`
COURSEGUIDE IN
BASIC ELECTRONICS (LECTURE)
ME 2251
Department of Electronics Engineering
SCHOOL of ENGINEERING and ARCHITECTURE
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SCHOOL OF ENGINEERING
AND ARCHITECTURE
ME 2251
COURSE LEARNING OUTCOMES
BASIC ELECTRONICS
(LECTURE)
1. Describe operation of semiconductor
materials; their applications; and, solve
basic problems involving diodes
2. Describe the elements of a regulated
power supply; analyze and solve
problems related to the different parts
of a regulated power supply
3. Describe the theory of operation of
Bipolar Junction Transistor (BJT) and
Field-Effect Transistor (FET): types,
construction; output characteristic
curves; biasing techniques; and,
demonstrate dc and ac analysis of the
different BJT and FET amplifier
configurations.
4. Describe and analyze operational
amplifiers.
2
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CONTENTS
PAGE
Title Page
………………………………1
Course Learning Outcomes
………………………………2
Course Introduction
………………………………7
Course Guide
………………………………7
Course Study Guide
………………………………8
Course Study Schedule
………………………………11
Evaluation
………………………………14
Contact Information of Course Facilitator
………………………………15
MODULE 1 SEMICONDUCTORS
………………………………16
UNIT 1 Semiconductor Theory and
PN Junction
………………………………16
ENGAGE
………………………………16
EXPLORE
………………………………17
EXPLAIN
………………………………23
ELABORATE
………………………………27
EVALUATE
………………………………27
UNIT 2 Clippers
………………………………30
ENGAGE
………………………………30
EXPLORE
………………………………30
EXPLAIN
………………………………32
ELABORATE
………………………………40
EVALUATE
………………………………40
UNIT 3 Clampers
………………………………41
ENGAGE
………………………………41
EXPLORE
………………………………41
3
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EXPLAIN
………………………………43
ELABORATE
………………………………49
EVALUATE
………………………………49
CLASSWORK
ASSIGNMENT
MODULE 2 REGULATED POWER SUPPLY
………………………………50
………………………………50
………………………………51
BLOCK DIAGRAM
………………………………51
UNIT 1 Rectifiers
………………………………52
ENGAGE
………………………………52
EXPLORE
………………………………53
EXPLAIN
………………………………69
ELABORATE
………………………………75
EVALUATE
………………………………75
UNIT 2 Capacitor Filter
………………………………76
ENGAGE
………………………………76
EXPLORE
………………………………76
EXPLAIN
………………………………82
ELABORATE
………………………………83
EVALUATE
………………………………83
UNIT 3 Zener Diode Voltage Regulator
………………………………84
ENGAGE
………………………………84
EXPLORE
………………………………84
EXPLAIN
………………………………85
ELABORATE
………………………………87
EVALUATE
………………………………87
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CLASSWORK
………………………………89
MIDTERM QUIZ
………………………………89
MIDTERM EXAM
………………………………90
MODULE 3 BIPOLAR JUNCTION TRANSISTOR
………………………………91
UNIT 1 Fundamentals of
Bipolar Junction Transistor
………………………………92
ENGAGE
………………………………92
EXPLORE
………………………………92
EXPLAIN
………………………………92
ELABORATE
………………………………93
EVALUATE
………………………………93
Unit 2 BJT AMPLIFIERS
………………………………94
ENGAGE
………………………………94
EXPLORE
………………………………94
EXPLAIN
………………………………94
ELABORATE
………………………………94
EVALUATE
………………………………95
CLASSWORK
………………………………96
FINAL QUIZ
MODULE 4 FIELD-EFFECT TRANSISTOR
………………………………96
………………………………97
UNIT 1 Fundamentals of
Field-Effect Transistor (FET)
………………………………97
ENGAGE
………………………………97
EXPLORE
………………………………97
EXPLAIN
………………………………98
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ELABORATE
………………………………98
EVALUATE
………………………………98
UNIT 2 FET AMPLIFIERS
………………………………99
ENGAGE
………………………………99
EXPLORE
………………………………99
EXPLAIN
………………………………100
ELABORATE
………………………………101
EVALUATE
………………………………101
MODULE 5 OPERATIONAL AMPLIFIERS
………………………………102
ENGAGE
………………………………102
EXPLORE
………………………………102
EXPLAIN
………………………………106
ELABORATE
………………………………109
EVALUATE
………………………………109
CLASSWORK
………………………………110
ASSIGNMENT
………………………………110
FINAL EXAM
………………………………110
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COURSE INTRODUCTION
This two-unit course covers introductory topics on devices and circuits that are
building blocks for various electronic applications. Topics that will find direct application
here are electrostatics, capacitors, DC circuit analysis, and Kirchhoff’s Laws. Most of the
illustrative problems cover basic circuit implementations. Other variations are left for the
student to study and apply the approach and method presented in the modules.
COURSE GUIDE
MODULES
To ensure that you will demonstrate the above cited course learning
outcomes, this course is divided into the following:
MODULE 1: SEMICONDUCTORS. This module covers basics of solid-state
technology. In particular, this module presents the properties of a PN junction
which is the basic building block of active devices such as solid-state diodes
and transistors. Also included are some common applications of diodes.
MODULE 2: REGULATED POWER SUPPLY. This module will introduce you the main
parts of a regulated power supply. Each part will be discussed separately, and
this includes operation, circuit analysis and basic design calculations.
MODULE 3: BIPOLAR JUNCTION TRANSISTOR (BJT). This module will introduce you
to an active semiconductor device that is used in electronic analog and digital
circuits. The topics are divided into two units: fundamentals of BJT and BJT
amplifiers. Unit 1 describes the characteristics and biasing of a BJT; and, unit 2
presents the three different types of amplifiers that utilize the BJT as the main
active device. It covers dc and ac analysis of BJT amplifiers, particularly on the
performance measures.
MODULE 4: FIELD-EFFECT TRANSISTOR (FET). This module presents another active
device that can perform functions similar to those of the bipolar junction
transistor. The topics are divided into two units: The topics are divided into two
units: fundamentals of FET and FET amplifiers. Unit 1 describes the
characteristics and biasing of an FET; and, unit 2 presents the three different
types of amplifiers that utilize the FET as the main active device. It covers dc
and ac analysis of FET amplifiers, particularly on the performance measures.
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MODULE 5: OPERATIONAL AMPLIFIERS. This module deals with the fundamentals
of operational amplifiers: types, application and analysis of basic circuits.
Review the course study guide and study schedule for your guidance. You can
go over the specific parts of the module through our course site in Google
Classroom.
COURSE STUDY GUIDE
The modules were prepared for you to learn diligently, intelligently, and
independently. Aside from meeting the content and performance standards of this
course in accomplishing the given activities, you will be able to learn other invaluable
learning skills which you will be very proud of as a responsible student. There are
numerous topics included in this course so it is imperative that you keep your focus
and not be overwhelmed. Your course facilitator will guide you on how to go about
the modules. In this course, you will explore and learn to understand, analyze and
solve problems in basic electronics. You are therefore encouraged to:
1. Download any of the following references for this course:
a. Boylestad, R.L. (2006). Electronic Devices and Circuit Theory. 9TH Ed.
Prentice Hall International, Inc.
b. Grob,B. (2004). Basic Electronics. 9th Ed. New York; McGraw-Hill.
c. Malvino, A. P. (2002). Electronic Principles. 6th Ed. New York: McGraw- Hill.
d. Cathey, J. J. (2002). Scahum’s outline of theory and problems of
electronic devices and circuits. New York: McGraw-Hill.
e. Floyd, T.L. (2001). Electronic Fundamentals: Circuits, devices and
Applications. 5th Ed. Upper Saddle River, New Jersey: Prentice Hall.
f. Godse, A.P., Bakshi, U.A. (2009). Basic Electronics. Pune India; Technical
publications Pune.
2. Schedule and manage your time to read and understand every part of the
module. Please note that in anticipation of any internet connection problems that
may occur, and to be able to cooperate with the government in observing the
health protocols, this online course will be delivered asynchronously.
3. Study and plan how you can manage to do the activities of this course in
consideration of your other modules from other courses. Be very conscious with
the study schedule. Post it at a conspicuous place so that you will always be
reminded of it.
4. Log in to the course site at least twice a week and as scheduled to keep abreast
of important announcements, discussions, and other class activities. Check the
STREAM page every time you log in for any announcements.
5. Dedicate at least six (6) hours per week for the lessons and assignments
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6. Follow the task schedule. Works pile up quickly if you will not do them on time;
cramming can increase your stress level and this, more often than not, is counterproductive.
7. Read and understand the assessment tools provided before you start doing your
tasks. Do not settle with the low standards, aim for the highest standards.
8. Browse and read the different materials even prior to doing the tasks in the
module. However, you need to ensure that you will not miss any part of the module
so that you will not miss to accomplish any required activity.
9. Participate in the course discussions and consultations that will be conducted
using Google Meet and Google Hangouts. If you will be using mobile app of
Google Meet and Google Hangouts, stay logged in so you can engage in the
discussion anytime and anywhere. If you are using the desktop app, regularly log
in to stay in the discussion.
10. Note that our Google Classroom is a virtual learning environment, not a social
networking site. Use recent and appropriate ID photo on your profile page.
11. Always remember that the discussions are academic in nature, which means that
relevant academic conventions apply.
a. Read and analyze the contributions made by your classmates in the
discussion forum. Respond appropriately and courteously. Always use
proper language.
b. Be polite and respectful. Do not be rude and do not make remarks that
may be construed as personal attack.
c. Do not post lengthy contributions. Go directly to the point and express it
concisely. Stay with the topic at hand.
d. Protect your privacy. Ponder before you post. If you wish to share something
private, do it by email or private chat.
12. Work independently and honestly in accomplishing the requirements. Welcome
challenges as these would hone your analytical abilities.
13. Always remind yourself of deadlines. Read in advance. Try to anticipate possible
conflicts between your personal schedule and the course schedule, and make
the appropriate adjustments. Inform your facilitator, through any means, about
any concern that could affect your participation and performance in the course
(e.g. cause of delays or "absences" or "silences" of more than a week's duration,
etc.).
14. Raise any query that you may have regarding the learning materials through the
comments section in our google classroom. You may also request to schedule a
face-to-face online meeting/forum with the faculty for clarifications that are
difficult to raise as a simple comment or question.
15. Always remember you are the learner; hence, you do the modules on your own.
Your family members, friends and companions at home can support you but the
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activities must be done by you. As Louisan, you always need to demonstrate the
core values of competence, creativity, social involvement and Christian spirit.
Using the Learning Modules
To be able to help you build, your own understanding from experiences and new
ideas, the modules in this course are designed based on the 5E Instructional Model
(Engage, Explore, Explain, Elaborate, and Evaluate). The following icons will help
you find some of the most critical ideas in the learning modules.
This part of the module are activities to pique your interest and get
you personally involved in the lesson, while pre-assessing your prior
understanding and familiarity with related concepts
This part of the module presents the main lesson
This part of the module contains further explanations and/or
illustrative problems
This part of the module contains activities that will allow you
to use your new knowledge and continue to familiarize
yourselves with related situations and applications
This icon is placed at the end of each module to remind you of
graded classworks like assignments, quizzes, and examinations
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COURSE STUDY SCHEDULE
Below are the details in the conduct of this course arranged in chronological order
with the corresponding topic learning outcomes and activities designed for you
to undergo the five stages of the 5E learning model.
Week
MODULE 1
UNIT 1
1st
UNIT 2
1st
UNIT 3
1st
MODULE 2
UNIT 1
2nd
TOPIC LEARNING OUTCOMES
ACTIVITIES
SEMICONDUCTORS
Semiconductor Theory and PN Junction
TLO 1: Describe a PN-junction
Engage: Assess familiarization with
diode; identify the terminals,
electronic devices and components; review
construction, operation, and
on conductors and insulators
equivalent circuits
Explore: Differentiate P and N type
materials; characteristic curve of a PN
junction or Diode; equivalent circuits of a
diode
Explain: Analyze and solve sample problem
on application of diode approximations or
equivalent circuits
Elaborate: Practice on Problem Set
Evaluate: Assignment/Quiz/Exam;
conceptual questions and problem solving
Clippers
TLO 2: Describe the operation,
Engage: Review of related topics in
analyze and draw the output
electrostatics and DC circuits
waveform corresponding to an
Explore: Describe and analyze basic
input waveform for a diode
clipping circuits and draw output
wave-shaping circuit
waveforms
Explain: Illustrate analysis of clipping circuit
and draw output waveform
Elaborate: Practice on Problem Set
Evaluate: Conceptual questions; analysis of
circuit and drawing of output waveform
Clampers
TLO 2: Describe the operation,
Engage: Review of capacitor action
analyze and draw the output
Explore: Describe and analyze basic
waveform corresponding to an
clamping circuits and draw output
input waveform for a diode
waveforms
wave-shaping circuit
Explain: Illustrate analysis of clamping circuit
and draw output waveform
Elaborate: Practice on Problem Set
Evaluate: Conceptual questions; analysis of
circuit and drawing of output waveform
Evaluative Assessment
Assignment on Module 1
REGULATED POWER SUPPLY
Rectifiers
TLO 3: Describe the elements
Engage: Review of diode biasing and
and operation of a regulated
approximations
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power supply and present
design procedures involving the
rectifier, filter, and the regulator.
