICE methods in this book can be downloaded at the author’s website www.engr.uky.edu/cathey. Errata for this book and selected evaluation versions of PSpice are also available at this website.

The book is written with the assumption that the user has some prior or companion exposure to SPICE methods in other formal course work. If the user does not have a ready reference to SPICE analysis methods, the three following references are suggested (pertinent version of PSpice is noted in parentheses):

1. SPICE: A Guide to Circuit Simulation and Analysis Using PSpice, Paul W. Tuinenga, Prentice-Hall, Englewood Cliffs, NJ, 1992, ISBN 0-13-747270-6 (PSpice 4).

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2. Basic Engineering Circuit Analysis, 6/e, J. David Irwin and Chwan-Hwa Wu, John Wiley & Sons, New York, 1999, ISBN 0-471-36574-2 (PSpice 8).

3. Basic Engineering Circuit Analysis, 7/e, J. David Irwin, John Wiley & Sons, New York, 2002, ISBN 0-471-40740-2 (PSpice 9).

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CHAPTER 1

Circuit Analysis: Port Point of View

1

1.1 Introduction

1

1.2 Circuit Elements

1

1.3 SPICE Elements

2

1.4 Circuit Laws

3

1.5 Steady-State Circuits

4

1.6 Network Theorems

4

1.7 Two-Port Networks

8

1.8 Instantaneous, Average, and RMS Values

13

CHAPTER 2

Semiconductor Diodes

30

2.1 Introduction

30

2.2 The Ideal Diode

30

2.3 Diode Terminal Characteristics

32

2.4 The Diode SPICE Model

33

2.5 Graphical Analysis

35

2.6 Equivalent-Circuit Analysis

38

2.7 Rectifier Applications

40

2.8 Waveform Filtering

42

2.9 Clipping and Clamping Operations

44

2.10 The Zener Diode

46

CHAPTER 3

Characteristics of Bipolar Junction Transistors

70

3.1 BJT Construction and Symbols

70

3.2 Common-Base Terminal Characteristics

71

3.3 Common-Emitter Terminal Characteristics

71

3.4 BJT SPICE Model

72

3.5 Current Relationships

77

3.6 Bias and DC Load Lines

78

3.7 Capacitors and AC Load Lines

82

CHAPTER 4

Characteristics of Field-Effect Transistors and Triodes 103

4.1 Introduction

103

4.2 JFET Construction and Symbols

103

4.3 JFET Terminal Characteristics

103

v

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4.4 JFET SPICE Model

105

4.5 JFET Bias Line and Load Line

107

4.6 Graphical Analysis for the JFET

110

4.7 MOSFET Construction and Symbols

110

4.8 MOSFET Terminal Characteristics

110

4.9 MOSFET SPICE Model

111

4.10 MOSFET Bias and Load Lines

114

4.11 Triode Construction and Symbols

115

4.12 Triode Terminal Characteristics and Bias

115

CHAPTER 5

Transistor Bias Considerations

136

5.1 Introduction

136

5.2

b

Uncertainty and Temperature Effects in the BJT

136

5.3 Stability Factor Analysis

139

5.4 Nonlinear-Element Stabilization of BJT Circuits

139

5.5

Q

-Point-Bounded Bias for the FET

140

5.6 Parameter Variation Analysis with SPICE

141

CHAPTER 6

Small-Signal Midfrequency BJT Amplifiers

163

6.1 Introduction

163

6.2 Hybrid-Parameter Models

163

6.3 Tee-Equivalent Circuit

166

6.4 Conversion of Parameters

167

6.5 Measures of Amplifier Goodness

168

6.6 CE Amplifier Analysis

168

6.7 CB Amplifier Analysis

170

6.8 CC Amplifier Analysis

171

6.9 BJT Amplifier Analysis with SPICE

172

CHAPTER 7

Small-Signal Midfrequency FET and Triode Amplifiers 200

7.1 Introduction

200

7.2 Small-Signal Equivalent Circuits for the FET

200

7.3 CS Amplifier Analysis

201

7.4 CD Amplifier Analysis

202

7.5 CG Amplifier Analysis

203

7.6 FET Amplifier Gain Calculation with SPICE

203

7.7 Graphical and Equivalent Circuit Analysis of Triode

Amplifiers

205

CHAPTER 8

Frequency Effects in Amplifiers

226

8.1 Introduction

226

8.2 Bode Plots and Frequency Response

227

8.3 Low-Frequency Effect of Bypass and Coupling Capacitors

229

8.4 High-Frequency Hybrid-

BJT Model

232

8.5 High-Frequency FET Models

234

8.6 Miller Capacitance

235

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CHAPTER 9

Operational Amplifiers

258

9.1 Introduction

258

9.2 Ideal and Practical OP Amps

258

9.3 Inverting Amplifier

259

9.4 Noninverting Amplifier

260

9.5 Common-Mode Rejection Ratio

260

9.6 Summer Amplifier

261

9.7 Differentiating Amplifier

262

9.8 Integrating Amplifier

262

9.9 Logarithmic Amplifier

263

9.10 Filter Applications

264

9.11 Function Generators and Signal Conditioners

264

9.12 SPICE Op Amp Model

265

CHAPTER 10

Switched Mode Power Supplies

287

10.1 Introduction

287

10.2 Analytical Techniques

287

10.3 Buck Converter

289

10.4 Boost Converter

290

10.5 Buck-Boost Converter

292

10.6 SPICE Analysis of SMPS

294

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1

Circuit Analysis: Port

Point of View

1.1. INTRODUCTION

Electronic devices are described by their nonlinear terminal voltage-current characteristics. Circuits containing electronic devices are analyzed and designed either by utilizing graphs of experimentally measured characteristics or by linearizing the voltage-current characteristics of the devices. Depending upon applicability, the latter approach involves the formulation of either small-perturbation equations valid about an operating point or a piecewise-linear equation set. The linearized equation set describes the circuit in terms of its interconnected passive elements and independent or controlled voltage and current sources; formulation and solution require knowledge of the circuit analysis and circuit reduction principles reviewed in this chapter.

1.2. CIRCUIT ELEMENTS

The time-stationary (or constant-value) elements of Fig. 1-1(a) to (c) (the resistor, inductor, and capacitor, respectively) are calledpassive elements, since none of them can continuously supply energy to a circuit. For voltagevand currenti, we have the following relationships: For the resistor,

v¼Ri or i¼Gv ð1:1Þ

whereRis itsresistancein ohms (), andG1=Ris itsconductancein siemens (S). Equation (1.1) is known asOhm’s law. For the inductor,

v¼L di

dt or i¼

1 L

ðt

1v d ð1:2Þ

whereLis itsinductancein henrys (H). For the capacitor,

v¼1 C

ðt

1i d or i¼C

dv

dt ð1:3Þ

whereC is its capacitancein farads (F). IfR, L, and Care independent of voltage and current (as well as of time), these elements are said to be linear: Multiplication of the current through each by a constant will result in the multiplication of its terminal voltage by that same constant. (See Problems 1.1 and 1.3.)

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The elements of Fig. 1-1(d) to (h) are calledactive elementsbecause each is capable of continuously supplying energy to a network. Theideal voltage sourcein Fig. 1-1(d) provides a terminal voltagevthat is independent of the currentithrough it. Theideal current sourcein Fig. 1-1(e) provides a currentithat is independent of the voltage across its terminals. However, thecontrolled(ordependent)voltage source

in Fig. 1-1(f) has a terminal voltage that depends upon the voltage across or current through some other element of the network. Similarly, thecontrolled(ordependent)current sourcein Fig. 1-1(g) provides a current whose magnitude depends on either the voltage across or current through some other element of the network. If the dependency relation for the voltage or current of a controlled source is of the first degree, then the source is called alinearcontrolled (or dependent) source. Thebatteryordc voltage sourcein Fig. 1-1(h) is a special kind of independent voltage source.

