Physics: Lecture Notes

PHYSICS
LECTURE NOTES
PHYS 395
ELECTRONICS
c D.M. Gingrich
University of Alberta
Department of Physics
1999
Preface
Electronics is one of the fastest expanding fields in research, application development and
commercialization. Substantial growth in the field has occured due to World War II, the
invention of the transistor, the space program, and now, the computer industry. The research
grants are high, jobs are available and there is much money to be made in areas related to
electronics. With the beginning of the “information superhighway” and computerized video
coming to your home, it is hard to imagine that electronics will not continue to expand in the
future. Electronics is everywhere in our lives.
It is difficult for the practicing engineer to stay informed of the most recent developments in
electronics. What is taught in this course could well be out of date by the time you actually go to
use it. However the physical concepts of circuit behavour will be largely applicable to any future
development.
The approach to electronics taken in this course will be a mixture of physical concepts and
design principles. The course will thus appear more qualitative and wordy compared to other
physics courses. Nevertheless, it is hoped that this course will become a useful tool for your
future physics laboratories and research.
We can not begin to scratch the surface of the field of electronics in a one term course.
Rather than cover a few topics in detail you will be exposed to most of the concepts and areas of
design. The knowledge you gain will hopefully allow you to communicate with design engineers
and technicians to enable them to design and build the electronics you require. You should also
be equipped to pursue any area of electronics that may interest you in the future. This will
include reading more detailed texts, the component data sheets and manuals. As well as,
understanding the popular literature, including manuals for your stereo, computer, etc.. But
above all I hope you find electronics interesting and enjoyable.
Contents
1 Direct Current Circuits 6

1.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.1.1 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.1.2 Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.1.3 Resistance and Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 The Schematic Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2.1 Electromotive Force (EMF) . . . . . . . . . . . . . . . . . . . . . . . 8
1.2.2 Ground . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Kirchoff’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3.1 Series and Parallel Combinations of Resistors . . . . . . . . . . . . . 10
1.3.2 Voltage Divider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3.3 Current Divider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.3.4 Branch Current Method . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.3.5 Loop Current Method . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.4 Equivalent Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.4.1 Thevenin’s and Norton’s Theorems . . . . . . . . . . . . . . . . . . . 15
1.4.2 Determination of Thevenin and Norton Circuit Elements . . . . . . . 15
1.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2 Alternating Current Circuits 21
2.1 AC Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1.1 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.1.2 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2 Circuit Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.2.1 RC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.2.2 RL Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.2.3 LC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.2.4 RCL Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.3 Sinusoidal Sources and Complex Impedance . . . . . . . . . . . . . . . . . . 28
2.3.1 Resistive Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.3.2 Capacitive Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.3.3 Inductive Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.3.4 Combined Impedances . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.4 Resonance and the Transfer Function . . . . . . . . . . . . . . . . . . . . . . 33
2.5 Four-Terminal Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
1
CONTENTS 2
2.6 Single-Term Approximations of H . . . . . . . . . . . . . . . . . . . . . . . . 40
2.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3 Filter Circuits 44
3.1Filters and Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.2Log-Log Plots and Decibels . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.3Passive RC Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.3.1 Low-Pass Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.3.2 Approximate Integrater . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.3.3 High-Pass Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.3.4 Approximate Differentiator . . . . . . . . . . . . . . . . . . . . . . . . 48
3.4 Complex Frequencies and the s-Plane . . . . . . . . . . . . . . . . . . . . . . 49
3.4.1 Poles and Zeros of H . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.5 Sequential RC Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.6 Passive RCL Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.6.1 Series RCL Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.7 Amplifier Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.7.1 One-, Two- and Three-Pole Amplifier Models . . . . . . . . . . . . . 58
3.7.2 Amplifier with Negative Feedback . . . . . . . . . . . . . . . . . . . . 59
3.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
4 Diode Circuits 64
4.1Energy Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
4.2The PN Junction and the Diode Effect . . . . . . . . . . . . . . . . . . . . . 65
4.2.1 Current in the Diode . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.2.2 The PN Diode as a Circuit Element . . . . . . . . . . . . . . . . . . . 67
4.2.3 The Zener Diode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
4.2.4 Light-Emitting Diodes . . . . . . . . . . . . . . . . . . . . . . . . . . 68
4.2.5 Light-Sensitive Diodes . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.3 Circuit Applications of Ordinary Diodes . . . . . . . . . . . . . . . . . . . . 69
4.3.1 Power Supplies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.3.2 Rectification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.3.3 Power Supply Filtering . . . . . . . . . . . . . . . . . . . . . . . . . . 71
4.3.4 Split Power Supply . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
4.3.5 Voltage Multiplier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4.3.6 Clamping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4.3.7 Clipping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.3.8 Diode Gate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.3.9 Diode Protection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
5 Transistor Circuits 81
5.1 Bipolar Junction Transistors . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.1.1 Transistor Operation (NPN) . . . . . . . . . . . . . . . . . . . . . . . 82
5.1.2 Basic Circuit Configurations . . . . . . . . . . . . . . . . . . . . . . . 83
CONTENTS 3
5.1.3 Small-Signal Models . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
5.1.4 Ideal and Perfect Bipolar Transistor Models . . . . . . . . . . . . . . 87

5.1.5 Transconductance Model . . . . . . . . . . . . . . . . . . . . . . . . . 87
5.2 The Common Emitter Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . 88
5.2.1 DC Biasing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
5.2.2 Approximate AC Model . . . . . . . . . . . . . . . . . . . . . . . . . 89
5.2.3 The Basic CE Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . 89
5.2.4 CE Amplifier with Emitter Resistor . . . . . . . . . . . . . . . . . . . 90
5.3 The Common Collector Amplifier . . . . . . . . . . . . . . . . . . . . . . . . 94
5.4 The Common Base Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . 94
5.5 The Junction Field Effect Transistor (JFET) . . . . . . . . . . . . . . . . . . 95
5.5.1 Principles of Operation . . . . . . . . . . . . . . . . . . . . . . . . . . 95
5.5.2 Small-Signal AC Model . . . . . . . . . . . . . . . . . . . . . . . . . . 96
5.6 JFET Common Source Amplifier . . . . . . . . . . . . . . . . . . . . . . . . 97
5.7 JFET Common Drain Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . 99
5.8 The Insulated-Gate Field Effect Transistor . . . . . . . . . . . . . . . . . . . 100
5.9 Power MOSFET Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
5.10 Multiple Transistor Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.10.1 Coupling Between Single Transistor Stages . . . . . . . . . . . . . . . 102
5.10.2 Darlington and Sziklai Connections . . . . . . . . . . . . . . . . . . . 102
5.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
6 Operational Amplifiers 105
6.1 Open-Loop Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

6.2 Ideal Amplifier Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 107
6.2.1 Non-inverting Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . 107
6.2.2 Inverting Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
6.2.3 Mathematical Operations . . . . . . . . . . . . . . . . . . . . . . . . 110
6.2.4 Active Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
6.2.5 General Feedback Elements . . . . . . . . . . . . . . . . . . . . . . . 114
6.2.6 Differential Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . 115
6.3 Analysis Using Finite Open-Loop Gain . . . . . . . . . . . . . . . . . . . . . 122
6.3.1 Output Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
6.3.2 Input Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
6.3.3 Voltage and Current Offsets . . . . . . . . . . . . . . . . . . . . . . . 127
6.3.4 Current Limiting and Slew Rate . . . . . . . . . . . . . . . . . . . . . 127
6.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
7 Digital Circuits 131
7.1 Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

7.1.1 Binary, Octal and Hexadecimal Numbers . . . . . . . . . . . . . . . . 131
7.1.2 Number Representation . . . . . . . . . . . . . . . . . . . . . . . . . 133
7.2 Boolean Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
7.3 Logic Gates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
7.4 Combinational Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
CONTENTS 4
7.4.1 Combinational Logic Design Using Truth Tables . . . . . . . . . . . . 137
7.4.2 The AND-OR Gate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

7.4.3 Exclusive-OR Gate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
7.4.4 Timing Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
7.4.5 Signal Race . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
7.4.6 Half and Full Adders . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
7.5 Multiplexers and Decoders . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
7.6 Schmitt Trigger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
7.7 The Data Bus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
7.8 Two-State Storage Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
7.9 Latches and Un-Clocked Flip-Flops . . . . . . . . . . . . . . . . . . . . . . . 148
7.9.1 Latches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
7.9.2 RS and RS Flip-Flops . . . . . . . . . . . . . . . . . . . . . . . . . . 148
7.10 Clocked Flip-Flops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
7.10.1 Clocked RS Flip-Flop . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
7.10.2 D Flip-Flop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
7.10.3 JK Flip-Flop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
7.11 Dynamically Clocked Flip-Flops . . . . . . . . . . . . . . . . . . . . . . . . . 151
7.11.1 Master/Slave or Pulse Triggering . . . . . . . . . . . . . . . . . . . . 151
7.11.2 Edge Triggering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
7.12 One-Shots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
7.13 Registers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
7.13.1 Data Registers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
7.13.2 Shift Registers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
7.13.3 Counters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
7.13.4 Divide-by-N Counters . . . . . . . . . . . . . . . . . . . . . . . . . . 155
7.14 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
8 Data Acquisition and Process Control 157
8.1 Transducers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

8.2 Signal Conditioning Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
8.2.1 De-bouncing the Mechanical Switch . . . . . . . . . . . . . . . . . . . 158
8.2.2 Op Amps for Gain, Offset and Function Modification . . . . . . . . . 158
8.2.3 Sample-and-Hold Amplifiers . . . . . . . . . . . . . . . . . . . . . . . 159
8.2.4 Gated Charge-to-Voltage Amplifier . . . . . . . . . . . . . . . . . . . 159
8.2.5 Comparator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
8.3 Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
8.3.1 Application to Interval Timers . . . . . . . . . . . . . . . . . . . . . . 159
8.4 Digital-to-Analog Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . 161
8.4.1 Current Summing and IC Devices . . . . . . . . . . . . . . . . . . . . 161
8.4.2 DAC Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
8.5 Analog-to-Digital Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . 161
8.5.1 Parallel-Encoding ADC (flash ADC) . . . . . . . . . . . . . . . . . . 162
8.5.2 Successive-Approximation ADC . . . . . . . . . . . . . . . . . . . . . 162
8.5.3 Dual-Slope ADC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
CONTENTS 5
8.6 Time-to-Digital Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
8.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

9 Computers and Device Interconnection 169
9.1 Elements of the Microcomputer . . . . . . . . . . . . . . . . . . . . . . . . . 169

9.1.1 Microprocessor and Microcomputer . . . . . . . . . . . . . . . . . . . 169
9.1.2 Functional Elements of the Computer . . . . . . . . . . . . . . . . . . 169
9.1.3 Mechanical Arrangement . . . . . . . . . . . . . . . . . . . . . . . . . 170
9.1.4 Addressing Devices on the Bus . . . . . . . . . . . . . . . . . . . . . 172
9.1.5 Control of the Bus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
9.1.6 Clock Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
9.1.7 Random Access Memory . . . . . . . . . . . . . . . . . . . . . . . . . 172
9.1.8 Read-Only Memory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
9.1.9 I/O Ports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
9.1.10 Interrupts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
9.2 8-, 16-, or 32-Bit Busses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
Chapter 1
Direct Current Circuits
These lectures follow the traditional review of direct current circuits, with emphasis on two-
terminal networks and equivalent circuits. The idea is to bring you up to speed for what is to
come. The course will get less quantitative as we go along. In fact, you will probably find the
course gets easier as we go.
1.1 Basic Concepts

Direct current (DC) circuit analysis deals with constant currents and voltages, while al-ternating
current (AC) circuit analysis deals with time-varying voltage and current signals whose time
average values are zero. Circuits with time-average values of non-zero are also important and
will be mentioned briefly in the section on filters. The DC circuit compo-nents considered in this
course are the constant voltage source, constant current source, and resistor. Electronics also
deals with charge Q, electric E and magnetic B fields, as well as, potential V . We will not be
concerned with a detailed description of these quantities but will use approximation methods
when dealing with them. Hence electronics can be considered as a more practical approach to
these subjects. For the details look at your classical physics and quantum mechanics courses.
1.1.1 Current
The fundamental quantity in electronics is charge and at its basic level is due to the charge
properties of the fundamental particles of matter. For all intensive purposes it is the electron (or
lack of electrons) that matter. The role of the proton charge is negligible.
The aggregate motion of charges is called current I
I= dq
, (1.1)
dt
where dq is the amount of positive charge crossing a specified surface in a time dt. Be aware
that the charges in motion are actually negative electrons. Thus the electrons move in the
opposite direction to the current flow.
The SI unit for current is the ampere (A). For most electronic circuits the ampere is a rather
large unit so the mA unit is more common.
6
CHAPTER 1. DIRECT CURRENT CIRCUITS 7
1.1.2 Potential Difference

It is often more convenient to consider the electrostatic potential V rather than electric field E as
the motivating influence for the flow of electric charge. The generalized vector properties of E
are usually unimportant. The change in potential dV across a distance dr in an electric field is
dV = −E · dr. (1.2)
A positive charge will move from a higher to a lower potential. The potential is also referred
to as the potential difference or, incorrectly, as just voltage:
V dV. (1.3)
V =V21=V2−V1= 2
V1
Remember that current flowing in a conductor is due to a potential difference between its
ends. Electrons move from a point of less positive potential to more positive potential and the
current flows in the opposite direction.
The SI unit of potential difference is the volt (V).
1.1.3 Resistance and Ohm’s Law

For most materials
V ∝I; V =RI, (1.4)
where V = V2 −V1 is the voltage across the object, I is the current through the object, and R is a
proportionality constant called the resistance of the object. Resistance is a function of the
material and shape of the object, and has SI units of ohms (Ω). It is more common to find units
of kΩ and MΩ. The inverse of resistivity is conductivity.
Resistor tolerances can be as bad as ±20% for general-purpose resistors to ±0.1% for
ultra-precision resistors. Only wire-wound resistors are capable of ultra-precision
applications.
The concept of current through and potential across are key to the understanding of and
sounding intelligent about electronics.
Now comes the most useful visual tool of this course.
1.2 The Schematic Diagram

The schematic diagram consists of idealized circuit elements each of which represents some
property of the actual circuit. Figure 1.1 shows some common circuit elements encountered in
DC circuits. A two-terminal network is a circuit that has only two points of interest, say
A and B.
+
V I R
(a) (b) (c)
Figure 1.1: Common circuit elements encountered in DC circuits: a) ideal voltage source, b)
ideal current source and c) resistor.
1.2.1 Electromotive Force (EMF)

Charge can flow in a material under the influence of an external electric field. Eventually the
internal field due to the repositioned charge cancels the external electric field resulting in zero
current flow. To maintain a potential drop (and flow of charge) requires an external energy
source, ie. EMF (battery, power supply, signal generator, etc.). We will deal with two types of
EMFs:
The ideal voltage source is able to maintain a constant voltage regardless of the current it
must put out (I → ∞ is possible).
The ideal current source is able to maintain a constant current regardless of the voltage
needed (V → ∞ is possible).
Because a battery cannot produce an infinite amount of current, a model for the behavior of a
battery is to put an internal resistance in series with an ideal voltage source (zero resistance).
Real-life EMFs can always be approximated with ideal EMFs and appropriate combinations of
other circuit elements.
1.2.2 Ground
A voltage must always be measured relative to some reference point. It is proper to speak of the
voltage across an electrical component but we often speak of voltage at a point. It is then
assumed that the reference voltage point is ground.
Under strict definition, ground is the body of the earth. It is an infinite electrical sink. It can
accept or supply any reasonable amount of charge without changing its electrical characteristics.
It is common, but not always necessary, to connect some part of the circuit to earth or
ground, which is taken, for convenience and by convention, to be at zero volts. Frequently, a
common (or reference) connection of the metal chassis of the instrument suffices. Sometimes
there is a common reference voltage that is not at 0 V. Figure 1.2 show some common ways of
depicting grounds on a circuit diagram.
When neither a ground nor any other voltage reference is shown explicitly on a schematic diagram, it is useful
for purposes of discussion to adopt the convention that the bottom line on a circuit is at zero potential.
(a) (b) (c)
Figure 1.2: Some grounding circuit diagram symbols: a) earth ground, b) chassis ground and c)
common.
1.3 Kirchoff ’s Laws

The conservation of energy and conservation of charge when applied to electrical circuits are
known as Kirchoff’s laws.
Conservation of energy – zero algebraic sum of the voltage drops Vi around a closed
circuit loop (imaginary loop)
Vi = 0. (1.5)
i
Conservation of charge – zero algebraic sum of the currents Ik flowing into a point (total
charge in, equals total charge out)
Ik = 0. (1.6)
k
When applying these laws to solve for circuit unknowns we will find the following defini-
tions useful:
• an element is an impedance (resistance) or EMF (ideal voltage source or ideal current

source),
• a node is a point where three or more current-carrying elements are connected,
• a branch is one element or several in series connecting two adjacent nodes, and
• an interior loop is a circuit loop not subdivided by a branch.
Using these definitions we can apply Kirchoff’s laws to a circuit to solve for the unknown
quantities. The general procedure is:
1. define the currents and voltages on a diagram,

2. apply Kirchoff’s laws to loops and nodes,
3. write down a set of linear algebraic equations, and
4. solve for the unknowns.
But before we look at general circuits lets consider how simple resistors add.
1.3.1 Series and Parallel Combinations of Resistors

Circuit elements are connected in series when a common current passes through each element.
The equivalent resistance Req of a combination of resistors Ri connected in series is given by
summing the voltage drops across each resistor.
V = Vi = I R i, (1.7)
i i
R = (1.8)
eq R i.
i
If Rj Rk, where Rk are all the other resistors than Req ≈ Rj ; the largest resistor wins. Circuit
elements are connected in parallel when a common voltage is applied across each
element. The equivalent resistance Req of a combination of resistors Ri connected in parallel
is given by summing the current through each resistor
V
I = Ii = , (1.9)
i i Ri
1 I 1
= = , (1.10)
R V R
eq i i
R
R = i i . (1.11)
eq i R
j=i j
If Rj Rk, where Rk are all the other resistors than Req ≈ Rj ; the smallest resistor wins. The
following “divider” circuits are useful combinations of resistors. Believe it or not,
they are a super useful concept that will often be used in one form or another; learn it.
1.3.2 Voltage Divider
A
+
R1
V I R R I
in A
in 1 2 out
V
R2 out
B B
(a) (b)
Figure 1.3: Divider circuits: a) voltage divider and b) current divider.
Consider the voltage divider shown in figure 1.3a. The voltage across the input source is
Vin = (R1 +R2)I and the voltage across the output between terminals A and B is Vout = R2I.
The output voltage from the voltage divider in thus
V = R V . (1.12)
out 2 in
R1+R2
Example: Determine an expression for the voltage V2 on the voltage divider in
figure 1.4.
R1
+
V2
V
R2 R3
Figure 1.4: Example voltage divider.
We take the bottom line in figure 1.4 to be at ground and define the current
flowing between V2 and ground to be I. Ohm’s law gives
V2 = I R23, where R23 = R2R3 . (1.13)
R2+R3
Applying Kirchoff ’s voltage law for the input source gives
V = I R, where R = R1 + R23. (1.14)

Combining the above two results and solving for V 2 leads to
R R2R3
23
V2 = V= R2+R3
R R V (1.15)
R R1 +
2 3
R2+R3
= R2R3 V. (1.16)
R1R2 + R2R3 + R3R1
1.3.3 Current Divider

Consider the current divider shown in figure 1.3b. The source current is divided between the two
resistors and is given by Iin = I1 + I2 = V /R1 + V /R2 . The voltage at the output is V = IoutR2 .
The output current from the current divider is thus
I = R1 Iin. (1.17)
out
R1+R2
Example: Determine an expression for the current I3 through the resistor R3
in the circuit shown in figure 1.5.
The current I is divided amongst the three resistors and hence we use our expres-sion for resistors in
parallel
I R1 R2 R3 I3
Figure 1.5: Example current divider.
V = R123I = R1R2R3 I
. (1.18)
R1R2 + R2R3 + R3R1
where V is the common voltage across the three parallel The current
resistors. through R3 is thus
V R1R2
I3 = = I. (1.19)
R3 R1R2 + R2R3 + R3R1
Now lets consider some general approaches to solving for unknowns in circuits.
1.3.4 Branch Current Method

Use both of Kirchoff’s laws. But be aware that an arbitrary application of Kirchoff ’s two
equations will not always yield an independent set of equations. But the following
approach will probably work.
1. Label the current in each branch (do not worry about the direction of the actual current).
2. Use only interior loops and all but one node.
3. Solve the system of algebraic equations.
1.3.5 Loop Current Method

This method is also referred to as the mesh loop method. The independent current variables are
taken to be the circulating current in each of the interior loops.
1. Label interior loop currents on a diagram.
2. Obtain expressions for the voltage changes around each interior loop.
3. Solve the system of algebraic equations.
Depending on the problem, it may ultimately be necessary to algebraically sum two loop currents in order to
obtain the needed interior branch current for the final answer.
Lets consider the example of the Wheatstone bridge circuit shown in figure 1.6. We wish to
calculate the currents around the loops. The three currents are identified as: Ia the clockwise
current around the large interior loop which includes the EMF, Ib the clockwise current around
the top equilateral triangle, and Ic the clockwise current around the bottom equilateral triangle.
The voltage loop expressions for the three current loops are
100 ohm
R2
=
90
=
ohm
R1
+ R5 = 2 ohm
V=25V A B
R3
110 ohm
=
80
=
ohm
R4
Figure 1.6: Loop method for the Wheatstone bridge circuit.
V = R1(Ia − Ib) + R3(Ia − Ic) (1.20)
0 = R1(Ib − Ia) + R2Ib + R5(Ib − Ic) (1.21)
0 = R3(Ic − Ia) + R5(Ic − Ib) + R4Ic. (1.22)

Collecting terms containing the same current gives
V = Ia(R1 + R3) − IbR1 − IcR3 (1.23)
0 = −IaR1 + Ib(R1 + R2 + R5) − IcR5 (1.24)
0 = −IaR3 − IbR5 + Ic(R3 + R4 + R5). (1.25)

If the values for the parameters shown in the diagram are used, the current values can be found
by solving the set of simultaneous equations to give
Ia = 0.267A, Ib = 0.140A, and Ic = 0.113A. (1.26)

Moreover, if we number the individual currents through each resistor using the same scheme as
we have for each component (current through R1 is I1, R2 has I2, etc.) and identify I0 as the
current out of the battery, then
I0 = Ia = 0.267A (1.27)
I1 = Ia − Ib = 0.127A (1.28)
I2 = Ib = 0.140A (1.29)
I3 = Ia − Ic = 0.154A (1.30)
I4 = Ic = 0.113A (1.31)
I5 = Ib − Ic = 0.027A. (1.32)
These are the same currents that would be found using only Kirchoff’s equations; however,
here we had to handle only three simultaneous equations instead of six.
Example: Use the loop current method to determine the voltage developed
across the terminals AB in the circuit shown in figure 1.7.
A
R R
+
V1 R R
+
V2
Figure 1.7: Example circuit for analysis using the loop current method.
Consider the clockwise current loop I A through the two resistors and the two
potentials. Similarly consider the clockwise current I B around the other
internal loop consisting of the three resistors and V 2 . Kirchoff ’s law gives
V1 − IA(2R) + IB R − V2 = 0 loop A. (1.33)
V2 − IB (3R) + IAR = 0 loop B. (1.34)
Solving the above two equations for the unknown loop currents IA and IB gives
V1 − V 2 = 2RIA − RIB (1.35)

V2 = −RIA + 3RIB , (1.36)
V1−V2 = 2R −R IA I
A
(1.37)
,
≡R
V2 − R 3R IB I B
(1.38)
1 3/5 1/5
R −1 =
R 1/5 2/5 ,
1 3 1 1 3 2
(1.39)
IA = R 5 (V1 −V2)+ 5 V2 = R 5 V1 − 5 V2
1 1 2 1 1 1 (1.40)
IB = R 5 (V1 −V2)+ 5 V2 = R 5 V1 + 5 V 2 .
The voltage across AB is given simply by
(1.41)
VAB = IBR
1 (1.42)
= (V1 + V2).
5
1.4 Equivalent Circuits

Equivalent circuits is often the hardest concept and most numerically intensive in the course.
Learning them well could make a difference on your midterm exam. Look in several books until
you find the explanation you understand best.
Since Ohm’s law and Kirchoff’s equations are linear, we can replace any DC circuit by a
simplified circuit. Just like a combination of resistors and Ohm’s law could give an equivalent
resistor, a combination of circuit elements and Kirchoff’s laws can give an equivalent circuit.
Two possibilities are shown in figure 1.8.
1.4.1 Thevenin’s and Norton’s Theorems

A Thevenin equivalent circuit contains an equivalent voltage source VTh in series with an
equivalent resistor RTh. A Norton equivalent circuit contains an equivalent current source IN in
parallel with an equivalent resistor RN.
1.4.2 Determination of Thevenin and Norton Circuit Elements

One approach to determine the equivalent circuits is:
1. Thevenin – calculate the open-circuit voltage VAB = VTh.

R A
Th
A
+
I N RN
V
Th
B B
(a) (b)
Figure 1.8: Thevenin and Norton equivalent circuits.
2. Norton – calculate the short-circuit current between A and B; IN.