UNIT 2
2nd
Filters
TLO 3: Describe the elements
and operation of a regulated
power supply and present
design procedures involving the
rectifier, filter, and the regulator.
UNIT 3
3rd
Zener Diode Voltage Regulator
TLO 3: Describe the elements
and operation of a regulated
power supply and present
design procedures involving the
rectifier, filter, and the regulator.
MODULE 3
UNIT 1
4th
UNIT 2
September
4th
Explore: Describe and analyze rectifying
circuits: half-wave and full-wave rectifiers
Explain: Solve sample problems on the
basic rectifiers
Elaborate: Practice on Problem Set
Evaluate: Assignment/Quiz/ExamConceptual questions and problem solving
on the three types of rectifiers
Engage: Review of capacitor action
Explore: Describe and analyze capacitor
filter in a power supply
Explain: Analyze and solve sample problem
on capacitor filter
Elaborate: Practice on Problem Set
Evaluate: Assignment/Quiz/ExamConceptual questions and problem solving
on capacitor filter
Engage: Open discussion on the need for a
stable supply; review on zener diode
Explore: Describe and analyze zener
voltage regulator
Explain: Illustrate analysis and solution to
problem on zener voltage regulation
Elaborate: Practice on Problem Set
Evaluate: Assignment/Quiz/ExamConceptual questions and problem solving
on zener voltage regulation
Evaluative Assessment
Midterm Quiz on Module 2
Midterm Exam on Modules 1 and 2
BIPOLAR JUNCTION TRANSISTOR (BJT)
Fundamentals of Bipolar Junction Transistor
TLO 4: Describe the types,
Engage: Review of PN junction
construction, operation,
Explore: Differentiate NPN and PNP
applications, and biasing
transistors; equivalent circuits of BJT; discuss
techniques of Bipolar Junction
characteristic curves for BJTs, biasing for
Transistors.
NPN and PNP transistors and operating
regions of BJT
Explain: Illustrate analysis on biasing and
operating regions of BJTs
Elaborate: Practice on Problem Set
Evaluate: Assignment/Quiz/ExamConceptual questions on BJTs
BJT AMPLIFIERS
TLO 5: Analyze and solve
Engage: Review of operating regions of BJT
problems on small signal BJT
Explore: Describe and analyze different
amplifiers.
types of biasing circuits for BJT
Explain: Illustrate analysis and solution to
problem on of BJT biasing circuits
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Elaborate: Practice on Problem Set on BJT
biasing circuits
Evaluate: Assignment/Quiz/ExamConceptual questions and problem solving
on BJT biasing circuits
5th
MODULE 4
UNIT 1
5th
UNIT 2
5th
MODULE 5
6th
Evaluative Assessment
Finals Quiz on Module 3
FIELD-EFFECT TRANSISTOR (FET)
Fundamentals of Field-Effect Transistor
TLO 6: Describe the types,
Engage: Review of characteristics and
construction, operation,
applications of BJT
applications, and biasing
Explore: Discuss construction of different
techniques for Field Effect
types of FETs; biasing techniques; and,
Transistor (FET)
applications
Explain: Illustrate analysis and solution to
problem on biasing of FET
Elaborate: Practice on Problem Set
Evaluate: Assignment/Quiz/ExamConceptual questions and problem solving
on FET
FET AMPLIFIERS
TLO 7: Analyze and solve
Engage: Review biasing of FET
problems on small signal FET
Explore: Discuss DC and AC equivalent
amplifiers.
circuits and performance measures and
characteristics of the three configurations of
FET amplifiers
Explain: Illustrate analysis and solution to
problem on DC and AC analysis of FET
amplifier
Elaborate: Practice on Problem Set
Evaluate: Assignment/Quiz/ExamConceptual questions and problem solving
on CS amplifiers
OPERATIONAL AMPLIFIERS
TLO 8: Describe and analyze
Engage: Review applications of BJT and FET
basic operational amplifiers
Explore: Discuss DC and AC equivalent
circuits and performance measures and
characteristics of the three configurations of
FET amplifiers
Explain: Illustrate analysis and solution to
problem on DC and AC analysis of FET
amplifier
Elaborate: Practice on Problem Set
Evaluate: Assignment/Quiz/ExamConceptual questions and problem solving
on CS amplifiers
Evaluative Assessment
Assignment on Module 4 Unit 2; Module 5
Final Exam on Modules 3 and 4
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EVALUATION
TO PASS THE COURSE, YOU MUST:
1.
2.
3.
4.
Read all course readings and answer the pre-assessment (or ENGAGE) questions
Take all quizzes and examinations
Submit all assignments
Obtain at least 50% of the total score points in each of the graded assignments,
quizzes and major exams
5. Complete the minimum online participation as required by the University
ASSESSMENT ACTIVITIES
Formative Assessment
Formative assessments such as the ENGAGE questions aim to enhance and
deepen your understanding of the course.
Summative Assessment
Assignments, Quizzes and Examination
Graded assignments will be posted one week before its due date. Schedule for
quizzes, Midterm and Final Individual Examination will be posted at least three days
ahead of time. You will be given tests to be accomplished within the given
timeframe and you are not allowed to edit their answers once submitted. Written
test contains combination of conceptual and problem solving questions.
All submissions are automatically time stamped and recorded. The honor pledge
or undertaking shall always be a part of all requirements submitted online.
DATES OF SUBMISSION:
MIDTERM CLASSWORKS: not later than March 20, 2021
FINAL CLASSWORKS: not later than April 17, 2021
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Rubric for Assignment (Problem Solving, Derivation, etc.)
Criteria
Weight of Criterion
Solution
50%
Presentation
50%
Full Credit
Partial Credit
All relevant
figures/diagrams are
shown and properly
labeled
Solution is presented in a
logical/orderly manner;
All working equations are
presented; equations are
properly written; all
relevant details are shown;
Final answer has
appropriate unit
Figure is properly drawn;
Equations are properly
written;
Work is neat
Incomplete figure;
Some answers are
given without the
complete solution;
Figure is not
properly drawn;
Equations are not
properly written;
Work is untidy
Rubric for Problem Solving in Quizzes and Examinations
Criteria
Weight of Criterion
Figure
30%
Solution
60%
Presentation
10%
Full Credit
Partial Credit
Appropriate visual
representation of the
situation as described in
the problem; properly
labeled
Solution is presented in a
logical/orderly manner; All
working equations and
presented; equations are
properly expressions; all
relevant details are shown;
answers are correct with
the proper unit
Figure is properly drawn;
Equations are properly
written;
Work is neat
Incomplete figure
Some results (final
and/or) are given
without the
complete solution
Figure is not
properly drawn;
Equations are not
properly written;
Work is untidy
CONTACT INFORMATION OF COURSE FACILITATOR
Department of Electronics Engineering
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MODULE 1
SEMICONDUCTORS
This module covers several topics on the PN junction of solid-state diode. Topics include
semiconductor theory on P-type and N-type materials, PN junction construction,
characteristics, biasing, and equivalent circuits. Illustrative problems on the application
of diode equivalent circuits (or approximations) are also included. Common specialpurpose diodes and their applications are also briefly discussed.
At the end of the module, you should be able to:
TLO 1: Describe a PN-junction diode; identify the terminals, construction, operation, and
equivalent circuits
TLO 2: Describe the operation, analyze and draw the output waveform corresponding to
an input waveform for a diode wave-shaping circuit
LEARNING OUTCOMES:
a. Differentiate P type and N type semiconductor materials
b. Describe the V-I characteristic curve of a diode
c. Solve problems applying various equivalent circuits of diode
UNIT 1: SEMICONDUCTOR THEORY AND PN JUNCTION
1. List electronic devices/equipment that you use or are familiar with.
2. Compare conductors and insulators in terms of a) the number of valence
electrons b) the ease of electron passage
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HISTORY
(Source:www.tel.com/museum/exhibition/history/)
1904
Two-electrode vacuum tube is invented; attributed to Thomas Alva Edison and John Ambrose Fleming
1946
World’s first general-purpose computer ENIAC (Electronic Numerical Integrator and Computer) is
announced; it was the largest electronic machine ever made-18000 vacuum tubes and weighing about 30
tons, occupied a 160 sq m room.
1948
Junction-type transistor is invented; attributed to William Bradford Shockley
1955
Japan’s first transistor radio (Sony) is released
1957
Esaki diode is invented; attributed to Leo Esaki
1959
Kilby patent on the integrated circuit is filed; attributed to Jack St. Kilby (Texas Instruments)
1965
Moore’s Law is announced: related to trends in computer manufacturing industries and integration rate;
attributed to Gordon Moore (co-founder of Intel)
1971
Intel 4004 is released: first single chip microprocessor; attributed to Masatoshi Shima
1977
World’s first personal computer Apple II is released
1980
Flash memory is invented; attributed to Fujio Masuoka
1983
Family Computer is released by Nintendo
1991
Carbon nanotube is discovered by Sumio Iijima
1993
First practical blue LED developed by Shuji Nakamura
1995
Sharp releases LCD TV
2002
Earth Simulator (supercomputer) records fastest processing speed
2004
Graphene (semiconductor material) creation experiments prove successful; Andre Geim and Konstantin
Novoselov
2007
Apple announces iPhone; attributed to Steve Jobs
2010
Apple announces iPad; attributed to Steve Jobs
2011
Japanese Super computer, the K computer achieves world’s fastest processing speed
2013
Google Glass beta test starts VR (Virtual Reality)/AR (Augmented Reality)- technologies for making
computer-generated environments
2015
Internet of Things: various object embedded with microprocessors connected via the Internet; pioneered
by Ken Sakamura
2016
AI (Artificial Intelligence): AlphaGo AI program developed by DeepMind defeated the human world
champion in the game Go
17
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Types of semiconductor materials:
1. Intrinsic Semiconductor: a material that is made of pure semiconductor
element (e.g. Si, Ge)
2. Extrinsic Semiconductor: a material that is made of combination of a
semiconductor element and non-semiconductor element (referred to as
impurities)
When a semiconductor material is combined with impurities, the material is said
to be doped, and the process is called doping.
Types of Doped or Extrinsic Semiconductor:
1. N-type: pure semiconductor combined with a pentavalent or
valence 5 element or donor (e.g. Arsenic); results in excess of
electrons
2. P-type: pure semiconductor combined with a trivalent or
valence 3 element or acceptor (e.g. Boron); results in excess of holes
(deficiency of electrons or electron vacancies)
Types of charge carriers in a semiconductor:
a) Electrons: electrons in the conduction band
b) Holes: electron vacancies in the valence band
e
e
e
electron
e
e
e
electron
e
electron
hole
hole
e
e
e
e
hole
e
hole
Hole flow
Electron flow
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When a P-type and an N-type are fused, a PN junction results:
Anode
Cathode
A
K
Symbol
Operation
1. Without any potential applied
electrons
Depletion Region
Diffusion of electrons occurs spontaneously by which electrons will move from the N-type
towards the P-type material. This is due to the high concentration or high charge or high
electron density in the N-type material. The migrating electrons from the N-type leave the
boundary positively-charged. These electrons then fill up the vacancies (or holes) in the
P-type at the boundary between the two materials- a process called recombination. The
diffusion continues until the electric field at the boundary is strong enough to prevent
further migration of electrons. This is referred to as the equilibrium state at which a
potential barrier or depletion region (or layer) now exists at the boundary between the
P- and N- type materials. And, there is a barrier potential at the boundary, which must be
overcome before any charge carrier can pass through.
What happens when a potential difference is applied across the diode? Application of
potential to the electrodes, or application of voltage across the junction is referred to
as biasing. To bias means to apply voltage.
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2. Forward Bias
What happens if the Anode is at positive potential and the cathode at negative
potential? Or generally, what happens if the Anode is at a higher potential relative to the
Cathode?
If the applied potential difference is large enough to overcome the barrier potential,
electrons from the N-type can now move to the P-type and by recombination, travel
through the P-type and out to the positive end of the source. Under these conditions, the
PN junction is considered to forward-biased.
3. Reverse Bias
What happens when the Anode is at a negative potential and the cathode at positive
potential? Or generally, what happens if the Anode is at a lower potential relative to the
Cathode?
With this manner of applying the voltage, the potential barrier (or depletion region) is
reinforced. Electrons are repelled by the negative end of the source, they enter the Ptype, undergo recombination and finally arriving at the boundary. Electrons in the Ntype are attracted by the positive end of the source, leaving more electron vacancies
at the boundary. Under these conditions, the PN junction is considered to be reversebiased.