1.3. SPICE ELEMENTS

The passive and active circuit elements introduced in the previous section are all available in SPICE modeling; however, the manner of node specification and the voltage and current sense or direction are clarified for each element by Fig. 1-2. The universal ground node is assigned the number 0. Otherwise, the node numbersn1(positive node) andn2(negative node) are positive integers

i i i i i i

i L

+

_ L R

+

_ L

+

_

+

_

+

_

+

_

+

_

L L L L

+

_ V i

L C

(a) (b) (c) (d) (e) ( f ) (g) (h)

Fig. 1-1

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selected to uniquely define each node in the network. The assumed direction of positive current flow is from noden1 to noden2.

The four controlled sources—voltage-controlled voltage source (VCVS), current-controlled voltage source (CCVS), voltage-controlled current source (VCCS), and current-controlled current source (CCCS)— have the associated controlling element also shown with its nodes indicated bycn1(positive) andcn2(negative). Each element is described by anelement specification statementin the SPICE netlist code. Table 1-1 presents the basic format for the element specification statement for each of the elements of Fig. 1-2. The first letter of the element name specifies the device and the remaining characters must assure a unique name.

1.4. CIRCUIT LAWS

Along with the three voltage-current relationships (1.1) to (1.3), Kirchhoff’s laws are sufficient to formulate the simultaneous equations necessary to solve for all currents and voltages of a network. (We use the termnetworkto mean any arrangement of circuit elements.)

Kirchhoff’s voltage law(KVL) states thatthe algebraic sum of all voltages around any closed loop of a circuit is zero; it is expressed mathematically as

Xn

k¼1

vk¼0 ð1:4Þ

wherenis the total number of passive- and active-element voltages around the loop under consideration.

Kirchhoff’s current law(KCL) states thatthe algebraic sum of all currents entering every node (junc-tion of elements)must be zero; that is

Xm

k¼1

ik¼0 ð1:5Þ

wheremis the total number of currents flowing into the node under consideration.

Table 1-1

Element Name Signal Type Control Source Value

Resistor R:::

Inductor L::: H

Capacitor C::: F

Voltage source V::: AC or DCa Vb

Current source I::: AC or DCa Ab

VCVS E::: ðcn1;cn2Þ V/V

CCVS H::: V::: V/A

VCCS G::: ðcn1;cn2Þ A/V

CCCS F::: V::: A/A

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1.5. STEADY-STATE CIRCUITS

At some (sufficiently long) time after a circuit containing linear elements is energized, the voltages and currents become independent of initial conditions and the time variation of circuit quantities becomes identical to that of the independent sources; the circuit is then said to be operating in the

steady state. If all nondependent sources in a network are independent of time, the steady state of the network is referred to as thedc steady state. On the other hand, if the magnitude of each nondependent source can be written asKsinð!tþÞ, whereKis a constant, then the resulting steady state is known as thesinusoidal steady state, and well-known frequency-domain, or phasor, methods are applicable in its analysis. In general, electronic circuit analysis is a combination of dc and sinusoidal steady-state analysis, using the principle of superposition discussed in the next section.

1.6. NETWORK THEOREMS

Alinear network(orlinear circuit) is formed by interconnecting the terminals of independent (that is, nondependent) sources, linear controlled sources, and linear passive elements to form one or more closed paths. Thesuperposition theoremstates thatin a linear network containing multiple sources, the voltage across or current through any passive element may be found as the algebraic sum of the individual voltages or currents due to each of the independent sources acting alone, with all other independent sources deactivated.

An ideal voltage source is deactivated by replacing it with a short circuit. An ideal current source is deactivated by replacing it with an open circuit. In general, controlled sources remain active when the superposition theorem is applied.

Example 1.1. Is the network of Fig. 1-3 a linear circuit?

The definition of a linear circuit is satisfied if the controlled source is a linear controlled source; that is, ifis a constant.

Example 1.2. For the circuit of Fig. 1-3,vs¼10 sin!tV, Vb¼10 V,R1¼R2¼R3¼1, and¼0. Find currenti2 by use of the superposition theorem.

We first deactivateVb by shorting, and use a single prime to denote a response due tovsalone. Using the method of node voltages with unknownv20 and summing currents at the upper node, we have

vsv

0 2 R1

¼v

0 2 R2

þv

0 2 R3

Substituting given values and solving forv20, we obtain

v20¼13vs¼103sin!t

Then, by Ohm’s law,

i20¼ v20 R2

¼10

3sin!tA R3

R2

i3

i2

L2

Ls Vb

i1

=i1

R1

_ +

_ + _

+

_ +

(13)

Now, deactivatingvsand using a double prime to denote a response due toVbalone, we have

i300¼ Vb R3þR1kR2

R1kR2 R1R2 R1þR2

where

i300¼

10 1þ1=2¼

20 3 A so that

Then, by current division,

i200¼ R1 R1þR2

i300¼

1 2i

00 3 ¼

1 2

20 3 ¼

10 3 A Finally, by the superposition theorem,

i2¼i20þi200¼

10

3 ð1þsin!tÞA

Terminals in a network are usually considered in pairs. Aportis a terminal pair across which a voltage can be identified and such that the current into one terminal is the same as the current out of the other terminal. In Fig. 1-4, ifi1i2, then terminals 1 and 2 form a port. Moreover, as viewed to the left from terminals 1,2, networkAis a one-port network. Likewise, viewed to the right from terminals 1,2, networkBis a one-port network.

The´venin’s theoremstates thatan arbitrary linear, one-port network such as network A in Fig. 1-4(a) can be replaced at terminals 1,2 with an equivalent series-connected voltage sourceVThand impedanceZTh (¼RThþjXThÞas shown in Fig. 1-4(b). VThis the open-circuit voltage of networkAat terminals 1,2 and

ZThis the ratio of open-circuit voltage to short-circuit current of network A determined at terminals 1,2 with

network B disconnected. If network A or B contains a controlled source, its controlling variable must be in that same network. Alternatively, ZTh is the equivalent impedance looking into networkA through terminals 1,2 with all independent sources deactivated. If networkAcontains a controlled source,ZThis found as thedriving-point impedance. (See Example 1.4.)

Example 1.3. In the circuit of Fig. 1-5,VA¼4 V, IA¼2 A, R1¼2, and R2¼3. Find the The´venin

equivalent voltageVThand impedanceZThfor the network to the left of terminals 1,2.

Linear network

A

Network B

Network B

Network B

(a) (b)

2 1

2 1

(c) 2 1

i1 i2

+

_ VTh

ZTh

YN IN

Fig. 1-4

VA VB

RB

IA +

_

+

_

R1 R2 1

2

(14)

With terminals 1,2 open-circuited, no current flows throughR2; thus, by KVL,

VTh¼V12¼VAþIAR1¼4þ ð2Þð2Þ ¼8 V

The The´venin impedanceZThis found as the equivalent impedance for the circuit to the left of terminals 1,2 with the independent sources deactivated (that is, withVAreplaced by a short circuit, andIAreplaced by an open circuit):

ZTh¼RTh¼R1þR2¼2þ3¼5

Example 1.4. In the circuit of Fig. 1-6(a),VA¼4 V,¼0:25 A=V,R1¼2, andR2¼3. Find the The´venin

equivalent voltage and impedance for the network to the left of terminals 1,2.