3. RTh = RN = VTh/IN.
An alternative to step 2 is to short all voltage sources, open all current sources, and calculate the
equivalent resistance remaining between A and B. We will use the latter approach whenever
manageable. To see if you understand equivalent circuits so far, convince yourself that RTh =
RN.
Solution: From Thevenin’s theorem
VAB(open) = VTh (1.43)

VTh
IAB(short) = . (1.44)
R
Th
According to Notron’s theorem
VAB(open) = INRN (1.45)
IAB(short) = IN. (1.46)
Therefore
VTh = INRN and (1.47)

V
Th = IN. (1.48)
R
Th
⇒ RTh = RN. (1.49)
Lets now return to our Wheatstone bridge example shown in figure 1.6. We will calculate the
current through R5 by replacing the rest of the circuit by its Thevenin equivalent.
• R5 is removed and the open terminals are labeled V Th. The polarity assigned is
arbi-trary as will be verified in the calculations.
• The evaluation of VTh is performed using Kirchoff’s laws:
0 = 25 − (100 + 80)I1 (1.50)
0 = 25 − (90 + 110)I2 (1.51)
0 = 100I1 + VTh − 90I2 (1.52)
The result is VTh = −2.64 V. The minus sign means only that the arbitrary choice of
polarity was incorrect.
100 90
100 90
A B A B
80 110
80 110
A
RTh
+
V
Th R
5
B
Figure 1.9: Thevenin’s theorem applied to the Wheatstone bridge circuit.
• The voltage source is shorted out and RTh is calculated (figure 1.9):
(100)(80) (90)(110)
RTh = + = 93.94Ω (1.53)
100 + 80 90 + 110
Note that when the source is shorted out, the resistors that were in series (R1 and R3; R2
and R4) become parallel combinations.
• The network is assembled in series as shown in figure 1.9 and the current through
R5 is calculated.
I = V = −2.64 =− 0.027A (1.54)
Th
5
R + R 93.94 + 2
Th 5
Note that the numerical value of the current is the same as that in the preceding calculations, but
the sign is opposite. This is simply due to the incorrect choice of polarity of VTh for this
calculation. In fact, the current flow is in the same direction in both examples, as would be
expected.
Example: Find the Thevenin equivalent components V Th and RTh for the
circuit in figure 1.10.
R R
A
V
+
R
V
Figure 1.10: Example circuit for analysis using a Thevenin equivalent circuit.
Shorting the V ’s to find R T h gives two resistors in parallel, which are in

series with a third resistor:
RR 3
RTh = R + = R. (1.55)
2 R+R
The open circuit voltage gives VT h. For the open circuit no current flows from
the node joining the two resistors to A. A is thus at -V relative to this node.
V
Around the interior loop Vloop = 2 (cf. voltage divider).
Therefore
VAB = Vloop − V = − V . (1.56)
2
1.5 Problems
1. Find the current in each resister in the circuit shown below. VA = 2 V, VB = 4 V,
R1 = 100 Ω, R2 = 500 Ω and R3 = 600 Ω. Hint: writing down the loop-current
equations in terms of the symbols will give you most of the marks.
R2
R1
VA R3 VB
+ +
2. Determine the Thevenin equivalent circuit of the circuit shown below. Hint: determine
Vth and Rth.
A
300ohm
600ohm
600ohm 300ohm
+
10V
3. Consider the following circuit:
+
V
A
R
R
V +
V +
(a) What is the Thevenin equivalent voltage?

(b) What is the Thevenin equivalent resistance?

(c) For a variable load resistance placed externally between the terminals A and B, plot
the current through the load as a function of VAB . Label the intercepts on both axes.
4. Sketch the current through a load resistance as a function of VAB for the circuit shown
below. Label both intercepts and the slope.
2R
V R A B R
3R
5. Find the voltage, VL, across the 3 Ω load resistor for the circuit below by replacing the
remaining circuit by its Thevenin equivalent. Hint: You can check your answer by
direct analysis of the entire circuit.
6ohm 15A
->
1ohm A 2ohm B
+ | VL 3ohm
|
20V \/ 15A
b
Chapter 2
Alternating Current Circuits
We now consider circuits where the currents and voltages may vary with time (V = V (t), I = I(t)
(also Q = Q(t))). These lectures will concentrate on the special case in which the signals are
periodic, with time average values of zero ( v(t) = i(t) = 0). Circuits with these signals are
referred to as alternating current (AC) circuits. In general signals will have both DC and AC
properties (v(t) = VAC(t) + VDC). We will concentrate only on the AC components and assume
that the DC properties can be treated separately using the methods of the previous lectures.
The algebraic equations representing Kirchoff’s laws for DC circuits will take the form of
differential equations for AC circuits. So now is a good time to review your differential
equations and complex number theory because we will use it.
2.1 AC Circuit Elements

In physical terms, EMFs can be regarded as circuit elements which put energy into a circuit and
a resistor R as an element which removes energy from a circuit. The energy is dissipated in the
resistor as heat. In AC circuits we have the additional circuit elements, capacitance C and
inductance L, which store energy in electric and magnetic fields respectively. C and L are
referred to as reactive elements while R is a resistive element. All three of these element are
considered passive elements. We will encounter active circuit elements in the lectures to follow.
For simplicity we will ignore radiation that might be emitted by high frequency circuits.
2.1.1 Capacitance
The fundamental property of a capacitor is that it can store charge and hence electric field
energy. The capacitance C between two appropriate surfaces is defined by
Q
V= , (2.1)
C
where V is the potential difference between the surfaces and Q is the magnitude of the charge
distributed on either surface.
21
CHAPTER 2. ALTERNATING CURRENT CIRCUITS 22
In terms of current, I = dQ/dt implies
dV 1 dQ I
= = . (2.2)
dt C dt C
In electronics we take I = ID (displacement current). In other words, the current flow-ing from or
to the capacitor is taken to be equal to the displacement current through the capacitor. You
should be able to show that capacitors add linearly when placed in parallel.
There are four principle functions of a capacitor in a circuit.
1. Since Q and E can be stored a capacitor can be used as a (non-ideal) source of I and
V.
2. Since a capacitor passes AC current but not DC current it can be used to connect parts of a
circuit that must operate at different DC voltage levels.
3. A capacitor and resistor in series will limit current and hence smooth sharp edges in
voltage signals.
4. Charging or discharging a capacitor with a constant current results in the capacitor having
a voltage signal with a constant slope, ie. dV /dt = I /C = constant if I is a constant.
Some capacitors (electrolytic) are asymmetric devices with a polarity that must be hooked-up
in a definite way. You will learn this in the lab. The SI unit for capacitance is farad (F). The
capacitance in a circuit is typically measured in µF or pF. Non-ideal cir-cuits will have stray
capacitance, leakage currents and inductive coupling at high frequency. Although important in
real circuit design we will slip over these nasties at this point.
Capacitors can be obtained in various tolerance ratings from ±20% to ±0.5%.

Because of dimensional changes, capacitors have a high temperature dependence of
capacitance. A capacitor does not hold a charge indefinitely because the dielec-tric is
never a perfect insulator. Capacitors are rated for leakage, the conduction through
the dielectric, by the leakage resistance-capacitance product in MΩ · µF. High
temperature increases leakage.
2.1.2 Inductance
Faraday’s law applied to an inductor states that a changing current induces a back EMF that
opposes the change. Or
dI
V=V V = L . (2.3)
− BA dt
Where V is the voltage across the inductor and L is the inductance measured in henry (H).
The more common units encountered in circuits are µH and mH.
The inductance will tend to smooth sudden changes in current just as the capacitance smoothes sudden
changes in voltage. Of course, if the current is constant there will be no
induced EMF. So unlike the capacitor which behaves like an open-circuit in DC circuits, an
inductor behaves like a short-circuit in DC circuits.
Applications using inductors are less common than those using capacitors, but inductors are
very common in high frequency circuits. We will again skip over the unpleasantness – that non-
ideal inductors have some resistance and some capacitance.
Inductors are never pure inductances because there is always some resistance in and
some capacitance between the coil windings. When choosing an inductor
(occasionally called a choke) for a specific application, it is necessary to consider the
value of the inductance, the DC resistance of the coil, the current-carrying capacity
of the coil windings, the breakdown voltage between the coil and the frame, and the
frequency range in which the coil is designed to operate. To obtain a very high
inductance it is necessary to have a coil of many turns. The inductance can be further
increased by winding the coil on a closed-loop iron or ferrite core. To obtain as pure
an inductance as possible, the DC resistance of the windings should be reduced to a
minimum. This can be done by increasing the wire size, which of course, increases
the size of the choke. The size of the wire also determines the current-handling
capacity of the choke since the work done in forcing a current through a resistance is
converted to heat in the resistance. Magnetic losses in an iron core also account for
some heating, and this heating restricts any choke to a certain safe operating current.
The windings of the coil must be insulated from the frame as well as from each
other. Heavier insulation, which necessarily makes the choke more bulky, is used in
applications where there will be a high voltage between the frame and the winding.
The losses sustained in the iron core increases as the frequency increases. Large
inductors, rated in henries, are used principally in power applications. The frequency
in these circuits is relatively low, generally 60 Hz or low multiples thereof. In high-
frequency circuits, such as those found in FM radios and television sets, very small
inductors (of the order of microhenries) are frequently used.
2.2 Circuit Equations

Recall that voltage V is related to current I, via the passive DC circuit element resistance R, by
Ohm’s law V = I R. Analogously, the change in voltage and change in current are related to the
current and voltage, via the passive AC circuit elements C and L, by
dV I dI
= and V = L . (2.4)
dt C dt
Applying the above three equations, along with Kirchoff’s loop rule, to AC circuits results in
a set of differential equations. These differential equations are linear with constant coeffi-cients
and can easily be solved for Q(t), I(t), and V (t). In general the solutions will consist of a
transient response and a steady-state response. The transient response describes the return to equilibrium
after the EMFs change suddenly. The steady-state response describes the long term behaviour when the circuit is
driven by a sinusoidal source.
We will first consider the transient response. This will be one of the few times we consider
non-oscillating AC behaviour. Since Ohm’s law and Kirchoff’s laws are linear we can use
complex exponential signals and take real or imaginary parts in the end. This is not true for
power, since it is non-linear (product of signals).
2.2.1 RC Circuit
Consider the resistor R and capacitance C in the circuit loop in figure 2.1. Notice that there is no
source.
R C
Figure 2.1: RC circuit.
We start with a differential version of Kirchoff’s voltage law.

d Vi = dVi = 0. (2.5)
d i
i dt
t
When applied to our circuit
dVC + dV = 0, (2.6)
R
dt dt
where VC is the voltage drop across the capacitor and VR is the voltage drop across the resistor.
The change in the voltage drop across the capacitor is given by our previous expression,
dVC = I . (2.7)
dt C
The change in the voltage drop across the resistor can be obtained from Ohm’s law
VR=RI⇒ dVR dI
=R . (2.8)
dt dt
Substituting these changes in voltage into Kirchoff’s equation gives
I dI
+R = 0, (2.9)
C dt
where the current due to the flow of charge on or off the capacitor is the same as through the
resistor.
Now we need some initial conditions. Notice that although the capacitor behaves as an open circuit to DC,
current must flow to charge or discharge the capacitor. Lets take the
case where the capacitor is initially charged and then the circuit is closed and the charge is
allowed to drain off the capacitor (eg. closing a switch). The resulting current will flow through
the resistor.
Solving for the current we obtain
−t/RC (2.10)
I(t) = I0e ,
where I(t = 0) = I0 is the initial current given by Ohm’s law
I0 =V0 . (2.11)
R
Using a time dependent version of Ohm’s law we can solve for the voltage across the resistor
−t/RC −t/RC −t/τ (2.12)

V (t) = RI(t) = RI0e = V0e = V0e ,
where V (t = 0) ≡ V0 is the initial voltage across the capacitor and τ ≡ RC is the commonly
defined time constant of the decay. You should also be able to solve for the voltage across the
capacitor and charge on the capacitor.
For the case of an applying voltage VB being suddenly placed into the circuit (inserting a
battery) the capacitor is initially not charged and the voltage across the capacitor is
−t/τ (2.13)
V (t) = VB (1 − e ).
In the first case, current and voltage exponentially decay away with time constant τ when the
switch is closed. The charge flows off the capacitor and through the resistor. The energy initially
stored in the capacitor is dissipated in the resistor.
In the second case the capacitor charges to a voltage VB until no current flows and hence the
voltage drop across the resistor is zero. Energy from the battery is stored in the capacitor.
In both cases the characteristic RC time constant occurs. In general this is true of all resistor-
capacitor combinations and will be important throughout the course.
2.2.2 RL Circuit
The response of the RL circuit, shown in figure 2.2, is similar to that of the RC circuit.
There are however some significant differences.
R L
Figure 2.2: RL circuit.

If a battery is inserted into the circuit the current raises quickly from zero to some finite
value. The EMF generated in the inductor impedes the current flow until it is constant.
The expression for the current in the RL circuit is
i(t) = VB (1 (2.14)
R
− e−tR/L)
where the time constant is now
L
τ = . (2.15)
R
The voltage across the resistor is an increasing exponential unlike the RC circuit in which
the voltage across the resistor decreased exponentially. Likewise, the voltage across the inductor
decreases with time while in the RC circuit the voltage across the capacitor increased with time.
There are other initial conditions we could work with in this circuit but these can now be
worked out by the student.
2.2.3 LC Circuit
Lets now consider the LC circuit in figure 2.3 which has no resistive element.
L C
Figure 2.3: LC circuit.
Kirchoff’s voltage law applied to the loop is
VL+VC =0. (2.16)

Substituting our previous expressions for VL and VC gives
d Q
I
L + = 0. (2.17)
dt C
Using I = dQ/dt gives
L d2Q Q
2 + = 0. (2.18)
dt C
The circuit equation is second-order in Q and one possible solution is
Q(t) = Q0 cos(ωt + φ), (2.19)
where Q(t = 0) = Q0 is the initial charge on the capacitor and φ is an arbitrary phase constant.
Considering the cases of Q0 = Qmax, gives φ = 0. The angular frequency ω is totally determined
by the other parameters of the circuit
√ ω2 = 1/(LC) (2.20)
and ωr ≡ 1/ LC is the natural or resonance frequency of the circuit.

We can also solve for the current and voltage across the capacitor
Q(t) = Q0 cos(ωr t), (2.21)
I(t) = d = −Q0ωr sin(ωr t), (2.22)

Q
dt
= −I0 sin(ωr t) = I0 cos(ωr t + π/2), and (2.23)
V (t) = Q(t) =Q0 cos(ωr t) = V0 cos(ωr t). (2.24)
C C
Notice that unlike the transient current and voltage responses of the RC and RL circuits, the
LC circuit oscillates. The energy in the circuit is shared back and forth between the inductor and
capacitor.
2.2.4 RCL Circuit

Lets now consider the case of all three passive circuit elements in series, as in figure 2.4.
R L
Figure 2.4: RCL circuit.
Applying Kirchoff’s law around the loop and using I = dQ/dt gives
L dI Q
+RI+ = 0 and (2.25)
dt C
L d2 + R d +Q = 0. (2.26)
Q Q
2 dt C
dt
The solution will not only depend on the initial conditions but also the relative values of R,
C and L.
There are three possible solutions:
2 −t/τ
1. under damped (R < 4L/C): Ae cos(ωt + φ),
2 −t/τ1 −t/τ2
2. over damped (R > 4L/C): A1e + A2e , and
2 −t/τ
3. critically damped (R = 4L/C): (A1 + A2t)e .
RCL circuits have a variety of properties, especially when driven by sinusoidal sources,
which will not be investigated here. My aim is simply to expose you to the area and get on to
more interesting topics. Driven oscillating systems also appear in other areas of physics and
hopefully you will encounter them there. The detailed considerations lead to discussions on
resonance and quality-factor Q.
2.3 Sinusoidal Sources and Complex Impedance

We now consider current and voltage sources with time average values of zero. We will use
periodic signals but the observation time could well be less than one period. Periodic signals are
also useful in the sense that arbitrary signals can usually be expanded in terms of a Fourier series
of periodic signals. Lets start with
v(t) = V0 cos(ωt + φV ) and (2.27)

i(t) = I0 cos(ωt + φI ). (2.28)
Notice that I have now switched to lowercase symbols. Lowercase is generally used for AC
quantities while uppercase is reserved for DC values.
Now is the time to get into complex notation since it will make our discussion easier and is
encountered often in electronics. The above voltage and current signals can be written
v(t) = j(ωt+φ )
(2.29)
V0e V and
i(t) = j(ωt+φ ) (2.30)
I0e I .
To be cleaver we will define one EMF in the circuit to have φ = 0. In other words, we will
pick t = 0 to be at the peak of one signal. The vector notation is used to remind us that complex
numbers can be considered as vectors in the complex plane. Although not so common in physics,
in electronics we refer to these vectors as phasors. Hence you should now review complex
notation.
The presence of sinusoidal v(t) or i(t) in circuits will result in an inhomogeneous differen-tial
equation with a time-dependent source term. The solution will contain sinusoidal terms with the
source frequency.
The extension of Ohm’s law to AC circuits can be written as
v(ω, t) = Z(ω)i(ω, t), (2.31)

where ω is the source frequency. Z is a generalized resistance referred to as the impedance. We can cancel
out the common time dependent factors to obtain
v(ω) = Z(ω)i(ω) (2.32)

and hence you see the power of the complex notation. For a physically quantity we take the
amplitude of the real signal
|v(ω)| = |Z(ω)||i(ω)|. (2.33)

We will now examine each circuit element in turn with a voltage source to deduce its
impedance.
2.3.1 Resistive Impedance

Kirchoff’s voltage law for a voltage source and resistor is
v(t) − Ri(t) = 0. (2.34)

Trying the solutions
jωt and jωt (2.35)
i(t) = ie v(t) = ve
leads to
v = Ri ⇒ ZR=R. (2.36)
The impedance is equal to the resistance, as expected.
2.3.2 Capacitive Impedance

Kirchoff’s voltage law for a voltage source and capacitor is
v(t) − q(t) = 0. (2.37)

C
Or
dv(t) − i(t)
= 0. (2.38)
dt C
Solving this equation gives
jωv = i ⇒ ZC = 1 (2.39)
C jωC
For DC circuits ω = 0 and hence ZC → ∞. The capacitor acts like an open circuit (infinite resistance) in
a DC circuit.
2.3.3 Inductive Impedance

Kirchoff’s voltage law for a voltage source and inductor is
v(t) − L di(t)
= 0. (2.40)
dt
Solving this equation gives
v = jωLi ⇒ ZL = jωL. (2.41)
For DC circuits ω = 0 and hence ZL = 0. There is no voltage drop across an inductor in DC (zero
resistance).
2.3.4 Combined Impedances

We now know the impedance for each of our passive circuit elements:
ZR = R; ZL = jωL; ZC = −j/(ωC). (2.42)

The equivalent impedance of a circuit can be obtained by using the following rules for
combining impedances.
In series
Zeq = Z i. (2.43)
i
In parallel Zi
i . (2.44)
Zeq =
i Z
j=i j
Appealing to the complex notation we can write
Zeq = R + jX(ω), (2.45)
where R is the resistance and X is called the reactance (always a function of ω). For a
series combination of R, L and C
Zeq = R + jωL + 1 , (2.46)

jωC
= R + j ωL − 1 . (2.47)
ωC
√
(ωL − 1/ωC) = 0 gives a special frequency, ω = 1/ LC.
Example: An inductor and capacitor in parallel form the tank circuit shown in
figure 2.5.
L
A B
Figure 2.5: Tank circuit with inductor and capacitor.
1. Determine an expression for the impedance of this circuit.

The impedance of an inductor and a capacitor are
ZL = jωL; ZC = 1 . (2.48)
jωC
Combining the impedances in parallel gives
Z = 1 −1 1 1 −1
ZC ZL
Z = Z +Z = (2.49)
i i L C ZC +ZL
= (1/(jωC))(jωL) = −jL/C (2.50)
1/(jωC) + jωL ωL − 1/(ωC)
= jωL . (2.51)
2
1 − ω LC
√ ?
2. What is the impedance when ω = 1/ LC
Substituting this value for ω into the above result gives
L
L
j√ LC
j C
Z = LC = (2.52)
0
1− LC
→ ∞. (2.53)
Example: The tank circuit schematic shown in figure 2.6 results from the use
of a real inductor.
L R
A B
-> C ->
i i
Figure 2.6: Tank circuit with real inductor

.
1. Find an expression for the impedance of this circuit. The

impedance of an inductor, capacitor and resistor are
ZL = jωL; ZC = 1 ; ZR=R. (2.54)

jωC
The resistor and inductor are in series and this combination of impedance is in parallel with
the capacitor. Combining the impedances gives
1 1 −1
ZC (ZL + ZR)
1 −1
Z = = + = (2.55)
Z Z
i i ZL+ZR C ZC +ZR+ZL
= (1/(jωC))(jωL + R) (2.56)
1/(jωC) + R + jωL
= L/C − jR/(ωC) . (2.57)
R + j(ωL − 1/(ωC))
2. If L =√1H, R = 100Ω, and C = 0.01µF, what is the impedance when

ω = 1/ LC?
Substituting this value for ω into the above equation gives
L/C − jR L/C L −j L
Z= =RC C (2.58)
R + j( L/C −
L/C)
Substituting the numerical values for the inductance, resistance and

capaci-tance gives
Z = 1 −j 1 (2.59)
2 −8 −8
10 × 10 10
6 4
= (10 − j10 )Ω (2.60)
= (100 − j) × 104Ω. (2.61)
3. What is the impedance when ω is very small?
Z ≈ −jR/(ωC) (2.62)
−j/(ωC)
= R. (2.63)
4. What is the phase angle between the voltage v AB and i at resonance

5
and at ω = 10 rad/s?
Rationalizing the denominator of the impedance gives
[L/C − jR/(ωC)][R − j(ωL − 1/(ωC))]
Z= . (2.64)
2 2
R + [ωL − 1/(ωC)]
Taking the real and imaginary components gives

RL/C − R/(ωC)[ωL − 1/(ωC)]
Re[Z] = (2.65)
R2 + [ωL − 1/(ωC)]2
R/(C2ω2)
, (2.66)
=R 2
2
+ [ωL − 1/(ωC)] 2
−R /(ωC) − L/C[ωL − 1/(ωC)]

Im[Z] = . (2.67)
2 2
R + [ωL − 1/(ωC)]
The inverse tangent of the ratio of the imaginary to real parts is
2 2 2
φ = tan−1 −R /(ωC) − ωL /C + L/(ωC ) (2.68)
2 2
R/(C ω )
= tan −R Cω − CL ω
−1
2 2 3
+ Lω (2.69)
R
−1 ω 2 2 2
= tan (L − CR − CL ω ) . (2.70)
R
√
There is a resonance at ωL − 1/(ωC) = 0 ⇒ ω = 1/ LC
and hence
−1
tan 1 (L−CR2−L) (2.71)
φ = √
res R LC
= tan−1 R C (2.72)
− L
= − tan−1 R C
(2.73)
L
= − tan (10 )
−1 −2 (2.74)
≈ 0. (2.75)
At ω = 105 rad/s.
105 (2.76)
φ = tan−1 1 − 10−8(102)2 − 10−8(1)2(105)2 102
=
−1
tan [103(1 − 10
−4
− 102)] (2.77)
−1
≈ tan (−10 )
5 (2.78)
≈ −π/2. (2.79)
2.4 Resonance and the Transfer Function

Lets now consider putting a sinusoidal source in our series RCL circuit and consider the voltage across
one of the circuit elements. The resistor for example in figure 2.7
L
v(t)
A
v
R AB(t)
B
Figure 2.7: Driven series RCL circuit.
Applying Ohm’s law v(jω) = Ri(jω) across the resistor gives (cf. a voltage divider)
vAB (jω) = ZR v(jω), (2.80)

ZR+ZL+ZC
= R v(jω), (2.81)
R + j(ωL − 1/(ωC))
≡ H(jω)v(jω), (2.82)
where H(jω) is known as the transfer function in the frequency domain. We have changed
independent variables from ω to jω for convenience.
We define
H(jω) = vAB (jω) = R . (2.83)
v(jω)R + j(ωL − 1/(ωC))
H(jω) contains all the information needed to characterize the circuit. In exponential form
jφ(ω) (2.84)
H(jω) = H(jω)e ,
where
R
H(jω) = (2.85)
2 2
R + (ωL − 1/(ωC))
and
1/(ωC) − ωL
−1 . (2.86)
φ(ω) = tan R
√
H(jω) has a maximum (resonance) given by ωL − 1/(ωC) = 0. Or ω = 1/ LC ≡ ωr is the
resonant frequency.
Example: Consider the series LCR circuit (figure 2.8) driven by a voltage pha-
sor v(t) = v0 exp(jωt).
C L
S -> R
G v(t)
Figure 2.8: Driven series LCR circuit.
1. At an angular frequency such that ωL = 2R and 1/(ωC) = R, write the

current phasor in terms of v(t) and R.
1
v(t) is given by v(t) = Zi(t), where Z = ZC + ZL + ZR = jωC + jωL + R.
At ωL = 2R and ωC1 = R
⇒Z = R
j + 2jR + R = R(1 + j) (2.87)
= √
2 Rejπ/4. (2.88)
Therefore
v(t) jωt
v0e
i(t) = Z =√ 2 jπ/4 (2.89)
Re
v0
i(t) = √ R ej(ωt−π/4). (2.90)
2
2. At the instant when v(t) is exactly real, calculate the three phasors
repre-senting the voltage developed across the R, C, and L circuit
elements. v(t) is real at t = 0. Thus
vR = Ri(t = 0) (2.91)
v0
= √ (2.92)
2 e−jπ/4
And
v0
vL = ZLi(0) = jωLi(0) = j2R √ 2 R e−jπ/4 (2.93)
= √ 0 jπ/2 −jπ/4
2v e e (2.94)
√
= 2 (2.95)
v0ejπ/4
Also
i(0) e−jπ/2 v0
= ZC i(0) = jωC = e −jπ/4
(2.96)
vC 1/R √ 2R
v0
= √ (2.97)
2 e−j3π/4.
3. Algebraically and with a sketch on the complex plane, show that the
complex voltage sum around the closed loop is zero.
The three voltage phasors are
vR = v 0 v0 [cos( π/4) + j sin( π/4)] (2.98)
√ e−jπ/4 = √
v02 2 − −
= (2.99)
2 (1 − j)
vL = √ √ 2v0[cos(π/4) + j sin(π/4)] (2.100)
2 v0 e jπ/4 =
= v0(1 + j) (2.101)
vC = v0 v0 3π/4) + j sin( 3π/4)] (2.102)
√ e−j3π/4 = √ [cos(
2 2 − −
v
0
= (1 + j) (2.103)
− 2
Around the closed loop v = v − vR − v L − v this expression is zero
i i . If
at t = 0 it will be zero for all time. v
C v0 j v0 j
Therefore 0
− 2(1 − ) − (1 + )+
v0 (1 + j) = 0.
2
Im
vL
1
v
−0.5 1 Re
−0.5
vC vR
Figure 2.9: Complex voltage sum around the closed loop of the driven LCR circuit.
Example: Sketch simplified versions of the circuit shown in figure 2.10 that
would be valid at:
R C
2L
100R
S ->
Vs
G
R
Figure 2.10: Example LCR circuit.
1. ω=0;
ω=0⇒ZC→∞;ZL→0.
v vs
Figure 2.11: Example circuit for ω = 0.
2. very low frequencies but not ω = 0;

When ω is small (ω = 0) C and L are in parallel and
ZC ZL [1/(jωC)](jωL) jL/C
Zeq = = = (2.104)
≈ ZC +ZL 1/(jωC) + jωL 1/(ωC) − ωL
jωL = ZL. (2.105)
2L and 100R in parallel gives

