ELECTENG 101
Electrical and Digital Systems
Electrical and Digital Systems
ii
Department of Electrical, Computer, and Software
Engineering
The Department of Electrical, Computer, and Software Engineering
currently offers three of the nine specialisations in the Bachelor of
Engineering (Hons) degree:
Electrical and Electronic Engineering
Computer Systems Engineering
Software Engineering
Cover Photos:
Graduates of the Department are making significant contributions towards
the society of the future.
Top Left
There is considerable pressure for countries to move away from their
dependence on fossil fuels to a variety of alternative energy sources. Solar
and wind are two of these that are being extensively developed both in New
Zealand and worldwide.
Top Right
Robots are increasingly being used in our modern world. The photo shows
one of the Department’s robots being used as part of a research project into
the next generation of automated manufacturing.
Bottom
Driverless electric vehicles that sense their environment will soon be a
reality for all of us. These will be charged inductively using technology
developed in the Department over several decades.
Electrical and Digital Systems
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Learning Resources
Online resources (available on Canvas)
In addition to the weekly lectures, these online resources have been developed to
assist students in their learning:
•
Formative Assignments – The online assignments in this course are
designed for learning. The questions are mostly numeric in nature, and
different values are given each time you practice them. There are no penalties
for answering them wrong and you can repeat most questions until you are
satisfied you have integrated the concepts with the mathematical models
that describes them.
•
Multiple-choice Questions (MCQs) – A large bank of MCQs has been
developed, covering various aspects of the course material. These will be
made available progressively in line with the course delivery as a way for
self-assessments of the relevant concepts covered throughout the course.
•
Online Forum – An online forum will be available where students can
engage in anonymous discussions, debates, and collaborations over the
course content. This will be monitored closely to provide guidance, and to
be in tune with common misconceptions that arise throughout the course.
Recommended texts
There is no prescribed textbook for this course because most electrical engineering
textbooks either focus on calculations or assume you already know most of
electrical engineering. What you need is the bridge in between these two ends, and
that is what this coursebook has been written for. That being said, books are nice,
and one good text is
Allan R. Hambley, Electrical Engineering – principles and applications; Any edition,
Pearson Education.
It covers most (but not all) topics we will visit and provides a nice balance between
computation and explanation of ideas.
Some of the course notes are also based on the following sources:
•
James Diefenderfer, Principles of Electronic Instrumentation; Second
edition, Holt-Saunders International Editions, 1979
•
Tony R. Kuphaldt, Lessons in Electric Circuits, Volume 1 - DC
This is a free series of textbooks on the subjects of electricity and electronics:
http://www.ibiblio.org/kuphaldt/electricCircuits/index.htm
•
Austin Hughes, Electric Motors and Drives: Fundamentals, Types and
Applications, Chapter 1; Fourth edition, Elsevier, 2013
Electrical and Digital Systems
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I have long held an opinion, almost amounting to conviction, in common I
believe with many other lovers of natural knowledge, that the various forms
under which the forces of matter are made manifest have one common
origin; or, in other words, are so directly related and mutually dependent,
that they are convertible, as it were, one into another, and possess
equivalents of power in their action.
Michael Faraday (1791 – 1867)*
Acknowledgements
A number of past and present colleagues, in particular,
Dr. Mark Andrews,
Prof. Gerard Rowe
Dr. Chris Smaill,
Prof. Udaya Madawala, and
Dr. Bill Collis
have influenced and contributed to the notes presented herein. The work and
effort of all these individuals is here gratefully acknowledged.
Copyright Notice
Material which is clearly indicated as owned by the University of Auckland may
only be used for "not-for-profit" educational purposes or private research and
study in accordance with the Copyright Act 1994, provided that textual and
graphical content are not altered and that the University's ownership of the
material is acknowledged. The University reserves the right to withdraw this
permission at any time. Reposting of course materials online without permission
of the copyright owner can result in copyright infringement. This can occur when
students post material belonging to faculty staff onto websites which make a profit
from on-selling course information. The permission to reproduce copyright
protected material does not extend to any material on this site that is identified as
being the copyright of a third party or individual staff members or students.
Authorisation to reproduce such material must be obtained from the copyright
owners concerned.
Michael Faraday was an English scientist who contributed to the study of electromagnetism and
electrochemistry. Among his major discoveries are the principles underlying electromagnetic
induction.
*
Electrical and Digital Systems
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The ELECTENG 101 Concept Map
Fundamental Electrical Quantities
Section 2 - Fundamental Theories
of Electrical & Digital Systems
Charge
Current
Voltage
Power
Tools of Circuit Analysis
Methods
Behaviour and Quantification of Electrical Quantities
Voltage and Current
Theorems
Power Metrics
Terminal Characteristic of Components
Equivalent
Resistance
Thevenin s
Theorem
Voltage &
Current Division
Principles
Signal
Characterisation
behaviour governed
and quantitifed by
Section 2.1
lead to the
development of
Resistor
Capacitor
used to
describe
Algebraic
Systems
Decimal
Boolean Algebra
Binary
Truth Table
Root Mean
Square Value
Inductor
Apparent Power
Superposition
Section 2.4
Positional
Numeral
Systems
Sinusoids
Kirchhoff s Laws
Ideal Wire
Section 2.3
Average
Power
Reactive Power
Ohm s Law
Reactive Elements
Ideal Sources
OPAMP
Hexadecimal
Section 2.5
Node-Voltage
Analysis
Power Factor
Section 2.2
underpins the
operations of
facilitates rapid analysis
and understanding of
Section 3 & 4 – Applications of the
Fundamental Theories of Electrical & Digital
Systems
Mathematical Models
used to design
and model
Section 3.2
Systems of Monitoring, Analysis, and Response Technology (SMART)
Section 3.1
Section 3.3
Monitoring Technologies
Sensors
Analysis Technologies
Analog Signal
Conditioning
Circuits
Sensor Interface
Analog & Digital
Processing Circuits
Electrical Grid
Section 3.4
Section 4.1
Section 4.2
Response Technologies
Generation
Systems
Transmission &
Distribution
Systems
Actuators
Strain Gauge
Voltage Divider
analysed by
Loading Effects
Thermometer
Thermistor
Analog-toDigital
Converters
DC Motors
Wheatstone
Bridge
OPAMP circuits
delivered
through
powered by
controls
Three-Phase
Electricity
Transformer
Logic Gates &
Circuits
Resistive
Electrical and Digital Systems
Dr. William Lee ELECTENG 101
vi
2
1.1
“SMART” Engineering................................................................. 2
1.2
Signals and Systems ..................................................................... 3
1.3
Dealing with Complex Systems .................................................. 5
Fundamentals of Electrical and Digital Systems ...... 7
2.1
Circuit Concepts............................................................................ 7
2.1.1. Charge and Current ......................................................... 8
2.1.2. Voltage and Common Reference ................................. 12
Week 1 – 2
1
Table of Contents
Introduction .................................................................... 1
2.1.3. Resistance and Ohm’s law ............................................ 15
2.1.4. Voltage and Current Sources ....................................... 20
2.1.5. Power ............................................................................... 23
2.1.6. Kirchhoff’s Current and Voltage Laws ....................... 26
2.1.7. Problems.......................................................................... 30
2.2
Methods and Theorems of Circuit Analysis ........................... 33
2.2.2. Voltage and Current Division Principles ................... 40
2.2.3. Node-Voltage Analysis ................................................. 44
2.2.4. Superposition Theorem................................................. 49
Week 2 – 3
2.2.1. Equivalent Resistance .................................................... 34
2.2.5. Thevenin’s Theorem ...................................................... 53
2.2.6. Problems.......................................................................... 60
2.3
Number Systems ......................................................................... 63
2.3.1. Standard Bases ............................................................... 63
2.3.3. Representation Limits ................................................... 69
2.3.4. Problems.......................................................................... 73
2.4
Boolean Algebra .......................................................................... 75
2.4.1. Basic Boolean Operations ............................................. 75
Week 7 – 8
2.3.2. Conversion between Bases ........................................... 65
2.4.2. Boolean Identities........................................................... 78
2.4.3. Converting Truth Tables to Boolean Equations ........ 82
2.4.4. Problems.......................................................................... 84
2.5
Alternating Current (AC) Concepts ......................................... 87
2.5.2. Average and Root Mean Square (RMS) Value........... 91
2.5.3. Capacitor and Inductor ................................................. 95
Week 11
2.5.1. Sinusoidal Representation ............................................ 88
2.5.4. Power Characterisation ................................................. 98
2.5.5. Problems........................................................................ 109
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“SMART” Engineering ............................................. 111
3.1
Monitoring Technologies ......................................................... 113
3.1.1. Resistive Input Transducer ......................................... 114
3.1.2. Sensor Interface Circuitry ........................................... 119
Week 4 – 5
3
3.1.3. Problems........................................................................ 129
Analysis Technologies – Signal Conditioning ...................... 131
3.2.1. Equivalent Circuits and the Loading Effect ............. 133
3.2.2. Amplifiers ..................................................................... 136
3.2.3. Operational Amplifiers ............................................... 141
3.2.4. Practical OPAMP Circuits .......................................... 143
Week 5 – 6
3.2
Analysis Technologies – Decision Making ............................ 159
3.3.1. Logic Gates.................................................................... 160
3.3.2. Analog to Digital Conversion .................................... 167
3.3.3. Problems........................................................................ 177
3.4
Response Technologies ............................................................ 181
3.4.1. DC Motors ..................................................................... 182
3.4.2. Problems........................................................................ 189
Electricity Supply ....................................................... 191
4.1
Electricity Generation ............................................................... 193
4.1.1. Three-Phase Electricity ................................................ 194
4.1.2. Problems........................................................................ 201
4.2
Electricity Transmission and Distribution ............................ 203
Week 11 – 12
4
Week 11
3.3
Week 9 – 10
3.2.5. Problems........................................................................ 153
4.2.1. Transformers ................................................................ 204
4.2.2. Problems........................................................................ 210
Electrical and Digital Systems
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Electrical and Digital Systems
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Introduction
The study of electrical and digital systems is important because electrical
technology is utterly pervasive and permeates everyday life to the extent that we
are rarely conscious of its effects unless a device or system fails. Electrical
technology has transformed human civilisation, improved quality of life, extended
lifetimes, and has progressively reduced the burden of human and animal labour
throughout the world. The revolution that started over a century ago with the
telegraph and electric light bulb – and now gives us autonomous robots on other
planets – shows no sign of slowing down and seems to be limited only by the
imagination of engineers.
Physicists in the 19th century greatly expanded our understanding of electrical
phenomena, such as the origin of electrical charge, electric and magnetic fields and
how they interact. Consequently, we have very precise ways to measure quantities
such as current and voltage. We also understand how to move electrons efficiently
from one location to another, how to accumulate charge and regulate the flow of
electrons – all through the use of electrical circuits.
The simplest electrical circuits are constructed from components you may already
be familiar with: batteries (to supply current and voltage), wires (metal conductors
to deliver the current or voltage), resistors (to impede the flow of current and
generate desired voltage), inductors (coiled conductors to intensify the magnetic
field and store energy in the magnetic field), capacitors (flattened conductors that
concentrate an electric field and store energy in the electric field).
V
R
C
L
Because we are so very adept at controlling the voltages and currents in a circuit,
it turns out to be extremely useful to use electrical quantities as stand-ins (or
analogues) for other physical quantities (such as temperature, mechanical strain,
velocity, etc.) and manipulate the analogue quantity (that is, voltage or current)
instead of the physical quantity itself.
Note that this means we must have a mechanism for converting or transducing the
physical quantity into an electrical analogue (and possibly vice versa). A simple
example might be a “smart” heating system: A thermostat is a very simple
temperature sensor that outputs a voltage to indicate “too cold”, “about right”,
and “too hot”. This output voltage can then be used to control the voltage applied
to resistive heating elements producing heat to achieve the desired temperature
automatically without manual intervention.
Dr. William Lee
Page 1
This simple self-regulating heating system reveals the internal workings of almost
all modern electrical and digital systems that appears to have some perception of
automation and intelligence – they consist of technologies that
•
monitor the environment and convert the physical quantities of interest into
electrical analogues;
•
analyse the electrical analogues for contextual interpretations to make
informed decisions;
•
respond appropriately to the situation based on the gathered information to
achieve pre-determined purposes.
The practice of modern electrical, computer systems, and software engineering
therefore revolves around the study, design and engineering of these electrical and
digital Systems of Monitoring, Analysis, and Response Technologies.
1.1
“SMART” Engineering
While electrical systems can be extremely varied in nature, but a useful block
diagram that applies to a wide range of realistic systems is shown below.
Input 1
Systems of Monitoring, Analysis, and Response Technologies
Monitoring Technologies
Input 2
Input 3
Sensors
Sensor
Interface
Input 4
(feedback)
Analysis Technologies
Signal
Conditioning
Decision
Making
Output 1
Response Technologies
Visual
Display
Output 4
(feedback)
Actuators
Output 2
Output 3
Monitoring Technologies
The monitoring technologies sense (or measure or sample) the environment in
which the circuits exist. For certain inputs this may require transduction
(converting a physical quantity into an analogue electrical quantity) or it may take
direct electrical input (for example, from another circuit).
Analysis Technologies
The analysis technologies take the electrical output(s) from the monitoring
technologies and process them in some way. For example, an audio amplifier may
take the millivolt-level output from a microphone and amplify the voltage to
several tens of volts required by the audio speaker.
In more elaborate systems (typically digital systems) the circuitry may respond
differently depending on the state of one or more inputs and the current state of
the processing circuitry. The system is therefore making decisions based on the
current and past inputs. This is an extremely important concept and underpins all
computer-systems thinking.
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Dr. William Lee
Introduction
Response Technologies
The response technologies generate the action or response (if any) to the inputs as
determined by the processing circuitry. In its simplest form, the action might be to
turn on a light or display text on an LCD panel. More sophisticated actions might
entail detailed behaviour such as when a mobile robotic device has plotted a
trajectory through a cluttered environment.
Frequently, the action itself is sensed so that the system can determine that the
action has actually taken place or to the desired precision. The general notion of
an output being measured as another input to the system is called feedback and it
turns out to be a very powerful and important tool in electrical engineering.
For example, when a car’s cruise control is set to 80 km/h the cruise control
computer needs to know if the target speed of 80 km/h has been achieved so that
the car’s acceleration can be changed accordingly. Accelerating from 50 km/h to 80
km/h requires different amount of engine torque depending on whether the car is
travelling on the flat or up an incline.
1.2
Signals and Systems
In the previous discussions we have casually referred to “circuits”, “amplifiers”,
and “computer systems” without being very precise about the differences. There
are many ways to classify different types of circuits, together with their inputs and
outputs, but the most important distinction is between continuous-time and
discrete-time.
General Signals
A signal is the generic term applied to any quantity that changes with time.
Voltages and currents are therefore signals and it is often handy to refer to “the
signal” without necessarily specifying if it is a current or voltage. A physical
quantity prior to being sensed is also a signal. For example, we may refer to “the
temperature signal”.
Continuous-time Signals
A continuous-time signal is one in which the signal is defined at all possible times 1.
For example, the voltage signal
0F
𝑣(𝑡) = 100 cos(2𝜋50𝑡) V
is a 50 Hz sinusoid with a peak value of 100 V. We can substitute any value of 𝑡
into the expression for 𝑣(𝑡) and get a sensible value. A continuous-signal is also
commonly referred to as an analog signal.
Note that this doesn’t mean that the signal is continuous. Rather, the signal is defined at a continuum
of times and we could, in principle, measure or know the value of the signal at any instant of time.
1
Dr. William Lee
Page 3
Discrete-time Signals
A discrete-time signal is one in which the signal is defined only at certain instants
of time. For example, the voltage signal
𝑣[𝑛] = 100 cos (2𝜋50 ⋅
𝑛
)V
100
(where 𝑛 is an integer) traces out the shape of a 50 Hz sinusoid with a peak value
of 100 V. However, because 𝑛 is an integer, we only “know” the signal at the times
𝑡𝑛 = 𝑛⁄100 = 0.01𝑛 (that is, every 10 milliseconds). In a minor abuse of
terminology, we will refer to a discrete-time signal as a digital signal.
Digital signals arise in practice because we measure or sample an analog signal at
discrete instants in time to process them in a digital circuit 2. This is referred to as
analog to digital conversion or ADC 3.
1F
2F
v[n]
v(t)
100V
100V
0.01
0.02
0.03
t
0.01
0.02
0.03
t
−100V
−100V
Analog Signal
Digital Signal
General Systems
A system is the generic term applied to the mechanism that is acting on the inputs
to produce an output. A simple circuit containing two resistors connected as a
voltage divider is therefore a system. The term system can also be used quite
broadly. For example, a high-level computer program that scans a text file looking
for a search phrase is also a system in the sense that we have defined it.
• An analog system or analog circuit is any system that processes analog signals.
• A digital system or digital circuit is any system that processes digital signals.
What is the “original” signal doing between, say, 𝑡1 = 0.01 ms and 𝑡2 = 0.02 ms? The answer is: we
don’t know! However, it is an amazing fact that provided we take “enough” measurements we don’t
need to know the signal value between sampling instants.
3 We can also go back the other way. Starting with a stream of samples we can generate an analog
voltage. Typically, the analog voltage is held constant until the next sample is available, producing a
type of staircase effect.
2
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Dr. William Lee
Introduction
1.3
Dealing with Complex Systems
The design of electrical and digital systems is a well-understood process and as
time has progressed the complexity of these systems has increased immensely. The
microprocessor is one example where the complexity (defined in this case to be the
number of transistors, or switches, on a single chip) is now enormous. The original
Intel 4004 microprocessor released in 1971 contained 2,300 transistors. At the time
this was a massive achievement and this level of device density (or integration) was
unprecedented.
Since 2010 the leading-edge designs have never had fewer than a billion switches.
For example, the Intel 10-core Xeon released in 2012 has approximately
2,600,000,000 switches and even this isn’t the most highly integrated device on the
market. Transistor counts alone are not a very good measure of complexity since a
microprocessor has a very regular architecture with many duplicated structures.
Nevertheless, how are engineers supposed to deal with this level of detail?
The key to dealing with complex systems is to compartmentalise a complex system
into many smaller sub-systems (indeed, the overall system is designed by
combining many smaller systems in the first place). Each sub-system is simple
enough to be understood by an average engineer and at no time is it necessary to
grapple with all details of the complete system. This keeps everyone sane.
The circuit diagram below shows the internal structure of the LM741 operational
amplifier – one of the workhorses of analog circuit design. It looks rather
complicated and a bit intimidating, but the dotted colour boxes delineate subsystems that are actually easy to understand. By thinking about how the subsystems interact (and not their internal components) it’s a much easier task to
figure out how the entire circuit works. As you will soon begin to see throughout
this course that once we have figured out how a system functions, we can abstract
away the complexity, reducing it down to an equivalent circuit modelling its
external electrical behaviour that is easy to understand – we no longer need to
worry about why it behaves the way it does – and focus our attention purely on its
engineering implications
Dr. William Lee
Page 5
What we see physically:
Non-inverting Inverting
Input
Input
VS+
Output
1
4
3
2
5
6
7
Offset Null Offset Null
VS−
What is inside internally:
Non-inverting
Input, v +
Inverting
Input, v
VS+
4.5kΩ
39kΩ
30pF
Output, vO
25Ω
7.5kΩ
25Ω
50kΩ
1kΩ
Offset Null
1kΩ
5kΩ
50kΩ
50Ω
Offset Null
VS
What we use to model its electrical behaviour externally:
v+
RO
RIN
A(v+
vO
v
v
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Dr. William Lee
2
Fundamentals of Electrical and Digital Systems
The study of electrical and digital systems revolves around the manipulation and
interpretation of the electrical quantities voltage and current. We do so by cleverly
altering, suppressing, and amplifying these quantities through electrical circuits
containing electrical elements such as resistors, capacitors, inductors, etc. While
the manipulation is standardised (via circuits), the meaning and purpose of these
signals, however, depends completely on what we want them to be.
For example, the output voltage of a thermistor may be fed into a microcontroller
to drive a seven-segment display as a direct indication of the temperature, but it
could also be used as the input of a regulatory system to control the current
through its resistive heating elements thereby creating a self-regulated heater 4. In
both cases, the voltage signal (the electrical analog of temperature) is processed
universally via (different) circuits, but the former decides the on/off state of the
display LEDs, while the latter controls the magnitude of the heating current.
3F
On the other end of the spectrum, the lowest level of the Internet transports vast
quantities of numbers from one location to another in the form of voltages and
currents. The transmission and reception of these signals are uniformly achieved
via wired(less) transceivers (which are just electrical circuits), but the meaning of
those numbers is up to us – they could represent an image, an audio, an e-mail, or
an actual number, etc.
Knowledge of electrical technology is vital for all disciplines of engineering
because it is simply not possible to do any engineering without encountering
electrical principles. But in order for us to control, utilise, and ultimately engineer
electrical and digital systems, we must first develop an appreciation of (1) what
these underlying analogue quantities mean, (2) how they behave in an electrical
circuit, and (3) the tools necessary to analyse them.
2.1
Circuit Concepts
At the heart of every electrical circuit is a collection of electrons. These electrons
are harnessed and energised through various engineering methods (i.e., electrical
sources). By providing a conducting path (e.g., wires) toward other electrical
devices (e.g., resistors, capacitors, inductors, etc.), the resulting flow of electrons (i.e.,
current) enables the delivery of energy to these devices over time. Through careful
monitoring of the energy changes in the electrons (i.e., voltage) across these devices,
we are able exploit, and take advantage of, these devices’ electrical behaviours to
store energy, accumulate charges, control the current, regulate power, and in doing
so, produce useful circuits that perform meaningful tasks.
It is important that we understand, and become fluent with, the terminologies and
nomenclatures we use in describing the various constituents, quantities, and
The idea of transducing a physical quantity into an electrical analogue (and vice versa) forms the
basis behind all disciplines of engineering due to the pervasiveness of electrical technology.
4
Dr. William Lee
Page 7
Circuit Concepts
phenomena pertaining to an electrical circuit for they form the language of
everything we practice and do in electrical engineering – it is simply not possible to
do any engineering without a concise way of articulating what we wish to convey.
Section 2.1 Concept Map
Fundamental Electrical Quantities
Voltage and Current
Current
used to derive
Charge
behaviour
governed by
Resistor
underpins
Voltage
Power
Kirchhoff s Laws
Ideal Wire
Ohm s Law
Ideal Sources
Component Characteristics
Behaviour and Quantification of Electrical Quantities
2.1.1. Charge and Current
Intended Learning Outcomes
•
•
Be able to recognise, and make use of, the relationship between charge and
current.
Be able to recognise the difference between conventional current and electronflow current pertaining to the modelling of electric circuit behaviour.
Electrostatic force is a fundamental force in nature that we have attributed its
origin to electric charge (symbol 𝑞 or 𝑄), a physical property of matter that causes
all matters to experience (and exert) such a force when placed next to another.
Through experiments, we have discovered that there are two types of charges –
like ones repel, and opposites attract – which allowed us to define the notion of
positive and negative charge. While we now have developed a model that describes,
quite accurately, the behaviour of these charges in terms of the electrostatic force
they exert upon each other, namely 5,
4F
𝐹∝
𝑞1 𝑞2
,
𝑟2
what exactly are they is still a mystery to some extent today. We do know that the
source of negative charge is the electron, but we have not completely narrowed
down the origin of positive charge 6. Thus, based on what we can observe, we have
associated the ‘flow of charges’ to the movement of electrons.
5F
Surprisingly, no one knows what an electron is to date, but we only know how it
behaves. Its behaviour has led us to link what we do know in terms of the
conservation of energy to how and why an electrical circuit work the way it does
– by clever means of working against the electrostatic forces, physicists discovered
This is known as the Coulomb’s law which states that the electrostatic force 𝐹 between two charges
is proportional to the magnitudes of the charges (𝑞1 , and 𝑞2 ) and decreases with distance 𝑟.
6 We do know that protons have positive charge, but these are not the only subatomic particles to
possess it. Positrons which is similar to electrons in all characteristics except its behaviour is opposite
is known to exist but hard to observe.
5
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
ways to separate electrons from positive charges 7, and in doing so the electrons
gain energy which can be transported to electrical devices through conducting
materials for them to make use of.
6F
Electrons returns to its energy neutral state
1
2
Work done by the battery
in separating the charges
Electrons gains energy, and want
to return to a charge neutral state
5
Work done by the electrons in
flowing through, thereby
transferring energy to the device
Device
4
Conducting material
provides an alternative path
for the electrons to do so 3
The amount of charge a matter has is measured in Coulombs (symbol C), and the
smallest free-standing particle known to date that possess charge is the electron
which has approximately (−)1.6 × 10−19 C of charge. In other words, 1 C of charge
is equivalent to approximately 6.3 × 1018 electrons.
Current
A current (symbol 𝑖 or 𝐼) is the number of charges flowing through a particular
point in one second. In other words, it measures the rate at which electric charges
pass through a section of an electric circuit at any given time, and this is denoted
mathematically by the derivative of the total number of charges 𝑞 passing the given
section over time 𝑡,
𝑑𝑞
𝑖≝ .
𝑑𝑡
Currents thus have the units Coulombs per second (Cs −1 ), but commonly we use
Ampere (symbol A).
We are generally more interested in the rate at which charges flow (i.e., current)
through an electrical device at a given instance in time rather than the total number
of charges that have gone by the device up to that time because the rate gives us
an indication of the electrical ‘movement’ that is happening in the device and
therefore tells us if the device is ‘working’. This in turn reveals how fast energy is
being absorbed (or delivered) by the device, which we are usually more interested
in, as opposed to how much energy has the device received (or supplied) 8, this is
because all electrical devices require a continuous supply of energy to function.
7F
E.g., through electrochemical reactions such as that inside a battery, or through rotating a wire in a
magnetic field such as that of a generator.
8 The obvious exception is when we have limited total energy capacity such as when considering that
of a practical battery. Most of the time, however, when we are interested in the behaviour of electrical
devices, we would assume it can be powered indefinitely by an unlimited source of energy (e.g., from
the mains supply).
7
Dr. William Lee
Page 9
Circuit Concepts
Conventional Current
Historically, it was once believed that the flow of charges was attributed to positive
charge ‘carriers’, and consequently the conventional current actually described the
rate at which these positive charge carriers flow. Much of the inventions developed
in electrical engineering went on following such a notion, and regrettably we
ended up with theories and notations all based on the conventional current 9.
8F
While we now know this is incorrect – it is the electron flow that occurs in the
conducting material – fortunately since the movement of an electron in one
direction can be viewed relatively to be a displacement of a ‘lack of an electron’ in
the opposite, other than the direction of charge flow, adhering to the conventional
current had no impact on the theories developed in terms of how devices behave.
Therefore, given the pervasiveness of our innocent error, the conventional current
remains in use today in describing all electrical operations.
Conventional Current
A
Actual (electron) flow
I=4A
B
Device
D
C
We denote (conventional) current in circuits by using arrows along the wires in
the direction of which (positive) charge flows.
Sometimes we may also use a double-subscript notation to indicate the direction
the charges are flowing in the order of from and to. Thus, for the circuit shown
above, we can also denote the 4A clockwise current by
𝐼𝐴𝐵 = 𝐼𝐵𝐶 = 𝐼𝐶𝐷 = 𝐼𝐷𝐴 = 4 A.
Example 1.
A current of 0.5 A flows through the light bulb in a torch for 8
minutes. Find the total number of charges that passed through the
light bulb in this time. How many electrons are needed to carry this
charge?
(ANS: 240 Coulombs, ≈ 1.5 × 1021 electrons)
For example, the symbol for the diode, an electrical device that allows current in one direction but
blocks in the other, takes on an arrowhead that points in the direction of conventional current.
Similarly, the transistor, an integral building block of all digital logics, has also inherited such a
notation in its symbolic representation.
9
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 2.
(NAQTQO)
The total amount of charge, 𝑞(𝑡), that has
passed through a circuit element over
time is shown. The current through this
element at the instant 𝑡 = 2 s is
(a) 0 A
Example 3.
(b) 1 A
(c) 2 A
q(t)
2C
1C
1
2
3
t (s)
(d) 3 A
If the plot in Example 2 is that of the current, 𝑖(𝑡), through the circuit
element instead, what would be the total amount of charges that has
passed through the element at 𝑡 = 2 s ?
(ANS: 3 Coulombs)
Dr. William Lee
Page 11
Circuit Concepts
2.1.2. Voltage and Common Reference
Intended Learning Outcomes
•
•
Be able to relate and interpret the meaning of voltage in an electric circuit to
the energy characteristic of the underlying charges in the circuit.
Be able to represent the changes in the energy of the charges across electrical
components using the notion of voltage and voltage arrows.
A voltage (symbol 𝑣 or 𝑉) is a mathematical difference, it is the difference in energy
(of each unit of charge) between two points. As such, a voltage is also commonly
known as a potential difference and is measured in units of Joules per Coulomb (JC −1)
but more commonly we use Volt (symbol V).
Because it is a difference (i.e., the result of a subtraction) between two points, for a
voltage value in a circuit to make sense we need to know which of the two points
the difference is taken with respect to. For example, the figure below shows two
ways in which we can calculate the voltage across a resistor:
Each charge here has 9 J
of energy
Each charge here has 4 J
of energy
A
The voltage across the resistor is 9 − 4 = 5 V
OR
B
The voltage across the resistor is 4 − 9 = −5 V
•
when we say the voltage across the resistor is 5 V, we mean one end (in this
case, 𝐴) of the resistor has 5 J of energy more than the other (i.e., 𝐵), and
•
likewise, we can also say the voltage across the resistor is −5 V, implying one
end (in this case, 𝐵) of the resistor has 5 J of energy less than the other (i.e., 𝐴).
Both statements are correct, and such a paradox arises because voltage values
themselves do not tell us which of the two ends the difference is calculated with
respect to: all we know is that the energy (of each unit of charge) has changed by 5 J
between the two resistor ends, but not which end this change is measured from.
To rid of this ambiguity in defining voltages,
every voltage value must be accompanied by a voltage arrow 10 with the arrow tail
indicating the end the difference is taken with respect to11.
9F
This nomenclature makes sense because the result of this difference tells us how
much higher the arrowhead end is relative to its tail end12. The arrow therefore
points in the direction of the assumed potential rise associated with the difference.
An alternative is the use of positive/negative sign pair as used in defining voltage sources –
negative denotes the end the difference is taken with respect to.
11 In other words, the tail end is the subtrahend, and the head end is the minuend of the subtraction.
12 If 𝑥 − 𝑦 = 𝑑 then 𝑥 = 𝑦 + 𝑑, and 𝑥 is therefore higher than 𝑦 by 𝑑.
10
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
A
A
𝑉𝐵𝐴 = 4 − 9 = −5 V
𝑉𝐴𝐵 = 9 − 4 = 5 V
B
B
(a)
(b)
We say the potential at 𝐵 is −5 V
“higher” than the potential at 𝐴.
We say the potential at 𝐴 is 5 V This means the potential at 𝐵 is 5 V
higher than the potential at 𝐵.
lower than 𝐴, in agreement with (a).
Both cases now identify the voltage across the resistor unanimously – a negative
voltage simply means the true polarity (i.e., potential rise) is opposite to what was
assumed. We say “the voltage at 𝐴 with respect to 𝐵 is 𝑉𝐴𝐵 .”, and clearly 𝑉𝐴𝐵 =
−𝑉𝐵𝐴 . Notice voltage values are relative and not absolute – we no longer know the
exact potential at 𝐴 or 𝐵 given the voltage across the resistor 𝑉𝐴𝐵 or 𝑉𝐵𝐴 .
Common Reference or Ground
Often, voltages are defined or measured in a circuit with respect to a common
reference point (this could be across multiple components) which we denote as
“ground” in the circuit and is represented by the symbols 13:
10F
“Ground” is simply a symbolic indicator for a common reference point in a
circuit from which we can define voltages “at”14 various points in the circuit by
measuring them with respect to this reference point.
This reference is often “understood” and so we omit its subscript when indicating
voltages that are measured with respect to it:
3V
A
𝑽𝑨𝑫 𝐨𝐫 𝐬𝐢𝐦𝐩𝐥𝐲 𝑽𝑨
𝑉𝐴𝐵
B
7V
10 V
𝑽𝑩𝑫 𝐨𝐫 𝐬𝐢𝐦𝐩𝐥𝐲 𝑽𝑩
D
6V
𝑉𝐵𝐶
C
1V
𝑽𝑪𝑫 𝐨𝐫 𝐬𝐢𝐦𝐩𝐥𝐲 𝑽𝑪
The “triangle ground” is used as the ground for digital circuits while the “line segment ground” is
used for analog circuits. We won’t make such a distinction and will assume they mean the same.
14 Voltage is a relative quantity between two points, it makes no sense to say the voltage “at” a point
without an implied point of comparison.
13
Dr. William Lee
Page 13
Circuit Concepts
Thus, we commonly abuse the notation and say that the voltage “at” 𝐴 is 10 V, “at”
𝐵 is 7 V, and “at” 𝐶 is 1 V, and we can say so because those voltages are
understood to be relative to the node designated as ground:
𝑉𝐴 = 𝑉𝐴𝐷 = 10 V,
𝑉𝐵 = 𝑉𝐵𝐷 = 7 V,
𝑉𝐶 = 𝑉𝐶𝐷 = 1 V, and 𝑉𝐷 = 𝑉𝐷𝐷 = 0 V 15.
Notice the placement of ground is purely arbitrary and does not affect the actual
voltage across (nor the current through) the components – the circuit has not
changed, only the reference at which point voltages are measured relative to has.
If the ground in the previous circuit is placed at 𝐶 instead, then the value of the
voltage “at” each labelled points is now the difference (in energy of the charges)
between those points and the new reference point 𝐶, and these are
𝑉𝐴 = 𝑉𝐴𝐶 = 9 V,
Example 4.
(LAZOXL)
𝑉𝐵 = 𝑉𝐵𝐶 = 6 V,
𝑉𝐶 = 𝑉𝐶𝐶 = 0 V,
and 𝑉𝐷 = 𝑉𝐷𝐶 = −1 V.
The voltage across each of the two electrical
components 𝑅1 and 𝑅2 connected end-to-end is
shown. The voltage across the two components 𝑅1
and 𝑅2 , i.e., 𝑉𝐴𝐶 is
(a) −5 V
(b) −1 V
(c) 1 V
(d) 5 V
A
R1
V
B
R2
V
C
This is where the aphorism “ground is a point taken/defined to be 0 V” comes from. It is not 0 V
because we “defined” it per se, but because the voltage “at” ground is measured with respect to itself,
and so this difference must be zero (but the actual energy of the charges there may very well be nonzero).
15
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
2.1.3. Resistance and Ohm’s law
Intended Learning Outcomes
•
•
•
Be able to interpret and relate Ohm’s law to the energy changes in the charges
through a resistive conductor in terms of the voltage across, current through,
and the resistance of the material.
Be able to recognise the characteristics of an ideal conductor and infer the
implications of these characteristics on its behaviour in an electric circuit.
Be able to determine relevant electrical quantities pertaining to a resistor in an
electric circuit in terms of its voltage, current, and resistance.
When an electrical charge flow through a conducting material (e.g., metals), it must
necessarily lose energy as it repels and collides with other charges (of the material)
in between. The amount of energy lost is subject to how ‘difficult’ it is for the
charges to flow through the material which, intuitively, must depend on both the
chemical (e.g., stability of valence electrons) and physical (e.g., length, crosssectional area, temperature) properties of the conductor itself.
Larger cross-sectional area, less energy lost
Cu
valence electron loosely
held, less energy lost
Cu
Cu
Cu
Cu
Longer length, more energy lost
From empirical evidence it was observed that the amount of energy lost through a
conductor is directly proportional to the rate at which the charges are flowing
through it – the larger the amount of charge is forced through a conductor at a
given instance, the larger the amount of energy is lost by each charge as they
propagate through, and vice versa. Given our understanding of the meaning of
voltage and current we therefore have a very concise mathematical way of
describing this observation via the well-known Ohm’s law:
⏟
𝑣
= difference in energy across the conductor
=
𝑅
⏟
= constant of proportionality
×
⏟𝑖
= rate of charge flow
The constant of proportionality, 𝑅, in Ohm’s law is known as the resistance of the
material whose value provides a measure of its ability to impede or resist the flow
of charges (i.e., current), summarising its intrinsic and extrinsic properties.
From our understanding of voltage and current, a voltage only make sense if it is
accompanied by a voltage arrow, likewise a current only make sense with an
indicated direction. This means there must be an inherently assumed voltage
polarity and current direction associated with Ohm’s law (which the symbols
themselves cannot convey):
Ohm’s law takes the current to be in the direction of a voltage drop.
Dr. William Lee
Page 15
Circuit Concepts
Such a convention makes intuitive sense since charges can only lose energy
through a resistor 16. Note that this also means if the current is taken to be in the
direction of a voltage rise (or vice versa), then Ohm’s law still applies, but we must
include a negative sign to reflect reality:
1F
i
i
R
v = Ri
R
v = Ri
From its origin, the unit of resistance is thus Volts per Ampere (VA−1 ) but more
commonly we simply use Ohm (symbol Ω). Ohm’s law makes intuitive sense
because
•
For a fixed amount of charge passing through the material at a given
instance (current), a higher resistance in the material means a larger
amount of energy must be lost (voltage) by the charges as they traverse
through, and vice versa
•
For a fixed amount of energy lost in the charges passing through the
material (voltage), a higher resistance in the material means a smaller
amount of charge must be flowing through at that instance (current), and
vice versa
Resistor
While the resistance of a material might seem like a non-ideal consequence of the
underpinning physics governing the movement of charges, the ingenuity in
electrical engineering lies exactly in exploiting such an undesirable characteristic of
the real world and using it to our advantage in achieving useful purposes. Indeed,
through careful manufacturing methods, we can (and do) intentionally produce
conductors of specific resistances with the intention of using them to do
meaningful tasks. These ‘resistive’ conductors are also known as resistors, and we
represent them symbolically by:
R
R
Fixed Resistors
R
R
Variable Resistors
It is important that we understand how the voltage across and current through a
resistor behave for they are one of the fundamental building blocks of electrical
circuits, and familiarity with its behaviour will allow us to develop an appreciation
for how they can be used to design and engineer useful electrical circuits.
This is also known as the passive-sign convention and, in fact, is used by electrical engineers to define
the voltage-current behaviours (i.e., terminal characteristics) of all electrical elements (e.g., capacitors,
inductors, diodes, etc.) – they all, by default, taken to receive energy (from the charges).
16
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 5.
(NMHPNF)
Example 6.
(ZLPJPF)
The direction and magnitude of the current through
the resistor in the portion of the circuit shown is
(a)
1 A downwards
(b)
2 A upwards
(c)
2 A downwards
(d)
3 A upwards
2V
2Ω
6V
In the circuit shown, the current 𝑖 through the 2 Ω
resistor in the indicated direction is
(a) −3 A
(b) −1 A
(c) 1 A
(d) 2 A
V
i
2Ω
4V
Dr. William Lee
Page 17
Circuit Concepts
Ideal Wire
An ideal wire is a perfect conductor that has zero resistance, the most desirable
property that we would like an electrical wire to have – charges can freely flow
through an ideal wire without any loss in energy as Ohm’s law would suggest:
𝑅
⏟
×
=0 Ω
⏟
𝐼
=
irrespective of the size of current
0⏟V
.
no difference in 𝐸 𝐚𝐜𝐫𝐨𝐬𝐬 the ideal wire
Observations:
•
the difference in energy across an ideal wire is zero regardless of the amount
of charge flowing through it at any given instant in time. Said differently,
•
the charges anywhere along an ideal wire must have the same amount of
energy so that the difference between any two points along the wire is zero.
The implication of the observations above is that
the current through an ideal wire depends on the circuit that it is connected in, it
cannot be determined by simply examining the ideal wire on its own.
This make sense because irrespective of the current through the ideal wire, the
voltage across it is always zero (since 𝑅 = 0 Ω) and by looking at such an isolated
view there is insufficient information to determine the size of this current 17 . This
contrasts with that of a resistor – if the voltage across a resistor is zero, then the
current through the resistor must be zero (since 𝐼 = 0⁄𝑅 = 0 for any 𝑅 ≠ 0 Ω).