With terminals 1,2 open-circuited, no current flows throughR2. But the control variableVLfor the voltage-controlled dependent source is still contained in the network to the left of terminals 1,2. Application of KVL yields

VTh¼VL¼VAþVThR1

VTh¼

VA

1R1

¼ 4

1 ð0:25Þð2Þ¼8 V

so that

Since the network to the left of terminals 1,2 contains a controlled source,ZThis found as the driving-point impedanceVdp=Idp, with the network to the right of terminals 1,2 in Fig. 1-6(a) replaced by the driving-point source of Fig. 1-6(b) andVAdeactivated (short-circuited). After these changes, KCL applied at nodeagives

I1¼VdpþIdp ð1:6Þ

Application of KVL around the outer loop of this circuit (withVAstill deactivated) yields

Vdp¼IdpR2þI1R1 ð1:7Þ

Substitution of (1.6) into (1.7) allows solution forZThas

ZTh¼

Vdp

Idp

¼R1þR2

1R1

¼ 2þ3

1 ð0:25Þð2Þ¼10

Norton’s theorem states thatan arbitrary linear, one-port network such as network A in Fig. 1-4(a) can be replaced at terminals 1,2 by an equivalent parallel-connected current sourceINand admittanceYNas

shown in Fig. 1-4(c). INis the short-circuit current that flows from terminal 1 to terminal 2 due to network A, and YN is the ratio of short-circuit current to open-circuit voltage at terminals 1,2 with network B

disconnected. If network A or B contains a controlled source, its controlling variable must be in that same network. It is apparent thatYN 1=ZTh; thus, any method for determiningZTh is equally valid for findingYN.

R1 I1 Idp

Ldp R2

RL VL =VL VA

+

_

+

_ +

_ 1

2

1

2 a

(a) (b)

(15)

Example 1.5. Use SPICE methods to determine the The´venin equivalent circuit looking to the left through terminals 3,0 for the circuit of Fig. 1-7.

In SPICE independent source models, an ideal voltage source of 0 V acts as a short circuit and an ideal current source of 0 A acts as an infinite impedance or open circuit. Advantage will be taken of these two features to solve the problem.

Load resistorRLof Fig. 1-7(a) is replaced by the driving point current sourceIdpof Fig. 1-7(b). The netlist code that follows forms a SPICE description of the resulting circuit. The code is set up with parameter-assigned values forV1;I2, andIdp.

Ex1_5.CIR - Thevenin equivalent circuit .PARAM V1value=0V I2value=0A Idpvalue=1A V1 1 0 DC {V1value}

R1 1 2 1ohm

I2 0 2 DC {I2value} R2 2 0 3ohm

R3 2 3 5ohm

G3 2 3 (1,0) 0.1 ; Voltage-controlled current-source Idp 0 3 DC {Idpvalue}

.END

If bothV1andI2are deactivated by setting V1value=I2value=0, currentIdp¼1 A must flow through the The´venin equivalent impedanceZTh¼RTh so thatv3¼IdpRTh¼RTh. Execution of <Ex1_5.CIR> by a SPICE program writes the values of the node voltages for nodes 1, 2, and 3 with respect to the universal ground node 0 in a file <Ex1_5.OUT>. Poll the output file to findv3¼Vð3Þ ¼RTh¼5:75.

In order to determine VTh (open-circuit voltage between terminals 3,0), edit <Ex1_5.CIR> to set V1value=10V, I2value=2A, and Idpvalue=0A. Execute <Ex1_5.CIR> and poll the output file to find

VTh¼v3¼Vð3Þ ¼14 V.

Example 1.6. Find the Norton equivalent currentINand admittanceYNfor the circuit of Fig. 1-5 with values as given in Example 1.3.

The Norton current is found as the short-circuit current from terminal 1 to terminal 2 by superposition; it is

IN¼I12¼current due toVAþcurrent due toIA¼

VA

R1þR2

þ R1IA

R1þR2

¼ 4

2þ3þ

ð2Þð2Þ 2þ3¼1:6 A

(16)

The Norton admittance is found from the result of Example 1.3 as

YN¼ 1

ZTh

¼1

5¼0:2 S

We shall sometimes double-subscript voltages and currents to show the terminals that are of interest. Thus,V13is the voltage across terminals 1 and 3, where terminal 1 is at a higher potential than terminal 3. Similarly,I13is the current that flowsfromterminal 1toterminal 3. As an example,VLin Fig. 1-6(a) could be labeledV12(but notV21).

Note also that an active element (either independent or controlled) is restricted to its assigned, or stated, current or voltage, no matter what is involved in the rest of the circuit. Thus the controlled source in Fig. 1-6(a) will provideVLA no matter what voltage is required to do so and no matter what changes take place in other parts of the circuit.

1.7. TWO-PORT NETWORKS

The network of Fig. 1-8 is atwo-portnetwork ifI1¼I10andI2¼I20. It can be characterized by the four variablesV1;V2;I1, andI2, only two of which can be independent. If V1 andV2 are taken as independent variables and the linear network contains no independent sources, the independent and dependent variables are related by theopen-circuit impedance parameters(or, simply, thez parameters) z11;z12;z21;andz22through the equation set

V1¼z11I1þz12I2 ð1:8Þ

V2¼z21I1þz22I2 ð1:9Þ

Each of the zparameters can be evaluated by setting the proper current to zero (or, equivalently, by open-circuiting an appropriate port of the network). They are

z11¼ V1

I1

I2¼0

ð1:10Þ

z12¼ V1

I2

I1¼0

ð1:11Þ

z21¼ V2

I1

I2¼0

ð1:12Þ

z22¼ V2

I2

I1¼0

ð1:13Þ

In a similar manner, ifV1andI2are taken as the independent variables, a characterization of the two-port network via thehybrid parameters(or, simply, theh-parameters) results:

V1¼h11I1þh12V2 ð1:14Þ I2¼h21I1þh22V2 ð1:15Þ

Linear network

I1 I2

I1¢ I2¢

1 2

1¢ 2¢

+ +

V1 V2

_ _

(17)

Two of thehparameters are determined by short-circuiting port 2, while the remaining two parameters are found by open-circuiting port 1:

h11¼ V1

I1

V2¼0

ð1:16Þ

h12¼ V1 V2

I1¼0

ð1:17Þ

h21¼ I2 I1

V2¼0

ð1:18Þ

h22¼ I2 V2

I1¼0

ð1:19Þ

Example 1.7. Find thezparameters for the two-port network of Fig. 1-9. With port 2 (on the right) open-circuited,I2¼0 and the use of (1.10) gives

z11¼V1 I1

I2¼0

¼R1kðR2þR3Þ ¼ R1ðR2þR3Þ R1þR2þR3

Also, the currentIR2flowing downward throughR2 is, by current division,

IR2¼ R1 R1þR2þR3

I1

But, by Ohm’s law,

V2¼IR2R2¼ R1R2 R1þR2þR3

I1

Hence, by (1.12),

z21¼ V2

I1

I

2¼0

¼ R1R2

R1þR2þR3

Similarly, with port 1 open-circuited,I1¼0 and (1.13) leads to

z22¼ V2

I2

I1¼0

¼R2kðR1þR3Þ ¼

R2ðR1þR3Þ R1þR2þR3

The use of current division to find the current downward throughR1yields

IR1¼ R2 R1þR2þR3

I2

and Ohm’s law gives

V1¼R1IR1¼

R1R2 R1þR2þR3I2

I1 R3 I2

R1

V1 R2

(18)

Thus, by (1.11),

z12¼ V1

I2

I1¼0

¼ R1R2

R1þR2þR3

Example 1.8. Find thehparameters for the two-port network of Fig. 1-9. With port 2 short-circuited,V2¼0 and, by (1.16),

h11¼ V1

I1

V2¼0

¼R1kR3¼ R1R3 R1þR3

By current division,

I2¼ R1 R1þR3

I1

so that, by (1.18),

h21¼ I2 I1

V2¼0

¼ R1

R1þR3

If port 1 is open-circuited, voltage division and (1.17) lead to

V1¼ R1 R1þR3

V2

h12¼ V1 V2

I1¼0

¼ R1

R1þR3

and

Finally,h22is the admittance looking into port 2, as given by (1.19):

h22¼ I2 V2

I1¼0

¼ 1

R2kðR1þR3Þ

¼R1þR2þR3

R2ðR1þR3Þ

Thezparameters and thehparameters can be numerically evaluated by SPICE methods. In electron-ics applications, thezandhparameters find application in analysis when small ac signals are impressed on circuits that exhibit limited-range linearity. Thus, in general, the test sources in the SPICE analysis should be of magnitudes comparable to the impressed signals of the anticipated application. Typically, the devices used in an electronic circuit will have one or more dc sources connected to bias or that place the device at a favorable point of operation. The input and output ports may be coupled by large capacitors that act to block the appearance of any dc voltages at the input and output ports while presenting negligible impedance to ac signals. Further, electronic circuits are usually frequency-sensitive so that any set ofzorh parameters is valid for a particular frequency. Any SPICE-based evaluation of thezandhparameters should be capable of addressing the above outlined characteristics of electronic circuits.