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Diode Characteristic Curve
πΌπΉ : Forward Current
ππΉ : Forward Voltage
ππΎ : Knee Voltage
πΌπ
: Reverse Current
ππ
: Reverse Voltage
ππ΅π : Breakdown Voltage
πΌπΉ : Forward Current
: current through the junction when it is forward-biased
ππΉ : Forward Voltage
: potential difference across the junction; anode at higher potential relative to
cathode
ππΎ : Knee Voltage
: minimum required voltage to overcome the potential barrier
: also called the barrier potential
: the amount of forward-bias voltage at which the forward current starts to increase
rapidly
: nominal values- 0.3 V for Ge, 0.7 V for Si
πΌπ
: Reverse Current
: current through the junction when it is reverse-biased
: very small amount when the applied reverse-bias voltage is less than the breakdown value
ππ
: Reverse Voltage
: potential difference across the junction; anode at lower potential relative to
cathode
ππ΅π : Breakdown Voltage
: amount of reverse-bias voltage that causes a large amount of reverse current
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Diode Equivalent Circuits
In analyzing circuits that contain diodes, the diode has to be replaced with its
equivalent as determined by its state: whether it is forward- or reverse- biased.
1. First or Ideal Approximation
a. A forward-biased diode is equivalent to a closed switch, or short circuit
b. A reverse-biased diode is equivalent to an open switch, or open circuit
This approximation is used when dealing with simple or basic circuits and/or
when accuracy of results is not a major consideration.
2. Second Approximation
a. A forward-biased diode is equivalent to a dc voltage source (or cell) at a value
equal to the knee voltage or barrier potential: 0.3 V for Ge and 0.7 V for Si
(nominal values)
b. A reverse-biased diode is equivalent to an open switch or open circuit
This approximation is used when accuracy of results is desired.
3. Third Approximation
a. A forward-biased diode is equivalent to a dc voltage source (or cell) in series
with a resistor. Voltage of cell is equal to the knee voltage or barrier potential:
0.3 V for Ge and 0.7 V for Si (nominal values); resistor has resistance equal to
the bulk resistance of the junction. Bulk resistance, ππ΅ is the resistance of the Pand N-type materials.
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b. A reverse-biased diode is equivalent to a resistance, called the reverse
resistance, ππ
.
This approximation is used when a much higher degree of accuracy is
desired; when the external resistances (e.g. load) are comparable to the
diode resistances.
Method of Analysis:
Step 1- Determine whether diode is forward-biased or reverse-biased.
a. Identify which end of the source is connected (directly or indirectly) to
diode each electrode, or which electrode is at a higher (or lower)
potential; or,
b. Determine whether the conventional current would tend to pass from
Anode to Cathode or Cathode to Anode.
Anode to Cathode: forward-biased
Cathode to Anode: reverse-biased
Step 2- Replace the diode with the appropriate equivalent circuit.
Step 3- Compute for the required quantities.
Examples
1. Given:
Determine the current in R applying each of the three diode approximations.
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Solution:
Is the diode forward- or reverse-biased?
Anode is to the positive terminal of the voltage source; Cathode is to the
negative terminal; therefore, diode is forward-biased.
OR
Conventional current is from positive to negative terminal of the voltage source,
in the clockwise direction around the loop; it passes from Anode to Cathode.
Therefore, diode is forward-biased.
a) First Approximation
πΌ=
πΌ=
π
π
3π
12 πΊ
πΌ = 0.25 π΄
b) Second Approximation
By Kirchhoff’s Voltage Rule:
Summing voltages around the loop,
(+π) + (−ππ΅ ) + (−πΌπ
) = 0
3 π − 0.7 π − πΌ(12 πΊ) = 0
πΌ=
2.3 π
12 πΊ
πΌ ≅ 0.192 π΄
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c) Third Approximation
By Kirchhoff’s Voltage Rule:
Summing voltages around the loop,
(+π) + (−ππ΅ ) + (−πΌππ΅ ) + (−πΌπ
) = 0
3 π − 0.7 π − πΌ(2 πΊ) − πΌ(12 πΊ) = 0
πΌ=
2.3 π
2 πΊ + 12 πΊ
πΌ ≅ 0.164 π΄
2. Given
Determine the voltage across π
3 .
Solution:
Is the diode forward- or reverse-biased?
Anode is to the negative terminal of the voltage source; Cathode is to the
positive terminal; therefore, diode is reverse-biased.
OR
Conventional current is from positive to negative terminal of the voltage source;
it would tend to pass from Cathode to Anode. Therefore, diode is reverse-biased.
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Diode is Reverse-Biased
a) First and Second Approximation
π3 = πΌ3 π
3
πΌπ· = 0
πΌ1 = πΌ2 = πΌ3
The resistors are in series; so, by voltage
division:
π
3
π3 = πΈ
π
π = π
1 + π
2 +π
3
π
π
27 ππΊ
π3 = (5 π) (
)
15 ππΊ + 62 ππΊ + 27 ππΊ
π3 ≅ 1.298 π
b) Third Approximation
By current division:
πΌ3 = πΌπ
ππ
ππ
+ π
23
π
23 = 62 ππΊ + 27 ππΊ = 89 ππΊ
π
=
ππ
//π
23 = π
1
≅ 84.015 ππΊ
1
1
+
1500 ππΊ 89 ππΊ
π
π = 15 ππΊ + 84.015 ππΊ ≅ 99.015 ππΊ
πΌπ =
5π
99.015 ππΊ
πΌ3 = (0.051 ππ΄) (
πΌπ ≅ 0.051 ππ΄
1500 ππΊ
)
84.015 ππΊ + 27 ππΊ
πΌ3 ≅ 0.048 ππ΄
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Practice Problem
Given: π
1 = 2 ππΊ
π
2 = 1.2 ππΊ
π·1 : ππ, ππ΅ = 2 πΊ, ππ
= 220 ππΊ
π
3 = 6.8 ππΊ
πΈ = 10 π
π·2 : ππ, ππ΅ = 5 πΊ, ππ
= 560 ππΊ
Apply each of diode
approximations and determine:
a. Current through π·1
b. Voltage across π·2
c. Voltage across π
3
Quiz, Exam-Conceptual Questions and Problem Solving
SUPPLEMENTARY TOPIC: SPECIAL-PURPOSE DIODES
The ordinary PN junction can be operated either under forward-biased or reversebiased condition. In the forward-biased condition, the PN junction passes
significant amount of current from Anode to Cathode; while in the reverse-biased
condition, it passes a very small amount of current or ideally no current at all.
Because of this behavior, it is a component can act either as a conductor or as
an insulator depending on how the potentials are applied and to the electrode,
and also on how much the potential difference is.
The PN junction serves many functions simply by exhibiting conductor/insulator
properties. To cater to more varied applications with more complexities and
stringent requirements, modifications have been introduced to the basic PN
junction.
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Listed in the table below are some of these special diodes.
Diode Type
Symbol
Sample
Zener
(Source:
circuitspedia.com)
Schottky
(Source:
electron.com)
LED, LASER
Used for purposes similar to
those of the ordinary PN
junction when forward and
reverse biased; usually
operated in the
breakdown region for
voltage regulation
Similar functions as a PN
junction; P-type is replaced
with a metal; faster
switching times and for
high frequency
applications, better
sensitivity to low voltages
Used for lighting; when
forward-biased, it emits
light; different elements
used for emission of varied
wavelengths (different
colors)
(LightEmitting
Diode,
Light
Amplification
by
Stimulated
Emission of
Radiation)
Function
(Sources:
learn.adafruit.com
amazon.com
physicsworld.com
Photodiode
(Source:
elprocus.com)
Used in devices that
detect light; in motion
sensors; amount of current
generated is determined
by the intensity of light
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Sample Application:
Shown above can be a block diagram for a basic smoke detector.
A light source is placed on one wall of the room and the photodiode on the
opposite wall. Under normal situation (no smoke), light from the source
continuously strikes the photodiode and a significant amount of current is passing
through the photodiode. This condition keeps the alarm off. If there is smoke, the
light is blocked and does not reach the photodiode. With no light on the
photodiode, the current drops to a significantly low value. The control and alarm
circuits can be designed in such a way that the decrease in photodiode current
would cause the alarm to be activated (may be a sound alarm or optical alarmflashing lights).
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UNIT 2 Clippers or Clipping Circuits
1. What happens if there is potential difference between two points in a circuit?
2. Enumerate the condition(s) for a PN junction to a) block current b) pass
current
3. For each of the following situations, identify which point is at a higher
potential, and indicate the direction of the conventional current (whether A
B or B to A):
a) point A is at +25 V and point B is at +20 V
b) point A is at -15 V and B at -8 V
4. State Kirchhoff’s Voltage Law.
A Clipper is a diode circuit that effectively removes unwanted parts of an input
waveform. Only parts of an input waveform above or below a predetermined
reference are reproduced at the output.
Considering an input waveform with symmetrical positive and negative halfcycles:
A peak clipper removes upper portion of the positive half-cycles of the input
waveform, or the lower portion of the negative half-cycles. A combination peak
clipper, or simply a combination clipper removes upper portion of the positive halfcycles and the lower portion of the negative half-cycles. Clippers are also referred
to as limiters. A positive base limiter reproduces only the upper portion of an input
waveform. A negative base limiter reproduces only the lower portion of an input
waveform.
Method of Analysis:
Step 1- Consider the positive and negative cycles of the input separately.
For each half-cycle of the input determine the state of the diode, whether
forward-biased or reverse-biased.
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a. Identify which end of the voltage source is connected (directly or
indirectly) to each diode electrode, or which electrode is at a higher
(or lower) potential
OR
b. Determine whether the conventional current would tend to pass from
Anode to Cathode or Cathode to Anode:
Anode to Cathode: forward-biased
Cathode to Anode: reverse-biased
Step 2- Replace the diode with the appropriate equivalent circuit. For a basic
approach, forward-biased diode is replaced with a short circuit; a
reverse-biased diode is replaced with an open circuit.
Step 3- Write the expression for the output voltage based on the circuit. Use
applicable concepts or laws (e.g. Ohm’s Law, Kirchhoff’s Laws,
properties of series/parallel circuit, etc.)
Step 4- For a cycle under analysis, if output is equal to the input, copy the
waveform of the input (output waveform is a duplicate of the input
waveform).
Applications of Clippers
1. Used in reducing/minimizing the amplitude unwanted energy (e.g.
noise)
2. Used to limit the input to a device or circuit to avoid
overdriving/overloading
3. Used to change the input waveform to the desired shape
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Sample Circuits with Analyses
1. Negative Clipper: circuit that removes the negative going portion of the input
waveform; produces only the positive going portion cycles
During the positive half-cycles of the input, conventional current (from + to -)
would tend to go clockwise around the loop; from Cathode to Anode.
Therefore, diode is reverse-biased. Ideally, the diode is an open switch (open
circuit).
By KVL:
(+π£π ) + (−ππ
) + (−π£π ) = 0
But π = 0
π£π = π£π
This means that during the
+half-cycle of the input, the
output waveform is a duplicate
of the +half-cycle waveform of
the input.
During the negative half cycles of the input, the conventional
current would tend
to go counterclockwise; from Anode to Cathode. Therefore, the diode is
forward-biased. Ideally, the diode is a closed switch (short circuit).
π£0 = π£πΎπ΄
Since the diode is equivalent
to a short circuit, voltage
across it is zero: π£πΎπ΄ = 0 = π£0
During the negative halfcycles of the input, the
output waveform is a line.
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π£π
π£π
2. Positive Clipper: removes positive going portion of the input waveform;
produces only the negative-going portion
During the positive half-cycles of the input, the conventional current would tend
to go clockwise; from Cathode to Anode. Therefore, the diode is reverse-biased.
Ideally, the diode is an open switch (open circuit).
π£π = π£π
= ππ
But current is zero. Therefore,
π£π = π£π
= 0
During the positive halfcycles of the input, the
output waveform is a line.
This means that during the
During the negative half-cycles of the input, the conventional current would
positive half-cycles of the
tend to go counterclockwise; from Anode to Cathode. Therefore, the diode is
input, the output voltage is
forward-biased. Ideally, the diode is a closed switch (short circuit).
zero: π£0 = 0
π£π = π£π
= ππ
Also, R is across the input;
π£π
= π£π
Therefore, during the negative
half-cycles of the input, the
output waveform is a duplicate
of the negative half-cycle
waveform of the input
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π£π
π£π
3. Biased Positive Clipper
Assume π£πππ > πΈ
During which portion of the input would there be current around the loop and in
what direction would the conventional current tend to go?
There is current if there is potential difference between two points in a circuit.
Positive half-cycle:
During the positive half-cycle of the input, the Anode and Cathode of the diode
are at positive potential.
Between π‘0 and π‘1 during which the instantaneous input voltage is less than the
voltage E, the current would tend to go counterclockwise; that is, from Cathode
to Anode. Therefore, the diode is reverse-biased. Ideally it is an open switch
(open circuit). At π‘1 ,when π£π is exactly equal to E, there is no potential difference
between Anode and Cathode so the diode is still reverse biased.
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-
+
-
+
How is the π£π determined?
π£0 is across the two branches
Right branch:
Left branch:
π£0 = +π£π΄πΎ + πΈ
OR
π£0 = +ππ
+ π£π
Using expression for the left branch:
π=0
therefore,
π£0 = π£π
This means that during the positive half-cycles of the input, when the
instantaneous value of the input voltage is less than the value of E, the output
waveform is a duplicate of the input waveform.
Between π‘1 and π‘2 during which the instantaneous value of the input voltage is
greater than the value of E, the current would tend to go clockwise; that is, from
Anode to Cathode. Therefore, the diode is forward-biased. Ideally it is a closed
switch (short circuit).
How is the π£π determined?