12F
0V
Ideal wires
1A
5V
5V
5Ω
0V
It should be clear that the current through the ideal wires is 1 A, determined by
the circuit completed by the ideal wires, but not the wires themselves: The 5 V
source ensures that each unit of charge at the positive terminal carries 5 J of energy
more than the negative. The ideal wires enable the charges to travel to and from
the 5 Ω resistor without losing any energy which ensures that there is also a
voltage of 5 V across the resistor. By Ohm’s law, the current through the 5 Ω
resistor is 1 A which must have come through via the ideal wires.
While all real wires have some non-zero resistances, in practice these are negligibly
small 18 in comparison to the resistances of other circuit components, and we can
therefore approximate them to be ideal to facilitate rapid analysis. Of course, when
it does matter, we may simply model the real wire as a resistor.
13F
Attempting to apply Ohm’s law to an ideal wire will result in 𝐼 = 𝑉⁄𝑅 = 0⁄0 A an indeterminate
form. This is a whimsy way of the mathematics telling you there is insufficient information to deduce
the current by just examining the piece of ideal wire.
18 For example, a typical copper wire has a resistance of ≈ 5 mΩ per metre at room temperature.
17
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 7.
(RAXOTE)
Example 8.
(EKEBPD)
Dr. William Lee
The magnitude of the current 𝑖 through
the ideal wire in the portion of the circuit
shown is
(a)
0A
(b)
1A
(c)
2A
(d)
3A
The voltage, 𝑉𝐴𝐵 , across the open
switch in the circuit shown below is
(a)
0V
(b)
3V
(c)
6V
(d)
12 V
2Ω
i
3A
3V
3Ω
3Ω
12 V
A
B
VAB
3Ω
Page 19
Circuit Concepts
2.1.4. Voltage and Current Sources
Intended Learning Outcomes
•
•
Be able to recognise the characteristics of an ideal voltage and current source
and infer the implications of these characteristics on their behaviour in a
circuit.
Be able to determine voltage across and current through an ideal source given
knowledge of its surrounding electrical behaviour.
An electrical source is a device that usually 19 powers, or provides energy to, an
electrical circuit. The most ideal characteristic that we would like a source to have
is the ability to maintain its intended operating behaviour with zero variability and
be able to do so indefinitely. In other words, ideal sources are those that can supply
(or receive) infinite amount of energy without any deviation from their
specifications regardless of what they are connected to. Of course, such a
behaviour is purely fictional in practice – all real sources have finite energy
capacities and will necessarily deviate from their rated output over time. Not only
so, depending on what these sources are connected to, their output may also vary
due to internal imperfections, and electrical noise. We are nonetheless interested
in these fictitious ideal sources for several reasons:
14F
•
In most well-designed engineering applications, the sources are designed to
operate within their rated limits, we are thus less concerned by their finite
limitations within the scale in which they operate 20 and may therefore be
15F
assumed ideal
•
Ideal sources provide very good approximations to most practical
behaviours and facilitate rapid analysis that would otherwise result in
analytical inconvenience with minimal gain, and if a design requires a more
accurate description, we may simply refine our initial estimates through
more in-depth models should the need arise
•
One of the most powerful theorems in electrical engineering (see Sect. 2.2.5
later) allows us to reduce almost all circuits to simply an ideal source with a
resistor, abstracting the complexities away from the analysis when they are
not of the focus
There are two types of electrical sources that are commonly used to model, and
describe, almost all phenomena that supply energy: the ideal voltage source and the
ideal current source.
In circuits containing multiple sources, they could also receive energy from other source(s)
depending on the circuit configuration and arrangement. For example, the battery in your mobile
phone provides energy to the underlying circuitry but it also receives energy from the mains supply
when ‘plugged in’.
20 For example, the bench power supplies in the laboratory are practically ideal for the sort of smallscale electronics that we would connect to it. But, of course, their energy capacity is limited by the
reserve of hydro-electricity from the mains supply which, despite being ‘enormous’, is finite.
19
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
Ideal Voltage Source
An ideal voltage source maintains a constant voltage across its terminals
regardless of how much current is developed through it due to its surrounding
environment. In other words, an ideal voltage source will supply or draw
whatever amount of current necessary in order to maintain the specified voltage
across its terminals.
Supplying current
Drawing current
i
VS
i
VS
Ideal voltage source
Such a behaviour also means that
the current through an ideal voltage source depends on the circuit it is placed in.
Without looking at the circuit an ideal voltage source is in, it is not possible to
determine the current through it. This behaviour contrasts with that of a resistor –
no matter where a resistor is placed in, if the voltage across the resistor is known
then its current is known – and is analogous to that of an ideal wire – an ideal wire
is essentially an ideal voltage source of zero volt!
This idealised behaviour of an ideal voltage source does entail some practical
considerations:
•
If the terminals of an ideal voltage source are connected by an ideal wire (i.e.,
a “short-circuit” 21), an infinite current will result (a constant voltage over
16F
zero resistance), in practice the source will simply break
•
Similarly, if ideal voltage sources of different values are connected in
parallel, an infinite current will result, in practice a large amount of current
will result between the sources and energy is wasted in the sources until they
reach a common voltage value (if they do not break first)
Example 9.
(FPHSEN)
The circuit shown contains two ideal
sources. The value of the current 𝑖 is
(a)
0A
(b)
2A
(c)
4A
(d)
8A
i
8V
2Ω
8V
If a pair of terminals is connected together with zero (or very low) resistance in between, we say
the terminals are shorted, and we call such a connection across the terminals a short-circuit. Similarly,
if there is no electrical connection (infinite resistance) across a pair of terminals, we say the terminals
are open, and we call such a lack of connection across the terminals an open-circuit.
21
Dr. William Lee
Page 21
Circuit Concepts
Ideal Current Source
An ideal current source maintains a constant current through its terminals
regardless of how much voltage is developed across it due to its surrounding
environment. In other words, an ideal current source will supply or draw
whatever amount of energy to or from the charges (i.e., voltage) necessary in order
to maintain the specified current through its terminals.
supplying voltage
IS
v
drawing voltage
IS
v
Ideal current source
Such a behaviour of also means that
the voltage across an ideal current source depends on the circuit it is placed in.
Without looking at the circuit an ideal current source is in, it is not possible to
determine the voltage across it. This behaviour contrasts with that of a resistor –
no matter where a resistor is placed in, if the current through the resistor is known
then its voltage is known.
The idealised behaviour of an ideal current source also entails some practical
considerations:
•
If an ideal current source is not connected to anything (i.e., an “open-circuit”),
an infinite voltage will result (a constant current through infinite resistance),
in practice the source will no longer function as intended
•
Similarly, if ideal current sources of different values are connected in series,
an infinite voltage will result, in practice the sources will stop functioning as
intended
Example 10.
(LQDAEP)
Page 22
The circuit shown contains two ideal
sources. The value of the voltage 𝑉𝑅
across the 5Ω resistor is:
(a)
19 V
(b)
15 V
(c)
−11 V
(d)
−15 V
3A
4V
5Ω
VR
Dr. William Lee
Fundamentals of Electrical and Digital Systems
2.1.5. Power
Intended Learning Outcomes
•
•
Be able to interpret and relate the power of a circuit component to the
underlying energy characteristic of the charges in an electric circuit.
Be able to determine, and infer, the power of an electrical device based on
knowledge of its voltage and current characteristics
All electrical devices function in the same way – by their ability to do work on the
electrical charges flowing through them. The only difference between the different
devices lies in the nature of the work done (i.e., energy conversion) that they are
designed to do. For example, a battery does work by separating the charges
through chemical reactions – work is done by the battery (on the charges) in
isolating opposite charges, converting chemical potential energy into electric
potential energy. Likewise, a resistor does work by resisting the flow of charges –
work is done by the charges (on the resistor) in moving through, and in doing so,
electric potential energy is converted into heat.
Since all electrical devices require a continuous source of energy for them to
operate, we often use the rate at which electric charges gain (from the device) or
lose (to the device) electric potential energy as a measure for comparing between
devices of similar purpose. Given our knowledge of the meaning of voltage across
and current through a device, we therefore have a very intuitive expression for
quantifying this rate:
𝑝
⏟
≝
⏟
𝑣
×
=Δ𝐸 across device per 𝑄
=𝐸 supplied/received per sec.
⏟𝑖
=no. of 𝑄 through device per sec.
The product thus measures the amount of energy transferred per second to or from
the device. Dimensionally, this product also makes sense since the unit is
JC −1 × Cs −1 = Js −1, Joules per second. This quantity is commonly known to you as
the power (symbol: 𝑝 or 𝑃) of an electrical device, and we usually use the unit Watt
(symbol W) instead.
p10V = 2 10
= 20 W supplied
10 V
2A
2A
5Ω
10 V
p5Ω = 2 10
= 20 W received
In the circuit above, it should be clear that the power developed in both the 10 V
source and the 5 Ω resistor is 20 W (10 ⋅ 2 = 20 W). However, clearly one of these
components is supplying 20 W while the other is receiving 20 W of power:
Dr. William Lee
Page 23
Circuit Concepts
•
Ohm’s law dictates there must be a 2 A current through the 5 Ω resistor, in the
direction of the 10 V voltage drop – each Coulomb of charge loses 10 J of energy
to the resistor as they flow through – the resistor receives 20 W of power from
the charges
•
All of the 2 A current must have come from the 10 V source, in the direction of
the 10 V voltage rise – each Coulomb of charge gains 10 J of energy from the
source as they flow through – the source supplies 20 W of power to the charges
There is thus a clear ambiguity in the power quantity of a device – it does not
provide details as to whether the power calculated is supplied or received by the
device, which can only be deduced by examining the underlying energy
distribution of the charges across the device. This ambiguity occurs because we
have failed to define the polarity of the voltage and the direction of the current as
used in the power quantity (which the symbols themselves cannot convey).
To address this (as with Ohm’s law), electrical engineers have adopted the
standard (albeit arbitrary) convention 22:
17F
The power definition takes the current to be in the direction of a voltage drop.
Said differently, the power quantity, by default, conveys the amount of power
received/absorbed by the device of interest.
Device A
2A
2A
2A
10 V
𝑝=+ 𝑣 ⋅ 𝑖
= + (10) ⋅ (2)
= + 20 W
= 20 W received
Device A
V
𝑝=− 𝑣 ⋅ 𝑖
= − (−10) ⋅ (2)
= + 20 W
= 20 W received
Device B
2A
V
𝑝=+ 𝑣
⋅ 𝑖
= + (−10) ⋅ (2)
= − 20 W
= 20 W supplied
Device B
10 V
𝑝=− 𝑣 ⋅ 𝑖
= − (10) ⋅ (2)
= − 20 W
= 20 W supplied
In this way we therefore have a unanimous way of measuring and expressing
power of device – by measuring the voltage to be against the direction in which the
current is measured, the product of the measured quantities will tell you the
amount of power received by the device, and a negative result simply indicate
power is being supplied by the device.
The passive-sign convention, the same one used to define the terminal characteristics of all devices.
E.g., recall Ohm’s law takes the voltage polarity across a resistor to be against the current direction
through it (see footnote 16 in Sect. 2.1.3).
22
Page 24
Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 11.
(MDXGLT)
Example 12.
(DQAHHA)
Example 13.
The 5 V source in the circuit shown is
(a)
Supplying 5 W of power
(b)
Supplying 10 W of power
(c)
Receiving 10 W of power
(d)
Receiving 15 W of power
5Ω
5V
15 V
In the circuit shown, the 2 A source is
(a)
Receiving 10 W of power
(b)
Receiving 30 W of power
(c)
Supplying 30 W of power
(d)
Supplying 50 W of power
10 Ω
2A
5V
Think about how your answer would differ in Example 12 if the
10 Ω resistor is replaced by a 1 Ω resistor.
(ANS: Receiving 6 W of power)
Dr. William Lee
Page 25
Circuit Concepts
2.1.6. Kirchhoff’s Current and Voltage Laws
Intended Learning Outcomes
•
Be able to apply, and make use of, Kirchhoff’s laws to relate currents and
voltages in an electric circuit
Be able to incorporate, and integrate, Kirchhoff’s laws into problems of relating
unknown electrical quantities to known ones in an electric circuit.
•
Kirchhoff’s Current Law (KCL)
A current, by definition, is the number of charges passing through a given point in
a circuit in one second. Intuitively, the amount of charge arriving at any point in
one second must 23 equal the amount of charge leaving that point in one second –
this is the underpinning idea behind Kirchhoff’s Current Law (KCL).
18F
KCL:
The algebraic sum of the currents leaving any node24 in a circuit equals zero.
Such a statement make sense because a positive current entering a node in a circuit
is equivalent to the negative of the current leaving the node, hence by assuming all
currents to be leaving 25, the algebraic signs will take care of those entering the node.
19F
20F
node
node
i1 i3
i4
i2
i6
node
i7
i8
i5
𝑖1 + 𝑖2 = 𝑖3
or
−𝑖1 − 𝑖2 + 𝑖3 = 0
𝑖4 = 𝑖5
or
−𝑖4 + 𝑖5 = 0
Example 14. For the partial circuit shown, determine
(PMBGPQ) the indicated current 𝑖𝑜 .
(a) 2 A
(b) −10 A
(c) −6 A
𝑖6 + 𝑖7 + 𝑖8 = 0
or
−𝑖6 − 𝑖7 − 𝑖8 = 0
8A
4A
-2 A
(d) −2 A
io
If they are not equal then there would be, for example, an accumulation of charges at some point
along the conducting path, and consequently a depletion of charges at some other point in the circuit
by the same amount (law of conservation of charges). This separation of charges implies a force of
attraction (Coulomb’s law) along the conducting path which contradicts the stationarity of charges.
24 A node in an electrical circuit is a point at which two or more circuit elements are connected
together, either directly, or via ideal wires.
25 You are not restricted to this arbitrary convention – you may choose to sum currents entering or
equate those entering and leaving. You will obtain the same KCL equations.
23
Page 26
Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 15. Consider the part of a circuit shown. Write a
KCL equation for the node of potential 𝑉1.
I1
1Ω
V1
1Ω
+8 V
Example 16. Node 𝑋 has a potential of 𝑉 volts.
Write a KCL equation at the node 𝑋.
Dr. William Lee
6.8 Ω X 4.7 Ω
12 V
3.3 Ω
2A
Page 27
Circuit Concepts
Kirchhoff’s Voltage Law (KVL)
A voltage, by definition, is the difference in energy (of each unit of charge)
between two points. Such an understanding immediately suggests that the total
voltage rises around a loop 26 in a circuit must 27 equal to the total voltage drops in
the same loop – this is the underpinning idea behind Kirchhoff’s Voltage Law (KVL).
21F
KVL:
2F
The algebraic sum of voltage rises around a loop in a circuit equals zero.
Charges at 𝐶 have 3 J of energy
less than those at 𝐵
Charges at 𝐵 have 5 J of energy
more than those at 𝐴
B
5V
3V
0V
C
D
R
VR
Charges at 𝐶 and 𝐷 have
the same amount of energy
VB
A
Charges at 𝐴 have 0 J of energy
Clearly the voltage across the resistor 𝑅 is 2 V given knowledge of the energy
differences (i.e., voltages) around the loop “𝐴𝐵𝐶”. Applying KVL we indeed have
5 = 3 + 𝑉𝑅 𝐨𝐫 5 − 3 − 𝑉𝑅 = 0
⇒
𝑉𝑅 = 2V.
Similarly, from the voltages around loop “ 𝐴𝐶𝐷 ” we deduce that 𝑉𝐵 = 𝑉𝑅 , in
agreement with KVL
𝑉𝑅 = 0 + 𝑉𝐵 𝐨𝐫 𝑉𝑅 − 0 − 𝑉𝐵 = 0
⇒
𝑉𝐵 = 𝑉𝑅 .
Notice “𝐴𝐵𝐶𝐷” also form a loop and so we also have
5 = 3 + 0 + 𝑉𝐵 𝐨𝐫 5 − 3 − 0 − 𝑉𝐵 = 0 ⇒
𝑉𝐵 = 2 V as expected from above.
Example 17. In the circuit shown, the voltage 𝑣 as
(NYAKOY) indicated is
(a) 6 V
(b) 10 V
(c) 14 V
(d) 30 V
12 V
10 V
8V
v
A loop in an electrical circuit is a (hypothetical) closed path starting from a node and proceeding
through the circuit, eventually returning to the same node. A loop does not necessarily need to be a
physically ‘complete’ circuit.
27 Otherwise, this suggests that the charges, upon returning to the original point, have, for example,
gained energy, and over time more energy would be absorbed (by the charges) than is supplied,
violating the law of conservation of energy.
26
Page 28
Dr. William Lee
Fundamentals of Electrical and Digital Systems
VR
Example 18. In the circuit shown, the voltage 𝑉𝑅 as
(XMDSYR)
indicated is
12 V
(a) −12 V
(b) −4 V
(c) 4 V
8V
(d) 12 V
Example 19. Write a KVL equation for each of the
three loops in the circuit shown 28.
I1
I2
2Ω
4Ω
5V
3V
I3
23F
6Ω
Note that the resulting equations are dependent – each one is a linear combination of the other two.
We thus need more information if we are to solve for the currents, and in this case, the KCL equation
𝐼1 + 𝐼2 = 𝐼3 .
28
Dr. William Lee
Page 29
Circuit Concepts
2.1.7. Problems
1.
Find the voltage 𝑉1 and current 𝐼1 in the circuit shown.
5V
I1
V1
4.7 kΩ
3.3 kΩ
9V
2.
In your house, two 230 V light bulbs are turned on. Bulb A has a thinner
tungsten wire filament than Bulb B. Which one is likely to be brighter?
Explain your answer.
3.
A resistor is connected in series with a 2 A current source. If the resistor
receives 20 W of electrical power. Find the voltage 𝑣 across the current
source, and the resistance of the resistor.
2A
4.
R
Find the current (magnitude and direction) through the voltage source and
through the 2 Ω resistance. Find the powers developed for each of the two
sources. State whether each source is delivering power or receiving it.
3A
Page 30
10 V
3Ω
2Ω
Dr. William Lee
Fundamentals of Electrical and Digital Systems
5.
For the circuit shown below
R
20 V
2A
(a)
If the current source receives 20 W of power, determine the value of
the resistance 𝑅 and the power of the voltage source. State whether the
voltage source receives or supplies power.
(b)
Investigate the power characteristic of the current source when the
resistor is replaced with a 𝑅 = 10 Ω resistor.
(c)
Repeat Question 5(b) when 𝑅 = 15 Ω.
Dr. William Lee
Page 31
Circuit Concepts
Solutions to Problems
𝑉1 = 5 V, 𝐼1 = −0.15 mA.
Bulb B.
𝑣2A = 10 V polarity upwards, 𝑅 = 5 Ω.
𝑖10V = 1 A downwards, 𝑖2Ω = 2 A downwards,
voltage source absorbs 10W, current source delivers 30W.
5(a). 𝑅 = 5 Ω, the voltage source is delivering 40 W of power.
5(b). The power of the current source is 0 W, it is neither supplying nor receiving
power.
5(c). The current source is supplying 20 W of power.
1.
2.
3.
4.
Page 32
Dr. William Lee
Fundamentals of Electrical and Digital Systems
2.2
Methods and Theorems of Circuit Analysis
When we wish to analyse a circuit, what we mean is to determine everything there
is to know about the circuit. In other words, the general goal of circuit analysis is
to determine sufficient information about the circuit such that the current through,
and voltage across every element in a circuit can be easily deduced. All other
characteristics (e.g., power) can be inferred from the knowledge of these two
parameters.
Surprisingly, as it turns out, Kirchhoff’s laws, along with the terminal behaviour of
the circuit elements (e.g., resistors, capacitors, and inductors) are all that is needed
to characterise all electrical circuits containing them. Nonetheless, useful results
and ‘shortcuts’ have been developed over time to simplify and/or speed up the
analysis, thereby reducing the complexity of the task. Here, we present some of
the common tools and results.
Section 2.2 Concept Map
Fundamental Electrical Quantities
Voltage and Current
behaviour governed
and quantified by
Charge
Current
Voltage
Kirchhoff s Laws
Power
Resistor
Ideal Wire
used to develop
Tools of Circuit Analysis
Methods
Equivalent
Resistance
Theorems
used
by
Thevenin s
Theorem
Ohm s Law
Ideal Sources
Component Characteristics
Behaviour and Quantification of Electrical Quantities
Voltage &
Current Division
Principles
Superposition
Node-Voltage
Analysis
Dr. William Lee
Page 33
Methods and Theorems of Circuit Analysis
2.2.1. Equivalent Resistance
Intended Learning Outcomes
•
•
•
Be able to recognise and identify resistive networks that are connected in
series-, and parallel-connection.
Be able to re-draw, and heuristically simplify two-terminal resistive networks
consisting of series- and parallel-connections into their equivalent resistance
by inspection.
Be able to determine, interpret, and apply the notion of equivalent resistance
for two-terminal resistive networks in finding relevant electrical quantities in
an electric circuit.
When a load 29 is connected to a circuit, it will draw a certain amount of current
and voltage from the circuit depending on the internal constituents of the load. A
different load will therefore result in a different current or voltage being drawn
from the same circuit. If we are only interested in the terminal behaviour of the load 30,
and not its internal operations, then for all intents and purposes, we can replace
the load as perceived by the circuit with a resistance according to Ohm’s law – the
ratio of the voltage across the load to the current it draws from the circuit –
without affecting the circuit behaviour. This resistance is commonly referred to as
the equivalent resistance 31 of the load seen across its terminals.
24F
25F
26F
A
A
iL
Circuit
vL
iL
Load
Equivalent
vL
Circuit
Req ≝ 𝑣𝐿 ⁄𝑖𝐿
To
B
B
You may have come across this notion of equivalent resistance when ‘combining’
two resistances in series, or in parallel into a single resistance. It is important that
we understand the meaning of these ‘total’ resistances in relation to the original
circuit they result from, as we often exploit these as means to deduce, or investigate,
behaviours about the original circuit and not just as a way of ‘simplifying’ it.
A load is a device or circuit connected to the output of some other circuit.
This may be when we are interested in optimising and fine-tuning a circuit for which a fixed load
is connected to, or purely to simplify the subsequent analyses of relevant electrical quantities external
to the load.
31 Depending on the context, this is also known as the input resistance of the load.
29
30
Page 34
Dr. William Lee
Fundamentals of Electrical and Digital Systems
Equivalent Series Resistance
Components are said to be in series if and only if there is only one path for the
charges to flow through them.
In other words, we say two components are in series if they share an exclusive node
in which nothing else is connected to, and this should make it obvious why the
current through components in series must be the same – every charge that leaves
one component has no choice but to enter the other, a direct application of
Kirchhoff’s current law.
1Ω
7V
6Ω
i2
−𝑖1 + 𝑖2 = 0
1Ω
i1
7V
12Ω
4Ω
9Ω
3Ω
7Ω
5Ω
7Ω
5Ω
−𝑖3 + 𝑖4 = 0
i3
i4
When two resistors are connected in series, the total voltage across, and the current
through the resistors is equivalent to that of a single resistor whose resistance is
equal to the arithmetic sum of the two resistances:
Applying KVL around the loop, with
Ohm’s law we have
i
i
R1
v1
R2
v2
𝑣 = 𝑖𝑅
⏟1 + 𝑖𝑅
⏟2
= 𝑣1
v
⇒
𝑣 = 𝑖 (𝑅
⏟ 1 + 𝑅2 ) ,
= 𝑣2
=𝑅eq
v
Req = R1+R2
which shows the series resistor has an
equivalent resistance of
𝑅eq =
𝑣
= 𝑅1 + 𝑅2 .
𝑖
A direct generalisation of the result to 𝑁 resistors in series is thus
𝑁
𝑅eq = 𝑅1 + 𝑅2 + ⋯ + 𝑅𝑁 = ∑ 𝑅𝑘 .
𝑘=1
Notice this also reveals the fact that a series equivalent resistance is larger than any
of the resistors contributing to the equivalent, and in particular, 𝑅eq is larger than
the largest resistor in 𝑅1 , 𝑅2 , …, 𝑅𝑁 .
Example 20.
(LLRCHB)
Dr. William Lee
In the circuit shown, which of the resistors are in series?
(a)
1 Ω and 2 Ω.
(b)
1 Ω and 3 Ω.
(c)
3 Ω and 4 Ω.
(d)
1 Ω, 3 Ω, and 4 Ω.
5V
1Ω
3Ω
2Ω
4Ω
Page 35
Methods and Theorems of Circuit Analysis
Equivalent Parallel Resistance
Components are said to be in parallel if and only if they are connected together
at both of their ends.
In other words, we say two components are in parallel if they share both nodes,
and this should make it obvious why the voltage across components in parallel
must be the same – the difference in energy across the components is measured
across the same two nodes that they share, the voltage between the two ends of
each component must be the same, a direct application of Kirchhoff’s voltage law.
9Ω
1Ω
7V
6Ω
9Ω
3Ω
v9Ω
3Ω
12Ω
4Ω
𝑣3Ω − 𝑣9Ω = 0
v3Ω
7Ω
5Ω
When two resistors are connected in parallel, the voltage across, and the total
current through the resistors is equivalent to that of a single resistor whose
resistance is equal to the harmonic sum of the two resistances:
Applying KCL at the top node, with
Ohm’s law we have
i
v
i1
i2
R1
R2
𝑖=
𝑣
𝑣
+
𝑅
⏟1 𝑅
⏟2
= 𝑖1
⇒
= 𝑖2
i
1
𝑣 = 𝑖(
),
⏟1⁄𝑅1 + 1⁄𝑅2
=𝑅eq
Req =
v
1
1 / R1 + 1 / R2
which shows the parallel resistors have an
equivalent resistance of
𝑅eq =
𝑣
1
=
.
𝑖 1⁄𝑅1 + 1⁄𝑅2
A direct generalisation of the result to 𝑁 resistors in parallel is thus
𝑅eq =
1
1
= 𝑁
.
⁄
⁄
⁄
1 𝑅1 + 1 𝑅2 + ⋯ + 1 𝑅𝑁 ∑𝑘=1 1⁄𝑅𝑘
Notice this also reveals the interesting fact that a parallel equivalent resistance is
smaller than any of the resistors contributing to the equivalent, and in particular,
𝑅eq is smaller than the smallest resistor in 𝑅1 , 𝑅2 , …, 𝑅𝑁 .
Example 21.
(QTXCIT)
Page 36
In the circuit shown, which of the resistors are in parallel?
(a)
1 Ω and 2 Ω.
(b)
1 Ω and 3 Ω.
(c)
1 Ω and 4 Ω.
(d)
2 Ω and 3 Ω.
4Ω
5V
1Ω
3Ω
2Ω
Dr. William Lee
Fundamentals of Electrical and Digital Systems
9Ω
1Ω
7V
6Ω
1Ω
7V
6Ω
3Ω
12Ω 7Ω
5Ω
4Ω
2.25Ω
12Ω
4Ω
7V
6Ω
12Ω
4Ω
3.25Ω
6Ω
Has equivalent
terminal behaviour to
7V
7V
7Ω
7V
3.25Ω
3.25Ω
3.75Ω
6Ω
10Ω
It should be clear from its origins that the resulting equivalent resistance of a circuit
is equivalent to the original in terms of the behaviour across the terminals in which
the equivalent resistance is considered, but not the internal characteristics – in the
above example, clearly the equivalent resistance of 7 Ω is equivalent to the original
in terms of what the 7 V voltage source ‘sees’, but the current through, say, the 3 Ω
in the original circuit no longer exists in the equivalent circuit 32. We will explore
this point in the following circuit:
27F
iS
1Ω
i3
9V
v6Ω
6Ω
3Ω
Suppose we are interested in finding the indicated parameters 𝑖𝑆 , 𝑣6Ω , and 𝑖3Ω –
we will do so by interpreting the intermediate and eventual equivalent resistances
that result in relation to the original circuit. Consider the following sequence of
equivalent circuits that arise from combining the relevant resistances into their
equivalent:
iS
iS
1Ω
i3
9V
v6Ω
6Ω
Original Circuit
iS
1Ω
9V
3Ω
9V
v6Ω
1Ω+2Ω = 3Ω
6Ω||3Ω = 2Ω
Equivalent Circuit 1
Equivalent Circuit 2
The final equivalent circuit in this case simply reveals that the total current supplied by the source
(drawn by the circuit) is 3 A. Notice this will corresponds to the current through the 1 Ω resistor in
the original, but nothing else.
32
Dr. William Lee
Page 37
Methods and Theorems of Circuit Analysis
It is not immediately obvious from the original circuit what any of the interested
quantities would be. However, notice if our interest is solely on finding 𝑖𝑆 then we
can combine the two parallel resistances into a single equivalent resistance of 2 Ω
– the equivalence is in terms of the voltage across, and the current through its
terminals in place of the two parallel resistors in the original circuit. This make
sense since the total current through the parallel connection is 𝑖𝑆 and this is
reflected through the series connection of the 2 Ω equivalent resistance to the 1 Ω
in Equivalent Circuit 1. Notice 𝑖3Ω no longer exists, one cannot therefore deduce
this current directly from Equivalent Circuit 1.
Similarly, when we are only interested in finding 𝑖𝑆 , we can combine the two series
resistances in Equivalent Circuit 1 into a single equivalent resistance of 3Ω – the
equivalence is in terms of the voltage across, and the current through its terminals
in place of the two series resistor in Equivalent Circuit 1. This make sense since the
total voltage across the series connection is 9V and this is reflected through the
single 3 Ω equivalent connected directly across the 9 V source in Equivalent Circuit
2. Notice now 𝑣6Ω no longer exists, one cannot therefore deduce this voltage
directly from Equivalent Circuit 2.
From Equivalent Circuit 2, it is now obvious that 𝑖𝑆 is 3 A (Ohm’s law). This newly
deduced information thus enables us to make use of in Equivalent Circuit 1 and
deduce that 𝑣6Ω = 6 V. Finally, reverting to the original circuit with these new
details reveals that 𝑖3Ω = 2 A.
From this example, we see that the necessity in ‘simplifying’ a circuit, or a section
of it, into its equivalent resistance is contingent on what we want to use it for in the
subsequent analysis, and this relies on our ability to correctly interpret what the
‘equivalence’ is in relation to the original circuit – Sometimes replacing a circuit by
its equivalent resistance may trivialise the analysis, while sometimes it may not
directly lead to the quantities of interest but allows quick deduction of new details
about the original circuit for us to revert to and make use of. Other times it may
simply get us further away from what we wish to achieve if we arbitrarily replace
circuits by their equivalent resistances without planning.
It is worth noting that not all electrical connections are either in series or
parallel 33, and so, one should take care when simplifying connections into their
equivalent resistances – sometimes a circuit cannot be reduced into a single
equivalent resistance through series and parallel reductions.
28F
Other connections exist, such as the wye (or “Y”) and delta (or “Δ”) connections. The simplification
of these networks is beyond the scope of this course.
33
Page 38
Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 22. Determine the equivalent resistance
of the circuit seen by the voltage
source. Use the equivalent resistance
to find the current 𝑖𝑆 drawn from the
source (by the circuit), and hence the
current 𝑖2Ω through the 2 Ω resistor.
iS
6V
1Ω
3Ω
i2
2Ω
4Ω
(ANS: 𝑅eq = 3 Ω, 𝑖𝑆 = 2 A, 𝑖2Ω = 2/3 A)
Dr. William Lee
Page 39
Methods and Theorems of Circuit Analysis
2.2.2. Voltage and Current Division Principles
Intended Learning Outcomes
•
•
Be able to recognise the applicability of voltage and current division principles
in an electrical circuit.
Be able to incorporate, and integrate, the voltage and current division
principles into problems requiring finding electrical quantities in a circuit
Voltage Division Principle
The voltage division principle or voltage divider is a convenient method for
determining the voltage across resistive components when they are connected in
series. It is so-called because the result suggests that the voltage across a single
resistor equals the applied voltage (across all resistors) divided down by some
factor. The voltage divider is a direct consequence of KVL.
Applying KVL around the loop, together with Ohm’s law
we have
i
R1
𝑉 = 𝑖𝑅
⏟1 + 𝑖𝑅
⏟2
v1
= 𝑣1
V
R2
v2
1
𝑉.
𝑅1 + 𝑅2
⇒
𝑖=
and
𝑣2 = 𝑖𝑅2 = (
= 𝑣2
and so
𝑣1 = 𝑖𝑅1 = (
𝑅1
)𝑉
𝑅1 + 𝑅2
𝑅2
) 𝑉.
𝑅1 + 𝑅2
The voltage across a resistor in series with other resistors is directly proportional
to the total voltage applied across these resistors by the fractional contribution of
its resistance to the total sum of resistances.
A direct generalisation of the result to 𝑁 resistors in series is thus
𝑣𝑝 = (
𝑅𝑝
)𝑉
𝑅1 + 𝑅2 + ⋯ + 𝑅𝑁
for 𝑝 = 1,2, … , 𝑁.
Notice the result only holds when the resistors of interest are in series, since it
relies on the fact that the current through the resistors are equal. We emphasise
this in the following example.
Page 40
Dr. William Lee
Fundamentals of Electrical and Digital Systems
2 kΩ
Example 23. Find the output voltage, 𝑉𝑎𝑏 , for the
circuit shown when
12 V
a
6 kΩ
(a) terminals 𝑎-𝑏 is an open-circuit,
b
(b) a 3 kΩ resistor is placed across 𝑎-𝑏.
(ANS: 9 V polarity upwards, and 6 V polarity upwards)
8V
Example 24. The voltage 𝑣 at the centre node in the
(KEMJQF)
partial circuit shown is
(a) 0.5 V
(b) 1.5 V
(c) 3.5 V
3 kΩ
(d) 4.5 V
v
1 kΩ
2V
Example 25. For the circuit shown, the voltage across
(FDKNAN) the 6 Ω resistor is
3
3
6
3
21 V
(a) 0 V
Dr. William Lee
(b) 10.5 V
(c) 12 V
(d) 14 V
Page 41
Methods and Theorems of Circuit Analysis
Current Division Principle
The current division principle or current divider is a convenient method for
determining the current through resistive components when they are connected in
parallel. It is so-called because the result suggests that the current through a single
resistor equals the total current divided down by some factor. The current divider
is a direct consequence of KCL.
Applying KCL at the top node, together with Ohm’s law we
have
I
v
i1
i2
R1
R2
𝐼=
𝑣
𝑣
+
𝑅
⏟1 𝑅
⏟2
= 𝑖1
⇒
𝑣=
1
𝐼.
1⁄𝑅1 + 1⁄𝑅2
= 𝑖2
and so
𝑖1 =
𝑣
1⁄𝑅1
𝑣
1⁄𝑅2
=(
=(
) 𝐼 and 𝑖2 =
) 𝐼.
𝑅1
1⁄𝑅1 + 1⁄𝑅2
𝑅2
1⁄𝑅1 + 1⁄𝑅2
The current through a resistor in parallel to other resistors is directly
proportional to the current supplied to these resistors by the fractional
contribution of its conductance 34 to the total sum of conductances.
29F
A direct generalisation of the result to 𝑁 resistors in parallel is thus 35
30F
𝑖𝑝 = (
1⁄𝑅𝑝
)𝐼
1⁄𝑅1 + 1⁄𝑅2 + ⋯ + 1⁄𝑅𝑁
for 𝑝 = 1,2, … , 𝑁.
Notice the result only holds when the resistors of interest are in parallel, since it
relies on the fact that the voltage across the resistors are equal. We emphasise this
in the following example.
The reciprocal of resistance is also known as the conductance, 𝐺, with a unit of Siemens, S.
For the special case of 𝑛 = 2, you may have seen the equivalent expression 𝑖1 = (𝑅2 ⁄(𝑅1 + 𝑅2 ))𝐼
and 𝑖2 = (𝑅1 ⁄(𝑅1 + 𝑅2 ))𝐼 which simply falls out of algebraic manipulation of the expression shown.
However, such an expression does not generalise to 𝑛 > 2, while the expression shown does and
preserves similarity to that of the voltage divider.
34
35
Page 42
Dr. William Lee
Fundamentals of Electrical and Digital Systems
4 kΩ a
Example 26. Find the current through terminals 𝑎-𝑏
for the circuit shown when
(a) the terminals 𝑎-𝑏 is short-circuited,
9A
2 kΩ
b
(b) a 3 kΩ resistor is placed across 𝑎-𝑏.
(ANS: 3 A from 𝑎 to 𝑏, and 2 A from 𝑎 to 𝑏)
Example 27. For the circuit shown, the current
(GDDZSA) through the 6 Ω resistor is
3
3
6
3
6A
(a) 0 A
Dr. William Lee
(b) 1 A
(c) 2 A
(d) 3 A
Page 43
Methods and Theorems of Circuit Analysis
2.2.3. Node-Voltage Analysis
Intended Learning Outcomes
•
Be able to recognise and differentiate between nodes and essential nodes in a
circuit.
Be able to write down the governing node-voltage equations necessary to
completely characterise a circuit containing one non-reference essential node.
Be able to incorporate and integrate node-voltage analysis into problems
requiring finding electrical quantities in a circuit.
•
•
Not all circuits can be simplified into a series/parallel equivalent 36, consequently
methods such as the voltage or current dividers cannot be used in these cases, and
one must fall back to Kirchhoff’s laws. Consider the following circuit:
31F
i1
4Ω
i3
i2
2Ω
3Ω
8V
5A
It should be clear that there are no resistors in series or in parallel, and the circuit
cannot be reduced any further 37. To analyse the circuit, we could apply KVL and
KCL to get
32F
KVL:
8 + 4𝑖1 − 3𝑖2 = 0,
(1)
KCL:
𝑖1 + 𝑖2 + 𝑖3 = 0,
(2)
𝑖3 = −5 A,
(3)
solving which completes the characterisation of the circuit. While feasible, such a
general approach is clearly non-systematic, requiring careful planning of which
laws to use and when 38.
3F
On the other hand, if we can identify the voltages ‘at’ the essential nodes 39 (i.e., the
“node voltages”) of the circuit, then all other voltages and currents in the circuit
can be trivially deduced thereby completing the analysis – this is the rationale
behind the node-voltage analysis method.
34F
Node-voltage analysis is the use of KCL to find the voltages ‘at’ nodes by
expressing the currents through those nodes in terms of their voltages.
Recall not all connections are either in series or parallel, connections such as the “Y” or “Δ” exist.
At least not without using additional circuit tools such as Thevenin’s theorem, see Sect. 2.2.5.
38 For example, why didn’t we write out a KVL equation for the right loop instead? Because we do
not know the voltage across the current source, 𝑉5A , and would therefore end up with an additional
unknown if we chose to do so: 3𝑖2 − 2𝑖3 − 𝑉5A = 0, where the polarity of 𝑉5A is taken to be upwards.
39 An essential node is a point where three or more elements join.
36
37
Page 44
Dr. William Lee
Fundamentals of Electrical and Digital Systems
Node-voltage analysis is carried out in the following way:
1.
Select one of the essential nodes as the reference (ground) node, in which all
other node voltages will be calculated with respect to.
2.
Write a KCL equation for each of the remaining essential nodes in terms of
the unknown essential node voltages.
3.
Solve the node-voltage equations simultaneously.
Note that the reference node can be arbitrary in general, however, selecting an
appropriate one may speed up the analysis.
Case 1: Selecting Bottom Essential Node as Ground
Suppose we wish to analyse the circuit above via node-voltage analysis. We will
do so by selecting the reference node to be at various points in the circuit and
examine its implications.
There is only one unknown node-voltage (at the top essential node), and so
we only need one equation to solve the circuit. Applying KCL at this essential
node we have
𝑖1 + 𝑖2 + 𝑖3 = 0.
Expressing each of these currents in terms of this node-voltage (say, 𝑉1) using
Ohm’s law we have
𝑖1 =
𝑉1 − 8
,
4
𝑖2 =
giving the node-voltage equation
solving which we deduce
𝑉1 − 0
,
3
and
𝑖3 = −5 A,
𝑉1 − 8 𝑉1
+ − 5 = 0,
4
3
𝑉1 =
5 + 8⁄4
= 12 V.