Example 1.9. For the frequency-sensitive two-port network of Fig. 1-10(a), use SPICE methods to determine thez

parameters suitable for use with sinusoidal excitation over a frequency range from 1 kHz to 10 kHz.

Thezparameters as given by (1.10) to (1.13), when evaluated for sinusoidal steady-state conditions, are formed as the ratios of phasor voltages and currents. Consequently, the values of thezparameters are complex numbers that can be represented in polar form aszij¼zijffij.

For determination of thezparameters, matching terminals of the two sinusoidal current sources of Fig. 1-10(b) are connected to the network under test of Fig. 1-10(a). The netlist code below models the resulting network with parameter-assigned values forII1andII5. Two separate executions of <Ex1_9.CIR> are required to determine all

(19)

Ex1_9.CIR - z-parameter evaluation .PARAM I1value=1mA I5value=0mA I1 0 1 AC {I1value}

R10 1 0 1Tohm ; Large resistor to avoid floating node Ci 1 2 100uF

RB 2 3 10kohm VB 0 3 DC 10V R1 2 4 1kohm R2 4 0 5kohm C2 4 0 0.05uF Co 5 4 100uF I5 0 5 AC {I5value}

R50 5 0 1Tohm ; Large resistor to avoid floating node .AC LIN 11 10kHz 100kHz

.PROBE .END

The values of R10 and R50 are sufficiently largeð11012Þso thatII1¼IICiandII5¼IICo. If sourceII5 is

deactivated by setting I5value=0 and I1value is assigned a small value (i.e., 1 mA), thenz11andz21are determined

by (1.10) and (1.12), respectively. <Ex1_9.CIR> is executed and the probe feature of PSpice is used to graphically display the magnitudes and phase angles ofz11andz21in Fig. 1-11(a). Similarly,II1is deactivated andII5is assigned

a small value (I1value=0, I5value=1mA) to determine the values ofz12andz22by (1.11) and (1.13), respectively.

Execution of <Ex1_9.CIR> and use of the Probe feature of PSpice results in the magnitudes and phase angles of

z12andz22as shown by Fig. 1-11(b).

Example 1.10. Use SPICE methods to determine thehparameters suitable for use with sinusoidal excitation at a frequency of 10 kHz for the frequency-sensitive two-port network of Fig. 1-10(a).

Theh parameters of (1.16) to (1.19) for sinusoidal steady-state excitation are ratios of phasor voltages and currents; thus the values are complex numbers expressible in polar form ashij¼hijffij.

Connect the sinusoidal voltage source and current source of Fig. 1-10(c) to the network of Fig. 1-10(a). The netlist code below models the resulting network with parameter-assigned values forII1 and VV5. Two separate

(20)

Fig. 1-11

(a)

(21)

Through use of the .PRINT statement, both magnitudes and phase angles ofVV1, VV5,IICi, andIICoare written to <Ex1_10.OUT> and can be retrieved by viewing of the file.

Ex1_10.CIR - h-parameter evaluation .PARAM I1value=0mA V5value=1mV I1 0 1 AC {I1value}

R10 1 0 1Tohm ; Large resistor to avoid floating node Ci 1 2 100uF

RB 2 3 10kohm VB 0 3 DC 10V R1 2 4 1kohm R2 4 0 5kohm C2 4 0 0.05uF Co 5 4 100uF V5 5 0 AC {V5value} .AC LIN 1 10kHz 10kHz

.PRINT AC Vm(1) Vp(1) Im(Ci) Ip(Ci) ; Mag & phase of inputs .PRINT AC Vm(5) Vp(5) Im(Co) Ip(Co) ; Mag & phase of outputs .END

Set V5value=0 (deactivatesVV5) and I1value=1mA. Execute <Ex1_10.CIR> and retrieve the necessary

values ofVV1;IICi;andIICoto calculateh11andh21by use of (1.16) and (1.18).

h11¼

Vmð1Þ

ImðCiÞff ðVpð1Þ IpðCiÞÞ ffi 0:9091

0:001 ff ð0:028þ08Þ ¼909:1ff 0:028

h21¼

ImðCoÞ

ImðCiÞ ff ðIpðCoÞ IpðCiÞÞ ffi

9:08104

1103 ff ð1808þ08Þ ¼0:908ff 1808

Set V5value=1mV and I1value=0 (deactivatesII1). Execute <Exl_10.CIR> and retrieve the needed values of

V

V1;VV5;andIICoto evaluateh12andh22by use of (1.17) and (1.19).

h21¼

Vmð1Þ

Vmð5Þff ðVpð1Þ Vpð5ÞÞ ffi

9:08104

1103 ff ð0808Þ ¼0:908ff08

h22¼

ImðCoÞ

Vmð5Þ ff ðIpðCoÞ Vpð5ÞÞ ffi

3:15106

1103 ff ð84:7808Þ ¼3:1510 3

84:78

1.8. INSTANTANEOUS, AVERAGE, AND RMS VALUES

Theinstantaneous valueof a quantity is the value of that quantity at a specific time. Often we will be interested in the average value of a time-varying quantity. But obviously, the average value of a sinusoidal function over one period is zero. For sinusoids, then, another concept, that of the root-mean-square(orrms) value, is more useful: For any time-varying functionfðtÞwith periodT, theaveragevalue over one period is given by

F0¼ 1 T

ðt0þT

t0

fðtÞdt ð1:20Þ

and the corresponding rms value is defined as

F ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1

T ðt0þT

t0

f2ðtÞdt s

ð1:21Þ

(22)

Example 1.11. Since the average value of a sinusoidal function of time is zero, thehalf-cycleaverage value, which is nonzero, is often useful. Find the half-cycle average value of the current through a resistanceRconnected directly across a periodic (ac) voltage sourcevðtÞ ¼Vmsin!t.

By Ohm’s law,

iðtÞ ¼vðtÞ

R ¼

Vm R sin!t

and from (1.20), applied over the half cycle fromt0¼0 toT=2¼,

I0¼

1

ð

0 Vm

R sin!t dð!tÞ ¼

1

Vm

R½cos!t

!t¼0¼

2

Vm

R ð1:22Þ

Example 1.12. Consider a resistanceRconnected directly across a dc voltage sourceVdc. The power absorbed by Ris

Pdc¼ Vdc2

R ð1:23Þ

Now replaceVdcwith an ac voltage source,vðtÞ ¼Vmsin!t. Theinstantaneous poweris now given by

pðtÞ ¼v 2ðtÞ

R ¼

Vm2

R sin

2!t ð1:24Þ

Hence, theaverage powerover one period is, by (1.20),

P0¼

1 2

ð2

0 Vm2

R sin

2!t dð!tÞ ¼V 2

m

2R ð1:25Þ

Comparing (1.23) and (1.25), we see that, insofar as power dissipation is concerned, an ac source of amplitudeVmis equivalent to a dc source of magnitude

Vmffiffiffi 2

p ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1

T

ðT

0 v2ðtÞdt

s

V ð1:26Þ

For this reason, the rms value of a sinusoid,V¼Vm=

ffiffiffi

2

p

, is also called itseffectivevalue.