π£0 is across the two branches;
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Using the right: π£0 = +πΈ
This means that during the positive half-cycles of the input, when the
instantaneous value of the input voltage is greater than the value of E, the
output waveform is a horizontal line through +πΈ.
Between π‘2 and π‘3 during which the instantaneous value of π£π is again less than E,
the diode is reverse-biased and the output is a duplicate of the input. This
situation is similar to that between π‘π and π‘1
During the negative half-cycles of the input, the conventional current would
tend to go counterclockwise; from Cathode to Anode. Therefore, the diode is
reverse-biased. Ideally, it is an open switch (open circuit).
Using the left branch: π£π = +ππ
− π£π
But π = 0; so, π£π = −π£π
During the negative half-cycles, the output waveform is a duplicate of the
negative half-cycle waveform.
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The circuit can also be called a positive peak biased clipper.
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4. How should the circuit in no. 3 be modified in order to produce the output
shown below?
The output waveform is for a positive base limiter.
Analysis and Solution:
Circuit in no. 3
π£πππ has to be greater than πΈ; π£πππ > πΈ
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Time interval π‘π to π‘1 : +π£π < πΈ
The output is equal to the battery voltage: π£π = +πΈ,
with positive terminal at the top end and negative at the bottom end.
So, during this time interval, the diode has to be forward biased: the anode at a
higher potential relative to the cathode. This can only happen if the anode is
connected to the positive terminal of the battery.
Time interval π‘1 to π‘2 : +π£π > πΈ
The output is equal to input voltage; the output waveform is a reproduction of
the input waveform: π£π = +π£π
During this time interval, the diode has to be reverse-biased.
During the negative half-cycles, the diode is forward-biased; ideally a short
circuit. Therefore, the output is equal to the battery voltage:
π£π = +πΈ
Final Circuit:
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Practice on the following circuits:
Nos. 1 and 2: Sketch the output waveform. Show your complete analysis.
1.
2.
3.
Draw the circuit for the following
output. Show your complete
analysis.
Assignment, Quiz, Exam
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UNIT 3 CLAMPERS or Clamping Circuits
Describe the circuit operation when the switch is at position 1, then thrown to
position 2. Assume that initially capacitor voltage is zero.
Consider a waveform that has positive and negative-going cycles and
symmetrical about the horizontal axis.
Sample input waveform:
The average value or the dc component is zero.
If this input is fed to a clamper, a dc component is added to the input waveform;
or the dc level would change. This means that the waveform would be shifted
upward or downward without any change in its shape. The dc component could
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either be positive or negative. The line along which the positive and negative
going portions are symmetrical would be above or below zero.
A clamper is a circuit that preserves the waveform of the input but with a different
dc level.
Method of Analysis:
Step 1- Determine whether the diode is forward-biased during very first positive
half-cycle or very first negative half-cycle.
Step 2- Replace the forward-biased diode with a short circuit and determine the
effective voltage which the capacitor will be charged to. Add up the
input voltage and battery voltage (series-opposing or series-aiding).
Indicate which terminal of the capacitor is positively-charged and which
one is negatively-charged.
Step 3- After the capacitor has been fully charged, the diode will be reversedbiased and will remain in this state for all the subsequent cycles.
Step 4- Replace the diode with an open circuit. Write the expression for the
output voltage separately for the positive and negative half-cycles.
Step 4- Draw the output waveform. Take note the shape of the input is
preserved, but shifted above or below a certain level
Application of Clampers:
1.
2.
3.
4.
In sonar and radar testing
In voltage multipliers
To remove distortions in a circuit
In video processing
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Sample Circuits with Analyses
1.Positive Clamper
The capacitor is charged during the time that the diode is forward biased. This
occurs in the very first negative half-cycle of the input. Capacitor is charged to a
maximum voltage equal to the peak value of the input. The right plate charged
positive and the left plate charged negative and the capacitor retains the charge
for the subsequent cycles of the input.
ππΆ = π£πππ
The analysis that follows covers the action of the circuit after the capacitor has
been fully charged.
During the positive half-cycles of the input, the total voltage across the diode is
is equal to the series combination of the input voltage and the capacitor
voltage:
π£πΎπ΄ = +ππΆ + π£π
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This total voltage sets the Cathode at positive potential and the Anode at
negative potential. Therefore, the diode is reverse-biased. Ideally it is an open
circuit.
π£πΎπ΄ = +ππΆ + π£π
; π£π ≤ ππΆ
How is π£π determined? π£π is across the three branches:
Using the leftmost branch:
π£π = πππππππ‘ππ π£πππ‘πππ + πππ π‘πππ‘πππππ’π ππππ’π‘ π£πππ‘πππ
π£π = ππΆ + π£π
ππΆ = π£πππ ;
when π£π = 0,
π£π = ππΆ
;
when π£π = π£πππ
π£π = π£πππ + π£πππ = 2π£πππ
During the negative half-cycles of the input, the total voltage across the diode is
is equal to the series combination of the input voltage and the capacitor
voltage:
π£πΎπ΄ = +ππΆ − π£π
But since the capacitor remains at its maximum value (equal to the peak value
of the input), Cathode is still at positive potential and the Anode at negative
potential. The diode remains reverse-biased.
π£πΎπ΄ = +ππΆ − π£π
; π£π ≤ ππΆ
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π£π is across the three branches:
Using leftmost branch:
π£π = πππππππ‘ππ π£πππ‘πππ − πππ π‘πππ‘πππππ’π ππππ’π‘ π£πππ‘πππ
π£π = ππΆ − π£π
ππΆ = π£πππ ;
when π£π = 0,
π£π = ππΆ
;
when π£π = π£πππ
π£π = π£πππ − π£πππ = 0
In summary:
The capacitor charges during the very first negative cycle to a voltage ππΆ = π£πππ ;
left plate is negative and the right plate positive. Capacitor retains the charge
and the voltage between its plates.
For subsequent positive half-cycles: π£π = ππΆ + π£π
For subsequent negative half-cycles: π£π = ππΆ − π£π
Output voltage is always positive; that is, the waveform is above the zero line
and the input waveform is preserved.
π£ππππ = ππΆ − π£πππ = 0
π£ππππ₯ = ππΆ +π£πππ
Peak-to-peak output voltage= π£ππππ₯ − π£ππππ = 2π£πππ
Dc component= average value = ππΆ
2.Biased Clamper
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The capacitor is charged during the time that the diode is forward biased. This
occurs in the very first positive half-cycle of the input. Capacitor is charged to a
maximum voltage equal to the combination of the peak value of the input and
the battery voltage. The left plate charged positive and the right plate charged
negative and the capacitor retains the charge for the subsequent cycles of the
input.
π£πππ > πΈ
Capacitor voltage reaches
ππΆ = π£πππ − πΈ
The analysis that follows covers the action of the circuit after the capacitor has
been fully charged.
During the positive half-cycles of the input, the total voltage across the diode is
is equal to the series combination of the instantaneous input voltage, the
capacitor voltage and the dc voltage.
π£π΄πΎ = −ππΆ + π£π − πΈ = − (π£πππ − πΈ) +π£π − πΈ
π£π΄πΎ = − π£πππ +π£π
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This total voltage sets the Anode at negative potential and the Cathode at
positive potential. Therefore, the diode is reverse-biased. Ideally it is an open
circuit.
π£π is across the three branches;
In the leftmost branch: π£π = −ππΆ + π£π = − (π£πππ − πΈ) +π£π
π£π = − π£πππ + πΈ + π£π
when π£π = 0 ,
π£π = − π£πππ + πΈ
;
when π£π = π£πππ ,
π£π = πΈ
During the negative half-cycles of the input, the total voltage across the diode is
is equal to the series combination of the instantaneous input voltage, the
capacitor voltage and the dc voltage.
But since the capacitor remains at its maximum value (ππΆ = π£πππ − πΈ), Cathode is
still at positive potential and the Anode at negative potential. The diode remains
reverse-biased.
π£π is across the three branches;
In the leftmost branch: π£π = −ππΆ − π£π = − (π£πππ − πΈ) −π£π
π£π = − π£πππ + πΈ − π£π
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when π£π = 0 ,
π£π = − π£πππ + πΈ
;
when π£π = π£πππ ,
π£π = −2π£πππ + πΈ
peak-to-peak
In the positive half-cycle:
π£ππππ₯ = − π£πππ + πΈ + π£πππ
π£ππππ₯ = +πΈ
In the negative half-cycle:
π£ππππ = − π£πππ + πΈ− π£πππ
π£ππππ = − 2π£πππ + πΈ
Peak-to-peak output voltage = π£ππππ₯ − π£ππππ = 2π£πππ
DC component=Average value= − (π£πππ − πΈ)
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Practice on the following circuits:
Sketch the output waveform. Show your complete analysis.
1.
2.
Assignment, Quiz, Examination
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MODULE 2
Regulated Power Supply
This module covers the basic parts of a regulated power supply. The basic block diagram
is discussed, together with the output waveform for each block and the purpose of each
block. The Units are divided into the Rectifiers, Filters and Regulator. Each of these Units
cover the basic operation of the circuit, the waveforms for the different quantities,
derivations of the average and effective values, and basic design calculations. In
addition to these, a discussion and analysis of voltage multipliers are also included.
Block Diagram with Waveforms
Transformer: It is an electrical device that can either step-up (increase) or stepdown the ac input to the desired value
Rectifier: It is circuit that converts ac into pulsating dc. Pulsating dc voltage has a
varying value but does not change polarity; the current produced has constant
direction. It can either be a circuit that outputs half-wave or full-wave.
Filter: It is a component or a circuit that reduces the variation or smooths out the
pulsating dc.
Regulator: It is a circuit that keeps the output voltage at a desired value. The
output is maintained in spite of changes/variations in the rectifier and/or the load
resistance.
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Transformers
(center-tapped)
ππ : primary voltage; RMS or effective value is
usually given
Nominal Line voltage or outlet voltage= ππππ½πΉπ΄πΊ
ππ : secondary voltage
ππ : number of turns at the primary winding
ππ : number of turns at the secondary winding
Turns ratio:
ππ ππ πππ
ππ
=
=
ππ ππ ππ π
ππ
ππππ π£πππ’π ππ ππ = πππ
ππ √2
ππππ π£πππ’π ππ ππ = ππ π
ππ √2
ππ < ππ ; ππ < ππ : step-down transformer
ππ > ππ ; ππ < ππ : step-up transformer
Center-tapped:
ππ
ππ 1 = ππ 2 =
2
At the end of this module, you should be able to:
TLO 3: Describe the elements and operation of a regulated power supply and present
design procedures involving the rectifier, filter, and the regulator.
LEARNING OUTCOMES:
a. Identify and describe the functions and output waveforms of the basic
parts of regulated power supply.
b. Analyze and solve problems related to half-wave and full-wave rectifiers;
capacitor filters; and, zener voltage regulators.
UNIT 1 RECTIFIERS
1. Differentiate DC and AC voltage sources in terms of a) value of voltage, b)
direction of current produced, and c) waveform or graph of value versus time
2. Enumerate electronic/electrical devices/equipment that you are using or are
familiar with, and identify the type of source for each of these.
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HALF-WAVE RECTIFIER
The diode is forward-biased during the positive
half-cycles, and reverse-biased during the
negative half-cycles. The output is similar to
that of the negative clipper.
Positive half-cycles:
ππ· = ππΏ
π£πΏ = ππΏ π
πΏ ; also, π£πΏ = ππ
ππ·ππ = ππΏππ
π£π· = π£π΄πΎ = 0
When ππ = ππ ππ ,
π£πΏ = π£πΏππ = ππ ππ
π£πΏππ
ππΏππ =
π
πΏ
Negative half-cycles:
ππ· = ππΏ = 0
π£πΏ = ππΏ π
πΏ ; π£πΏ = 0
ππ·ππ = ππΏππ
π£π· = π£π΄πΎ = −ππ + π£πΏ = −ππ
When ππ = ππ ππ ,
π£π· = π£π·ππ = −ππ ππ
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Determine average value, ππΏππ£π and effective value, ππΏπ
ππ of load voltage,
Average value is the same as the DC value; effective value is the same as RMS
(Root Mean Square) and AC value.
For the average value, imagine that the waveform for one cycle from 0 to 2π is
pressed down until a rectangle is formed. The height of this rectangle corresponds
to the average or dc value of the voltage. The exact expression for this quantity
can be obtained by calculus.
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π΄π£πππππ ππππ’π = π·πΆ ππππ’π =
πππ‘ππ ππππ πππ πππ ππ¦πππ π΄ π
=
π΅ππ π
π΅
π
ππΏππ£π =
∫0 π£πΏππ π ππππ‘πππ‘
ππΏππ£π =
ππΏππ£π =
π½π³πππ =
ππ³ππ
π£πΏππ
2π
2π − 0
π£πΏππ
2π
π
(−πππ ππ‘)
[− cos(π) − (−πππ 0)] =
0
π£πΏππ
2π
ππ³ππ = ππππ
[−(−1) − (−1)]
π°π³πππ =
π
π½π³πππ
πΉπ³
π°π«πππ = π°π³πππ
Effective Value = AC Value = RMS Value
Step 1- Square the function
Step 2- Get the Mean or Average Value
Step 3- Extract the Square Root
Step 1:
2
π£πΏ2 = (π£πΏππ π ππππ‘)
ππ‘
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Step 2:
ππππ =
π΄π
π΅ππ π
π
ππππ =
∫0 π£2πΏππ π ππ2 ππ‘πππ‘
2π − 0
π’π π π‘πππππππππ‘πππ πππππ‘ππ‘π¦:
ππππ =
π£πΏ2ππ
2π
π ππ2 ππ‘ =
π
1
2
(1 − πππ 2ππ‘)
1
∫ (1 − πππ 2ππ‘)πππ‘
0
2
π
ππππ =
ππππ =
π£πΏ2ππ
1
[ππ‘ − π ππ2ππ‘]
4π
2
0
π£πΏ2ππ
1
1
{[π − sin(π)] − [0 − sin(0)]}
4π
2
2
ππππ =
π£πΏ2ππ
4π
[π] =
π£πΏ2ππ
4
Step 3:
π£πΏ2ππ
√
ππΏπ
ππ = √ππππ =
4
π½π³πΉπ΄πΊ =
ππ³ππ
π·π³πππ = π°π³πΉπ΄πΊ π½π³πΉπ΄πΊ
π
ππ³ππ = ππππ
Minimum Average Diode Current Rating
πΌπ·ππ£π = πΌπΏππ£π
Minimum Peak Diode Current Rating
ππ·ππ = ππΏππ
ππππππ’π πΌπ·ππ£πππ‘π =
ππππππ’π ππ·ππππ‘π =
πΌπ·ππ£π
0.8
ππ·ππ
0.8
ππ·ππ = πΌπ·ππ£π π
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Minimum Peak Reverse Diode Voltage Rating
π£π·ππ = π£πΏππ
ππππππ’π π£π·ππππ‘π =
π£π·ππ
0.8
π£πΏππ = ππΏππ£π π
Transformer Turns Ratio
ππ πππ
ππ
=
ππ ππ π
ππ
ππ π
ππ =
ππ ππ
√2
ππ ππ = π£πΏππ
Minimum Average Secondary Current Rating
πΌπ ππ£π = πΌπΏππ£π
ππππππ’π πΌπ ππ£πππ‘π =
πΌπ ππ£π
0.8
FULL-WAVE RECTIFIER
1. Center-Tapped Full-Wave Rectifier
ππ : total secondary voltage
ππ 1 : voltage between top of
secondary to center tap
ππ 2 : voltage between bottom of
secondary to center tap
ππ = πππ + πππ
πππ = πππ =
-
ππ
π
+
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During the positive half-cycles of secondary
voltage, top of secondary is positive with
respective to the bottom of the secondary.
Top of the secondary is positive with respect
to the center tap; bottom is negative with
respect to the center tap.
Current in the upper loop would tend to go
clockwise; from anode to cathode of π·1 .
π«π : forward- biased
Current in the lower loop would tend to go
clockwise as well; from cathode to anode of
π·2 .
π«π : reverse- biased
π·1 : ideally short
ππ·1 = ππΏ
ππ·ππ = ππΏππ
πππ π,
π£πΏ = ππ 1 =
ππ³ is positive with respect to ground.
When ππ 1 = ππ 1ππ
π£πΏ = π£πΏππ = ππ 1ππ
ππ
2
π£π·1 = π£π΄πΎ = 0
π·2 : ideally open
ππ·2 = 0
π£π·2 = π£π΄πΎ = −ππ
When ππ = ππ ππ
During negative half-cycles of secondary
voltage, top of secondary is negative with
respective to the bottom of the secondary.
Top of the secondary is negative with
respect to the center tap; bottom is positive
with respect to the center tap.
Current in the upper loop would tend to go
counterclockwise; from cathode to anode
of π·1 .
π«π : reverse- biased
Current in the lower loop would tend to go
counterclockwise as well; from anode to
cathode of π·2 .
π«π : forward- biased
π£πΏ = ππΏ π
πΏ
π£π·2 = π£π·2ππ = −ππ ππ
π·1 : ideally open
ππ·1 = 0
π£π·1 = π£π΄πΎ = −ππ
When ππ = ππ ππ
π£π·1 = π£π·1ππ = −ππ ππ
π·2 : ideally short
ππ·2 = ππΏ
ππ·ππ = ππΏππ
π£πΏ = ππΏ π
πΏ
πππ π,
π£πΏ = ππ 2
ππ³ is positive with respect to ground.
When ππ 2 = ππ 2ππ
π£πΏ = π£πΏππ = ππ 2ππ =
ππ ππ
2
π£π·1 = π£π΄πΎ = 0
There is current in πΉπ³ during both positive and negative half-cycles of the secondary
voltage, and it does not change direction (right to left in the figures). The resulting
ππ·2 = ππΏ
voltage across πΉπ³ is always positive with respect to ground for both half-cycles of the
π£π·2 = π£π΄πΎ = −ππ
secondary voltage.
When ππ = ππ ππ
π£π·2 = π£π·2ππ = −ππ ππ
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Determine average value, ππΏππ£π and effective value, ππΏπ
ππ of load voltage,
Average value is the same as the DC value; effective value is the same as RMS
(Root Mean Square) and AC value.
For the average value, imagine that the waveform for one cycle from 0 to π is
pressed down until a rectangle is formed. The height of this rectangle corresponds
to the average or dc value of the voltage. The exact expression for this quantity is
determined by calculus.
π΄π£πππππ ππππ’π = π·πΆ ππππ’π =
πππ‘ππ ππππ πππ πππ ππ¦πππ π΄ π
=
π΅ππ π
π΅
π
ππΏππ£π =
∫0 π£πΏππ π ππππ‘πππ‘
ππΏππ£π =
π−0
π£πΏππ
π
π
(−πππ ππ‘)
0
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ππΏππ£π =
π½π³πππ =
π£πΏππ
π
[− cos(π) − (−πππ 0)] =
πππ³ππ
ππ³ππ =
π
π£πΏππ
π
[−(−1) − (−1)]
ππππ
π°π³πππ =
π
π½π³πππ
πΉπ³
The area for one cycle for the diode current waveform is half of the area for that
of the load current; therefore,
π°π«πππ =
π°π³πππ
π
ππ«ππ = ππ³ππ
Effective Value = AC Value = RMS Value
Step 1- Square the function
Step 2- Get the Mean or Average Value
Step 3- Extract the Square Root
Step 1:
2
π£πΏ2 = (π£πΏππ π ππππ‘)
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Step 2:
π΄π
π΅ππ π
ππππ =
π
ππππ =
∫0 π£2πΏππ π ππ2 ππ‘πππ‘
π−0
π’π π π‘πππππππππ‘πππ πππππ‘ππ‘π¦:
ππππ =
π£πΏ2ππ
π
π ππ2 ππ‘ =
π
1
2
(1 − πππ 2ππ‘)
1
∫ (1 − πππ 2ππ‘)πππ‘
0
2
π
ππππ =
ππππ =
π£πΏ2ππ
1
[ππ‘ − π ππ2ππ‘]
2π
2
0
π£πΏ2ππ
1
1
{[π − sin(π)] − [0 − sin(0)]}
2π
2
2
ππππ =
π£πΏ2ππ
2π
[π] =
π£πΏ2ππ
2
Step 3:
π£πΏ2ππ
ππΏπ
ππ = √ππππ = √
2
π½π³πΉπ΄πΊ =
ππ³ππ
π·π³πππ = π°π³πΉπ΄πΊ π½π³πΉπ΄πΊ
ππ³ππ =
√π
Minimum Average Diode Current Rating
πΌπΏ
πΌπ·ππ£π = ππ£π
2
ππππππ’π πΌπ·ππ£πππ‘π =
ππππ
π
πΌπ·ππ£π
0.8
Minimum Peak Diode Current Rating
ππ·ππ = πΌπ·ππ£π π
ππππππ’π ππ·ππππ‘π =
ππ·ππ
0.8
Minimum Peak Reverse Diode Voltage Rating
π£π·ππ = 2π£πΏππ
ππππππ’π π£π·ππππ‘π =
π£π·ππ
0.8
ππΏ π
π£πΏππ = ππ£π
2
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Transformer Turns Ratio
ππ π
ππ =
ππ πππ
ππ
=
ππ ππ π
ππ
ππ ππ
√2
ππ ππ = 2π£πΏππ
Minimum Average Secondary Current Rating
ππππππ’π πΌπ ππ£πππ‘π =
πΌπΏ
πΌπ ππ£π = ππ£π
2
πΌπ ππ£π
0.8
1. Bridge Type Full-Wave Rectifier
a
c
b
d
During the positive half-cycles of secondary
voltage, top of secondary is positive with
respective to the bottom of the secondary.
Current would tend to go clockwise,
separate at the node a. It would go from A to
K of π·1 but K to A of π·2 . At node b, current
would separate: K to A of π·4 and through π
πΏ .
At node c, current from π
πΏ would from A to K
of π·3 (π·2 is reverse-biased). This current would
go back to the sources (π·4 is reverse-biased).
π«π : forward- biased π«π : forward- biased
π«π : reverse- biased π«π : reverse- biased
ππ·2 = ππ·4 = 0
π£πΏ = ππΏ π
πΏ
ππ·1 = ππ·3 = ππΏ
ππ·ππ = ππΏππ
With π·1 and π·3 as short-circuits, π
πΏ is across the source;
so, π£πΏ = ππ
π£πΏππ = ππ ππ
ππ³ is positive with respect to ground.
π£π·1 = π£π·3 = 0
π£π·2 = π£π·4 = π£π΄πΎ = −ππ
π£π·2ππ = π£π·4ππ = −ππ ππ
Anode of π·2 is connected to the bottom of the
secondary via the short-circuit π·3 . Cathode of π·2 is
connected to the top of the secondary.
Anode of π·4 is connected to the bottom of the
secondary. Cathode of π·4 is connected to the top of
the secondary via the short-circuit π·1 .
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a
c
b
d
During the negative half-cycles of secondary
voltage, top of secondary is positive with
respective to the bottom of the secondary.
Current would tend to go counterclockwise,
separate at the node d. It would go from A to
K of π·4 but K to A of π·3 . At node b, current
would separate: K to A of π·1 and through π
πΏ .
At node c, current from π
πΏ would from A to K
of π·2 (π·3 is reverse-biased). This current would
go back to the source (π·1 is reverse-biased).
π«π : reverse- biased π«π : reverse- biased
π«π : forward- biased π«π : forward- biased
ππ·1 = ππ·3 = 0
ππ·2 = ππ·4 = ππΏ
π£πΏ = ππΏ π
πΏ
ππ·ππ = ππΏππ
With π·2 and π·4 as short-circuits, π
πΏ is across the
source; so, π£πΏ = ππ
π£πΏππ = ππ ππ
ππ³ is positive with respect to ground.
π£π·2 = π£π·4 = 0
π£π·1 = π£π·3 = π£π΄πΎ = −ππ
π£π·1ππ = π£π·3ππ = −ππ ππ
Anode of π·3 is connected to the top of the
secondary via the short-circuit π·2 . Cathode of
π·3 is connected to the bottom of the
secondary.
Anode of π·1 is connected to the top of the
secondary. Cathode of π·1 is connected to the
bottom of the secondary via the short-circuit
π·4 .
There is current in πΉπ³ during both positive and negative half-cycles of the secondary
voltage, and it does not change direction (right to left in the figures). The resulting
voltage across πΉπ³ is always positive with respect to ground for both half-cycles of the
secondary voltage.
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Determine average value, ππΏππ£π and effective value, ππΏπ
ππ of load voltage,
Average value is the same as the DC value; effective value is the same as RMS
(Root Mean Square) and AC value.
For the average value, imagine that the waveform for one cycle from 0 to π is
pressed down until a rectangle is formed. The height of this rectangle corresponds
to the average or dc value of the voltage. The exact equation for this quantity
can be obtained by calculus.