1⁄4 + 1⁄3
The result says the voltage at the top node is 12 V higher than the bottom node.
Dr. William Lee
Page 45
Case 2: Selecting Top Essential Node as Ground
Methods and Theorems of Circuit Analysis
There is only one unknown node-voltage (at the bottom essential node), and
so we only need one equation to solve the circuit. Applying KCL at this
essential node we have
−𝑖1 − 𝑖2 − 𝑖3 = 0.
Expressing each of these currents in terms of this node-voltage (say, 𝑉2 ) using
Ohm’s law we have 40
35F
𝑖1 =
0 − (𝑉2 + 8)
,
4
𝑖2 =
0 − 𝑉2
,
3
giving the node-voltage equation:
−(
solving which we deduce
and 𝑖3 = −5 A,
−𝑉2 − 8
0 − 𝑉2
)−(
) − (−5) = 0,
4
3
𝑉2 =
−5 − 8⁄4
= −12 V.
1⁄4 + 1⁄3
Case 3: Selecting an Arbitrary Node as Ground
The result is consistent with before – the voltage at the bottom node is 12 V lower
than the top node.
If we select, for example, the non-essential node between the 5 A source and
the 2 Ω resistor as our reference, we now have two unknown essential node
voltages at the top (say, 𝑉3 ) and bottom (say, 𝑉4 ) since it is not immediately
obvious what the voltages at these nodes are with respect to the chosen
reference.
Applying KCL at the top and bottom essential nodes respectively, we have
𝑉3 − (𝑉4 + 8) 𝑉3 − 𝑉4 𝑉3
+
+ =0
4
3
2
⇒
𝑉3 − (𝑉4 + 8)
𝑉3 − 𝑉4
)+5 = 0
)−(
4
3
⇒
13𝑉3 − 7𝑉4 = 24
and
−(
−𝑉3 + 𝑉4 = −12,
from which 𝑉3 = −10 V and 𝑉4 = −22 V.
The result is, again, consistent to previous analyses – the voltage 𝑉3 at the top node is
12 V higher than the bottom node.
Remember there is an inherent polarity and direction assumed in Ohm’s law 𝑣 = 𝑖𝑅: the polarity
of 𝑣 and direction of 𝑖 are opposite. This means for example, given the direction of 𝑖1 , the polarity
across the 4Ω resistor must be upwards, and so the voltage across it, with the top node assumed to be
at a higher potential, is (0 − (𝑉2 + 8)) volts.
40
Page 46
Dr. William Lee
Fundamentals of Electrical and Digital Systems
Notice in all three cases, the calculated node voltages are necessarily different since
they are with respect to different points in the circuit. However, the voltage across
and current through all elements in the circuit are identical in all three cases – they
must, since the circuit has not changed, only the way we have chosen to analyse it
has.
Example 28. Determine the current through the 4 Ω
resistor by means of node-voltage
analysis
3Ω
6Ω
4Ω
14 V
6V
(ANS: 11/9 A downwards)
Dr. William Lee
Page 47
Methods and Theorems of Circuit Analysis
Example 29. Find the voltage at node 𝐴.
7V
A
4Ω
8V
3Ω
2Ω
9V
(ANS: 107/13 V)
Example 30. A correct node-voltage equation for the node 𝑉 is:
(KSIANR)
𝑉
𝑉 𝑉+2
+
+
=0
(a)
𝑅1 𝑅2
𝑅3
(b)
Page 48
V
𝑉 𝑉−2 𝑉+2
+
+
=0
𝑅1
𝑅2
𝑅3
(c)
𝑉+2 𝑉 𝑉+2
+
+
=0
𝑅1
𝑅2
𝑅3
(d)
𝑉 𝑉−2 𝑉
+
+
=0
𝑅1
𝑅2
𝑅3
R3
2V
R2
R1
2V
Dr. William Lee
Fundamentals of Electrical and Digital Systems
2.2.4. Superposition Theorem
Intended Learning Outcomes
•
•
Be able to recognise and re-draw the equivalent of a circuit containing
independent sources when they are set to zero.
Be able to incorporate the notion of superposition in finding electrical
quantities in a circuit containing multiple independent sources.
When tackling complex engineering problems, a common strategy is to reduce, or
break, the problem down into manageable sub-problems. By solving these ‘simpler’
problems separately, the solution to the original problem is found upon
integrating the results together. The superposition theorem is a powerful tool in
circuit analysis that demonstrates such a philosophy for solving complex circuits
in an intuitive way. The theorem says:
Any linear 41 response (i.e., voltage or current) in a circuit containing multiple
sources is equal to the algebraic sum of the contributions to that response due to
each sources acting individually with all other sources set to zero.
36F
To set a voltage source to zero we replace it with a short-circuit, to set a current
source to zero we replace it with an open-circuit 42.
37F
The proof of the theorem is beyond the scope of the course, but it relies on the so-called linearity
property of the circuit response of interest. Roughly, a response 𝑦 = 𝑓(𝑥) due to some input 𝑥 is linear
if it satisfies the property 𝑓(𝑎𝑥1 + 𝑏𝑥2 ) = 𝑎𝑓(𝑥1 ) + 𝑏𝑓(𝑥2 ). One can easily check that the voltage and
current behaviour of a resistor (𝑣 = 𝑅𝑖), a capacitor (𝑖 = 𝐶 𝑑𝑣⁄𝑑𝑡 ), and an inductor (𝑣 = 𝐿 𝑑𝑖 ⁄𝑑𝑡) are
all linear. The power quantity 𝑝 = 𝑣𝑖 of a component on the other hand, is not (try expressing 𝑝 as a
function of either 𝑣 or 𝑖 for any of the components above, you’ll find that the resulting expression
fails to satisfy the linearity property), and so superposition cannot be used to find the power
developed across a component.
42 Why? Because an ideal wire (i.e., short-circuit) has zero voltage across it (but potentially non-zero
current through it). Likewise, a lack of electrical path (i.e., open-circuit) has zero current through it
(but potentially non-zero voltage across it).
41
Dr. William Lee
Page 49
Methods and Theorems of Circuit Analysis
Consider the circuit shown below, we will determine the voltage 𝑉𝑋 across the 3Ω
resistor by superposition.
4Ω
2Ω
VX
3Ω
8V
Response due to 8V Source
First, we consider the contribution of the
8 V source to the response 𝑉𝑋 . Setting the
current source to zero, the circuit looks like:
Response due to 5A Source
5A
Second, we consider the contribution of the
5 A source to the response 𝑉𝑋 . Setting the
voltage source to zero, the circuit looks like:
4Ω
2Ω
VX
3Ω
8V
The voltage across the 3 Ω resistor can be found via, for example, the voltage
divider
3
24
𝑉𝑋(8V) =
⋅8=
V, polarity 𝐮𝐩𝐰𝐚𝐫𝐝𝐬.
3+4
7
4Ω
2Ω
VX
3Ω
5A
The voltage across the 3 Ω resistor can be found via, for example, the current
divider and Ohm’s law:
𝑖3Ω =
1⁄3
20
⋅5=
A
1⁄3 + 1⁄4
7
⇒
𝑉𝑋(5A) = 3𝑖3Ω =
60
V, polarity 𝐮𝐩𝐰𝐚𝐫𝐝𝐬
7
By the superposition theorem, the actual voltage 𝑉𝑋 across the 3 Ω resistor with
both sources in effect is therefore the algebraic sum of these two responses
𝑉𝑋 = 𝑉𝑋(8V) + 𝑉𝑋(5A) =
24 60
+
= 12 V polarity upwards.
7
7
Notice the result is consistent with what we have deduced via node-voltage
analysis from the previous section.
Page 50
Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 31. By means of superposition 43, find the
current (magnitude and direction)
through the 3 Ω resistor.
38F
6Ω
3Ω
6Ω
6V
12 V
(ANS: 1.5 A upwards)
As a further practice, try determining the current by using node-voltage analysis and convince
yourself that you must obtain the same answer.
43
Dr. William Lee
Page 51
Methods and Theorems of Circuit Analysis
Example 32. By means of superposition 44, find the
voltage across the 4 Ω resistor, make
sure to indicate its polarity.
39F
2Ω
4Ω
4 A 12 A
2Ω
12 V
(ANS: 14 V polarity upwards)
Example 33. [Optional] In the circuit shown, 𝑅1 receives 3 W of power from each
(LZDHZZ)
of the sources when they are considered separately (by setting the
other to zero). The power absorbed by 𝑅1 when both sources are in
effect is:
R1
(a) 3 W
(b) 6 W
(c) 9 W
(d) 12 W
V1
R2
V2
As before, try finding the current of interest via node-voltage analysis to convince yourself of the
equivalence between the methods. Note that in this case, you will need to solve a minimum of two
node-voltage equations simultaneously as there are at least two unknown essential node voltages.
44
Page 52
Dr. William Lee
Fundamentals of Electrical and Digital Systems
2.2.5. Thevenin’s Theorem
Intended Learning Outcomes
•
•
•
Be able to identify and distinguish between the terms ‘circuit’ and ‘load’ as
relevant in the context of defining the terminal characteristics across a pair of
terminals in an electrical system.
Be able to apply, and provide relevant interpretation to, Thevenin’s theorem in
the context of representing the terminal behaviour of an electric circuit.
Be able to incorporate, and make use of, Thevenin’s theorem in analysing
resistive electrical systems.
Often, when we connect a load (e.g., mobile phone, speaker) to a circuit (e.g.,
power supply, amplifier), we are more interested in what the voltage and current
supplied (by the circuit) to the load are, as opposed to how the circuit is able to
produce them. In other words, we are usually more concerned with the terminal
characteristics of the circuit (its output voltage and output current) as seen by the
load over the details of the circuit’s internal operations that give rise to the
observed responses.
When we are just interested in the terminal behaviour of a circuit, the Thevenin’s
theorem is a convenient tool that ‘hides’ the complexity of the circuit behind its
terminals, thereby considerably simplifies the analysis of the external surroundings
connected to the circuit. The theorem says:
The external behaviour of any linear 45 two-terminal circuit at its terminals can
be modelled by an equivalent circuit consisting of a voltage source 𝑉𝑇𝐻 in series
with a resistance 46 𝑅𝑇𝐻 , where 𝑉𝑇𝐻 is the open-circuit voltage across the
terminals, and 𝑅𝑇𝐻 is the equivalent resistance looking into the circuit from the
terminals when independent sources are set to zero 47.
40F
41F
42F
i
A
Any circuit
comprising
resistances and
electrical sources
RTH
Thevenin’s
v
B
i
A
Load
Equivalent
VTH
v
Load
B
Thevenin’s theorem thus allows us to effectively analyse the behaviour of a circuit
from its terminals subject to different external environments without needing to
re-analyse the entire system internally due to the external change.
The proof of Thevenin’s theorem relies on the superposition theorem which in itself requires the
linearity property of the responses in the circuit.
46 Thevenin’s theorem also applies to steady-state responses due to AC excitations in circuits
containing reactive components (inductors or capacitors). In these cases, the resistance is generalised
to an impedance which consists of a series resistance and reactance. This is, of course, outside the scope
of this course.
47 Remember to set a voltage source to zero we replace it with a short-circuit, while to set a current
source to zero we replace it with an open-circuit, see Sect. 2.2.4.
45
Dr. William Lee
Page 53
Methods and Theorems of Circuit Analysis
Consider the circuit shown below, we will determine the voltage 𝑉𝑋 across the 3 Ω
resistor via Thevenin’s theorem.
4Ω
2Ω
3Ω
VX
8V
5A
Thevenin Voltage
By Thevenin’s theorem, we can replace the circuit seen by the 3 Ω resistor by a
series voltage source 𝑉𝑇𝐻 and resistance 𝑅𝑇𝐻 without affecting the behaviour
experienced by the 3 Ω . To do so, we disconnect the 3 Ω from the circuit, and
determine (1) the open-circuit voltage across the pair of terminals the 3 Ω was
connected to, and (2) the equivalent resistance of the circuit as seen from the open
terminals.
To find the open-circuit voltage 𝑉𝐴𝐵 it should
be clear that, with the 3 Ω disconnected, the
current through the 4 Ω must be 5 A, and so
by KVL we have 48
43F
4Ω
A
2Ω
VTH
8V
B
5A
𝑉𝐴𝐵 = 8 + 4 ⋅ 5 = 28 V,
this is the Thevenin voltage 𝑉𝑇𝐻 , i.e., 𝑉𝑇𝐻 = 𝑉𝐴𝐵 = 28 V.
To find the equivalent resistance seen from
the terminals 𝐴-𝐵 we ask the question 49 – if a
charge is to flow from terminal 𝐴, through
the circuit, to terminal 𝐵 (or vice versa), what
resistive network (and hence the equivalent
resistance) will the charge see?
Thevenin Resistance
4F
4Ω
A
RTH
2Ω
B
Clearly, in this case since there is no conducting path along the 2Ω branch,
the charge has to flow through the 4 Ω in order to go from terminal 𝐴 to
terminal 𝐵 and so immediately,
𝑅𝑇𝐻 = 4 Ω,
this is the Thevenin resistance.
Alternatively, apply node-voltage analysis at node 𝐴 with node 𝐵 selected as ground to obtain the
node-voltage equation −5 + (𝑣 − 8)⁄4 = 0 ⇒ 𝑣 = 28 V as expected.
49 Recall the formal approach in calculating the equivalent resistance across a pair of terminals is to
apply a notional voltage (current) across it and determine the current (voltage) drawn from the source
(by the circuit). The ratio of the two (Ohm’s law) then gives the equivalent resistance of the circuit.
In this particular case, applying, arbitrarily, an 8 V across the terminals A-B, the current drawn is 2 A
and thus 8⁄2 = 4 Ω as expected.
48
Page 54
Dr. William Lee
Fundamentals of Electrical and Digital Systems
The Thevenin equivalent of the circuit seen by the 3 Ω resistor thus consists of a
voltage source of 28 V in series with a resistance of 4 Ω. This means we can replace
the circuit that the 3 Ω is connected to by this equivalent without affecting the
voltage across and current through the 3 Ω
4Ω
A
i
4Ω
8V
28 V
5A
B
i
Thevenin’s
2Ω
3Ω
VX
A
VX
3Ω
Equivalent
B
and this trivially reduces the problem of finding 𝑉𝑋 to that of a voltage divider
𝑉𝑋 =
3
⋅ 28 = 12 V
3+4
which is consistent with the prior analyses via node-voltage analysis and
superposition theorem.
Thevenin s Theorem
External environment of interest
i
4Ω
VX
8V
A
4Ω
2Ω
A
2Ω
VTH
3Ω
5A
B
8V
B
+
A
4Ω
RTH
5A
2Ω
B
Internal circuit
Open-Circuit Voltage seen from
Terminals of Interest
Resistance seen from Terminals
of Interest
=
Has an equivalent
external behaviour to
RTH
4Ω
i
28V
VX
B
Dr. William Lee
A
A
3Ω
VTH
Thevenin Equivalent Circuit
of the internal circuit
B
Page 55
Methods and Theorems of Circuit Analysis
Example 34. Determine the Thevenin equivalent of
the circuit seen by the 3 Ω resistor, and
use it to find the current 𝑖3Ω through it.
9V
4Ω
i3Ω
6Ω
3Ω
(ANS: 𝑉TH = 5.4 V polarity upwards, 𝑅TH = 2.4 Ω, 𝑖3Ω = 1 A)
Notice when applying Thevenin’s theorem, we have designated where the ‘load’ is
which consequently dictated what the ‘internal circuit’ (to be replaced by its
Thevenin equivalent) is. The choice of where the external environment is in a
circuit is purely decided by what we are interested in.
Page 56
Dr. William Lee
Fundamentals of Electrical and Digital Systems
For example, if we are interested in determining the voltage across the 4 Ω resistor
instead, we would designate the load to be the 4 Ω and replace the circuit it ‘sees’
by its Thevenin equivalent:
A
4Ω
VY
2Ω
Thevenin s
5A
Equivalent
3Ω
B
8V
RTH
A
VY
VTH
4Ω
B
Thevenin Voltage
Internal circuit
To find the open-circuit voltage 𝑉𝐴𝐵 across
the pair of terminals the 4Ω was connected
to, we note that the current through the 3 Ω
must be 5 A, and so by KVL we have
A
2Ω
VTH
3Ω
B
5A
8V
𝑉𝐴𝐵 = 3 ⋅ 5 − 8 = 7 V,
Thevenin Resistance
this is the Thevenin voltage 𝑉𝑇𝐻 , i.e., 𝑉𝑇𝐻 = 𝑉𝐴𝐵 = 7 V.
Since there is no conducting path through
the 2 Ω, it should be clear that the equivalent
resistance seen by a charge flowing from
terminal 𝐴 to 𝐵 is
A
2Ω
RTH
3Ω
B
𝑅𝑇𝐻 = 3 Ω
which is the Thevenin resistance as seen
from terminal 𝐴-𝐵.
Replacing the circuit seen by the 4 Ω by its Thevenin equivalent
A
VY
4Ω
B
8V
2Ω
Thevenin s
5A
Equivalent
3Ω
A
VY
3
4Ω
7V
B
trivially reduces the problem down to, once again, a voltage divider
𝑉𝑌 =
4
⋅ 7 = 4 V,
3+4
and is consistent with the prior analysis since by KVL 𝑉𝑌 = 𝑉𝑋 − 8 = 12 − 8 = 4 V
as expected.
Dr. William Lee
Page 57
Methods and Theorems of Circuit Analysis
Example 35. Determine the Thevenin equivalent of
the circuit seen by the 6 Ω resistor, and
use it to find the current 𝑖6Ω through it.
4Ω
9V
i6Ω
6Ω
3Ω
(ANS: 𝑉TH = 27⁄7 V polarity upwards, 𝑅TH = 12⁄7 Ω, 𝑖6Ω = 0.5 A)
Example 36. Find the Thevenin equivalent of the
circuit as seen across the terminals a-b,
and use this equivalent circuit to
determine how much current would
pass through a 3 Ω resistor if connected
across the terminals a and b.
Page 58
2Ω
3A
5Ω
A
8Ω
B
Dr. William Lee
Fundamentals of Electrical and Digital Systems
(ANS: 𝑉TH = 24 V, 𝑅𝑇𝐻 = 13 Ω, 𝑖3Ω = 1.5 A from 𝐴 to 𝐵)
It is important to emphasise that, as seen from all of the examples, the Thevenin
equivalent circuit only preserves the external behaviour of the original circuit at
the terminals interest, but not its internal behaviour – with no load connected,
there is no current in the Thevenin equivalent circuits, yet a non-zero current
clearly exists in all of the replaced circuits 50.
45F
Finally, it is worth noting that while any linear two-terminal circuit has a Thevenin
equivalent, not all Thevenin equivalents have a determinant form. For instance,
attempting to find the Thevenin equivalent circuit seen by the 2 Ω resistor in the
circuit presented on Page 54 will result in 𝑉𝑇𝐻 = ∞, and 𝑅𝑇𝐻 = ∞ for the Thevenin
parameters, and consequently any attempts to characterise the voltage across, and
current through the 2Ω using this Thevenin equivalent will yield an indeterminant
value of ∞/∞! Note that this does not mean that the electrical behaviour across the
2 Ω is undefined (e.g., there is clearly a 5 A current through it) but rather it says
that the particular method of analysis has failed to incorporate all of the available
information (in the circuit) which would otherwise uniquely characterise the
behaviour of interest 51.
46F
This notion of preserving the external equivalence is no different to, for example, the conversion of
two series resistances into an equivalent ‘total’ resistance – the equivalence of this total resistance to
the original two is in terms of the external voltage across the two resistors, and current through them.
One can no longer identify, say, the internal voltage across one of the series resistances from this total
resistance.
51 This is because in this particular situation, irrespective of what the current source value is, we end
up with the same Thevenin equivalent circuit of 𝑉𝑇𝐻 = 𝑅𝑇𝐻 = ∞. The resulting equivalent circuit
therefore fails to incorporate the current source value pertaining to this circuit that would otherwise
uniquely determine the effect on the 2 Ω.
50
Dr. William Lee
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Methods and Theorems of Circuit Analysis
2.2.6. Problems
1.
In the circuit shown, four resistances are connected to a voltage source.
A
8.2 Ω
5.6 Ω
3.9 Ω
12 V
2.2 Ω
Find the current through the 2.2 Ω resistor by
(a)
(b)
(c)
2.
finding the supply current and making use of the current division
principle,
making use of node-voltage analysis to determine the voltage at node
𝐴 with respect to the bottom rail, and
making use of Thevenin’s theorem to replace the circuit seen by the
2.2 Ω by its Thevenin’s equivalent.
Consider the circuit show below.
1Ω
A
4Ω
6V
2V
Determine the voltage at node 𝐴 by making use of
(a)
(b)
(c)
3.
the voltage division principle,
node-voltage analysis, and
superposition theorem.
Consider the circuit shown below.
A
3Ω
6V
4Ω
5Ω
2A
Determine the voltage at node 𝐴 by making use of
(a)
(b)
(c)
node-voltage analysis,
superposition theorem, and
Thevenin’s theorem (to replace the circuit seen by the 4 Ω resistor).
Hence, or otherwise, find the voltage across the 2 A current source.
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
4.
Consider the circuit shown below.
4Ω
3A
12 V
6Ω
Determine the current through the 6 Ω resistance by making use of
(a)
(b)
(c)
5.
node-voltage analysis,
superposition theorem, and
Thevenin’s theorem (to replace the circuit seen by the 6 Ω resistor).
Consider the circuit shown below.
3Ω
15 Ω
6Ω
10 V
3A
3Ω
Determine the current through the 10 V source by making use of
(a)
(b)
6.
superposition theorem, and
Thevenin’s theorem (to replace the circuit seen by the 10 V source).
Determine the current through the 2 V source in the circuit shown.
Dr. William Lee
2V
2Ω
1Ω
3A
Page 61
Methods and Theorems of Circuit Analysis
Solutions to Problems
1.
𝐼2.2Ω = 10⁄27 ≈ 0.37 A.
1(a). 𝐼supply = 10⁄9 A to be divided between 3.9 Ω and (5.6 + 2.2) Ω.
1(b). 𝑉𝐴 = 26⁄9 V across (5.6 + 2.2) Ω, then use Ohm’s law.
1(c). The Thevenin equivalent seen by the 2.2Ω is 𝑉𝑇𝐻 ≈ 3.87 V and 𝑅𝑇𝐻 ≈ 8.24 Ω.
2.
𝑉𝐴 = 2.8 V.
2(a). 𝑉1Ω = 0.8 V and 𝑉4Ω = 3.2 V, apply KVL.
2(b). Directly 𝑉𝐴 = 2.8 V.
2(c). 𝑉𝐴(2V) = 1.6 V, 𝑉𝐴(6V) = 1.2 V.
3.
𝑉𝐴 = 48⁄7 ≈ 6.857 V, 𝑉2A = 118⁄7 ≈ 16.857 V polarity upwards.
3(a). Directly 𝑉𝐴 ≈ 6.857 V, apply KVL to find 𝑉2A.
3(b). 𝑉𝐴(6V) = 24⁄7 V, 𝑉𝐴(2A) = 24⁄7 V.
𝑉2𝐴(6V) = 24⁄7 V, 𝑉2𝐴(2𝐴) = 94⁄7 V.
3(c). The Thevenin equivalent seen by the 4Ω is 𝑉𝑇𝐻 = 12 V and 𝑅𝑇𝐻 = 3 Ω.
The Thevenin equivalent seen by the 2A is 𝑉𝑇𝐻 = 24⁄7 V and 𝑅𝑇𝐻 = 47⁄7 Ω.
4.
𝐼6Ω = 0 A.
4(a). 𝑉6Ω = 0 V, then use Ohm’s law.
4(b). 𝐼6Ω(12V) = 1.2 A upwards, 𝐼6Ω(3A) = 1.2 A downwards.
4(c). The Thevenin equivalent seen by the 6Ω is 𝑉𝑇𝐻 = 0 V and 𝑅𝑇𝐻 = 4 Ω.
5.
𝐼10V = 22⁄9 ≈ 2.44 A upwards.
5(a). 𝐼10V(10V) = 40⁄9 A upwards, 𝐼10V(3A) = 2 A downwards.
5(b). The Thevenin equivalent seen by the 10V is 𝑉𝑇𝐻 = 4.5 V and 𝑅𝑇𝐻 = 2.25 Ω.
6.
𝐼2V = 0 A.
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
2.3
Number Systems
A number system is a systematic way to represent numbers. Each digit's value
depends on the digit itself, its position, and the base of the system. For example,
consider the number 32110 , where the subscript represents the base. The base, in
this case, is 10, indicating the decimal number system we are familiar with. The
number can be expanded as follows:
Base
32110 = 3 × 102 + 2 × 101 + 1 × 100
Position
Besides the familiar representation, there are other number systems in use. Let us
explore some common ones. In particular, we are interested in the binary number
system, which only contains two possible digits - 0 and 1. It is very powerful as it
is used in many computer processors and digital applications.
Section 2.3 Concept Map
Mathematical Models
Positional Numeral Systems
Hexadecimal
conversion
between
Decimal
conversion
between
conversion
between
Binary
conversion
between
Other Bases
2.3.1. Standard Bases
Intended Learning Outcomes
•
•
Be able to understand and reason that ordering and symbols used to represent
numbers are abstract concepts and can be arbitrarily chosen.
Be able to acquire the knowledge to identify and differentiate various number
systems, particularly those relevant to computer applications, and effectively
expand numbers using different number systems.
While counting objects in daily life, we often count from 0 to 9 and so on. We are
used to the decimal number system, which is among many approaches to order and
count objects in daily life as sets of 10 (decimal) and using 10 unique symbols to
represent them. Here, 10 is the base of the decimal number system. However, it is
possible to count differently. We could count using sets of 8, and 8 unique symbols
or even just sets of 2, with 2 unique symbols. Depending on the different bases
possible, there are different number systems.
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Number Systems
Suppose we want to count the people before us in a queue, as shown below:
Let us count using different number systems.
I. Base: 10
II. Base: 16
Unique symbols: 0,1,2,3,4,5,6,7,8,9
Unique symbols:
0,1,2,3,4,5,6,7,8,9, A, B, C, D, E, F
- called the decimal number system.
III. Base: 3
Unique symbols: 0,1,2
- called the hexadecimal number system.
IV. Base: 3
Unique symbols: A, B, C
- called the ternary number system.
Note that the ordering and symbols used above are arbitrary and can be defined
according to your preference to create a number system. In our everyday life,
we use the decimal number system (base-10) to count things like people in a
queue. This system uses digits from 0 to 9 to represent quantities.
The “people” we counted can be treated as entities. Examples of entities that can
be counted are days, alphabets, trees and discrete amplitude levels of a signal,
among others. In base-10, we count from 0 to 9 before the number rolls over (like
counting from 0 to 9 on our fingers). For instance, 10 means there are 10 entities
present. Now, invent your number system and count the people in the queue:
Good luck with making it as popular as the decimal number system!
V. Finally, let us use the binary number system to count the people in the queue:
Base: 2, Unique symbols: 0, 1
We discussed a few bases and their corresponding number systems (unique
symbols list). The decimal number system (base-10) is commonly used in daily life.
The binary number system (base-2) and hexadecimal number system ( base-16) are
common in computer applications.
In the binary number system, the numeral at each position is called a binary digit or
bit. For the hexadecimal system, the numeral at each position is called a hex digit.
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Dr. Jesin James
Fundamentals of Electrical and Digital Systems
In a number, the left-most digit is the most significant digit, while the right-most
digit is the least significant digit. When performing computations, it is crucial to
ensure the accuracy of the most significant digit, while errors in the least significant
digit have a minimal impact on overall system performance. E.g., consider the
difference of 5.23 vs 5.22, and 5.23 vs 6.23.
Most significant digit in the binary number system becomes most significant bit
(MSB), and least significant digit becomes least significant bit (LSB).
A number may be compact in one base but very long (possibly infinitely long) in
another. Consider Example (f). In base-16, the number 𝐴. 10𝐵 is compact, but the
same number represented in base-10 is 10.065185546875, which is rather long.
Hence, the choice of the number system will depend on the application.
Other Bases
Octal numbers (base-8) were commonly used in early computer development,
with each digit ranging from 0 to 7. 32-bit data paths in modern microprocessors
would make base-32 numbers potentially attractive. However, the catch is trying
to devise unique symbols for the “digits” from 132 to 3132 , not to mention the
problem of memorising them.
2.3.2. Conversion between Bases
Intended Learning Outcomes
•
Be able to recognize patterns and relationships between numbers in different
bases, enabling conversion of a number between different bases.
Engineers frequently encounter the need to convert numbers between bases when
working with computer systems. In this section, we will explore the methods and
techniques used to perform these conversions.
Converting a number in the decimal number system to any base
Consider a decimal number that has both an integer and a fractional part, for
example: 14.62510. The integer and fractional parts are converted separately to the
other base and then combined. You may find following steps useful to convert a
number from decimal system (base-10) to any other number system (new base)52.
Integer part conversion
Step 1 Divide the decimal number to be converted by the value of the new base.
Step 2 Get the remainder from Step 1 as the rightmost digit (LSD) of the new
base number.
Step 3 Divide the quotient of the previous divide by the new base.
Step 4 Record the remainder from Step 3 as the new base number's next digit
(to the left).
Repeat steps 3 and 4, getting remainders from right to left until the quotient
becomes 0 in Step 3. The last remainder thus obtained will be the Most Significant
Digit (MSD) of the new base number.
52
After some practice, these conversions will become intuitive.
Dr. Jesin James
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Number Systems
Fractional part conversion
Step 1 Multiply the number to be converted by the value of the new base.
When you multiply, the result will have an integer part and a fractional
part.
Step 2 Get the integer part from Step 1 as the new base system's leftmost digit
(MSD).
Step 3 Multiply the fractional part of the previous multiplication by the new
base.
Step 4 Record the integer part from Step 3 as the new base number's next digit
(to the right).
Repeat Steps 3 and 4, getting integer parts from left to right, until the fractional
part becomes 0 in Step 3. The last integer part thus obtained will be the LSD of the
new base number.
Example 37. Convert 14.62510 to binary.
Integer part = 14
Step
Operation
Result
Remainder
Integer part
Fractional part
Fractional part = 0.625
Step
Operation
Once the conversion is completed, remember to include the base as the suffix.
(ANS: 1110.1012 )
Example 38. Convert 9.7510 to hexadecimal.
Integer part = 9
Step
Page 66
Operation
Result
Remainder
Dr. Jesin James
Fundamentals of Electrical and Digital Systems
Fractional part = 0.75
Step
Operation
Integer part
Fractional part
(ANS: 9. 𝐶16 )
Converting a number in any base to decimal number system
To convert a number from any base to the decimal number system, we expand the
number by general number system representation. This includes multiplying each
digit of the number by its corresponding digit-weighting factor and then summing
up all the product terms in the expansion to obtain the decimal number.
Example 39. Convert 1.1012 and 𝐵𝐸716 to decimal.
(ANS: 1.62510 , 304710 )
Conversion between bases
With the knowledge to convert a number in decimal number system to any base and
any base to decimal number system, we can convert a number between any two bases.
Example 40. Convert 𝐴. 416 to binary.
(ANS: 1010.01002 )
Dr. Jesin James
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Number Systems
Converting a number between binary and hexadecimal number systems
Converting between base-16 and base-2 is straightforward because 16 = 24 ,
meaning each hex digit corresponds to exactly four bits. To convert from
hexadecimal to binary, concatenate the binary representation of each hex digit.
Conversely, to convert from binary to hexadecimal453
7 , group the bits in sets of 4 on
either side of the binary point.
First, let us find the binary equivalent of all the unique hexadecimal symbols:
Hex
Binary
0
1
2
3
4
5
6
7
Hex
Binary
8
9
A
B
C
D
E
F
Knowing the binary equivalent of each of the unique symbols in the hexadecimal
number system, we can now convert a number between binary and hexadecimal
number systems.
Example 41. Convert
(a) 𝐵𝐸716 to binary.
(b) 1110100102 to hex.
(c) 𝐴. 416 to binary.
(d) 101.11111001012 to hex
(ANS: (𝑎)1011111001112 , (𝑏)1𝐷22
(𝑐)1010.01002 , (𝑑)5. 𝐹9416 )
53
Cheat sheet: 8421 can be used to convert from hexadecimal to binary quickly.
8 4
2
1
Sum
8×1+4×0+2×0+1×1= 9
1 0
0
1
So, if you want to find the binary representation for 9, find the combination of 1s and 0s under 8421
to get the sum as 9.
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Dr. Jesin James
Fundamentals of Electrical and Digital Systems
2.3.3. Representation Limits
Intended Learning Outcomes
•
Be able to develop the ability to critically assess the limitations of number
systems, understanding the implications of representation limits.
Be able to acquire the skills to mathematically analyse and quantify the impact
of having a fixed number of unique symbols in a number system, allowing for
a comprehensive understanding of the system's capabilities and constraints.
•
We will now focus our discussion on the binary number system and some practical
limitations to be considered while using the binary number system to represent
numbers. Using the binary number system, let us re-visit counting the people
(entities) in a queue. A few more people have joined the queue, as shown below:
Let us use the binary number system to count the people in the queue:
Base: 2, Unique symbols: 0, 1
From the above example, depending on the number of entities to count (represent),
we have to choose the number of bits needed to represent the entities. Hence, the
number of bits we choose imposes a limit on the number of entities we can represent
using the number system. This is called the representation limit.
Representation limit is considered while choosing computer processors for various
applications. Computer processors are designed with a fixed number of bits5054; and
8, 16, 32 and 64 - bit systems are common. For example, an 8 - bit computer
processor can work with numbers (called data) that are 8 bits wide.
Any pattern of 8 bits is generally called a byte. Half a byte or 4 bits is called a nibble. 16 bits is called
a word and 32 bits is called a long word.
54
Dr. Jesin James
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Number Systems
Suppose we need to represent the decimal number 12810 using 6 bits only.
Converting 12810 to binary,
Step
Operation
Result
Remainder
This is an 8 - bit representation. But we have only 6 bits to represent. In computer
memory, the bits are shifted cyclically from the left, with the MSB entering the
memory register first55. Story the binary representation of 12810 to a 6 - bit memory.
When the binary number is stored in the memory, and the binary number is longer
than the length of the memory available, then the memory locations get overwritten cyclically. This is called memory overflow and can cause an overflow error
or display a message “the number is too large”.
For the above example, the 6 bits in the memory would be:
Converting it back to decimal gives:
which is not the number we want to represent. This is a simple example to show
that there are limits on the numbers we can represent using the binary number
system456, depending on how many bits we have to represent the number.
Example 42. Calculate the largest number (considering positive integers only)
in the decimal number system that can be represented by an 8 bit computer processor and a 5 - bit computer processor?
The range of numbers that can be represented using 8 bits is 000000002 to
111111112. The range of numbers that can be represented using 5 bits is ………
to ………. In the decimal number system, this is:
Number of bits
8
Base-2 number
000000002
111111112
Base-10 number
5
(ANS: 25510 , 3110 )
Notice that this is only one strategy of storing bits in memory. Other approaches exist and effective
strategies of memory organisation is an interesting research area.
56Any number system for that matter.
55
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Dr. Jesin James
Fundamentals of Electrical and Digital Systems
Example 43. Complete the following table listing the largest decimal number
that can be represented by 𝑘 bits, where 𝑘 = 1, 2, 3, 8.
No. of bits (𝑘)
1
2
3
.
.
8
The largest number in binary
The largest number in decimal
We can use this information to calculate the number of bits needed to represent a
decimal number. 25510 is the largest number that can be represented using 8 bits
memory, and 3110 is the largest number that can be represented using 5 bits.
Also, 255 = 28 − 1, and 31 = 25 − 1. This observation can be generalised to:
The maximum decimal number 𝑁 that can be represented by 𝑘 bits is 2𝑘 − 1.
In the examples discussed above, the numbers started from 0 and went up to the
largest decimal number. For example, the numbers start from 0 up to 255. Hence,
256 different entities are represented by 8 bits.
Example 44. Complete the table listing the number of bits required to represent
the following decimal numbers.
Decimal number (𝑁)
0
1
2
3
4
5
6
7
8
9
Binary number
Number of bits (𝑘)
Consequentially, the number of bits needed to represent a number 𝑁 can be
determined by finding an integer 𝑘 such that:
𝑁 ≤ 2𝑘 − 1
Note that 25510 is the largest number that can be represented by 8 bits. So, any
number less than or equal to 25510 can also be represented by 8 bits. Hence, the
equation contains the ≤ sign. To calculate 𝑘,
𝑘 ≥ log 2 (𝑁 + 1),
as 𝑘 must be an integer, we use the ceiling function. Hence, 𝑘 = ⌈log 2 (𝑁 + 1)⌉ .
This equation means, to represent the decimal number 𝑁, at least 𝑘 bits are needed.
Dr. Jesin James
Page 71
Number Systems
Example 45. Calculate the largest positive integer that a 64-bit processor can
store. Represent your answer in the decimal number system.
(ANS: 264 − 1)
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Dr. Jesin James
Fundamentals of Electrical and Digital Systems
2.3.4. Problems
1.
Express the following binary numbers in decimal form.
(a) 101.1012 (b)
0111.112
(c)
1010.012
(d) 111.1112 (e)
1000.01012 (f)
10101.0112
2.
Express the following numbers in binary, octal, and hexadecimal forms.
(a) 17310
(b)
313.062510
(c)
112.510
(d) 65510
(e)
38.7510
3.
Convert the numbers to decimal form
(a) 3𝐹2. 𝐷16 (b)
75.368 (c)
FA5.616
(d)
725.38
4.
Consider the positive decimal integers range 0 through to 106
(a) Alteast, how many bits are needed to represent all the numbers?
(b) Can you represent all the numbers in the range using 25 bits?
(c) Can you represent all the numbers in the range using 19 bits?
(d) Let us generalise this. Consider the positive decimal integers range 0
through to 𝑁 . Alteast, how many bits are needed to represent all the
numbers in the range?
5.
Compute the decimal equivalents of the following binary numbers.
5
3
(a)
⏞2
1. 111
𝑁−1
⏞
1. 11111
2
(b)
(c)
1. ⏞
111 … 1112
𝐻𝑖𝑛𝑡: 𝑈𝑠𝑒 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑡𝑜 𝑁 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑎 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑝𝑟𝑜𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛:
𝑁−1
0
1
2
𝑥 +𝑥 + 𝑥 + ⋯+ 𝑥
𝑁−1
= ∑ 𝑥𝑛 =
𝑛=0
6.
1 − 𝑥𝑁
.
1−𝑥
Atleast, how many bits are required to represent the following entities?
(a) The seasons in Aotearoa New Zealand (Summer, Autumn, Winter, Spring)
(b) The number of days in a leap year
(c) All integer amplitudes of a sine wave that varies from [−5, 2], including
both the extreme amplitude values.
(d) All the colours in a rainbow
(e) The average number of steps walked by a person per day, known to be
16384. Note that this will be the memory size of the steps counter on your
smartwatch.
(f) All colours on a traffic light
7. For question 6, generally, let 𝑘 be the number of bits required to represent the
entities. Are all possible combinations of the bits (𝑘 bits) used to represent the
entities listed from (a) to (f)? If no, which combinations remain unused?
Dr. Jesin James
Page 73
Number Systems
Solutions to Problems
1(a). 5.62510 (b) 7.7510 (c) 10.2510 (d) 7.87510 (e) 8.312510 (f) 21.37510
2(a). 101011012 , 2558 , 𝐴𝐷16
(b)
1110000.12 , 160.48 , 70.816
(e) 100110.112, 46.68, 26. 𝐶16
100111001.00012 , 471.048 , 139.116
(c)
(d)
10100011112 , 12718 , 28𝐹16
3(a). 1010.812510 (b) 61.4687510 (c) 4005.37510 (d)
469.37510
4(a). 20 bits (b) Yes (c) No (d) ⌈log 2(𝑁 + 1)⌉
5(a). 1.875 (b) 1.96875 (c) 2(1 − 2−𝑁 )
6(a). 2 (you are trying to represent 4 entities: 0, 1, 2, 3)
6(b). 9 (you are trying to represent the number 366: 0 to 366, hence 367 entities)
6(c). 3 (you are trying to represent 8 entities: -5, -4, -3, -2, -1, 0, 1, 2)
6(d). 3 (you are trying to represent 7 entities (colours): 0 to 6)
6(e). 15 (you are trying to represent the number 16384: 0 to 16384, hence 16385
entities)
6(f). 2 (you are trying to represent 3 entities: 0, 1, 2)
7(a) All combinations of 2 bits (00, 01, 10, 11) are used to represent the 4 entities.