ZRZL (100R)(jω2L)
Zeq = = = jω2L. (2.106)
ZR+ZL 100R + jω2L
L
A
R
100R 2L
v vs
R
Figure 2.12: Example circuit for very low frequencies but not ω = 0.
3. very high frequencies but not ω = ∞;

ω large (ω = ∞) (note: 100R + R ≈ 100R).
C
A
R
100R 2L
v vs
R
Figure 2.13: Example circuit for very high frequencies but not ω = ∞.
4. ω=∞.
ω = ∞ ⇒ ZC = 1/jωC → 0; ZL = jωL → ∞.
Example: For the circuit shown in figure 2.15 plot |Zeq| as a function of fre-
6
quency over the range ω = 1 rad/s to ω = 10 rad/s.
The equivalent impedance for the three components in parallel is
Z = ZRZLZC (2.107)
eq
ZLZR + ZLZC + ZC ZR
= (R)(jωL)(1/jωC) (2.108)
1
(jωL)(R) + (jωL)( jωC ) + ( jωC1 )(R)
= RL/C (2.109)
L 1
C + jR(ωL − ω )
C
|Zeq | = RL/C . (2.110)

( LC 2 2
) + R (ωL −
1 2
ωC )
R 100R
v vs R
Figure 2.14: Example circuit for ω = ∞.
1kohm 10H 0.1microF
Figure 2.15: Example circuit with components in parallel.
Plugging in the numerical values gives

| Z eq | = 3 7 (2.111)
10 × 10 × 10
7
10 2 6 2
( −7 ) + 10 (10ω −
10
ω )
10
= 1011 (2.112)
6
1016 + 108(ω − ω )2
10
= 107 . (2.113)
6
10
108+(ω− ω )2
A table of values and its plot follows.
Table 2.1: Numerical values for example circuit.

log Z
ω(rad/s) log10 ω |Zeq | 10 | eq |
0 1
1(10 ) 0 10 1
10 1
1 102 2
2 √ 3
103 2 1/ 2 ×3 10 2.8
10 3 10 3
4 √ 3
105 4 1/ 2 ×2 10 2.8
10 5 10 2
6 1
10 6 10 1
2.5 Four-Terminal Networks

Our previous resonance circuit is an example of a two-terminal network. A source is present but
no load. A four-terminal network also has the source removed. The four-terminal network can be
described by a transfer function. A generic four-terminal network is shown in figure 2.17. Such a
circuit can be analyzed simply by considering it as a voltage divider. In general
H= Z2 . (2.114)
Z1+Z2
2.6 Single-Term Approximations of H

A circuit with a few components quickly leads to a complicated expression for the transfer
function. It is often sufficient, and of course, easier to work with approximations to the transfer
function.
Let H be the ratio of two polynomials
P(jω)
H(jω) = , (2.115)
Q(jω)
where
N M
i k
P(jω) = Pi(jω) and Q(jω) =Qk(jω) . (2.116)
i=0 k=0
If one term dominates in each polynomial
n and m (2.117)
P(jω) ≈ Pn(jω) Q(jω) ≈ Qm(jω) .
Thus
Figure 2.16: Plot of |Zeq| for example circuit.
H(jω) = n = Pn
Pn(jω) Pn
(jω)(n−m) = ω(n−m)ej(n−m)π/2 and (2.118)
Qm(jω)
m Q Q
m m
Pn (n−m)
|H(jω)| = ω . (2.119)
Qm
We define log ≡ log10 and plot
log |H(jω)| = log P + (n − m) log ω,
n
(2.120)
Qm
which is a straight line on a log-log plot with integer slope.
As an example, consider our RCL circuit:
H(jω) = R
(2.121)
R + j(ωL − 1/(ωC))
CHAPTER 2. ALTERNATING CURRENT CIRCUITS log |Hhigh(jω)|
log(ωr /Q) − lo
which
has a
A Z C
1 slope
of −1.
Z
2
B D
Figure 2.17: Generic four-terminal network.
= 1
1 + j(ωL/R − 1/(ωRC))
= 1
1 + j(ωQ/ωr − ωr Q/ω)
= 1 ,
1 + jQ(ω/ωr )(1 − (ωr /ω)2)

1 ωrL
where ωr = √ and Q = R .
LC
At low frequencies ω/ωr → 0 and

H (jω) =
1 ≈
jω
low
1 − jQωr /ω Qωr
ω
|Hlow(jω)| = .
Qωr
On a log-log plot
log |Hlow(jω)| = log(1/(Qωr )) + log ω,

which has a slope of +1.
At high frequencies ωr /ω → 0 and
H (jω) = 1 ≈ −jωr
high
1 + jQω/ωr Qω
|Hhigh(jω)| =ωr .
Q
ω
On a log-log plot
42
(2.122)
(2.123)
(2.124)
(2.125)
(2.126)
(2.127)
(2.128)
(2.129)
(2.130)
2.7 Problems
1. Consider the following circuit:
Input Output
(a) What is the impedance of the circuit?

(b) For the input v(t) = v exp(jωt), what current flows through the capacitor?
(c) What is the phase difference between the voltage across the capacitor (VC ) and the
applied voltage? Be sure to specify whether VC leads or follows the applied voltage.
2. Consider a circuit consisting of an inductor (L), a capacitor (C) and a resistor (R) in series
jωt
with a voltage source v(t) = v0e . Let ω = 1 krad/s, R = 1 kΩ, L = 1 H and C=10µF.
(a) What is the magnitude of the current through the circuit?

(b) What is the phase of the current relative to the source voltage?
(c) What is the magnitude of the voltage drop across the inductor?
(d) What is the phase of the voltage across the inductor relative to the source voltage?
(e) At what frequency is the amplitude of the voltage across the resistor largest?
3. Derive the transfer function for the four-terminal network shown below.
R1
Vin C1
R2 C2 Vout
Chapter 3
Filter Circuits
Lets now apply our knowledge of AC circuits to some practical applications. We will first look
at some simple passive filters (skipping active filters) and then an amplifier model. Again we
will rely on complex variables.
3.1 Filters and Amplifiers

Simplistically, filters and amplifiers can be considered as four-terminal networks described by a
transfer function as follows:
vout(jω) = H(jω)vin(jω). (3.1)
Figure 3.1 shows some ideal transfer functions. If H(jω) = H ≡ A is a real constant then we
call the network an ideal amplifier. If H(jω) = Θ(j(ω − ω0)) is a heavyside step function we refer
to the circuit as an ideal low-pass filter, and if H(jω) = 1 − Θ(j(ω − ω0)) an ideal high-pass filter.
3.2 Log-Log Plots and Decibels

A log-log plot of a circuit’s transfer function can be a useful qualitative tool to allow us to
understand most of the important features of filter and amplifier circuits. The commonly used
decibel unit will be defined. Although unappealing to the physicist this unit is still in wide spread
use in electronics. Lets start.
If P1 and P2 are two powers, we define the decibel as
P2 V 2 V
2
dB ≡ 10 log P = 10 log V 2 = 20 log V . (3.2)

2
1 1 1
where we have used P ∝ V 2.

The decibel is a property of the network and not the signals. Hence we can make use of any
convenient signals in defining decibel. If two constant equal amplitude sources, |vin(jω1)| = |
vin(jω2)|, are applied to a four-terminal network we may write
44
CHAPTER 3. FILTER CIRCUITS 45
H H
1 1
(a) (b)
omega c omega omega c omega

H H H
A 1 1
(c)
(d) slope = -1 (e)
omega omega slope = +1

omega c
omega c omega
Figure 3.1: a) Ideal amplifier, b) Ideal low-pass filter, c) ideal high-pass filter, d) low-pass filter
and e) high-pass filter.
|vout(jω1)| = |H(jω1)||vin(jω1)|, (3.3)

|vout(jω2)| = |H(jω2)||vin(jω2)| and (3.4)
v2 H(jω2)
= . (3.5)
out
v1 H(jω1)
Therefore
H(jω2)
dB = 20 log . (3.6)
H(jω1)
Using the approximation procedure of a previous lecture
n n (3.7)
|H(jω)| ∝ |(jω) | = ω ,
and if ω1 and ω2 are not too different
dB = 20 log ω2
n = n20 log ω .
2
n
(3.8)
ω1 ω1
By definition an octave interval is when ω2 = 2ω1 and hence
dB/octave = 20n log10(2) = 6.02n ≈ 6n. (3.9)
Likewise for a decade interval ω2 = 10ω1 and

dB/decade = 20n log10(10) = 20n. (3.10)
3.3 Passive RC Filters

We will now use our passive circuit elements to design some filter circuits. Inductors are not
very good devices and hence we will concentrate on the use of resistors and capacitors.
3.3.1 Low-Pass Filter

Figure 3.2 shows one possible low-pass filter. The circuit is essentially a frequency-sensitive
voltage divider. At high frequencies the output behaves as if it is shorted while at low
frequencies the output appears as an open circuit.
R H
A C
corner
C 1
slope = -1
-1
B D omega
Figure 3.2: RC low-pass filter omega c omega
Mathematically we have
1/(jωC)
v = (3.11)
R + 1/(jωC) vin.
out
H(jω) ≡ vout = 1 .
v (3.12)
1 + jωRC
in
The approximations are
ω → 0 ⇒ H(jω) → Hlow = 1. (3.13)

1
ω → ∞ ⇒ H(jω) → Hhigh =jωRC , (3.14)
1 ω−1.
|Hhigh| = RC (3.15)
At the corner
|Hhigh| = |Hlow| ⇒ 1 = 1. (3.16)
RCωc
Therefore
ωc = 1 (3.17)
R
C
is the corner frequency of the filter. At the corner frequency
H(jωc) = 1 = 1 = 1−j , (3.18)
1+j 2
1 + jωcRC
1
|H(jωc)| = √ . (3.19)
2
√
We say that the output is down by 1/ 2 at the corner frequency.
3.3.2 Approximate Integrater

The low-pass filter acts as an approximate integrater at high frequencies. Assume
jωt (3.20)
vin(t) = ve
and integrate to obtain
v
jωt jωt
vout = v e dt = e + vout(t = 0). (3.21)
jω
The DC term is unimportant and may be dropped to obtain
1
v = (3.22)
out j vin.
ω
We define
H v 1
integrate ≡ out (3.23)
v =j .
in
ω
For a low-pass filter at high frequencies ω ωc and
1 1 (3.24)
H = = H .
high jωRC R integrate
C
Thus the low-pass filter integrates at high frequencies but also attenuates the signal by 1/
(RC).
C
A C H
1
omega
R slope = 1
D omega
omega c
B
Figure 3.3: RC high-pass filter.
3.3.3 High-Pass Filter

Figure 3.3 shows one possible high-pass filter. Mathematically we can write
H(jω) = R = jωRC . (3.25)
R + 1/(jωC) 1 + jωRC
At low and high frequencies
Hlow = jωRC and Hhigh = 1. (3.26)
At the corner frequency ω = ωc we have

|Hlow| = |Hhigh| (3.27)
and therefore
1. (3.28)
ωc =
RC
3.3.4 Approximate Differentiator

A high-pass filter acts as an approximate differentiator at low frequencies. Consider
jωt (3.29)
vin = ve
and differentiate to obtain
v = dvin = jωvejωt = jωvin. (3.30)
out
dt
We define
1H .
H = jω = low
differentiate (3.31)
RC
Again the filter attenuates the signal by 1/(RC).
Example: Write the transfer function H(jω) for the network in figure 3.4 and
from it find:
1. the corner frequency,
Treating the circuit like a voltage divider, the transfer function is
1/(jωC) = 1 . (3.32)
H(jω) =
1/(jωC) + jωL 1−
ω2LC
−1
2
ω LC .
1= 1 2 1
For ω ≈ 0 ⇒ H(jω) → 1.
2
ωC LC =LC .
⇒ ωC
For large ω ⇒ H(jω) →
For the corner
frequency Therefore
ωC = √ 1 = 1 = 1 × 103rad/s. (3.33)
LC
(1
×
10−6)1 /2
1 H
1 microF
Figure 3.4: Four-terminal network without resistance.
2. the value of |H| at the corner

frequency. At the corner frequency
H(jωC ) = 1 → ∞. (3.34)
1−1
3. How many degrees of phase shift are introduced by this network just
below and just above the corner frequency?
Since H(jω) is always real there is no phase shift.
3.4 Complex Frequencies and the s-Plane

We will now consider s-plane techniques. Not because we will use them, but more to under-
stand some of the common electronics terminology.
We can enhance the usefulness of the transfer function H(jω) by transforming to a complex
frequency. Define the complex variable s such that
s = σ + jω, (3.35)
where σ is an inverse time constant.
Our exponential function now becomes

st σt jωt
f (t) = Ae = Ae e (3.36)
and we have a rich set of cases
ω=0, σ>0, exponential growth,
eσt
ω=0, σ<0, e−|σ|t decay,
ω=0, σ=0, ejωt oscillation,
ω=0, σ>0, eσ ejωt growing oscillations and
ω=0, σ<0, decaying oscillations.
e−|σ|ejωt
So we can not only describe oscillatory behavior but transient responses as well.
3.4.1 Poles and Zeros of H

As before, consider expanding the transfer function as the ratio of two polynomials
H(s) = P(s)
. (3.37)
Q(s)
If an are the roots of P(s) and bm are the roots of Q(s) we can write
H(s) = A (s − a1)(s1 − a2) . . . (s − an) , (3.38)
(s − b1)(s − b2) . . . (s − bm)

where A is a real constant, an are zeros of H and bm are poles (infinities) of H. Knowledge of an
and bm determines H(s) everywhere.
Lets now look at our two filter circuits. For a low-pass filter
H(s) = 1 1/(RC)
= (3.39)
1 + sRC s + 1/(RC)
and the filter has one pole at −1/(RC). For a high-pass filter
H(s) = sRC s
= (3.40)
1 + sRC s + 1/(RC)
and it has one pole at −1/(RC) and one zero at 0. We refer to these two types of filters as single-
pole filters.
There is a general rule that there must be at least as many reactive elements as poles. Based
on the location of the poles we are able to deduce the general response properties of the filter.
We will not do this here.
Example: If a transfer function has poles at p1 = (−1, 2) and p2 = (−1, −2)

and a zero at (0,0), as shown in figure 3.5,
+5
−5 +5
−5
Figure 3.5: Poles and zeros in the complex plane.
1. sketch |H(jω)| on the interval 0 ≤ ω < 10.

The transfer function is given by
H(s) = |s − (0, 0)| = ω (3.41)

|s − (−1, 2)||s − (−1, −2)| |s − p1 ||s − p2|
Plugging in values for ω gives the table 3.1.
Table 3.1: Numerical values of the transfer function.

ω |s − (−1, 2)| |s − (−1, −2)| H(s)
0 2.24 2.24 0
1 1.41 3.16 0.224
2 1.00 4.12 0.485
3 1.41 5.10 0.416
4 2.24 6.08 0.294
5 3.16 7.07 0.224
6 4.12 8.06 0.180
7 5.10 9.06 0.152
8 6.08 10.05 0.131
9 7.07 11.05 0.115
10 8.06 12.04 0.103
2. If |H(j10)| = 1, what is the approximate value of |H| at its highest point?

If |H(j10)| = 1 then Hmax at ω = 2 is Hmax = 0.485. Therefore
0.485 ×1=4.7. (3.42)
0.103
Figure 3.6: The transfer function from the table above.
3.5 Sequential RC Filters

Single-pole filters are rather limited (6 dB/octave slope). For better band-pass and band-reject
filters we require more poles and zeros and thus more reactive circuit elements. A simple
solution is to connect two or more single-pole RC filters in sequence. If filter H 2 draws no
current from filter H1, the transfer function for the combined filter is
H = H1H2(= H2H1). (3.43)

One way to do this is to choose a large impedance for H 2. Hence H has more poles and zeros
than H1 or H2.
The corner frequency for a high-pass filter is ωH = 1/(RC) and the transfer function may
be written as
H = jωRC = jω/ωH . (3.44)
high
1 + jωRC 1 + jω/ωH
For a low-pass filter the corner frequency is ωL = 1/(RC) and
1 = 1
H = . (3.45)
low 1 + jωRC 1 + jω/ωL
We may build a two-section low-pass filter by requiring R2 > R1 and 1/C2 > 1/C1, as shown
in figure 3.7, so that
R
A 1 R
2 C H
1
slope = -1
C C
1 2
slope = -2
D w omega
B L1 w L2
Figure 3.7: Two-section low-pass filter.
1
Hlow = . (3.46)
(1 + jω/ωL1 )(1 + jω/ωL2 )
A special case occurs when R1C1 = R2C2 ⇒ ωL1 = ωL2 and we obtain one corner frequency but
−2
the slope of the filter is ω .
The results are similarly for a two-section high-pass filter.
A band-pass filter can be built from one low-pass filter and one high-pass filter, as shown in
figure 3.8. The order of the filter sections does not matter as long as the impedance rule is
obeyed.
R1 C2
A C H
1 slope = -1
slope = +1
C1 R2
D omega
w H w L
B
Figure 3.8: Band-pass filter.

H = jω/ωH . (3.47)
pass
(1 + jω/ωH )(1 + jω/ωL)

If ωL ≤ ωH we have only two straight regions and the band-pass frequency range degenerates to
zero.
Example: Show that the magnitude of the transfer function
H = jω/ωH (3.48)
pass
(1 + jω/ωH )(1 + jω/ωL)
falls off −6 dB/octave at both the low- and high-frequency extremes.
For small ω (ie. ω ωH and ω ωL) H ≈ jω/ωH ; |H(jω)| = ω/ωH ;
log10 H = log10 ω + C .
Thus
d(log10 |H|) = 1 ⇒ 6db/octave.
(3.49)
d(log10 ω)
For large ω (ie. ω ωH and ω ωL) H ≈ −jωL/ω; |H(jω)| = ωL/ω; |H(jω)| =
− log10 ω + C .
And thus
d(log10 |H|) −⇒−
= 1 6dB/octave.
(3.50)
d(log10 ω)
3.6 Passive RCL Filters

Sequential RC filters always have real poles and hence smooth rounded corners. To improve
these filters we introduce an inductor (which is good for high frequencies).
3.6.1 Series RCL Circuit

Consider the RCL circuits as shown in figure 3.9. Each has a low-frequency and high-frequency
approximation. Considering the band-reject filter (figure 3.6d) we obtain for the transfer function
H = 1/(jωC) + jωL (3.51)

R + 1/(jωC) + jωL
= 1 − ω2LC . (3.52)
(1 − ω2LC) + jωRC
The approximations are:
ω→0; H→HL=1 and (3.53)
ω 1. (3.54)
→∞; H H=
→ H √
We notice a zero in the transfer function at ω0 = 1/ LC. In the low-medium frequency
range
1 ∝ ω−1,
ω<ω0; H→HLM= (3.55)
jωRC
for high-medium frequencies
ω > ω ; H→ H = −ω2LC = jωL ∝ ω+1. (3.56)
0 HM jωRC R
Solving for the corner frequencies we have
L R C R
A C A C
(a) C (b) L
D B D
B
C R
L C
A A C
R C
(c)
(d)
D L
B
B D
Figure 3.9: LCR filters: a) low-pass, b) high-pass, c) band-pass and d) band-reject.
1 = 1/|jωRC| ⇒ ω1 = 1/(RC), (3.57)
1 = |jωL/R| ⇒ ω2 = R/L and (3.58)
√ √ .
ω0 = 1/ LC = ω1ω2 (3.59)
Example: Sketch |H(jω)| for the LCR circuit shown in figure 3.10 for the two
conditions R = 0.5 L/C and R = 2 L/C. In each case, determine the values of |
H| at ω = 0, ∞, and ωc, and label these points on the sketches.
The transfer function is
R + jωL
H(jω) = 1 . (3.60)
R + jωL + jωC
For ω small
H(jω) ≈ R
= jωRC (3.61)
1/(jωC)
|H(jω)| = ωRC. (3.62)
C L
Figure 3.10: LCR circuit with two components across the output.
For large ω: H(jω) = 1.

For the corner frequency: 1 = ωC RC → ωC = 1/(RC).
For R = 0.5 L/C =1 L ,
√
2
ω 2
H = 2 LC; ωC = √ .
low LC
R + 2j
H(jωC ) = L/C R + 4jR
= (3.63)
R + 2j L/C − j/2 L/C R + 4jR − jR
= 1 + 4j = (1 + 4j)(1 − 3j) = 13 + j (3.64)
1 + 3j 1+9 10
|H(jωC )| = 132 + 12 = 170 = 1.30. (3.65)

102 10
For R = 2 L/C,
√ 1
Hlow = 2ω LC; ωC = √ LC
2
R + j/2
H(jωC ) = L/C = 2 + j/2 (3.66)
R + j/2 L/C − 2j L/C 2 + j/2 − 2j
4 + j(4 + j)(4 + 3j) 13 + 16j
= = = (3.67)
4 − 3j 16+9 25
2 2 √
|H(jωC )| = 13 + 16 = 425 = 0.825. (3.68)
252 25
Figure 3.11 is a sketch of the transfer functions.
Example:
1. Write an expression for the transfer function of the circuit shown in fig-
ure 3.12.
ZCL = (1/(jωC))(jωL) = jωL (3.69)

1/(jωC) + jωL 1−
ω2CL
Figure 3.11: Sketch of the transfer functions for the above circuit.
jωL 1
H(jω) = 2
1−ω CL
jωL
= 2
(3.70)
R+ 2
1−ω CL 1 − jR/(ωL)(1 − ω CL)
= 1 . (3.71)
1R
1 + j ωRC − ω L
2. What phase shift is introduced by this filter at very small and very large
frequencies?
1 −j −1
For large ω H(jω) ≈ jωRC = RC ω
1 →φH=− π . (3.72)
H = ω e −1 −jπ/2
high RC 2
For small ω H(jω) ≈ 1 = jω L
(−j/ω)(R/L) R
L π
H = φL=+ . (3.73)
low R 2
ωejπ/2 →
3. On a log-log scale, sketch |H(jω)| and the phase shift as a function of ω .
2 R 1
For the corner frequency ω RC = L →ωC= √ LC .
C
H(jωC ) = 1; φC = 0.
-> ->
Vin R Vout
L
C
Figure 3.12: Circuit with components in parallel at the output.
3.7 Amplifier Model

Enough of filters. Lets now look at the simple amplifier model in figure 3.14
vout(jω) = A(jω)vin(jω), (3.74)

vout(s) = A(s)vin(s). (3.75)
Notice that vout is with respect to ground while vin is a voltage difference. For a typical
6
operational amplifier |H| ≈ 10 at ω = 0. As ω → ∞, A(ω) < A(ω = 0) due to internal
capacitance, ie. the amplifier behaves like a low-pass filter.
A(s) = A0Hlow(s), (3.76)
where Hlow is the transfer function of a low-pass filter.
3.7.1 One-, Two- and Three-Pole Amplifier Models

The simplest amplifier has
A
0
A1(s) = , (3.77)
1 + s/sa
where sa is a real positive number (corner frequency). There is a single pole at −sa on the
negative real axis. This means that the impulse response will decay exponentially without
ringing.
If we cascade multiple (three) single-pole amplifiers together
A3(s) = A0 . (3.78)
(1 + s/sa)(1 + s/sb)(1 + s/sc)
There will still be no oscillation to an impulse.
Figure 3.13: Transfer function and phase shift for the above circuit.
3.7.2 Amplifier with Negative Feedback

Feedback is a widely used technique to improve the characteristics of an imperfect amplifier.
A generalized amplifier with negative voltage feedback is shown in figure 3.15.
The overall transfer function (closed-loop gain) can be written as
H(s) = v .
3
(3.79)
v1
Realizing that
v2 = v1 − v4; v2 = v3 and v4 = Fv3.
(3.80)
A(s)
we may write
H(s) = A .
(3.81)
1+AF
If |AF| 1 ⇒ H(s) = 1/F. The transfer function is now independent of the amplifier gain. F is the
transfer function of a stable resistive network which means that H will also be stable.
v + A
in - v
out
Figure 3.14: A simple amplifier model.
+ v3 = A v2
v2 A
- v3
+ v1
v4 F v3
v4 = F v3
Figure 3.15: Amplifier with negative voltage feedback.

3.8 Problems
1. In the following circuit, the input signal is v = v0 exp(jωt), and the components have been
chosen such that L = R/ω and C = 3/(ωR). The output is at the terminals AB.
A
R
C
v(t) S L
(a) What is the transfer function H(jω)?

(b) Find the current in the circuit.
(c) What is the voltage drop across each element of the circuit?
(d) Show algebraically that at any instant the potential difference around the circuit is
zero.
(e) At time t = N π/ω, where N = 0, 2, 4, ... make a sketch showing the voltage across
each element in the complex plane and show that the vector sum of the voltage drops
is equal to the voltage supplied.
(f) Write an expression for |H(jω)|.
(g) What is the limit of |H(jω)| as ω goes to zero?
(h) What is the limit of |H(jω)| as ω goes to infinity?
(i) What is the corner frequency?
(j) What is |H(jω)| at the corner frequency?
(k) Make a sketch showing the characteristics of |H(jω)| on a log-log plot. Label the
slope of the curve where possible, the corner frequency, and the value of |H(jω)| at
the corner frequency.
(l) Describe the high and low frequency behavior in dB/octave.
2. (a) Draw a passive LCR low-pass filter and write down the transfer function of your four-
terminal network.
(b) Determine approximations to the transfer function and filter corner frequency(s).
√
(c) Write the resonance frequency, ωr = 1/ LC, in terms of the corner frequency(s).
3. Write down the transfer function, H(jω), for the network shown below, and from it find:
(a) the corner frequency(s) and

(b) the value(s) of |H(jω)| at the corner frequency(s).
(c) Sketch |H(jω)| and the voltage phase-shift as a function of ω .
(d) What type of filter is this?
Vin R Vout
L
4. (a) What is the transfer function for the following circuit?
R R R
input C C C output
(b) Describe the frequency response at low and high frequencies?