From this point on, unless an explicit statement is made to the contrary, all currents and voltages in the frequency domain (phasors) will reflect rms rather than maximum values. Thus, the time-domain voltage

vðtÞ ¼Vmcosð!tþÞwill be indicated in the frequency domain asVV¼Vj, whereV¼Vm=

ffiffiffi

2

p

.

Example 1.13. A sinusoidal source, a dc source, and a 10resistor are connected as shown by Fig. 1-12. If

vs¼10 sinð!t308ÞV andVB¼20 V, use SPICE methods to determine the average value ofiðI0Þ, the rms value of iðIÞ, and the average value of powerðP0Þsupplied toR.

The netlist code below describes the circuit. Notice that the two sources have been combined as a 10 V sinusoidal source with a 20-V dc bias. The frequency has been arbitrarily chosen as 100 Hz as the solution is independent of frequency.

(23)

Ex1_13.CIR - Avg & rms current, avg power vsVB 1 0 SIN(20V 10V 100Hz 0 0 -30deg) R 1 0 10ohm

.PROBE .TRAN 5us 10ms .END

The Probe feature of PSpice is used to display the instantaneous values ofiðtÞandpRðtÞ. The running average and running RMS features of PSpice have been implemented as appropriate. Both features give the correct full-period values at the end of each full-period of the source waveform. Figure 1-13 shows the marked values asI0¼2:0 A,

I¼2:1213 A, andP0¼45:0 W.

Solved Problems

1.1 Prove that the inductor element of Fig. 1-1(b) is a linear element by showing that (1.2) satisfies the converse of the superposition theorem.

Leti1andi2be two currents that flow through the inductors. Then by (1.2) the voltages across the inductor for these currents are, respectively,

v1¼L di1

dt and v2¼L

di2

dt ð1Þ

(24)

Now supposei¼k1i1þk2i2, wherek1 andk2are distinct arbitrary constants. Then by (1.2) and (1),

v¼L d

dtðk1i1þk2i2Þ ¼k1L di1

dt þk2L di2

dt ¼k1v1þk2v2 ð2Þ

Since (2) holds for any pair of constantsðk1;k2Þ, superposition is satisfied and the element is linear.

1.2 IfR1¼5,R2¼10,Vs¼10 V, andIs¼3 A in the circuit of Fig. 1-14, find the currentiby using the superposition theorem.

WithIsdeactivated (open-circuited), KVL and Ohm’s law give the component ofidue toVsas

i0¼ Vs R1þR2

¼ 10

5þ10¼0:667 A

WithVsdeactivated (short-circuited), current division determines the component ofidue toIs:

i00¼ R1 R1þR2Is¼

5

5þ103¼1 A

By superposition, the total current is

i¼i0þi00¼0:667þ1¼1:667 A

1.3 In Fig. 1-14, assume all circuit values as in Problem 1.2 except thatR2¼0:25i. Determine the currentiusing the method of node voltages.

By (1.1), the voltage-current relationship forR2is

vab¼R2i¼ ð0:25iÞðiÞ ¼0:25i2

so that i¼2 ffiffiffiffiffiffivab

p

(1)

Applying the method of node voltages ataand using (1), we get

vabVs

R1

þ2 ffiffiffiffiffiffivab

p

Is¼0

Rearrangement and substitution of given values lead to

vabþ10 ffiffiffiffiffiffivab

p

25¼0

Lettingx2¼vaband applying the quadratic formula, we obtain

10

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð10Þ24ð25Þ

q

2 ¼2:071 or 12:071

The negative root is extraneous, since the resulting value ofvabwould not satisfy KVL; thus,

vab¼ ð2:071Þ2¼4:289 V and i¼22:071¼4:142 A

R1

R2 Is

Vs +

_

a

i

b

(25)

Notice that, because the resistanceR2is a function of current, the circuit is not linear and the superposition

theorem cannot be applied.

1.4 For the circuit of Fig. 1-15, find vab if (a) k¼0 and (b) k¼0:01. Do not use network theorems to simplify the circuit prior to solution.

(a) Fork¼0, the currentican be determined immediately with Ohm’s law:

i¼ 10

500¼0:02 A

Since the output of the controlled current source flows through the parallel combination of two 100- resistors, we have

vab¼ ð100iÞð100k100Þ ¼ 1000:02ð100Þð100Þ

100þ100¼ 100 V ð1Þ (b) Withk6¼0, it is necessary to solve two simultaneous equations with unknownsiandvab. Around the

left loop, KVL yields

0:01vabþ500i¼10 ð2Þ Withiunknown, (1) becomes

vabþ5000i¼0 ð3Þ

Solving (2) and (3) simultaneously by Cramer’s rule leads to

vab¼

10 500 0 5000

0:01 500 1 5000

¼

50,000

450 ¼ 111:1 V

1.5 For the circuit of Fig. 1-15, use SPICE methods to solve for vab if (a) k¼0:001 and (b) k¼0:05.

(a) The SPICE netlist code fork¼0:001 follows:

Prb.1_5.CIR Vs 1 0 DC 10V R1 1 2 500ohm

E 2 0 (3,0) 0.001 ; Last entry is value of k F 0 3 Vs 100

R2 3 0 100ohm RL 3 0 100ohm .DC Vs 10 10 1 .PRINT DC V(3) .END

Execute <Prb1_5.CIR> and poll the output file to findvab¼Vð3Þ ¼ 101 V.

500 W

100 W 100i

10 V kLab Lab RL = 100 W

+

_

+

_

+

_ c

i

a

d 0 b

2

1 3

(26)

(b) Edit <Prb1_5.CIR> to set k¼0:05, execute the code, and poll the output file to find

vab¼Vð3Þ ¼ 200 V.

1.6 For the circuit of Fig. 1-16, findiLby the method of node voltages if (a) ¼0:9 and (b) ¼0.

(a) Withv2andvabas unknowns and summing currents at nodec, we obtain

v2vs

R1 þv2

R2

þv2vab

R3

þi¼0 ð1Þ

But i¼vsv2

R1 (2)

Substituting (2) into (1) and rearranging gives

1 R1 þ 1 R2 þ 1 R3 v2 1 R3 vab¼

1

R1

vs ð3Þ

Now, summation of currents at nodeagives

vabv2 R3

iþvab

RL

¼0 ð4Þ

Substituting (2) into (4) and rearranging yields

1

R3

R1

v2þ

1

R3

þ 1

RL

vab¼

R1

vs ð5Þ

Substitution of given values into (3) and (5) and application of Cramer’s rule finally yield

vab¼

2:1 0:1vs

0:1 0:9vs

2:1 1

0:1 1:1

¼

1:9vs

2:21¼0:8597vs

and by Ohm’s law,

iL¼

vab

RL¼

0:8597vs

10 ¼0:08597vs A

(b) With the given values (including¼0Þsubstituted into (3) and (5), Cramer’s rule is used to find

vab¼ 3 vs

1 0

3 1

1 1:1

¼

vs

2:3¼0:4348vs

i

=i

R3 = 1 9 R1 = 1 9

R2 = 1 9 RL = 10 9

(27)

TheniLis again found with Ohm’s law:

iL¼

vab

RL

¼0:4348vs

10 ¼0:04348vs A

1.7 If V1¼10 V,V2¼15 V,R1¼4, andR2¼6in the circuit of Fig. 1-17, find the The´venin equivalent for the network to the left of terminalsa;b.