7(b) All combinations of 9 bits are not used to represent the 367 (0 to 366) entities.
Only 00000000002 to 1011011102 are used. The combinations from
1011011112 to 11111111112 remain unused.
7(c) All combinations of 3 bits are used to represent the 8 entities:
000, 001, 010 … . 111.
7(d) All combinations of 3 bits are not used to represent the 7 entities. 111 remains
unused.
7(e) All combinations of 15 bits are not used to represent the 16385 (0 to 16384)
entities. Only 0000000000000002 to 1000000000000002 are used. The
combinations from 1000000000000012 to 1111111111111112 remain
unused.
7(f) All combinations of 2 bits are not used to represent the 3 entities. The
combination 11 remains unused.
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Dr. Jesin James
Fundamentals of Electrical and Digital Systems
2.4
Boolean Algebra
The binary number system is widely used in computer processing. Unlike the
algebra of real numbers, which uses ten unique symbols to represent numbers, the
binary system has only two symbols: 0 and 1. In Boolean algebra, we manipulate
these binary symbols, which have logical significance rather than numerical meaning.
Section 2.4 Concept Map
Mathematical Models
Positional
Numeral Systems
Algebraic
Systems
Decimal
Binary
underpins
the study of
Boolean Algebra
Hexadecimal
equivalence between
Other Bases
Truth Table
2.4.1. Basic Boolean Operations
Intended Learning Outcomes
•
•
•
•
Be able to understand the logical significance of binary mathematical symbols
and their role in representing information in computer systems.
Be able to compare and contrast the operations performed by different Boolean
operations, such as AND, OR, and NOT, on binary mathematical symbols.
Be able to create truth tables, which show the output values for different input
combinations, for any Boolean operation to analyse its behaviour and logic..
Be able to derive any Boolean operation using the basic Boolean operations to
build complex logical expressions.
For any algebraic system, we need to define the mathematical symbols and
manipulations that can be performed on those symbols. In the binary number
system, the mathematical symbols used are binary 1 (HIGH) and binary 0 (LOW).
We shall discuss some manipulations that are developed for these mathematical
symbols, which constitute the basic operations in Boolean Algebra.
Consider two binary variables, 𝑋 and 𝑌 which can only take the values 0 or 1.
Truth Table:
To examine all possible combinations of inputs 𝑋 and 𝑌, we use a truth table. A
truth table exhaustively lists all input-output pairs for a logic function or circuit
(you will learn logic circuits later), providing a clear understanding of how output
variables relate to input variables. In a truth table with 𝑁 inputs there are 2𝑁 rows.
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Boolean Algebra
AND Operation
The AND operation results in HIGH only when both inputs are HIGH.
While constructing the truth table, first, we write all possible combinations of
inputs 𝑋 and 𝑌. Executing the AND operation, we can complete the truth table.
𝑋
𝑌
𝑍 =𝑋⋅𝑌
AND is represented with a dot (⋅). The truth table can be summarised as:
1, when
𝑋⋅𝑌 ={
0, when
OR Operation
The OR operation results in HIGH when at least one of the inputs is HIGH.
Representing this information as a truth table:
𝑋
𝑌
𝑍 =𝑋+𝑌
OR is represented with a + sign. The truth table can be summarised as:
0, when
𝑋+𝑌 = {
1, when
AND and OR are binary operators that require two input variables.
NOT / COMPLEMENT Operation
NOT is a unary operator that performs negation operation – basically negating
HIGH to LOW, and LOW to HIGH. Representing this as a truth table:
𝑋
𝑍=𝑋
Symbolically, NOT is represented with a ̅ sign above the variable. The truth table
can be summarised as:
1, when
𝑋={
0, when
All other Boolean operations can be expressed in terms of these fundamental
operations – AND, OR, NOT, also called basic operations.
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Dr. Jesin James
Fundamentals of Electrical and Digital Systems
XOR (Exclusive OR) Operation
This is an operation derived from the basic operations. XOR operations result
HIGH only when the inputs are different. Representing this using a truth table:
𝑋
𝑌
𝑍 = 𝑋⊕𝑌
XOR is represented with a ⊕ sign. The truth table can be summarised as:
1, when
𝑋⊕𝑌 = {
0, when
XOR can be represented using the basic Boolean operations:
𝑋⊕𝑌 =𝑋⋅𝑌+𝑋⋅𝑌
Example 46. Prove: 𝑋 ⊕ 𝑌 = 𝑋 ⋅ 𝑌 + 𝑋 ⋅ 𝑌 using truth tables
(ANS: Construct the truth table of LHS and RHS, and compare. )
Example 47. Two-input XNOR Boolean operation results HIGH only when the
inputs are the same.
a. Establish a relationship between XNOR and XOR operations.
b. Represent XNOR in terms of the basic Boolean operations.
(ANS: a)𝑋 XNOR 𝑌 = 𝑋 ⊕ 𝑌, b) 𝑋 XNOR 𝑌 = 𝑋 ⋅ 𝑌 + 𝑋 ⋅ 𝑌)
Dr. Jesin James
Page 77
Boolean Algebra
2.4.2. Boolean Identities
Intended Learning Outcomes
•
•
•
Be able to derive basic Boolean identities by analysing and understanding truth
tables, recognizing relationships between input and output values.
Be able to apply Boolean identities to simplify Boolean expressions, reducing
complexity and improving interpretability.
Be able to draw analogies between real algebra and Boolean algebra,
understanding how operators and their precedence operate in both systems.
In the previous section, we compared truth tables to demonstrate the equivalence
of Boolean expressions. However, relying solely on truth tables can be timeconsuming, especially as the number of variables increases. To simplify Boolean
manipulations, we use Boolean identities, which are relationships that allow us to
express expressions in more useful ways. These identities help make Boolean
equations more concise, similar to how recognizing trigonometric identities
simplifies algebraic equations. For example, when manipulating an algebraic
equation, it is useful to recognise that 𝑠𝑖𝑛(𝐴) 𝑐𝑜𝑠(𝐵) − 𝑐𝑜𝑠(𝐴) 𝑠𝑖𝑛(𝐵) can be written
as 𝑠𝑖𝑛(𝐴 − 𝐵), since it usually simplifies the resulting equations. Similarly, we will
use Boolean identities to make Boolean equations as compact as possible. Basic
Boolean identities can be derived from the truth tables. Let us explore some
Boolean identities using a combined truth table for AND, OR, and NOT operations.
(a)
𝑋
(b)
𝑌
(c)
𝑍 =𝑋⋅𝑌
(d)
𝑍 =𝑋+𝑌
(e)
𝑍=𝑋
(i)
(ii)
(iii)
(iv)
Observing the above table, from columns (a), (b) and (c), we can see that:
𝑋⋅0=
(Rows (i), (iii))
𝑋⋅1=
(Rows (ii), (iv))
𝑋⋅𝑋 =
(Rows (i), (iv))
𝑋⋅𝑋 =
(Rows (ii), (iiii))
Summarising these into Boolean identities:
Identity
𝑋. 0 = 0
1
𝑋. 1 = 𝑋
2
𝑋. 𝑋 = 𝑋
3
4
𝑋. 𝑋 = 0
Page 78
Observation
ANDing with 0 obliterates 𝑋
ANDing with 1 has no effect
ANDing with itself has no effect
ANDing with its complement obliterates 𝑋
Dr. Jesin James
Fundamentals of Electrical and Digital Systems
By considering different possibilities of 𝑋 and the impact of the AND operator, we
can understand various identities. For instance, consider 𝑋 ⋅ 𝑋 . When 𝑋 =
1, then 𝑋 = 0, giving 1 ⋅ 0 = 0. When 𝑋 = 0, there is still a 0 in the expression,
causing the AND operator to return 0 as the result. Regardless of the value of
𝑋 (0 or 1), there will always be a 0 in the 𝑋 ⋅ 𝑋 expression, resulting in 0.
From the truth table on the previous page, examining columns (a), (b), and (d), we
can observe that:
Summarising these into Boolean identities:
Identity
Observation
𝑋+0=
5
ORing with 0
𝑋
+
1
=
6
ORing with 1
𝑋
+
𝑋
=
7
ORing with itself
8
ORing with its compliment
𝑋+𝑋 =
From the truth table on the previous page, examining columns (a) and (e): What
happens if we complement (e)? It becomes the same as (a). So, 𝑋 = 𝑋
Identity
Observation
9
The complement of a complement negates itself
𝑋=𝑋
Further examination of the truth table and introducing a new variable can lead to
the derivation of additional identities which can be proven using truth tables57.
10
11
12
13
14
15
16
17
Identity
𝑋+𝑌 =𝑌+𝑋
𝑋. 𝑌 = 𝑌. 𝑋
(𝑌
𝑋 + + 𝑍) = (𝑋 + 𝑌) + 𝑍
𝑋. (𝑌. 𝑍) = (𝑋. 𝑌). 𝑍
𝑋. (𝑌 + 𝑍) = 𝑋. 𝑌 + 𝑋. 𝑍
𝑋 + 𝑌. 𝑍 = (𝑋 + 𝑌). (𝑋 + 𝑍)
̅̅̅̅̅
𝑋. 𝑌 = 𝑋 + 𝑌
̅̅̅̅̅̅̅̅
𝑋 + 𝑌 = 𝑋. 𝑌
Observation
OR is commutative
AND is commutative
OR is associative
AND is associative
AND distributes over OR
OR distributes over AND
De Morgan’s law I
De Morgan’s law II
De Morgan’s laws are transformational laws that help transform the OR operation
to AND or the AND operation to OR, as required.
We can construct analogies between the Boolean operators and the operators in
real algebra. We can see that the AND behaves similar to ………… and OR behaves
similar to ………….. This analogy helps to understand the order of precedence of
Boolean operations. NOT is performed first, followed by AND, and then OR.
Consider 2-variables or 3-variables truth tables for each of the identities 10 to 17 and pick the rows
in the truth table that help you prove the identities, just like we did for identities 1 to 9.
57
Dr. Jesin James
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Boolean Algebra
Example 48. Prepare the truth table for the Boolean expression 𝑋. (𝑌 + 𝑍), and
prove that AND operation distributes over OR operation
Example 49. Simplify the following to have the fewest Boolean operations:
(𝐴. 𝐴)
(a) ̅̅̅̅̅̅̅̅
(b) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
(𝐴 ⋅ 𝐵) ⋅ (𝐴 ⋅ 𝐵)
(c) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
(𝐴 ⋅ 𝐴) ⋅ (𝐵. 𝐵)
̅̅̅̅̅̅̅̅
(d) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
𝐴 + 𝐴 ⋅ ̅̅̅̅̅̅̅̅
𝐵+𝐵
(e) 𝐴 + 𝐴. 𝐵
(f) 𝐴 + ̅̅̅̅
𝐵𝐴
Page 80
Dr. Jesin James
Fundamentals of Electrical and Digital Systems
(𝐴 + 𝐵)
(g) 𝐴 + 𝐵 + ̅̅̅̅̅̅̅̅̅̅
(ANS: (a)𝐴, (b)𝐴 + 𝐵, (c)𝐴 + 𝐵, (d)𝐴 + 𝐵, (e)𝐴, (f)1, (g)𝐴. 𝐵)
Dr. Jesin James
Page 81
Boolean Algebra
2.4.3. Converting Truth Tables to Boolean Equations
Intended Learning Outcomes
•
Be able to develop the ability to evaluate Boolean expressions by utilising truth
tables and applying Boolean identities, enabling the simplification and
optimisation of logical expressions.
By now, you understand that both truth tables and Boolean algebraic equations
can be used to represent Boolean operations. But when do we choose to use either of
these approaches? Consider a scenario where you want to design two physically
separate switches controlling an alarm. The intended behaviour is as below:
• When both switch 𝐴 and switch 𝐵 are OFF – The alarm should turn ON
• When only switch 𝐵 is ON – The alarm should turn OFF
• When only switch 𝐴 is ON – The alarm should turn ON
• When both switch 𝐴 and switch 𝐵 are ON – The alarm should turn ON
The truth table is a good approach to represent this logic
(consider active HIGH5 logic58 – ON = 1, and OFF = 0):
Switch 𝐴
Switch 𝐵
Alarm
Now, we will explore the conversion of information from a truth table into a
Boolean algebraic expression: Consider the following truth table replacing switch
𝐴 with 𝐴 , switch 𝐵 with 𝐵 and alarm with 𝑍.
𝐴
𝐵
𝑍
(1)
<<
(2)
(3)
<<
(4)
<<
Careful examination shows that 𝑍 = 1 for row (1) OR row (3) OR row (4). Now
looking at row (3), we see that row (3) represents the condition 𝐴 = 1 AND 𝐵 = 0,
which can also be written as 𝐴 = 1 AND 𝐵 = 1. Similarly, row (4) represents the
condition 𝐴 = 1 AND 𝐵 = 1. Combining the conditions for the three rows gives:
𝑍 =𝐴⋅𝐵+𝐴⋅𝐵+𝐴⋅𝐵
This is the sum of products equation representing the truth table. This is not
necessarily the most compact representation. We use Boolean identities to simplify:
𝑍 =𝐴⋅𝐵+𝐴⋅𝐵+𝐴⋅𝐵
=
=
=
58
There is also active-LOW where HIGH is represented by 0, and LOW is represented by 1.
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Dr. Jesin James
Fundamentals of Electrical and Digital Systems
Notice that we had quite a bit of work to do because there were three terms in our
expression. However, an inspection of the truth table shows that there is only one
row with 𝑍 = 0. We can derive an expression for the output using zero-rows.
From row (2) of the truth table, we see that 𝑍 = 𝐴 ⋅ 𝐵. Inverting both sides, we
discover a much shorter route to the solution:
𝑍 = 𝐴. 𝐵
=
=
Example 50. (a) Write a Boolean expression for 𝑆 in the following truth table in
terms of 𝐴, 𝐵, 𝐶, and determine the number of Boolean operations
in the expression.
(b) Simplify the expression to have the minimum number of
Boolean operations.
𝐴
0
0
0
0
1
1
1
1
𝐵
0
0
1
1
0
0
1
1
𝐶
0
1
0
1
0
1
0
1
𝑆
0
0
0
0
0
1
1
1
(ANS: (a)10 operations needed, (b)Simplified to 𝑆 = 𝐴(𝐵 + 𝐶), 2 operations needed)
Dr. Jesin James
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Boolean Algebra
2.4.4. Problems
1.
Prepare a truth table for the Boolean expressions
(a)
(b)
(c)
𝑆 = 𝐴𝐵 + 𝐵𝐶
𝑍 =𝐶⋅𝐴⋅𝐵
𝑍 = (𝐴 ⊕ 𝐵) ⋅ 𝐶
2.
Use a truth table to verify that ̅̅̅̅̅̅
𝐴 ⋅ 𝐵 ≠ 𝐴 ⋅ 𝐵.
3.
Prove the identity 𝐴 + 𝐴 ⋅ 𝐵 = 𝐴 + 𝐵 using Boolean algebra.
4.
Simplify the expressions to obtain the minimum number of Boolean
operations:
(a) 𝐴 + 𝐴 ⋅ 𝐵
(b) 𝐴 + 𝐴 ⋅ 𝐵
Observing the identity in Problem 3 and the results in Problem 4 a, b, use
words to generalise the identity in Problem 359?
5.
Simplify the following Boolean expressions to obtain the minimum number
of Boolean operations:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
6.
𝑍 = ̅̅̅̅̅̅̅̅̅̅
(𝐴 + 𝐴)
𝑍 = 𝐶 + (𝐷𝐴𝐵𝐶)
𝑍 = 𝐴(𝐴 + (𝐴𝐵))
𝑍 = 𝐴 + 𝐵𝐴
𝑍 = 𝐴𝐵 + 𝐴 + 𝐵
𝑍 = ̅̅̅̅̅̅̅̅̅
𝐴 + 𝐵𝐶
𝑍 = 𝐴(𝐴 + 𝐵 + 𝐶)
𝑍 = ̅̅̅̅̅̅̅̅
𝐴+𝐵
𝐹 = 𝐴 ⋅ 𝐵 ⋅ 𝐶 + 𝐴 ⋅ (𝐵 + 𝐶)
𝐷 =𝐴⋅𝐵⋅𝐶+𝐴⋅𝐵
𝐸 = (𝐴 + 𝐵) ⋅ (𝐴 + 𝐴 ⋅ 𝐵)
𝐷 = (𝐴 + 𝐵) ⋅ (𝐴 + 𝐶)
An XOR operation with 𝑛 inputs gives a true (i.e. a HIGH or 1) output when
the number of true inputs is odd.
Prepare the truth table for
a) a 3-input XOR operation,
b) a 3-input XNOR operation.
c) express 3-input XOR and 3-input XNOR operation using basic Boolean
operations only.
This problem is a good example to not associate the identities to the exact variable name. Rather
the variables represent logical states which can be 0 or 1 in Boolean algebra.
59
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Dr. Jesin James
Fundamentals of Electrical and Digital Systems
Dr. Jesin James
Page 85
Boolean Algebra
Solutions to Problems
3.
The proof of this identity is straightforward using a truth table but requires
a trick when using Boolean algebra.
𝐴+𝐴⋅𝐵 = 𝐴
⏟
⋅ 1 + 𝐴 ⋅ 𝐵 = 𝐴 ⋅ (𝐵
⏟ + 1) + 𝐴 ⋅ 𝐵 = 𝐴 ⋅ 𝐵 + 𝐴 + 𝐴 ⋅ 𝐵
𝐵+1=1
𝐴⋅1=𝐴
= 𝐴 + 𝐵 ⋅ (𝐴
⏟+ 𝐴) = 𝐴 + 𝐵
𝐴+𝐴=1
a) 𝐴 + 𝐵, b) 𝐴 + 𝐵
Observation: if we have the first variable ORed to the negation (NOT) of the
first variable ANDed by a second variable, we can replace the entire expression
with just first variable OR second variable. The impact of the negated first variable
can be ignored.
4.
5
𝑍 = 0 (b).
𝑍=𝐶
𝑍 = 1 (f).
𝑍 = 𝐴𝐵𝐶
𝐹 = 𝐴 ⋅ (𝐵 + 𝐶)
𝐸=𝐵
(a).
(e).
(i).
(k).
(c).
(g).
(j).
(l).
𝑍=𝐴
(d).
𝑍=𝐴
(h).
𝐷 = 𝐴 ⋅ (𝐵 + 𝐶)
𝐷 =𝐴+𝐵⋅𝐶
𝑍=𝐴
𝑍 = 𝐴𝐵
6.
𝐴
0
0
0
0
1
1
1
1
𝐵
0
0
1
1
0
0
1
1
𝐶
0
1
0
1
0
1
0
1
𝐴⊕𝐵⊕𝐶
0
1
1
0
1
0
0
1
𝐴 ⊕ 𝐵 ⊕ 𝐶 (XNOR)
1
0
0
1
0
1
1
0
c) from the truth table:
𝐴 ⊕ 𝐵 ⊕ 𝐶 = 𝐴 𝐵 𝐶 + 𝐴 𝐵 𝐶 + 𝐴 𝐵 𝐶 + 𝐴𝐵𝐶
𝐴 ⊕ 𝐵 ⊕ 𝐶 = 𝐴 𝐵 𝐶 + 𝐴 𝐵 𝐶 + 𝐴 𝐵 𝐶 + 𝐴𝐵𝐶 = 𝐴 𝐵 𝐶 + 𝐴 𝐵𝐶 + 𝐴𝐵 𝐶 + 𝐴𝐵𝐶
Page 86
Dr. Jesin James
Fundamentals of Electrical and Digital Systems
2.5
Alternating Current (AC) Concepts
Our studies of circuit behaviours thus far have been largely limited, in scope, to a
specific “class” of inputs – constant, or direct current (DC), signals. These are signals
in which the polarity of the voltage, and direction of the current do not change (e.g.,
a battery), and consequently, the responses of a circuit to these signals must also
be time-invariant. In practice however, all interesting phenomena vary with time
to convey some sort of intelligible information (e.g., a musical piece, a video
recording, or the voltage output of an ECG implant, etc.), and naturally, in order
for us to design and create electronics to interact with these phenomena, we need
to extend our understanding of circuits to how they would behave when excited
by time-varying, or alternating current (AC), signals.
While it may appear insurmountable at first – signals come in practically infinite
number of forms and realisations – one of the great advances of the 19th century
was the work of Joseph Fourier who discovered that any practical signal can be
expressed as a sum of sinusoids, and in doing so, paved the path for an unifying
way of representing arbitrary signals of all shapes and forms using just a set of
“basic” building-blocks – the humble sinusoids.
The ability to represent arbitrary signals as a combination of sinusoids means that
if we are able to understand how a circuit would respond to a single sinusoidal
input, then we have a way to analyse the circuit subject to any arbitrary AC inputs
– all analyses can be reduced to a sum of individual sinusoidal analysis via the
superposition theorem (see Sect. 2.2.4) – we can therefore restrict our study of circuit
behaviours to just sinusoidal sources 60 without loss of generality!
52F
Section 2.5 Concept Map
Mathematical Models
Signal Characterisation
Fundamental Electrical Quantities
time-varying nature
described by
Current
Sinusoids
Other TimeVarying
Functions
constitute
used to derive
Charge
Voltage
provides rationale
for understanding
Power
behaviour can be summarised by
time-varying nature
characterised by
Root Mean
Square Value
a convenient metric
for calculating
Voltage and Current
Equivalent
Resistance
Terminal Characteristic of Components
Superposition
Voltage &
Current Division
Principles
power
behaviour
OPAMP
quantified by
Ideal Wire
Capacitor
Node-Voltage
Analysis
Tools of Circuit Analysis
Apparent Power
Ohm s Law
the ratio gives
Inductor
Ideal Sources
Theorems
Reactive Power
vectorially
sums to
Kirchhoff s Laws
Thevenin s
Theorem
Methods
Resistor
Average
Power
Power Factor
Reactive Elements
Power Metrics
Behaviour and Quantification of Electrical Quantities
This study is also important since the generation, transmission, distribution, and consumption of
electrical energy from our mains supply are all in sinusoidal forms due to the ease at which they can
be produced (to be discussed in Sect. 4).
60
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Alternating Current (AC) Concepts
2.5.1. Sinusoidal Representation
Intended Learning Outcomes
•
Be able to represent and model signals as a sinusoidal function in time and
make relevant interpretation about the underpinning signals they represent.
A sinusoid is a periodic signal 61 that varies sinusoidally with time. Mathematically,
we can express a sinusoidal signal using either the sine or cosine function. Here, we
will arbitrarily take the cosine convention, and define a sinusoid to be of the form
53F
𝑥(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜙) = 𝐴 cos(2𝜋𝑓𝑡 + 𝜙),
where
𝐴 (amplitude)
is the magnitude (≥ 0) of the maximum deviation from the
average value of 𝑥(𝑡).
𝜔 (angular frequency)
is the rate (in rad s −1 ) at which the signal ‘travels’ and is
related to the frequency, f (in Hz), and period, T (in second),
by 𝜔 = 2𝜋𝑓 = 2𝜋/𝑇.
𝜙 (phase)
is the angle (in ° or rad) that the waveform has shifted from
a standard cosine, this corresponds to a time shift of 𝜙/𝜔
along the time-axis.
𝑣(𝑡)
𝑇 = 20 ms
𝑣(𝑡) = 5 cos(2𝜋50𝑡 − 45∘ )
5
𝑡 (ms)
𝐴=5
−5
𝜙
𝜋/4
=
= 2.5 ms
2𝜋𝑓 2𝜋50
Note that the phase 𝜙 is often expressed in degrees, however, the unit of 𝜔𝑡 is in
radians62. Thus, in order to evaluate, say 𝑣(5ms) = 5 cos(0.5𝜋 − 45∘ ), the argument
of the cosine must be converted to the same unit via the unit conversions:
𝜋
180°
1° =
rad, and
1 rad =
180
𝜋
A periodic signal is one that repeats its value in regular intervals. Specifically, a signal 𝑥(𝑡) is said
to be periodic if there is a non-zero constant 𝑇 such that 𝑥(𝑡 + 𝑇) = 𝑥(𝑡) for all time 𝑡. The number 𝑇
is known as the period of the signal, and the smallest of such a number is known as the fundamental
period.
62 Simply because we are preconditioned to be able to interpret and visualise the meaning of degrees
in comparison to the more mathematically convenient counterpart, radians.
61
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
Thus, we can evaluate 𝑣(5ms) as
𝑣(5ms) = 5 cos (0.5𝜋
⏟ − 45∘ ) = 5 cos(45∘ ) =
=90∘
or equivalently,
5
√2
= 5 cos (0.5𝜋 − ⏟
45∘ ) = 5 cos(0.25𝜋) =
=0.25𝜋
V
5
√2
V,
depending on your (calculator’s!) preference.
Example 51. A sinusoidal current 𝑖(𝑡) has a maximum measured amplitude of
10 A. The current passes through one complete cycle in 1 ms and
the magnitude of the current at time 𝑡 = 0 was measured to be 5
A and decreasing. Determine
(a)
(b)
(c)
(d)
the frequency of the current in Hz and rad s −1 ,
an expression for 𝑖(𝑡) using the cosine function,
the instantaneous value at 𝑡 = 2⁄3 ms, and
the first time after 𝑡 = 0 that 𝑖(𝑡) = −5 A
(ANS: 1 kHz and 2𝜋1000 rad s −1 , 𝑖(𝑡) = 10 cos(2𝜋1000𝑡 + 60∘ ) A, 5 A, 1/6 ms)
Since trigonometric functions are periodic over 360∘ , their representations are
therefore non-unique and we have infinite ways of representing the same signal, e.g.,
𝑣(𝑡) = 5 cos(2𝜋50𝑡 − 45∘ ) = 5 cos(2𝜋50𝑡 − 45∘ ± 360∘ 𝑛) for 𝑛 = 0,1,2, ….
To manage this rather chaotic situation, the agreed convention is to
express the phase between −180∘ < 𝜙 ≤ 180∘ , the so-called principal values.
In doing so, all sinusoids can thus be uniquely represented by a corresponding one
whose phase is within the principal values.
Dr. William Lee
Page 89
Alternating Current (AC) Concepts
It should also be clear that in order for us to make meaningful comparisons
between sinusoids (of the same frequency), they must be expressed in the same
form. For example, if the sinusoidal voltage across, and current through a
component is 𝑣(𝑡) = 5 cos(2𝜋50𝑡 − 45∘ ) and 𝑖(𝑡) = −2 sin(2𝜋50𝑡 + 165∘ )
𝑣(𝑡) = 5 cos(2𝜋50𝑡 − 45∘ ) V
𝑖(𝑡) = −2 sin(2𝜋50𝑡 + 165∘ ) A
𝑡 (ms)
45∘ 60∘
105∘
it is not immediately obvious in its existing form how far ‘behind’ or ‘ahead’ (in
phase, or time) the current 𝑖(𝑡) is relative to the voltage 𝑣(𝑡). But when expressed
in the same form, this comparison becomes immediately obvious since the phase
are now expressed with reference to the same trigonometric function (cosine):
𝑖(𝑡) = −2 sin(2𝜋50𝑡 + 165∘ ) = 2 cos(2𝜋50𝑡 − 105∘ ) A
revealing that the voltage is ahead (or ‘leads’) of the current through the load by
−45∘ − (−105∘ ) = 60∘ . Equivalently, we can say that the current is ahead of the
voltage by −105 − (−45) = −60∘ which is identical to the former statement as the
negative implies that it is behind (or ‘lags’) the voltage by 60∘ .
Convince yourself of the following equivalence:
sin(𝑥) = cos(𝑥 − 90∘ ) ,
and
− cos(𝑥) = cos(𝑥 ± 180∘ ).
Example 52. Calculate the phase difference between the two voltages
𝑣1 (𝑡) = −10 cos(2𝜋50𝑡 + 50∘ ) and 𝑣2 (𝑡) = 12 sin(2𝜋50𝑡 − 10∘ ),
and state which sinusoid is leading in time.
(ANS: 𝑣2 leads 𝑣1 by 30∘ or 5⁄3 ms)
Page 90
Dr. William Lee
Fundamentals of Electrical and Digital Systems
2.5.2. Average and Root Mean Square (RMS) Value
Intended Learning Outcomes
•
Be able to characterise common AC periodic signals in terms of their rootmean-square values
While sinusoidal signals can be described and modelled precisely using
trigonometric functions, unlike DC signals these functions are rather awkward to
work with, and compare between, due to their time-varying nature. Often, we
need not know the specific value of the AC signal at any given point in time but
are usually content with a descriptive measure that captures its central behaviour,
allowing a quick way to discern between the various AC signals – some sort of
measure of average.
While the standard arithmetic average is one of such measures, it is not very
informative when it comes to quantifying AC, and in particular, sinusoidal signals.
Recall the average, or the mean, of a continuous periodic signal 𝑥(𝑡) is
mathematically given by
𝑋AVG =
1 𝑡0 +𝑇
∫
𝑥(𝑡) 𝑑𝑡
𝑇 𝑡=𝑡0
or in words: we ‘sum up’ (i.e., integrate) the signal over its period and divide by
(i.e., take the average over) the period.
Notice the standard average when applied to an arbitrary sinusoidal signal of the
form 𝑥(𝑡) = 𝐴 cos(2𝜋𝑓𝑡 + 𝜙) will always be zero and therefore does not provide
any meaningful information about them:
Example 53. Show that the average of an arbitrary sinusoidal signal of the form
𝑥(𝑡) = 𝐴 cos(2𝜋𝑓𝑡 + 𝜙) is zero.
Dr. William Lee
Page 91
Alternating Current (AC) Concepts
A commonly used measure in place of the normal average is the root mean square
(rms or RMS) value. Much like the arithmetic average, the RMS value is just another
type of average that provides some measure of central tendency of the signals.
The RMS value of a signal 𝑥(𝑡), as its name suggests, is the
square-root of the mean (average) of the square of the signal over its period 𝑇.
Mathematically, this is captured concisely by
𝑡0 +𝑇
1
2
𝑋RMS = √ ∫ (𝑥(𝑡)) 𝑑𝑡.
𝑇
𝑡=𝑡0
As we will see later on, the RMS value 63 conveniently allows us to do power
calculations due to AC excitations in a familiar and meaningful way.
5F
The RMS value is also sometimes referred to as the effective value because it turns out to be the value
of a DC current (or voltage) that effectively delivers the same amount of average power (more on what
this means later on) to a resistor as the underpinning AC current that has this RMS value.
63
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 54. Apply the definition of RMS value to show that the RMS value of an
arbitrary sinusoidal signal of the form 𝑥(𝑡) = 𝐴 cos(2𝜋𝑓𝑡 + 𝜙) is
𝑋RMS = 𝐴/√2.
Hint: you may find the identity cos(2𝐴) = 2 cos2(𝐴) − 1 useful.
From the above example we see that the RMS value of a sinusoid does not depend
on either its frequency or the phase shift. It should also be obvious that
the RMS value of a signal necessarily depends on the (shape and size of the) signal
itself and is therefore different depending on the underlying signal for which
the RMS value is to be calculated for.
For example, we have seen that the RMS value of a standard sinusoid relates to its
amplitude by a factor of √2, however, this is not necessarily true for a square wave
as demonstrated in the following example.
Dr. William Lee
Page 93
Alternating Current (AC) Concepts
Example 55. Determine the average and the RMS
value of a square-wave signal of
period 𝑇 that is ‘ON’ for a fraction 𝐷
of the period, and ‘OFF’ for the
remainder of the period.
x(t)
A
T
DT
2T
t
(ANS: 𝑋AVG = 𝐴𝐷, 𝑋RMS = 𝐴√𝐷)
When in doubt, apply the definition of RMS value and see where the result leads!
Example 56. Determine the average and the RMS
value of the sawtooth waveform
shown.
x(t)
A
T
2T
t
(ANS: is 𝑋AVG = 𝐴/2, 𝑋RMS = 𝐴/√3)
Page 94
Dr. William Lee
Fundamentals of Electrical and Digital Systems
2.5.3. Capacitor and Inductor
Intended Learning Outcomes
•
Be able to recognise the time-domain terminal characteristic of capacitors and
inductors, and their implications towards their behaviours under DC, and
under sinusoidal excitations.
Capacitor
A capacitor is an electrical component that consists of two conductive ‘plates’
separated by a non-conductive (i.e., insulating) region (e.g., air, ceramic, plastic,
etc.) – there is no direct conducting path through a capacitor. Capacitors are
presented symbolically by
C
C
Fixed Capacitor
Polarised Capacitor
+
C
Variable Capacitor
When a voltage is applied across a capacitor, positive charges are accumulated
onto one of the plates which in turn attract (repel) equal number of negative
(positive) charges from the other plate. This displacement of equal but opposite
charges on the two separated plate establishes a force of attraction (i.e., Coulomb’s
law, see Sect. 2.1.1) between them. As more charges are displaced, the attraction
force increases until it balances out with the force underpinning the voltage
applied across the capacitor – equilibrium is reached with no more movement of
charges.
From the description of its operation, it should become intuitive that the number
of positive (or negative) charges accumulated on the plate is directly proportional
to the voltage applied across the capacitor:
𝑞 = 𝐶𝑣,
where the constant of proportionality is known as the capacitance of the capacitor
characterising its “capacity” to “store” charge (energy).
In order to relate the current (rate of flow of charge) to the capacitor to the voltage
across it, we can simply take the derivative of the governing behaviour above to
get
𝑑𝑣
𝑖=𝐶 ,
𝑑𝑡
the terminal characteristic of the capacitor. The terminal characteristic says in order
for a current to be present “through” the capacitor, the voltage across the capacitor
must be changing, as otherwise the capacitor will simply be at its equilibrium state.
Dr. William Lee
Page 95
Alternating Current (AC) Concepts
Capacitors are therefore not particularly interesting at DC (although there are
certainly uses for them!) since it will simply “charge” to its balanced state at which
𝑑𝑣 ⁄𝑑𝑡 = 0 ⇒ 𝑖 = 0, and behaves like an open-circuit. On the other hand, when an
AC, sinusoidal, voltage 𝑣(𝑡) = 𝐴 cos(2𝜋𝑓𝑡 + 𝜙𝑣 ) is applied across a capacitor we
see that 64
56F
𝑖(𝑡) = 𝐶
𝑑𝑣(𝑡)
= 𝐶 ⋅ (−2𝜋𝑓 ⋅ 𝐴 sin(2𝜋𝑓𝑡 + 𝜙𝑣 ))
𝑑𝑡
=
2𝜋𝑓𝐶 ⋅ 𝐴
⏟
(𝜙𝑣 + 90∘ ))
cos (2𝜋𝑓𝑡 + ⏟
amplitude of 𝑖(𝑡)
phase of 𝑖(𝑡)
which says that the resulting
current “through” the capacitor leads the voltage across it by exactly 90∘
since (𝜙𝑣 + 90∘ ) − 𝜙𝑣 = 90∘ .
Inductor
An inductor (or a coil) is another electrical component that has the ability to store
energy. It consists of a conductive wire wound in a spiral shape around a magnetic
core (ceramic, ferrite, iron, etc.) – there is thus a direct conducting path through an
inductor. Inductors are presented symbolically by
L
L
L
Fixed Inductor
Iron-cored Inductor
Variable Inductor
A current through a wire creates a continuous circular magnetic field around it –
the larger the current, the stronger this magnetic field becomes. When the wire is
in the shape of a coil, the particular geometry allows the induced magnetic field
within the coil to ‘link’ together, creating a resultant flux 𝜙 inside the coil along
one direction 65. If the coil consists of 𝑁 turns, then this resultant flux is reinforced
𝑁 times as the current ‘flows’ through the turns. The ‘total’ flux, 𝜆, is therefore
directly proportional to the current 𝑖 through the coil:
57F
𝜆(𝑡) = 𝑁𝜙(𝑡) = 𝐿𝑖(𝑡)
where the constant of proportionality 𝐿 is known as the inductance 66 of the inductor
characterising its ability to ‘induce’ magnetic field.
58F
Recall the derivative of 𝑦 = cos(𝑎𝑥) is 𝑑𝑦⁄𝑑𝑡 = −𝑎 sin(𝑎𝑥).
This direction is commonly determined via the so-called right-hand rule.
66 The inductance depends on the number of coil windings, the shape of the inductor, but also the
magnetic core the coil is on, etc. These are factors that determines the ‘strength’ of the resultant
magnetic field inside the inductor.
64
65
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
In order to relate the voltage across the inductor to the current through it, we make
use of Faraday’s law 67 and take the derivative of the above to get
59F
𝑑𝜆(𝑡)
𝑑𝑖(𝑡)
=𝐿
𝑑𝑡
𝑑𝑡
⇒
𝑣(𝑡) = 𝐿
𝑑𝑖(𝑡)
𝑑𝑡
the terminal characteristic of the inductor. The terminal characteristic says in order
for a voltage to be present across an inductor, the current through the inductor
must be changing, as otherwise the flux through the inductor will be constant.
Like the capacitor, the inductors are therefore not particularly interesting at DC,
since they will simply have uniform magnetic flux through them (due to constant
current) at which 𝑑𝑖 ⁄𝑑𝑡 = 0 ⇒ 𝑣 = 0, and behaves like a short-circuit. On the other
hand, when an AC, sinusoidal current 𝑖(𝑡) = 𝐴 cos(2𝜋𝑓𝑡 + 𝜙𝑖 ) is applied across an
inductor we see that
𝑣(𝑡) = 𝐿
𝑑𝑖(𝑡)
= 𝐿 ⋅ (−2𝜋𝑓 ⋅ 𝐴 sin(2𝜋𝑓𝑡 + 𝜙𝑖 ))
𝑑𝑡
=
2𝜋𝑓𝐿 ⋅ 𝐴
⏟
amplitude of 𝑣(t)
(𝜙𝑖 + 90∘ ))
cos (2𝜋𝑓𝑡 + ⏟
phase of 𝑣(𝑡)
which says that the result voltage across the inductor leads the voltage across it by
exactly 90∘ . Equivalently, this says the
current through the inductor lags the voltage across it by exactly 90∘
since 𝜙𝑖 − (𝜙𝑖 + 90∘ ) = −90∘ .
Faraday’s law states that any change in the magnetic flux in a coil will induce a voltage
proportional to the rate of change of the flux to oppose the current creating the flux.
67
Dr. William Lee
Page 97
Alternating Current (AC) Concepts
2.5.4. Power Characterisation
Intended Learning Outcomes
•
•
•
Be able to differentiate, relate, and interpret between average power, apparent
power, and instantaneous power of an underlying device driven by sinusoidal
excitations.
Be able to relate, and make connections, between the power factor of a device
to its power characteristics, and provide relevant interpretation towards its
equivalent electrical component composition.
Be able to determine the relevant power characteristics of a device subject to
sinusoidal excitations.
Power is perhaps the most important quantity in all electric utilities, electronics,
and communication systems as the underlying operating principles of these
systems all revolve around the transmission, conversion, and utilisation of power.
As such, every industrial and household electrical devices has a power rating – the
nominal power required by the device.
Conveniently, sinusoidal sources are also the predominant means of providing
electric power due to the ease at which they are produced, it is therefore important
that we understand, and be able to interpret, the power characteristics of a device
under sinusoidal excitations.
Instantaneous Power
Recall the power of an electrical device is, by definition, the product of the voltage
across and the current through that device. When a device is driven by a constant
source (i.e., in a DC circuit), its power must therefore be constant during its
operations (why?). On the other hand, if a device is excited by a time-varying
source (i.e., in an AC circuit), then the voltage 𝑣(𝑡) and current 𝑖(𝑡) of the device
must be time-varying, and so must its power:
𝑝(𝑡) = 𝑣(𝑡)𝑖(𝑡).
When the power of a device is a function of time, we often prepend it with the term
“instantaneous” and refer to it as the instantaneous power of the device to be
explicit of its time-dependent nature: it describes the rate at which energy is
absorbed68 by a device at any instant of time.
i(t)
Device
v(t)
Recall the voltage and current in the power definition are taken to be in opposing directions as per
the passive-sign convention. The product therefore gives the amount of power absorbed or received by
the device. If the product is negative (at certain instances in time) then it means the device is delivering
or supplying power at those instances (see Sect. 2.1.5).
68
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
Our interest in the instantaneous power revolves around sinusoidal voltage and
current signals of the general form
𝑣(𝑡) = 𝑉𝑝 cos(𝜔𝑡 + 𝜙𝑣 )
and
𝑖(𝑡) = 𝐼𝑝 cos(𝜔𝑡 + 𝜙𝑖 )
for some constants 𝑉𝑝 , 𝐼𝑝 , 𝜙𝑣 , and 𝜙𝑖 . Equivalently, since the signals are identically
periodic (of the same frequency), the zero-time frame of reference is therefore
arbitrary (as the relative ‘position’ of the signals do not change), so we lose nothing
by selecting the zero-time instant to be at (one of) the peak of the voltage signal:
new reference
𝑣(𝑡)
𝑖(𝑡)
𝑡
𝜙𝑣 𝜙𝑖 − 𝜙𝑣
𝜙𝑖
In doing so, we can conveniently consider the signals
𝑣(𝑡) = 𝑉𝑝 cos(𝜔𝑡)
𝑖(𝑡) = 𝐼𝑝 cos(𝜔𝑡 + (𝜙𝑖 − 𝜙𝑣 ))
and
to describes the exact same situation 69. The (instantaneous) power 𝑝(𝑡) through an
arbitrary device due to a sinusoidal excitation is thus given by
61F
𝑝(𝑡) = 𝑉𝑝 cos(𝜔𝑡) ⋅ 𝐼𝑝 cos(𝜔𝑡 + (𝜙𝑖 − 𝜙𝑣 ))
= 𝑉𝑝 𝐼𝑝 cos(𝜔𝑡) cos(𝜔𝑡 + (𝜙𝑖 − 𝜙𝑣 )) .
While the above expression provides a precise description of the power of a device
at any given time, it is not a useful description for its behaviour. For example, here
is a plot of the power across a device and its anteceding excitations:
𝑣(𝑡)
4V
𝑣(𝑡) = 4 cos(2𝜋50𝑡) V
𝑖(𝑡)
𝑖(𝑡) = 3 cos(2𝜋50𝑡 − 60∘ ) A
3A
20 ms
−4 V
𝑡
−3 A
20 ms
𝑡
𝜋/3
= 3.33ms
2𝜋50
𝑝(𝑡)
9W
−3 W
offset = 3 W
𝑡
20 ms
𝑝(𝑡) = 12 cos(2𝜋50𝑡) cos(2𝜋50𝑡 − 60∘ ) W
The only reason we have done so is to simplify the mathematics in the discussions to come. All of
the results still hold if we do not make this shift in the time reference, but the mathematics become
unnecessarily messy.
69
Dr. William Lee
Page 99
Alternating Current (AC) Concepts
It should be evident from the plot that none of the salient features of the resulting
instantaneous power (e.g., its peaks, troughs, period, and offset) is immediately
obvious from looking at its mathematical expression. Its description (as it stands)
therefore does not provide any intuition for its underlying behaviour.
To reconcile the apparent discrepancy between the mathematical description 𝑝(𝑡)
and its graphical representation, we can re-write it equivalently as70
𝑝(𝑡) = 𝑉𝑝 𝐼𝑝 cos(𝜔𝑡) cos(𝜔𝑡 + (𝜙𝑖 − 𝜙𝑣 ))
1
1
= 𝑉𝑝 𝐼𝑝 cos(𝜙𝑣 − 𝜙𝑖 ) + 𝑉𝑝 𝐼𝑝 cos(2𝜔𝑡 − (𝜙𝑣 − 𝜙𝑖 )).
⏟
⏟
2
2
=constant
=time−varying
Notice by deliberately breaking 𝑝(𝑡) down into a much more amicable form, we
can now deduce that the instantaneous power due to sinusoidal excitations
generally consists of
•
a constant component whose value depends on the phase difference between
𝑣(𝑡) and 𝑖(𝑡),
•
a time-varying component that varies sinusoidally at twice the frequency of
the underpinning voltage and current, and
•
may be negative for a portion of each cycle – power is delivered by the device
(to the circuit it is connected in) due to capacitors and/or inductors.
Returning to the previous example, by writing out the power in the more
meaningful form
𝑝(𝑡) = 12 cos(2𝜋50𝑡) cos(2𝜋50𝑡 − 60∘ ) = 3 + 6 cos(2𝜋100𝑡 − 60∘ ) W,
it should now be apparent what the salient graphical features of 𝑝(𝑡) that eluded
its previous form are.
Despite this neat algebraic workaround, the instantaneous power still presents
itself as mathematical burden to convey on a day-to-day basis. Clearly, we need a
more coherent way (or ways) of summarising the useful information contained
within the instantaneous power of a device:
we need a (few) descriptive metric(s) that we can infer from the instantaneous
power in order to communicate the largest amount of information about the
nature of the underlying device as simply as possible without the mathematical
baggage that comes with writing out the instantaneous power.
70
Use the trigonometric identity 2 cos(𝐴) cos(𝐵) = cos(𝐴 − 𝐵) + cos(𝐴 + 𝐵).
Page 100
Dr. William Lee
Fundamentals of Electrical and Digital Systems
Average Power
One of the most intuitive summarising metrics for a time-varying phenomenon is
its average, the simplest indicator of central behaviour. The average of the
instantaneous power of a device over (one of) its period is therefore an obvious
candidate for a useful measure. If the voltage and current of the device are
sinusoidal, then it should be evident from the earlier argument that this average is
1 𝑡0 +𝑇
𝑃AVG = ∫
𝑝(𝑡) 𝑑𝑡
𝑇 𝑡=𝑡0
1 𝑡0 +𝑇 1
1 𝑡0 +𝑇 1
= ∫
𝑉𝑝 𝐼𝑝 cos(𝜙𝑣 − 𝜙𝑖 ) 𝑑𝑡 + ∫
𝑉𝑝 𝐼𝑝 cos(2𝜔𝑡 − (𝜙𝑣 − 𝜙𝑖 )) 𝑑𝑡
𝑇
𝑇
⏟ 𝑡=𝑡0 2
⏟ 𝑡=𝑡0 2
1
= 𝑉𝑝 𝐼𝑝 cos(𝜙𝑣 −𝜙𝑖 )
2
=0 (why?)
and so
1
𝑃AVG or 𝑃 ≝ 𝑉𝑝 𝐼𝑝 cos(𝜙𝑣 − 𝜙𝑖 ).
2
We therefore denote this quantity as the average power of a device, a number that
provides some central description about the power behaviour of the device.
The unit of the average power is the usual Watt (W).
Example 57. A 150 Ω resistive lamp is connected to a sinusoidal voltage source
of 230√2 V in amplitude. Find an expression for the power
absorbed by the lamp over time, and how much power it absorbs
on average.
(ANS: 𝑝(𝑡) =
Dr. William Lee
2116
3
cos 2(2𝜋50𝑡) W =
1058
3
+
1058
3
cos(2𝜋100𝑡) W, 𝑃 =
1058
3
W)
Page 101
Alternating Current (AC) Concepts
For the purpose of power calculations, the RMS value makes it useful appearance
here, as it allows us to re-write the average power expression in a more nostalgic
form in comparison to the DC scenario:
𝑉𝑝
𝐼𝑝
1
𝑃 = 𝑉𝑝 𝐼𝑝 cos(𝜙𝑣 − 𝜙𝑖 ) =
⋅
⋅ cos(𝜙𝑣 − 𝜙𝑖 ) = 𝑉RMS 𝐼RMS cos(𝜙𝑣 − 𝜙𝑖 ).
2
⏟
⏟
√2
√2
=𝑉RMS
=𝐼RMS
The RMS values are so useful in practice that, unless specified otherwise, values
quoted in AC context are the RMS values not maximum values. Here, we will make
the distinction by including an RMS label in its unit where appropriate. So, for
example, the New Zealand mains voltage is 230 V(rms), and this corresponds to a
sinusoid with an amplitude of 230√2 V.
Example 58. A resistive heater draws, on average, 1.5 kW of power from the
NZ mains supply – 230 V(rms). Determine the
(a) peak value of the current drawn by the heater, and
(b) the equivalent resistance of the heater.
(ANS: 150√2/23 A, 529/15 Ω)
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
While the average power metric provides an easy-to-understand measure about
the power behaviour of a device but, it turns out that, it is only ‘half’ of the picture.
This is because the average power metric comes with some caveats:
Inability to Distinguish between Capacitive and Inductive Devices
Consider two devices where one is purely capacitive and the other purely inductive
(i.e., both contain no resistors). The phase difference between the (sinusoidal)
voltage and current of these devices must be 𝜙𝑣 − 𝜙𝑖 = ±90∘ (why?). If we are to
compute the average (of the) power of the two devices we will discover, in both
cases, that
1
𝑃 = 𝑉𝑝 𝐼𝑝 cos (𝜙
⏟𝑣 − 𝜙𝑖 ) = 0 W,
2
∘
=±90
the average power metric therefore tells us nothing about what the nature of the
underlying devices is. By extension, this means that a capacitive and an inductive
device (containing non-zero resistance) could both produce the same non-zero
average power but there is no way for us to differentiate between them.
Misleading Notion of Zero Electrical Movements
When the average power is zero, it can engender a false notion of zero power (and
therefore zero voltage and current) when clearly there are still non-zero electrical
movements in the devices as seen from the instantaneous power:
1
1
1
(𝜙𝑣 − 𝜙𝑖 )) = ± 𝑉𝑝 𝐼𝑝 sin(2𝜔𝑡) ≠ 0.
𝑝(𝑡) = 𝑉𝑝 𝐼𝑝 cos (𝜙
⏟𝑣 − 𝜙𝑖 ) + 𝑉𝑝 𝐼𝑝 cos(2𝜔𝑡 − ⏟
2
2
2
∘
∘
⏟
⏟
=±90
=0
=±90
=± sin(2𝜔𝑡)
The resulting 𝑝(𝑡) shows that a (purely) capacitive or inductive device still draws
current and voltage, but in a way such that half of the time power is being absorbed
(stored) by the device, while the same amount of power is delivered (extracted)
from the device during the other half of the cycle. The average (power) is therefore,
unsurprisingly, zero because energy is alternately exchanged between the device
and the circuit (source) driving it in equal ‘volume’ – the energy never leaves the
circuit to do anything ‘useful’, instead, remains in electrical form indefinitely.
For this very reason, the average power is also commonly referred to as the real
power or the active power because any non-zero value of it is an indicative
measure of the amount of power that is being converted from electrical to nonelectrical form (e.g., heat for a resistor) – it measures the amount of power that
leaves the circuit to do something ‘useful’.
The average power, 𝑃, therefore undermines the richness of information about a
device contained within its instantaneous power 𝑝(𝑡) – there are other useful
information we can extract from the instantaneous power that the average power
alone does not capture.
Dr. William Lee
Page 103
Alternating Current (AC) Concepts
Power Factor
Another useful metric that complements the average power metric is the phase
angle difference of the device voltage and current, i.e., 𝜙𝑣 − 𝜙𝑖 . This metric is
useful because it directly provides an indication of the nature of the device (that
the average power fails to capture):
−90∘
⏟
≤
𝜙𝑣 − 𝜙𝑖
≤
capacitive
current 𝐥𝐞𝐚𝐝𝐬 voltage
⏟∘
90
.
inductive
current 𝐥𝐚𝐠𝐬 voltage
It also plays a role in determining the average power, as the cosine of this angle
difference appears as a factor that ‘controls’ the size of the average power:
⏟
0
≤ cos(𝜙𝑣 − 𝜙𝑖 ) ≤
⏟
1
.
resistive
𝐚𝐥𝐥 of 𝑝(𝑡) leaves the circuit
capacitive 𝐨𝐫 inductive
𝐧𝐨𝐧𝐞 of 𝑝(𝑡) leaves the circuit
The latter seems to give us a measure of the ‘proportion’ of the (instantaneous)
power of the device that is being converted into other (non-electrical) forms. We
therefore denote the cosine of the phase angle difference between the device
voltage and current as the power factor or p.f. and the phase angle difference as
the power factor angle:
p. f. ≝ cos ( ⏟
𝜙𝑣 − 𝜙𝑖 ).
≝ p.f. angle
Notice the nature of the device cannot be determined from the value of the power
factor71. As such, power factors must be specified with the accompanying terms
leading or lagging to indicate if the device current leads or lags the device voltage:
•
If the device current lags the device voltage (i.e., 𝜙𝑖 < 𝜙𝑣 or 𝜙𝑣 − 𝜙𝑖 > 0), the
device is inductive, and it has a lagging p.f.
•
If the device current leads the device voltage (i.e., 𝜙𝑖 > 𝜙𝑣 or 𝜙𝑣 − 𝜙𝑖 < 0),
the device is capacitive, and it has a leading p.f.
Because cosine is an even function. Recall a function 𝑓(𝑥) is even if it satisfies 𝑓(−𝑥) = 𝑓(𝑥) and is
odd if 𝑓(−𝑥) = −𝑓(𝑥) . Thus, cosine is even because cos(−𝑥) = cos(𝑥) and sine is odd because
sin(−𝑥) = − sin(𝑥). Returning to our context, a phase angle difference of 𝜙𝑣 − 𝜙𝑖 = ±60∘ will both
result in a power factor of cos(±60∘ ) = 0.5 and so we cannot tell if the current leads (capacitive) or
lags (inductive) the voltage from the numerical value of the power factor.
71
Page 104
Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 59. A voltage of 𝑣(𝑡) = 100 cos(5𝑡) V is applied across a load and the
current was measured to be 𝑖(𝑡) = −4 sin(5𝑡 − 30∘ ) A (in the
direction of a voltage drop). Determine
(a) the instantaneous power of the load,
(b) the power factor of the load,
(c) the average power absorbed/delivered by the load,
(d) state whether the load is inductive or capacitive.
(ANS: 𝑝(𝑡) = −400 cos(5𝑡) sin(5𝑡 − 30∘ ) W, 0.5 leading, 100 W, capacitive)
Dr. William Lee
Page 105
Alternating Current (AC) Concepts
Apparent Power
Specifying the power factor (or the power factor angle) along with the average
power provided additional information on the nature (i.e., capacitive, inductive,
or resistive) of the device. However, it still fails to address the misleading notion
of zero instantaneous power: a device could do very little ‘useful’ work (i.e., low
average power) but still draws an appreciable amount of current due to reactive
elements. Returning, once again, to the extreme case when the device is purely
reactive (i.e., 𝜙𝑣 − 𝜙𝑖 = ±90∘), both the average power and the power factor are
zero, yet there is still a non-zero (and possibly large) voltage and current through
the device.
On the other hand, a closer look at the average power expression shows that it
appears to be just a ‘portion’ of (as quantified by the power factor) some
underlying encompassing fixed quantity:
1
𝑆 = 𝑉𝑝 𝐼𝑝 = 𝑉RMS 𝐼RMS ,
2
that is, 𝑃 = 𝑆 cos(𝜙𝑣 − 𝜙𝑖 ). Graphically, this can be visualised by the so-called
power triangle:
𝑃
𝜙𝑣 − 𝜙𝑖
𝑆
or
𝜙𝑣 − 𝜙𝑖
𝑆
𝑃
depending on whether the device is inductive (former) or capacitive (latter)72. The
quantity 𝑆 is called the apparent power 73 and solves this problem, for it gives an
overall measure of the size 74 of the load: regardless of the size of the average power
or the power factor, this apparent power quantity is zero if and only if the device
voltage or current is zero. Thus, even if the device does no ‘useful’ work, the
apparent power still provides an indication of the ‘size’ of power required by the
device for it to function. The apparent power therefore gives an indicative measure
of the total power ‘capacity’ required by the device (from the perspective of the
source) irrespective of whether the power supplied by the source is being drawn
to do anything useful75.
6F
67F
The unit of apparent power is volt-ampere (VA).
Note that the average power 𝑃 can also be positive or negative depending on whether the device
is a source or a load. In the discussions thus far, we have assumed that all devices are loads, that is,
they receive, rather than provide, power.
73 It is so called because it seems apparent that the power should be the voltage-current product, by
analogy with DC resistive circuits.
74 We say load 𝐴 is ‘bigger’ than load 𝐵 if 𝐴 draws more current, or draws more power, than 𝐵.
75 What is the ‘opposite’ side of the power triangle, 𝑆 sin(𝜙 − 𝜙 )? That is known as the reactive power
𝑣
𝑖
of the device and is the ‘dual’ to the average power: it quantifies the ‘proportion’ of the
(instantaneous) power retained in electrical form due to reactive elements. As such, it is also known
as phantom or wattless power, power that does no useful work. This is beyond the scope of the course.
72
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Dr. William Lee
Fundamentals of Electrical and Digital Systems
Example 60. A single-phase fan draws 1 kW of active power when plugged
into the NZ mains supply. The fan current was measured to lag
the voltage by 45∘ . Determine
(a) the expression for the current 𝑖(𝑡) supplied by the source,
(b) the power factor, and the apparent power of the fan.
(c) By connecting a carefully chosen capacitor in parallel to the
fan, it is possible to alter the current supplied by the source
to be in phase with the voltage supplied while delivering the
same amount of real power to the fan. Determine the
expression for this new current 𝑖(𝑡) , and by comparing
with (a), explain why it is desirable to do so.
(ANS: 𝑖(𝑡) =
Dr. William Lee
200
23
cos(2𝜋50𝑡 − 45∘ ) A, p. f. =
1
√2
lagging, 𝑖(𝑡) =
100√2
23
cos(2π50t + 0∘ ) A)
Page 107
Alternating Current (AC) Concepts
Example 61. [Optional] The capacitance value, 𝐶, of the capacitor required for
part (c) in Example 60 can be determined as follows.
(a) Show that the current through the capacitor is
𝑖𝐶 (𝑡) = −230√2(2𝜋50)𝐶 sin(2𝜋50𝑡) A,
(b)
Since the supply current 𝑖(𝑡) = 𝑖fan (𝑡) + 𝑖𝐶 (𝑡), where 𝑖fan (𝑡)
is the current found in part (a) of Example 60. By summing
the two currents and show that in order for the resulting
𝑖(𝑡) to be in phase with the supply voltage, 𝑣(𝑡) =
230√2 cos(2𝜋50𝑡) V , the capacitance must be 𝐶 ≈
60.172 μF.
Hint: use the identity 𝑐𝑜𝑠(𝐴 − 𝐵) = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵.
Page 108
Dr. William Lee
Fundamentals of Electrical and Digital Systems
2.5.5. Problems
1.
A sinusoidal voltage signal is described by 𝑣(𝑡) = 50 cos(30𝑡 + 10∘ ) V,
determine its amplitude, phase, period, frequency (in Hz), and its
instantaneous value at 𝑡 = 10 ms.
2.
Express the following sinusoidal signals using the standard cosine
representation:
(a)
(b)
(c)
𝑣(𝑡) = 10 sin(2𝜋50𝑡 + 30∘ ) V,
𝑖(𝑡) = −9 sin(8𝑡 + 100∘ ) A, and
𝑝(𝑡) = −20 cos(2𝜋100𝑡 + 45∘ ) W.
3.
Describe the following sinusoidal signal using the standard cosine
representation:
4.
Determine the phase angle difference between the two current signals
𝑖1 = −4 sin(377𝑡 + 120∘ ) A
and
𝑖2 = 5 cos(377𝑡 − 65∘ ) A,
and state whether 𝑖1 leads or lags 𝑖2 .
5.
Determine the average and the RMS value of the voltage waveform shown.
6.
The voltage across a 10 Ω resistor is given by 𝑣(𝑡) = 50 cos(200𝑡) V.
Determine the peak and average power absorbed by the resistor.
7.
The voltage across, and current through a load is 𝑣(𝑡) = 160 cos(50𝑡) V and
𝑖(𝑡) = −33 sin(50𝑡 − 30∘ ) A (measured in the direction of a voltage drop).
Determine (a) the instantaneous power of the load, (b) the average power
absorbed by the load, and (c) whether the load is capacitive or inductive.
Dr. William Lee
Page 109
Alternating Current (AC) Concepts
8.
A domestic appliance is supplied by an AC 230 V(rms) supply. If the
appliance draws 150 W of average power at a lagging power factor of 0.7,
determine the peak value of the current drawn by the appliance.
9.
An electric heater draws 1.5 kW of average power from a 230 V(rms), 50 Hz
electricity supply. The heater can be considered as a purely resistive
appliance. Determine
(a) the resistance of the heater,
(b) the peak and RMS value of the current drawn by the heater,
(c) the power factor and the phase angle difference between the voltage
across and current through the heater, and
(d) the apparent power supplied by the electricity supply.
Solutions to Problems
1.
𝐴 = 50 V, 𝜙 = 10∘ , 𝑇 ≈ 0.21 s, 𝑓 ≈ 4.77 Hz, 𝑣(10m) ≈ 44.48 V.
2(a). 𝑣(𝑡) = 10 cos(2𝜋50𝑡 − 60∘ ) V.
2(b). 𝑖(𝑡) = 9 cos(8𝑡 − 170∘ ) A.
2(c). 𝑝(𝑡) = 20 cos(2𝜋100𝑡 − 135∘ ) W.
3.
𝑣(𝑡) = 100 cos(100𝜋𝑡) V.
4.
𝑖1 lags 𝑖2 by 85∘ .
5.
𝑉AVG = 6 V, 𝑉RMS ≈ 7.48 V.
6.
peak power = 250 W, average power 𝑃 = 125 W.
7(a). 𝑝(𝑡) = −5280 cos(50𝑡) sin(50𝑡 − 30∘ ) W.
7(b). 𝑃 = 1320 W
7(c). Capacitive.
8.
𝑖peak ≈ 1.32 A.
9(a). 𝑅 ≈ 35.27 Ω
9(b). 𝑖peak ≈ 9.22 A.
9(c). p. f. = 1, 𝜙𝑣 − 𝜙𝑖 = 0∘ .
9(d). 𝑆 = 1500 VA.
Page 110
Dr. William Lee
3
“SMART” Engineering
In general, the phenomena of interest to engineers are not inherently electrical.
However, the instruments used by engineers to interpret and act upon these
phenomena are electrical. This occurs because of the great ease with which
electrical signals may be used, interpreted, and recorded. For instance, a Chemical
engineer monitoring (and controlling) the temperature in a kiln will almost
inevitably turn to electronic instrumentation to carry out this process.
An input transducer (or commonly referred to as a sensor) is a device which
converts a non-electrical signal into an electrical signal. This signal is usually so
small that it cannot be directly recorded or displayed on instrumentation or
directly input to a computer-based data acquisition system. The output of the
transducer is passed through a signal conditioning circuit which provides
amplification and may sometimes perform other signal conditioning functions.
Often, we may wish to take some action based upon our interpretation of the
transducer’s signal. If we wish to control the operation of some non-electrical plant,
we must convert the electrical signal into a non-electrical signal. To do this we use
an output transducer (which is more commonly referred to as an actuator).
These new terms are best illustrated by using the familiar example of a sound
system composed of a turntable, preamplifier, amplifier, and speaker. The
cartridge of the turntable is the input transducer. It converts the mechanically
inscribed sound on the record into a weak electrical signal. The preamplifier
amplifies (i.e., increases the amplitude of) this signal, and the amplifier provides
further (higher power) amplification in order to produce a sufficiently powerful
signal to drive the voice coil of the speaker. Thus, the pre-amplifier and the
amplifier perform signal conditioning. The speaker is the output transducer since
it converts the electrical signal into a non-electrical (sound) signal.
The kinds of measurements a Civil engineer makes are markedly different from
those made by a Chemical and Materials engineer or by a Biomedical engineer.
However, it is really only the type of transducer (input and output) that is used that
differs between the various fields of engineering. Once some sort of electrical
signal is obtained to represent the physical quantity being measured, it is
processed by signal conditioning devices (such as amplifiers) and displayed on
meters or acquired by computer-based data acquisition systems that are common
to all engineering discipline. Thus, the underlying instrumentation systems in
which every modern engineer use to deal with the physical world is the same, and
it is therefore crucial that we foster an appreciation for how these systems of
monitoring, analysis, and response technologies work.
Dr. William Lee
Page 111
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Dr. William Lee
“SMART” Engineering
3.1
Monitoring Technologies
Many texts categorise transducers by the measurand (e.g., temperature, pressure,
displacement, etc.), an alternative, electrical engineering, approach is to categorise
based upon means of operation. In other words, instead of classifying a transducer
by the type of non-electrical signal it monitors, it is more meaningful to categorise
the transducer by the underlying electrical parameter whose change in magnitude
relates to the non-electrical signal of interest. This way of classification provides a
natural scheme for organising all transducers by the five common electrical
analogues: voltage, current, resistance, capacitance, and inductance.
We will primarily focus on the operations of resistive transducers and examine
some of the electronics necessary to interface them with subsequent analysis and
response systems. The remaining types of transducer all operates via the same
principle, the only difference is the electrical property that is used as the analogue
to the non-electrical signal of interest and the underlying mechanisms for doing so.
Section 3.1 Concept Map
Fundamental Electrical Quantities
Behaviour and Quantification of Electrical Quantities
Voltage and Current
behaviour governed
and quantified by
Charge
Current
Voltage
Component Characteristics
Power
Resistor
Kirchhoff s Laws
Ideal Wire
Tools of Circuit Analysis
Methods
Theorems
Equivalent
Resistance
Thevenin s
Theorem
Ohm s Law
lead to the
development of
Ideal Sources
Voltage &
Current Division
Principles
Superposition
Node-Voltage
Analysis
underlying physics
informs the operations of
Strain Gauge
Voltage Divider
used by
means of
Thermometer
shortcomings led
the invention of
Thermistor
can generalise
to understand
Wheatstone
Bridge
Other
Other
Transducers
Other Interfaces
Resistive
Sensors
Sensor Interface
Monitoring Technologies
facilitates the
analysis and design of
Loading Effects
output
behaviours
highlight
Analog Signal
Conditioning
Circuits
Analysis Technologies
SMART Engineering
Dr. William Lee
Page 113
Monitoring Technologies
3.1.1. Resistive Input Transducer
Intended Learning Outcomes
•
•
Be able to describe, and relate, the operation of resistive transducers to the
governing physical quantities.
Be able to determine the relevant electrical and physical quantities pertaining
to a resistive transducer within its nominal operating limits.
Resistive input transducers are devices in which a change in their resistances is
related to the non-electrical parameter of interest. These transducers are simply
resistive materials designed and constructed in clever arrangements to exploits the
change in their physical properties due to variation in the non-electrical quantity
being sensed. They operate by variation in length (𝑙), area (𝐴), or resistivity (𝜌),
parameters that affects the resistance 𝑅 of a material by 76
70F
𝑅=𝜌⋅
𝑙
𝐴
(unit: Ω)
Naturally, the phenomenon of interest should only change one of these parameters
in a resistive transducer so that a proper interpretation can be made from the
corresponding change in resistance.
Strain Gauge
Structural engineers (a sub-discipline within the field of Civil engineering) may be
concerned with monitoring of the structural health of load-bearing members in
large infrastructure such as bridges and high-rise buildings. The transducer used
for such monitoring is known as a strain gauge. The strain gauge is an example of
a resistive input transducer whose change in resistance is related to changes in
length. Small changes (which are indicative of strain in a member under
investigation) may be accurately measured in this manner. These devices, one of
which is shown below, are the size of a postage stamp or smaller.
Tension increases resistance
Insensitive to
lateral force
The strain gauge is affixed to the member to be tested via an adhesive coating
supplied on the back of the gauge. As the member under test (or subject to longterm monitoring) is stretched or compressed along the indicated direction, changes
Can you link this relationship to the intuitive description of the underlying physics of operation of
a resistor as given in Sect. 2.1.3?
76
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Dr. William Lee
“SMART” Engineering
in the length of the long axis of the gauge produce a directly proportional change
in the resistance. This is described quantitatively by
Δ𝑅
=𝐺⋅
𝑅𝑜
Δ𝑙
( )
𝑙𝑜
⏟
= strain, 𝜖
where the constant of proportionality, 𝐺, is the gauge factor which is specific to the
material and the manufacturing process of the device 77.
71F
A complicating feature is that the resistance of the gauge also depends on its
temperature (see Resistance Thermometers), introducing a potential confounding
factor. To overcome this difficulty, a second strain gauge (of properties similar to
the first and held at the same temperature, but not attached to the test member) is
used commonly used as a reference in the interfacing circuitry to the first to
compensate for any temperature variation during the measurement (see the
Wheatstone Bridge later on in Sect. 3.1.2).
77
For example, the commercially popular constantan gauge has a gauge factor of 2.
Dr. William Lee
Page 115
Monitoring Technologies
Resistance Thermometers
One of the ways in which we can measure temperature is to make use of the fact
that the value of any resistance is temperature dependent – what appeared to be a
nuisance for the strain gauge is now exploited to an advantage. Specifically, the
resistivity of the material changes with temperature according to the approximate
linear relationship
𝜌(𝑇) = 𝜌0 (1 + 𝛼(𝑇 − 𝑇0 ))
𝑅(𝑇) = 𝑅0 (1 + 𝛼(𝑇 − 𝑇0 ))
⇒
where 𝜌0 is the resistivity at temperature 𝑇0 , and 𝛼 is the temperature coefficient of
the resistive material 78. This means the change in resistance of the material can be
used as a direct surrogate to measure the change in temperature from 𝑇0
72F
Δ𝑅
= 𝛼 ⋅ (Δ𝑇).
𝑅0
ΔR/Ro
Nickel
6
Copper
5
4
Platinum
3
2
1
0
200
400
600
800
1000
ΔT (˚C)
Generally, the resistance thermometer is constructed by winding platinum wire
around a former, the longer the length of the wire, the more sensitive the
thermometer is to changes in temperature 79. Platinum is commonly used for its
linear temperature dependency over other materials despite its modest coefficient
in comparison to others, and is the standard from −190 ∘ C to 660 ∘ C.
73F
Glass coating
Ceramic rod
Platinum coil
Temperature coefficients of some common materials are: nickel (0.0067), copper (0.0043), silver
(0.0041), and platinum (0.0039). Note that in general 𝛼 also varies as a function of temperature, but
over limited range it is approximately constant.
79 A longer wire necessarily means a larger resistance, a change in temperature will thus give rise to
a larger measurable difference in the resistance.
78
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Dr. William Lee
“SMART” Engineering
Care must be taken to avoid 𝑖 2 𝑅 heating (or Ohmic heating), which becomes
significant as the current becomes too large and would invariably confound the
measurement through self-heating.
It is worth noting that the resistivity 𝜌 of a material is not only a function of
temperature, but also subject to a variety of other physical phenomena such as
pressure, light intensity, magnetic forces, etc. Resistive transducers can thus be
made to detect numerous other non-electrical quantities. Notice the only difference
is the mechanism in which the material property is changed.
Dr. William Lee
Page 117
Monitoring Technologies
Thermistor
Another common resistive input transducer used for temperature sensing is the
thermistor. The active material of a thermistor is a type of semiconductor 80 (e.g.,
germanium, silicon, metallic oxides, etc.) whose resistance changes with
temperature. What makes thermistors particular attractive over the resistance
thermometer is their relatively large temperature dependence.
74F
R/Ω
10 6
Thermistor (2252 Ω at 25˚C
10 5
Resistance Thermometer
(100 Ω at 0˚C
10 4
10 3
10 2
T / ˚C
10 1
0
100
200
300
While the response is a non-linear, exponentially decaying, function of
temperature
𝑅(𝑇) = 𝑅0 𝑒
1 1
𝛽( − )
𝑇 𝑇0
where 𝛽 is a material specific constant, and 𝑅0 is the known resistance at a given
temperature 𝑇0 , over a small range of temperatures a linear approximation is valid.
In general thermistors are used over the temperature range from 0 ∘ C to 100 ∘ C,
although higher temperature measurements are possible.
Having a large sensitivity means the physical size of thermistors can thus be small
(typically 1 to 2 mm) and be made cheap in comparison to their counterpart. Once
again, care must be taken to avoid 𝑖 2 𝑅 heating which is especially detrimental for
the thermistors due to their highly negative temperature coefficient leading to
thermal runaway – an increase in temperature due to self-heating would decrease
the resistance and thereby further increase the current until catastrophic failure
unless a current-controlling resistance is also in the circuit.
2mm
A semiconductor is a material that acts as both an insulator and a conductor depending on the
temperature and the potential difference applied across it.
80
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Dr. William Lee
“SMART” Engineering
3.1.2. Sensor Interface Circuitry
Intended Learning Outcomes
•
•
Be able to determine, and provide specifications for, the relevant circuitry
necessary to interface a resistive transducer to subsequent electrical systems.
Be able to anticipate, and provide the specifications for, the necessary signal
conditioning circuitry, given the characteristics of both the transducer circuitry
and the subsequent circuitry
In order for transducers to be useful, we must have accurate means of monitoring
the value, or at the very least the changes in the value, of their electrical properties
as the non-electrical quantities of interest vary. For a resistive transducer, this
means we need electronic circuits that can facilitate the measurement of miniscule
changes in resistance, but also flexible enough to allow the sensor to be interfaced
directly for use with subsequent control systems for automatic analysis and
response without manual intervention.
The Voltage Divider
The most primitive but intuitive circuit for determining the resistance of a resistive
transducer is just a basic voltage divider with a known resistor 𝑅 in series with the
transducer of unknown value 𝑅𝑥 , supplied by a known constant voltage supply 𝑉𝑆 :
Rx
A
VS
vOUT
R
B
Standard analysis shows that the output voltage 𝑣OUT is related to the transducer
resistance 𝑅𝑥 by
𝑣OUT = 𝑣𝐴 − 𝑣𝐵 = (
𝑅
) ⋅ 𝑉𝑆
𝑅 + 𝑅𝑥
⇒
𝑅𝑥 = (
𝑉𝑆
𝑣OUT
− 1) ⋅ 𝑅.
Thus, by measuring the output voltage 𝑣OUT over time, we can directly infer what
the transducer resistance is, and hence changes in the underpinning physical
quantity being monitored.
In practice it may not be necessary to determine the actual resistance, instead, we
may simply use the output voltage as an analogue to 𝑅𝑥 as it changes (inverse)
proportionally to 𝑅𝑥 . In this way, the variation in the output tracks the changes in
resistance thereby providing a useable voltage representation of measurand. This
voltage can then be used by subsequent electrical or digital systems that make use
of the sensed physical quantity (e.g., to regulate the temperature, to trigger a
warning indicator, or simply to display the measurand in a meaningful way).
Dr. William Lee
Page 119
Monitoring Technologies
Example 62. A Pt100 sensor81 is a platinum thermometer with a resistance of
100 Ω at 0 ∘ C and a temperature coefficient of 0.0039/∘ C. The sensor
is to be used to monitor the temperature between 0 to 95 ∘ C. Design
a voltage-divider interface with a 5 V battery such that the output
will vary nominally about 2.5 V at 25 ∘ C, and determine the largest
output swing possible over the monitored range.
(ANS: 0.4 V)
“Pt” is the symbol for platinum, and “100” stands for the resistance in ohms at 0 ∘ C. A Pt1000 sensor
therefore has a resistance of 1000 Ω at 0 ∘ C. Pt100 is most commonly used.
81
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Dr. William Lee
“SMART” Engineering
Despite its simplicity, the voltage-divider interface comes with many weaknesses
and sources of error:
Power Supply Variation
The output voltage is a function of the supply voltage, which has inherent noise,
imprecisions, and will likely drift over time leading to inaccurate measurements
and error in the inferred value of 𝑅𝑥 .
Temperature Drift
When the measurand of the sensor 𝑅𝑥 is unrelated to temperature (e.g., strain), any
temperature-related drift in 𝑅𝑥 will inevitably confound the measurement.
Idle DC Offset
Changes in the output of the divider (due to variation in 𝑅𝑥 ) is a small variation
about a large offset (DC) voltage. If this output voltage is to be used by subsequent
systems, it necessarily requires a modest amount of amplification (typically 10 to
1000 times), but this means the meaningless offset will also be amplified by the
same amount and likely become incompatible with the following systems, unless
the offset is removed.
Dr. William Lee
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Monitoring Technologies
Example 63. A resistance thermometer 𝑅𝑥 is placed in a modified voltage
divider interface circuit shown so that the output voltage 𝑣𝑆 can be
used to monitor the temperature of a kiln electrically. From prior
testing, the resistance of 𝑅𝑥 is 1000 Ω at the nominal operating
temperature of the kiln, and it varies between 600 Ω and 1500 Ω
for the sort of temperature variations expected in the kiln.
R1
Rx
5V
250 Ω
(a)
(b)
vS
Explain why the correlation between the output voltage 𝑣𝑆
and the temperature of the kiln is negative, i.e., an increase
in 𝑣𝑆 corresponds to a decrease in temperature, vice versa.
Find the resistance of 𝑅1 if, for the expected temperature
variations in the kiln, the maximum output voltage of the
sensor circuit is to be 2 V.
(ANS: 1000 Ω)
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Dr. William Lee
“SMART” Engineering
The Wheatstone Bridge
The shortcomings of the simple voltage-divider interface can be mitigated by
modifying it to contain an additional divider in parallel, and taking the output to
be across the two branches instead,
R1
R2
A
VS
vOUT
R3
Rx
B
giving rise to the topology commonly referred to as the Wheatstone Bridge.
By placing the transducer 𝑅𝑥 in one ‘arm’ of the bridge, standard analysis shows
that the output 𝑣OUT is given by
𝑣OUT = 𝑣𝐴 − 𝑣𝐵 = (
𝑅3
𝑅𝑥
−
) ⋅ 𝑉𝑆 .
𝑅1 + 𝑅3 𝑅2 + 𝑅𝑥
This clever design alleviates the aforementioned weaknesses of the voltage divider
interface in the following ways:
Immunity to Power Supply Variation
To measure the unknown resistance 𝑅𝑥 , we can adjust 82 the resistor values 𝑅2 and
𝑅3 with a known 𝑅1 until 𝑣OUT = 0 (at which we say the bridge is balanced) and can
deduce immediately
𝑅3
𝑅𝑥
−
=0
𝑅1 + 𝑅3 𝑅2 + 𝑅𝑥
76F
⇒
𝑅𝑥 =
𝑅2 𝑅3
.
𝑅1
Notice via this method the inferred value of 𝑅𝑥 is independent of the supply
voltage and only depends on the resistor ratio.
To do so, we usually use a coarser variable resistor for 𝑅2 and a finer variable resistor for 𝑅3 .
Therefore, by varying 𝑅2 we get a ballpark estimate for 𝑅𝑥 while 𝑅3 provides the fine tuning
necessary to narrow down the value.
82
Dr. William Lee
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Monitoring Technologies
Example 64. A Wheatstone bridge has 𝑅1 a fixed 1 kΩ resistor, while 𝑅3 can be
adjusted in 1-Ω steps from 0 Ω to 1100 Ω, and 𝑅2 can be selected
to be 1 kΩ, 10 kΩ, 100 kΩ, or 1 MΩ.
(a)
If the bridge is balanced with 𝑅3 = 732 Ω and 𝑅2 = 10 kΩ,
what is the value of the resistive transducer 𝑅𝑥 ?
(b)
What is the largest value of 𝑅𝑥 for which the bridge can be
balanced?
(c)
Suppose that 𝑅2 = 1 MΩ. What is the increment between
values of 𝑅𝑥 for which the bridge can be precisely
balanced?
(ANS: 7320 Ω, 1.1 MΩ, 1 kΩ increment)
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Dr. William Lee
“SMART” Engineering
Temperature Drift Compensation
When using the output voltage 𝑣OUT as an analogue for the measurand, if we wish
to make the measurements independent of temperature (e.g., when sensing strain),
we can decouple this dependency from 𝑣OUT by replacing 𝑅3 with an identical
gauge to the active one, 𝑅𝑥 , but mounted in a strain-free position. Both gauges will
now track in parallel with variations in temperature minimising its effect, leaving
changes in 𝑣OUT largely due to changes in strain.
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Monitoring Technologies
Variable DC Offset Control
Since 𝑣OUT is taken as a difference across the two branches, its variation can be (but
not necessarily) made to vary about zero volt – we set the values of the known
resistors so that at the nominal 𝑅𝑥 , the bridge is balanced. In doing so, the variation
in the output voltage directly represents the physical quantity without the offset.
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Dr. William Lee
“SMART” Engineering
Example 65. A strain gauge 𝑅𝑥 is attached to a structural member, and placed
in a Wheatstone bridge as shown in order to monitor the strain of
the structural member electrically through the output voltage 𝑣𝑆 .
It is known from prior testing that the unstrained resistance of 𝑅𝑥
is 1000 Ω, and it varies between 900 Ω and 1100 Ω for the sort of
strains the member is designed to withstand.
Wheatstone Bridge
1 kΩ
R1
5V
vs
4 kΩ
(a)
(b)
Structural member
Rx
Determine the resistance of 𝑅1 if it is desired to have 𝑣𝑆 vary
nominally about an unstrained value of 0 V.
What is the expected output voltage range of the interface
under normal operation?
(ANS: 250 Ω, 0.1610 V)
Dr. William Lee
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Monitoring Technologies
Practical Considerations
While the variation in the output voltage of the sensor interface circuitry is
representative of the underpinning physical quantity being monitored, in practice
this variation is extremely small and cannot be used directly to drive or control
any subsequent systems. For example, strain measurements rarely involve
quantities larger than a few milli-strain (i.e., Δ𝑙/𝑙 is in the ballpark of 10−3 ),
resulting in only a fraction of a percent change in the resistance of a strain gauge.
Similarly, the temperature coefficients of resistive thermometers are typically very
small, between 2 × 10−5 and 2 × 10−2 per ∘ C leading to almost unmeasurable
change in output voltage. While the thermistors have a larger sensitivity (typically
−4% to −7% of 𝑅 per ∘ C) the corresponding changes in the output voltage is still
miniscule in comparison to the sort of magnitudes required by typical electronics.
Further complications arise in the form of unavoidable offset (as previously
mentioned) inherent to the operation of the chosen interface circuitry, but more
noticeably, the undesirable effect the subsequent systems have on the output of
the sensor circuitry once connected – the loading effect (see Sect. 3.2.1 later on) –
causing the voltage to deviate from its expected value.
We must therefore have some sort of electronics that amplify the output voltage of
the sensing circuit, remove the unwanted effects inherent to the sensing circuit
where appropriate, and minimise the impact the subsequent systems have on the
output of the sensor circuit before it can be interpreted and used subsequently. The
signal conditioning circuitry used to alleviate these undesirable artefacts form the
basis behind the analysis technologies to be discussed in Sect. 3.2.
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Dr. William Lee
“SMART” Engineering
3.1.3. Problems
1.
Briefly explain (with the aid of suitable diagrams if necessary) the operation
of a resistive transducer which is based on variation in length, and
temperature. Describe how a resistive pressure sensor might operate based
on your understanding of the physical properties of resistive transducers.
2.
Poorly ventilated indoor spaces pose a risk for airborne transmission of
respiratory viruses. A simple method of assessing the adequacy of
ventilation is to monitor the carbon dioxide (𝐶𝑂2 ) levels. A resistive sensor,
𝑅𝑥 , that is positively correlated to 𝐶𝑂2 concentration is to be placed in a
Wheatstone bridge interface as shown so that the output voltage, 𝑣𝑂 , can be
used to track 𝐶𝑂2 level of a room.
R1
R3
VS
vO
Rx
R2
(a)
(b)
3.
Briefly explain how a resistive 𝐶𝑂2 sensor might operate based on your
understanding of resistance and the properties that affect it.
Briefly explain what the correlation is between the output voltage of
this bridge circuit and the 𝐶𝑂2 level of the room it is monitoring.
The output 𝑣S (𝑡) of a voltage divider interface used to host an input
transducer produces a signal in the range of 5 mV to 10 mV. The input 𝑣𝐸 (𝑡)
of the subsequent electronics accepts voltages in the range of 0 V to 5 V. To
maximise the resolution of this system, a signal conditioning circuit is to be
placed in between to ‘scale’ the sensor output to occupy the entire input
range of the electronics.
Voltage Divider
Sensor Interface
vS(t)
vIN
Signal
Conditioning
Circuit
vOUT
vE(t)
Subsequent
Electronics
This can be achieved by designing the signal conditioning circuit to perform
either of the two following transformations:
(1)
𝑣𝑂𝑈𝑇 = 𝑎 ⋅ 𝑣𝐼𝑁 − 𝑏
OR
(2)
𝑣𝑂𝑈𝑇 = 𝛼 ⋅ (𝑣𝐼𝑁 − 𝛽)
for some constants 𝑎, 𝑏 and 𝛼, 𝛽 (that can be set by the circuit parameters of
the conditioning circuit) so that for 5 mV ≤ 𝑣𝐼𝑁 ≤ 10 mV, the output of the
signal conditioning circuit is to be 0 V ≤ 𝑣𝑂𝑈𝑇 ≤ 5 V.
(a)
(b)
Determine the required constants 𝑎 , 𝑏 and 𝛼, 𝛽 of the two possible
signal conditioning circuits that will satisfy the requirements.
Out of the two solutions above, briefly explain which one is preferable.
Dr. William Lee
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Monitoring Technologies
Solutions to Problems
1.
See Coursebook.
2(a). Any causal link between how a gaseous compound can lead to a change in
the resistance through one or more of its underlying parameters.
2(b). The output voltage is negatively correlated to 𝐶𝑂2 .
3(a). 𝑎 = 1000, 𝑏 = 5 V and 𝛼 = 1000, 𝛽 =5 mV.
3(b). Solution (2).
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Dr. William Lee
“SMART” Engineering
3.2
Analysis Technologies – Signal Conditioning
Once a non-electrical signal of interest, such as speech signal, image, temperature,
or humidity levels, is captured and converted to an electrical signal using an input
transducer, it is now the engineer’s job to analyse it. Analysis can firstly involve
signal conditioning, where the acquired analog signal is manipulated to meet the
requirements of the next processing stage.
Signal conditioning encompasses various operations that may be necessary
depending on the specific characteristics of the captured signal. Sometimes the
captured signal may be too small in amplitude, or all the values acquired have
negative amplitude, or the signals are captured from multiple transducers that
need to be combined into one signal for further analysis. In such scenarios, the
acquired signal should be amplified, unwanted values removed or attenuated, the
signal shifted to required amplitude levels or signals from multiple transducers
added or subtracted to obtain the signal to be analysed. One or more of these signal
conditioning methods can be conducted to obtain a signal that is usable for further
analysis. All these operations on the signal form to the signal conditioning stage of
SMART technologies.
The signal that undergoes conditioning can then be stored as it is in a data
acquisition system or can be used to make some decisions. The analog signal is
converted to digital for both acquisition and decision making. The decision-making
uses other analysis technology, such as logic gates, which we will discuss in the
next chapter. Finally, based on the decisions made, response technologies are
employed to convey the decisions as physical signals, such as light, sound, or
motion.
In this section, we will discuss amplification, a signal conditioning process. We
will also discuss about an analog circuit block called operation amplifier (OPAMP)
that is commonly used for signal conditioning.
Before we go into the details of amplifier configuration, we shall discuss a model
for amplifiers, as we commonly do in electrical engineering. This model simplifies
the circuit by replacing it with an equivalent circuit for analysis purposes.
Additionally, we will discuss the impact of connecting an electric load to a circuit
and its influence on the model we have created. This model will then be extended
to specific amplifier types in the subsequent sections.
Dr. Jesin James
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Analysis Technologies – Signal Conditioning
Section 3.2 Concept Map
Fundamental Electrical Quantities
Behaviour and Quantification of Electrical Quantities
Voltage and Current
behaviour governed
and quantified by
Charge
Current
Voltage
Component Characteristics
Power
Resistor
Kirchhoff s Laws
Ideal Wire
Tools of Circuit Analysis
Methods
Theorems
Equivalent
Resistance
Superposition
Ohm s Law
used to model the
external behaviour of
Ideal Sources
OPAMP
used to understand the
ideal behaviour of
Voltage &
Current Division
Principles
Thevenin s
Theorem
Node-Voltage
Analysis
facilitates the
analysis and design of
Strain Gauge
Voltage Divider
Loading Effects
higlighted
Thermometer
OPAMP circuits
Wheatstone
Bridge
Thermistor
output
behaviours
Other
Transducers
Other
Other Interfaces
Resistive
Sensors
motivated the
development of
Sensor Interface
Monitoring Technologies
Analog Signal
Conditioning
Circuits
Analysis Technologies
SMART Engineering
Page 132
Dr. Jesin James
“SMART” Engineering
3.2.1. Equivalent Circuits and the Loading Effect
Intended Learning Outcomes
•
Be able to analyse sensor circuits and represent them using equivalent circuit
models, allowing for simplified analysis and design considerations.
Be able to interpret the effect of connecting multiple circuits one after the other
applying the knowledge of circuit analysis and loading.
Be able to determine appropriate circuit design strategies to minimise the effect
of loading.
•
•
Equivalent Circuits
When dealing with complex electrical systems like sensor circuits and amplifier
circuits, simplifying their representation is crucial for analysis. Imagine a sensor
circuit that measures something (let's call it 𝑉SENSOR ) and has some internal
resistance (𝑅SENSOR ). Instead of dealing with the entire complex circuit, we use a
simpler model such as the Thevenin equivalent circuit83. This model includes a
voltage source (𝑉SENSOR) and a resistor (𝑅SENSOR ), which together mimic how the
sensor behaves without all the detailed complexity.
+
RSENSOR
VSENSOR
Thevenin equivalent
circuit of the input
sensor
VIN
Output voltage of the
sensor is the input voltage
VIN of the next stage
Let us try this approach for one of the sensor interface circuits you came across in
Section Error! Reference source not found..
Example 66. Represent the following sensor circuit by its Thevenin equivalent
as seen by the output terminals 𝑎 − 𝑏.
Rx
a
VS
R
b
Notice that the Thevenin equivalent circuit is only one approach to represent the sensor by its
equivalent circuit. There are other approaches such as Norton’s theorem which can be used.
83
Dr. Jesin James
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Analysis Technologies – Signal Conditioning
𝑉 𝑅
𝑆
(ANS: 𝑣TH = 𝑉SENSOR = 𝑅+𝑅
, 𝑅TH = 𝑅SENSOR = 𝑅 ||𝑅𝑋 )
𝑋
Also, refer to Problem 1 at the end of this Chapter to practice finding the equivalent
circuit for another sensor interface circuit.
The source voltage 𝑉SENSOR represents the measurand while the resistance
𝑅SENSOR (the Thevenin resistance) accounts for the resistance of the sensor. The
equivalent circuit can be seen as an abstraction of the real circuit. In the circuit
shown above, the sensor's output is not connected to any other circuits. Due to this,
the current in the circuit is 0, making 𝑉IN = 𝑉SENSOR . In real applications, the
voltage measured by the sensor will then be passed to a signal conditioning circuit
such as an amplifier. This is called loading the sensor circuit or adding a load to the
sensor circuit, which we will discuss below.
Loading in Electrical Circuits
If an electrical circuit has an output terminal - a pair of terminals producing an
electrical signal - any circuit connected to this terminal is called a load. For instance,
when you connect an amplifier to a music player, the amplifier is the load. In
sensor circuits, the device that measures something (like a strain gauge or
thermometer) is the source, and any circuit connected to its output terminals is a
load. We say that the circuit connected to the output terminal loads or is loading
the sensor.
Why It Matters:
Understanding loading is important because when you connect another circuit to
the output of a sensor, like an amplifier, it affects how the sensor operates. This
additional circuitry adds its own resistance (let's call it load resistance 𝑅IN ) which
changes how much current can flow through the sensor. Now, let us connect this
load resistance to the output of the equivalent sensor circuit.
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Dr. Jesin James
“SMART” Engineering
+
RSENSOR
VSENSOR
Thevenin equivalent
circuit of the input
sensor
RIN
VIN
Output voltage to the
amplifier from the sensor,
and the input voltage of
the amplifier stage
RIN represents the input
resistance of the next
amplifier stage
The circuit will behave differently when a load is connected. What is 𝑉IN in this
case? As the load is connected, current flows through the circuit and voltage drops
in the resistors 𝑅SENSOR and 𝑅IN .
𝑅
Applying the voltage-divider principle: 𝑉IN = 𝑉SENSOR 𝑅 + 𝑅IN
IN
SENSOR
Our goal is to ensure that the voltage delivered by a sensor reaches the connected
circuit (load) as fully as possible. When we look at the equation above, we see that
the voltage 𝑉SENSOR from the sensor gets divided between the sensor’s own
resistance (𝑅SENSOR ) and the resistance of whatever is connected to it (𝑅IN ).
Because of this division, the voltage seen by the load (𝑉IN ) is less than the voltage
produced by the sensor ( 𝑉SENSOR ). This happens because some current flows
through the circuit, causing this voltage drop. This effect is known as loading, and
it is something we want to minimize. For circuit designers, minimizing loading or
minimise the effect of load means ensuring that as much of the sensor's voltage as
possible reaches the load without being reduced. This is crucial for the accurate
operation of sensors and other devices like amplifiers.
Example: Household Electrical Appliances
Think of your home's electrical outlets. They provide a steady voltage (like the
sensor’s voltage). When you plug in appliances (the load), they collectively create
a load on the power supply. If a high-power appliance with low resistance (like a
heater) is turned on, it reduces the overall resistance in the circuit. As a result, other
appliances (like lights) might dim because they receive less voltage.
Designing to Minimize Loading
To minimize loading effects, engineers design the load (such as an amplifier
connected to a sensor) with a high input resistance (𝑅IN ) compared to the sensor’s
output resistance (𝑅SENSOR ). This way, the load draws less current from the sensor,
allowing more of the sensor's voltage to pass through to the load.
For amplifier designers, this means ensuring the amplifier has a much higher input
resistance than the sensor’s output resistance. This design choice helps maintain
the accuracy and performance of the sensor’s measurements by reducing the
unwanted voltage drop caused by loading. By understanding and managing
loading effects, engineers can ensure that the amplifier has the least loading effect
on the sensor, or in other words, the amplifier loads the sensor the least.
Dr. Jesin James
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Analysis Technologies – Signal Conditioning
3.2.2. Amplifiers
Intended Learning Outcomes
•
Be able to evaluate and recommend design considerations for amplifiers by
considering the impact of loading and conducting thorough circuit analysis
The output voltage of the sensor interface circuit may be small, so an amplifier is
connected to amplify it. An amplifier magnifies electrical quantities such as voltage
or current. Let us develop a model that effectively mimics the amplifier behaviour.
What is “seen” by the input terminal of the amplifier?
As discussed in Section 3.2.1, looking into the input terminals of the amplifier
(load) from the sensor circuit (source), we see a finite resistance. This resistance
represents the effective resistance seen by the sensor circuit when connected to
the amplifier - called the input resistance 𝑅IN . This is shown in the figure below.
Now, what is “seen” by the output terminal of the amplifier?
Let the input voltage to an amplifier be 𝑣IN . At the output terminals of the amplifier,
the amplified signal 𝑣OUT is produced as a multiple of the input signal 𝑣IN . This
𝑣
amplification factor 𝐴 is defined as 𝐴 = 𝑣OUT . To model the amplifier's output
IN
behavior accurately, we use a Thevenin equivalent circuit, consisting of a voltage
source representing 𝑣OUT and an output resistance 𝑅OUT . This 𝑅OUT is the
equivalent resistance of the amplifier as seen by the output terminal.
vIN
Amplifier Block
vIN
RIN
vOUT
ROUT
A·vIN
vOUT
Amplifier Block
Interaction with Load
The amplifier’s output is connected to subsequent stages in the system, such as
signal conditioning circuits or actuators like loudspeakers. These stages
collectively form the load for the amplifier. The load presents a resistance 𝑅LOAD
when viewed from the amplifier's output terminals, influencing signal
transmission. Similar to analysing loads in the context of sensors (Section 3.2.1),
this resistance 𝑅LOAD is depicted in the figure below:
vIN
Amplifier Block
vIN
RIN
ROUT
A·vIN
Amplifier Block
Page 136
vOUT Load
vOUT
RLOAD
Input resistance of load
Dr. Jesin James
“SMART” Engineering
Example 67. Consider the circuit given below:
RS
VS
RA
RL
a) Find the equivalent resistance as seen by the voltage source 𝑉S and resistor
𝑅S together.
b) Find the Thevenin equivalent circuit as seen by 𝑅L .
𝑅
A
(ANS: a) 𝑅A || 𝑅L , b) 𝑅TH = 𝑅A || 𝑅S, 𝑉TH = 𝑉S 𝑅 +𝑅
)
A
Dr. Jesin James
S
Page 137
Analysis Technologies – Signal Conditioning
Similar to the above example where we found the equivalent circuit representation
of the circuit given, let us now find the equivalent circuit representation of a circuit
consisting of a sensor, amplifier and load.
Equivalent Circuit Representation
To facilitate analysis and design, each component in the sensor-amplifier-load
system can be represented by its equivalent circuit model. This approach simplifies
complex systems into manageable components, aiding engineers in optimising
performance and troubleshooting issues effectively.
Sensor
vIN
Amplifier Block
Gain = A
vOUT
Load
Notice that the above figure can be considered as the equivalent circuit of a typical
SMART technology, consisting of monitoring (sensor), analysis (amplifier) and
response (for example, a loudspeaker represented by its load resistance).
Amplifier Design Considerations
Based on our understanding of amplifiers with regards to their equivalent circuit
and loading, let us discuss some considerations while designing amplifiers.
Design Parameters
Motivation
Consideration
𝑅IN : Input resistance
𝑅IN should be large to ensure that
most of the voltage from the previous
stage (source) reaches the amplifier's
input. This minimizes the loading effect
on the source.
𝐴: Amplifier gain or No- The amplifier must amplify the
load voltage or Open- output
signal. With no load
loop gain
connected: 𝑣OUT = 𝐴. 𝑣IN .
𝐺:Circuit gain or Closed With a load connected, the circuit gain
𝑣
loop gain
is given by: 𝐺 = 𝑣OUT
IN
𝑅OUT : Output Resistance
Page 138
𝑅OUT is in series with 𝐴. 𝑣IN and affects
how efficiently the amplifier delivers
power to the load ( 𝑅LOAD ). A small
𝑅OUT minimizes power loss as heat
within the amplifier, ensuring that
𝑣OUT closely matches 𝐴. 𝑣IN .
Dr. Jesin James
“SMART” Engineering
Example 68. A voltage amplifier has a gain of 100, input resistance of 1 MΩ,
and output resistance of 10 Ω. The output terminal of the
amplifier is connected to a loud speaker that has an equivalent
resistance of 50 Ω as seen by the amplifier. Calculate the circuit
gain 𝐺 for the amplifier circuit, and check if it meets the amplifier
design considerations.
(ANS: 𝐺 = 83.3)
Example 69. For the following circuit choose an appropriate value for the
amplifier gain 𝐴 to ensure that when 𝑣S = 20 mV, then 𝑣OUT =
6 V. Also, what is the overall gain of the circuit?
500
vs
Rs
20mV
Amplifier Block
vIN
RIN
2k
25
ROUT
A·vIN
RLOAD vOUT
75
= 6V
(ANS: 𝐴 = 500, Overall gain = 300)
Dr. Jesin James
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Analysis Technologies – Signal Conditioning
It is quite common for amplifiers to be cascaded. That is, one amplifier's output
feeds into another amplifier's input.
Example 70. Consider 3 amplifiers cascaded together. Find expressions for:
a) Circuit gains of individual amplifier circuits.
b) Circuit gain of the cascaded amplifier circuit
𝑣
𝑣
IN
2
𝑣
𝑣
(ANS: 𝐺1 = 𝑣 2 , 𝐺2 = 𝑣3 , 𝐺3 = OUT
, Circuit gain of cascaded circuit = 𝑣OUT )
𝑣
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3
IN
Dr. Jesin James
“SMART” Engineering
3.2.3. Operational Amplifiers
Intended Learning Outcomes
•
Be able to extend the understanding of amplifiers to derive and comprehend
the ideal properties of an OPAMP.
The operational amplifier, or OPAMP for short, is the workhorse and fundamental
building block of many analog circuits. Modern OPAMPs possess characteristics
that make circuit design and analysis easier. The simplicity of analysis results from
one important concept—the virtual earth principle—which is covered later on.
An OPAMP is a difference amplifier, as shown in the diagram below.
v+
+
-
v-
AA
vOUT
𝑣OUT = 𝐴 ⋅ (𝑣 + − 𝑣 − )
• The terminal labelled “+” is called the non-inverting terminal.
We will use 𝑣 + to refer to the voltage at the non-inverting terminal.
• The terminal labelled “−” is called the inverting terminal.
We will use 𝑣 − to refer to the voltage at the inverting terminal.
• The gain A is called the open-loop gain of the OPAMP.
The OPAMP needs a power supply and a ground connection to operate, but we will
omit them from subsequent circuit diagrams to reduce clutter.
Vcc+
v+
v-
+
-
A
Vcc-
vOUT
This is a more detailed diagram of the
OPAMP with power supply. Typically,
𝑉CC+ = +15 V and 𝑉CC− = −15 V.
These are called the rail voltages and
refer to the maximum and minimum
voltages that an OPAMP can handle. It's
important to ensure that the input and
output voltages of the OPAMP stay
within these voltages to prevent
distortion or damage to the OPAMP.
This is a picture of a common OPAMP:
the LM74184. It has eight pins, not all of
which are used.
You have already seen the internal schematic of an OPAMP in Section 1.3. This
circuit has 20 transistors and 11 resistors, but thanks to Thevenin’s theorem, we
can simplify it down to a few of resistors and a voltage source!
84
Source: https://robotist.co/product/lm741-741-ic-op-amp-operational-amplifier-ic/
Dr. Jesin James
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Analysis Technologies – Signal Conditioning
OPAMP Model
v+
Δv
v-
v+
+
-
vOUT
AA
ROUT
+
RIN
vIN
v-
OPAMP circuit symbol
A·(v +-v -)
vOUT
-
Internal Model of an OPAMP
Looking at the schematic in Section 1.3 we need a simple model to understand how
an OPAMP behaves. An OPAMP, like any amplifier, can be represented using an
equivalent model we've discussed before.
The input resistance of the OPAMP is represented by a resistor 𝑅IN . Consider the
input terminal with 𝑣 + and 𝑣 − . A previous stage connected the input terminal
(such as a sensor), “sees” a finite resistance.
We model the OPAMP using a Thevenin equivalent circuit. This includes a voltage
source 𝐴(𝑣 + − 𝑣 − ) and a resistor 𝑅OUT in series. It accurately models the potential
difference available at the input of next stage to which the OPAMP is connected
(load) and the current available to the load.
Now, we will use OPAMP in circuits to form OPAMP circuits. OPAMP has been
designed keeping amplifier design considerations in mind to ensure that efficient
signal amplification happens. We will discuss some of the (cool) OPAMP properties:
Properties of an Ideal OPAMP
Property
𝐴: Open-loop gain
𝑅IN : Input resistance
𝑅OUT : Output
resistance
Amplifier Design Consideration
The amplifier must amplify the
output signal.
To ensure that the OPAMP does not
load the previous stage.
To ensure that the next stage will not
load the OPAMP.
Ideal OPAMP characteristics
Real OPAMPs can only approximate the properties of an ideal OPAMP. For example,
Typical Parameters
for the LM741 OPAMP.
Page 142
Property
𝐴
𝑅IN
𝑅OUT
Value
200000
2 MΩ
75 Ω
Dr. Jesin James
“SMART” Engineering
3.2.4. Practical OPAMP Circuits
Intended Learning Outcomes
•
Be able to utilize the knowledge of circuit analysis and ideal OPAMP properties
to analyse, design, and implement various OPAMP topologies that meet specific
circuit requirements.
Practical OPAMP circuits consists of an OPAMP with an assortment of other
components, most commonly resistors connected around the OPAMP. Each unique
arrangement of circuit elements is called a topology. Each topology behaves
differently and has certain advantages and disadvantages. As the OPAMP is now
part of a circuit, we will analyse both OPAMP parameters and OPAMP circuit
parameters, as below:
𝐴
OPAMP parameters
(refer to Properties of
an Ideal OPAMP table)
𝑅IN
𝐺
OPAMP circuit
parameters
𝑅IN(CCT)
- The voltage gain of the OPAMP or open-loop
gain.
- OPAMP manufacturer defines this.
- The input resistance of the OPAMP.
- OPAMP manufacturer defines this.
- The voltage gain of the circuit, circuit gain or
closed-loop gain.
- When the OPAMP is becomes part of a circuit,
then the voltage gain of the entire circuit needs to
be considered. 𝐺 = 𝑣OUT /𝑣IN .
- The input resistance of the circuit.
- This is based on the circuit components.
Different OPAMP circuits exist. In the context of this course, we will only discuss
OPAMP circuits with negative feedback.
OPAMP Circuits with Negative Feedback:
Let's examine the given OPAMP circuit, which includes components like resistors,
voltage sources, ground, and input-output terminals. It's important to note that
certain components may not be explicitly shown in the circuit diagram, but they
are assumed to be present. To fully represent the circuit, we will include all
connections, including the voltage sources, DC power supply, and ground85
R2
vIN
R1
-
vOUT
+
In all the topologies that we discuss, try to think of all the connections associated with the
simplified topology circuit.
85
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Analysis Technologies – Signal Conditioning
Inverting Amplifier
Now let us analyse the same OPAMP topology by obtaining expressions for the
voltage gain of the circuit.
R2
vIN
R1
-
vOUT
+
The gain indicates that this topology will amplify the signal and invert it. Hence,
this OPAMP topology is called an Inverting Amplifier.
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Dr. Jesin James
“SMART” Engineering
Now, that analysis was long and quite complicated! Let us learn an idea that will
make the analysis easier.
Virtual Earth Principle (VEP)
Consider the following fragment of an OPAMP circuit.
Assume that 𝑅IN = ∞ and 𝐴 = 105 .
If 𝑣OUT is some reasonable value,
+
v
for example, 𝑣OUT = 1 V
+
𝑣
1𝑉
vOUT
Δv
AA
Then 𝛥𝑣 = OUT = 5 = 10 𝜇𝑉,
v-
-
𝐴
10
Therefore,
Hence, we can take 𝑣 + and 𝑣 − as
essentially the same potential, even
though they have no ohmic connection!
There is nothing special about the value 𝑣OUT = 1 V. If A is large enough, then
𝛥𝑣 → 0 (𝛥𝑣 tends to 0), regardless of the value of 𝑣OUT . Also, 𝑣 + and𝑣 − will
always be at approximately the same potential. The VEP states that the OPAMP
does what is necessary to keep the input terminals at the same potential.
In OPAMP circuits with negative feedback, the virtual earth principle plays a
crucial role. This principle states that using negative feedback from the output
to the inverting input of the OPAMP minimizes the voltage difference between
its inverting and non-inverting input terminals. VEP simplifies OPAMP circuit
analysis by maintaining nearly equal voltages at both input terminals.
Applying positive feedback, where the output connects to the non-inverting
input, amplifies the voltage difference between input terminals. This breaks the
virtual earth principle and can lead to unstable or oscillatory circuit behavior.
In the context of this course, we will focus solely on OPAMP circuits with negative
feedback, where the virtual earth principle holds.
How exactly does the virtual earth principle help us analyse circuits?
Simply because the two inputs must follow each other, it reduces the mathematical
complexity.
Dr. Jesin James
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Analysis Technologies – Signal Conditioning
Now let us analyse the same inverting amplifier by employing virtual earth
principle.
Based on properties of the ideal
OPAMP and observing the
circuit:
R2
vIN
R1
-
vOUT
+
𝐴 and 𝑅IN are provided by the OPAMP manufacturer. Depending on the circuit, 𝐺
and 𝑅IN(CCT) can be determined.
Observations
The gain of the OPAMP is 𝐴 = ∞ , and yet the overall circuit gain is finite and equal
to 𝐺 = −𝑅2 /𝑅1 . The external components are responsible for this by a mechanism
called negative feedback—something we do not have time to discuss in this course
formally. The upshot is that the resistive components completely determine the
circuit gain 𝑅1 and 𝑅2 .
Even though R IN(OPAMP) = ∞ the external component determines the circuit input
resistance 𝑅1 . This is probably the biggest drawback of this topology because it is
difficult to have a large value of 𝑅1 and a large value for G without making 𝑅2 too
large. Remember that the input resistance of an amplifier should be high to ensure
that all voltage from the previous stage reaches the amplifier. Also, a large circuit
gain is good for obtaining high amplification from the amplifier. If one is made
high in this circuit, it will reduce the other.
Circuit gain 𝐺 = −𝑅2 /𝑅1 means that this topology will amplify the signal and
invert it. Hence, this OPAMP topology is called an Inverting Amplifier.
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Dr. Jesin James
“SMART” Engineering
Non-inverting Amplifier
Let us analyse the following OPAMP topology by obtaining expressions for the
voltage gain and input resistance of the circuit.
vIN
+
vOUT
-
R2
R1
Observations
The circuit gain G is determined by the external components 𝑅1 and 𝑅2 . Note that
the feedback goes from the output to the inverting input terminal—just like the
inverting amplifier. It is common practice to fix one of the resistors in the above
equation, and vary the other to obtain the required circuit gain 𝐺.
The input resistance of the circuit is ∞. This means that the input resistance of the
amplifier stage is large; hence, the previous sensor stage connected to the amplifier
will not be loaded. The choice of the resistors completely controls the circuit gain.
Circuit gain 𝐺 = 1 + 𝑅2 /𝑅1 means that this topology will amplify the signal and
invert it. Hence, this OPAMP topology is called a Non-inverting Amplifier.
Dr. Jesin James
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Analysis Technologies – Signal Conditioning
Inverting Summing Amplifier
R1
v1
RF
R2
v2
+
vOUT
Let us analyse the above OPAMP topology by obtaining an expression for the output
voltage of the circuit. Since this circuit has two inputs and one output, we will
compute the output voltage as a function of those inputs, rather than attempt to
develop an expression for the gain.
Observations
• The output voltage 𝑣OUT is the inverted (negative) weighted sum of the inputs
𝑣1 and 𝑣2 .
•
This topology sums up multiple inputs to generate the output voltage; hence,
it is called an Inverting Summing Amplifier.
•
Note that there can be more than two inputs, and the circuit will still multiply
the inputs with a gain and to generate the output voltage.
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Dr. Jesin James
“SMART” Engineering
Differential Amplifier
R2
v1
R1
v2
R3
-
vOUT
+
R4
Let us analyse the circuit above to obtain an expression for the output voltage of
the circuit. The analysis of this topology is more difficult because we cannot
determine 𝑣 − or 𝑣 + by simple inspection—all we know is that 𝑣 − = 𝑣 +.
Since this circuit has two inputs and one output, we will compute the output
voltage as a function of those inputs, rather than attempt to develop an expression
for the gain.
.
Observations
• The output voltage 𝑣OUT is the weighted difference of the inputs 𝑣1 and 𝑣2 of the
form 𝛼 ⋅ 𝑣2 − 𝛽 ⋅ 𝑣1 Hence, this topology is called a Differential Amplifier.
𝑅
If we choose 𝑅3 = 𝑅1 and 𝑅4 = 𝑅2 then, 𝑣OUT = 𝑅2 ⋅ (𝑣2 − 𝑣1 ). This is called a
1
Difference Amplifier because the output is an amplified version of the difference
𝑣2 − 𝑣1 , rather than the more general weighted difference of the form.
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Analysis Technologies – Signal Conditioning
Example 71.
(a) Design an OPAMP circuit with a voltage gain of – 20, and an
input resistance of 10 kΩ.
(b) Suppose the input signal supplied to the circuit is 𝑣IN (𝑡) =
0.5 sin (2𝜋. 1000𝑡), then draw the output signal from this
circuit.
Note: It is usual to select resistances to be no less than 1kΩ so that we do not draw
too much current from the power supply or the OPAMP.
(ANS: Need inverting amplifier topology, 𝑅1 = 10 kΩ, 𝑅2 = 200 kΩ)
Example 72. Find 𝑖 (mA) and 𝑣OUT (V) for the following circuit. Based on your
observation of the circuit, are the current and voltage values
expected?
R2
2k2
vIN
3V
-
R1
5k6
+
vOUT
i
R3
15k
(ANS: 𝑖 = −0.5357 mA, 𝑣OUT = −1.18 V)
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Dr. Jesin James
“SMART” Engineering
Example 73.
(a) For the circuit shown below, determine an expression for
the output voltage in terms of the input voltage and the
other component values.
(b) Suppose the input to this circuit is 𝑣IN (𝑡) =
0.5 sin (2𝜋. 1000𝑡), plot the output signal 𝑣OUT (𝑡).
R2
vIN
R1
3k
VREF
6k
-
vOUT
+
1V
(ANS: 𝑣OUT = −2𝑣IN + 3)
Dr. Jesin James
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Analysis Technologies – Signal Conditioning
Observations
•
This circuit is called a Level Shifter because it changes the DC level of the input
signal 𝑣IN . This is not the only level shifter circuit possible. Check out Problems
6, 14 for other examples of level shifters.
•
Notice that in this case, the scaling happens first (𝑣IN is scaled by a factor) and
then shifted by a DC value determined by 𝑉REF and the resistors.
Example 74. Determine an expression for the overall circuit voltage gain 𝐺.
R2
vIN
R1
-
+
R4
R3
-
+
vOUT
R1 = R 2 = 2 kΩ
R 3 = 1 kΩ
R 4 = 8 kΩ
The OPAMPs are assumed to be ideal, so the output resistance of each is zero. This
means that the second OPAMP does not load the first, and the overall gain is simply
the product of the two individual sub-circuit gains.
−𝑅
−𝑅
3
1
(ANS: 𝐺 = 𝑅 4 . 𝑅 2)
We conclude our discussion of OPAMPs here. We considered ideal OPAMPs and
used them for signal conditioning applications such as amplification, summing,
difference and level shifting. The engineer chooses the required signal
conditioning depending on the input signal obtained from sensors, and then an
OPAMP circuit is designed to meet the requirements of the next stage. Corrections
will be made to account for the non-ideal behaviour of OPAMPs.
Applications of OPAMPs:
Some of the more common applications are audio and video preamplifiers and
buffers, voltage and current regulators, digital to analog converters, analog
computers and analog telephone systems to amplify voltage level drop due to the
resistance of copper phone lines.
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“SMART” Engineering
3.2.5. Problems
1.
Represent the following sensor circuit by its Thevenin equivalent circuit as
seen by the output terminals 𝑎 − 𝑏 . (This problem is related to the first
example in this chapter)
R1
R2
a
VS
vSENSOR
R3
2.
Rx
b
Consider an amplifier with gain 𝐴 = 20, input resistance 𝑅IN = 50 Ω and
output resistance 𝑅OUT = 8 Ω. An input voltage 𝑣IN = 100 mV is connected
across the input terminals of the amplifier, and no load is connected across
the output terminal of the amplifier.
(a) Calculate the input current through 𝑅IN .
(b) Calculate the output voltage 𝑣OUT.
3.
A load is connected to the output of the amplifier, as shown below.
(a) Calculate the output voltage 𝑣OUT if 𝑣IN = 2 V and hence, determine
the circuit gain 𝐺 = 𝑣OUT /𝑣IN .
(b) Why is the circuit gain G different to the amplifier gain A?
ROUT
vIN
4.
RIN
A·vIN
RL
𝑅IN = 50 Ω
𝑅OUT = 8 Ω
𝑅L = 8 Ω
𝐴 = 20
vOUT
When the output resistance of an amplifier is not zero, it is clear that all of
the generated power is not delivered to the load, and some power must be
wasted in the amplifier. (Ever wondered why your audio amplifier at home
is warm to the touch?)
Consider just the output portion of an amplifier shown in the circuit below.
ROUT
A·vIN
(a)
(b)
RL
vL
Derive the equation for power delivered to the load.
Given that we cannot remove the effect of 𝑅OUT, a natural question is
to ask: what value of 𝑅L results in the maximum possible power
delivered to the load R L ? Assume that 𝐴𝑣IN = 50 V and 𝑅OUT = 8 Ω.
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Analysis Technologies – Signal Conditioning
Calculate the power delivered to 𝑅L for 𝑅L = 4, 8, 16 Ω. What value
of 𝑅L gives maximum power transfer to the load?
Prove that maximum power is delivered to the load when the load
resistance is equal to the output resistance of the amplifier
(c)
5.
Given below are the properties of four electronic devices. Which of the
following is most likely an OPAMP?
A. 𝑅IN = 10 MΩ, 𝑅OUT = 70 Ω, 𝐴 = 10
B. 𝑅IN = 1 MΩ, 𝑅OUT = 10 MΩ, 𝐴 = 100000
C. 𝑅IN = 10 MΩ, 𝑅OUT = 70 Ω, 𝐴 = 100000
D. 𝑅IN = 1 MΩ, 𝑅OUT = 1 MΩ, 𝐴 = 100000
6.
Determine an expression for the output voltage in terms of the input voltage
and the other component values. Notice that the circuit shown below is a
type of level shifter.
vIN
+
vOUT
-
𝑉REF = −1 V
𝑅1 = 2 kΩ
𝑅2 = 6 kΩ
R2
R1
VREF
7.
A cascade of two amplifiers is shown in the circuit below. Determine an
expression for the overall circuit voltage gain 𝐺 = 𝑣OUT /𝑣IN and calculate
its numerical value.
R4
vIN
+
R3
R1
8.
vOUT
-
-
+
𝑅1 = 2 kΩ
𝑅2 = 10 kΩ
𝑅3 = 1 kΩ
𝑅4 = 8 kΩ
R2
The circuit below is a simplified form of a digital to analog converter (DAC).
The inputs 𝐴, 𝐵, 𝐶 and 𝐷 are assumed to be high or low, i.e., 1 V or 0 V.
A
1k
B
2k
C
4k
D
Page 154
8k
8k
+
1k
1k
-
vOUT
+
Dr. Jesin James
“SMART” Engineering
(a) Write down an expression for the output voltage in terms of the input
voltages and the resistor values.
(b) If the inputs are 𝐴𝐵𝐶𝐷 = 0000, what is the output voltage?
(c) If the inputs are 𝐴𝐵𝐶𝐷 = 1111, what is the output voltage?
(d) If the inputs are 𝐴𝐵𝐶𝐷 = 1011, what is the output voltage?
9.
Design the below circuit to obtain 𝑣OUT (𝑡) = −3(𝑣IN (𝑡) − 2).
R2
vIN
R1
v2
R3
-
vOUT
+
R4
10.
Analyse the below circuit using superposition theorem to find the output
voltage of the circuit. Notice the similarity of this circuit with the inverting
summing amplifier.
R1
v1
RF
R2
v2
+
11.
An OPAMP circuit is shown below. Calculate the load voltage 𝑣𝐿 and the
load current 𝑖𝐿 .
VF
iF
iB
VB
12.
vOUT
R
+
vL
iL
RL
𝑉F = 1 V
𝑉B = 4 V
𝑅 = 3 kΩ
𝑅L = 1 kΩ
Design an operational amplifier circuit that has a voltage gain of +20.
Suppose the input signal supplied to the circuit is 𝑣IN (𝑡) =
0.5 sin (2𝜋. 1000𝑡), then draw the output signal from this circuit.
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Analysis Technologies – Signal Conditioning
13.
Obtain the expressions for the currents marked in the below circuit based
on the voltages and resistors already labelled?
vIN
+
iX
io vOUT
iY
R2
R1
Suppose the input to this circuit is 𝑣IN (𝑡) = 0.5 sin (2𝜋. 1000𝑡), plot the
output signal 𝑣OUT (𝑡). Notice the similarity of this circuit to the differential
amplifier
14.
2k
vIN
1k
2V
4k
-
vOUT
+
2k
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“SMART” Engineering
Solutions to Problems
1.
𝑅
𝑅
3
𝑋
𝑣TH = 𝑣SENSOR = 𝑉𝑠 (𝑅 +𝑅
− 𝑅 +𝑅
) , 𝑅TH = 𝑅SENSOR = 𝑅2 ||𝑅𝑋 + 𝑅1 ||𝑅3
3
1
𝑋
2
2(a). 2 mA
2(b). 2 V
Drawing an amplifier model like below will be useful for the analysis:
ROUT
vIN
100 mV
RIN
A·vIN
vOUT
3(a). 𝑣OUT = 20 V, 𝐺 = 10
3(b). 𝐺 ≤ 𝐴 because the resistors always attenuate the value of open-loop gain 𝐴.
𝑅
4(a). 𝑃𝐿 = (𝐴 ⋅ 𝑣IN )2 ⋅ (𝑅 +𝑅𝐿 )2 .
𝐿
OUT
4(b).
𝑹𝑳
𝑷𝑳
4Ω
69.44 W
8Ω
78.12 W
16 Ω
69.44 W
Maximum power is delivered to the load when 𝑅𝐿 = 8 Ω.
4(c). Think about how you can calculate the maxima (or minima) of a function.
5.
C
𝑅1
6.
𝑣OUT = 4𝑣IN + 3 V, (Also, 𝑣 − = 𝑉REF + 𝑅 +𝑅
⋅ (𝑣OUT − 𝑉REF ))
1
7.
8(a).
2
𝑅 𝑅 +𝑅
8 2+10
𝐺 = − 𝑅4 ⋅ 1𝑅 2 = − 1 ⋅ 2 = −48
3
1
A
B
C
D
𝑣OUT = 8 × ( 1 + 2 + 4 + 8 ) = 8A + 4B + 2C + D
Note that if the inputs ABCD are treated as binary digits (where A is the MSB
and D is the LSB), then 𝑣OUT is the numerical value of the 4-bit word ABCD as
a voltage.
8(b). 0 V
8(c). 15 V 8(d). 11 V
9.
The output of this circuit is of the form 𝑣OUT (𝑡) = 𝛼 ⋅ 𝑣2 − 𝛽 ⋅ 𝑣IN = −𝛽 ⋅
𝑣IN + 𝛼 ⋅ 𝑣2 . We need to design for 𝑣OUT (𝑡) = −3(𝑣IN (𝑡) − 2), which can be
written as 𝑣OUT (𝑡) = −3𝑣IN (𝑡) + 6. Now, we need to choose the resistor
values and voltage source 𝑣2 to match the 𝑣OUT (𝑡) = −3𝑣IN (𝑡) + 6
expression.
One possible design is 𝑅1 = 1 kΩ, 𝑅2 = 3 kΩ, 𝑅3 = 3 kΩ, 𝑅4 = 1 kΩ, 𝑣2 = 6 V.
Try out different possibilities by changing the signs of the terms in the output
expression such as 𝑣OUT (𝑡) = −3𝑣IN (𝑡) − 6, 𝑣OUT (𝑡) = 3𝑣IN (𝑡) + 6 etc.
10.
11.
12.
13.
14.
𝑅
𝑅1
𝑅
𝑅2
𝑣OUT = − ( 𝐹 . 𝑣1 + 𝐹 . 𝑣2 ). Find out the output due to the voltage source 𝑣1
alone by setting the other voltage source to 0. Then, find out the output due
to the voltage source 𝑣2 alone by setting the other voltage source to 0 .
Combine the out voltages employing superposition theorem,.
v𝐿 = 3 V, 𝑖𝐿 = 3 mA. Notice that the currents 𝑖F and 𝑖B are zero as they are
both input currents to the OPAMP.
Need non-inverting amplifier topology. An option: 𝑅1 = 1 kΩ, 𝑅2 = 19 kΩ.
Note: Is 19 𝑘𝛺 available in E12 resistor series? If no, how can you change the design
to choose resistors from E12 series only?
𝑣 −𝑣
𝑣
−𝑣
𝑖0 = 0, 𝑖Y = IN 𝑅 OUT , 𝑖X = OUT𝑅 IN
2
2
The output signal will be an inverted sine wave centered at 2 V , and
amplitude varying from 3 V to 1 V.
Dr. Jesin James
Page 157
Analysis Technologies – Signal Conditioning
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Dr. Jesin James
“SMART” Engineering
3.3
Analysis Technologies – Decision Making
All the analysis technologies we discussed so far take the analog signal
(continuous-amplitude, continuous-time signal) as the input. However, most of
the processing of the signals in computers and other digital technology uses digital
signals (discrete-amplitude discrete-time signals). In the next sections, we will
discuss some approaches to analysing (or processing) digital signals. We will also
discuss how analog signals are converted to digital signals by analog to digital
conversion.
Section 3.3 Concept Map
Fundamental Electrical Quantities
Charge
Current
Mathematical Models
Voltage
Power
Positional
Numeral
Systems
Algebraic
Systems
Decimal
Boolean Algebra
Tools of Circuit Analysis
Methods
Theorems
Equivalent
Resistance
Superposition
Behaviour and Quantification of Electrical Quantities
Voltage and Current
Component Characteristics
Hexadecimal
Truth Table
Resistor
Voltage &
Current Division
Principles
Binary
Other Bases
Kirchhoff s Laws
Ideal Wire
Thevenin s
Theorem
Ohm s Law
Node-Voltage
Analysis
Ideal Sources
Strain Gauge
Voltage Divider
Loading Effects
Wheatstone
Bridge
OPAMP circuits
Thermometer
OPAMP
Other
Transducers
design requirements
governed by the input
requirements of
Sensors
Sensor Interface
output can be
interpreted
and processed by
Logic Gates &
Circuits
Other Interfaces
Resistive
used to design
and understand
Analog-toDigital
Converters
Thermistor
Other
underpins the
operations of
Analog Signal
Conditioning
Circuits
Analog & Digital
Processing Circuits
Analysis Technologies
Monitoring Technologies
SMART Engineering
Dr. Jesin James
Page 159
Analysis Technologies – Decision Making
3.3.1. Logic Gates
Intended Learning Outcomes
•
•
•
•
Be able demonstrate the ability to apply the principles of Boolean operations
to analyse and understand the operation of basic logic gates.
Be able to design and construct derived gates by combining and modifying
basic logic gates to achieve specific logic functions.
Be able to apply the principles of Boolean algebra, to simplify complex logic
gate operations and optimise circuit design.
Be able to utilise logical design techniques to design logic circuits that
effectively solve a variety of digital logic problems and perform specific
functions, such as arithmetic operations, and control functions.
We discussed how an analog signal is acquired and processed. Now, we will shift
our focus to digital signals. Digital signals consist of a sequence of 1s and 0s, and
manipulating these signals is essential for digital circuits. Boolean algebra
provides the tools needed to handle digital circuits. These circuits consist of logic
gates, which are basic electronic circuits capable of processing binary numbers.
Logic gates are the building blocks of digital systems, including microprocessors
and microcontrollers, performing logic functions critical to their operation.
Basic Gates
The basic Boolean operators and gates have the same names and carry out
corresponding logical operations.
AND Gate
The AND gate performs the Boolean AND operation, producing a HIGH output
only if all inputs are HIGH. It can be demonstrated as a switch-lamp circuit.
Switch-lamp circuit for AND gate
X
Y
Switch X
Switch Y
Battery
The truth table for AND gate
Z
Lamp
𝑿
𝒀
𝒁=𝑿⋅𝒀
0
0
1
1
0
1
0
1
0
0
0
1
From the above circuit, only when switches 𝑋 and 𝑌 are ON (CLOSED) will the lamp
𝑍 turn ON (HIGH, meaning LIGHT UP). The truth table of AND gate is the same
as that of the Boolean AND operator. The symbol for AND gate is given below,
where 𝑍 = 𝑋 ⋅ 𝑌 (two inputs) and 𝑍 = 𝑊 ⋅ 𝑋 ⋅ 𝑌 (three inputs) . It is easy to
generalize the AND gate to many more inputs.
X
Y
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Z
W
X
Y
Z
Dr. Jesin James
“SMART” Engineering
OR Gate
OR gate performs the Boolean OR operation. The output of an OR gate is HIGH if
any of its inputs are HIGH. Let us draw a switch-lamp circuit for the OR gate.
Switch-lamp circuit for OR gate
The truth table for OR gate
𝑿
𝒀
𝒁 = 𝑿+𝒀
0
0
1
1
0
1
0
1
0
1
1
1
From the above circuit, when switch X or Y is ON (CLOSED), the lamp Z will turn
ON (HIGH). The truth table of the OR gate is the same as that of the OR operator.
The symbol for the OR gate is given below, where 𝑍 = 𝑋 + 𝑌 (two inputs). It is easy
to generalize the OR gate to many more inputs.
X
Y
Z
NOT Gate
NOT Gate performs the Boolean NOT operation. The output of the NOT gate is
the inverse (or complement) of the input.
Switch-lamp circuit for NOT gate
The truth table for NOT gate
𝑿
𝒁=𝑿
1
0
0
1
From the above circuit, when switch X is OFF (OPEN), the lamp Z will turn ON
(meaning LIGHT UP). The truth table of the NOT gate is the same as that of the
Boolean NOT operator. The symbol for the NOT gate is given below, where 𝑍 = 𝑋.
X
Z
An inversion bubble is used to denote NOT gate to avoid clutter in digital circuits.
The inversion bubble is a circle that denotes that the variable has been inverted.
Example 75.
(a) Prepare a truth table for this circuit.
(b) Draw a compact version of this
circuit using the inversion bubble.
X
Z
Y
Looking at the circuit, output 𝑍 is HIGH when either 𝑋 is LOW, or 𝑌 is HIGH.
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Derived Gates
Using the basic gates as building blocks, more gates are designed to implement
particular logical operations. These are called derived gates.
XOR (Exclusive OR) Gate
The output of an XOR gate is HIGH if its inputs are different. XOR gate operation
is defined as 𝑍 = 𝑋 ⊕ 𝑌 = 𝑋𝑌 + 𝑋𝑌.
𝑿
𝒀
𝒁 = 𝑿⊕𝒀
0
0
X
0
1
Z
Y
1
0
1
1
Symbol
XOR gate truth table
NAND (NOT-AND) Gate
NAND gate is formed by inverting the output of the AND gate. The output of the
NAND gate is LOW if all its inputs are HIGH. The operation is defined as 𝑍 = 𝑋. 𝑌.
X
Y
Z
Symbol
𝑿
0
0
1
1
𝒀
𝒁 = 𝑿. 𝒀
0
1
0
1
NAND gate truth table
NOR (NOT-OR) Gate
NOR gate is formed by inverting the output of the OR gate. The output of a NOR
gate is LOW if any of its inputs are HIGH. The operation is defined as 𝑍 = 𝑋 + 𝑌.
X
Y
Z
Symbol
𝑿
0
0
1
1
𝒀
𝒁=𝑿+𝒀
0
1
0
1
NOR gate truth table
XNOR (or NOT-XOR) Gate
The XNOR gate is also called an equivalence gate because its output is HIGH when
both inputs are the same or equivalent.
𝑿
𝒀
𝒁 = 𝑿⊕𝒀
0
0
0
1
X
Z
1
0
Y
1
1
XNOR gate truth table
Symbol
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Designing a derived gate instead of using combinations of basic gates to achieve
the same operation makes it easy to mass-produce a particular derived gate, rather
than using basic gates and expecting the assembly line to connect it all together.
Implementing Logic Operations using Universal Gates
NOR and NAND gates are universal gates because they can be used to implement
all basic logic operations.
Example 76. For the following logic circuits, construct the truth table, the
equivalent Boolean algebraic expression and identify the type of
gate.
(a)
(b)
X
X
Y
Z
Truth table:
Truth table:
Boolean expression:
Boolean expression:
Type of gate:
Type of gate:
(c)
(d)
X
Z
X
Z
Z
Y
Y
Truth table:
Truth table:
Boolean expression:
Boolean expression:
Type of gate:
Type of gate:
(ANS: 𝑎) NOT gate 𝑏) AND gate 𝑐) OR gate 𝑑) AND gate )
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In the above example, the logic circuit for an operation was provided to you. If you
have a known logic operation and want to construct a logic circuit to perform that
operation, you can utilize NAND gates (or NOR gates) exclusively.
We will explore the process of building logic circuits using only NAND gates. The
aim is to express all terms of the Boolean expression in the form 𝑍 = 𝑋. 𝑌. This can
be done by observation and rearrangement. Another approach is to take the
complement of the expression twice and explore the resulting expression to find
possible terms of the form 𝑍 = 𝑋. 𝑌. When you take complement twice, its effect
nullifies but introduces the NOT operator you need for NAND operation786
8 .
Example 77. Draw the circuits for the following Boolean expressions using
NAND gates only.
(a) 𝐶 = ̅̅̅̅̅̅
𝐴⋅𝐵+𝐴⋅𝐵
(b) 𝑍 = 𝐴 ⋅ 𝐵 + 𝐴 ⋅ 𝐵
86
You may find using the De-Morgan’s laws multiple times useful in these cases.
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Designing Logic Circuits
Now, let us implement more complex logic circuits using the logic gates.
Example 78. Implement the following digital logic using logic gates:
0
+0
=0
0
+1
=1
1
+0
=1
1
+1
=1
The requirement is similar to decimal addition but defined differently for the case
1 + 1. Note here that the inputs are single bits, and the output is also one bit.
Let us write the truth table for this Let us derive a Boolean expression for
representation:
truth table.
0 + 0 = 02
0 + 1 = 12
1 + 0 = 12
1 + 1 = 12
Logic circuit:
(ANS: Assuming output is 𝑆, and inputs are 𝐴 and 𝐵. 𝑆 = 𝐴 + 𝐵.)
Example 79. Implement the following digital logic using logic gates:
0
+0
= 00
0
+1
= 01
1
+0
= 01
1
+1
= 10
We can see that the above requirement is similar to decimal addition but defined
differently for the case 1 + 1. Note here that the inputs are single bits, but the
output requires two bits. This is called a binary half adder.
Let us write the truth table for this Let us derive a Boolean expression for
representation:
truth table.
0 + 0 = 02
0 + 1 = 12
1 + 0 = 12
1 + 1 = 102
Logic circuit:
(ANS: Assuming outputs are 𝑆 𝑎𝑛𝑑 𝐶, and inputs are 𝐴 and 𝐵. 𝑆 = 𝐴⨁𝐵, 𝐶 = 𝐴𝐵.)
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Example 80. Implement the following digital logic using logic gates:
Gate 2
Sensor 2
Sensor 1
Lane 1
Gate 1
Sensor 3
Lane 2
Gate 3
Suppose an engineering system designed to monitor digital signals from three
piezoelectric sensors at different locations for traffic control to a recycling facility,
as shown above. Below is the solution proposed by an engineer:
There are three gates and there are two lanes to enter the facility. Lane 1 is for
waste collection from industrial vehicles. Lane 2 is for staff entry. All industrial
vehicles come through Gate 1 and enter the facility through Gate 2. All staff enter
through Gate 3. Sensor 1 detects the vehicle at Gate 1, Sensor 2 weighs the vehicle
at Gate 2, and Sensor 3 detects the vehicle at Gate 3. Sensor 2 will produce HIGH
output when the weight is within a threshold. Other sensors produce high output
when a vehicle is detected.
Gate 1 opens when an industrial vehicle is detected, Gate 3 opens when a staff
vehicle is detected, Gate 2 opens for the industrial vehicle if its weight is within a
threshold and no staff vehicle is detected.
Let us write the truth table for this representation:
Let us derive a Boolean expression from the above truth table.
Now let us draw a logic circuit for the Boolean expressions above:
(ANS: If outputs are 𝐺1 , 𝐺2 , 𝐺3 𝑎nd inputs are 𝑆1 , 𝑆2 , 𝑆3 . 𝐺1 , = 𝑆1 , 𝐺2 = 𝑆2 𝑆3 , 𝐺3 = 𝑆3 )
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3.3.2. Analog to Digital Conversion
Intended Learning Outcomes
•
•
•
Be able to understand the steps involved in analog to digital conversion.
Be able to justify the selection of sampling rate and bit rate for analog to digital
conversion based on specific requirements and considerations.
Be able to design an analog to digital converter (ADC) for range of the physical
signal being converted, considering factors such as resolution, and range.
Analog to digital conversion is performed by an analog-to-digital converter (ADC).
The main steps in analog to digital conversion are:
Sampling – converting continuous-time to discrete-time
Sampling involves measuring the analog signal only at periodic time points. This
ensures that the time axis is discretised, as the signal value is captured only at
certain time instants. Consider the analog signal (continuous-time continuousamplitude signal) shown in the figure below. It is sampled at intervals of 𝑇, which
is the time period in seconds. By sampling, we capture the signal only at certain
time points, but there is no restriction on the amplitude values captured. Hence,
we obtain a discrete-time continuous-amplitude signal.
Figure: Analog to digital conversion- Steps
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Amplitude (V)
Example 81. Consider the continuous-time signal given below. Draw sampled
version of this signal sampled at the rate of:
a) 1 sample in 2 seconds
b) 1 sample in 1 second
2
1.5
1
0.5
0
10
Time (s)
What is the sampling frequency for all the cases?
a) 1 sample in 2 seconds
b) 1 sample in 1 second
(ANS: 𝑎) 0.5 Hz, 𝑏) 1 Hz)
The rate at which a continuous-time signal is sampled to obtain a discrete-time
signal is called the sampling rate or sampling frequency (𝒇𝒔 ). In signal processing
research, you will often see sampling rate quoted as an important parameter in
reports and interpretation of results. So, why is sampling rate so important?
To accurately capture the shape of a signal, it is desirable to have closely spaced
samples. In applications like speech signal recording, where the digital signals are
converted back to analog for human listening, a higher number of samples leads
to better-quality reconstructed speech signals. However, this also increases storage
requirements and processing demands. Therefore, the sampling rate ( 𝑓𝑠 ) is
determined based on the variation of the analog signal to capture adequate
variations for effective reconstruction. The variation of the analog signal is
measured by its frequency components87, and the sampling rate is defined as the
reciprocal of the sampling interval (𝑇):
The unit of sampling frequency is:
Based on research, the lowest sampling rate required to be able to effectively reconstruct a signal
has been calculated. This is called the Nyquist sampling rate. The Nyquist sampling rate is two times
the highest frequency component in the input signal.
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Example 82. A speech recording session captures a speech signal from a
speaker at a sampling rate of 44.1 𝑘𝐻𝑧. The recording session was
conducted for 30 minutes uninterruptedly. How many samples
were captured during the recording session? Also, what is the
time interval between two given samples?
10
Amplitude
5
0
-5
-10
-15
-20
-25
0
10
20
Samples
30
40
50
(ANS: 79.3 × 106 samples in 30 mins. Interval between 2 samples = 2.26 × 10−5 s)
Example 83. The speech signal given below is sampled at 44.1 𝑘𝐻𝑧, and a total
of 83,790 samples are captured. What is the total time required to
capture these samples?
15000
Amplitude
10000
5000
0
-5000
-10000
0
2000
4000
Samples
6000
8000
(ANS: 1.9 s)
Quantisation and Encoding – converting continuous-amplitude to
discrete-amplitude
The next step in analog to digital conversion is discretising the amplitude through
quantisation. The signal's amplitude is divided into zones, and each sample's
amplitude is approximated to one amplitude in a given zone. This process is called
quantisation. In the figure (previous page), the amplitude is divided into eight
zones (0 to 7), and each zone is assigned a 3-bit binary code word to represent the
approximate value of each sample. This process is called encoding.
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Amplitude zone assignment scheme:
Many schemes are followed to divide the amplitude range into amplitude zones
for discretisation. The one we will follow is to divide the amplitude range equally.
For example, if the total amplitude varies from 𝑉MIN to 𝑉MAX , and if the width of
each amplitude zone is:
𝑉
−𝑉
MAX
MIN
∆ = Total number
, then the amplitude zones are:
of amplitude zones
0 𝑉 to 2 𝑉
𝑉MIN to 𝑉MAX
𝑉MIN ≤ 𝑣 < 𝑉MIN + ∆
𝑉MIN + ∆ ≤ 𝑣 < 𝑉MIN + ∆ + ∆
𝑉MIN + 2∆ ≤ 𝑣 < 𝑉MIN + 2∆ + ∆
v(t)
222222222222
Amplitude (V)
Zone #
Zone 0:
Zone 1:
Zone 2:
….
Last zone:
2
1.5
1
0.5
0
𝑉MAX − ∆≤ 𝑣 < 𝑉MAX
22222222222222
10
Time (s)
Notice the upper limit of every zone is not included within a zone’s voltage range88.
Choosing the number of bits:
Each amplitude zone is encoded using a 𝑘 − bit code word. Using 𝑘 bits, we can
represent 𝑁 amplitude zones, where 𝑁 = 2𝑘 − 1 is the largest number that can be
represented using 𝑘 bits89. The zone numbering would then be 0, 1, 2 … 𝑁.
How many amplitude zones can be represented using all 𝒌 – bit code words?
The number of amplitude zones is related to the number of bits 𝑘 in a code word.
Hence, for every 𝑘 given in the table, the number of amplitude zones are:
No. of bits All amplitude
(𝒌)
zones
3
Maximum number
of amplitude zones
Encoded zone numbers
4
Notice that column 3 of the table gives the maximum number of amplitude zones
that can be represented using 𝑘 bits. So, a lower number of amplitude zones can
also be represented using the same 𝑘 bits.
If the upper limit of the voltage range needs to be included in the amplitude zones, we can either
modify the zone assignment scheme or add an additional zone to accommodate it. This can be
achieved by including the upper limit in the last zone (include 2 V in last zone) or creating a separate
zone for values greater than or equal to the upper limit (for ≥ 2 V) and allocating bits accordingly.
89 This is the expression for the number of bits used to represent a number.
88
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Consider the same continuous-time signal as the previous
example. After sampling the signal at 1 Hz, illustrate the
quantisation of the sampled signal to:
a) 2 amplitude zones
2
b) 4 amplitude zone
1.5
What is the resolution in each
1
case?
Amplitude (V)
Example 84.
0.5
0
10
Time (s)
a) 2 amplitude zones
b) 4 amplitude zone
(ANS: 𝑎) 1𝑉, 𝑏)0.5 𝑉)
In ADC, the number of amplitude zones used is the second important
consideration after the sampling frequency. The resolution of an ADC refers to the
smallest increment of amplitude that is treated as an amplitude zone, or the
smallest amplitude that the ADC can measure. To accurately represent the signal's
amplitude, lower resolution is needed, which requires a larger number of zones,
leading to longer code words, increased memory, and processing requirements.
In most music applications, a sampling frequency of 44.1 kHz is considered
optimal, meaning 44,100 samples are captured per second. A sample rate of 48 kHz
is common for music or audio intended for video. In terms of the number of bits
used, 16-bit or 24-bit are commonly employed in audio systems.
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Choosing an Analog to Digital Converter:
In engineering instrumentation, we need to choose ADCs specifications that match
our signal requirements. Let us understand this using an example.
What do we need? (Signal Requirements)
Imagine we have an analog signal with the following characteristics:
• It varies every 1 millisecond (1 ms).
• The voltage ranges from 0 V to +2 V.
• The smallest change in voltage we need to detect is 0.5 mV.
What do we have? (Available ADC Specifications):
We need to refer to the ADC
datasheets for various parameters
such as bit resolution, sampling rate,
and reference voltage. While all
IN
parameters should be considered, we
vIN
will focus on a few parameters in the
context of this course.
VREF
REF
ADC
GND
DN-1
.
.
.
D2
D1
D0
1. Reference voltage: The ADC compares the input analog voltage against the
reference voltage 𝑉REF to determine the digital output. If 𝑉REF 90 is 2.5 𝑉, the ADC
converts any voltage within the range 0 𝑉 to 2.5 𝑉 into digital data.
To ensure accurate representation of the analog signal, the entire analog signal
range should ideally fall within the range of 𝑽𝐑𝐄𝐅 . If analog signal voltage
exceeds 𝑉REF, the ADC may saturate, leading to inaccurate digital values.
2. Bit resolution - This determines the number of discrete levels (or digital steps)
an ADC can represent.
A higher bit rate means the ADC can discern smaller changes in analog voltage,
providing finer digital representation. A 13-bit ADC can represent 213 = 8192
discrete levels. This means it can differentiate between smaller changes in
voltage compared to an 8-bit ADC, which can represent only 28 = 256 levels.
3. Sampling rate - The number of samples the ADC takes per second
Nyquist-Shannon Theorem: According to this theorem, to accurately
reconstruct a signal from its samples, the sampling rate must be at least twice
the highest frequency component of the signal (Nyquist rate). Insufficient
sampling rate can distort the reconstructed signal.
It is possible to have negative reference voltages and reference voltages with a range varying from
a positive value to a negative value. The principle of ADC conversion remains same in these cases.
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From these ADCs, which would be suitable for the designed specifications?
ADC Parameters ADC 1 ADC 2 ADC 3 ADC 4
12
12
14
14
Bit resolution
16 kHz 8 kHz
8 kHz 16 kHz
Sampling rate
1
V
2
V
4V
1V
Reference voltage
Digital signals offer key advantages over analog signals. After noise is added to an
analog signal, it is usually impossible to determine the precise amplitude of the
original signal. By contrast, after noise is added to a digital signal, we can still
determine the logic values – provided that the noise amplitude is not too large.
Also, digital signals require lower memory and processing than analog signals.
Example 85.
A temperature sensor with a temperature range of 0 ℃ ≤
temperature < 32 ℃ over a voltage range 1 V ≤ voltage <
2.6 V placed in a chemical factory is recording temperature
variations at the rate of 1 sample per 5 minutes.
To design an ADC for these readings,
(a) how many bits are required to detect a change of 0.25 ℃ in temperature?
(b) what is the minimum sampling frequency required?
Consider that a 1 Hz sampling frequency 7-bit ADC with a reference voltage of 4 V
was chosen . What is the encoded ADC output if the sensed temperature is:
(c) 31 ℃
(d) 8 ℃
(e) 41 ℃
Consider that a 1 Hz sampling frequency 7-bit ADC with a reference voltage of 3 V
was chosen . What is the encoded ADC output, if the sensed temperature is:
(f) 31 ℃
(g) 8 ℃
(h) 41 ℃
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(ANS: 𝑎) 7 bits, b) 0.003 Hz 𝑐) 10100012 , 𝑑) 01011002 , 𝑒)11000012 ,
𝑓)11011002 , 𝑔) 01110112 , ℎ) 11111112)
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Consider an analog signal with voltage 𝑣 that varies from 0 V ≤
𝑣 < 5 V.
a) If a 2 − bit ADC with reference voltage 4 𝑉 is used to digitise the analog
signal, what is the binary output for input voltage = 2.1 𝑉 ?
b) If a 2 − bit ADC with reference voltage 2 − 4 𝑉 is used to digitise the
analog signal, what is the binary output for input voltage = 2.1 𝑉 ?
c) If a 2 − bit ADC with reference voltage 4 𝑉 is used to digitise the analog
signal, what is the binary output for input voltage = 4.2 𝑉 ?
d) If a 2 − bit ADC with reference voltage 8 𝑉 is used to digitise the analog
signal, what is the binary output for input voltage = 4.2 𝑉 ?
e) If a 3 − bit ADC with reference voltage 8 𝑉 is used to digitise the analog
signal, what is the binary output for input voltage = 4.2 𝑉 ?
Example 86.
(ANS: 𝑎) 102 , 𝑏)102 , 𝑐)112 , 𝑑)102 , 𝑒) 1002 )
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Time to Pause and Think About Data
As we conclude our discussion on ADCs, it is time to pause and think about what
we have learned. A transducer captures a physical signal (such as speech and
images). The analog signal is then converted to a digital signal using an ADC. The
output of the ADC is a stream of 1s and 0s, commonly referred to as data. We are
living in a data-driven world, where our images, speech, location, and opinions
(only to name a few) are collected and used to develop models and train the
human-computer interaction systems we interact with.
We as engineers need to be mindful of the data we collect or the data given to us
by others for technology development. Often the data is collected from individuals,
communities or ecosystems that are nurtured and maintained by communities.
This data is then stored and analysed by multiple people across multiple devices
and may even reach different parts of the world. In the journey of the data from
the source of its collection to technology development, two major issues have
caused concern to individuals and communities sharing their data:
•
•
Can the privacy and confidentiality of the data be maintained?
Who has ownership of the data collected?
There are many accounts of communities being disadvantaged due to the data they
shared (examples of instances where the data was used to spread negative traits of
the community) or the benefits of the data do not reach back to the community that
provided the data in the first place (examples of large technology developers using
data from communities, building technology based on it and not sharing the
profits to the community). Such instances are more common for indigenous
communities, whose knowledge is much sought after.
All is not grim as there is a growing realization in the communities, their leaders
and technology developers regarding the need to protect the collected data. Near
2020, the concept of data sovereignty was introduced, which refers to the
understanding that data is subject to the laws of the nation within which it is
collected. How this can be implemented is still a discussion in progress91.
Māori regard data as taonga (a unique resource), and the researchers/technology
developers who collect/use their data are kaitiaki (guardians) of the data. There is
a shift from the Western view of ownership to guardianship. There are many ways
in which Māori organisations are trying to make sure that the data collected from
the Māori community will benefit the community. Here we have discussed data
only from a Māori perspective. Different communities will have their view on data.
Ultimately, it is also up to us as technology developers to make sure that we are
guardians of the data entrusted to us and that the benefits reach back to the
community that shared the data.
A great example of these discussions is happening at the University of Auckland which aspires to
be a Māori Data Sovereignty organisation according to the university’s 2030 Vision Taumata Teitei.
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3.3.3. Problems
1.
Develop expressions and a truth table for the circuits below. Simplify the
expressions to have the minimum number of Boolean operations.
(a)
(b)
W
X
X
Y
Z
Z
Y
(c)
(d)
W
X
Z
X
Y
Z
W
X
Z
Y
(e)
(f)
W
X
Z
Y
Y
2.
The Boolean expression for XOR gate is 𝑍 = 𝑋 ⊕ 𝑌. But it can also be written
as 𝑍 = 𝑋. 𝑌 + 𝑋𝑌 . Draw a circuit with basic gates to implement the XOR gate.
3.
The expression for XNOR gate is 𝑍 = 𝑋 ⊕ 𝑌. But it can also be written as 𝑍 =
𝑋. 𝑌 + 𝑋. 𝑌 . Draw a circuit with basic gates to implement the XNOR gate.
4.
Implement the simplified expressions obtained in Problem 1 using NAND
gates only. For Question 1(f), assume you have a 3-input NAND gate.
5.
Implement Question 1(f) using only 2-input NAND gates.
6.
Consider two logic variables 𝑋and 𝑌 that produce an output 𝑍 based on the
logical operation " − "defined below. Implement a logic circuit for the below
logical operation using NAND gates only:
0
−0
=0
7.
0
−1
=0
1
−0
=1
1
−1
=0
Consider a house with two motion sensors and an alarm switch. One sensor
is at the entrance (S1), and the other is inside the house (S2). If S1 is ON, then
the calling bell in the house should ring. If S2 is ON, then the burglar alarm
should ring. There is also a switch for the alarm that can be turned OFF or
Dr. Jesin James
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Analysis Technologies – Decision Making
ON to activate the alarm. If the switch is OFF (LOW), then the alarm should
not ring even if S1 detects motion. Design a logic circuit to implement this.
8.
An audio recorder is used to capture bird sounds. One recording session is
24 hours. The recording is conducted at 44.1 𝑘𝐻𝑧 sampling frequency and
using a 16-bit binary system. What size storage device (in gigabytes – GB) is
required for this recording?
9.
Using a Carbon Monoxide sensor, the concentration of Carbon Monoxide in
the atmosphere is monitored in the range of 0 to 100 ppm (excluding 100
ppm). The voltage range of the sensor is 0.45 ≤ 𝑣 < 4.35 𝑉. If a concentration
resolution of 2 ppm is desired, what is the resolution of the ADC that will be
required?
10.
Consider an analog signal ranging from 0 𝑉 ≤ Amplitude ≤ 4 𝑉.
(a) Quantise the analog signal using a 2-bit ADC, with reference voltage =
4 V, dividing the amplitude range into equal amplitude zones. Is the upper
limit 4 V included in this ADC scheme?
(b) How can the coding scheme be modified to include the upper limit 4 V?
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Dr. Jesin James
“SMART” Engineering
Solutions to Problems
1(a). 𝑍 = (𝑊 ⋅ 𝑋) + 𝑌
1(d). 𝑍 = 𝑋𝑌 + 𝑋𝑌
2.
1(b).
1(e).
𝑍 = ̅̅̅̅̅̅̅̅
𝑋+𝑌
𝑍 =𝑊⋅𝑋
1(c).
1(f).
𝑍 = (𝑊 + 𝑋). 𝑌
𝑍 =𝑊⋅𝑋⋅𝑌
X
Y
Z
3.
X
Y
Z
4(a). 𝑍 = 𝑊. 𝑋. 𝑌
4(b).
𝑍 =𝑋⋅𝑌
4(c).
𝑍 = ̅̅̅̅̅̅̅̅̅
𝑊. 𝑋. 𝑌
4(d). 𝑍 = 𝑋. 𝑌. 𝑋. 𝑌
4(e).
𝑍 = 𝑊. 𝑋
4(f).
𝑍 = 𝑊. 𝑋. 𝑌
5.
To implement 𝑍 = 𝑊 ⋅ 𝑋 ⋅ 𝑌 using 2-input NAND gates only:
We need to implement 𝑊 ⋅ 𝑋 using the NAND gate only. 𝑊 ⋅ 𝑋 represents
AND gate. So, implementing 𝑊 ⋅ 𝑋 first, and then inverting it with a NAND
gate (implemented as NOT gate) will give us 𝑊 ⋅ 𝑋.
Then 𝑊. 𝑋 and 𝑌 can be input into a NAND gate to produce 𝑊 ⋅ 𝑋 ⋅ 𝑌. This is
how a 3-input NAND gate can be implemented using 2-input NAND gates.
Too obtain 𝑍 = 𝑊 ⋅ 𝑋 ⋅ 𝑌, we pass 𝑊 ⋅ 𝑋 ⋅ 𝑌 through another NAND gate.
W
X
6.
7.
8.
Y
W.X.Y
𝑍 = 𝑋 ⋅ 𝑌, using NAND only, implement 𝑍 = 𝑋. 𝑌
𝑋 = Calling bell, 𝑌 = Burglar alarm, 𝐴 = Door sensor, 𝐵 = Inside sensor, 𝐶 =
Alarm switch
𝑋 = 𝐴, 𝑌 = 𝐵𝐶.
Sampling frequency of 44.1 kHz means 44.1 × 103 samples are captured in
1 second. Also, each sample is represented by 16 bits, as a 16 − bit binary
system is used. You may need this information for the unit conversion: 8
bits = 1 byte, GB = giga byte = 1 × 109 bytes.
Answer = 7.62 GB
9.
78 mV
10(a). No
10(b). If equal zones are not required, then final zone can be 3𝑉 ≤ Amplitude ≤ 4𝑉.
Another approach is to use an ADC with a higher reference voltage.
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Analysis Technologies – Decision Making
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“SMART” Engineering
3.4
Response Technologies
We have familiarised ourselves with monitoring and analysis technologies in our
journey to being acquainted with SMART Engineering. What remains is to
familiarise ourselves with response technologies. Once we have monitored nonelectrical signals, converted them to electrical signals (via input transducers), and
analysed the electrical signals, it is then required to respond to the monitored
signal based on the analysis conducted. Some common responses are alarm sounds,
lights turning on, and motors rotating (to open or close). The above three cases
convert an electrical signal into a non-electrical signal. In the case of alarm, the
electrical signal is converted to a sound signal. For the light, the conversion is to a
light signal. In the case of the motor, the electrical signal is converted to motion
(mechanical signal). As briefly mentioned in the introduction to SMART
Engineering, an output transducer or actuator is used to convert this conversion
from an electrical signal to a non-electrical signal. There are many output
transducers such as an electric lamp, LED (Light Emitting Diode), heater, motor
and loudspeakers. Even without a detailed understanding of the working of all
output transducers mentioned above, you must be able to say the output signal in
each case.
We will only focus on the operations of an electromechanical output transducer.
The electric machines that convert electrical energy into mechanical energy are
called electric motors.
Section 3.4 Concept Map
Fundamental Electrical Quantities
Charge
Current
Mathematical Models
Voltage
Power
Positional
Numeral
Systems
Algebraic
Systems
Decimal
Tools of Circuit Analysis
Methods
Theorems
Equivalent
Resistance
Superposition
Boolean Algebra
Behaviour and Quantification of Electrical Quantities
Binary
Voltage and Current
Component Characteristics
Hexadecimal
Resistor
Voltage &
Current Division
Principles
Truth Table
Other Bases
Kirchhoff s Laws
Thevenin s
Theorem
Node-Voltage
Analysis
Ideal Wire
Inductor
Ohm s Law
Reactive Elements
Ideal Sources
OPAMP
facilitates the analysis
and design of
Strain Gauge
Voltage Divider
Loading Effects
Wheatstone
Bridge
OPAMP circuits
Thermometer
Thermistor
Analog-toDigital
Converters
Other
Logic Gates &
Circuits
DC Motors
Other
Transducers
Other Interfaces
Resistive
Sensors
Sensor Interface
Monitoring Technologies
Analog Signal
Conditioning
Circuits
Analog & Digital
Processing Circuits
Analysis Technologies
output
controls
Actuators
Response Technologies
SMART Engineering
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Response Technologies
3.4.1. DC Motors
Intended Learning Outcomes
•
•
•
Be able to apply the understanding of the fundamental laws of
electromagnetism to determine the direction of rotation of DC motors.
Be able to create an accurate electrical model of a PMDC motor considering its
working, and utilise circuit analysis techniques to analyse its performance.
Be able to develop the knowledge to analyse the efficiency of a PMDC motor
considering factors such as input power, output power and losses.
An electric machine operates under the principle of electromagnetism, which states
that an electric current generates a magnetic field. This magnetic field is the region
around a magnet or a moving electric charge where the force of magnetism acts.
Conventionally, the magnetic field direction is outward from the north pole and
into the south pole of a magnet.
Faraday's law tells us that a changing magnetic field near a wire (conductor)
induces a voltage in the wire, but only if there is a closed loop circuit for the electric
charges to flow through. Ampere's law states that an electric current flowing
through a wire creates a magnetic field around the wire. The direction of this
induced current is such that it opposes the magnetic field that caused it. The righthand rule helps determine the direction of the magnetic field created92.
These laws are the foundation of how all electric machines work. For example,
electric motors use these principles to convert electricity into mechanical motion.
Generally, an electric machine (motor) has a stationary member called a stator, and
a moving member called a rotor. However, the movement can also be linear, which
is called a linear motor. Applications of DC motors include electric shavers,
elevators, steel mills, rolling mills, locomotives, and excavators.
Building a Motor with an Electromagnet
Consider a current-carrying coil wrapped around a conductor to form an
electromagnet near a permanent magnet (figure below). This electromagnet has the
freedom to rotate. By reversing the direction of the current in the electromagnet,
its poles (north and south) can switch (see figure), leading to attraction or repulsion
with the permanent magnet and causing rotational motion.
N
N
S
+
+
S
As shown in the figures (next page), two permanent magnets are placed on either
side of the electromagnet to make this stronger. The direction of current through
the coil is alternated to ensure that the iron core rotates continuously. The
Let the fingers of your right hand point in the direction of the current flow (𝐼). Orient the palm of
your hand, so that as you curl your fingers, you can sweep them over to point into the direction of
north pole of the magnet field.
92
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“SMART” Engineering
permanent magnets are not reversed. The changes are made to the polarity of the
electromagnet only. This is the basic principle of working of an electric motor.
N
N
S
S
+
+
Figure: Magnetic poles attracting
Figure: Magnetic poles repelling
Permanent Magnet DC (PMDC) Motor
The figure below (left) shows a PMDC)motor, which is similar in design to the
simple setup with an iron core and permanent magnet we discussed earlier.