π΄π£πππππ ππππ’π = π·πΆ ππππ’π =
πππ‘ππ ππππ πππ πππ ππ¦πππ π΄ π
=
π΅ππ π
π΅
π
ππΏππ£π =
∫0 π£πΏππ π ππππ‘πππ‘
ππΏππ£π =
ππΏππ£π =
π½π³πππ =
π£πΏππ
π
π−0
π£πΏππ
π
π
(−πππ ππ‘)
0
[− cos(π) − (−πππ 0)] =
πππ³ππ
ππ³ππ = ππππ
π
π£πΏππ
π
[−(−1) − (−1)]
π°π³πππ =
π½π³πππ
πΉπ³
The area for one cycle for the diode current waveform is half of the area for that
of the load current; therefore,
π°π«πππ =
π°π³πππ
π
ππ«ππ = ππ³ππ
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Effective Value = AC Value = RMS Value
Step 1- Square the function
Step 2- Get the Mean or Average Value
Step 3- Extract the Square Root
Step 1:
2
π£πΏ2 = (π£πΏππ π ππππ‘)
Step 2:
ππππ =
π΄π
π΅ππ π
π
ππππ =
∫0 π£2πΏππ π ππ2 ππ‘πππ‘
π−0
π’π π π‘πππππππππ‘πππ πππππ‘ππ‘π¦:
ππππ =
π£πΏ2ππ
π
π
π ππ2 ππ‘ =
1
2
(1 − πππ 2ππ‘)
1
∫ (1 − πππ 2ππ‘)πππ‘
0
2
π
ππππ =
ππππ =
π£πΏ2ππ
1
[ππ‘ − π ππ2ππ‘]
2π
2
0
π£πΏ2ππ
1
1
{[π − sin(π)] − [0 − sin(0)]}
2π
2
2
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ππππ =
π£πΏ2ππ
2π
[π] =
π£πΏ2ππ
2
Step 3:
π£πΏ2ππ
√
ππΏπ
ππ = √ππππ =
2
π½π³πΉπ΄πΊ =
ππ³ππ
√π
ππ³ππ = ππππ
Minimum Average Diode Current Rating
πΌπΏ
πΌπ·ππ£π = ππ£π
2
π
(π½π³πΉπ΄πΊ )
π·π³πππ = π°π³πΉπ΄πΊ π½π³πΉπ΄πΊ =
πΉπ³
ππππππ’π πΌπ·ππ£πππ‘π =
πΌπ·ππ£π
0.8
Minimum Peak Diode Current Rating
ππ·ππ = πΌπ·ππ£π π
ππππππ’π ππ·ππππ‘π =
ππ·ππ
0.8
Minimum Peak Reverse Diode Voltage Rating
π£π·ππ = π£πΏππ
ππππππ’π π£π·ππππ‘π =
π£π·ππ
0.8
ππΏ π
π£πΏππ = ππ£π
2
Transformer Turns Ratio
ππ πππ
ππ
=
ππ ππ π
ππ
ππ π
ππ =
ππ ππ
√2
ππ ππ = π£πΏππ
Minimum Average Secondary Current Rating
πΌπ ππ£π = πΌπΏππ£π
ππππππ’π πΌπ ππ£πππ‘π =
πΌπ ππ£π
0.8
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Sample Problems:
Given: Rectifier connected across the secondary of a transformer with turns ratio,
ππ : ππ of 10:1.
Determine all relevant quantities for the half-wave and full-wave rectifiers. Let π
πΏ = 2 ππΊ
Solution:
a) Half-wave rectifier
ππΏππ£π =
π£πΏππ
π
1
ππ π
ππ = (220 π) ( ) = 22 π
10
ππ ππ = (22 π)√2 ≅ 31.113 π = π£πΏππ
ππΏππ£π =
π£πΏππ = ππ ππ
31.113 π
π
ππ ππ = ππ π
ππ √2
ππ π
ππ = πππ
ππ (
ππ
)
ππ
πππ
ππ = ππππ π£πππ‘πππ = 220 π
ππΏππ£π ≅ 9.904
πΌπΏππ£π =
πΌπΏππ£π =
ππΏππ£π
π
πΏ
9.904 π
2 π₯103 πΊ
ππΏππ =
ππΏππ =
π£πΏππ
π
πΏ
31.113 π
2 π₯103 πΊ
πΌπΏππ£π = 4.952 π₯10−3 π΄
ππΏππ ≅ 15.557π₯10−3 π΄
πΌπ·ππ£π = πΌπΏππ£π
ππ·ππ = ππΏππ
πΌπ·ππ£π = 4.952 π₯10−3 π΄
ππ·ππ ≅ 15.557π₯10−3 π΄
π£π·ππ = ππ ππ
π£π·ππ ≅ 31.113 π
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b) Center-tapped Full-Wave Rectifier
ππΏππ£π =
ππΏππ£π =
2π£πΏππ
π£πΏππ =
π
2(15.557 π)
π
ππΏππ£π = 9.904 π
9.904 π
2 π₯103 πΊ
πΌπ·ππ£π =
πΌπΏππ£π
2
4.952 π₯10
2
From part a): ππ ππ ≅ 31.113 π
31.113 π
π£πΏππ =
2
ππΏππ =
πΌπΏππ£π = 4.952 π₯10−3 π΄
πΌπ·ππ£π =
2
π£πΏππ ≅ 15.557 π
π£πΏππ
ππΏππ =
π
πΏ
ππΏ
πΌπΏππ£π = ππ£π
π
πΏ
πΌπΏππ£π =
ππ ππ
15.557 π
2 π₯103 πΊ
ππΏππ ≅ 7.779π₯10−3 π΄
ππ·ππ = ππΏππ
−3
π΄
πΌπ·ππ£π = 2.476π₯10−3 π΄
ππ·ππ = 7.779π₯10−3 π΄
π£π·ππ = ππ ππ
π£π·ππ ≅ 31.113 π
c) Bridge Type Full-Wave Rectifier
ππΏππ£π =
2π£πΏππ
π£πΏππ = ππ ππ
π
2(31.113 π)
ππΏππ£π =
π
From part a): ππ ππ ≅ 31.113 π
ππΏππ£π ≅ 19.807 π
πΌπΏππ£π =
πΌπΏππ£π =
ππΏππ£π
π
πΏ
19.807 π
2 π₯103 πΊ
πΌπΏππ£π ≅ 9.904π₯10−3 π΄
ππΏππ =
ππΏππ =
π£πΏππ
π
πΏ
31.113 π
2 π₯103 πΊ
ππΏππ = 15.557π₯10−3 π΄
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πΌπΏππ£π
2
πΌπ·ππ£π =
ππ·ππ = ππΏππ
ππ·ππ = 15.557π₯10−3 π΄
9.904 π₯10−3 π΄
πΌπ·ππ£π =
2
πΌπ·ππ£π = 4.952π₯10
−3
π£π·ππ = ππ ππ
π΄
π£π·ππ = 31.113 π
1. Design a rectifier fed from 220-V line through a step-down transformer. The load
requires an average voltage of 12 V and average power of 5 W.
Solution:
ππΏππ£π = 12 π
ππΏππ£π = 5 π
a) Half-Wave Rectifier
π£πΏππ = (12 π)π
ππΏππ£π = πΌπΏπ
ππ ππΏπ
ππ
ππΏπ
ππ =
π£πΏππ
π£πΏππ ≅ 37.699 π
2
ππΏπ
ππ =
π£πΏππ = ππΏππ£π π
πΌπΏπ
ππ =
ππΏππ
2
πΌπΏππ£π =
ππΏππ
π
(37.699 π)
≅ 18.850 π
2
πΌπΏπ
ππ =
5π
= 0.265 π΄
18.850 π
ππΏππ = 2(0.265 π΄) = 0.53 π΄
πΌπΏππ£π =
ππΏππ
π
=
0.53 π΄
= 0.169 π΄
π
πΌπ·ππ£π = πΌπΏππ£π
ππππππ’π πΌπ·ππ£πππ‘π =
ππππππ’π πΌπ·ππ£πππ‘π =
πΌπ·ππ£π
0.8
0.169 π΄
0.8
π΄ππππππ π°π«ππππππ ≅ π. πππ π¨
ππ·ππ = ππΏππ = 0.53 π΄
ππ·ππ
ππππππ’π ππ·ππππ‘π =
0.8
ππππππ’π ππ·ππππ‘π =
0.53 π΄
0.8
ππππππ’π ππ·ππππ‘π ≅ 0.663 π΄
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π£π·ππ = π£πΏππ = 37.699 π
ππππππ’π π£π·ππππ‘π =
ππ πππ
ππ
=
ππ ππ π
ππ
π£π·ππ
0.8
ππ π
ππ =
ππ ππ
ππππππ’π π£π·ππππ‘π =
ππππππ’π π£π·ππππ‘π = 47.124 π
ππ ππ = π£πΏππ = 37.699 π
√2
ππ π
ππ =
ππ πππ
ππ
220 π
=
=
ππ ππ π
ππ 26.657 π
37.699 π
0.8
37.699 π
√2
≅ 26.657 π
ππ 8
≅
ππ 1
ππππππ’π πΌπ ππ£πππ‘π =
πΌπ ππ£π = πΌπΏππ£π = 0.169 π΄
ππππππ’π πΌπ ππ£πππ‘π =
πΌπ ππ£π
0.8
0.169 π΄
0.8
ππππππ’π πΌπ ππ£πππ‘π = 0.211 π΄
b) Center-Tapped Full-Wave Rectifier
ππΏππ£π = πΌπΏπ
ππ ππΏπ
ππ
ππΏπ
ππ =
π£πΏππ
ππΏππ
√2
πΌπ·ππ£π =
πΌπΏππ£π =
2ππΏππ
π
=
ππΏππ£π π
2
πΌπΏππ£π =
(12 π)π
2
π£πΏππ ≅ 18.850 π
√2
π£πΏππ =
πΌπΏπ
ππ =
π£πΏππ =
ππΏπ
ππ =
2ππΏππ
π
πΌπΏππ£π
2
2(0.53 π΄)
= 0.337 π΄
π
πΌπΏ
0.337
πΌπ·ππ£π = ππ£π =
= 0.169 π΄
2
2
(18.850 π)
πΌπΏπ
ππ =
√2
≅ 13.329 π
5π
= 0.375 π΄
13.329 π
ππΏππ = √2(0.375 π΄) = 0.53 π΄
ππππππ’π πΌπ·ππ£πππ‘π =
ππππππ’π πΌπ·ππ£πππ‘π =
πΌπ·ππ£π
0.8
0.169 π΄
0.8
π΄ππππππ π°π«ππππππ ≅ π. πππ π¨
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ππ·ππ = ππΏππ = 0.53 π΄
ππ·ππ
ππππππ’π ππ·ππππ‘π =
0.8
0.53 π΄
0.8
ππππππ’π ππ·ππππ‘π =
π΄ππππππ ππ«πππππ ≅ π. πππ π¨
π£π·ππ = 2π£πΏππ = 2(18.850 π) = 37.7π
ππππππ’π π£π·ππππ‘π =
ππ πππ
ππ
=
ππ ππ π
ππ
π£π·ππ
0.8
ππ π
ππ =
ππ ππ
ππππππ’π π£π·ππππ‘π =
π΄ππππππ ππ«πππππ = ππ. πππ π½
ππ ππ = 2(π£πΏππ ) = 37.7 π
√2
ππ π
ππ =
ππ πππ
ππ
220 π
=
=
ππ ππ π
ππ 26.658 π
37.7 π
0.8
37.7 π
√2
≅ 26.658 π
ππ 8
≅
ππ 1
πΌπΏ
0.337 π΄
πΌπ ππ£π = ππ£π =
= 0.169 π΄
2
2
πΌπ
ππππππ’π πΌπ ππ£πππ‘π = ππ£π
0.8
ππππππ’π πΌπ ππ£πππ‘π =
0.169 π΄
0.8
ππππππ’π πΌπ ππ£πππ‘π = 0.211 π΄
c) Bridge Type Full-Wave Rectifier
ππΏππ£π = πΌπΏπ
ππ ππΏπ
ππ
ππΏπ
ππ =
π£πΏππ
ππΏππ
√2
πΌπ·ππ£π =
ππΏππ£π π
2
πΌπΏππ£π =
πΌπΏππ£π
2
(12 π)π
2
π£πΏππ ≅ 18.850 π
√2
π£πΏππ =
πΌπΏπ
ππ =
π£πΏππ =
ππΏπ
ππ =
2ππΏππ
π
(18.850 π)
πΌπΏπ
ππ =
√2
≅ 13.329 π
5π
= 0.375 π΄
13.329 π
ππΏππ = √2(0.375 π΄) = 0.53 π΄
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πΌπΏππ£π =
2ππΏππ
π
=
2(0.53 π΄)
= 0.337 π΄
π
πΌπΏ
0.337
πΌπ·ππ£π = ππ£π =
= 0.169 π΄
2
2
ππ·ππ = ππΏππ = 0.53 π΄
ππ·ππ
ππππππ’π ππ·ππππ‘π =
0.8
ππππππ’π πΌπ·ππ£πππ‘π =
ππππππ’π πΌπ·ππ£πππ‘π =
ππππππ’π π£π·ππππ‘π =
ππ πππ
ππ
=
ππ ππ π
ππ
0.8
ππ π
ππ =
ππ ππ
ππππππ’π ππ·ππππ‘π =
0.53 π΄
0.8
π΄ππππππ ππ«πππππ ≅ π. πππ π¨
ππππππ’π π£π·ππππ‘π =
18.850 π
0.8
π΄ππππππ ππ«πππππ ≅ ππ. πππ π½
ππ ππ = π£πΏππ = 18.850 π
√2
ππ πππ
ππ
220 π
=
=
ππ ππ π
ππ 13.329 π
0.169 π΄
0.8
π΄ππππππ π°π«ππππππ ≅ π. πππ π¨
π£π·ππ = π£πΏππ = 18.850 π
π£π·ππ
πΌπ·ππ£π
0.8
ππ π
ππ =
18.850 π
√2
≅ 13.329 π
ππ 16
≅
ππ
1
πΌπ ππ£π = πΌπΏππ£π = 0.337 π΄
ππππππ’π πΌπ ππ£πππ‘π =
πΌπ ππ£π
0.8
ππππππ’π πΌπ ππ£πππ‘π =
0.337 π΄
0.8
ππππππ’π πΌπ ππ£πππ‘π = 0.421 π΄
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Problem Set:
1. Derive the formula for the RMS value for the following
a)
b)
2.Design a rectifier fed from 220-V line through a step-down transformer. The load
requires an average voltage of 9 V and average power of 2.5 W. Use each of the
three circuits: half-wave, center-tapped full-wave and bridge type full-wave.