With terminalsa;bopen-circuited, only loop currentIflows. Then, by KVL,

V1IR1¼V2þIR2

I¼V1V2 R1þR2

¼1015

4þ6 ¼ 0:5 A so that

The The´venin equivalent voltage is then

VTh¼Vab¼V1IR1¼10 ð0:5Þð4Þ ¼12 V

Deactivating (shorting) the independent voltage sourcesV1andV2gives the The´venin impedance to the left

of terminalsa;bas

ZTh¼RTh¼R1kR2¼ R1R2 R1þR2

¼ð4Þð6Þ

4þ6¼2:4

VTh andZThare connected as in Fig. 1-4(b) to produce the The´venin equivalent circuit.

1.8 For the circuit and values of Problem 1.7, find the Norton equivalent for the network to the left of terminalsa;b.

With terminalsa;bshorted, the component of currentIabdue toV1alone is

Iab0 ¼V1 R1

¼10

4 ¼2:5 A

Similarly, the component due toV2alone is

Iab00¼

V2 R2¼

15 6 ¼2:5 A Then, by superposition,

IN¼Iab¼I

0

abþI

00

ab¼2:5þ2:5¼5 A Now, withRThas found in Problem 1.7,

YN¼ 1

RTh

¼ 1

2:4¼0:4167 A

INandYNare connected as in Fig. 1-4(c) to produce the Norton equivalent circuit.

Iab

R1 R2

V2 V1

R3

Vab I

a

b +

+

_ +

_

_

(28)

1.9 For the circuit and values of Problems 1.7 and 1.8, find the The´venin impedance as the ratio of open-circuit voltage to short-circuit current to illustrate the equivalence of the results.

The open-circuit voltage is VTh as found in Problem 1.7, and the short-circuit current isIN from Problem 1.8. Thus,

ZTh¼

VTh

IN

¼12

5 ¼2:4

which checks with the result of Problem 1.7.

1.10 The´venin’s and Norton’s theorems are applicable to other than dc steady-state circuits. For the ‘‘frequency-domain’’ circuit of Fig. 1-18 (wheresis frequency), find (a) the The´venin equivalent and (b) the Norton equivalent of the circuit to the right of terminalsa;b.

(a) With terminalsa;b open-circuited, only loop current IðsÞflows; by KVL and Ohm’s law, with all currents and voltages understood to be functions ofs, we have

I¼ V2V1 sLþ1=sC

Now KVL gives

VTh¼Vab¼V1þsLI¼V1þ

sLðV2V1Þ sLþ1=sC ¼

V1þs2LCV2 s2LCþ1

With the independent sources deactivated, the The´venin impedance can be determined as

ZTh¼sLk 1 sC¼

sLð1=sCÞ sLþ1=sC¼

sL s2LCþ1

(b) The Norton current can be found as

IN¼VTh ZTh

¼

V1þs2LCV2 s2LCþ1

sL s2LCþ1

¼V1þs2LCV2

sL

and the Norton admittance as

YN¼ 1

ZTh¼

s2LCþ1

sL

1.11 Determine thezparameters for the two-port network of Fig. 1-19.

ForI2¼0, by Ohm’s law,

Ia¼ V1

10þ6¼

V1

16

sL

+

_

+

_ a

b IL(s)

I(s) Load

V1(s) V2(s) 1 sC

(29)

Also, at nodeb, KCL gives

I1¼0:3IaþIa¼1:3Ia¼1:3

V1

16 ð1Þ

Thus, by (1.10),

z11¼ V1

I1

I2¼0

¼16

1:3¼12:308

Further, again by Ohm’s law,

Ia¼

V2

6 ð2Þ

Substitution of (2) into (1) yields

I1¼1:3 V2

6 so that, by (1.12),

z21¼ V2

I1

I2¼0

¼ 6

1:3¼4:615

Now withI1¼0, applying KCL at nodeagives us

I2¼Iaþ0:3Ia¼1:3Ia ð3Þ

The application of KVL then leads to

V1¼V2 ð10Þð0:3IaÞ ¼6Ia3Ia¼3Ia¼3I2

1:3 so that, by (1.11),

z12¼V1 I2

I1¼0

¼ 3

1:3¼2:308

Now, substitution of (2) in (3) gives

I2¼1:3Ia¼1:3

V2

6 Hence, from (1.13),

z22¼ V2

I2

I1¼0

¼ 6

1:3¼4:615

1.12 Solve Problem 1.11 using a SPICE method similar to that of Example 1.9.

The SPICE netlist code is

b a

I1 I2

I1 I2

+

V1

_

+

V2 VB = 0

_ 10 W

6 W 0.3Ia

Ia

0

2 1

3

+ _

(30)

Prbl_12.CIR z-parameter evaluation .PARAM I1value=1mA I2value=0mA I1 0 1 AC {I1value}

F 1 0 VB 0.3 R1 1 2 10ohm

VB 2 3 0V ; Current sense R2 3 0 6ohm

I2 0 2 AC {I2value}

.DC I1 0 1mA 1mA I2 1mA 0 1mA ; Nested loop .PRINT DC V(1) I(I1) V(2) I(I2)

.END

A nested loop is used in the .DC statement to eliminate the need for two separate executions. As a consequence, data is generated forI1¼I2¼1mA andI1¼I2¼0, which is extraneous to the problem.

Execute <Prb1_12.CIR> and poll the output file to obtain data to evaluate thezparameters by use of (1.10) to (1.13).

z11¼ V1

I1

I2¼0

¼Vð1Þ

IðI1Þ

IðI2Þ¼0

¼1:23110

2

1103 ¼12:31

z12¼ V1

I2

I1¼0

¼Vð1Þ

IðI2Þ

IðI1Þ¼0

¼2:30810

3

1103 ¼2:308

z21¼ V2

I1

I2¼0

¼Vð2Þ

IðI1Þ

IðI1Þ¼0

¼4:61510

3

1103 ¼4:615

z22¼ V2

I2

I

1¼0

¼Vð2Þ

IðI2Þ

IðI1Þ¼0¼4:61510 3

1103 ¼4:615

1.13 Determine thehparameters for the two-port network of Fig. 1-19.

ForV2¼0;Ia0; thus,I1¼V1=10 and, by (1.16),

h11¼ V1

I1

V2¼0

¼10

Further,I2¼ I1and, by (1.18),

h21¼ I2 I1

V2¼0

¼ 1

Now,Ia¼V2=6. WithI1¼0, KVL yields

V1¼V210ð0:3IaÞ ¼V210ð0:3Þ V2

6 ¼ 1 2V2 and, from (1.17),

h12¼ V1 V2

I1¼0

¼0:5

Finally, applying KCL at nodeagives

I2¼Iaþ0:3Ia¼1:3

V2

6 so that, by (1.19),

h22¼ I2 V2

I1¼0

¼1:3

(31)

1.14 Use (1.8), (1.9), and (1.16) to (1.19) to find thehparameters in terms of thezparameters.

SettingV2¼0 in (1.9) gives

0¼z21I1þz22I2 or I2¼ z21 z22

I1 ð1Þ

from which we get

h21¼ I2 I1

V2¼0

¼ z21

z22

Back substitution of (1) into (1.8) and use of (1.16) give

h11¼ V1

I1

V2¼0

¼z11

z12z21 z22

Now, withI1¼0, (1.8) and (1.9) become

V1¼z12I2 and V2¼z22I2

so that, from (1.17),

h12¼ V1 V2

I

1¼0

¼z12 z22

and, from (1.19),

h22¼ I2 V2

I1¼0

¼ I2

z22I2

¼ 1

z22

1.15 The hparameters of the two-port network of Fig. 1-20 areh11¼100;h12¼0:0025;h21¼20, andh22¼1 mS. Find the voltage-gain ratio V2=V1.