Arrows represent the current direction. In a PMDC motor, there are curved
permanent magnets that create a magnetic field. The electromagnet in this motor
is formed by a metal loop called the armature. For simplicity, the figure below
(right) shows a single coil in this armature. This coil acts as an electromagnet with
a north and south pole, as illustrated in the figure93.
Armature coil
Armature coil
S
N
+
Permanent magnet
+
Permanent magnet
Permanent magnet
Permanent magnet
The direction of rotation of the armature
The armature in a PMDC motor spins because of the attraction and repulsion
between unlike and like poles of magnets. To determine the direction of rotation,
we use the right-hand rule. Looking at the left side of the armature, where the
current flows from bottom to top (fingers point upward), and the magnetic field
from the permanent magnets curls clockwise (fingers curl clockwise), the
resulting force pushes downward. On the right side, applying the same rule
shows the force pushing upward. Therefore, the armature rotates in an anticlockwise direction, as shown in the left figure below.
N
S N
Permanent magnet
+
Permanent magnet
Figure: Unlike poles face each other. It
may cause the armature to stop rotating if
the polarity is not changed.
Permanent magnet
S
+
Permanent magnet
Figure: Direction of electric current is
reversed, causing the direction of
armature’s magnetic field to change.
Note that in the motor, we cannot “see” the north and south poles of the electromagnet. As the
direction of current through the armature changes, the direction of the magnetic field also changes.
93
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Response Technologies
To keep the armature spinning continuously, the current direction through the
armature is reversed, changing the magnetic field's direction. This reversal ensures
the armature keeps rotating. In the figures above, we focused on the perpendicular
sides of the armature relative to the permanent magnet's magnetic field. On the
parallel sides, where the magnetic field and current are parallel, no significant
force acts, allowing smooth rotation.
Changing the current direction
Changing the current direction in the armature is achieved through a commutator
system using slip-rings and carbon brushes, as shown below. Here is how it works:
•
•
•
•
•
Current flows from the battery to the brush, then to the commutator segment.
It passes through the armature coil, returns through the other commutator
segment, the brush, and back to the battery.
As the armature turns further, the commutator ring also turns.
When the armature reaches the next position, the brushes switch between
commutator segments.
This brief switch reverses the current direction through the armature. By
changing the current direction, the magnetic field direction changes, ensuring
continuous motor rotation.
This process of commutation, where the brushes and commutator change
connections, allows for continuous movement and efficient operation of the motor.
B
C
Armature coil
C
N
A +
D
Permanent magnet
Permanent magnet
D
S
+
B
A
Permanent magnet
2
Commutator
2
Brush
Brush
1
Permanent magnet
1
Slip Ring
Figure: Commutator in a PMDC Motor
Example 87.
What is the direction of rotation of the armature for the PMDC
motor design given below:
Armature coil
Permanent magnet +
Permanent magnet
(ANS: Clockwise direction)
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Dr. Jesin James
“SMART” Engineering
Induced Voltage
The two conductors (two sides of the armature coil) rotate, sweeping past the
magnetic field. Based on Faraday's law, this is the case of a current-carrying
conductor moving in a magnetic field, which will induce a voltage in the conductor.
This voltage or EMF (electromotive force) acts opposite the applied voltage and is known
as Back EMF. In the case of the PMDC, a voltage is induced in each conductor with
an equal magnitude but opposite polarity.
The above section briefly described the working of a PMDC motor. There are many
other types of motor configurations that use the principle of electromagnetism. The
details are out of the scope of this course. Now, we will move on to how a motor
is modelled electrically for analysis purposes.
Electric Model of a PMDC Motor
To electrically model the PMDC model, we consider the following:
Te
ωr
B
I
C
EAB
B
ECD
A
Permanent magnet
D
+
VDC
Permanent magnet
Magnetic field 𝑩 – Magnetic field produced by the motor’s permanent magnet.
Electric current 𝑰 – Current through the armature coil of the motor.
Induced Voltage 𝑬 − The induced voltage (or EMF) acts opposite the applied
voltage and is known as Back EMF. In the case of the PMDC, a voltage is induced
in each conductor with an equal magnitude but opposite polarity, as shown in the
figure above. Hence, the total induced voltage, 𝐸 = 𝐸AB + 𝐸CD .
Since the induced voltage in each of these two conductors is proportional to the
angular speed 𝜔𝑟 , the total induced voltage 𝐸 is represented in terms of 𝜔𝑟 as:
𝐸 = 𝐸AB + 𝐸CD = 𝑘𝑒 𝜔𝑟
where 𝑘𝑒 is the back EMF constant of the motor. The unit of 𝑘𝑒 is
on the design of the motor.
and depends
Armature resistance 𝑹𝑨 - The resistance of the motor’s coil or armature winding.
Electromagnetic Torque 𝑻𝒆 − In rotational motion, torque is the force responsible
for causing an object to rotate. In the case of an electric motor, the electromagnetic
torque 𝑇𝑒 is produced due to the interaction between the magnetic field 𝐵 and
electric current 𝐼. Since the torque is proportional to the current 𝐼 in the conductor,
it is represented in units of newton-meters (Nm) by:
𝑇𝑒 = 𝑘𝑡 𝐼
where 𝑘𝑡 is the torque constant of the motor. It has units of Nm/A (newton-meters
per ampere) and depends on the design of the motor.
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Response Technologies
Rotor Speed 𝝎𝒓 − Assume that the rotor rotates at 𝑁𝑟 revolutions-per-minute
(rpm). The speed at which the rotor is spinning is called the rotor speed. The
angular speed of the rotor, 𝜔𝑟 can then be expressed in radians per second by:
𝜔𝑟 = 2𝜋 × frequency, where frequency = number of revolutions per second
𝑁𝑟 revolutions per 𝑚𝑖𝑛𝑢𝑡𝑒 𝑁𝑟
=
60
60
𝑁𝑟
Hence,
𝜔𝑟 = 2𝜋
radians/second
60
Note: Both 𝑘𝑒 and 𝑘𝑡 are equal in value but have different units. The inductance of
the coil is ignored for simplicity.
frequency =
A PMDC motor can therefore be represented by a simple model or equivalent
circuit incorporating all the properties we discussed above.
Te
ωr
I
B
EAB
Te
ωr
B
ECD
A
Permanent magnet
RA
C
E
VDC
D
+
VDC
Permanent magnet
When the motor is spinning at constant rotor speed, we can apply KVL as:
𝑉DC = 𝐼𝑅𝐴 + 𝐸 = 𝐼𝑅𝐴 + 𝑘𝑒 𝜔𝑟
Using this simplified model of the PMDC, an analysis of the PMDC motor circuit
can be conducted.
The Efficiency of a PMDC Motor
Electric motors convert electrical power 𝑃IN into mechanical power 𝑃M to perform
tasks like rotating machinery or moving vehicles.
The efficiency of a motor indicates how
well it converts input electrical power
into useful mechanical power,
minimising power loss 𝑃LOSS .
PIN
Electric
Motor
POUT = PM
PLOSS
Input Electrical Power (𝑷𝐈𝐍 ): This is the total electrical power supplied to the
motor from a DC voltage source, given by:
𝑃IN = 𝑉DC 𝐼
where, 𝑉DC is the DC voltage and 𝐼 is the current through the motor.
Mechanical Power Output (𝑷𝐌 ): This is the power produced by the motor in the
form of rotational motion. For rotational motion94, 𝑊 = 𝑇𝑒 𝜃, where 𝑇 is the torque
and 𝜃 is the angle of rotation.
In equation form, Work 𝑊 equals force times distance, or 𝑊 = 𝐹𝑑, where 𝐹 is the linear force in
N, and 𝑑 is the distance in m. Work has a rotational analog. To relate a linear force acting for a certain
distance with the idea of rotational work, you relate force to torque 𝑇 (its angular equivalent) and
distance to angle 𝜃. This gives, 𝑊 = 𝑇𝜃.
94
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“SMART” Engineering
For electric motors with a rotary motion, mechanical power:
𝑊
𝑇𝑒 𝜃
𝑃M =
=
= 𝑇𝑒 𝜔𝑟
𝑡
𝑡
where, 𝜔𝑟 (radians/second) is the angular speed. It is therefore clear from the
above equation that for any given mechanical power output 𝑃M , if speed is high,
then torque is low and vice versa.
Power Loss (𝑷𝐋𝐎𝐒𝐒): Some of the input power is lost due to factors like armature
resistance 𝑅A , which dissipates power as heat. This loss is given by:
𝑃LOSS = 𝐼 2 𝑅A
Now, the relationship between all the power elements discussed above is:
𝑃IN = 𝑃LOSS + 𝑃M
𝐼 2 𝑅A + 𝐸𝐼 = 𝑃LOSS + 𝑃M
We know that 𝑃LOSS = 𝐼 2 𝑅A. Hence, 𝑃M =
𝑃M represents the power absorbed by the back EMF 𝐸 to produce the
electromagnetic torque, 𝑇𝑒 due to the conversion of electrical energy to mechanical
energy. As a result of 𝑇𝑒 , the motor rotates at an angular frequency 𝜔𝑟 , converting
the power absorbed by 𝐸 into mechanical power, 𝑃M .
Hence, 𝑃M = 𝐸𝐼 = 𝑇𝑒 𝜔𝑟
Efficiency of a PMDC Motor:
If the losses are only due to armature resistance, then the efficiency 𝜂 of the motor
as a percentage can be determined as:
𝜂=
𝑃OUT
𝑃M
𝑃IN − 𝑃LOSS
%=
%=
%
𝑃IN
𝑃IN
𝑃IN
If the powers 𝑃OUT (or 𝑃LOSS ) and 𝑃IN are obtained, and then efficiency can be
calculated.
In summary, the efficiency of a PMDC motor is a measure of how effectively it
converts electrical energy into mechanical energy. It's influenced by factors such
as the motor's design, load conditions, and any losses due to resistance. Higher
efficiency motors convert more of the input power into useful work, making them
desirable for various applications.
Dr. Jesin James
Page 187
Response Technologies
Example 88.
A PMDC commutator motor with armature resistance of 0.5 𝛺,
supplied by a 24 𝑉 DC battery, is used in an electric scooter.
The back emf of the motor is 22 𝑉 when running at 1500 𝑟𝑝𝑚.
a) Draw the electric model of the motor.
b) Calculate the efficiency of the motor.
(ANS: 𝑎) 𝐼 = 4 A, 𝑇𝑒 = 0.56 Nm, 𝑏) 𝜂 = 91.6 %)
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Dr. Jesin James
“SMART” Engineering
3.4.2. Problems
1.
An electric motor runs at 𝜔𝑚 = 50𝜋 rad/s when connected to a load of torque
= 100 𝑁𝑚. If the motor's efficiency is 80 %, calculate the average electric
input power to the motor in 𝑘𝑊.
2.
A car window is opened and closed using a PMDC commutator motor,
which is supplied by a 12 𝑉 DC power source. Both opening and closing of
the window are performed at constant motor speeds of 60 𝑟𝑝𝑚 and 80 𝑟𝑝𝑚,
respectively. The motor has an armature resistance of 0.4 𝛺 and draws 1 𝐴 of
current when closing the window.
(a)
(b)
(c)
3.
Draw the electrical model of the PMDC model while opening and
closing the motor using the information provided.
Calculate the efficiency of the motor while opening the window.
Calculate the efficiency of the motor while closing the window.
A cordless drill is driven by a PMDC commutator motor, which is supplied
by a 12 𝑉 DC power source. The motor has 4 𝑊 of power loss in the
armature resistance when it is delivering 20 𝑊 of output power at 200 𝑟𝑝𝑚
to tighten a screw. If there are no other losses, determine the following:
(a)
(b)
Draw the electrical model of the PMDC model using the information
provided.
Calculate the efficiency of the motor
Dr. Jesin James
Page 189
Response Technologies
Solutions to Problems
1.
19.635 kW
2.(a) Opening: 𝑉DC = 12 𝑉, 𝑅A = 0.4 Ω, 𝐼 = 8.25 𝐴,
𝑟𝑎𝑑
𝐸 = 8.7 𝑉,
𝜔𝑟 = 6.283
, 𝑇𝑒 = 11.418 𝑁𝑚
𝑠
Closing: 𝑉DC = 12 𝑉, 𝑅A = 0.4 Ω, 𝐼 = 1 𝐴,
𝑟𝑎𝑑
𝐸 = 11.6 𝑉,
𝜔𝑟 = 8.377
, 𝑇𝑒 = 1.384 𝑁𝑚
𝑠
2.(b) 𝜂 = 72.46 %
2.(c) 𝜂 = 96.61 %
3. (a) 𝑉DC = 12 𝑉, 𝑅A = 1 Ω, 𝐼 = 2 𝐴, 𝐸 = 10 𝑉, 𝜔𝑟 = 20.94
𝑟𝑎𝑑
, 𝑇𝑒 = 0.96 𝑁𝑚
𝑠
3.(b) 𝜂 = 83.33 %
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Dr. Jesin James
4
Electricity Supply
All engineers either directly or indirectly work with electrical energy and there is
thus important information to be gained from understanding how electrical
energy is supplied and delivered within our society. While it may not be necessary
for all engineers to understand, in detail, every aspect of our electricity supply
network, every engineer must have at least some basic understandings behind how
electric power is generated and transmitted so that meaningful questions can be
raised in situations where it matters.
For example, in establishing a new residential area, structural engineers are
required to assess the suitability of ground structures for power poles as well as
the pole itself. Mechanical engineers may be called for to assist with the traffic light
pole design, while traffic engineers might be called for to identify traffic flows to
optimise power poles and transformers placements. Environmental engineers may
be tasked to investigate the impacts and pollutants related to power generation
and distribution systems, while biomedical engineers might be working with
electrical and computer systems engineers in developing life-saving and lifesustaining equipment in case of accidents and emergencies associated with the
newly established electrical network.
Every country has an electric power system or network, called the power grid, by
which electricity is delivered from the producers (power stations) to the consumers.
The basic structure of a power grid consists of electrical generators (modelled as
voltage sources) connected to industrial and household electrical loads (modelled
using circuit components) by means of electrical transformers and transmission lines.
Due to the ease at which sinusoidal voltages can be produced, the generation,
transmission, distribution, and consumption of electrical energy from our mains
supply are thus predominantly in AC form.
Electrical (Power) Grid
Generator
Transformer
Transmission lines
Transformer
Industrial Consumers
Electrical Loads
Residential Consumers
Dr. William Lee
Page 191
Electricity Generation
Page 192
Dr. William Lee
Electricity Supply
4.1
Electricity Generation
Our analyses thus far have been focused on single-phase circuits: systems consist of
a supply connected through a pair of wires to a load. Circuits or systems that
contain multiple AC sources operating at the same frequency but at fixed phases
apart are known as polyphase systems. In particular, the three-phase system is by far
the most prevalent to date. Nearly all electric power is generated and distributed
in three-phase, and when a single-phase input is required (e.g., household
appliances), they are taken from the three-phase system instead.
Section 4.1 Concept Map
Fundamental Electrical Quantities
Charge
Voltage
Current
Power
Behaviour and Quantification of Electrical Quantities
Voltage and Current
Mathematical Models
Power Metrics
Positional
Numeral
Systems
Average
Power
Decimal
time-varying nature
characterised by
Kirchhoff s Laws
Algebraic
Systems
Boolean Algebra
Binary
Component Characteristics
Tools of Circuit Analysis
Methods
Equivalent
Resistance
Hexadecimal
Resistor
Ideal Sources
Truth Table
Superposition
Other Bases
Ideal Wire
Voltage &
Current Division
Principles
Reactive Power
Theorems
OPAMP
Apparent Power
Signal Characterisation
Ohm s Law
Thevenin s
Theorem
Power Factor
Capacitor
Node-Voltage
Analysis
Sinusoids
Root Mean
Square Value
Inductor
Reactive Elements
Generation
Systems
Single-Phase
Electricity
powered by
facilitates the
analysis and
design of
drawn from
Three-Phase
Electricity
Monitoring Technologies
Analysis Technologies
SMART Engineering
Dr. William Lee
Response Technologies
Electrical Grid
Page 193
Electricity Generation
4.1.1. Three-Phase Electricity
Intended Learning Outcomes
•
•
Be able to describe, and relate, the advantages of a three-phase system to its
practical implications in power systems.
Be able to determine the relevant electrical quantities pertaining to a balanced
three-phase four-wire system.
There are very good reasons for the choice of polyphase, and in particular, threephase systems. As we will shortly see, the sum of the power output of an 𝑛-phase
( 𝑛 ≥ 2 ) system is constant, resulting in uniform power transmission and less
vibrations in machinery. The three-phase systems are also favoured due to its costefficiency trade-off: a three-phase system can provide three times the power
throughput of a single-phase for only one additional wiring. This maximum copper
utilisation cannot be achieved with a greater or fewer number of phases!
The fundamental operating principle behind a generator is identical to that of a
motor, both are based on Faraday’s laws which, in this context, says that an electrical
(mechanical) force can be created through some mechanical (electrical) motion of
(in) a conductor under the presence of a magnetic field. The only difference
between the two electric machines is the nature of the causality between the
motions:
•
In the case of a motor, an external electrical source of excitation generates a
current which, under the influence of a magnetic field, induces a rotational
motion and thereby converting electrical energy to mechanical.
•
In the case of a generator, an external mechanical source of excitation
generates a rotational motion which, under the influence of a magnetic field,
induces a current and thereby converting mechanical energy to electrical.
Thus, in the former (i.e., motor) case, the current is the cause of the rotational
motion in the coil and the coil acts as a load driven by the external electrical source
supplying the current. In the latter (i.e, generator) case, the current is the effect due
to the rotational motion of the coil and the coil acts as a source (of electrical energy)
made possible by some external mechanism creating the rotational motion.
Load
N
Source
Source
Load
S
Motor
N
S
Generator
In practice, the rotational motion of the generator comes from the magnet instead
of the coil for obvious reasons.
Page 194
Dr. William Lee
Electricity Supply
Load
Load
a
a
a
a
vaa (t)
N
N
S
S
t=0
t = 3T/4
T/2
Load
a
a
T
t
Load
a
S
S
N
N
t = T/4
t = 2T/4
Three-phase voltages are often produced via a generator whose cross-sectional
view is shown. The generator consists of a rotating magnet (a.k.a., the rotor)
surrounded by a stationary winding (a.k.a., the stator). Three separate phase coils
with terminals 𝑎-𝑎′ , 𝑏-𝑏 ′ , and 𝑐-𝑐 ′ are physically placed 120∘ apart around the
stator. Terminals 𝑎 and 𝑎′ , for example, stand for one of the ends of the coils going
into and the other end coming out of the page. As the rotor rotates, its magnetic
field alternately “cuts” the three coils and induces sinusoidal voltages in the coils
according to Faraday’s law 95. If the coils are identical but placed 120∘ apart, then
81F
the induced voltages are equal in magnitude and frequency 𝜔 and are only out
of phase with each other (by 120∘ ), the voltages are said to be balanced.
a
Stator
3-phase output
Rotor
v 𝑣 (𝑡) 𝑣 (𝑡) 𝑣 (𝑡)
𝑐𝑛
𝑏𝑛
𝑎𝑛
b
a
c'
b'
c
𝑣𝑎𝑛 (𝑡)
S
N

c
120
a'
𝑣𝑏𝑛 (𝑡)
b
120 240
ωt
𝑣𝑐𝑛 (𝑡)
n
Mathematically, taking the voltage generated in the 𝑎-phase coil to be the zerophase reference, the generated voltages can therefore be expressed as
𝑣𝑎𝑛 (𝑡) = 𝑉𝑝 cos(𝜔𝑡 + 0∘ ) V,
and
𝑣𝑏𝑛 (𝑡) = 𝑉𝑝 cos(𝜔𝑡 − 120∘ ) V,
𝑣𝑐𝑛 (𝑡) = 𝑉𝑝 cos(𝜔𝑡 + 120∘ ) V.
The principle of operation of the generator should make clear that the different sources of electricity
generation (i.e., renewable, fossil fuels, nuclear, etc.) differ simply in the mechanisms by which the
rotational movement of the rotor is achieved.
95
Dr. William Lee
Page 195
Electricity Generation
Here is a somewhat remarkable geometric result that, as we shall see, underpins
the practical benefits of a three-phase system:
Example 89. Show that the sum of the (instantaneous) voltages generated from
a balanced three-phase generator is zero at all times, that is,
𝑣𝑎𝑛 (𝑡) + 𝑣𝑏𝑛 (𝑡) + 𝑣𝑐𝑛 (𝑡) = 0.
𝐴+𝐵
𝐴−𝐵
) cos (
) useful.
2
2
Hint: you may find the identity cos(𝐴) + cos(𝐵) = 2 cos (
Given its construction, the generator can be therefore modelled as three voltage
sources, along with the electrical loads, a typical balanced three-phase system can
therefore be represented as the following circuit 96.
82F
RS
a
Rl
A
iA(t)
n
Rn
N
iN(t)
RS
b
Rl
B
iB(t)
RS
c
Rl
C
iC(t)
van(t)
vcn(t)
RL
vbn(t)
Three-Phase Generator
Transmission Lines
Three-Phase or Three
Single-Phase Load(s)
Note that we have omitted the transformers to decouple, and simplify, the
discussions in order to develop an initial appreciation for the advantages of a
three-phase system but without jeopardising a basic understanding of the
calculations involved.
Note that there are two ways of interconnecting the separate phase windings of the generator (and
the electrical load): either as the WYE (𝑌)-connection shown or a DELTA (Δ)-connection. Without loss
of generality, and for simplicity, we only consider WYE-connected source and load as the remaining
combinations (i.e., 𝑌-to-Δ, Δ-to-𝑌, Δ-to-Δ) can all be equivalently reduced, and hence analysed under
this 𝑌-to-𝑌 configuration.
96
Page 196
Dr. William Lee
Electricity Supply
Cost Efficiency
The economic benefit of a three-phase system becomes apparent when one
attempts to analyse the three-phase circuit. To that end, notice in each phase of the
circuit the source (Thevenin) resistance 𝑅𝑆 , the line resistance 𝑅𝑙 , and the load
(equivalent) resistance 𝑅𝐿 are in series. Furthermore, their sum in each of the
phases are equal since the circuit is balanced. To quantify the currents and voltages
we could apply node-voltage analysis (KCL) at node 𝑁:
𝑣𝑁 (𝑡) 𝑣𝑁 (𝑡) − 𝑣𝑎𝑛 (𝑡) 𝑣𝑁 (𝑡) − 𝑣𝑏𝑛 (𝑡) 𝑣𝑁 (𝑡) − 𝑣𝑐𝑛 (𝑡)
+
+
+
= 0,
𝑅𝑛
𝑅𝑆 + 𝑅𝑙 + 𝑅𝐿
𝑅𝑆 + 𝑅𝑙 + 𝑅𝐿
𝑅𝑆 + 𝑅𝑙 + 𝑅𝐿
from which
(
1
3
1
+
) 𝑣𝑁 (𝑡) =
(𝑣 (𝑡) + 𝑣𝑏𝑛 (𝑡) + 𝑣𝑐𝑛 (𝑡))
𝑅𝑛 𝑅𝑆 + 𝑅𝑙 + 𝑅𝐿
𝑅𝑆 + 𝑅𝑙 + 𝑅𝐿 ⏟ 𝑎𝑛
=0V!
and hence 97
83F
𝑣𝑁 (𝑡) = 0 V !
The implication of this result is significant since it shows that (in a balanced system)
the current through the neutral wire is zero, effectively an open circuit, and can
be removed without affecting the load currents or voltages 98.
84F
Thus, we can deliver three times the amount of power of a single-phase system
(which requires two wires) for only one additional wiring.
In practice, the phases are unlikely to be perfectly balanced (especially if each
phase is used to power different single-phase systems) and there will be a small,
but non-zero, voltage at 𝑣𝑁 (𝑡). The neutral line is therefore needed, however, since
the current capacity is much lower than the live lines, it can be much thinner with
lower insulation requirements and, therefore, cheaper.
Remarkably, this result still holds even if there are reactive elements in (a more realistic model of)
the system, but the analysis of which is beyond the scope of this course.
98 Can you see why, from an analytical point of view, we can also replace the neutral wire with a
short circuit between node 𝑛 and 𝑁 without affecting the circuit behaviour?
97
Dr. William Lee
Page 197
Electricity Generation
Example 90. A balanced 3-phase 4-wire generator has an output resistance of
0.2 Ω per phase and an internal voltage of 120 V(rms) / 50 Hz per
phase. The generator feeds a balanced 3-phase 4-wire load with
an equivalent resistance of 39 Ω per phase via transmission lines
each with a resistance of 0.8 Ω. By taking the 𝑎-phase internal
voltage of the generator as the reference, sketch the equivalent
circuit of this power system, and use it to determine
(a) an expression for the line current through the 𝑎-phase of the
system,
(b) the RMS value of the voltages across each phase of the load,
(c) the RMS value of the voltage across any pair of terminals of
the load.
𝐴+𝐵
𝐴−𝐵
) sin (
) useful.
2
2
Hint: you may find the identity cos(𝐴) − cos(𝐵) = −2 sin (
(ANS: 𝑖𝐴 (𝑡) = 3√2 cos(2𝜋50𝑡) A, 117 V(rms), 117√3 V(rms))
Page 198
Dr. William Lee
Electricity Supply
Power Output
Another major advantage of three-phase systems, compared with single-phase
systems, is that its total power is constant rather than pulsating. The total
instantaneous power 𝑝𝑇 (𝑡) delivered by a three-phase generator, to a three-phase
load is simply the sum of the instantaneous power supplied by the generator in
each phase of the system. If the system is balanced, we have
𝑝𝑇 (𝑡) = 𝑣𝑎𝑛 (𝑡)𝑖𝐴 (𝑡) + 𝑣𝑏𝑛 (𝑡)𝑖𝐵 (𝑡) + 𝑣𝑐𝑛 (𝑡)𝑖𝐶 (𝑡)
with
𝑖𝐴 (𝑡) = 𝐼𝑝 cos(𝜔𝑡 − 𝜃) V,
𝑖𝐵 (𝑡) = 𝐼𝑝 cos(𝜔𝑡 − 𝜃 − 120∘ ) V,
and
𝑖𝐶 (𝑡) = 𝐼𝑝 cos(𝜔𝑡 − 𝜃 + 120∘ ) V,
where 𝜃 = 𝜙𝑣 − 𝜙𝑖 is the phase angle difference between the 𝑎-phase generator
voltage 𝑣𝑎𝑛 (𝑡) and current 𝑖𝐴 (𝑡).
Thus, evaluating 𝑝𝑇 (𝑡) we discover that
𝑝𝑇 (𝑡) = 𝑉𝑝 𝐼𝑝 cos(𝜔𝑡) cos(𝜔𝑡 − 𝜃)
+ 𝑉𝑝 𝐼𝑝 cos(𝜔𝑡 − 120∘ ) cos(𝜔𝑡 − 𝜃 − 120∘ )
+ 𝑉𝑝 𝐼𝑝 cos(𝜔𝑡 + 120∘ ) cos(𝜔𝑡 − 𝜃 + 120∘ )
1
1
1
= 3 ( 𝑉𝑝 𝐼𝑝 cos(𝜃)) + 𝑉𝑝 𝐼𝑝 ∑ cos(2𝜔𝑡 − 𝜃 + 120∘ 𝑛)
2
2
⏟
𝑛=−1
1
= 3 ( 𝑉𝑝 𝐼𝑝 cos(𝜙𝑣 − 𝜙𝑖 )) W !
2
=0!
the total instantaneous power developed in a balanced three-phase circuit does
not change with time: 𝑝𝑇 (𝑡) is a constant, whose value is simply three times the
average power developed in each phase.
A consequence of this fact is that machineries (e.g., motors) that are designed to
accept three-phase power will have less vibration and constant, rather than
fluctuating, mechanical outputs.
Dr. William Lee
Page 199
Electricity Generation
Example 91. A 3-phase 4-wire balanced load is connected to a 425 V(rms) 3phase distribution line. The load draws a current of 9 A(rms) from
each line. The power factor of the load was measured to be 0.96
lagging. Determine the instantaneous power received by the
three-phase load and its average.
(ANS: 𝑝𝑇 (𝑡) = 11.016 kW, 𝑃average = 11.016 kW)
Page 200
Dr. William Lee
Electricity Supply
4.1.2. Problems
1.
A single-phase load on a particular 3-phase 4-wire power system draws a
current of 50 A(rms) at 235 V(rms) with a lagging power factor of 0.80.
Determine the value of the average power supplied to the load.
2.
At a location in a particular 3-phase 4-wire power system network, the
voltage measured between a line and the neutral is 225 V(rms). Determine
the value of the peak voltage measured between any two lines at that
location.
3.
A resistive load is connected between one line and the neutral of a 3-phase
system. The (average) power it receives in this situation is 20 kW. What
power would it receive if it were connected between two lines of the same
system? Assume the resistance of the load does not change.
4.
A 3-phase balanced load is connected to a 400 V(rms) 3-phase 4-wire
distribution line. The power received by the load is 15 kW.
(a) Find the current supplied by each phase if the power factor is 1.
(b) Find the current supplied by each phase if the power factor is 0.85.
(c) Based on your findings in (a) and (b), give one reason why factories
and other large electricity consumers strive to have power factors as
close to 1.00 as possible.
Dr. William Lee
Page 201
Electricity Generation
Solutions to Problems
1.
9.4 kW
2.
≈ 551.14 V
3.
60 kW
4(a). 12.5 A(rms)
4(b). 14.71 A(rms)
4(c). Lower power factor requires higher current and therefore thicker wires to
deliver the same amount of power. Energy charges are higher for energy
supplied at a lower power factor.
Page 202
Dr. William Lee
Electricity Supply
4.2
Electricity Transmission and Distribution
The most efficient power stations are usually the largest. It makes sense to run
these large power stations 24 hours a day, even if all the power they generate is
not needed locally. Since sources of raw energy are often located far from load
centres, most of the power generated by these power stations must therefore be
transmitted over large distances to densely populated areas where electrical
energy is in high demand.
For example, much of New Zealand’s generating capacity is in the south of the
South Island while much of our population is in the north half of the North Island.
Generators may also be sited far from large population centres for environmental
or safety reasons (e.g., nuclear power stations).
Early power systems used relatively low voltages for safety and because the
devices they ran (mostly electric lamps) required relatively low voltages. The
problem with delivering a large amount of power at a low voltage is that the
current must be large (why?) and consequently the amount of power lost becomes
significant over long distances, in addition to thicker, and therefore more
expensive, cable requirements. To improve the economy associated with power
transmission and distribution, voltages are, therefore, “stepped” up during
transmission, and “stepped” back down for distribution. This is accomplished
using electrical transformers.
Section 4.2 Concept Map
Fundamental Electrical Quantities
Charge
Voltage
Current
Power
Behaviour and Quantification of Electrical Quantities
Voltage and Current
Mathematical Models
Power Metrics
Positional
Numeral
Systems
Average
Power
Decimal
time-varying nature
characterised by
Algebraic
Systems
Kirchhoff s Laws
Boolean Algebra
Binary
Component Characteristics
Tools of Circuit Analysis
Methods
Equivalent
Resistance
Hexadecimal
Resistor
Ideal Sources
Truth Table
Superposition
Other Bases
Ideal Wire
Voltage &
Current Division
Principles
Reactive Power
Theorems
OPAMP
Apparent Power
Signal Characterisation
Ohm s Law
Thevenin s
Theorem
Power Factor
Capacitor
Node-Voltage
Analysis
Root Mean
Square Value
Sinusoids
Inductor
Reactive Elements
Single-Phase
Electricity
facilitates the
analysis and design of
drawn from
Three-Phase
Electricity
Generation
Systems
Monitoring Technologies
Analysis Technologies
SMART Engineering
Dr. William Lee
delivered via
the use of
Transformer
Transmission &
Distribution
Systems
Response Technologies
Electrical Grid
Page 203
Electricity Transmission and Distribution
Example 92. 1 MW of power is delivered to a resistive load through a cable of
1 Ω. Determine the power lost in the cable if the power
(a) is delivered at 230 V(rms), and
(b) the power is delivered at 220 kV(rms).
(ANS: 18.90 MW, 20.66 W)
4.2.1. Transformers
Intended Learning Outcomes
•
Be able to interpret, and make use of, relevant transformer ratings for
determining electrical quantities in circuits containing transformers.
The transformer is the most-commonly used electrical machine, one of the most
useful electrical devices ever invented. Transformers can raise or lower voltage or
current in an AC circuit 99. They can also be used to isolate one circuit from another
and enable us to transmit electrical energy over great distances, distribute it
conveniently, and use it safely in factories and homes.
85F
An ideal transformer is simply constructed from two separate inductors (coils)
placed either superimposed on, or nearby, one another, on a magnetic core as
shown, and its principle of operation is governed by Faraday’s law.
Primary Side
Transformer
Secondary Side
Iron Core
RS
vS(t)
i1(t)
Φ(t)
v1(t)
i2(t)
v2(t)
Load
Primary Winding Secondary Winding
N1 turns
N2 turns
The ability for transformers to step up or step down voltage and distribute power economically is
one of the major reasons for generating AC rather than DC voltages in power systems.
99
Page 204
Dr. William Lee
Electricity Supply
When the primary side of the transformer is excited by a sinusoidal voltage 𝑣𝑆 (𝑡),
a magnetic flux 𝜙(𝑡) is induced in the primary coil of the transformer as a result of
the (sinusoidal) current 𝑖1 (𝑡). By Faraday’s law, this flux 𝜙(𝑡) is related to the
voltage 𝑣1 (𝑡) across the primary coil, and the current 𝑖1 (𝑡) through it, by
𝑣1 (𝑡) = 𝑁1
𝑑𝜙(𝑡)
𝑑𝑡
(= 𝐿1
𝑑𝑖1
).
𝑑𝑡
Due to the magnetic core, which act as an ideal ‘conductor’ of magnetic fields, all
the magnetic flux 𝜙(𝑡) induced by the primary coil will flow through, and intersect,
the secondary coil. A voltage 𝑣2 (𝑡) is therefore induced on the secondary coil and,
according to Faraday’s law, is given by
𝑣2 (𝑡) = 𝑁2
𝑑𝜙(𝑡)
𝑑𝑡
(= 𝐿2
𝑑𝑖2
).
𝑑𝑡
Since the flux 𝜙(𝑡) (and its rate of change) through both coils is identical (it is the
same flux), we therefore deduce that the induced voltage 𝑣2 (𝑡) on the secondary
side of the transformer is related to the antecedent voltage 𝑣1 (𝑡) on the primary
side of the transformer by
𝑣1 (𝑡) 𝑁1
= ,
𝑣2 (𝑡) 𝑁2
the ratio between the number of turns on the primary and secondary side. This
ratio is commonly referred to as the turns ratio of the transformer, and determines
the size of the output (secondary) voltage for a given input (primary) excitation.
The transformer is said to be a step-up transformer if 𝑁2 > 𝑁1 and a step-down
transformer if 𝑁2 < 𝑁1 (why?).
Notice since there are no resistive elements in the ideal transformer, the total
power ‘entering’ the transformer must equal the total power ‘leaving’ it. In other
words, the transformer is a constant power device, that is,
𝑣1 (𝑡)𝑖1 (𝑡) = 𝑣2 (𝑡)𝑖2 (𝑡)
from which we can also deduce (by re-arrangement) that the induced secondary
current 𝑖2 (𝑡) of a transformer relates to its primary driving current 𝑖1 (𝑡) by
𝑖1 (𝑡) 𝑁2
= ,
𝑖2 (𝑡) 𝑁1
the reciprocal of its turns ratio.
Dr. William Lee
Page 205
Electricity Transmission and Distribution
Once we have understood a transformer’s terminal behaviour, we often abstract
away its electromagnetic complexities (when they not of the focus), and represent
the transformer symbolically by
i1
i2
v1
v2
N1 : N2
Transformer Specifications
An ideal transformer can operate across all voltages and currents, and thus the
turns ratio (i.e., 𝑁1 /𝑁2 ) is all that is needed to completely describe its terminal
behaviour. In practice, however, like all electrical equipment, transformers can
only behave as intended when operating under certain rated conditions. These
conditions are usually provided by the manufacturer in the form of transformer
maximum power handling capacity (think apparent power) and the nominal primary
and secondary voltage at which the maximum power deliverable was measured.
Example 93. A single-phase transformer with the following ratings:
40 kVA, 2.4 kV(rms) / 200 V(rms), and 50 Hz.
is connected between a 120 V(rms) utility supply and a load with
an equivalent resistance of 100 Ω. Determine
(a) the power received by the load,
(b) the current supplied by the utility supply,
(c) the equivalent resistance of the transformer and the load as
seen by the utility supply, and
(d) the smallest load resistance that the transformer can handle
with the given utility supply without exceeding its ratings.
Page 206
Dr. William Lee
Electricity Supply
(ANS: 1 W, 8.33 mA, 14.4 kΩ, 50 mΩ)
It should be clear from Example 93 that the ratings of a transformer need to be
interpreted carefully especially when the transformer is used in conditions outside
of its tested conditions: The power rating only specifies the maximum power that
the transformer can handle, but the actual power in operation depends on the
circuit that the transformer is placed in. Likewise, the voltage (and hence current)
ratings simply refer to the nominal voltage (current) values that the transformer
was tested at for measuring the maximum power handling capability, but the
actual operating voltage (current) is contingent on the actual circuit in effect.
Dr. William Lee
Page 207
Electricity Transmission and Distribution
Example 94. A 30 kVA, 11 000 / 230 V(rms) 50 Hz single-phase transformer for
rural application is connected on the primary across a singlephase of a 3-phase 4-wire feeder. The secondary supplies the
electrical load of a farm. At a particular time, the farm’s load is 25
kVA at 220 V(rms). Determine
(a) the primary voltage of the transformer, and
(b) the primary current of the transformer.
(ANS: 10.52 kV(rms), 2.38 A(rms))
Page 208
Dr. William Lee
Electricity Supply
Example 95. A source delivers 12 W of (average) power to a 300 Ω load
through an ideal transformer. If the equivalent resistance of the
transformer and the load as seen by the source is 7500 Ω ,
determine
(a) the turns ratio of the transformer, and
(b) the source voltage and the current it supplies.
(ANS: 5, 300 V(rms), 40 mA(rms))
Dr. William Lee
Page 209
Electricity Transmission and Distribution
4.2.2. Problems
1.
A single-phase transformer is rated 50 kVA, 33 kV(rms) / 230 V(rms). At a
certain time, it is actually operating at 45 kVA with a primary voltage of 30
kV (rms). Determine
(a) the primary current, and
(b) the secondary voltage.
2.
A resistive load of 300 Ω is connected to the secondary windings of a
transformer rated 230 V(rms) / 50 V(rms). Determine the equivalent
resistance as seen by the source looking into the primary of the transformer.
3.
A certain step-up transformer has 1000 turns in the primary and 2000 turns
in the secondary. Why would it not be practical to make it with just 10 turns
in the primary and 20 in the secondary?
4.
A voltage source 𝑉𝑆 is to be connected to a resistive load 𝑅𝐿 = 10 Ω by a
transmission line having a resistance 𝑅line = 10 Ω as shown below. In part
(a), no transformers are used. In part (b), a step-up transformer is used at the
sending end of the line, and a step-down transformer is used at the load end.
For each case, calculate the (average) power delivered by the source, the
power dissipated in the line resistance, the power delivered to the load, and
the efficiency, defined as the power delivered to the load as a percentage of
the source power.
Rline = 10 Ω
RL = 10 Ω
VS = 100 V(rms)
(a)
Rline = 10 Ω
RL = 10 Ω
VS = 100 V(rms)
1 : 10
10 : 1
(b)
Solutions to Problems
1(a). 1.5 A(rms)
1(b). 209 V(rms)
2.
6348 Ω
3.
For the required secondary voltage, (a) the required flux may be too large
and exceeds the core limit, and (b) .the required primary current may be too
large to be feasible.
4(a). 𝑃𝑆 = 500 W, 𝑃line = 250 W, 𝑃𝐿 = 250 W, Efficiency = 50 %.
4(b). 𝑃𝑆 = 990.1 W, 𝑃line = 9.803 W, 𝑃𝐿 = 980.3 W, Efficiency = 99.01%.
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Dr. William Lee