(c) Sketch the magnitude |H(jω)| on a log-log plot. (label slopes, the corner fre-
quency(s), and |H(jωC )|)
−1
(d) What is the signal attenuation for ω = 3(RC) .
5. (a) Show that the transfer function of a single-pole RC filter drops by 3 db at the corner
frequency. This is often refereed to as the 3 db down-point.
3
(b) Design a bandpass RC filter with the transfer function shown below. ω1 = 10 rad/s
and ω2 = 105 rad/s are the 3 db down-points of the RC filter sec-tions. Choose
impedances so that the first section is not much affected by the loading of the second
section.
(c) Are ω1 and ω2 the 3 db down-points of a bandpass filter?
Chapter 4
Diode Circuits
So far we have only considered passive circuit elements. Now we will consider our first reactive
circuit element, the diode. Hopefully things will start to get a little more interesting.
Along with the diode there is the transistor, which will be discussed in future lectures. These
circuit elements are commonly made from a semiconductor basic material. Semicon-ductors
along with the passive circuit elements are often integrated onto a single electronics chip. If a
hundred or so transistors are on a chip it is referred to as an integrated circuit or IC. Large-scale
integrated circuits, LSI, contain thousands of transistors. And today even very-large-scale
integrated circuits, VLSI, exist, with hundreds of thousands of transistors.
Now lets talk physics.
4.1 Energy Levels

The physical principles of semiconductor devices can be understood by considering quantum
energy levels in the material. I will just give a sketchy view here and define some terminology.
Valance electrons are the electrons outside the closed shells of an atom. In silicon and
germanium there are four valence electrons, arsenic has five and gallium three. The large
number of valance electrons favour these elements as semiconductors.
To sustain an electric current, a material must have charge carriers that are free to move.
There is some probability that an atom may eject a valence electron which is then free to move
in the material. The conductivity of a material is thus a function of the number of free charge
carriers per unit volume. Based on these probability densities it is common to divide materials
into three categories: conductors, semiconductors and insulators.
In crystals atoms interact and bind by sharing valence electrons. The wave function is no
longer associated with a single atom but extends over the entire crystal. One effect of the
interaction between the atoms is that the otherwise degenerate energy levels split into closely
spaced levels. Since the number of atoms is large, it is common to refer to this set of levels as a
continuous energy band.
In solid materials there usually exist a valance band which is an energy region where the
states are filled or partially filled by valence electrons. The conduction band is defined to be
the lowest unfilled energy band. So our three materials can be characterized by their band
structure. An insulator has the valence and conduction band well separated. A
64
CHAPTER 4. DIODE CIRCUITS 65
semiconductor has the valence band close to the conduction band – separated by about a 1 eV
gap. Conductors on the other hand have the conduction and valence bands overlapping.
The interesting property of a semiconductor is that thermally excited electrons can move
from the valence band to the conduction band and conduct current. Silicon and germanium have
thermally excited electrons at room temperature and hence their common use in diodes and
transistors.
When an electron has been excited into the conduction band, the hole left behind in the
valence band is also free to move through the crystal. A quantum mechanical treatment of this
effect puts the hole on an approximately equal footing with the electron. Temperature causes the
thermal generation of electron-hole pairs. One of the components of the pair will add a little to
the majority charge carriers. The other component of the electron-hole pair will become the
minority charge carrier. Minority charge carriers limit ideal performance and increase with
increasing temperature.
A common method for generating even more charge carriers in a semiconductor is by
doping. That is, replacing a few atoms of the base material with atoms of a different element.
These impurities will contribute an excess electron or hole which is loosely bound and hence can
be excited into the conduction band by thermal energy. In N-type semiconductors the majority of
free charge carriers are negative, while in a P-type semiconductor the majority are positive.
4.2 The PN Junction and the Diode Effect

By joining a P-type and N-type semiconductor together we can make a diode (figure 4.1)
Slope
is E
E V Electron
P−TYPE N−TYPE
Energy
+ −
++ −−
++ − 0
++ −−
+ −
−
Majority Deplection Majority Good Poor Good
Carriers Region Carriers Conductor Conductor Conductor
Figure 4.1: PN junction diode.

Initially both semiconductors are totally neutral. The concentration of positive and negative
carriers are quite different on opposite sides of the junction and the thermal energy-powered
diffusion of positive carriers into the N-type material and negative carriers into the P-type
material occurs. The N-type material acquires an excess of positive charge near the junction and
the P-type material acquires an excess of negative charge. Eventually diffuse charges build up
and an electric field is created which drives the minority charges and eventually equilibrium is
reached. A region develops at the junction called the depletion region. This region is essentially
un-doped or just intrinsic silicon.
To complete the diode conductor, leads are placed at the ends of the PN junction.
4.2.1 Current in the Diode

The behaviour of a diode depends on its polarity in the circuit (figure 4.2). If the diode is
reverse biased (positive potential on N-type material) the depletion region increases. The only
charge carriers able to support a net current across the PN junction are the minority carriers and
hence the reverse current is very small. A forward-biased diode (positive poten-tial on P-type
material) has a decreased depletion region; the majority carriers can diffuse across the junction.
The voltage may become high enough to eliminate the depletion region entirely.
+ +
I r I f
P N P N
a) b)
Figure 4.2: Diode circuit connections: a) reversed biased and b) forward biased.
An approximation to the current in the PN junction region is given by (shown in fig-ure 4.3a)
V /V (4.1)
I = I0(e T − 1),
where both I0 and VT are temperature dependent. This equation gives a reasonably accurate
prediction of the current-voltage relationship of the PN junction itself – especially the tem-
perature variation – and can be improved somewhat by choosing I0 and VT empirically to fit
a particular diode. However, for a real diode, other factors are also important: in particular, edge
effects around the border of the junction cause the actual reverse current to increase slightly with
reverse voltage, and the finite conductivity of the doped semi-conductor ulti-mately restricts the
forward current to a linear increase with increasing applied voltage. A better current-voltage
curve for the real diode is shown in the figure 4.3b.
I I
V/V
I (e T−1)
0
slope = 1/Rf
V VPN V
slope = 1/Rr
a) b)
Figure 4.3: Current versus voltage a) in the PN junction region and b) for an actual PN diode.
Various regions of the curve can be identified: the linear region of forward-biasing, a non-
linear transition region, a turn-on voltage (VP N ) and a reverse-biased region. We can assign a
dynamic resistance to the diode in each of the linear regions: Rf in the forward-biased region and
Rr in the reverse-biased region. These resistances are defined as the inverse slope of the curve:
1/R = ∆I /∆V . The voltage VP N , represents the effective voltage drop across a forward-biased
PN junction (the turn-on voltage). For a germanium diode, VP N is approximately 0.3 V, while
for a silicon diode it is close to 0.6 V.
4.2.2 The PN Diode as a Circuit Element

Diodes are referred to as non-linear circuit elements because of the above characteristic curve.
For most applications the non-linear region can be avoided and the device can be modeled by
piece-wise linear circuit elements. Qualitatively we can just think of an ideal diode has having
two regions: a conduction region of zero resistance and an infinite resistance non-conduction
region. For many circuit applications, this ideal diode model is an adequate representation of an
actual diode and simply requires that the circuit analysis be separated into two parts: forward
current and reverse current. Figure 4.4 shows a schematic symbol for a diode and the current-
voltage curve for an ideal diode.
A diode can more accurately be described using the equivalent circuit model shown in figure
4.5. If a diode is forward biased with a high voltage it acts like a resistor (Rf ) in series with a
voltage source (VP N ). For reverse biasing it acts simply as a resistor (Rr ). These
I
If
conduction
region
V
non−conduction
a) region
b)
Figure 4.4: a) Schematic symbol for a diode and b) current versus voltage for an ideal diode.
approximations are referred to as the linear element model of a diode.

VPN
Rf Rr
+ -
If Ir
VPN Rf
ideal
diodes
Rr
Figure 4.5: Equivalent circuit model of a junction diode.
4.2.3 The Zener Diode

There are several other types of diodes beside the junction diode. As the reverse voltage
increases the diode can avalanche-breakdown (zener breakdown). This causes an increase in
current in the reverse direction. Zener breakdown occurs when the electric field near the
junction becomes large enough to excite valence electrons directly into the conduction band.
Avalanche breakdown is when the minority carriers are accelerated in the electric field near
the junction to sufficient energies that they can excite valence electrons through collisions.
Figure 4.6 shows the current-voltage characteristic of a zener diode, its schematic symbol and
equivalent circuit model in the reverse-bias direction. The best zener diodes have a breakdown
voltage (VZ ) of 6-7 V.
4.2.4 Light-Emitting Diodes

Light-emitting diodes (LED) emit light in proportion to the forward current through the diode. LEDs are low voltage
devices that have a longer life than incandescent lamps. They
I
Rf
V
R Z VZ
+
Z
V
I Z
Rr I Z
a) b) c)
RZ
Figure 4.6: a) Current versus voltage of a zener diode, b) schematic symbol for a zener diode and
c) equivalent circuit model of a zener diode in the reverse-bias direction.
respond quickly to changes in current (10 MHz). LEDs have applications in optical-fiber
communication and diode lasers. They produce a narrow spectrum of coherent red or infrared
light that can be well collimated.
As an electron in the conduction band recombines with a hole in the valence band, the
electron makes a transition to a lower-lying energy state and releases energy in an amount equal
to the band-gap energy. Normally the energy heats the material. In an LED this energy goes into
emitted infrared or visible light.
4.2.5 Light-Sensitive Diodes

If light of the proper wavelength is incident on the depletion region of a diode while a reverse
voltage is applied, the absorbed photons can produce additional electron-hole pairs. Photo-diodes
or photocells can receive frequency-modulated light signals. LEDs and photodiodes are often
used in optical communication as receiver and transmitter respectively.
4.3 Circuit Applications of Ordinary Diodes

Lets briefly discuss some applications of ordinary diodes. For many circuits only the basic diode
effect is of any significance and these circuits can be analyzed under the assumption that the
diode is an ideal device.
4.3.1 Power Supplies

Batteries are often shown on a schematic diagram as the source of DC voltage but usually the actual DC voltage
source is a power supply. A more reliable method of obtaining DC
power is to transform, rectify, filter and regulate an AC line voltage. Power supplies make use of
simple circuits which we will discuss presently.
DC power supplies are often constructed using a common inexpensive three-terminal
regulator. These regulators are integrated circuits consisting of several solid state devices and are
designed to provide the desirable attributes of temperature stability, output current limiting and
thermal overload protection.
In power supply applications it is common to use a transformer to isolate the power supply
from the 110 V AC line. A rectifier can be connected to the transformer secondary to generate a
DC voltage with little AC ripple. The object of any power supply is to reduce the ripple which is
the periodic variation in voltage about the steady value.
4.3.2 Rectification
Figure 4.7 shows a half-wave rectifier circuit. The signal is exactly the top half of the input
voltage signal, and for an ideal diode does not depend at all on the size of the load resistor.
V0
S Vs R
G
Figure 4.7: Half-wave rectifier and its output waveform.
The rectified signal is now a combination of an AC signal and a DC component. Generally, it

is the DC part of a rectified signal that is of interest, and the un-welcomed AC component is
described as ripple. It is desirable to move the ripple to high frequencies where it is easier to
remove by a low-pass filter.
When diodes are used in small-signal applications – a few volts – their behaviour is not
closely approximated by the ideal model because of the PN turn-on voltage. The equivalent
circuit model can be used to evaluate the detailed action of the rectifier under these condi-tions.
During the part of the wave when the input is positive but less than the PN turn-on voltage, the
model predicts no loop current and the output signal voltage is therefore zero. When the input
exceeds this voltage, the output signal becomes proportional to vs − VP N , or about 0.6 V lower
than the source voltage.
The diode bridge circuit shown in figure 4.8 is a full-wave rectifier. The diodes act to route the current from
both halves of the AC wave through the load resistor in the same direction, and the voltage developed across the
load resistor becomes the rectified output signal. The diode bridge is a commonly used circuit and is available as a
four-terminal component in a number of different power and voltage ratings.
Figure 4.8: Full-wave rectifier and its output waveform.
4.3.3 Power Supply Filtering

The rectified waveforms still have a lot of ripple that has to be smoothed out in order to generate
a genuine DC voltage. This we do by tacking on a low-pass filter. The capacitor value is chosen
in order to ensure small ripple, by making the time constant for discharging much longer than the
time between re-charging.
4.3.4 Split Power Supply

Often a circuit requires a power supply that provides negative voltage as well as positive voltage.
By reversing the direction of the diode and the capacitor (if it is polarized), the half-wave
rectification circuit with low-pass filter provides a negative voltage. Similarly, reversing the
direction of the diodes and capacitor in the full-wave rectified supply produces a negative
voltage supply. A split power supply is shown in figure 4.9.
+V
G GROUND
-V
Figure 4.9: Split power supplies.

4.3.5 Voltage Multiplier

A voltage multiplier circuit is shown in figure 4.10. We can think of it as two half-wave rectifier
circuits in series. During the positive half-cycle one of the diodes conducts and charges a
capacitor. During the negative half-cycle the other diode conducts negatively to charge the other
capacitor. The voltage across the combination is therefore equal to twice the peak voltage. In this
type of circuit we have to assume that the load does not draw a significant charge from the
capacitors.
Figure 4.10: Voltage doubler circuit.
4.3.6 Clamping
When a signal drives an open-ended capacitor the average voltage level on the output termi-nal
of the capacitor is determined by the initial charge on that terminal and may therefore be quite
unpredictable. Thus it is necessary to connect the output to ground or some other reference
voltage via a large resistor. This action drains any excess charge and results in an average or DC
output voltage of zero.
A simple alternative method of establishing a DC reference for the output voltage is by using
a diode clamp as shown in figure 4.11. By conducting whenever the voltage at the output
terminal of the capacitor goes negative, this circuit builds up an average charge on the terminal
that is sufficient to prevent the output from ever going negative. Positive charge on this terminal
is effectively trapped.
C V0
S ->
Vs
G
Figure 4.11: Diode clamp circuit and its output waveform.

4.3.7 Clipping
A diode clipping circuit can be used to limit the voltage swing of a signal. Figure 4.12 shows a
diode circuit that clips both the positive and negative voltage swings to references voltages.
S ->
G Vs +V1 -V2
+
Figure 4.12: Diode clipping circuit and its output waveform.
4.3.8 Diode Gate

Diodes can also be used to pass the higher of two voltages without affecting the lower. A nifty
example is shown in figure 4.13 The 12 V battery does nothing until the power fails; then it takes
over without interruption.
+12V
to
+15V
+15V
- +
12V
Figure 4.13: Diode OR gate.
4.3.9 Diode Protection

An important use of diodes is to suppress the voltage surge present when an inductive load is
switched out of a circuit – inductive surge suppression. With inductors it is not possible to turn off the current
suddenly since the inductor will try to keep the current flowing when the switch is opened. A diode in a DC circuit
or back-to-back zener diodes in an AC circuit can be used to shunt the inductor and prevent it from conducting.
Example: For each circuit in figure 4.14 sketch the output voltage as a function of
time if vs(t) = 10 cos(2000πt) V. Assume that the circuit elements are ideal.
1 kohm
1 kohm
S ->
Vs
S -> 1 kohm S -> Vz = 6 V
Vs G Vs
G G B
C
A
Vz = 6 V
0.1 microF
S ->
1 kohm S ->
G Vs Vs
G
D
E
Figure 4.14: Circuits with a single ideal diode.
The forward and reverse biased approximations for the circuit in figure 4.14a
are shown in figure 4.15 and the output voltage is sketched in figure 4.20a.
1 kohm 1 kohm
S -> v=0 S -> v=Vs

G Vs G Vs
FORWARD BIAS REVERSE BIAS
Figure 4.15: Single diode circuit a).
The forward and reverse biased approximations for the circuit in figure 4.14b
are shown in figure 4.16 and the output voltage is sketched in figure 4.20b.
The forward and reverse biased approximations for the circuit in figure 4.14c
are shown in figure 4.17 and the output voltage is sketched in figure 4.20c.
S -> 1 kohm v=Vs S -> 1 kohm v=0

G Vs G Vs
B
Figure 4.16: Single diode circuit b).
1 kohm
1 kohm
+ S -> v=0
S -> Vz = 6 V
G Vs v>Vz Vs
G
Figure 4.17: Single diode circuit c).
The forward and reverse biased approximations for the circuit in figure 4.14d
are shown in figure 4.18 and the output voltage is sketched in figure 4.20d.
The forward and reverse biased approximations for the circuit in figure 4.14e
are shown in figure 4.20 and the output voltage is sketched in figure 4.20e.
Vz = 6 V
S -> 1 kohm v=Vs S -> 1 kohm v<-Vz
G Vs G Vs
D
Figure 4.18: Single diode circuit d).
Example: Assuming that the diodes in the circuit below are ideal, write
expres-sions for the voltage at points A and B.
Consider when the current flows in the clockwise direction (figure 4.22). V A =
VB = V0 in the steady state because the charge just builds (it has nowhere to
“drain” to).
Consider when the current flows in the anti-clockwise direction (figure 4.23).
VA = V0 cos ωt, since this is just the output terminal of the voltage source. V B =
V0 in the steady state, again since the charge cannot go anywhere.
Thus by superposition
V = V + V0 cos ωt (4.2)
A 0
V = V + V0 = 2V0. (4.3)
B 0
0.1 microF 0.1 microF
S -> Vs(0)=10V S -> v=0

Vs Vs
G G
FOWARD BIAS REVERSE BIAS
Figure 4.19: Single diode circuit e).
Figure 4.20: Sketch of the output voltage as a function of time.

A B
C
S
C
G
V0 cos(omega t)
Figure 4.21: A circuit with two ideal diodes.
B
C
S
C
G
V0 cos(omega t)
Figure 4.22: Current clockwise.
A
B
C
S C
G
V0 cos(omega t)
Figure 4.23: Current anti-clockwise.

4.4 Problems
1. (a) Make a sketch showing the current through an ideal diode as a function of the applied
voltage. Also sketch the current through a real signal diode as a function of voltage.
(b) Make a sketch showing the current drawn through a Zener diode as a function of the
applied voltage. Show how to determine the forward resistance (Rf ), the reverse
resistance (Rr ) and the Zener resistance (RZ ) from your sketch. Label the voltages
VP N and VZ .
2. Sketch the expected output waveforms when
(a) a 100 Hz sine wave with a peak voltage of 5 V, and

(b) a 100 Hz square wave with a peak-to-peak voltage of 10 V
are applied separately to each of the circuits below.
A 6.3V B
C D
5V
3V
3. The Zener diode in the following circuit is characterized by VP N = 0.6 V and |VZ | = 4.8
V. Terminal B is at ground and there is no external load resistor.
A
10kohm
+
13.2V
B
(a) What is the voltage at terminal A in the above circuit?

(b) What is the current through the Zener? (What reasonable approximation makes this
straightforward?).
(c) If the Zener is dissipating 1.0 µW (in heat), what can we conclude to be the effective
impedance of the diode for this situation?
4. The effects of a Zener diode in a circuit can be treated analytically by using an equiv-alent
circuit model of the reverse-biased condition. In the low voltage region, before breakdown,
the Zener diode can be treated like any other reverse-biased diode. How-ever, the Zener
diode is normally operated in the breakdown region.
(a) Write down an equivalent linear-circuit model in the breakdown voltage region.
(b) Replace the Zener diode in the following voltage reference circuit by your equiv-
alent circuit model.
Vs R VL
I Iz IL RL
(c) Determine the contribution of the of Zener diode to this voltage reference circuit by
calculating the elements of a Thevenin equivalent circuit representation.
(d) Show that for small Zener diode effective resistance the Thevenin equivalent volt-age
is close to the Zener breakdown voltage and thus is insensitive to changes in the
source voltage.
(e) Show that for small Zener diode effective resistance, the Thevenin equivalent resistor
gives the voltage source a reasonably small output impedance.
(f) The combined results show a voltage reference that is insensitive to voltage changes
in the original EMF and to changes in the load current. State the as-sumptions under
which this result is valid.
Chapter 5
Transistor Circuits
The circuits we have encountered so far are passive and dissipate power. Even a transformer that
is capable of giving a voltage gain to a circuit is not an active element. Active elements in a
circuit increase the power by controlling or modulating the flow of energy or power from an
additional power supply into the circuit.
Transistors are active circuit elements and are typically made from silicon or germanium and
come in two types. The bipolar transistor controls the current by varying the number of charge
carriers. The field effect transistor (FET) varies the current by varying the shape of the
conducting volume.
Before starting we will define some notation. The voltages that are with respect to ground are
indicated by a single subscript. Voltages with repeated letters are power supply voltages. And
voltages between two terminals are indicated by a double subscript.
5.1 Bipolar Junction Transistors

By placing two PN junctions together we can create a bipolar transistor. In a PNP transistor the
majority charge carriers are holes and germanium is favoured for these devices. Silicon is best
for NPN transistors where the majority charge carriers are electrons.
The thin and lightly doped central region is known as the base (B) and has majority charge
carriers of opposite polarity to those in the surrounding material. The two outer regions are
known as the emitter (E) and the collector (C). Under the proper operating conditions the
emitter will emit or inject majority charge carriers into the base region, and because the base is
very thin, most will ultimately reach the collector. The emitter is highly doped to reduce
resistance. The collector is lightly doped to reduce the junction capacitance of the collector-base
junction.
The schematic circuit symbols for bipolar transistors are shown in figure 5.1. The arrows on
the schematic symbols indicate the direction of both IB and IC . The collector is usually at a
higher voltage than the emitter. The emitter-base junction is forward biased while the collector-
base junction is reversed biased.
81
CHAPTER 5. TRANSISTOR CIRCUITS 82
C C
B B
E E
a) b)
Figure 5.1: a) NPN bipolar transistor and b) PNP bipolar transistor.
5.1.1 Transistor Operation (NPN)

If the collector, emitter, and base of an NPN transistor are shorted together as shown in figure
5.2a, the diffusion process described earlier for diodes results in the formation of two depletion
regions that surround the base as shown. The diffusion of negative carriers into the base and
positive carriers out of the base results in a relative electric potential as shown in figure 5.2b.
B
+V C E
N P N
C E
B
a) Depletion b)
Region
Figure 5.2: a) NPN transistor with collector, base and emitter shorted together, and b) voltage
levels developed within the shorted semiconductor.
When the transistor is biased for normal operation as in figure 5.3a, the base terminal is
slightly positive with respect to the emitter (about 0.6 V for silicon), and the collector is
positive by several volts. When properly biased, the transistor acts to make IC IB . The depletion
region at the reverse-biased base-collector junction grows and is able to support the increased
electric potential change indicated in the figure 5.3b.
For a typical transistor, 95% to 99% of the charge carriers from the emitter make it to
V
V
CB EB C
V
CE
IC + B +
IB
E
N P N
V B
C E CB BE V
V > V
IC>IB CE BE
Figure 5.3: a) NPN transistor biased for operation and b) voltage levels developed within the
biased semiconductor.
the collector and constitute almost all the collector current IC . IC is slightly less than IE and we
may write α = IC /IE , where from above α = 0.95 to 0.99.
The behaviour of a transistor can be summarized by the characteristic curves shown in figure
5.4. Each curve starts from zero in a nonlinear fashion, rises smoothly, then rounds a knee to
enter a region of essentially constant IC . This flat region corresponds to the condition where the
depletion region at the base-emitter junction has essentially disappeared. To be useful as a linear
amplifier, the transistor must be operated exclusively in the flat region, where the collector
current is determined by the base current.
A small current flow into the base controls a much larger current flow into the collector. We
can write
IC = βIB = hF E IB , (5.1)
where β is the DC current gain and hF E is called the static forward-current transfer ratio. From
the previous definition of α and the conservation of charge, IE = IC + IB , we have
β= α
. (5.2)
1−α
For α = 0.99 we have β = 99 and the transistor is a current amplifying device.
5.1.2 Basic Circuit Configurations

In any transistor circuit design you must supply a DC bias current and voltage to operate in the
linear region of the characteristic curve. The DC operating point is defined by the values of IB ,
IC , VBE and VCE . You must also obtain the proper AC operation.
The transistor is a three-terminal device that we will use to form a four-terminal circuit. Small voltage changes
in the base-emitter junction will produce large current changes in
knees I
constant IC B
I
C IB3
IB2
IB1
non−linear region
V
CE V
PN V
BE
Figure 5.4: Characteristic curves of an NPN transistor.
the collector and emitter, whereas small changes in the collector-emitter voltage have little effect
on the base. The result is that the base is always part of the input to a four-terminal network.
There are three common configurations: common emitter (CE), common collector (CC) and
common base (CB), as shown in figure 5.5.
C C C E C
B
B
E E B B
E
b) c)
a)
Figure 5.5: Transistor basic circuit configurations: a) common emitter (CE), b) common
collector (CC) and c) common base (CB).
The operating characteristics of the different circuit configurations are shown in table 5.1.
5.1.3 Small-Signal Models

A simple transistor model is given by IC = hF E IB . A more general model capable of describing
the family of characteristic curves is given by
IC = IC (IB , VCE ), (5.3)

Characteristic CE CC CB
power gain yes yes yes
voltage gain yes no yes
current gain yes yes no
input resistance 3.5 kΩ 580 kΩ 30 Ω
output resistance 200 kΩ 35Ω 3.1 MΩ
voltage phase change yes no no
Table 5.1: Operating characteristics of different transistor circuit configurations.
where IC is a transistor-dependent function.

For AC analysis only time changes are important and we may write
dIC =∂IC dIB + ∂IC dVCE , (5.4)
dt ∂IB dt ∂VCE dt
where the partial derivatives are evaluated at a particular IB and VCE – the operating point. hf e ≡
∂IC /∂IB |IB,VCE is the forward current transfer ratio and describes the vertical spacing ∆IC /∆IB
between the curves. The output admittance (inverse resistance) is hoe ≡ ∂IC /∂VCE |IB,VCE and
describes the slope ∆IC /∆VCE of one of the curves as it passes through the operating point.
Using these definitions we may write
dIC = hf e dIB + hoe dVCE . (5.5)
dt dt dt
The input signal VBE is also related to IB and VCE , and a similar argument to the above
gives
dVBE = hie dIB + hre dVCE , (5.6)
dt dt dt
where hie is the input impedance and hre is the reverse voltage ratio.
The differential equations are linear only in the limit of small AC signals, where the h
parameters are effectively constant. The h parameters are in general functions of the variables IB
and VCE . We arbitrarily picked IB and VCE as our independent variables. We could have picked
any two of VBE , VCE , IB and IC . Because the current and voltage variables are mixed the h
parameters are known as hybrid parameters.
In general the current and voltage signals will have both DC and AC components. The time
derivatives involve only the AC component and if we restrict ourselves to sinusoidal AC signals,
we may replace the time derivatives by the signals themselves (using complex
notation). Our hybrid equations become
iC = hf eiB + hoevCE and (5.7)
v = h i + h v . (5.8)
BE ie B re CE
The hybrid parameters are often used as the manufacturer’s specification of a transistor, but
there are large variations between samples. Thus one should use the actual measured parameters
in any detailed calculation based on this model. Table 5.2 shows typical values for the hybrid
parameters.
Symbol Name Typical Value

h
fe forward current ratio 102
h
ie input impedance 2×103 Ω
h −5 −1
oe output admittance 2 × 10 Ω
h −4
re reverse voltage ratio 10
Table 5.2: Hybrid parameters.
The relationship between the voltages and currents for a transistor in the common emitter
configuration is shown in figure 5.6.
vC C
vBE = vB - vE
vCE = vC - vE
iC
B vB
iC = hfe iB
+ hie vCE
Rs iB RL
S
VS
G
vE vE
E E
Figure 5.6: Transistor in the common emitter configuration.
We now make a few approximations to our hybrid parameter model to get an intuitive feel
for how transistors behave in circuits. The voltage across the load resistor is
vE − vC = iC RL, (5.9)
vC − vE = −iC RL and (5.10)

v = −iC RL. (5.11)
CE
Substituting this into our first hybrid equation gives

iC = h i . (5.12)
fe B
1 + hoeRL
If hoeRL 1 (good to about 10%) we can write
iC = hf eiB , (5.13)
which is the AC equivalent of IC = βIB = hF E IB . Similarly, using the second hybrid

equation gives
v Rh
CE L fe
v =− h Rh h . (5.14)
BE ie − L f e ie
If RLhf ehre 1 (good to about 10%) we have

v h
CE fe
v = − h RL, (5.15)
BE ie
which is the AC voltage gain.
5.1.4 Ideal and Perfect Bipolar Transistor Models

By ignoring hoe and hre we can define a simplified AC model for the transistor (perfect
transistor) that is independent of the circuit configuration:
iC = hf eiB and (5.16)

v = hieiB . (5.17)
BE
There is no PN voltage drop in these equations since it is a DC effect.