By Ohm’s law,I2¼ V2=RL, so that (1.15) may be written

V2

RL

¼I2¼h21I1þh22V2

Solving forI1 and substitution into (1.14) give

V1¼h11I1þh12V2¼

ð1=RLþh22Þ h21

V2h22þh12V2

which can be solved for the voltage gain ratio:

V2 V1¼

1

h12 ðh11=h21Þð1=RLþh22Þ¼

1

0:0025 ð100=20Þð1=2000þ0:001Þ¼ 200

Two-port network

Vs V1

I1

I1

I2

I2 +

_ +

_

V2 RL = 2 kW 1 kW

+

_

(32)

1.16 Determine the The´venin equivalent voltage and impedance looking right into port 1 of the circuit of Fig. 1-20.

The The´venin voltage isV1of (1.8) with port 1 open-circuited:

VTh¼V1jI1¼0¼z12I2 ð1Þ

Now, by Ohm’s law,

V2¼ RLI2 ð2Þ

But, withI1¼0, (1.9) reduces to

V2¼z22I2 ð3Þ

Subtracting (2) from (3) leads to

ðz22þRLÞI2¼0 ð4Þ

Since, in general,z22þRL6¼0, we conclude from (4) thatI2¼0 and, from (1),VTh¼0. Substituting (2) into (1.8) and (1.9) gives

V1¼z11I1þz12I2¼z11I1 z12 RL

V2 ð5Þ

and V2¼z21I1þz22I2¼z21I1 z22

RLV2 (6)

V1is found by solving forV2and substituting the result into (5):

V1¼z11I1 z12z21 z22þRL

I1

ThenZThis calculated as the driving-point impedanceV1=I1:

ZTh¼

Vdp

Idp ¼ V1

I1 ¼z11 z12z21 z22þRL

1.17 Find the The´venin equivalent voltage and impedance looking into port 1 of the circuit of Fig. 1-20 ifRL is replaced with a current-controlled voltage source such thatV2¼I1, whereis a constant.

As in Problem 1.16,

VTh¼V1jI

1¼0¼z22I2

But ifI1¼0, (1.9) and the defining relationship for the controlled source lead to V2¼I1¼0¼z22I2

from whichI2¼0 and, hence,VTh¼0.

Now we letV1¼Vdp, so thatI1¼Idp, and we determineZThas the driving-point impedance. From (1.8), (1.9), and the defining relationship for the controlled source, we have

V1¼Vdp¼z11Idpþz12I2 ð1Þ

V2¼Idp¼z21Idpþz22I2 ð2Þ

Solving (2) forI2and substituting the result into (1) yields

Vdp¼z11Idpþz12

z21 z22

Idp

from which The´venin impedance is found to be

ZTh¼

Vdp

Idp

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1.18 The periodic current waveform of Fig. 1-21 is composed of segments of a sinusoid. Find (a) the average value of the current and (b) the rms (effective) value of the current.

(a) BecauseiðtÞ ¼0 for 0 !t< , the average value of the current is, according to (1.20),

I0¼

1

ð

Imsin!t dð!tÞ ¼

Im

½cos!t!t¼¼

Im

ð1þcosÞ

(b) By (1.21) and the identity sin2x¼1

2ð1cos 2xÞ,

I2¼1

ð I

2

msin2ð!tÞdð!tÞ ¼

Im2 2

ð

ð1cos 2!tÞdð!tÞ

¼I

2

m 2 !t

1 2sin 2!t

!t¼

¼I

2

m

2 þ 1 2sin 2

I¼Im

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

þ1 2sin 2

2

s

so that

1.19 Assume that the periodic waveform of Fig. 1-22 is a current (rather than a voltage). Find (a) the average value of the current and (b) the rms value of the current.

(a) The integral in (1.20) is simply the area under thefðtÞcurve for one period. We can, then, find the average current as

I0¼1

T 4

T

2þ1

T

2

¼2:5 A

(b) Similarly, the integral in (1.21) is no more than the area under thef2ðtÞcurve. Hence,

I¼ 1

T 4

2T

2þ1

2T

2

1=2 ¼4:25 A

i, A Im

0 α p p + α 2p ωt

Fig. 1-21

T T

L, V

4

1

1 2

0 3T

2

t

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1.20 Calculate the average and rms values of the currentiðtÞ ¼4þ10 sin!tA.

SinceiðtÞhas period 2, (1.20) gives

I0¼

1 2

ð2

0

ð4þ10 sin!tÞdð!tÞ ¼ 1

2½4!t10 cos!t

2 !t¼0¼4 A

This result was to be expected, since the average value of a sinusoid over one cycle is zero. Equation (1.21) and the identity sin2x¼1

2ð1cos 2xÞprovide the rms value ofiðtÞ:

I2¼ 1

2

ð2

0

ð4þ10 sin!tÞ2dð!tÞ ¼ 1

2

ð2

0

ð16þ80 sin!tþ5050 cos 2!tÞdð!tÞ

¼ 1

2 66!t80 cos!t 50

2 sin 2!t

2

!t¼0

¼66

so that I¼pffiffiffiffiffi66¼8:125 A:

1.21 Find the rms (or effective) value of a current consisting of the sum of two sinusoidally varying functions with frequencies whose ratio is an integer.

Without loss of generality, we may write

iðtÞ ¼I1cos!tþI2cosk!t

where k is an integer. Applying (1.21) and recalling that cos2x¼1

2ð1þcos 2xÞ and cosxcosy¼ 1

2½cosðxþyÞ þcosðxyÞ, we obtain

I2¼ 1

2

ð2

0

ðI1cos!tþI2cosk!tÞ2dð!tÞ

¼ 1

2

ð2

0 I12

2 ð1þcos 2!tÞ þ

I22

2 ð1þcos 2k!tÞ þI1I2½cosðkþ1Þ!tþcosðk1Þ!t

( )

dð!tÞ

Performing the indicated integration and evaluating at the limits results in

I¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I2 1 2 þ I2 2 2 r

1.22 Find the average value of the power delivered to a one-port network withpassive sign convention

(that is, the current is directed from the positive to the negative terminal) ifvðtÞ ¼Vmcos!tand iðtÞ ¼Imcosð!tþÞ.

The instantaneous power flow into the port is given by

pðtÞ ¼vðtÞiðtÞ ¼VmImcos!tcosð!tþÞ

¼1

2VmIm½cosð2!tþÞ þcos By (1.20),

P0¼

1 2

ð2

0

pðtÞdt¼Vm

4 Im

ð2

0

½cosð2!tþÞ þcosdð!tÞ

After the integration is performed and its limits evaluated, the result is

P0¼ VmIm

2 cos¼

Vmffiffiffi 2

p Imffiffiffi 2

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Supplementary Problems

1.23 Prove that the capacitor element of Fig. 1-1(c) is a linear element by showing that it satisfies the converse of the superposition theorem. (Hint: See Problem 1.1.)

1.24 Use the superposition theorem to find the currentiin Fig. 1-14 ifR1¼5;R2¼10;Vs¼10 cos 2tV, and

Is¼3 cosð3tþ=4ÞA. Ans: i¼0:667 cos 2tþcosð3tþ=4ÞA

1.25 In Fig. 1-23, (a) find the The´venin equivalent voltage and impedance for the network to the left of terminalsa;b, and (b) use the The´venin equivalent circuit to determine the currentIL.