Assignment, Quiz, Exam
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UNIT 2 CAPACITOR FILTER
1. Sketch the graph of capacitor voltage versus time for a
a. charging capacitor
b. discharging capacitor
2. Based on the graph in no.1a), express the capacitor voltage after one time
constant, in terms of the charging voltage.
3. Based on the graph in no.1b), express the capacitor voltage after one time
constant, in terms of the initial voltage
Ripple factor, ππ is a figure of merit used on power supply filters. It is a measure of
the smoothness of the output, and how close the output waveform is to that of
dc.
Ripple: Variations in the output above and below the average value; the average
value refers also to the dc component of the output or load voltage.
ππππππππππ‘ππ =
π
ππ π£πππ’π ππ ππππππ π£πππ‘πππ
ππ£πππππ π£πππ’π ππ ππππ π£πππ‘πππ
ππ =
πππ
ππ
ππΏππ£π
equation 1
Ideally, ππ = 0; that is, the ripple must be as small as possible
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Consider the following load voltage waveform:
For a sawtooth ripple:
πππ
ππ =
ππππ
equation 2
√3
In the waveform above:
π£πΏππ = ππΏππ£π + ππππ
In equation 1:
πππ
ππ = ππ ππΏππ£π
ππππ
√3
= ππ ππΏππ£π
ππππ = √3ππ ππΏππ£π
equation 3
substitute in equation 2
substitute in equation 3
π£πΏππ = ππΏππ£π + √3ππ ππΏππ£π
Equation 4
ππ³ππ = π½π³πππ [π + √πππ ]
This formula is used for
solving the peak load
voltage when the ripple is
approximated
as
a
sawtooth wave.
The ripple must be as small as possible so that the average value will approach
the maximum (peak) value.
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1. Half-wave Rectifier with Capacitor Filter
Assume that the capacitor is initially uncharged and ππ = 0. During the first positive
half-cycle, from π‘0 to π‘1 , the anode is at positive potential and cathode at zero
potential (the current would tend to go in the clockwise direction). The diode is
forward biased and the capacitor will charge with the upper terminal positivelycharged and the lower terminal negatively-charged. The capacitor voltage will
reach a maximum voltage equal to the peak value of the source:
ππ = ππ ππ .
From π‘1 to π‘2 , when ππ < ππ , cathode at higher potential relative to the anode;
diode reverse-biased; capacitor discharges through π
πΏ and ππ decreases.
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From π‘2 to π‘3 (negative half-cycle), cathode still at higher potential relative to the
anode, diode remains reverse biased and capacitor continues to discharge
through π
πΏ and capacitor voltages decreases further.
From π‘3 to π‘4 , (second positive half-cycle) when the source voltage is greater than
the capacitor voltage, the anode is at a higher potential relative to the cathode;
diode is forward-biased. The capacitor recharges up to a maximum voltage equal
to the peak value of the source voltage.
In subsequent cycles, the diode periodically goes through forward-bias and
reverse-bias states and the capacitor undergoes charging and discharging,
respectively.
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2. Full-wave Rectifier with Capacitor Filter
For half-wave and full-wave rectifier with capacitor
π = πΆπ
πππ’ππ‘πππ 5
πΆ=
βπ: change in the voltage
πΌπ‘
π
ππ
π = πΌπ‘ πππ’ππ‘πππ 6
πΆ=πΌ
βπ‘
βπ
πππ’ππ‘πππ 7
βπ =peak-to-peak of ripple= 2(peak value of ripple)
βπ = 2ππππ
equation 8
Using equation 1 (page 86) and equation 2 (page 87),
βπ = 2ππ ππΏππ£π √3
βπ‘:time interval
βπ‘ = πβππππππ π‘πππ + πππ πβππππππ π‘πππ = π‘π + π‘π
In the graph, time for one cycle or period, π = π‘π + π‘π = βπ‘
But π‘π βͺ π‘π
so, π ≅ π‘π
and
1
π=π
π: frequency
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πΌ = πΌπΏππ£π =
ππΏππ£π
π
πΏ
Back to equation 7,
ππΏ
1
1
πΆ = ( ππ£π ) ( )
π
πΏ
π 2ππ ππΏππ£π √3
Capacitance:
πͺ=
Ripple factor:
π
ππ =
π√πππ ππΉπ³
πΌπ·ππ£π = πΌπΏππ£π (HW)
ππ·ππ = πΌπΏππ£π [1 + 2π√
π
π√πππΉπ³ πͺ
πΌπ·ππ£π =
1
√3ππ
πΌπΏππ£π
2
ππΉπ = 2ππ»π
(FW)
π£π·ππ ≅ 2ππ ππ
]
Design requirements:
In design, minimum required Capacitance,
πΆπππ =
1
2√3ππ
πΏ ππ
Minimum voltage rating,
ππππππ‘π =
ππΆπππ₯
0.8
ππΆπππ₯ = π£πΏππ
Minimum average secondary current,
πΌπ ππ£πππ‘π =
πΌπ ππ£π
0.8
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Sample Problems:
Given: Full-wave rectifier; transformer secondary voltage = 20 πππ
Capacitance 220 ππΉ; average load voltage=25 V
Determine the a) ripple factor
Solution: π = 120 π»π§
ππ ππ = ππ π
ππ √2
b) average load current
ππΏππ£π = 25 π
πΆ = 220π₯10−6 πΉ
π£πΏππ = 28.284 π
ππ ππ = (20 π)√2 = 28.284 π
28.284 π
− 1)
25
ππ =
√3
ππ = 0.076
π£πΏππ = ππΏππ£π [1 + ππ √3]
(
28.284 π = 25 π[1 + ππ √3]
π
πΏ =
ππ π
ππ = 20 π
1
πΌπΏππ£π =
2√3(0.076)(120 π»π§)(220π₯10−6 πΉ)
ππΏππ£π
π
πΏ
πΌπΏππ£π =
25 π
143.877 πΊ
πΌπΏππ£π = 0.174 π΄
π
πΏ = 143.877 πΊ
Design a half-wave rectifier with a capacitor filter that will deliver to the load an
average voltage is 12 V at 250 mA. Ripple should not be higher than 4%.
Solution: π = 60 π»π§
ππΏππ£π = 12 π
πΌπΏππ£π = 250 ππ΄
ππ = 0.04
π£πΏππ = ππΏππ£π [1 + ππ √3]
π£πΏππ = 12.831 π
π£πΏππ = (12 π)[1 + (0.04)√3]
ππ·ππ = πΌπΏππ£π [1 + 2π√
1
ππ √3
ππ·ππ = (0.250 π΄) [1 + 2π√
]
1
(0.04)√3
πΌπ·ππ£π = πΌπΏππ£π = 250 ππ΄
ππππππ’π πΌπ·ππ£πππ‘π =
π£π·ππ = 2π£πΏππ
ππ·ππ = 6.218 π΄
250 ππ΄
0.8
π£π·ππ = 2(12.831 π)
πππ ππ·ππππ‘π =
]
ππππππ’π ππ·ππππ‘π =
6.218 π΄
0.8
ππ·ππ
0.8
πππ ππ·ππππ‘π = 7.773 π΄
ππππππ’π πΌπ·ππ£πππ‘π =
πΌπ·ππ£π
0.8
πππ πΌπ·ππ£πππ‘π = 312.5 ππ΄
π£π·ππ = 25.662 π
min ππππ‘π =
25.662 π
0.8
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ππππππ’π π£π·ππππ‘π =
πΆπππ =
π£π·ππ
πππ π£π·ππππ‘π = 32.078 π
0.8
1
π
πΏ =
2√3(0.04)(60 π»π§)(48 πΊ )
ππΏππ£π
πΌπΏππ£π
πΆπππ = 2.506π₯10−3 πΉ
ππππππ’π ππππ‘π =
π
πΏ =
12 π
0.250 π΄
π
πΏ = 48 πΊ
π£πΏππ
πππ ππππ‘π =
0.8
(12.831 π)
0.8
πππ ππππ‘π = 16.039 π
ππ πππ
ππ
=
ππ ππ π
ππ
ππ ππ = π£πΏππ = 12.831 π
ππ π
ππ =
ππ ππ
√2
ππ π
ππ =
ππ ππ
√2
=
12.831 π
√2
ππ
220 π
=
ππ 9.073 π
ππ 24
≅
ππ
1
ππ π
ππ = 9.073 π
ππππππ’π πΌπ ππ£πππ‘π =
πΌπ ππ£π = πΌπΏππ£π = 250 ππ΄
πΌπ ππ£π
0.8
πππ πΌπ ππ£πππ‘π =
250 ππ΄
0.8
πππ πΌπ ππ£πππ‘π = 312.5 ππ΄
1. Following the analysis applied to the half-wave rectifier with capacitor filter, show
how to arrive at the output waveform for a center-tapped full-wave with filter
capacitor. See output waveform on page 90.
2. Given: Half-wave rectifier; transformer secondary voltage = 32 V; Capacitance
4700 μF; average load voltage 18 V. Determine ripple factor and average load
current
3. Design a half-wave rectifier with a capacitor filter that will deliver to the load an
average voltage is 15 V at 280 mA. Ripple should not be higher than 5%.
Assignment, Quiz, Exam
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UNIT 3 ZENER VOLTAGE REGULATOR
1. Describe/Enumerate devices or equipment that require stable supply.
2. Describe operation of diode in the forward-bias and reverse-bias regions
For the zener diode to maintain a constant
voltage across the load, it has to be set in
the breakdown region. A reverse bias is
applied and the amount of potential
difference must be sufficient for the reverse
current to surge. At one end of this voltage
regulation region, there is a minimum
amount of reverse current that will keep the
diode in the breakdown region. This is
denoted as πΌπ§πππ in the graph, and typically
20% of the current rating of the diode. At
the other end of the voltage regulation
region, there is a maximum allowable
reverse current to keep the diode from
being damaged. This is denoted as πΌπ§πππ₯
in the graph, and typically 80% of the
current rating of the diode.
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π½π³ = π½π
When
unloaded:
πΌπΏπππ = 0
ππππ − πΌπ π
π πππ₯ − ππ§ = 0
ππππ₯ − πΌπ π
π πππ − ππ§ = 0
πΌπ = 0.2πΌπ§ + πΌπΏπππ₯
πΉππππ =
πΌπ = 0.8πΌπ§ + πΌπΏπππ
π½πππ − π½π
π. ππ°π + π°π³πππ
πΉππππ =
π½πππ − π½π
π. ππ°π
ππππ − ππ§
ππππ₯ − ππ§
>
0.2πΌπ§ + πΌπΏπππ₯
0.8πΌπ§
πΌπ§ = πΌπ§max ππ‘π = ππππππ’π ππππππ‘π‘ππ π£πππ’π
π°ππ¦ππ± πππ
>
π°π³πππ
π½
−π½
π. π [π½ πππ − π½π ] − π. π
πππ
π
minimum π·πππ = π°ππ¦ππ± πππ
π½π
Sample Problem
Given: Output from filter varies from 9 to12 V; average output voltage at 6 V at a
current of 100 mA.
Solution: ππΏ = 6 π
πΌπΏπππ₯ = 100 ππ΄
ππππ = 9 π
ππππ₯ = 12 π
Solve for minimum permitted value of zener current which is the same as the
maximum rated value of the zener current.
ππ§ = ππΏ = 6 π
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π°ππ¦ππ± πππ
>
π°π³πππ
πΌπ§max ππ‘π >
π½
−π½
π. π [π½ πππ − π½π ] − π. π
πππ
π
100 ππ΄
9π−6π
0.8 [12 π − 6 π ] − 0.2
πΌπ§max ππ‘π = 500 ππ΄
min π·πππ = (0.500 π΄)(6 π)= 3 W
minimum π·πππ = π°ππ¦ππ± πππ
π½π
Assuming 6 V, 3W zener diode is available.
π°ππ¦ππ± πππ
= πΌπ§ = 500 ππ΄
Solve for the series resistance
πΉππππ =
π½πππ − π½π
π. ππ°π
π
π πππ =
12 π − 6 π
0.8(500 ππ΄)
π
π πππ = 15 πΊ
πΉππππ =
π½πππ − π½π
π. ππ°π + π°π³πππ
π
π πππ₯ =
Assuming that available diode is 6 V, 5W:
πΉππππ =
π½πππ − π½π
π. ππ°π
9π−6π
0.2(500 ππ΄) + 100 ππ΄
π
π πππ₯ = 15 πΊ
Use standard value: π
π = 15 πΊ
πππ‘π 5 π
πΌπ§ = πΌπ§max ππ‘π =
=
= 0.833 π΄
ππ§
6π
12 π − 6 π
π
π πππ =
0.8(0.833 π΄)
π
π πππ = 9 πΊ
πΉππππ =
π½πππ − π½π
π. ππ°π + π°π³πππ
π
π πππ₯ =
9π−6π
0.2(0.833 π΄) + 0.100 π΄
π
π πππ₯ = 11 πΊ
Use standard value: π
π = 10 πΊ ππ 11 πΊ
7.5 πΊ ≤ π
π ≤ 10 πΊ
Standard Resistance Values (5% tolerance)
1.0
1.1
1.2
1.3
1.5
1.6
1.8
2.0
2.2
2.4
2.7
3.3
3.6
3.9
4.3
4.7
5.1
5.6
6.2
6.8
7.5
8.2
3.0
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Given: The input to a voltage regulator circuit varies from 11.4 to 12.2 V. The load voltage
is to be maintained at 9.1V with maximum current of 95 mA.
a) Which of the following zener diode can be used?