Since iC is typically 100 times larger than iB we can make the approximation iE ≈ iC . An
ideal transistor is defined such that hie = 0 and hence vB = vE . When the effects of hie cannot be
ignored, we can use the perfect transistor model as described by the two equations above. On a
circuit diagram hie can be added directly to the ideal transistor symbol.
The ideal transistor model has the following two working rules:
1. The base and emitter are at the same AC voltage (vB = vE ). They differ only by a constant
DC potential VP N .
2. The collector current is equal to the emitter current and proportional to the base current (iE
= iC , IE = IC , iC = hf eiB and IC = hF E iB ).
5.1.5 Transconductance Model

The transconductance model provides an alternative description of transistor operation. The
forward transconductance, which has units of inverse resistance, is defined as gm = hf e/hie.
Using the perfect transistor model we can write
iC = gm(vB − vE ) = gmvBE . (5.18)

Since the transconductance is used to describe field effect transistors, it is sometimes conve-nient to apply the
same parameter to bipolar transistors.
We now turn to the description of some simple amplifiers that use a single bipolar tran-sistor.
Our goal will be to estimate the voltage gains, current gains, input impedances, output
impedances and corner frequencies of the amplifiers. The characteristics of a perfect amplifier
are as follows:
• infinite input impedance (the amplifier draws no current),

• zero output impedance,
• infinite voltage and current gain, and
• stable, linear, etc.
5.2 The Common Emitter Amplifier

The common emitter configuration is the most versatile of the three. It has low input impedance,
moderate output impedance, voltage gain and current gain. The input and output are often
capacitively coupled. Before performing an AC analysis we will discuss DC biasing.
5.2.1 DC Biasing
DC biasing is setting up a circuit to operate a transistor at a desired operating point on its
characteristic curve. Three bias networks for the common emitter amplifier are shown in figure
5.7.
In figure 5.7a the only path for DC bias current into the base is through RB . VCC is a power
supply voltage which is generally greater than 10 V such that VP N can be ignored. The DC
voltage at the collector should be large enough to provide at least a 2 V drop between collector
and emitter and clearly must be less than VCC . In the absence of other circuit requirements, a
convenient algebraic choice for VC is VCC /2. DC circuit analysis results in the following
relative sizes of the two resistors:
RB = 2hF E RC . (5.19)
Although the circuit works reasonably well, the fact that hF E is quite variable among
samples leads to a bad design. A well-designed circuit should have an operating point that is less
dependent on this parameter.
Figure 5.7b shows a network with the base-biasing resistor connected to the collector instead
of VCC . RF acts as a negative feedback resistor since it feeds the collector current back into the
base. Analysis gives
RF = hF E RC . (5.20)
Therefore a change in hF E has only half the effect of the previous design.
A more common bias stabilization technique employs a series resistor between the emitter
and ground. This circuit has about the same sensitivity to changes in hF E as the previous circuit.
Vcc
Rc Vcc
RB
IB Vc Rc
VB~0 Ic = hFE IB I1 RB1
Vc
a) Ic
VB~VE
Vcc
VE
Rc IR = Ic + IB I2 RB2 RE
RF ~ Ic
IB Vc c)
VB~0 Ic = hFE IB
b)
Figure 5.7: Bias circuits for the common emitter amplifier.
A further improvement can be made by introducing a second base-bias resistor as shown in

figure 5.7c. The bias voltage is determined almost entirely by the two bias resistors. These
biasing methods can also be used for the common collector and common base configurations.
5.2.2 Approximate AC Model

The circuit shown in figure 5.8a is the basic common emitter amplifier using the simplest biasing
method. Because it is constant, the power supply voltage VCC is an AC ground indistinguishable
from the normal ground of the circuit. We can therefore relocate the upper end of RB and RC to
the common ground line as shown in figure 5.8b. The transistor symbol is ideal and hie is shown
explicitly as the input impedance and hence iS = iB .
5.2.3 The Basic CE Amplifier

We can now use the AC equivalent circuit to calculate the AC voltage gain between the base and
collector. The base voltage is developed across the input resistor hie and vB = hieiB . The collector
voltage can be similarity expressed as the voltage drop across the resistor RC : 0 − vC = RC hf eiB .
Eliminating iB , we can write the amplifier voltage transfer function between the base and collector as
v =− h R . (5.21)
C fe C
v
B h
ie
Vcc
Rc
vc
C
RB vc
Rs C vB hfeiB
B hie B'
vB B ic Vcc
Rc
S vs is iB
iB
is
E RB
G
a) b)
Figure 5.8: a) Basic CE amplifier and b) AC equivalent circuit drawn using an ideal transistor
symbol with hie shown explicitly.
o
The minus sign indicates that the voltage signal at the collector is 180 out of phase with the
signal at the base.
The input impedance to this amplifier circuit is just the parallel combination of RB and hie,
and since hie is usually much smaller than RB , the input impedance generally reduces to just the
input impedance of the transistor itself, namely, hie. The circuit output impedance is the collector
resistance RC .
The high-frequency operation of the common emitter amplifier is limited by the parasitic
capacitance between the collector and base. This capacitance provides a path by which the large
and inverted signal at the collector drives a feedback current into the base. The base-to-collector
voltage gain of this amplifier looks like a low-pass filter.
5.2.4 CE Amplifier with Emitter Resistor

The CE amplifier is often constructed with an emitter resistor RE as shown in figure 5.9. This
resistor provides a form of negative feedback that can be used to stabilize both the DC operating
point and the AC gain. It can be shown that the voltage transfer function across the transistor is
vC =−A=− h R . (5.22)
fe C
vB h + h R
ie fe E
If hf eREhie, the gain becomes independent of the hybrid parameters:

R
vC =−A=− C
. (5.23)
vB RE
Because it is unaffected by variations in the hybrid parameters, this result is valid even for a
large-amplitude signal.
The input resistance can be shown to be Rin = hf eRE . Since hie is small, it can usually be
neglected whenever the emitter impedance is more than a few hundred ohms.
vB hie vc
iC
iB vE Rc
RE
Figure 5.9: CE amplifier with emitter resistor.
For small input signals it is often desirable to retain the large voltage gain of the basic CE
amplifier even though an emitter resistor is used for DC stability. This can be done if a large
capacitor CE is used to bypass the AC signal around the emitter resistor. CE shorts out the
emitter bias resistor RE for AC signals. The magnitude of the resulting transfer function is
similar to a high-pass filter, with RE setting the gain (A = RC /RE ) at low frequencies and the
capacitor maximizing the gain (A = hf eRC /hie) at high frequencies.
Example: Determine values for RB and RC that will yield the operating point
shown in the circuit in figure 5.10 (VCC = +20 V ) under the assumption that:
1. VBE = 0.
VBE=0⇒VB=0.
V I
IB= CC= C.
R h
B FE
Therefore
R = h V =100 × 20 = 2 × 106 Ω
CC
B FE −3 (5.24)
IC 1 × 10
= 2M Ω. (5.25)
RC = VCC − VC = 20−10 = 104Ω

−3
(5.26)
IC 10
= 10kΩ. (5.27)
2. VBE = 0.6 V.
VBE = 0.6V ⇒ VB = 0.6V .
V −V I
I = CC B = C .
B
RB hF E
Therefore
RB = h V V = 100 × (20 − 0.6)
FE CC − B (5.28)
−3
IC 1 × 10
= 1.94M Ω. (5.29)
RC = VCC − VC (5.30)
IC
= 10kΩ. (5.31)
3. Repeat both calculations for hF E = 80.

For hF E = 80, RC = 10kΩ remains unchanged.
80×20
VBE = 0 ⇒ RB = = 1.6MΩ .
−3
1×10
⇒ 80(20−0.6)
V = 0.6V = = 1.55M Ω .
BE 1×10−3
Vcc=+20V
Ic=1mA
RC
RB
Vc=10V
hFE=100
Figure 5.10: Example transistor circuit with a single bias transistor.
Example:
1. If hF E = 100, determine an expression for the base-biasing resistor R B
that will result in a DC operating point V C = VCC /2 for the circuit in figure
5.11. Now hF E = 100. The voltage drop across the bias resistor is V CC =
RB I1 + 100RI2. Charge conservation (I1 = I2 + IB ) gives
VCC = RB(I2 + IB ) + 100RI2 (5.32)

= RBIB + (RB + 100R)I2. (5.33)
The voltage at the base is VB = 100RI2. Neglecting the VP N voltage drop

gives
VB ≈ VE = R(IC + IB). (5.34)
IC+IB IC
EliminOAating VB gives I2 = 100 . Using IC = hF E IB ⇒ IB = h gives
FE
I2 = 1 + 1/hF E IC .
(5.35)
100
Substituting equation 5.35 into equation 5.33 gives
VCC = RB + (RB + 100R) 1 + 1/hF E IC . (5.36)
h 100
FE
The voltage drop across the collector resistor is

VCC − VC = 5RIC (5.37)
VCC − VCC /2 = VCC /2 = 5RIC . (5.38)

Therefore
10R = RB + (RB + 100R) 1 + 1/hF E (5.39)
h
FE 100
= 0.01RB + 0.0101RB + 1.01R (5.40)
RB = 8.99 R = 447R (5.41)
0.0201
= 447R. (5.42)
2. If the circuit is built with the R B just found, what will be the operating
point if hF E = 50?
VCC−VC .
If RB = 447R and hF E = 50, VCC − VC = 5RIC ⇒ IC = 5R
Substitution into equation 5.36 gives
5RVCC = RB + (RB + 100R) 1 + 1/hF E (5.43)
h 100
VCC − VC FE
5 447 1 + 1/50
= + (447 + 100) (5.44)
1 − VC /VCC 50 100
= 14.5194 (5.45)
V = 1− 5 V (5.46)
C CC
14.5194
= 0.66VCC . (5.47)
Vcc
Ic
5R
I1
RB
Vc=Vcc/2
IB
I2
100R
R
Figure 5.11: Example transistor circuit with two bias resistors.

5.3 The Common Collector Amplifier

The common collector amplifier is also called the emitter follower amplifier because the output
voltage signal at the emitter is approximately equal to the voltage signal input on the base. The
amplifier’s voltage gain is always less than unity, but it has a large current gain and is normally
used to match a high-impedance source to a low-impedance load: the amplifier has a large input
impedance and a small output impedance.
A typical common collector amplifier is shown in figure 5.12.
Vcc
vB
iB
vE
RE
Figure 5.12: Basic common collector amplifier.
The voltage gain can be written as

vE = h R . (5.48)
fe E
vB h + h R
ie fe E
The gain is thus in phase and slightly less than unity. The output impedance of the CC amplifier
can be substantially less than the output impedance of the driving signal source.
5.4 The Common Base Amplifier

The common base amplifier is also known as the grounded base amplifier. This amplifier can
produce a voltage gain but generates no current gain between the input and the output sig-nals. It
is normally characterized by a very small input impedance and an output impedance like the
common emitter amplifier. Because the input and output currents are of similar size, the stray
capacitance of the transistor is of less significance than for the common emitter amplifier. The
common base amplifier is often used at high frequencies where it provides more voltage
amplification than the other one-transistor circuits.
A common base circuit is shown in figure 5.13. Above the corner frequency the capacitor
between base and ground on the circuit provides an effective AC ground at the transistor’s base.
The voltage gain is given by

v =h R , (5.49)
C fe C
vE h
ie
which is the same as the CE amplifier except for the lack of voltage inversion.
ic
vE vc
Rc
iB
RE RB1 Vcc
RB2 CB
Figure 5.13: Basic common base amplifier.
The input impedance looking into the emitter (Rin = hie/hf e) is quite small. The output
impedance of this circuit is never greater than RC .
Because of its high-frequency response and small input impedance, this circuit is often used
to receive high-frequency signals transmitted via a coaxial cable. For this purpose the input
impedance of the amplifier is adjusted to match the distributed impedance of the coaxial cable –
usually 50 to 75 Ω.
5.5 The Junction Field Effect Transistor (JFET)

Bipolar junction transistors have low input impedance, small high-frequency gain and are non-
linear when |VCE | < 2 V. The input impedance is naturally restricted by the forward-biased
base-emitter junction. There are always problems due to the main charge carriers passing
through the region where the majority carriers are of opposite polarity.
The junction field effect transistor (JFET) overcomes some of the problems of the bipolar
junction transistor. JFETs come in two types: N-channel and P-channel, and are shown in figure
5.14.
The designation refers to the polarity of the majority charge carriers in the bar of semi-
conductor that connects the drain terminal D to the source terminal S. Since the channel is
formed from a single-polarity (unipolar) material, its resistance is a function only of the ge-
ometry of the conducting volume and the conductivity of the material. The JFET operates with
all PN junctions reverse-biased so as to obtain a high input impedance into the gate.
5.5.1 Principles of Operation

Figure 5.15 shows an N-channel JFET with DC bias voltage applied. Just as for a simple diode,
the depletion region grows as the reverse bias across the PN junction is increased, thereby
constricting the cross section of the conducting N-channel material and increasing the resistance
of the channel. The major current ID in the channel is caused by the applied voltage between the
drain and source, VDS , and is controlled by the applied voltage between the gate and source, VGS .
D D
P
N D D
G G
P P G
N N G
S S
S S
N−Channel P−Channel
Figure 5.14: Basic geometry and circuit symbols of JFETS.
The JFET has two distinct modes of operation: the variable-resistance mode, and the pinch-
off mode. In the variable-resistance mode the JFET behaves like a resistor whose value is
controlled by VGS . In the pinch-off mode, the channel has been heavily constricted with most of
the drain-source voltage drop occuring along the narrow and therefore high-resistance part of the
channel near the depletion regions.
The characteristic curves of a typical JFET are shown in figure 5.16. At small values of VDS
(in the range of a few tenths of a volt), the curves of constant VGS show a linear relationship
between VDS and ID . This is the variable-resistance region of the graph. As VDS increases, each
of the curves of constant VGS enters a region of nearly constant ID . This is the pinch-off region,
where the JFET can be used as a linear voltage and current amplifier. At VGS = 0 the current
through the JFET reaches a maximum known as IDSS , the current from Drain to Source with the
gate Shorted to the source. If VGS goes positive for this N-channel JFET, the PN junction
becomes conducting and the JFET becomes just a forward-biased diode.
5.5.2 Small-Signal AC Model

The JFET characteristic curves can be described by an equation of the form
ID = ID(VGS , VDS ), (5.50)
where the function varies with the particular transistor. This expression yields the AC relationship
I
D
D
+
G P V
DS
P
V
GS
+ N
ID
Figure 5.15: An N-channel JFET with DC bias voltages applied.
iD = ∂ID vGS + ∂ID vDS , (5.51)
∂VGS ∂VDS
where the AC currents and voltages are complex but the partial derivatives evaluate to real
numbers. In the pinch-off region the curves of constant VGS are essential flat (∂ID /∂VDS = 0)
and allow the equation to be written as
iD = gmvGS , (5.52)
where gm = ∂ID /∂VGS is the transconductance.
5.6 JFET Common Source Amplifier

The common source configuration for a FET is similar to the common emitter bipolar tran-sistor configuration, and
is shown in figure 5.17. The common source amplifier can provide both a voltage and current gain. Since the input
resistance looking into the gate is extremely large the current gain available from the FET amplifier can be quite
large, but the voltage gain is generally inferior to that available from a bipolar device. Thus FET amplifiers are most
useful with high output-impedance signal sources where a large current gain is the pri-mary requirement. The
source by-pass capacitor provides a low impedance path to ground
ID
IDSS VGS1=0V
VGS2=−0.5V
VGS3=−1.0V
V
DS
Figure 5.16: Characteristic curves of a typical N-channel JFET.
for high frequency components of vDS and hence AC signals will not cause a swing in the bias
voltage.
VDD
RD
vG vD
RG
Rs
Figure 5.17: JFET common source amplifier.
Since the FET gate current is small we can make the approximations iS = iD and vS = −vGS :
the source is positive with respect to the gate for reverse-bias. Since at low frequencies we can
ignore the capacitor the source voltage is given by
vS = RS iS = RS iD . (5.53)
Using the transconductance equation we can write

vS = gmRS vG . (5.54)
1 + gmRS
With the approximation iS = iD we can write the drain voltage as

vD = − vS RD
. (5.55)
RS
The voltage gain is thus
vD = −A=− gmRD .
(5.56)
vG 1 + gmRS
If gmRS1, this reduces to
vD =−A=− R .
D (5.57)
v R
G S
5.7 JFET Common Drain Amplifier

The common drain FET amplifier is similar to the common collector configuration of the bipolar
transistor. A general common drain JFET amplifier, self-biased, is shown in fig-ure 5.18. This
configuration, which is sometimes known as a source follower, is characterized by a voltage gain
of less than unity, and features a large current gain as a result of having a very large input
impedance and a small output impedance.
VDD
vG
vs
RG
Rs
Figure 5.18: JFET common drain amplifier.
Using AC circuit analysis
iD = gmvGS and (5.58)

vS = gm(vG − vS ). (5.59)
RS
The voltage gain between the gate and source is
vS gmRS
= . (5.60)
If gmRS 1, vS = vG and we have a voltage follower.
5.8 The Insulated-Gate Field Effect Transistor

The insulated-gate FET, also known as a metal oxide semiconductor field effect transistor
(MOSFET), is similar to the JFET but exhibits an even larger resistive input impedance due to
the thin layer of silicon dioxide that is used to insulate the gate from the semiconductor channel.
This insulating layer forms a capacitive coupling between the gate and the body of the transistor.
The consequent lack of an internal DC connection to the gate makes the device more versatile
than the JFET, but it also means that the insulating material of the capacitor can be easily
damaged by the internal discharge of static charge developed during normal handling.
The MOSFET is widely used in large-scale digital integrated circuits where its high input
impedance can result in very low power consumption per component. Many of these circuits
feature bipolar transistor connections to the external terminals, thereby making the devices less
susceptible to damage.
The MOSFET comes in four basic types, N-channel, P-channel, depletion and enhance-ment.
The configuration of an N-channel, depletion MOSFET is shown in figure 5.19a. Its operation is
similar to the N-channel JFET discussed previously: a negative voltage placed on the gate
generates a charge depleted region in the N-type material next to the gate, thereby reducing the
area of the conduction channel between the drain and source. How-ever, the mechanism by
which the depletion region is formed is different from the JFET. As the gate is made negative
with respect to the source, more positive carriers from the P-type material are drawn into the N-
channel, where they combine with and eliminate the free negative charges. This action enlarges
the depletion region towards the gate, reducing the area of the N-channel and thereby lowering
the conductivity between the drain and source. For negative applied gate-source voltages the
observed effect is much like a JFET, and gm is also about the same size.
However, since the MOSFET gate is insulated from the channel, positive gate-source
voltages may also be applied without losing the FET effect. Depending on the construction
details, the application of a positive gate-source voltage to a depletion-type MOSFET can repel
the minority positive carriers in the depleted portion of the N-channel back into the P-type
material as discussed below, thereby enlarging the channel and reducing the resistance. If the
device exhibits this behaviour, it is known as an enhancement-depletion MOSFET.
A strictly enhancement MOSFET results from the configuration shown in figure 5.19b.
Below some threshold of positive gate-source voltage, the connecting channel of N-type ma-
terial between the drain and source is completely blocked by the depletion region generated by
the PN junction. As the gate-source voltage is made more positive, the minority positive carriers
are repelled back into the P-type material, leaving free negative charges behind. The effect is to
shrink the depletion region and increase the conductivity between the drain and source.
5.9 Power MOSFET Circuits

Traditionally, MOSFET devices have had the drain-to-source current confined to a thin pla-nar volume of silicon
lying parallel to the gate. The limited cross-sectional area of material
D D
N
N
U G P U
G
N P N
N
N
S S
G D U D
G U
a) S b) S
Figure 5.19: a) Depletion or depletion-enhancement type MOSFET and b) enhancement type

MOSFET.
thus available for conduction effectively limits the power-handling capability of MOSFET
devices to less than 1 W. More recently, new designs and manufacturing techniques have been
developed to produce a more complicated, three-dimensional gate structure. These transistors are
identified by various manufactures as HEXFET, VMOS, or DMOS, depend-ing on the geometry
of the gate structure: respectively hexagonal, V-shaped, or D-shaped. They feature power
dissipations exceeding 100 W and excellent high-frequency operation. In contrast to the normal
MOSFET, these devices have a much larger forward transconduc-tance. These devices thus
feature very high current gain at both high frequency and high power, a combination that is hard
to obtain with traditional bipolar power transistors.
5.10 Multiple Transistor Circuits

Integrated circuit technology has largely eliminated the need to fabricate multiple transistor circuits from discrete
transistors.
5.10.1 Coupling Between Single Transistor Stages

Quite often the single-transistor amplifier discussed in the previous lectures does not provide
enough gain for an application, or more often, it does not combine gain with the desired input
and output impedance characteristics. Perhaps the most obvious solution (but generally not the
best) is to connect several single-transistor amplifiers, or stages, in tandem one after the other.
Because of DC biasing considerations, it is usually not practical to connect the output of one
stage directly to the input of another; some kind of coupling device must be used that permits a
change in the DC level between two stages.
A possible circuit features capacitor coupling between single-transistor stages. The inter-
stage coupling capacitor allows the following transistor to be DC biased without concern for the
DC component of the driving signal. Thus the DC analysis of the two transistor stages is
separated into two distinct problems.
5.10.2 Darlington and Sziklai Connections

Two bipolar transistors in either the Darlington or Sziklai connection can be used as a single,
high gain, transistor. These circuits have AC current gains on the order of h2f e, providing high
amplification in a single stage.
The arrangement of two NPN transistors shown in figure 5.20a is known as the Darlington
connection. The three unconnected terminals, the base of Q1 and the collector and emitter of Q2,
behave like a single transistor with current gain hf e1hf e2 and an overall base to emitter DC
voltage drop of 2VP N . Two transistors in the Darlington connection can be purchased in a
single three-terminal package.
A similar circuit result can be obtained with the Sziklai connection of figure 5.20b. Because
it uses one transistor of each polarity (NPN and PNP), this connection is also known as the
complementary Darlington. The combination again results in a three-terminal device that
behaves like a single high-current-gain transistor. The overall circuit behaves like a transistor of
the same polarity as Q1.
C
C
B Q1 Q2
iB hfe iB 2 hfe iB 2
hfe iB B Q1 hfe iB
Q2
a) b)
E E
Figure 5.20: a) The Darlington connection of two transistors to obtain higher current gain and b) the Sziklai
connection of an NPN and PNP transistor to obtain the equivalent of a high-current-gain NPN.
5.11 Problems
1. Consider the common-emitter amplifier shown below with values VCC = 12 V, R1 = 47
kΩ, R2 = 12 kΩ, RC = 2.7 kΩ, RE = 1 kΩ, and a forward-bias voltage drop across the
base-emitter junction of 0.7 V. Calculate approximate values for VB , VE , IC , VC , and the
voltage gain. Using the measured h parameters hie = 3600 Ω and hf e = 150, calculate the
input impedance, base current, and voltage gain. Compare your calculated voltage gain
with your approximation.
V
CC
R1 RC
V
B VC
IC
VE
R
2 R
E
Chapter 6
Operational Amplifiers
The operational amplifier (op-amp) was designed to perform mathematical operations. Al-
though now superseded by the digital computer, op-amps are a common feature of modern
analog electronics.
The op-amp is constructed from several transistor stages, which commonly include a
differential-input stage, an intermediate-gain stage and a push-pull output stage. The dif-ferential
amplifier consists of a matched pair of bipolar transistors or FETs. The push-pull amplifier
transmits a large current to the load and hence has a small output impedance.
The op-amp is a linear amplifier with Vout ∝ Vin. The DC open-loop voltage gain of a typical
op-amp is 103 to 106. The gain is so large that most often feedback is used to obtain a specific
transfer function and control the stability.
Cheap IC versions of operational amplifiers are readily available, making their use popular in
any analog circuit. The cheap models operate from DC to about 20 kHz, while the high-
performance models operate up to 50 MHz. A popular device is the 741 op-amp which drops off
6 dB/octave above 5 Hz. Op-amps are usually available as an IC in an 8-pin dual, in-line
package (DIP). Some op-amp ICs have more than one op-amp on the same chip.
Before proceeding we define a few terms:
linear amplifier – the output is directly proportional to the amplitude of input signal.
open-loop gain, A – the voltage gain without feedback (≈ 105).

closed-loop gain, G – the voltage gain with negative feedback (approximation to H(jω)).
negative feedback – the output is connected to the inverting input forming a feedback loop
(usually through a feedback resistor RF ).
6.1 Open-Loop Amplifiers

Figure 6.1a shows a complete diagram of an operational amplifier. A more common version of
the diagram is shown in figure 6.1b, where missing parts are assumed to exist. The inverting
input means that the output signal will be 1800 out of phase with the input applied to this
terminal. On the diagram V++ ≡ +VCC = +15 V (DC) and V−− ≡ −VCC = −15 V (DC).
105
CHAPTER 6. OPERATIONAL AMPLIFIERS 106
VCC is typically, but not necessarily, ±15 V. The positive and negative voltages are necessary to
allow the amplification of both positive and negative signals without special biasing.
ground or
V++ common
+
non-inverting
V+
input
output Vout
inverting V-
input
V-- b)
+
a)
Figure 6.1: a) Complete diagram of an operational amplifier and b) common diagram of an

operational amplifier.
For a linear amplifier (cf. a differential amplifier) the open-loop gain is
vout = A(jω)(v+ − v−). (6.1)
The open-loop gain can be approximated by the transfer function

A(jω) = A0Hlow(jω), (6.2)
where A0 is the DC open-loop gain and Hlow is the transfer function of a passive low-pass filter.
We can write
A(jω) = A , (6.3)
0
1 + jω/ωc
5
where A0 ≈ 2 × 10 and fc ≈ 5 Hz.
Two conditions must be satisfied for linear operation:
1. The input voltage must operate within the bias voltages:

−VCC /A0 ≤ (v+ − v−) ≤ +VCC /A0.
2. For no clipping the output voltage swing must be restricted to
−VCC ≤ vout ≤ +VCC .
6.2 Ideal Amplifier Approximation

The following are properties of an ideal amplifier, which to a good approximation are obeyed by
an operational amplifier:
1. large forward transfer function,
2. virtually nonexistent reverse transfer function,
3. large input impedance, Zin → ∞ (any signal can be supplied to the op-amp without loading
problems),
4. small output impedance, Zout → 0 (the power supplied by the op-amp is not limited),
5. wide bandwidth, and
6. infinite gain, A → ∞.
If these approximations are followed two rules can be used to analyze op-amp circuits:
Rule 1: The input currents I+ and I− are zero, I+ = I− = 0 (Zin = ∞).
Rule 2: The voltages V+ and V− are equal, V+ = V− (A = ∞).
To apply these rules requires negative feedback.