Ans: ðaÞ VTh¼V1I2R2;ZTh¼R1þR2; ðbÞ IL¼ ðV1I2R2Þ=ðR1þR2þRLÞ

1.26 In the circuit of Fig. 1-18, V1¼10 cos 2tV;V2¼20 cos 2tV;L¼1 H;C¼1 F, and the load is a 1-

resistor. (a) Determine the The´venin equivalent for the network to the right of terminalsa;b. (b) Use the The´venin equivalent to find the load currentIIL. (Hint: The results of Problem 1.10 can be used here with

s¼j2.) Ans: ðaÞ VVTh¼23:333ff08V;ZTh¼ j0:667; ðbÞ IIL¼19:4ff33:698A:

1.27 In Fig. 1-24, find the The´venin equivalent for the bridge circuit as seen through the load resistorRL.

Ans: VTh¼VbðR2R3R1R4Þ=ðR1þR2ÞðR3þR4Þ;ZTh¼R1R2=ðR1þR3Þ þR2R4=ðR2þR4Þ

1.28 Suppose the bridge circuit in Fig. 1-24 is balanced by lettingR1¼R2¼R3¼R4¼R. Find the elements of

the Norton equivalent circuit. Ans: IN¼0;YN¼1=R

1.29 Use SPICE methods to determine voltagevabfor the circuit of Fig. 1-24 ifVb¼20 V,RL¼10,R1¼1, R2¼2,R3¼3, andR4¼4. (Netlist code available at author download site.)

Ans: vab¼Vð2;3Þ ¼1:538 V

R2

I2 RL

IL R1

V1 a

b

_ +

Fig. 1-23

+

+

_ _

Vb

RL

Lab R1

R2

R3

R4

a b

0 1

2 3

(36)

1.30 For the circuit of Fig. 1-25, (a) determine the The´venin equivalent of the circuit to the left of terminalsa;b, and (b) use the The´venin equivalent to find the load currentiL.

Ans: ðaÞ VTh¼120 V;ZTh¼20; ðbÞ iL¼4 A

1.31 Apply SPICE methods to determine load currentiLfor the circuit of Fig. 1-25 if (a) the element values are as shown and (b) the VCCS has a value of 0.5vabwith all else unchanged. (Netlist code available at author

download site.) Ans: ðaÞ iL¼4 A; ðbÞ iL¼ 6 A

1.32 In the circuit of Fig. 1-26, letR1¼R2¼RC¼1and find the The´venin equivalent for the circuit to the right of terminalsa;b (a) ifvC¼0:5i1and (b) ifvC¼0:5i2.

Ans: ðaÞ VTh¼0;ZTh¼RTh¼1:75; ðbÞ VTh¼0;ZTh¼RTh¼1:667

1.33 Find the The´venin equivalent for the network to the left of terminalsa;bin Fig. 1-15 (a) ifk¼0, and (b) ifk¼0:1. Use the The´venin equivalent to verify the results of Problem 1.4.

Ans: ðaÞ VTh¼ 200 V;ZTh¼RTh¼100; ðbÞ VTh¼ 250 V;ZTh¼RTh¼125

1.34 Find the The´venin equivalent for the circuit to the left of terminalsa;bin Fig. 1-16, and use it to verify the results of Problem 1.6. Ans: VTh¼12ð1þÞvs;ZTh¼RTh¼12ð3Þ

1.35 An alternative solution for Problem 1.3 involves finding a The´venin equivalent circuit which, when con-nected across the nonlinearR2¼0:25i, allows a quadratic equation in currentito be written via KVL. Find

the elements of the The´venin circuit and the resulting current.

Ans: VTh¼25 V;ZTh¼RTh¼5;i¼4:142 A

1.36 Use (1.10) to (1.15) to find expressions for thezparameters in terms of thehparameters.

Ans: z11¼h11h12h21=h22;z12¼h12=h22;z21¼ h21=h22;z22¼1=h22

30 V +

_

+ a

_

b

Lab RL = 10 W 10 W

5 W

0.25Lab

iL

0 2 1

Fig. 1-25

Ls

LC RC

R2

R1 i1 i2

a

b +

_

+

_

(37)

1.37 For the two-port network of Fig. 1-20, (a) find the voltage-gain ratioV2=V1in terms of thezparameters,

and then (b) evaluate the ratio, using theh-parameter values given in Problem 1.15 and the results of Problem 1.36. Ans: ðaÞ z21RL=ðz11RLþz11z22z12z21Þ; ðbÞ 200

1.38 Find the current-gain ratioI2=I1for the two-port network of Fig. 1-20 in terms of thehparameters. Ans: h21=ð1þh22RLÞ

1.39 Find the current-gain ratioI2=I1for the two-port network of Fig. 1-20 in terms of thezparameters. Ans: z21=ðz22þRLÞ

1.40 Determine the The´venin equivalent voltage and impedance, in terms of thezparameters, looking right into port 1 of the two-port network of Fig. 1-20 ifRLis replaced with an independent dc voltage sourceVd, connected such thatV2¼Vd. Ans: VTh¼z12Vd=z22;ZTh¼ ðz11z22z12z21Þ=z22

1.41 Find the The´venin equivalent voltage and impedance, in terms of thehparameters, looking right into port 1 of the network of Fig. 1-20 ifRLis replaced with a voltage-controlled current source such thatI2¼ V1,

where >0 and thehparameters are understood to be positive.

Ans: VTh¼0;ZTh¼ ðh11h22h12h21Þ=ðh22þh12Þ

1.42 Determine the driving-point impedance (the input impedance with all independent sources deactivated) of the two-port network of Fig. 1-20. Ans: ðz11RLþz11z22z12z21Þ=ðz22þRLÞ

1.43 Evaluate thezparameters of the network of Fig. 1-16.

Ans: z11¼2;z12¼1;z21¼þ1;z22¼2

1.44 Find the currenti1in Fig. 1-3 if¼2;R1¼R2¼R3¼1;Vb¼10 V, andvs¼10 sin!tV.

Ans: 2 A

1.45 For a one-port network with passive sign convention (see Problem 1.22), v¼Vmcos!tV and

i¼I1þI2cosð!tþÞA. Find (a) the instantaneous power flowing to the network and (b) the average

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30

Semiconductor Diodes

2.1. INTRODUCTION

Diodes are among the oldest and most widely used of electronic devices. Adiodemay be defined as a near-unidirectional conductor whose state of conductivity is determined by the polarity of its terminal voltage. The subject of this chapter is thesemiconductor diode, formed by the metallurgical junction of p-type andn-type materials. (Ap-type material is a group-IV elementdopedwith a small quantity of a group-V material;n-type material is a group-IV base element doped with a group-III material.)

2.2. THE IDEAL DIODE

The symbol for thecommon, orrectifier,diodeis shown in Fig. 2-1(a). The device has two terminals, labeledanode(p-type) andcathode(n-type), which makes understandable the choice ofdiodeas its name. When the terminal voltage is nonnegative (vD0), the diode is said to beforward-biasedor ‘‘on’’; the positive current that flows ðiD0) is calledforward current. When vD<0, the diode is said to be

reverse-biasedor ‘‘off,’’ and the corresponding small negative current is referred to asreverse current.

The ideal diode is a perfect two-state device that exhibits zero impedance when forward-biased and infinite impedance when reverse-biased (Fig. 2-2). Note that since either current or voltage is zero at any instant, no power is dissipated by an ideal diode. In many circuit applications, diode forward voltage drops and reverse currents are small compared to other circuit variables; then, sufficiently accurate results are obtained if the actual diode is modeled as ideal.

Theideal diode analysis procedureis as follows:

Step 1: Assume forward bias, and replace the ideal diode with a short circuit.

Step 2: Evaluate the diode currentiD, using any linear circuit-analysis technique.

Step 3: IfiD0, the diode is actually forward-biased, the analysis is valid, and step 4 is to be omitted. iD

Anode Cathode

+ _

D

LD

(a)

Fig. 2-1

References

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