Diode A: 9.1 V, 1 W; Diode B: 9.1 V, 2 W; Diode C: 9.1 V, 3 W; Diode D: 9.1 V, 5 W
List all standard resistance values that can be used.
Assignment, Quiz, Exam
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SUPPLEMENTARY TOPIC: VOLTAGE MULTIIPLIERS
As shown in the previous Units, a transformer is used to step-down the ac outlet
voltage when a low supply voltage is needed. If a higher supply voltage is
needed, then, a step- up transformer is used instead. However, for many
applications the required ratings correspond to bulky and thus relatively costly
transformer. An alternative to this is the voltage multiplier. It consists basically of
diodes and capacitors which are much smaller in size and generally are very
much less expensive.
n-stage Voltage Multiplier
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MODULE 3
Bipolar Junction Transistors
This module is divided into two parts: Fundamentals of Bipolar Junction Transistors (BJT)
and BJT Amplifiers. The first part covers comparison between the two types of BJT,
characteristic curves, operating regions; and biasing circuits. The second part covers the
description and performance measures of the three BJT amplifier configurations. Two ac
parameters, the h-parameters and r-parameter are both applied in the ac analysis of the
amplifiers. Illustrative calculations are also included.
Bipolar Junction Transistor is a solid-state counterpart of the vacuum tube triode. It is an
active device, which, just like the diode, can alter the applied signal waveform, whether
in its value or shape. BJT is a building block in many electronic circuits, devices and
equipment. This module focuses on the BJT function as an amplifier. Any device can be
properly utilized if its properties, characteristics and behavior under certain conditions are
well understood. This particular topic is wide-ranging but due to the limited number of
hours allotted, only basic concepts and principles relating to BJTs are included.
At the end of this module, you should be able to:
TLO 4: Describe the types, construction, operation, applications, and biasing
techniques of Bipolar Junction Transistors.
TLO 5: Analyze and solve problems on small signal BJT amplifiers.
LEARNING OUTCOMES:
a. Differentiate NPN and PNP Bipolar Junction Transistors
b. Compare the BJT biasing circuits
c. Solve problems on BJT biasing circuits
d. Apply ac equivalent circuit of BJT in determining performance measures
such as gains, input and output resistances.
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UNIT 1 FUNDAMENTALS OF BIPOLAR JUNCTION TRANSISTOR (BJT)
Describe the following:
a) two biasing methods of a PN junction and corresponding amount of current
that results
b) PN junction input and output characteristic curves
Watch the PowerPoint Presentation.
Sample transistors
(Source: techbook.co.in.)
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Problem Set
1.
Determine:
a) Transistor terminal voltages
b) Transistor junction voltages
2.
Draw the Load Line
Assignment, Quiz, Exam
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UNIT 2 BJT AMPLIFIERS
Describe the following:
c) two biasing methods of a PN junction and corresponding amount of current
that results
d) PN junction input and output characteristic curves
Watch the PowerPoint Presentation.
Problem Set
1.
Determine:
a) dc transistor terminal
voltages
b) βππ
′
c) π
ππ
d) π£π
π
e) π΄π π = ππΏ
π
f)
π
π′ (to the left of the
load)
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2.
Si transistor; β=50
Determine:
g) dc transistor terminal
voltages
h) βππ
′
i) π
ππ
j) π£π
π
k) π΄π π = ππΏ
π
l)
π
π′ (to the left of the load)
Assignment, Quiz, Examination
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MODULE 4
Field-Effect Transistors
In this module, the fundamentals of field-effect transistors are discussed: the two major
types, the construction, schematic symbols; biasing circuits and amplifier configurations.
Based on the discussion presented here, you should be able to have a general idea on
the basic differences between BJT and FET and the advantages and disadvantages one
has over the other; and for which purpose or application one is preferred over the other.
At the end of this module, you should be able to:
TLO 6: Describe the types, construction, operation, applications, and biasing
techniques of Field-Effect Transistors.
TLO 7: Analyze and solve problems on small signal FET amplifiers.
LEARNING OUTCOMES:
a. Differentiate the types of FETs
b. Analyze and solve problems on FET biasing circuits
c. Describe FET amplifier configurations common source, common drain and
common gate
d. Analyze and solve problems on FET amplifiers
UNIT 1 FUNDAMENTALS OF FIELD-EFFECT TRANSISTOR (FET)
Describe the operating regions and applications of bipolar junction transistors.
Watch PowerPoint presentation.
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Sample Transistors
(Source: alibaba.com)
1. Consider the voltage divider circuit for biasing an n-channel D-MOSFET.
Given the following quantities: πΌπ·ππ = 6 ππ ; ππ = −3 π; Supply ππ·π· = 18 π;
π
1 = 110 ππΊ, π
2 = 10 ππΊ, π
π· = 1.8 ππΊ, π
π = 750 πΊ . Determine πΌπ· and ππ·π
2. Consider the voltage divider circuit for biasing an n-channel E-MOSFET.
Given the following quantities: πΌπ·ππ = 3 ππ ππ‘ 10 π; πππ» = 5 π; Supply ππ·π· = 40 π;
π
1 = 22 ππΊ, π
2 = 18 ππΊ, π
π· = 3 ππΊ, π
π = 820 πΊ . Determine πΌπ· and ππ·π
Assignment, Quiz, Examination
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UNIT 2 FET AMPLIFIERS
Differentiate the biasing for n-channel and p-channel of:
a) JFET b) D-MOSFET c) E-MOSFET
Performance Measures for FET Amplifiers
Quantity
Common Source
Common Drain
Common Gate
π
ππ
∞
∞
1
ππ
π΄π£
−ππ π
π
1 + ππ π
π
ππ π
π
1 + ππ π
π
ππ π
π
π
π
∞
1
ππ
∞
π
π′
π
π
π
π β₯ π
π
π
π
ππ = π‘ππππ πππππ’ππ‘ππππ
ππ = πππ (1 −
ππΊπ
ππΊπππΉπΉ
)
πππ =
2πΌπ·ππ
|ππΊπππΉπΉ |
πππ : maximum value for ππ (when ππΊπ = 0)
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Method of Analysis of Amplifiers
DC Analysis
AC Analysis
1. Replace the ac source with a
short circuit
1. Replace the dc source with a
short circuit
2. Replace capacitors with open
circuits
2. Replace capacitors with short
circuits
3. Apply the derived biasing
circuits equations to solve for
unknown currents and voltages
3. Identify the amplifier
configuration by locating the
input and output terminals
4. Solve for unknown ac quantities
using the appropriate
equations for the amplifier
configuration
Watch the PowerPoint Presentation
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3. Given:
πΌπ· = 3 ππ΄
ππΊπππΉπΉ = −4 π
Determine
a) the ππ
b) Input resistance of the
amplifier
c) Input resistance seen
by the source (to the right
of πΆ1 )
d) Voltage gain, π΄π£
e) Output resistance
including π
πΏ
Assignment, Quiz, Examination
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MODULE 5
Operational Amplifiers
This module covers fundamentals of operational amplifier (op amp): characteristics of
ideal op amp, voltage gain and method of analysis, and applications. Only the basic
circuits are dealt with in the sample problems, and adopting ideal op amp approach.
At the end of this module, you should be able to:
TLO 8: Describe and analyze basic operational amplifiers
LEARNING OUTCOMES:
a. Describe basic operational amplifiers and their functions
b. Describe characteristics of ideal op amp
c. Solve basic op amp circuits
How does a cellphone able to perform numerous functions and store a large
amount of information within such a small device?
Operational Amplifier (Op Amp)
An operational amplifier (op amp) is a commonly used analog IC, intended to be used
with external passive devices (e.g. resistors, capacitors) basically functions as a voltage
amplifier.
An Integrated Circuit (IC) also called microelectronic circuit: miniaturized
(microscopic in size) active and passive components are fabricated together and
interconnected on a semiconductor base (called a substrate). Substrate is also known as
wafer: thin slice of Si; circular and ranges from 4 to 18 inches in diameter and
approximately 1 mm thick. Each wafer is cut into dies (or microchips or simply chips)
which are packaged individually
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An Analog or Linear Op Amp processes continuous signal: value of voltage varies
with time; voltage values within a range (theoretically infinity). This type of Op Amp
may be found in AF (audio frequency) amplifier, RF (radio frequency) amplifier;
power amplifier, operational amplifier, multiplier, voltage regulator, microwave
amplifier, and RF receiver.
Characteristics of Ideal Op-amp:
1.
2.
3.
4.
5.
Infinite open- loop gain
Infinite input impedance, and so zero input current (π
ππ = ∞; input current is zero)
Zero input offset voltage (πππ’π‘ = 0 when πππ = 0; πππ+ = πππ− )
Infinite bandwidth (constant gain at all input frequencies: from 0 π‘π ∞ π»π§)
Zero output impedance (π
ππ’π‘ = 0)
These features simplify the analysis of circuits with op- amps.
Basic Functions:
1. An op-amp is usually used in combination with another amplifier to increase the
level of very weak signals.
2. Can be used to filter out noises and other unwanted signals
3. Can be used in computers to perform logical and arithmetical operations
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Industrial applications in automotive, instrumentation, aerospace, etc.
(Source: ti.com)
Basic Circuits
1. Non-inverting Amplifier
πΊπππ =
πππ’π‘
π
2
=1+
πππ
π
1
Generally, π
2 is chosen to
be greater than π
1
Used when a high impedance is required
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2. Inverting Amplifier
πΊπππ =
πππ’π‘
π
2
=−
πππ
π
1
Used when a phase inversion is desired
3. Voltage Follower
πππ’π‘ = πππ
Essentially a non-inverting amplifier circuit where π
2 = 0 (short-circuit) and π
1 = ∞
(open-circuit). Used to provide isolation between two stages (acts as a buffer), or
for impedance matching.
4. Differential Amplifier
If π
1 = π
3 , π
2 = π
4:
πΊπππ =
πππ’π‘
π
2
=−
πππ
π
1
πππ = πππ2 − πππ1
Used for amplifying the difference between two input signals (e.g. in volume
control circuits, in stabilizing amplifiers, in automatic gain control circuits)
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5. Summing Amplifier
Based on the non-inverting amplifier:
πΊπππ =
πππ’π‘ = − (πππ1
πππ’π‘
π
2
=−
πππ
π
1
π
π
π
π
π
π
+ πππ2 +πππ3 )
π
1
π
2
π
3
πππ’π‘ = −π
π (πππ1
1
1
1
+ πππ2 +πππ3 )
π
1
π
2
π
3
Example 1
Determine πππ’π‘
Solution:
Input is at the positive or non-inverting input terminal: πππ = πππ+
Output voltage,
πππ’π‘ = πππ (1 +
π
2
)
π
1
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In the circuit, π
2 = 10 ππΊ and π
2 = 5 ππΊ;
By Voltage Division:
18 ππΊ
πππ = 15 π (
)=9π
18 ππΊ + 12 ππΊ
πππ’π‘ = 9 π (1 +
10 ππΊ
)
5 ππΊ
π½πππ = ππ π½
Example 2
Determine πππ’π‘
Solution:
For two inputs at the inverting input:
πππ’π‘ = −π
π (πππ1
1
1
+ πππ2 )
π
1
π
2
In the circuit: π
π = 5 ππΊ ; πππ1 = 4 π , π
1 = 2 ππΊ ; πππ2 = 6 π π
2 = 1 ππΊ
1
1
πππ’π‘ = −(5 ππΊ) [(4 π) (
) + (6 π) (
)]
2 ππΊ
1 ππΊ
π½πππ = −ππ π½
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Example 3
Determine πΌ and Vout
Solution:
The given circuit is similar to a
differential op amp; but,
π
1 ≠ π
3 , π
2 ≠ π
4
So, the given formula cannot be
directly applied.
Since ideally, πΌππ = 0, the 12 ππΊ and 24 ππΊ resistors have the same current
To solve for πΌ:
By KVL: 8 π − πΌ(12 ππΊ) − (πππ− ) = 0
Ideally, πππ− = πππ+
By voltage division:
10 ππΊ
πππ+ = 5 π (
)=2π
10 ππΊ + 15 ππΊ
πππ− = 2π
So,
πΌ=
To solve for πππ’π‘ :
8π−2π
12 ππΊ
π° = π. π ππ¨
By KVL
8 π − πΌ(12 ππΊ) − πΌ(24 ππΊ) − πππ’π‘ = 0
πππ’π‘ = 8 π − (0.5 ππ΄)(12 ππΊ + 24 ππΊ)
π½πππ = −ππ π½
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1.
Determine πΌ and πππ’π‘
2.
a) What is the required
value of π
for a
voltage gain of 10
b) Determine πππ’π‘
Assignment, Quiz, Exam
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