Feedback is used to control and stabilize the amplifier gain. The open-loop gain is too large
to be useful since noise will causes the circuit to clip. Stabilization is obtained by feeding the
output back into the input (closed negative feedback loop). In this way the closed-loop gain does
not depend on the amplifier characteristics.
6.2.1 Non-inverting Amplifiers

Figure 6.2 shows a non-inverting amplifier, sometimes referred to as a voltage follower.
Vin
Vout
Figure 6.2: Non-inverting, unity-gain amplifier.

Applying our rules to this circuit we have
V+ = V− ⇒ Vin = Vout. (6.4)
I+ = I− = 0 ⇒ Rin = ∞. (6.5)
The amplifier gives a unit closed-loop gain, G(jω) = 1, and does not change the sign of the input
signal (no phase change).
This configuration is often used to buffer the input to an amplifier since the input resis-tance
is high, there is unit gain and no inversion. The buffer amplifier is also used to isolate a signal
source from a load.
Often a feedback resistor is used as shown in figure 6.3.
Vin
Vout
I=0
RF
RI
Figure 6.3: Non-inverting amplifier with feedback.
For this circuit

V+ = V− ⇒ Vin = RI V .
out (6.6)
RI +RF
The gain is
V R + R R
out = I F =1+ F and (6.7)
V R R
in I I
RF
G(jω) = 1+ . (6.8)
RI
(6.9)
6.2.2 Inverting Amplifiers

An inverting amplifier is shown in figure 6.4. Analysis of the circuit gives
V V V V
in − − = − − out . (6.10)
RI RF
Since V+ = V− = 0 (V− is at virtual ground),
V V
in = − out . (6.11)
RI RF
The gain is
RF
Vin
RI
Vout
Figure 6.4: Inverting amplifier.
V = − RF and (6.12)
out
V R
in I
R
G(jω) = − F . (6.13)
RI
(6.14)
The output is inverted with respect to the input signal.
A sketch of the frequency response of the inverting and non-inverting amplifiers are shown
in figure 6.5.
OPEN−LOOP
H GAIN A(jω)
A0
APPROXIMATE CLOSE−LOOP
GAIN G(jω)
G
ACTUAL CLOSE−LOOP
GAIN H(jω)
ω
OPEN−LOOP ωc ωΙ
BANDWIDTH
Figure 6.5: Inverting and non-inverting amplifier frequency response.
The input impedance of the inverting amplifier is Rin = Vin/I. Since Vin − I RI = 0 we have
Rin = RI .
A better circuit for approximating an ideal inverting amplifier is shown in figure 6.6 The extra resistor is a
current bias-compensation resistor. It reduces the current bias by eliminating non-zero current at the inputs.
RF
RI
RB
Figure 6.6: Inverting amplifier with bias compensation.
6.2.3 Mathematical Operations

Current Summing Amplifier
Consider the current-to-voltage converter shown in figure 6.7. Applying our ideal amplifier rules
gives
RF
Vin
Vout
Figure 6.7: Current-to-voltage converter.
V+ = V− = 0 ⇒ 0 − Vout = I RF . (6.15)
Therefore Vout = −I RF and the circuit acts as a current-to-voltage converter.

Figure 6.8 shows several current sources driving the negative input of an inverting am-
plifier. Summing the current into the node gives
V1
R1
V2
R2 RF
V3 Vout
R3
Figure 6.8: Current summing amplifier.

V1 V V V
2 3 out
+ + =− . (6.16)
R1 R2 R3 RF
Therefore
Vout = − R V1 − RF V2 − RF V3. (6.17)
F
R1 R2 R3
If R1 = R2 = R3(≡ R), we have
R
Vout = − F
R (V1 + V2 + V3), (6.18)

and the output voltage is proportional to the sum of the input voltages.
For only one input and a constant reference voltage
Vout = − RF V RF
in −
V , (6.19)
RI RR ref
where the second term represents an offset voltage. This provides a convenient method for
obtaining an output signal with any required voltage offset.
Differentiation Circuit
To obtain a differentiation circuit we replace the input resistor of the inverting amplifier with a
capacitor as shown in figure 6.9.
C R
Vin
Vout
Figure 6.9: Differentiation circuit.
Replacing RI with ZC = 1/(jωC) in the voltage gain gives

Vout R
G(jω) = =− = −jωRC, (6.20)
V Z
in C
or
Vout = −jωRCVin = −RC dVin .
(6.21)
dt
Using dV /dt = I /C gives
d(Vin − 0) =1 0 − Vout
(6.22)
dt CR
and thus the same result as above.
The frequency response is shown in figure 6.10.
H
G=RCω
A0
RC
ω
ωc
ωI
Figure 6.10: Differentiation circuit frequency response.
Integration Circuit
Integration is obtained by reversing the resistor and the capacitor as shown in figure 6.11.
The capacitor is now in the feedback loop.
C
R
Vin
Vout
Figure 6.11: Integration circuit.

Analysis gives
G(jω) = V = Z = −1 (6.23)
out C
V jωRC
in − R
or
V = V = −1 Vindt (6.24)
out − in
.
jωRC RC
Using dV /dt = I /C gives
d(0 − Vout) = 1 Vin − 0 . (6.25)
dt C R
and thus the same result as above.
A0
1
G=−−−
RCω
ωI ωc ω
Figure 6.12: Integration circuit frequency response.
The frequency response is shown in figure 6.12.

We can combine the above inverting, summing, offset, differentiation and integration circuits
to build an analog computer that can solve differential equations. However, today, the
differentiators and integrators are mainly used to condition signals.
6.2.4 Active Filters

Filters often contain embedded amplifiers between passive-filter stages as shown in fig-ure 6.13.
Figure 6.13: Buffer amplifier as part of active filter.
These filters have a limited performance since the poles are still real and hence the knees are
not sharp. For example, a three stage high-pass filter with buffer amplifiers has a transfer
function
G(jω) = jω/ωcjω/ωcjω/ωc = −j(ω/ωc)3 , (6.26)
1 + jω/ωc 1 + jω/ωc 1 + jω/ωc (1 + jω/ωc)3
and only drops 18 db/octave.

For complex poles we must use either integrators or differentiators. Consider figure 6.14.
RI
Vin
Vout
Figure 6.14: Active filter with complex poles.
The closed-loop gain is

G = −ZRC = − R/jωC = −R . (6.27)
R RI (R + 1/jωC) RI (1 + jωRC)
I
By exchanging the input resistor for a capacitor we can change between a low-pass and high-
pass filter.
6.2.5 General Feedback Elements

The feedback elements in an operation amplifier design can be more complicated than a simple
resistor and capacitor. An interesting feedback element is the analog multiplier as defined in
figure 6.15.
Vx(t) Vz(t)=Vx(t)Vy(t)
Vy(t) MULT
Figure 6.15: Five-terminal network that performs the multiplication operation on two voltage
signals.
The multiplier circuit itself can be thought of as another op-amp with a feedback resistor whose
value is determined by a second input voltage. Multiplication circuits with the ability to handle
input voltages of either sign (four-quadrant multipliers) are available as integrated circuits and
have a number of direct uses as multipliers. But when used in a feedback loop around an
operational amplifier, other useful functional forms result.
The circuit of figure 6.16 gives an output that is the ratio of two signals, whereas the circuit of figure 6.17
yields the analog square-root of the input voltages.
V1(t) V1(t)
V3=-----
V2(t) MULT V2(t)
Figure 6.16: A multiplier as part of the feedback loop that results in the division operation.
v2(t)=sqrt(v1(t))
V1(t)
MULT
Figure 6.17: A multiplier as part of the feedback loop that results in the square-root opera-tion.
6.2.6 Differential Amplifiers

The differential amplifier amplifies the difference between two input signals (−) and (+). This
amplifier is also referred to as a differential-input single-ended output amplifier. It is a precision
voltage difference amplifier, and forms the central basis of more sophisticated instrumentation
amplifier circuits.
A differential amplifier is shown in figure 6.18. The voltage V3 is given by (cf. voltage
divider)
R1 R2
V1
V3
V3 Vout
V2
R1 R2
Figure 6.18: Differential amplifier.

V3 = R 2 V2 , (6.28)
R1+R2
and thus
V1−V3 =V V (6.29)
3 − out
R1 R2
leads to
V = R2 (V2 − V1). (6.30)
out
R1
The differential amplifier is usually limited in its performance by the low input impedance of
2R1. Two buffer amplifiers are commonly added to remove this limitation and form the simple
instrumentation amplifier in figure 6.19.
V1
Vout
V2
Figure 6.19: Instrumentation amplifier.
Example: If the open-loop gain curve in figure 6.20 describes the amplifier
shown, write an expression for Vout when Vin = 10 cos(1000t) mV.
By definition
Vout = A(jω)(V+ − V−). (6.31)
Since V− is at ground and V+ = Vin, we have Vout = A(jω)Vin.

A
A(jω) = 0 , (6.32)
1 + jω/ωC
5
where A0 = 2 × 10 and ωC = 25 rad/s (4 Hz).
Since Vin = 10 cos(1000t) mV ⇒ ω = 1000 rad/s.
Therefore
A= 2×105 = 2×105 = 2×105 (1 − 40j), (6.33)
1 + j1000/25 1 + 40j 1+402

A = 2×105 2×10
5 = 1/2 104, (6.34)
| | √ ≈ 40 ×
1 + 402
−1 −40 = −88.6o(−1.55 rad), (6.35)
φ = tan
1
|A(jω)|
5
A0=2X10
5
10 GAIN 0o Vin
Vout
A
o
PHASE −90
o
−180
ωc=25rad/s (4Hz)
ω=2πf 6 (Hz)
1 10
Figure 6.20: Open loop gain curve and amplifier.
and
V = (0.5 × 104) × 10 cos(1000t) × 10−3e−j1.55 (6.36)

out
(6.37)
= 50ej1000te−j1.55
= 50 cos(1000t − 1.55) (6.38)
≈ 50 cos(1000t) V. (6.39)
Example: The variable resistor in the circuit in figure 6.21 provides a DC

offset to the output voltage; a circuit of this type is often used to provide a
zero output for some specific but nonzero input signal. What range of input
voltage V0 can be zeroed out by this circuit?
The current flowing from the input to the output is
I = Vin − V− or I = V V . (6.40)
−− out
10 kΩ 10 kΩ
Therefore
Vin − V− = V V (6.41)
− − out
V = (6.42)
out 2V− − Vin
≈ 2V+ − Vin. (6.43)
10kΩ 15 = 15 .
V+ ranges from 0 to 10kΩ+100kΩ 11
Therefore Vout ranges from Vout = −Vin to Vout = 3011 − Vin.
For Vout = 0 ⇒ Vin = 0 to 0 = 30/11 − Vin or Vin = 30/11 = 2.7 V.
10kohm
10kohm
Vin
100kohm Vout
10kohm
+15V
Figure 6.21: Circuit with DC offset.
The range of input singals that can be zeroed is
0 ≤ Vin ≤ 2.7 V. (6.44)
Example: A two-input current summing amplifier can be used to shift the DC

level of an AC signal. For the circuit shown below, determine the average
value of the output signal if the input is Vin = 2 sin(6000t) V.
20kohm
100kohm
-15V
Vin
10kohm Vout
Figure 6.22: Two-input current summing amplifier.
The current sum into the node is

15V V V V V V
− − − + −in − = −− out . (6.45)
100kΩ 10kΩ 20kΩ
since V− is at virtual ground (V+ = 0).
15V Vin
(6.46)
Vout = 20kΩ −
(6.47)
= 3V − 2Vin.
V = 3V − 2 Vin . (6.48)
out
since Vin = 2 sin(6000t)V ⇒ Vin = 0.

Therefore Vout = 3V .
Example: Assuming ideal operational amplifiers, determine G(jω) for each of

the circuits below. Then determine the single-term approximations to the
transfer function at various frequencies, sketch the resulting straight-line
approximation to |G|, and label all magnitudes, corners and slopes.
C
V1
R2 V2
R1
V1 V2 C R
b)
a)
V1
V2 V1 V2
R2
C
R R1 C
d)
c)
Figure 6.23: Amplifiers with a feedback capacitor.
a) Since V− is at virtual ground

V1−0 =0−V2 , (6.49)
R1 ZF
where R2
ZF = R2/jωC = (6.50)
R2 + 1/jωC 1 + jωR2C
V2
G(jω) = = −Z F
(6.51)
V1 R1
= − R2 1 . (6.52)
R1 1 + jωR2C
ω → 0 ⇒ G = −R2/R1; |G| = R2/R1.

ω → ∞ ⇒ G = −1/jR1Cω; |G| = (1/R1C)ω−1.
R
2 1 1
R1 = R1CωC R2C .
⇒ωC=
Figure 6.24: Straight-line approximation to |G| for amplifier a).
b)
V2−V− = V−−0
(6.53)
R ZC
V2−V+ ≈ V+ (6.54)
R ZC
V2−V1 = V1 . (6.55)
R Z
C
Therefore V2 = V1(R/ZC + 1) = V1(1 + jωCR)

V2
G(jω) = = 1 + jωRC. (6.56)
V
1
ω →0⇒G=1;|G|=1.
ω → ∞ ⇒ G = jωRC; |G| = (RC)ω+1.
RCωC = 1 ⇒ ωC = 1/(RC).
Figure 6.25: Straight-line approximation to |G| for amplifier b).
c)
V2−V− = V−−0 (6.57)
ZC R
V2−V+ ≈ V+ (6.58)
Z
C R
V2−V1 = V1 . (6.59)
Z R
C
Therefore
Z
V2 = C V1+V1 = 1 + 1 V1 (6.60)
R jωRC
G(jω) = 1 + 1 . (6.61)
jωRC
−1
ω → 0 ⇒ G = 1/(jωRC); |G| = 1/(RC)ω .
ω →∞⇒G=1;|G|=1.
1 = 1/(ωC RC) ⇒ ωC = 1/(RC).
Figure 6.26: Straight-line approximation to |G| for amplifier c).
d)
V2−V− = V−−0
R (6.62)
ZC +R2 1
V2−V1 = V1 (6.63)
1/jωC + R2 R1
Therefore
V2 = 1/jωC + R2 V1+V1 (6.64)
R1
G(jω) = 1 + R2 + 1 (6.65)
R1 jωR1C
−1
ω → 0 ⇒ G = 1/(jωR1C); |G| = 1/(R1C)ω .
ω →∞⇒G=1+R2/R1; |G|=1+R2/R1.
1 1
= 1 +R →ωC= 2
.
R1CωC R1 CR1+CR2
6.3 Analysis Using Finite Open-Loop Gain

The infinite gain approximation is very useful but a more complete description is required if we are to understand
the limitations of the op-amp. Real op-amps have a large but finite
Figure 6.27: Straight-line approximation to |G| for amplifier d).
input impedance, small but non-zero output impedance and large but finite open-loop gain. They
also have voltage and current asymmetries at the inputs. We will analyze some circuits using an
finite open-loop gain and consider output impedance, input impedance, and voltage and current
offsets.
6.3.1 Output Impedance

An op-amp will in general have a small resistive output impedance from the push-pull output
stage. We will model the open-loop output impedance by adding a series resistor R0 to the
output of an ideal op-amp as shown in figure 6.28.
V1 R0 Vout
Vin
Figure 6.28: Real, current-limiting operational amplifier partially modeled by an ideal am-plifier
and an output resistor.
Assuming no current into the input terminals (unloaded), and hence no current through
R0, we have V1 = Vout = V(open). Using the open-loop transfer function V1 = A(jω)(Vin− Vout)
we obtain
A(jω)
V(open) = Vin. (6.66)
1 + A(jω)
Shorting a wire across the output gives Vout = 0 and hence
I(short) = V1 (6.67)
R0
= A(jω)
Vin. (6.68)
R0
Using the standard definition for the impedance gives
Z = V(open) = R0 . (6.69)
out
I(short) 1 + A(jω)
If A ≈ A0 1 than Zout ≈ R0/A0, which is small as required by our infinite open-loop gain
approximation.
We can now draw the impedance outside the feedback loop and use
A(jω) = A = A (6.70)
0 0
1 + jω/ωc 1+
s/ωc
to obtain
Z R0ωc + R0s
out
= . (6.71)
ωc(1 + A0) + s
The circuit can now be modeled by a resistor R0/A0 in series with an inductor R0/(A0ωc) all
in parallel with another resistor R0 (three passive components) as shown in figure 6.29. Students
should convince themselves of this.
A
V=---Vin R0/A0 R0/(A0omegaC)
1+A Vout
Vin Zout
R0
C
Figure 6.29: An equivalent circuit for a 741-type operational amplifier.
If the op-amp is used to drive a capacitive load, the inductive component in the output impedance could set
up an LCR resonant circuit which would result in a slight peaking of the transfer function near the corner frequency
as shown in figure 6.30
H(jω)
−12 db/octave
ω
ω
c
Figure 6.30: The overall transfer function when the amplifier drives a capacitive load.
6.3.2 Input Impedance

When calculating the output impedance we still assumed an infinite input impedance. In this
section we will calculate the finite input impedance assuming a zero output impedance. We
consider a model that assumes an internal resistor RT connecting the inverting and non-inverting
input terminals of the op-amp as shown in figure 6.31
Consider an inverting amplifier and remove the input resistor so that the input impedance can
be calculated directly at the amplifier’s input terminals.
I2
I1 RF
IT -
V1
RT
Vout
Figure 6.31: Model for calculating the input impedance of the inverting amplifier.
The input impedance is defined by

Zin = V1 (6.72)
I1
and the current at the summing junction is
I1 = V + I2. (6.73)
1
R
T
The current through the feedback resistor is
I2 = V V (6.74)
1 − out
R
F
and the output voltage is related to V1 by the open-loop gain

Vout = A(jω)(0 − V1). (6.75)
The resulting input impedance is thus

Zin = RT R F . (6.76)
RF + RT (1 + A(jω))
For large A
Zin = R . (6.77)
F
A(jω)
The closed-loop input impedance is thus small and almost independent of the large RT of the
operational amplifier.
Now consider the non-inverting amplifier shown in figure 6.32.
V1
+
Vout
I1
RT
I2 R2
-
V2
V2
I1 I1+I2
R1
Figure 6.32: Model for calculating the input impedance of the non-inverting amplifier.
The student should calculate the input impedance by recognizing that I1 is much less than I2,
since RT is much greater than R1 or R2. Your result should be
Zin = RT A(jω) + G(jω)

, (6.78)
G(jω)
where G(jω) is the closed-loop gain of the amplifier. Notice that in contrast to the low input
impedance for the inverting amplifier, the non-inverting amplifier exhibits a closed-loop input
impedance that is much larger than the open-loop value RT .
6.3.3 Voltage and Current Offsets

Since op-amps are generally DC coupled, there will appear a nonzero output even when the
inputs are grounded or connected to give no input signal. The voltage offset is the result of
slightly different transistors making up the differential input stage. The voltage offset can be
reduced by using an externally-adjustable bias resistor (voltage offset null circuitry).
The current offsets at the inverting and non-inverting input terminals are usually base
currents into two identical bipolar transistors. Thus their difference can be expected to be much
less than either base current alone. Using this fact the student should be able to explain the
reason for having an extra resistor between the non-inverting input and ground for the inverting
amplifier. The resistor should have a value equal to the input resistor and feedback resistor in
parallel.
We define the following:
output offset voltage – The voltage at the output when the input voltage is zero (input
terminals grounded).
common mode voltage – The voltage at the output when the voltage at the inverting and non-
inverting inputs are equal.
common mode rejection ratio (CMRR) – The ratio of the op-amp gain when operating in
differential mode to the gain when operating in common mode.
common mode rejection (CMR) – The ability to respond to only differences at the input
terminals: CMR ≡ 20 log10(CMRR).
6.3.4 Current Limiting and Slew Rate

The presence of resistance at the output of the op-amp limits the current that the amplifier can
deliver into a load, as shown in figure 6.33. Current limiting is a nonlinear property that
invalidates the two normal approximation rules. When an op-amp is driven into a current-
limiting condition it goes into saturation and becomes a constant current source. For a large load
the output signal will be voltage-limited. A similar breakdown of the rules occurs when the
amplifier is driven into voltage-limited operation.
The op-amp performance can be demonstrated by applying a step function to the input and
observing the output response, as shown in figure 6.33b. The actual output will have a finite
slope (slew rate) and overshoot the final voltage value. It then approaches the final voltage either
exponentially or with some damped ringing. The slew rate and overshoot are nonlinear effects.
The settling time after amplifier saturation is defined as the time between the edge of the applied step function
and the point where the amplifier output settles to within some stated percentage of the target voltage value.
CURRENT VOLTAGE
IDEAL SETTLING
LIMITED
V LIMITED OUTPUT TIME
out SIGNAL
V
LINEAR ACTUAL
OUTPUT
OPERATION SLEW RATE
|Z
load| (V/µs)
Figure 6.33: a) Voltage-limited and current-limited operational regions for an operational amplifier and b)
definition of slew rate and settling time for an operational amplifier.
6.4 Problems
1. Consider the circuit below. (You may assume that the op-amps are ideal.)
C1
R2
R1
Vin C2
Vout
(a) Write an expression for the transfer function G(ω). Express your result in terms of
the amplitude of the output and the phase relative to the input. Let R1 = 1 kΩ,
R2 = 100 Ω, C1 = 100 µF and C2 = 1 µF. Do not simplify the algebra.
(b) What are the (real) zeroes in the transfer function, if any?
(c) What are the (real) poles in the transfer function if any?
(d) Sketch the transfer function as a function of ω on a log-log plot. Your sketch should
show the slope of |G(ω)| in the large and small ω limits, the corner frequencies, and
the value of |G(ω)| at the corner frequencies.
(e) Describe the dependence of the output on frequency at small and large frequencies in
dB/octave.
2. (a) Write the two rules for the analysis of circuits which utilize “ideal” op-amps.
(b) Write an expression for the potential Vout for the following circuit.
Z2
Z2
Vin Z1
Vout
Z1
(c) Let Z1 = 1 kΩ and Z2 = 10 kΩ. Sketch a log-log plot showing the function G(ω) for
the above circuit assuming that a general purpose op-amp such as the 741 is used.
(d) The 741 op-amp has a corner frequencies of 4 Hz, DC open-loop gain of 2 × 105 and
a fall off at high frequency of 6 dB/octave. What is the frequency domain over which
the amplifier defined in part (c) will have constant gain? What is the gain of the
amplifier in this frequency domain?
(e) Suppose now that the impedance Z1 is replaced with a capacitor with a capaci-tance
C (Z2 = 10 kΩ). For frequencies much greater than 4 Hz, the 741 op-amp will
attenuate the signal if the product RC is greater than some maximum value. For a
frequency of 50 kHz, what is the maximum value for C for which the op-amp [A(ω)]
will not attenuate the output signal at high frequencies? Hint: If you sketch A(ω)
and G(ω), you will be able to see the constraint on G(ω).
3. Set up an operational amplifier circuit to solve the equation

d2x dx
+5 + 7x + 3 = 0.
dt2 dt
2 2
Hint: the input is d x/dt .
4. Draw a schematic diagram of a circuit for which Vout = ln(Vin). Specify the limitations on
the input voltage range, if any.
Chapter 7
Digital Circuits
Analog signals have a continuous range of values within some specified limits and can be
associated with continuous physical phenomena.
Digital signals typically assume only two discrete values (states) and are appropriate for any
phenomena involving counting or integer numbers.
While we were mostly interested in voltages and currents at specific points in analog circuits,
we will be interested in the information flow in digital circuits.
The active elements in digital circuits are either bipolar transistors or FETs. These transistors
are permitted to operate in only two states, which normally correspond to two output voltages.
Hence the transistors act as switches.
Before starting we will first review number systems and Boolean algebra.
7.1 Number Systems

The two digital states can be given various names: ON/OFF, true/false, high/low, 1/0, etc.. The 1
and 0 notation naturally leads to the use of binary (base 2) numbers. Octal (base 8) and
hexadecimal (base 16) numbers are also used since they provide a condensed number notation.
Decimal (base 10) numbers are not of much use in digital electronics.
7.1.1 Binary, Octal and Hexadecimal Numbers

Consider a decimal number with digits a b c. We can write abc as
2 1 0 (7.1)
abc10 = a × 10 + b × 10 + c × 10 .
Similarly, in the binary system a number with digits a b c can be written as
2 1 0 (7.2)
abc2 = a × 2 + b × 2 + c × 2 .
Each digit is known as a bit and can take on only two values: 0 or 1. The left most bit is the
highest-order bit and represents the most significant bit (MSB), while the lowest-order bit is the
least significant bit (LSB).
131
CHAPTER 7. DIGITAL CIRCUITS 132
Conversion from binary to decimal can be done using a set of rules, but it is much easier to
use a calculator or tables (table 7.1).
Table 7.1: Decimal, binary, hexadecimal and octal equivalents.
Decimal Binary Hex Octal

00 00000 00 00
01 00001 01 01
02 00010 02 02
03 00011 03 03
04 00100 04 04
05 00101 05 05
06 00110 06 06
07 00111 07 07
08 01000 08 10
09 01001 09 11
10 01010 0A 12
11 01011 0B 13
12 01100 0C 14
13 01101 0D 15
14 01110 0E 16
15 01111 0F 17
16 10000 10 20
The eight octal numbers are represented with the symbols 0, . . . , 7, while the 16 hexadec-
imal numbers use 0, . . . , 9, A, . . . , F .
In the octal system a number with digits a b c can be written as
2 1 0 (7.3)
abc8 = a × 8 + b × 8 + c × 8 ,
while one in the hexadecimal system is written as

2 1 0 (7.4)
abc16 = a × 16 + b × 16 + c × 16 .
A binary number is converted to octal by grouping the bits in groups of three, and converted
to hexadecimal by grouping the bits in groups of four. Octal to hexadecimal conversion, or visa
versa, is most easily performed by first converting to binary.
Example: Convert the binary number 1001 1110 to hexadecimal and to decimal.
100111102 = 9E(HEX) (7.5)

= 27+24+23+22+21 (7.6)
= 158(DECIMAL). (7.7)
Example: Convert the octal number 1758 to hexadecimal.
1758 = 0011111012 (7.8)

= 07D(HEX). (7.9)
Example: Convert the number 146 to binary by repeated subtraction of the

largest power of 2 contained in the remaining number.
14610 = 128+16+2 (7.10)

7 4 1 (7.11)
= 2 +2 +2
= 100100102. (7.12)
Example: Devise a method similar to that used in the previous problem and
convert 785 to hexadecimal by subtracting powers of 16.
78510 = 3×162+16+1 (7.13)

= 311(HEX). (7.14)
7.1.2 Number Representation

We define the following
word: a binary number consisting of an arbitrary number of bits.
nibble: a 4-bit word (one hexadecimal digit).
byte: an 8-bit word.
We often use the expressions 16-bit word (short word) or 32-bit word (long word) depending on
the type of computer being used. Most fast computers today actually employ a 64-bit word at the
hardware level.
If a word has n bits it can represent 2n different numbers in the range 0 to 2 n −1. Negative
numbers are usually represented by the so called 2’s complement notation. To obtain the 2’s
compliment of a number first take the complement (invert each bit) and then add 1. All the
negative numbers will have a 1 in the MSB position, and the numbers will now range from −2n−1
to 2n−1 − 1. The electronic advantages of the 2’s complement notation becomes evident when addition is
performed. Convince yourself of this advantage.
7.2 Boolean Algebra

The binary 0 and 1 states are naturally related to the true and false logic variables. We will find
the following Boolean algebra useful. Consider two logic variables A and B and the result of
some Boolean logic operation Q. We can define
Q≡AANDB≡A·B. (7.15)
Q is true if and only if A is true AND B is true.
Q≡AORB≡A+B. (7.16)
Q is true if A is true OR B is true.

(7.17)
Q≡NOTA≡A.
Q is true if A is false.
A useful way of displaying the results of a Boolean operation is with a truth table. We will
make extensive use of truth tables later. If no “–” is available on your text processor or circuit
drawing program an “N ” can be used, ie. A ≡ N A .
We list a few trivial Boolean rules in table 7.2.
Table 7.2: Properties of Boolean Operations.
A·0 = 0
A+0 = A
A·1 = A
A+1 = 1
A·A = A
A+A = A
A·A = 0
A+A = 1
The Boolean operations obey the usual commutative, distributive and associative rules of
normal algebra (table 7.3).
We will also make extensive use of De Morgan’s theorems (table 7.4).
7.3 Logic Gates

Electronic circuits which combine digital signals according to the Boolean algebra are referred to
as logic gates; gates because they control the flow of information. Positive logic is an
electronic representation in which the true state is at a higher voltage, while negative logic has
the true state at a lower voltage. We will use the positive logic type in this course.
In digital circuits all inputs must be connected.
Table 7.3: Boolean commutative, distributive and associative rules.
A = A
A·B = B·A
A+B = B+A
A·(B+C) = A·B+A·C
A·(B·C) = (A·B)·C
A+(B+C) = (A+B)+C
A+A·B = A
A·(A+B) = A
A·(A+B) = A·B
A+A·B = A +B
A+A·B = A+B
A+A·B = A+B
Table 7.4: De Morgan’s theorems.
A·B = A+B
A+B = A·B
Logic circuits are grouped into families, each with their own set of detailed operating rules.
Some common logic families are:
RTL: resistor-transistor logic,
DTL: diode-transistor logic,
TTL: transistor-transistor logic,
NMOS: N-channel metal-oxide silicon,
CMOS: complementary metal-oxide silicon and
ECL: emitter-coupled logic.
The ECL is very fast. The MOS features very low power consumption and hence is often used in
LSI technology. The TTL is normally used for small-scale integrated circuit units.
The schematic symbols of the basic gates and the logic truth tables are shown in figure 7.1. The
open circle is used to indicate the NOT or negation function and can be replaced by an inverter
in any circuit. A signal is negated if it passes through the circle. Any logic operation can be
formed from NAND or NOR gates or a combination of both. We also commonly have gates
with more than two inputs. Inverter gates can be formed by applying
the same logic signal to both inputs of an NOR or NAND gate.
AND
A B Q
A
0 0 0
Q=A·B
B
Q 0
1 0
AND 1 0 0
1 1 1
NAND
A B Q
A Q 0 0 1
B
Q=A·B 0 1 1
NAND 1 0 1
1 1 0
OR
A B Q
A Q=A+B 0 0 0
B Q 0 1 1
OR 1 0 1
1 1 1
NOR
A B Q
A 0 0 1
Q Q=A+B
B 0 1 0
NOR 1 0 0
1 1 0
NOT
A Q
Q=A
A NA OR 0 1
1 0
INVERT
Figure 7.1: Symbols and truth tables for the four basic two-input gates: a) AND, b) NAND, c) OR, d) NOR and e) the
inverter.
7.4 Combinational Logic

We will design some useful circuits using the basic logic gates, and use these circuits later on as
building blocks for more complicated circuits.
We describe the basic AND, NAND, OR or NOR gates as being satisfied when the inputs
are such that a change in any one will change the output. A satisfied AND or NOR gate has a
true output, whereas a satisfied NAND or OR gate has a false output. We sometimes identify the
input logic variables A, B, C, etc. with an n-bit number ABC . . ..
7.4.1 Combinational Logic Design Using Truth Tables

The following steps are a useful formal approach to combinational problems:
1. Devise a truth table of the independent input variables and the resulting output quan-tities.
2. Write Boolean algebra statements that describe the truth table.

3. Reduce the Boolean algebra.
4. Mechanize the Boolean statements using the appropriate logic gates.
Consider the truth table that defines the OR gate. Using the lines in this table that yield a true
result gives.
Q = A·B+A·B+A·B (7.18)
= A·B+A·B+A·B+A·B (7.19)
= A·(B+B)+B·(A+A) (7.20)
= A+B (7.21)
Since Q is a two-state variable all other input state combinations must yield a false. If the truth
table had more than a single output result, each such result would require a separate equation. An
alternative is to write an expression for the false condition.
Q =A·B (7.22)
Q =A·B (7.23)
Q = A+B (7.24)
= A+B (7.25)
7.4.2 The AND-OR Gate

Some logic families provide a gate known as an AND-OR-INVERT or AOI gate (figure 7.2).
Q = A·B+C·D (7.26)
Q = A·B+C·D (7.27)
B AoB
Q
C CoD
Figure 7.2: The basic AND-OR-INVERT gate.
7.4.3 Exclusive-OR Gate

The exclusive-OR gate (EOR or XOR) is a very useful two-input gate. The schematic symbol
and truth table are shown in figure 7.3.
XOR (EOR)
A B Q
A
Q=A⊕B 0 0 0
Q 1 0 1
B
EOR
0 1 1
1 1 0
Figure 7.3: The schematic symbol for the exclusive-OR gate (EOR or XOR) and its truth table.
From the truth table
Q = A·B+A·B (7.28)
Q = A·B+A·B (7.29)
and we can draw the mechanization directly from the truth table (figure 7.4).
7.4.4 Timing Diagrams

Normally signals flip from one logic state to another. The time it takes the signal to move
between states is the transition time tt, where the time is measured between 10% and 90% of the
signal levels. Delays within the logic elements result in a propagation (pulse) delay tpd, where
the time is measured between 50% of the input signal and 50% of the output response. Definitions of
the transition time and propagation delay are shown in figure 7.5.
NA NAoB
A
NAoB+NBoA
B
NB NBoA
Figure 7.4: A mechanization of the exclusive-OR directly from the truth table.
A
90% 50%
10%
V
t
tt 50%
Q
t
pd
Figure 7.5: The transition time of the input and output signals, and the propagation delay through
a gate.
7.4.5 Signal Race

Signal racing is the condition when two or more signals change almost simultaneously. The
condition may cause glitches or spikes in the output signal as shown in figure 7.6. The effects of
these glitches can be eliminated by using synchronous timing techniques. In synchronous timing
the glitches are allowed to come and go, and the logic state changes are initiated by a timing
pulse (clock pulse).
7.4.6 Half and Full Adders

From basic gates, we will develop a full adder circuit that adds two binary numbers. Consider
adding two 2-bit binary numbers X1X0 and Y1Y0. X0 + Y0 = C1Z0, where C1 is the carry bit. The
truth table for all combinations of X0 and Y0 is shown in table 7.5. From the truth table
Z0 = X0·Y0+X0·Y0 =X0⊕Y0 (7.30)

C = X ·Y
0 0
(7.31)
1
B
GLITCH
Q
a) t
B
NA
∆t1 INVERTER DELAY
NB ∆t2 INVERTER DELAY + WIRE

DIFFERENCE DELAY
EXPANDED GLITCH
Q
b)
Figure 7.6: a) A timing diagram for the EOR circuit. b) An expanded view of the glitch shows it
to be caused by a signal race condition.
The mechanization of these two equation is shown in figure 7.7.

This circuit is known as the half adder. It can not handle the addition of any two arbitrary
numbers because it does not allow the input of a carry bit from the addition of two previous
digits. A circuit that can handle these three inputs can perform the addition of any two binary
numbers.
The truth table for three input variables is shown in figure 7.8.
From the truth table
C2 = C1·X1·Y1+C1·X1·Y1+C1·X1·Y1+C1·X1·Y1 (7.32)
= X ·Y +C ·X +C ·Y (7.33)
1 1 1 1 1 1
This is known as majority logic. And a majority detector is shown in figure 7.9
Z1 = C1·X1·Y1+C1·X1·Y1+C1·X1·Y1+C1·X1·Y1 (7.34)
= C1 ⊕ (X1 ⊕ Y1) (7.35)
Table 7.5: The binary addition of two 2-bit numbers. The 20 column.
X Y Z C
0 0 0 1
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
A
S
B
A S
HALF
ADDER
B C
Figure 7.7: A mechanization of the half adder using an EOR and an AND gate.
The following device (figure 7.10) is known as a full adder and is able to add three single
bits of information and return the sum bit and a carry-out bit.
The circuit shown in figure 7.11 is able to add any two numbers of any size. The inputs are
X2X1X0 and Y2Y1Y0, and the output is C3Z2Z1Z0.
C1 X1 Y1 Z1 C2
0 0 0 0 0 Cin
0 0 1 1 0 S
HALF
0 1 0 1 0 ADDER
A S1
0 1 1 0 1 Cout
HALF
1 0 0 1 0 ADDER
B C1
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
Figure 7.8: The binary addition of two 2-bit numbers. The 21 column.
C1 C1oX1
X1
C2
C1oY1
Y1
X1oY1
Figure 7.9: A mechanization of the majority detector.
X1 Z1
Y1 SUM
MAJORITY C2
C1
CARRRY-IN DETECTOR
CARRY-OUT
Figure 7.10: The full adder mechanization.
X2 Y2 X1 Y1 X0 Y0
C2 C1 0
FULL FULL FULL

ADDER ADDER ADDER
C3 Z2 C2 Z1 C1 Z0
Figure 7.11: A circuit capable of adding two 3-bit numbers.

Example: If the input to the circuit in figure 7.12 is written as a number ABCD,
write the nine numbers that will yield a true Q.
A AoB
B
C CoD Q=AoB+CoD+NAoCoND
D
NA
C
ND NAoCoND
Figure 7.12: A typical logic function.
Table 7.6: The truth table for the typical logic function example.
ABCD A·B C·D A·C·D Q

0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 1 0 0 0 1 1
0 0 1 1 0 1 0 1
0 1 0 0 0 0 0 0
0 1 0 1 0 0 0 0
0 1 1 0 0 0 1 1
0 1 1 1 0 1 0 1
1 0 0 0 0 0 0 0
1 0 0 1 0 0 0 0
1 0 1 0 0 0 0 0
1 0 1 1 0 1 0 1
1 1 0 0 1 0 0 1
1 1 0 1 1 0 0 1
1 1 1 0 1 0 0 1
1 1 1 1 1 1 0 1
ABCD = (2, 3, 6, 7, 11, 12, 13, 14, 15) gives Q true.
Example: Using the 2’s complement convention, the 3-bit number ABC can
represent the numbers from -3 to 3 as shown in table 7.7 (ignore -4). Assuming
that A, B, C and A, B, C are available as inputs, the goal is to devise a circuit
that will yield a 2-bit output EF that is the absolute value of the ABC number.
You have available only two- and three-input AND and OR gates.
1. Fill a truth table with the ABC and EF bits.

The truth table is shown in table 7.7.
Table 7.7: Truth table with for the ABC and EF bits.
Value A B C E F
0 0 0 0 0 0
1 0 0 1 0 1
2 0 1 0 1 0
3 0 1 1 1 1
-1 1 1 1 0 1
-2 1 1 0 1 0
-3 1 0 1 1 1
-4 1 0 0
2. Write a Boolean algebra expression for E and for F .
E = A·B·C+A·B·C+A·B·C+A·B·C (7.36)
= A·(B·C+B·C)+A·(B·C+B·C) (7.37)
= A·B·(C+C)+A·(B⊕C) (7.38)
= A·B+A·(B⊕C). (7.39)
F = A·B·C+A·B·C+A·B·C+A·B·C (7.40)
= A·(B·C+B·C)+A·(B·C+B·C) (7.41)
= A·C·(B+B)+A·C·(B+B) (7.42)
= A·C+A·C (7.43)
= (A+A)·C (7.44)
= C. (7.45)
3. Mechanize these expressions.

The mechanized expressions are shown in figure 7.13.
Example: Suppose that the 2-bit binary number AB must be transmitted

between devices in a noisy environment. To reduce undetected errors
introduced by the transmission, an extra bit P is often included to add
redundancy to the informa-tion. Assume that P is set true or false as needed
to make an odd number of true bits in the resulting 3-bit number ABP . When
the number is received, logic circuits are required to generate an error signal
E whenever the odd number of bits condition is not met.
1. Develop a truth table of E in terms of A, B and P .

The truth table is shown in table 7.8.
A
E
B
C
Figure 7.13: Mechanization for the ABC and EF bits.
2. Write a Boolean expression for E as determined directly from the truth table.
E =A·B·P +A·B·P +A·B·P +A·B·P (7.46)

3. Using De Morgan’s theorem twice, reduce this expression to one EOR and
one N EOR operation. (This is very similar to the half-adder problem.)
E = A·(B·P +B·P)+A·(B·P +B·P) (7.47)

= A·B⊕P +A·(B⊕P) (7.48)
= A⊕(B⊕P) (7.49)
Table 7.8: Truth table for E in terms of A, B and P.

A B P E
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 0
A E
B
P
Figure 7.14: Mechanization for E.
7.5 Multiplexers and Decoders

Multiplexers and decoders are used when many lines of information are being gated and passed
from one part of a circuit to another.
Multiplexing is when multiple data signals share a common propagation path. Time
multiplexing is when different signals travel along the same wire but at different times. These
devices have data and address lines, and usually include an enable/disable input. When the
device is disabled the output is locked into some particular state and is not effected by the inputs.
Shown in figure 7.15 is a 4-line to 1-line multiplexer.
S0
NA0oNA1oS0
S1
A0oNA1oS1
S2
NA0oA1oS2
S3
A0oA1oS3
A0
A1
Figure 7.15: 4-line to 1-line multiplexer.
A decoder de-multiplexes the signals back onto several different lines. Shown in figure 7.16
is a binary-to-octal decoder (3-line to 8-line decoder).
Decoders (octal decoder) can also convert a 3-bit binary number to an output on one of eight
lines. Hexadecimal decoders are 4-line to 16-line devices. When the decoder is disabled the
outputs will be high. A decoder would normally be disabled while the address lines were
changing to avoid glitches on the output lines.
A0
A1
ND0=NA0oNA1oNA2
A2
ND1=A0oNA1oNA2
ND2=NA0oA1oNA2
ND3=A0oA1oNA2
ND4=NA0oNA1oA2
ND5=A0oNA1oA2
ND6=NA0oA1oA2
ND7=A0oA1oA2
Figure 7.16: Octal decoder.
7.6 Schmitt Trigger

A noisy input signal to a logic gate could cause unwanted state changes near the voltage
threshold. Schmitt trigger logic reduces this problem by using two voltage thresholds: a high
threshold to switch the circuit during low-to-high transitions and a lower threshold to switch the
circuit during high-to-low transitions. Such a trigger scheme is immune to noise as long as the
peak-to-peak amplitude of the noise is less than the difference between the threshold voltages. A
gate with the Schmitt trigger feature has a small hysteresis curve drawn inside the gate symbol.
Schmitt triggers are mostly used in inverters or simple gates to condition slow or noisy signals
before passing them to more critical parts of the logic circuit.
7.7 The Data Bus

A bus is a common wire connecting various points in a circuit; examples are the ground bus and
power bus. The data bus carries digital information. A data bus is usually a group of parallel
wires connecting different parts of a circuit with each individual wire carrying a different logic
signal. The data bus is connected to the inputs of several gates and to the outputs of several
gates. You cannot connect directly the outputs of normal gates. For this purpose three-state
output logic is commonly used but will not be discussed here.
A data bus line may be time multiplexed to serve different functions at different times.
At any time only one gate may drive information onto the bus line but several gates may receive
it. In general, information may flow on the bus wires in both directions. This type of bus is
referred to as a bidirectional data bus.
7.8 Two-State Storage Elements

Analog voltage storage times are limited since the charge on a capacitor will eventually leak
away. The problem of discrete storage reduces to the need to store a large number of two-state
variables. The four commonly used methods are: 1) magnetic domain orientation, 2) presence or
absence of charge (not amount of charge) on a capacitor, 3) presence or absence of an electrical
connection and 4) the DC current path through the latches and flip-flops of a digital circuit. We
will discuss only the latter.
7.9 Latches and Un-Clocked Flip-Flops

It is possible using basic logic gates to build a circuit that remembers its present condition.
It is also possible to build counting circuits. The basic counting unit is the flip-flop (FF).
7.9.1 Latches
All latches have two inputs: data and enable/disable. And typically Q and Q outputs. A ones-
catching latch can be built as shown in figure 7.17.
When the control input C is false, the output Q follows the input D, but when the con-trol input
goes true, the output latches true as soon as D goes true and then stays there independent of
further changes in D.
One of the most useful latches is known as the transparent latch or D-type latch. The
transparent latch is like the ones-catching latch but the input D is frozen when the latch is
disabled. The operation of this latch is the same as that of the statically triggered D flip-flop
discussed below.
7.9.2 RS and RS Flip-Flops

The RS flip-flop (RSFF) is the result of cross-connecting two NOR gates as shown in fig-ure
7.18. The RS inputs are referred to as active ones.
The ideal flip-flop has only two rest states, set and reset, defined by QQ = 10 and QQ = 01,
respectively.
A very similar flip-flop can be constructed using two NAND gates as shown in figure 7.19.
The RS inputs are now active zeros.
These FFs are often referred to as the set/reset type and are un-clocked.
D D
D+(CoQ)
C
CoQ
Figure 7.17: An AND-OR gate used as a “ones catching” latch and its timing diagram.
S R Q Q S NQ
0 0 no change RS flip-flop S Q
0 1 0 1
1 0 1 0 R NQ
1 1 undefined R Q
Figure 7.18: The RS flip-flop constructed from NOR gates, and its circuit symbol and truth table.
7.10 Clocked Flip-Flops

A clocked flip-flop has an additional input that allows output state changes to be synchro-nized
to a clock pulse.
7.10.1 Clocked RS Flip-Flop

We first consider the static clocked (level-sensitive) RS flip-flop shown in figure 7.20. The
symbol x in the following tables represents either the binary state 0 or 1.
The first five lines in the truth table give the static input and output states. The last four lines
show the state of the outputs after a complete clock pulse p.
NS
Q
S R Q Q
0 0 undefined NS Q
0 1 1 0 RS flip-flop
1 0 0 1 NR NQ
1 1 no change NR
NQ
Figure 7.19: The RS flip-flop constructed from NAND gates, and its circuit symbol and
truth table.
S R
C C C Q Q
x x 0 no change
0 0 1 no change
0 1 1 0 1 Sc
Sc Q
1 0 1 1 0 S Q
1 1 1 undefined C C
0 0 p no change R NQ Rc
NQ
0 1 p 0 1 Rc
1 0 p 1 0
1 1 p undefined
Figure 7.20: The clocked RS flip-flop can be constructed from an RS flip-flop and two
additional gates, the schematic symbol for the static clocked RSFF and its truth table.
7.10.2 D Flip-Flop
The D flip-flop avoids the undefined states in the RSFF truth table by reducing the number of
input options (figure 7.21).
D C Q Q D Sc Q D Q
x 0 no change C C C
0 p 0 1 Rc NQ NQ
1 p 1 0
Figure 7.21: Statically triggered D flip-flop (transparent latch) mechanized with clocked RS, and
the schematic symbol and its truth table.
The statically clocked DFF is also known as a transparent latch.
7.10.3 JK Flip-Flop
The JKFF simplifies the RSFF truth table but keeps two inputs (figure 7.22). The toggle state is
useful in counting circuits. If the C pulse is too long this state is undefined and hence the JKFF
can only be used with rigidly defined short clock pulses.
J K C Q Q DEV1
Sc Q
0 0 p no change J
S
J Q
0 1 p 0 1 C
C C
1 0 p 1 0 K NQ
K
Rc R Q
1 1 p toggle
Figure 7.22: The basic JK flip-flop constructed from an RS flip-flop and gates, and its schematic
symbol and truth table.
7.11 Dynamically Clocked Flip-Flops

We distinguish two types of clock inputs.
static clock input – a clock input sensitive to the signal level and
dynamic clock input – a clock input sensitive to signal edges.
7.11.1 Master/Slave or Pulse Triggering

We can simulate a dynamic clock input by putting two flip-flops in tandem, one driving the other
in a master/slave arrangement as shown in figure 7.23. The slave is clocked in a complementary
fashion to the master.
MASTER
SLAVE
J
S Q
S Q
C
K R NQ
R NQ
Figure 7.23: An implementation of the master/slave flip-flop.
This arrangement is still pulse triggered. The data inputs are written onto the master flip-flop
while the clock is true and transferred to the slave when the clock becomes false. The
arrangement guarantees that the QQ outputs of the slave can never be connected to the slave’s
own RS inputs. The design overcomes signal racing (ie. the input signals never catch up with
the signals already in the flip-flop). There are however a few special states when a transition can
occur in the master and be transferred to the slave when the clock is high. These are known as
ones catching and are common in master/slave designs.
7.11.2 Edge Triggering

Edge triggering is when the flip-flop state is changed as the rising or falling edge of a clock signal passes through a
threshold voltage (figure 7.24). This true dynamic clock input is
insensitive to the slope or time spent in the high or low state.
DELAY
∆tD
D
∆tD
Figure 7.24: A slow or delayed gate can be used to convert a level change into a short pulse.
Both types of dynamic triggering are represented on a schematic diagram by a special

symbol near the clock input (figure 7.25). In addition to the clock and data inputs most IC flip-
flop packages will also include set and reset (or mark and erase) inputs. The additional inputs
allow the flip-flop to be preset to an initial state without using the clocked logic inputs.
J K C S R Q Q J J S
J
0 0 ↓ 1 1 no change Q Q Q
0 1 ↓ 1 1 0 1 C C
C
1 0 ↓ 1 1 1 0 NQ NQ NQ
1 1 ↓ 1 1 toggle K K K
x x x R
0 1 1 0
x x x 1 0 0 1
a) b) c)
Figure 7.25: The schematic symbols for a) a positive edge-triggered JKFF, b) a negative (falling)
edge-triggered JKFF and c) a negative edge-triggered JKFF with set and reset inputs.
7.12 One-Shots
The one-shot (also called a monostable multivibrator) is essentially an unstable flip-flop. When a one-shot is set by
an input clock or trigger pulse, it will return to the reset state on its own accord after a fixed time delay. Hence a
one-shot is able to generate a pulse of a particular width following an input pulse. One-shots are often used in
pairs with the output of the first used to trigger the second. Unfortunately the time relationship between the
signals becomes excessively interdependent and it is better to generate signal transitions synchronized with the
circuit clock.
7.13 Registers
Registers are formed from a group of flip-flops arranged to hold and manipulate a data word
using some common circuitry. We will consider data registers, shift registers, counters and
divide-by-N counters.
7.13.1 Data Registers

The circuit shown in figure 7.26 uses the clocked inputs of D flip-flops to load data into the
register on the rising edge of a LOAD pulse.
D0 D1 D2
D Q D Q D Q
C C C
LOAD
Figure 7.26: A data register using the clocked inputs to D-type flip-flops.
It is also possible to load data and still leave the clock inputs free (figure 7.27). The loading
process requires a two-step sequence. First the register must be cleared, then it can be loaded.
LOAD
D0 D1 D2
S S S
D Q D Q D Q
C R Q C R Q C R Q
NCLEAR
Figure 7.27: A more complicated data-loading technique leaves the clocked inputs free but
requires a clear-load pulse sequence.
7.13.2 Shift Registers

A simple shift register is shown in figure 7.28. A register of this type can move 3-bit parallel
data words to a serial-bit stream. It could also receive a 3-bit serial-bit stream and save it
for parallel use.
D D Q D Q
A
D Q
C C
C
CLOCK
Figure 7.28: A 3-bit shift register constructed with D flip-flops.
If A is connected back to D the device is known as a circular shift register or ring counter. A
circular shift register can be preloaded with a number and then used to provide a repeated pattern
at Q.
7.13.3 Counters
There are several different ways of categorizing counters:
1. binary-coded decimal (BCD) versus binary,
2. one direction versus up/down and
3. asynchronous ripple-through versus synchronous.
Counters are also classified by their clearing and preloading abilities. The BCD type count is
decimal, and is most often used for displays. In the synchronous counter each clock pulse is fed
simultaneously or synchronously to all flip-flops. For the ripple counter, the clock pulse is
applied only to the first flip-flop in the array and its output is the clock to the second flip-flop,
etc.. The clock is said to ripple through the flip-flop array.
Shown in figure 7.29 is a binary, ripple-through, up counter.
COUNT ENABLE
S S S
J Q J Q J Q
COUNT C C C
K
R Q
K R Q K R Q
Figure 7.29: A 3-bit ripple counter constructed from JK flip-flops.
Because of pulse delays, the counter will show a transient and incorrect result for short time
periods. If the result is used to drive additional logic elements, these transient states may lead to
a spurious pulse. This problem is avoided by the synchronous clocking scheme shown in figure
7.30. All output signals will change state at essentially the same time.
COUNT ENABLE ENABLE

OUT
S
S J Q
J Q
C J S
C Q
K R Q
K R Q C
K R Q
COUNT
Figure 7.30: A 3-bit synchronous counter.
7.13.4 Divide-by-N Counters

A common feature of many digital circuits is a high-frequency clock with a square wave output.
If this signal of frequency f drives the clock input of a JKFF wired to toggle on each trigger, the
output of the flip-flop will be a square wave of frequency f /2. This single flip-flop is a divide-
by-2 counter. In a similar manner an n flip-flop binary counter will yield an output frequency
that is f divided by 2n.
7.14 Problems
1. Using only two-input NOR gates, show how AND, OR and NAND gates can be made.
2. The binary addition of two 2-bit numbers (with carry bits) looks like the following:
C2 C1
X1 X0
+ Y1 Y0
Z2 Z1 Z0
(a) Write a truth table expressing the outputs C2 and Z1 as a function of C1, X1 and
Y1.
(b) Write an algebraic statement in Boolean algebra describing this truth table.
(c) Implement this statement using standard (AND, OR, EX-OR and inverter gates).
3. If the 3-bit binary number ABC represents the digits 0 to 7:
(a) Make a truth table for A, B, C and Q, where Q is true only when an odd number of
bits are true in the number.
(b) Write a statement in Boolean algebra for Q.
(c) Convert this equation to one that can be mechanized using only two XOR gates.
Draw the resulting circuit.
4. You need to provide a logic signal to control an experiment. The experiment is con-trolled
by the four signals A, B, C and D, which make up the data word ABCD. The control line
Q should be set high only if this data word takes on the values 1, 3, 5, 7, 11 or 13.
(a) Write a truth table for this function.

(b) Using boolean algebra, write an expression indicating when Q is true. Your
statement should include one logical statement for each of the six true conditions,
each separated by the OR function. Therefore this statement should utilize a number
of AND/NAND functions, and five OR statements.
(c) Simplify this statement so that it can be implemented using two two-input AND
gates, one OR gate, one exclusive OR gate and one logical inverter. The expression
has the form Q = ( ) · {( ) + ( ) · [( ) ⊕ ( )]}, where ( ) stands for any of the four input
signals or their logical inverses.
(d) Implement this simplified function using logic gates.
Chapter 8
Data Acquisition and Process Control
The purpose of most electronic systems is to measure or control some physical quantity.
The system will need to acquire data from the environment, process this data and record it.
Acting as a control system it will also have to interact with the environment.
The flow of information in a typical data acquisition system (DAQ) can be described as
follows.
1. The input transducers measure some property of the environment.
2. The output from the transducers is conditioned (amplified, filtered, etc.).
3. The conditioned analog signal is digitized using an analog-to-digital converter (ADC).
4. The digital information is acquired, processed and recorded by the computer.
5. The computer may then modify the environment by outputting control signals.
6. The digital control signals are converted to analog signals using a digital-to-analog
converter (DAC).
7. The analog signals are conditioned (eg. amplified and filtered) appropriately for an output
transducer.
8. The output transducer interacts with the environment.
8.1 Transducers
Electrical systems are only able to respond to voltage and current signals in the electrical
domain: amplitude, frequency, phase and time constant. An input transducer is necessary to
convert a signal from its domain of origin (non-electrical) into the electrical domain.
A transducer may generate an electrical signal by varying one of the following: voltage,
current, resistance, capacitance, self-inductance, mutual-inductance, VP N , VZ , hf e or gm The
most fundamental transducers respond to temperature, electromagnetic radiation intensity, force,
displacement or chemical concentration. If coupled to the time domain these devices can be used
to measure any physical or chemical quantity. Examples of input transducers
157
CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL 158
are: a radio antenna, a photo-diode, a phototube, a piezoelectric crystal, a thermocouple, a Hall

effect device, a mechanical switch, a strain gauge, an ionization chamber, etc..
The output transducer transfers signals out of the electrical domain and into the do-main that
can be perceived by one of the five human senses. A substantial amount of power is usually
required to transfer information out of the electrical domain. Examples of output transducers are
the motor, cathode-ray tube, loudspeaker, light-emitting diode, radio-frequency transmitter, etc.
8.2 Signal Conditioning Circuits

Signal conditioning occurs in the interface between the transducers and the electrical circuit. A
low-level signal amplifier and a low-pass filter are common signal conditioners after the input
transducer. The output signal is usually conditioned by a low-pass filter and some type of power
amplifier.
8.2.1 De-bouncing the Mechanical Switch

The pushbutton or toggle switch is a simple form of data entry into a digital system. However, a
−1
problem occurs since the normal human reaction time is about 10 s and digital electronics
−8
responds to times of the order of 10 s. Thus any unnoticed mechanical contact bounce of a few
milliseconds will be seen as several distinct switch closures by a digital system. We may de-
bounce the mechanical switch by using an RC circuit and Schmitt trigger logic or a flip-flop
latch. The latter design requires a break-before-make action, which means that during the throw
there is a time when the common is connected to neither terminal.
8.2.2 Op Amps for Gain, Offset and Function Modification

An operational amplifier can be used to provide the following signal conditioning:
1. increase the amplitude of the signal,
2. filter the signal,
3. decrease the signal output impedance or
4. provide a variable gain and offset control.
The latter is most useful for calibrating a transducer’s output signal.

The dynamic range of a signal from an input transducer may be too large to process through
the DAQ system (eg. the ADC is often the limiting factor). One can use a linear amplifier and
choose to overflow or reduce the overall gain. The latter approach will cause a loss in precision.
Another approach is to use a nonlinear amplifier, such as one with a logarithmic gain, Vout =
log(Vin).
8.2.3 Sample-and-Hold Amplifiers

The purpose of the sample-and-hold amplifier is to freeze an analog voltage at the instant the
HOLD command is issued and make that analog voltage available for an extended period. Figure
8.1 shows various ways of converting three analog input signals to digital signals for acquisition
by a single digital n-bit bus.
8.2.4 Gated Charge-to-Voltage Amplifier

The gated charge-to-voltage amplifier is designed specifically as an integrating amplifier to
measure the area under a narrow pulse. Its capacitor must be discharged before a new sample can
be taken. If the initial charge on the capacitor is zero, then the output voltage from the amplifier
follows the gate signal. This sampling amplifier is normally used with pulsed signals when the
area under the signal is of primary interest. The entire signal is integrated and the output is
insensitive to the details of the signal shape.
If the signal pulse rides on a relatively constant but nonzero offset voltage, the effect of the
offset can be determined by generating a gate when no signal pulse is present. The resulting
output voltage is known as a pedestal and can be subtracted from the data signal at a later point
in the system.
8.2.5 Comparator
The comparator is used to provide a digital output indicating which of two analog input voltages
is larger. It is a single bit analog-to-digital converter. The comparator is very similar to an
operational amplifier but has a digital true/false output. Since the comparator is basically an
amplifier, the op-amp schematic symbol is used, but to avoid confusion the symbol C may be
inserted inside the op-amp symbol.
8.3 Oscillators
Often the need arises for some type of repetitive signal to serve as a timing reference for various
logic or control functions. This need is served by a constant-frequency square wave oscillator.
8.3.1 Application to Interval Timers

With increasing system complexity, the need may arise for several repetitive timing signals with different periods.
If each timing signal is obtained from a separate oscillator, the signals will have a random and variable phase
relationship. They will be asynchronous and may lead to glitches. A better technique is to use one high-frequency
oscillator with a short period and from it derive all longer-period signals. If the longer period is a multiple of two of
the clock period a counter with flip-flops can be used. If the period is not a multiple of two of the clock period a
divide-by-N counter can be used.
A S/H n−bit
V1
ADC n
A S/H n−bit digital
V2
ADC n MUX n
A S/H n−bit
V3
ADC n
A S/H analog n−bit
V1
A S/H
V2
MUX ADC n
A S/H
V3
A analog n−bit
V1
A S/H
V2
MUX ADC n
A
V3
Figure 8.1: Three schemes for multiplexing several analog signals down to one digital input
path. The notation S/H indicates sample-and-hold, ADC means analog-to-digital converter,
MUX means multiplexer, and the /n symbol across a line indicates n digital signals.
8.4 Digital-to-Analog Conversion

The process of converting a number held in a digital register to an analog voltage or current is
accomplished with a digital-to-analog converter (DAC). The DAC is a useful interface between
a computer and an output transducer.
8.4.1 Current Summing and IC Devices

DACs are normally switched current devices designed to drive the current-summing junction of
an operational amplifier. Figure 8.2 shows a simple 4-bit DAC.
Vref
Im Im/2 Im/4 Im/8
Iref Vref
Im = ----
8R 16R 2R
2R 4R
R
MSB A B C D
I
Vout
Figure 8.2: The current DAC uses the summing input to an op-amp to yield a voltage output.
LSB and MSB refer to the least and most significant bits of a binary number.
Each current is proportional to the value of a bit position in a binary number.

The design requires a number of different precisely defined resistor values. We can im-prove
the circuit by replacing it with a circuit that requires fewer distinct resistor values.
8.4.2 DAC Limitations

The output of a DAC can only assume discrete values. The relationship between the input binary
number and the analog output of a perfect DAC is shown in figure 8.3.
Common DAC limitations are an anomalous step size between adjacent binary numbers, non-
monotonic behaviour, or a zero output.
8.5 Analog-to-Digital Conversion

The analog-to-digital converter (ADC) is used to convert an analog voltage to a digital number (figure 8.4).
ANALOG
OUTPUT ANOMALOUS NON−MONOTONIC
PERFECT STEP SIZE BEHAVIOUR

DAC
+1/2 LSB error −3/2 LSB error
000001010011100101110111 000001010011100101110111 000001010011100101110111
DIGITAL INPUT DIGITAL INPUT DIGITAL INPUT

a) b) c)
Figure 8.3: Output signals from DACs showing a) the ideal result, and b) a differential
nonlinearity or c) non-monotonic behavior, both caused by imperfectly matched resistors.
8.5.1 Parallel-Encoding ADC (flash ADC)

The parallel-encoding or flash ADC design provides the fastest operation at the expense of high
component count and high cost (figure 8.5).
The resistor network sets discrete thresholds for a number of comparators. All comparators with
thresholds above the input signal go false while those below go true. Then digital encoding logic
converts the result to a digital number.
8.5.2 Successive-Approximation ADC

The successive-approximation ADC is the most commonly used design (figure 8.6). This design
requires only a single comparator and will be only as good as the DAC used in the circuit.
The analog output of a high-speed DAC is compared against the analog input signal. The digital
result of the comparison is used to control the contents of a digital buffer that both drives the
DAC and provides the digital output word.
The successive-approximation ADC uses fast control logic which requires only n compar-
isons for an n-bit binary result (figure 8.7).
8.5.3 Dual-Slope ADC

The limitations associated with the DAC in a successive-approximation ADC can be avoided by using the analog
method of charging a capacitor with a constant current; the time required to charge the capacitor from zero to the
voltage of the input signal becomes the digital output. When charged by a constant current the voltage on a
capacitor is a linear function
DIGITAL DAC A
S/H ADC PROCESSOR
ANALOG ANALOG
IN OUT
CONTROL
LOGIC
Figure 8.4: A generalized hybrid and digital circuit by which input analog data can be
transmitted, stored, delayed, or otherwise processed as a digital number before re-conversion
back to an analog output.
of time and this characteristic can be used to connect the analog input voltage to the time as
determined by a digital counter.
8.6 Time-to-Digital Conversion

It is possible to digitize relatively long time intervals by incrementing a counter with a repetitive
signal derived from an oscillator. For short time intervals the time-to-digital converter (TDC)
circuit shown in figure 8.8 can be used.
The resulting voltage on the capacitor is proportional to the time between the START and STOP
pulses as shown in figure 8.9.
Vin
Vref
R
C
R
C
R DEV1
C
Q7
Q6 EN
R Q5
Q4
C Q3 S2
Q2 S1
Q1 S0
R Q0
3Vref encoding
C
----- logic
8
R
C
R
Vref
C
----
8
R
Figure 8.5: A 3-bit, parallel-encoding or flash ADC.
+
V
IN C
−
Analog Output
V DAC
REF Digital Inputs
START
Register
CONVERSION HIGH/LOW
CONTROL LOGIC
CONVERSION Output Buffer
COMPLETE CLOCK
Figure 8.6: The block diagram of an 8-bit successive-approximation ADC.

XXX
MSB
TEST
0XX 1XX
MSB−1 MSB−1
TEST TEST
00X 01X 10X 11X
MSB−2 MSB−2 MSB−2 MSB−2

TEST TEST TEST TEST
000 001 010 011 100 101 110 111
Figure 8.7: The bit-testing sequence used in the successive approximation method.
RESET
START DEV1
ADC DIGITAL
S
D Q OUTPUT
C R Q
STOP
-Vref
Figure 8.8: A time-to-digital converter. The circuit is shown with the switches in the signal-
holding position after a STOP pulse and with RESET false.
RESET
START
STOP
∆t
VC V C=
REF ∆t
V
RC t
0
Figure 8.9: TDC timing signals.

8.7 Problems
1. Design an inverting op-amp circuit that has potentiometers for gain and offset.
2. Assuming that the diode shown in the circuit below exactly follows the equation I =
V /V −7
I0(e T − 1) with I0 = 10 A and VT = 50 mV, sketch Vout versus Vin over the input range
−2 V to +2 V. Show the scale on both axes.
10kohm
Vin
Vout
Figure 8.10: Operational amplifier with diode in feedback.
3. Using two 1020-type DACs and two op-amps, design a circuit whose analog output is
proportional to the product of two digital numbers.
4. How many comparators are needed to build an 8-bit flash encoder?
5. Using a TDC, devise an experiment and show a complete block diagram of a laboratory
system to determine the speed of a bullet.
6. Consider the bad design of the synchronous counter shown below.

Q0 Q1 Q2 Q3
S S S
J Q J Q J Q
S
J Q
K R Q C C
K R Q K R Q C
K R Q
CLOCK
(a) What must the JK inputs of the first flip-flop be connected to for the circuit to
“count”?
(b) What must the SR inputs be connected to for the flip-flop output to be well defined?
(c) Determine the truth table and decimal output for the synchronous counter.
7. (a) Using clocked D-latchs, design a divide-by-4 ripple through counter.

(b) Using clocked JK flip-flops, design a divide-by-4 synchronous counter.
(c) For the synchrnonous divide-by-4 up counter, add appropriate gating so that it may
be made to count down by using a mode control signal.
8. (a) In block diagram form, draw the simplest circuit that can multiplex 16 parallel lines,
transmit serially and then demultiplex back into 16 lines. You should build your
multiplexer and demultiplexer using parallel-to-serial and serial-to-parallel shift
registers, and anything else you think you need.
(b) How may lines between the multiplexer and demultiplexer are needed?
9. Prepare a comparison table of the three different types of ADCs presented in class. Take
into account complexity, resolution, accuracy and speed.
10. An ideal TTL buffer produces an output of either 0 V or 5 V for input voltages of 0 V and
> 0 V respectively. Using ideal TTL buffers and ideal op-amps, design a 4-bit digital-to-
analog converter that can produce voltages in the range of 0 to 5 V.
11. A ramp signal generator is a useful device that gives an output voltage that increases in
fixed steps with increasing time. Design a ramp signal generator circuit using 4-bit binary
counters and an 8-bit DAC.
Chapter 9
Computers and Device

Interconnection
These lectures will deal with the interface between computer software and electronic instru-
mentation.
9.1 Elements of the Microcomputer

We will treat the microprocessor as just another IC chip on a circuit board. Our stripped-down
version will result in a computer suitable to specialized, single-process or single-circuit
applications.
In a typical research laboratory arrangement, the computer would be used to control
equipment, acquire data from the measurement apparatus, perform mathematical operations on
the data and make these available on a display device for operator review.
9.1.1 Microprocessor and Microcomputer

The key element of a microcomputer is the LSI microprocessor chip, whose circuitry can
acquire, interpret, and execute a sequence of logical and arithmetic instructions. Although most
computer programming is done in a high level language, the processor itself deals only with
binary numbers that represent operation codes (move data, add, subtract, compare, jump, etc.)
and addresses (routing information for the data flow) of registers, memory locations or
input/output data ports.
Beside the microprocessor, a complete computer requires a power supply, memory, in-
terface circuits to provide ports to external devices, and input/output devices such as a keyboard,
display, DACs and ADCs, and magnetic or other (optical) data storage.
9.1.2 Functional Elements of the Computer

The functional elements of a computer are the
• central processing unit (CPU),
169
CHAPTER 9. COMPUTERS AND DEVICE INTERCONNECTION 170
• random access storage (memory), and
• input/output to external devices (I/O).
The sub-units of the CPU are
• instruction decode and CPU control,
• control of addressing for memory and I/O ports,
• data transfer control,
• data and address registers and
• arithmetic logic unit.
To keep track of the CPU steps, the processor maintains a special register, known as the
program counter. The program counter points to (contains the address of) the next instruction
to be executed. The instruction itself specifies
• the operation to be performed,
• the processor registers to be used and
• possibly data (or a pointer to data in memory).
The CPU will typically perform the following execution cycle:
1. use the program counter to fetch the next instruction.
2. decode the instruction and fetch data from memory into internal registers as required,
3. perform the instruction and put the result in another internal register and
4. set status bits in the status register as required.
The various functional units of the computer are connected by one or more multi-wire digital
buses which pass data, addresses, and control information between the units as shown in figure
9.1.
9.1.3 Mechanical Arrangement

Machines with the lowest cost and highest reliability are generally those with the fewest
mechanical connectors and socketed components, while those with more connectors and sockets
are more easily expanded and maintained. One design is to have all circuits on mechanically
equivalent boards that plug into a common bus, this is the most modular arrangement and is
easily maintained. An alternative design is the single-board computer which places the CPU and
a substantial portion of the memory and I/O interface circuits onto a single circuit board. This
design is more compact and less expensive.
E
Instruction
X
decode and
T
CPU control
E Additional
R MEMORY I/O CPU
I
N ADDRESS
N
T Control A
L
E
I/O DEVICE
R
B
N DATA
U
A Control
S
L
B CPU
U Registers
S
ALU
Figure 9.1: A typical computer design showing two multi-wire buses, an internal bus connect-ing functional units
within the CPU and an external bus for connecting additional computer subsystems.
9.1.4 Addressing Devices on the Bus

The address determines the destination or source of information. Since the wires of a bus are
common to all functional units, each unit will see all the data placed on the bus lines. The
address lines are used within a receiving unit to determine if available information should be
processed or ignored. Each data repository on a common bus will have a unique address.
When the CPU needs to transfer data between itself and a particular location, it imple-ments
a sequence of signals as specified by the read or write operation protocol for the bus. The range
of numbers that can be represented by the available address lines (wires) on a bus is known as
the address space. A range of numbers is used mostly to access information from memory and
is thus known as the memory address space. Some processors assign a few of these memory
addresses to other input/output devices. A feature known as memory-mapped I/O.
9.1.5 Control of the Bus

The information flow on the computer bus is time-multiplexed to allow different functional units
to use the same bus lines at different times. We will assume that only one device tries to write
(drive) a given bus line at any time. A special unit, known as a bus master, has the
responsibility for controlling the other units, which are correspondingly known as the bus
slaves. If several units are capable of becoming bus masters these units must arbitrate amongst
themselves to determine which is to have control of the bus for a given interval of time. Often a
direct memory access (DMA) unit is allowed to communicate with a slave memory without
going through the CPU bus master. This allows memory access at a higher speed than having to
go through the intermediate CPU bus master.
9.1.6 Clock Lines

The changes of state of all bused signals are synchronized to one or more clock signals, which
are distributed to all functional units on the bus.
Read and write operations between the CPU and memory units are the most common. One
approach is for the CPU to set the address lines and data lines and then blindly issue the control
pulses, assuming that the slave unit will respond as needed within the allowed time. A more
elegant but slightly slower technique uses a handshake which requires an acknowledgment
response from the slave before proceeding.
9.1.7 Random Access Memory

The random-access memory (RAM) supports both read and write operations. Integrated circuit
RAM comes in two main types:
static RAM in which a single bit of memory is simply a digital flip-flop and requires only
continuous power to maintain its state.
dynamic RAM in which a bit of memory is a storage capacitor in either the charged or
discharged condition. The term dynamic refers to the need to periodically renew or refresh
the slowly discharging capacitor.
Compared to dynamic memory, static memory has the following advantages. It is simpler to
use, about ten times faster and more reliable. On the other hand, it is more expensive, consumes
more power and requires more physical space. Because of power consumption in an IC the
largest static RAM is 16K bits. The largest dynamic RAM is 260K and hence is used for normal
applications while the static RAM is used for special fast applications within the same computer.
Both types of RAM are volatile, meaning that stored information is lost when power is removed
from the chip. Some computer designs provide a limited amount of non-volatile read/write RAM
storage by using special low-power (and slower) dynamic memories powered by re-chargeable
batteries.
9.1.8 Read-Only Memory

Another form of non-volatile random access storage is the read-only memory (ROM). Here a
single memory bit is nothing more than a connection that is either open or closed. The most
common ROM types are known as field-programmable (as opposed to factory pro-grammable).
This programming process consists of stepping through all the bits and setting the necessary
ones by burning open the fuse-like material associated with that bit.
There are many varieties of field-programmable ROM units:
PROM – programmable read-only memory,

EPROM – erasable PROM (using ultraviolet light), and
EEPROM (E2PROM) – electrically erasable PROMs.

The most common uses of ROM memory in a computer are to provide initialization such as
memory tests and disk bootstrapping.
9.1.9 I/O Ports

Input and output ports are the pathways by which the CPU communicates with the world outside
the computer. The pathway may be either
• 1-bit wide (bit-serial),

• 8-bit wide (byte-serial), or
• 16- to 32-bit wide (word-serial).
9.1.10 Interrupts
Real-time applications require the computer to respond immediately to an external stimulus. The hardware
interrupt can be used to suspend the current sequence of instructions, perform a specific and usually short I/O
task, then return to the original sequence of instructions.
9.2 8-, 16-, or 32-Bit Busses

A microprocessor can be characterized by the width of its internal and external buses. The width
of the internal bus determines the largest number that can be transferred or processed in a single
clock cycle. The width of the external bus determines the number of memory addresses
accessible by the processor. A wider bus is faster and more convenient for the programmer. A
narrower bus is cheaper and requires simpler external hardware with fewer wires and
connections.

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PHYSICS

1 Direct Current Circuits 6

2.1 AC Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1Filters and Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

5.1 Bipolar Junction Transistors . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.1.4 Ideal and Perfect Bipolar Transistor Models . . . . . . . . . . . . . . 87

6.1 Open-Loop Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

7.1 Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

7.4.2 The AND-OR Gate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

8.1 Transducers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

8.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

9.1 Elements of the Microcomputer . . . . . . . . . . . . . . . . . . . . . . . . . 169

1.1 Basic Concepts

1.1.2 Potential Diﬀerence

1.1.3 Resistance and Ohm’s Law

V ∝I; V =RI, (1.4)

1.2 The Schematic Diagram

(a) (b) (c)

1.2.1 Electromotive Force (EMF)

(a) (b) (c)

1.3 Kirchoﬀ ’s Laws

• an element is an impedance (resistance) or EMF (ideal voltage source or ideal current

1. define the currents and voltages on a diagram,

1.3.1 Series and Parallel Combinations of Resistors

1.3.2 Voltage Divider

Figure 1.3: Divider circuits: a) voltage divider and b) current divider.

Figure 1.4: Example voltage divider.

V = I R, where R = R1 + R23. (1.14)

1.3.3 Current Divider

Figure 1.5: Example current divider.

1.3.4 Branch Current Method

2. Use only interior loops and all but one node.

3. Solve the system of algebraic equations.

1.3.5 Loop Current Method

1. Label interior loop currents on a diagram.

3. Solve the system of algebraic equations.

Figure 1.6: Loop method for the Wheatstone bridge circuit.

V = R1(Ia − Ib) + R3(Ia − Ic) (1.20)

0 = R1(Ib − Ia) + R2Ib + R5(Ib − Ic) (1.21)

0 = R3(Ic − Ia) + R5(Ic − Ib) + R4Ic. (1.22)

V = Ia(R1 + R3) − IbR1 − IcR3 (1.23)

0 = −IaR1 + Ib(R1 + R2 + R5) − IcR5 (1.24)

0 = −IaR3 − IbR5 + Ic(R3 + R4 + R5). (1.25)

Ia = 0.267A, Ib = 0.140A, and Ic = 0.113A. (1.26)

V1 − IA(2R) + IB R − V2 = 0 loop A. (1.33)

V2 − IB (3R) + IAR = 0 loop B. (1.34)

V1 − V 2 = 2RIA − RIB (1.35)

1.4 Equivalent Circuits

1.4.1 Thevenin’s and Norton’s Theorems

1.4.2 Determination of Thevenin and Norton Circuit Elements

1. Thevenin – calculate the open-circuit voltage VAB = VTh.

Figure 1.8: Thevenin and Norton equivalent circuits.

2. Norton – calculate the short-circuit current between A and B; IN.

Solution: From Thevenin’s theorem

VAB(open) = VTh (1.43)

VTh = INRN and (1.47)

⇒ RTh = RN. (1.49)

• The evaluation of VTh is performed using Kirchoﬀ’s laws: