Unit 1 Physics Foundation, Circuit Elements, KVL & KC Unit 2

Unit 1

Physics Foundation, Circuit

Elements, KVL & KCL

Unit 2

Analysis Techniques

Unit 3

Op Amps & Two-Ports

ROSE-HULMAN INSTITUTE OF TECHNOLOGY ECE 203 DC Circuits Winter 09 - 10

Page 1 of 3

Course Information and Syllabus Description: ECE 203 DC Circuits 3R‐3L‐4C F, W, S Prerequisite: MA 111 and PH 112 Definition of voltage, current, energy and power. Ohm’s Law. Non‐ideal voltage and current sources. Measurement of voltage, current and resistance. Kirchhoff’s Laws. Circuit simplification by series and parallel reduction. Thevenin, Norton and Maximum Power Theorems. Superposition Theorem. Mesh and Nodal Analysis. Two‐Port Circuits. Operational Amplifiers. Integral laboratory. You must get a C or better in ECE 203 to take ECE 204 – AC Circuits Instructor: Dr. Carlotta A. Berry Moench, D‐211 812‐877‐8657 berry123@rose‐hulman.edu Textbook: J.W. Nilsson & S.A. Riedel, Electric Circuits, 8th edition, Prentice Hall, 2008. What is expected of You: First and foremost, professional work is the norm in this course. All of your written work and your conduct in class are to be at the level of one who is studying a profession—the profession of engineering. This means a number of things:

1. Your work is neatly done in a professional manner, using formats specified.

2. Your work is honestly done. You are encouraged to discuss course material with classmates to help each other understand and assimilate the concepts. Nevertheless, distinguish between helping someone understand concepts and providing them with specific answers. You are expected to work individually on homework without reference to others’ work.

3. Your work is done on time. As a rule of thumb, expect to put in eight hours per week outside of class doing homework, reading the text, and studying.

4. You are attentive and engaged in the lecture (i.e. not sleeping, reading the newspaper, surfing the web, doing homework for other courses, disturbing others with electronics).

Grading:

Grades will be assigned at the end of the quarter based on the grade weights and grading scale shown below: Midterm Test I 12% A 90 – 100 Midterm Test II 12% B+ 85 – 89 Midterm Test III 12% B 80 ‐ 84 Final Exam 24% C+ 75 ‐ 79 Homework 10% C 70 ‐ 74 Laboratory assignments 15% D+ 65 ‐ 69 Lab Practical Tests 10% D 60 ‐ 64 Quizzes 5% F Below 60

ECE 203 DC Circuits Winter 09 - 10

CAB Page 2 of 3

Independent of point totals:

• You must satisfactorily complete each of the eight lab projects in order to receive a passing grade in the course

Examinations:

In this course, examinations make up 70% of the grade and warrant careful preparation. Examination questions will be based on the lecture material, textbook, homework, and laboratory work. The three midterm tests will be fifty minutes in duration during the regular class meeting time. Midterms and final examinations will be closed book and closed notes. You will be provided with an excerpt from the NCEES FE Reference handbook that includes some formulas. It is your responsibility to memorize or derive any formula missing from this document.

Quizzes:

There will be weekly quizzes that involve solving short problems or answering questions on required reading. The purposes for these quizzes are:

• to give me feedback on the current level of understanding of the class

• to give you feedback on your current level of understanding

• to give you practice on problems similar to the exam format

• to encourage collaborative learning in the classroom

• to also take attendance

Homework:

The homework is intended to help you to understand the concepts presented in the course, and to provide you with practice in problem solving.

• Problem sets are due each Thursday in class before the bell rings at the beginning of class. Answers and solutions will be distributed using ANGEL.

• Homework turned in after the bell rings is late and will incur a 20% penalty.

• Homework turned in after 5 pm on the due date will not be accepted.

• Arrange to turn in homework early if you will be away for job interviews, athletic events, etc.

• The required format is described in the document on Angel and in the ECE Department’s guidelines and standards for writing assignments. It is your responsibility to make your methods and results clear to the grader. http://ece‐1.rose‐hulman.edu/ece/images/stories/files/ecewritingstandards.pdf

Laboratory Supplies:

Each student team must purchase an ECE203 kit with breadboard from the parts room before the first lab. Each individual student is required to purchase a laboratory notebook (10” x7‐7/8”, 80 sheets, 5x5 Quad Ruled, #26‐251, available at the bookstore). The format for notebook entries can be found under the ECE Department’s guidelines and standards for writing assignments. Additional specifications are provided in the ECE203 – DC Circuits Laboratory Manual

ECE 203 DC Circuits Winter 09 - 10

CAB Page 3 of 3

Prelabs:

Prelab exercises are due the day before lab at the beginning of class before the bell rings. Each student should do the pre‐lab in their lab notebook and make a photocopy to turn in. The solutions to the prelab may be presented at the start of the lab period. Any student that has not completed the prelab must do it at the beginning of lab for zero credit. This team must still finish the laboratory project within the allotted time.

Laboratory Notebooks:

Laboratory notebooks will be collected at the conclusion of each laboratory period. The laboratory notebook will be graded and both members of the team share the notebook grade. Each team member must alternate submission of the lab notebook as well as circuit building on a weekly basis.

Re‐grades:

All requests for re‐grades must be made in writing within one week of the return of the assignment or exam. The student should not make any marks on the document and must attach a memorandum that details a technical justification for the reason for the submission. It should be noted that based upon the request, the grade may increase, decrease or remain the same.

Attendance:

Regardless of whether formal attendance is taken, attendance at each class is expected. As a rule of thumb you should consider yourself seriously behind if you miss more than four classes in a four credit‐hour course. According to our Academic Rules and Procedures, “A student whose total absences in a course, excused or unexcused, exceed two per credit is liable to fail the course.” Eight absences in this course are grounds for failure. Missing an attendance check due to lateness may be counted as an absence.

If you miss a lab with an excused absence you need to make it up within 1 week without penalty. If you miss a lab without an excused absence, you need to make it up within 1 week and you will receive a grade of zero. If you come to lab more than 15 minutes late you need to complete the lab on your own.

Missed exams will not be made up. The final exam grade will be used to replace a missing test grade in the case of excused absences. An excused absence from an examination normally requires advance approval or formal documentation of an emergency. An examination that is missed for an unexcused reason will be given a grade of zero. Students are not excused from scheduled exams for intramural athletics or fraternity events.

Calculators & Computers:

You will need a calculator that can perform arithmetic with complex numbers (TI‐83 plus or better). You are encouraged to practice doing the homework with the same calculator you will use on the exam. It is important to learn to do simultaneous equations (or matrix) calculations with your calculator to be successful in this course. Maple can be used on the homework problems, but not in the exams.

Academic accommodation:

Those students with documented special needs may request extra time on timed tests. Students need to contact me at least 2 business days prior to each exam to make the necessary arrangements.

ROSE‐HULMAN INSTITUTE OF TECHNOLOGY WEDNESDAY LAB

ECE 203 DC Circuits Winter 2009 ‐ 10

CAB Rev. 10/15/09

Course Calendar

Class Day Date Topic Reading Due1‐1 M 11/30 Syllabus, Introduction, Circuit Analysis Overview 1.1‐ 1.3 1‐2 T 12/1 Voltage, current, power, energy, passive sign convention 1.4 – 1.6 Prelab 11‐L W 12/2 Lab 1. Introduction to Laboratory Techniques 1‐3 R 12/3 Sources, Resistors, Ohm’s Law, Power Calculations 2.1 – 2.3 HW 12‐1 M 12/7 Kirchhoff’s Laws 2.4 – 2.5 Quiz2‐2 T 12/8 Kirchhoff’s Laws 2.4 – 2.5 Prelab 22‐L W 12/9 Lab 2. Ohm’s Law, KVL, and KCL 2‐3 R 12/10 Resistors in parallel and series 3.1 – 3.2 HW 23‐1 M 12/14 Voltage and Current Divider 3.3 – 3.4 Quiz3‐2 T 12/15 Wheatstone Bridge and Delta‐Wye Equivalents 3.5 – 3.7 Prelab 33‐L W 12/16 Lab 3. Voltage and Current Divider HW 33‐3 R 12/17 Midterm Test 1 (up through Chapter 3)

WINTER BREAK (12/19/09 – 01/03/10)

4‐1 M 1/4 Node‐voltage method 4.1 – 4.2 4‐2 T 1/5 Node‐voltage method (special cases) 4.3 – 4.4 Prelab 44‐L W 1/6 Lab 4. Node‐voltage method 4‐3 R 1/7 Mesh‐current method 4.5 HW 45‐1 M 1/11 Mesh‐current method (special cases) 4.6 – 4.7 Quiz5‐2 T 1/12 Source Transformations 4.9 Prelab 55‐L W 1/13 Lab 5. Mesh‐current method 5‐3 R 1/14 Thevenin and Norton Equivalents 4.10 – 4.11 HW 56‐1 M 1/18 Thevenin and Norton Equivalents 4.10 – 4.11 Quiz6‐2 T 1/19 Maximum Power Transfer 4.12 6‐L W 1/20 Lab 6. Lab Practical Test I 6‐3 R 1/21 Superposition 4.13 HW 67‐1 M 1/25 Introduction to Operational Amplifiers 5.1 – 5.2 Quiz7‐2 T 1/26 Midterm Test 2 (up through Chapter 4 section 12) 7‐L W 1/27 Lab 7. Source Transformations, Thevenin and Norton Equivalents Prelab 77‐3 R 1/28 Inverting and summing op amps 5.3 – 5.4 HW 78‐1 M 2/1 Noninverting, difference, cascaded op amps 5.5 – 5.6 Quiz8‐2 T 2/2 More realistic op amps 5.7 Prelab 88‐L W 2/3 Lab 8. Operational Amplifiers 8‐3 R 2/4 Operational amplifier applications HW 89‐1 M 2/8 Terminal equations, two‐port parameters 18.1 – 18.2 Quiz9‐2 T 2/9 Midterm Test 3 (up through Chapter 5) Prelab 99‐L W 2/10 Lab 9. Inverting and Noninverting Op Amps 9‐3 R 2/11 Two – port parameters 18.2 HW 910‐1 M 2/15 Terminated two‐port circuits 18.3 Quiz10‐2 T 2/16 Cascaded two‐port networks 18.4 10‐L W 2/17 Lab 10. Lab Practical Test II 10‐3 R 2/18 Final Review, Course Evaluations HW 10

ROSE‐HULMAN INSTITUTE OF TECHNOLOGY FRIDAY LAB


CAB Rev. 10/18/09

Course Calendar

Class Day Date Topic Reading Due1‐1 M 11/30 Syllabus, Introduction, Circuit Analysis Overview 1.1‐ 1.3 1‐2 T 12/1 Voltage, current, power, energy, passive sign convention 1.4 – 1.6 1‐3 R 12/3 Sources, Resistors, Ohm’s Law, Power Calculations 2.1 – 2.3 HW 1, Prelab 11‐L F 12/4 Lab 1. Introduction to Laboratory Techniques2‐1 M 12/7 Kirchhoff’s Voltage and Current Laws 2.4 – 2.5 Quiz2‐2 T 12/8 Kirchhoff’s Voltage and Current Laws 2.4 – 2.5 2‐3 R 12/10 Resistors in parallel and series 3.1 – 3.2 HW 2, Prelab 22‐L F 12/11 Lab 2. Ohm’s Law, KVL, and KCL3‐1 M 12/14 Voltage and Current Divider 3.3 – 3.4 Quiz3‐2 T 12/15 Wheatstone Bridge and Delta‐Wye Equivalents 3.5 – 3.7 12/16 HW 3

3‐3 R 12/17 Midterm Test 1 (up through Chapter 3)3‐L F 12/18 Lab 3. Voltage and Current Divider Prelab 3

WINTER BREAK (12/19/09 – 01/03/10)

4‐1 M 1/4 Node‐voltage method 4.1 – 4.2 4‐2 T 1/5 Node‐voltage method (special cases) 4.3 – 4.4 4‐3 R 1/7 Mesh‐current method 4.5 HW 4, Prelab 44‐L F 1/8 Lab 4. Node‐voltage method5‐1 M 1/11 Mesh‐current method (special cases) 4.6 – 4.7 Quiz5‐2 T 1/12 Source Transformations 4.9 5‐3 R 1/14 Thevenin and Norton Equivalents 4.10 – 4.11 HW 5, Prelab 55‐L F 1/15 Lab 5. Mesh‐current method6‐1 M 1/18 Thevenin and Norton Equivalents 4.10 – 4.11 Quiz6‐2 T 1/19 Maximum Power Transfer 4.12 6‐3 R 1/21 Superposition 4.13 HW 66‐L F 1/22 Lab 6. Lab Practical Test I7‐1 M 1/25 Introduction to Operational Amplifiers 5.1 – 5.2 Quiz7‐2 T 1/26 Midterm Test 2 (up through Chapter 4)7‐3 R 1/28 Inverting and summing op amps 5.3 – 5.4 HW 7, Prelab 77‐L F 1/29 Lab 7. Source Transformations, Thevenin and Norton Equivalents8‐1 M 2/1 Noninverting, difference, cascaded op amps 5.5 – 5.6 Quiz8‐2 T 2/2 More realistic op amps 5.7 8‐3 R 2/4 Operational amplifier applications HW 8, Prelab 88‐L W 2/5 Lab 8. Operational Amplifiers9‐1 M 2/8 Terminal equations, two‐port parameters 18.1 – 18.2 Quiz9‐2 T 2/9 Midterm Test 3 (up through Chapter 5)9‐3 R 2/11 Two ‐ port parameters 18.2 HW 9, Prelab 99‐L W 2/12 Lab 9. Inverting and Noninverting Op Amps10‐1 M 2/15 Terminated two‐port circuits 18.3 Quiz10‐2 T 2/16 Cascaded two‐port networks 18.4 10‐3 R 2/18 Final Review, Course Evaluations HW 1010‐L F 2/19 Lab 10. Lab Practical Test II

ROSE‐HULMAN INSTITUTE OF TECHNOLOGY


CAB Rev. 10/15/09

Assigned Homework Problems

Homework Problems Due

HW1 Chapter 1 1.1, 1.9, 1.15, 1.30 P1 The following figure shows the power consumption of a certain household in one day. Calculate: a. The total energy consumed in kWh b. The average power per hour

12/3/09

HW2 Chapter 2 2. 1, 2.11, 2.18, 2.29 P2 Find R for the following circuit

12/10/09

HW3 Chapter 3 3.5, 3.10, 3.14, 3.52 P3 For the following Wheatstone bridge circuit, a. R is a 10 kΩ variable resistor, what is the range of Vab? b. For what value of R is the bridge balanced? c. If a 5 kΩ resistor is placed between terminals a and b, and R = 7.5 kΩ,

what is the current through the 5 kΩ resistor?

12/16/09

2

HW4 Section 4.1 – 4.4 4.6, 4.11, 4.17, 4.26 P4 Use the node‐voltage method to determine the gain, vo/vs for the following transistor amplifier circuit.

1/7/10

HW5 Section 4.5 – 4.9 4.32, 4.37, 4.47, 4.59 P5 The following circuit models a common‐emitter transistor amplifier. Find ix using source transformation.

1/14/10

HW6 Section 4.10 – 4.11 4.64, 4.66, 4.69, 4.77 P6 The Thevenin equivalent terminals a‐b of the linear network shown in the following figure can be determined by measurement. When a 10 kΩ resistor is connected to terminals a‐b, the voltage Vab is measured as 6 V. When a 30 kΩ resistor is connected to the terminals, Vab is measured as 12 V. Determine: a. The Thevenin equivalent at terminals a‐b b. Vab when a 20 kΩ resistor is connected across terminals a‐b

1/21/10

3

HW7 Section 4.12 – 4.13 4.81, 4.85, 4.90, 4.97 P7 For the following circuit, determine the value of R such that the maximum power delivered to the load is 5 mW.

1/28/10

HW8 Chapter 5 5.3, 5.11, 5.15, 5.28 P8 In the following op amp circuit, determine the current io.

2/4/10

HW9 Two‐Port Parameters 18.4, 18.8, 18.9, 18.11, 18.19

2/10/10

HW10 Terminated and Cascaded Two‐Port Networks 18.36, 18.37, 18.38. 18.39, P10 For the following two port circuit, if the h parameters are h11 = 16 Ω, h12 = 3 , h21 = ‐2, h22 = 0.01 S, find the power delivered to the 25Ω resistor.

2/18/10

+ 1V

_

+ 2V

_

1I

+ _ 10 V 25 Ω

Rate with respect to time

ELECTRICAL SYSTEMS

have components with 2 terminals that can be described in terms of current and/or voltage which are

Introduction

Have capacity to do work

ENERGY, w (J) Per unit charge

VOLTAGE, v (V)

CHARGE, q (C)

Have positive and negative

CURRENT, i (A)Separation yields

Flow yields

POWER, p (W)

LAW OF CONSERVATION OF

ENERGY

Can be neither created nor destroyed only transferred

IDEAL CIRCUIT ELEMENTS

PASSIVE SIGN CONVENTION

drawn such that current flows from the positive terminal (charge) to negative terminal (charge) obeys the

VOLTAGE DROP

VOLTAGE RISE

POWER DELIVERED

POWER ABSORBED

P = VI < 0

P = VI > 0

Flows from the (-) ve terminal to a (+)ve terminal

Flows from the (+) ve terminal to a (-)ve terminal

ECE203 – DC Circuits - C.A. BERRY 11/15/09

establishes voltage and current in a circuit without relying on voltage and current elsewhere in the circuit is an

establishes voltage and current in a circuit which depends on voltage and current elsewhere in the circuit is a

produces a constant current in a circuit

ACTIVE CIRCUIT ELEMENTS

PASSIVE CIRCUIT ELEMENTS

Model a device capable of generating electric energy

ELECTRICSOURCES

Model a device that cannot generate electric energy

RESISTORS

CAPACITORS

INDUCTORS

i = C dv/dt

v = L di/dt

V = iRImpede the flow of current

Store magnetic flux and induces voltage related to current

store electric charge and produce a current based upon electric field

DEPENDENT SOURCE

INDEPENDENT SOURCE

Convert non-electric energy to electric energy

IDEAL VOLTAGE SOURCE

VOLTAGE-CONTROLLED CURRENT SOURCE

IDEAL CURRENT SOURCE

CURENT-CONTROLLED VOLTAGE SOURCE

VOLTAGE-CONTROLLED VOLTAGE SOURCE

CURRENT-CONTROLLED CURRENT SOURCE

produces a constant voltage in a circuit

Produces a current based upon current elsewhere in the circuit

Produces a voltage based upon current elsewhere in the circuit

Produces a current based upon a voltage elsewhere in the circuit

Produces a voltage based upon voltage elsewhere in the circuit

IDEAL CIRCUIT ELEMENTS


states that current into a node equals current out of a node is

states that voltage rises equal voltage rises around a loop is

with sources that produce current or voltagethat is constant with time are

is usedto apply

is usedto apply

can be analyzed using can be simplified by

DC CIRCUITS

ELECTRICAL SYSTEMS

SUPERPOSITIONOHM’S LAWLAW OF

CONSERVATION OF ENERGY

COMBINING RESISTORS

SOURCE TRANSFORMATIONS ONE PORT CIRCUITS

THEVENIN/NORTON THEOREMS

MAXIMUM POWER TRANSFER

WYE DELTA TRANSFORMATIONSSERIESPARALLEL

in

in

using

KIRCHHOFF’S CURRENT LAW

VOLTAGE DIVIDERMESH-CURRENT METHODCURRENT DIVIDERNODE-VOLTAGE

METHOD

KIRCHHOFF’S VOLTAGE LAW


Operational Amplifiers

Infinite input impedance condition

that behave like voltage controlled voltage sources and can perform math operations are

Active Circuit Elements

Ideal op-amps have 2 constraints or assumptions

Virtual short condition

KCL

Op-amps are analyzed using the 2 assumptions and

Inverting SummingNon-inverting Difference

Can be used to find the voltage characteristics for several types of op-amps


More realistic model

Driven at one port and loaded at the other port are

PORTS

THEVENIN/NORTON EQUIVALENT

Can be described by the voltage and current characteristics at their

DC CIRCUITS

ONE PORT

TERMINATED

CASCADED

Signals can be fed into or extracted from either

TWO PORT CIRCUITS

Have separate ports for input and output described by

PARAMETERS (z, b, a, h, y, g)

Has an output port described by the

Two port circuits with two port circuits as inputs our outputs are

Described by the


Voltage, CurrentPower, Energy

Passive sign conventionSources

Resistors, Ohm’s LawKirchhoff’s Voltage LawKirchhoff’s Current Law

Resistors in series and parallelVoltage and Current Divider

Wheatstone BridgeDelta-Wye Equivalents

ECE203 DC Circuits Winter 2009/2010

C.A. Berry Lec1‐1.docx Page 1 of 2

Lecture 1‐1 Circuit Analysis Overview

Reading: 1.1 – 1.3

Objectives: • Be able to able to briefly and clearly explain static electricity and electrical circuit • Be able to use SI units and the standard prefixes for power of 10

**************************************An electric circuit is a mathematic model that approximates the behavior of an actual electric system.

A commonly used mathematical model for electric systems is a circuit model. The elements that comprise the circuit model are called ideal circuit components.

Circuit analysis is the tools applied to an electric circuit using mathematical techniques to predict the behavior of the circuit model and its ideal circuit components. The physical prototype is an actual electric system, constructed from actual electric components. In‐Class Activity: Watch the video from Jimmy Kimmel Live. What is the source? What is the resistance? What is the switch? Why do you think that Jimmy had to jump in the air in order to shock the people?



DC Circuit Water Analogy It is possible to understand a simple DC circuit by relating it to a water circuit.

Pressure, P drives water around the closed loop pipe at a certain volume flowrate, F

Voltage, V drives the charge around the closed circuit at a certain electric current, I

A pump takes water at a low pressure and does work on it to eject it at a higher pressure

A battery takes in charge at a lower voltage, does work on it and ejects it at a higher voltage

Constriction of the pipe represents resistance to water flow and causes a pressure drop

Resistance to electric current causes a voltage drop

The reservoir supplies water to the pipe and is a pressure reference

The ground supplies charge to the circuit and holds a reference voltage

A large pipe has very little resistance to flow A wire has very little resistance to charge flow http://hyperphysics.phy‐astr.gsu.edu/hbase/electric/watcir.html

Review the International System of Units and Standardized prefixes for powers of 10.



Lecture 1‐2 Voltage, current, power, energy, passive sign convention

Reading: 1.4 – 1.6 Objectives:

• Be able to briefly and clearly explain voltage, current, power, energy • Be able to write formulas for voltage, current, power • Be able to identify whether an element is delivering or absorbing power • Be able to explain the difference between power and energy • Be able to apply the law of conservation of energy to confirm that power balances for a

network • Be able to use the passive sign convention to calculate the power for an ideal basic

circuit element given its voltage and current

**************************************The separation of electric charge creates an electric force (voltage). Whenever positive and negative charges are separated, energy is expended. Voltage is the energy per unit charge created by this separation.

v = dqdw

v is voltage in volts (V), w is energy in Joules (J), q is charge in Coulombs (C) The motion of charge creates an electric fluid (current). The rate of charge flow is known as electric current.

i = dtdq

i is current in amperes (A), q is charge in Coulombs (C), t is time in seconds (s)

In‐Class Activity The current flowing into the positive terminal of an element is given by the following waveform. If the element is initially charged to 2 C, what is the total charge transferred to the element in 5 seconds?

-2-101234

0 1 2 3 4 5

i(t), A

time, s



An ideal basic circuit element has 3 attributes: (1) it has only 2 terminals, (2) it is described mathematically in terms of current and/or voltage, (3) it cannot be subdivided into other elements. A voltage drop is when current flows through an element from the positive to negative terminal on an element, otherwise it is a voltage rise. The passive sign convention states that when the direction of current flow through an element is in the direction of the voltage drop across the element, use a positive sign that relates the voltage to the current, otherwise use a negative sign. Concept Question: The charge flowing through a light bulb is shown in the figure on the right. Which of the following figures shows the current flowing through the light bulb? a) I b) II c) III d) none of the above

************************************************************************ The useful output of electrically‐based systems often is non‐electrical. This output is expressed in terms of power or energy. Power is the time rate of expending or absorbing energy.

p = dtdw

p is the power in Watts (W), w is the energy in Joules (J), t is the time in seconds (s)

The power associated with the flow or charge follows directly from the definition of voltage and current

p = dtdw

= dqdw

.dtdq

= vi

p is the power in Watts (W), v is the voltage in volts (V), i is the current in amperes (A)

i, A i, A i, A I II III

+ v -i

q, C

time, s



If the current flowing through an element is in the direction of the voltage drop across the element, p = + vi, otherwise it is p = ‐vi. If the numerical value of power is positive (p > 0), the element is absorbing power, otherwise the element is delivering power. The law of conservation of energy states that energy can neither be created nor destroyed only transferred. Therefore, the total power delivered to an electric circuit must be equal to the total power absorbed. In‐Class Activity: Figure 1 and 2 shows the current through and the voltage across a device. Sketch the power and find the total energy absorbed by the device for the period of 0<t<8 ms.

In‐Class Activity: In the following circuit, a) identify which elements are drawn to obey the passive sign convention, b) using the numerical values on the table, identify which elements are absorbing power and which are delivering power, c) confirm that the law of conservation of energy is satisfied. element Voltage (V) Current (A) a 8 ‐4 b 15 15 c ‐7 3.5 d 10 19 e 5 ‐18.5 f 2 ‐0.5

Figure 1

-15

-10

-5

0

5

10

15

0 1 2 3 4 5 6 7 8

time, ms

i, m

A

Figure 2

05

1015202530354045

0 1 2 3 4 5 6 7 8

time, ms

v, V

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5 6 7 8

time, ms

p, W

d f e

a

c b

+ vd _

- ve +

+ vf _

+ va -

+ vb - + vc -

ia

ib

id ie

ic

if



Lecture 1‐3 Sources, resistors, Ohm’s law, Power Calculations

Reading: 2.1 – 2.3

Objectives:

• Be able to explain Ohm’s law, power across a resistor, ideal voltage and current sources, active and passive circuit elements

• Be able to identify resistors in a network • Be able to solve for voltage, current and power in a resistive network • Be able to apply the law of conservation of energy to a resistive circuit

**************************************An electric source is a device that is capable of converting nonelectric energy to electric energy and vice versa. These sources can either absorb power or deliver power, generally by maintaining either voltage or current. Independent sources establish a voltage or current in a circuit without relying on voltage or currents elsewhere in the circuit. Dependent sources or controlled sources establish a voltage or current in a circuit whose value depends on the value of a voltage or current elsewhere in the circuit. Dependent sources are used to model transistors and operational amplifiers.

Element Description Graphic Independent Sources

Establishes a voltage or current in a circuit without relying on voltages and currents elsewhere in the circuit

ideal current source This circuit symbol is a circle and must include an arrow for the direction of the current and a value. is

ideal voltage source This circuit symbol is a circle and must include a polarity for the reference voltage and a value

vs Dependent or Controlled Sources

Establishes a voltage or current in a circuit which depends on a voltage or current elsewhere in the circuit.

current controlled current source (CCCS)

This circuit symbol is a diamond and it must include an arrow for the direction of the supplied current, a formula to calculate the supplied current from the controlling current, ix. is = βix (β is unit less)

βix

voltage controlled voltage source (VCVS)

This circuit symbol is a diamond and it must include a reference polarity for the supplied voltage, a formula to calculate the supplied voltage from the controlling voltage, vx. vs = μvx (μ is unit less)

μvx

+ _

+ _



voltage controlled current source (VCCS)

This circuit symbol is a diamond and it must include an arrow for the direction of the supplied current, a formula to calculate the supplied current from the controlling voltage, vx. is = αvx (α’s units are A/V)

αvx

current controlled voltage source (CCVS)

This circuit symbol is a diamond and it must include a reference polarity for the supplied voltage, a formula to calculate the supplied voltage from the controlling current, ix. vs = ρix (ρ’s units are V/A)

ρix

An active circuit element is one that models a device capable of generating electric energy A passive circuit element models physical devices that cannot generate electric energy. Resistors, Inductors, and Capacitors are examples of passive circuit elements. In‐Class Activity What value of vs is required in order for the following interconnection to be valid?

In‐Class Activity What value of α is required in order for the interconnection to be valid?

Resistance, R is the capacity of materials to impede the flow of current or the flow of electric charge. The circuit element used to model this behavior is the resistor. Many useful devices take advantage of resistance heating including stoves, toasters, irons, and space heaters.

+ _



Ohm’s law describes the voltage across a resistor, v = iR, where v is the voltage in volts, i is the current in amperes, and R is the resistance in ohms (Ω)

The reciprocal of resistance is conductance, G measured in Siemens (S) or mhos.

G = 1/R (S)

The power dissipated in a resistor is given by the following formulas

p = i2R = v2/R = v2G = i2/G (W) FE Preview: A 5 Ω resistor is placed in series with a varying current. Most nearly how much energy is dissipated by the resistor over the 4 s time interval shown?

a) 80 J b) 180 J c) 200 J d) 320 J

In‐Class Activity For the following circuit, confirm that it satisfies the law of conservation of energy.

In‐Class Activity For the following circuit, confirm that it satisfies the law of conservation of energy.

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 1 2 3 4 5 6

time, seconds

i(t),

Ampe

res



Lecture 2‐1 Kirchhoff’s Voltage and Current Laws

Reading: 2.4 – 2.5

Objectives:

• Be able to briefly and clearly explain Kirchhoff’s voltage and current laws • Be able to solve circuits for voltage, current, and power using KCL and KVL

**************************************

A node is a point where two or more circuit elements meet. Kirchhoff’s current law (KCL ) states that the algebraic sum of all the currents at any node in a circuit equals zero. (i.e. current in = current out)

i1 – i2 + i3 + i4 – i5 = 0

i1 + i3 + i4 = i2 + i5

A closed loop or path in a circuit is any closed path in a circuit through selected basic circuit elements and return to the original node with passing any intermediate node more than once. Kirchhoff’s voltage law states that the algebraic sum of the voltages around any closed path in a circuit equals zero. (i.e. voltage drops = voltage rises)

v1 + v4 = v2 + v3 + v5

‐v1 + v2+v3‐v4 +v5 = 0 Write 3 KCL and 3 KVL equations for the following circuit.



In‐Class Activity 1: Use KCL, to find all of the voltages and current in the following circuit. Finally, confirm that the circuit satisfies the law of conservation of energy.

element V, V I, A Pdel, W Pabs, W

5A 10A 2 Ω 6 Ω TOTAL



In‐Class Activity 2: Use KVL, to find all of the voltages and current in the following circuit. Finally, confirm that the circuit satisfies the law of conservation of energy.

element V, V I, A Pdel, W Pabs, W 12 V 6 V 2 Ω 4 Ω

12 Ω TOTAL



In‐Class Activity 3: Use KVL to find v1 and v2.



Lecture 2‐2 Kirchhoff’s Voltage and Current Laws

Reading: 2.4 – 2.5

Objectives:

• Be able to briefly and clearly explain Kirchhoff’s voltage and current laws

• Be able to solve circuits for voltage, current, and power using KCL and KVL

**************************************Recall that

• KVL states that ∑ = 0 (i.e. voltage rises = voltage drops)

• KCL states that ∑ = 0 (i.e. current in = current out)

In‐Class Activity 1: For the following circuit, find io and vo. Confirm that the circuit obeys the law of conservation of energy.


6 A io/4 8 Ω 2 Ω

TOTAL



In‐Class Activity 2: For the following circuit, find io.



In‐Class Activity 3: For the following circuit, find vx and vo. Confirm that the circuit obeys the law of conservation of energy.


35 V 2 vx 10 Ω 5 Ω

TOTAL



In‐Class Activity 4: For the following circuit, use KVL and KCL to find

a) i1, i2, i3 b) v1, v2, v3



Lecture 2‐3 Resistors in parallel and series

Reading: 3.1 – 3.2

Objectives:

• Be able to recognize resistors connected in series and in parallel and use the rules of combining resistors to yield equivalent resistance

**************************************

When two elements connect at a single node, they are said to be in series. Elements in series have the same current.

The equivalent resistance for resistors in series is the sum of the individual resistors. Therefore the equivalent resistance is always larger than the largest individual resistor.

Req = R1 + R2 + … + Rn When two elements connect at a single node pair, they are said to be in parallel. Elements in parallel have the same voltage.

The equivalent resistance is reciprocal of the sum of the individual conductances. Therefore the equivalent resistance is always smaller than the smallest individual resistor.

Req = 1

n21 R1

R1

R1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛+++ L

Geq = G1 + G2 + … + Gn

The special case for 2 parallel resistors is the product over the sum,

Req = (R1R2)/(R1 + R2)



Concept Question: In the following circuit, a 75‐W and 50‐W bulb are placed in parallel across a voltage source, which of the following statements is true a) the current through the 50‐W bulb is larger b) the current through the 75‐W bulb is larger c) the current through both bulbs is the same d) none of the above Concept Question: In the following figure, as more identical resistors are added to the parallel circuit between points P and Q, the total resistance between points P and Q a) increases b) decreases c) remains the same d) none of the above In‐Class Activity 1: For the following circuit,

a) Find Rab b) If a 10V source is placed across terminals a and b, what the is the power delivered by the

source?

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In‐Class Activity 3: For the following circuit,

a) Find Rab b) If a 25 V source is placed across terminals a and b, what is the voltage, Vdb?



Lecture 3‐1 Voltage and Current Divider

Reading: 3.3 – 3.4

Objectives:

• Be able to design simple voltage divider and current divider circuits

• Be able to apply current and voltage division to solve simple circuits

• Be able to derive the voltage divider using KVL

• Be able to derive the current divider using KCL

**************************************The voltage divider is used to calculate the voltage across several resistors when the voltage is supplied from a single source. In order to use the voltage divider, all of the elements must be in series.

Use KVL to find the voltage across R3: ‐v + v1 + v2 + v3 = 0 Use Ohm’s law to rewrite voltages across the resistors: ‐v + iR1 + iR2 + iR3 = 0 Solve for the current through the resistors: i = v/ (R1 + R2 + R3) Use Ohm’s law to find v3: v3 = iR3 = vR3/(R1 + R2 + R3)

Therefore the voltage divider rule is vn = vR

R

xx

n

∑

The current divider is used to calculate the current through several resistors when the current is supplied from a single source. In order to use the current divider rule, all of the elements must be in parallel.

Use KCL to find the current through R3: ‐i + i1 + i2 + i3 = 0 Use Ohm’s law to rewrite the currents across the resistors: ‐i + v/R1 + v/R2 + v/R3 = 0 Solve for the voltage across the resistors: v = i[1/R1 + 1/R2 +1/ R3]

‐1 = i/(G1 + G2 + G3) Use Ohm’s law to find i3: i3 = v/R3 = vG3 = iG3/(G1 + G2 + G3) = = i[1/R1 + 1/R2 +1/ R3]

‐1/ R3

Therefore the current divider rule is in = iRR

iGGG

G

n

eq

n21

n =++ L




a) use the voltage to find v1 and v2 b) use the current divider to find i1 and i2




a) use the current divider to find i1, i2 and v2 b) use the voltage to find v1



In‐Class Activity 3: For the above circuit,

a) Write a voltage divider relationship between the 10V source and Va, Vb, and Vc. b) Use the voltage divider to find Va, Vb, and Vc.

In‐Class Activity 4: For the above circuit,

a) Write a current divider relationship between Ia, Ib, and Ic b) Use the current divider to find Ib and Ic c) Write a current divider relationship to find Id d) Use the current divider to find Id



Lecture 3‐2 Wheatstone Bridge and Delta‐Wye Equivalents

Reading: 3.5 – 3.7

Objectives: • Be able to identify a Wheatstone Bridge and find the resistance to balance the bridge • Be able to analyze a Wheatstone bridge to find unknown voltages and currents • Be able to apply delta to Wye equivalents to solve simple circuits

**************************************The purpose of the Wheatstone Bridge is to make precise resistance measurement. One application of a Wheatstone Bridge is to place a strain gauge in one leg in order to measure

small changes in resistance in the range of 1 Ω to 1 MΩ to detect the strain on a device. A

detector called a galvanometer can be placed in the bridge to measure current in the μA range. The bridge is balanced when there is no current flowing through the galvanometer.

In‐Class Activity 1: Determine what value of resistance, Rx, causes the Wheatstone bridge to balance.

In‐Class Activity 2: For the following Wheatstone bridge, determine what value of resistance, Rx, causes the bridge to balance. If these are 25 mW resistors, do any of them exceed their power‐dissipating capacity?



In‐Class Activity 3: For the following Wheatstone bridge, determine the detector current, id if the voltage drop across the detector is negligible.

In‐Class Activity 4: For the following circuit, what is the bridge output voltage, vo?



Delta‐Wye Equivalents When the internal resistance of the galvanometer is modeled in the Wheatstone Bridge this creates a special configuration of resistances that is neither series nor parallel. Therefore this interconnection cannot be simplified by using series and parallel combinations. These special

configurations are referred to as Delta (Δ) or Wye (“Y”) interconnections and they can be transformed to equivalent representations.

Δ or π interconnection Y or T interconnection

Ra = R1 =

Rb = R2 =

Rc = R3 =

NOTE: Each resistor in the Δ network is the sum of all possible products of Y resistors taken two at a time divided by the opposite Y resistor.

The Y and Δ configurations are said to be balanced when the resistance in each branch is the same and RY = RΔ/3 or RΔ =3 RY

The following figure is a superposition of the Y and Δ configurations to aid in transforming from one to another.

What does equivalent mean? Equivalent means that both configurations have the same voltage and current characteristics at the terminals a, b, and c. If these 2 networks were in a black box and you measured voltage

and current, you could not state whether it was a Δ or Y interconnection in the box. However, you CANNOT state that the voltage across and current through the internal resistors is the same for both.



In‐Class Activity 5:

For the following circuit, determine the equivalent resistance at terminals a and b if all of the

resistors have a value of 30Ω.


a) use 2 different Δ‐to‐Y transformations to find the voltage v1. b) Use a Δ‐to‐Y transformation to find the power delivered by the 80V source.

Node-voltage methodMesh-current method

Source TransformationsThevenin and Norton

EquivalentsMaximum Power Transfer

Superposition



Lecture 4‐1 Node‐voltage method Reading: 4.1 – 4.2

Objectives:

• Be able to briefly and clearly explain in your own words node, node‐voltage method • Be able to identify nodes in a circuit • Be able to apply the node‐voltage method to calculate voltage, current, and power in an

electric circuit

**************************************A planar circuit is a circuit that can be drawn on a plane with no crossing branches. The node‐voltage method can be used on planar and non‐planar circuits. However, the mesh‐current method cannot be used on non‐planar circuits (see Figures 4.1 and 4.2 on page 94 in the textbook for examples of both). All of the terms used to describe a circuit are provided in the following table. node

a point where two or more circuit elements join

essential node

a node where 3 or more circuit elements join

path a trace of adjoining basic elements with no elements included more than once

branch

a path that connects two nodes

essential branch

a path which connects two essential nodes without passing through an essential node

loop a path whose last node is the same as the starting node

mesh

a loop that does not enclose any other loops



Recall that Kirchhoff’s current law states that the sum of the currents in and out of a node must equal zero. The node‐voltage law is based upon using KCL to analyze a circuit to independent simultaneous equations to solve for node voltages. The number of equations required is based upon the branches and nodes in the circuit. If there are n nodes in a circuit, then it is possible to derive n – 1 equations using KCL and b‐(n‐1) equations using KVL. In‐Class Activity: How many nodes and branches are in the following circuit? How many KCL and KVL equations are required to solve for the unknown branch currents?

Steps to implement the node‐voltage method

i. make a neat layout of the circuit with no branches overlapping ii. mark one of the essential nodes the reference node (typically the one with the most

branches near the bottom of the circuit) [note: this is the ground (0 V), all node voltages are with respect to this node]

iii. node voltages are defined as the voltage rise from the ground node to a non reference node

iv. label all of the node voltages on the circuit (v1, v2, v3, etc.) v. draw currents leaving each labeled node if it is not given vi. use Ohm’s law to write the KCL equations at each of the non‐reference labeled

nodes vii. Solve the simultaneous equations by using Cramer’s rule, matrix manipulation or

your calculator to find the node voltages viii. Finally complete the power table by using the derived values.



In‐Class Activity 1: For the above circuit, execute steps i – v of the node‐voltage method. Use the node‐voltage method to confirm that the circuit satisfies the law of conservation of energy. element V (V) I (A) Power (W)

delivered Power (W) absorbed

20mA 5 mA 10 mA 2 kΩ 4 kΩ 2 kΩ 4 kΩ 1 kΩ TOTAL



In‐Class Activity 2: Use the node‐voltage method to find the voltage across the current sources in the following circuit.



Lecture 4‐2 Node‐voltage method (special cases)

Reading: 4.3 – 4.4

Objectives:

• Be able to apply the node‐voltage method to circuits with voltage sources to solve for voltage, current, and power

**************************************

As seen from the prior lecture, when circuit includes dependent sources, the node voltage method must be supplemented with constraint equations imposed on the dependent source. When a voltage source is between two non‐reference nodes, the equation derivation is simplified because the voltage between the 2 essential nodes is constrained by the value of the source. This voltage source between two non‐reference nodes and any element in parallel with it is called a supernode. Properties of a supernode:

1. The voltage source inside the supernode provides a constraint equation needed to solve for the node voltages.

2. A supernode has no voltage of its own. 3. A supernode requires the application of both KCL and KVL.

To analyze the supernode:

1. perform KCL into and out of the supernode, (i.e., i4Ω + i3Ω+ i2Ω + i6Ω = 0) 2. perform KVL at the supernode (i.e., v2 – v1 = 3 V)

Finally, since the purpose of the node‐voltage method is to use KCL to find all unknown node‐voltages, it is also not necessary to perform KCL at the 7V source. Instead of labeling this node as v1 or v2, refer to it at “7V”.



In‐Class Activity 1: For the above circuit, a) outline how you would use the node‐voltage method to find all of the unknown voltages b) confirm that the circuit satisfies the law of conservation of energy

Element V (V) I (A) Power (W) delivered

Power (W) absorbed

7V 3V 4 Ω 3 Ω 4 Ω 6 Ω 2 Ω TOTAL



In‐Class Activity 2: For the above circuit, are there any supernodes? if so, what are they? Use the node‐voltage method to analyze the circuit and confirm that the circuit satisfies the law of conservation of energy.


Power (W) absorbed

10 V 5 V 2 Ω 4 Ω 6 Ω 8 Ω TOTAL



In‐Class Activity 3: For the following circuit, use the node‐voltage method to find v1, v2, v3, vo, and Io.



Lecture 4‐3 Mesh‐current method

Reading: 4.5

Objectives:

• Be able to briefly and clearly explain in your own words loop, mesh, mesh‐current and mesh‐current method

• Be able to identify loops and meshes in a circuit • Be able to apply the mesh‐current method to calculate voltage, current, and power in an

electric circuit

**************************************Recall that the mesh‐current method is only valid for non planar circuits. For the mesh‐current method, use KVL to describe a circuit in terms of b – (n – 1) independent simultaneous equations. A mesh current is the current that exists only in the perimeter of a mesh. Note that Figure 4.18 illustrates mesh currents and Figure 4.4 illustrates branch currents. It should be evident that it is not always possible to identify a mesh current in terms of a branch current. In order to implement the mesh‐current method,

1. Assign mesh current i1, i2, . . . . in to the n meshes in the circuit 2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms

of the mesh currents 3. Solve the resulting n simultaneous equations to get the mesh currents. 4. Determine the relationship between the mesh currents and branch currents in order

to find all the voltage and power in the circuit Similar to the analysis with dependent sources for the node‐voltage method, the existence of a dependent source must be supplemented by an appropriate constraint equation. Concept Question: In the following circuit, all of the light bulbs are identical. Knowing that the intensity of a light bulb is proportional to the power dissipated across it. When the switch is closed, which of the following statements is true? a) Bulb B is brighter b) Bulb B is dimmer c) Bulb B remains the same d) none of the above



In‐Class Activity 1: For the above circuit, use the mesh‐current method to find all of the mesh currents and confirm that it satisfies the law of conservation of energy.

Element V (V) I (A) Pdel (W) Pabs (W) 80 V 5 Ω 30 Ω 90 Ω 26 Ω 8 Ω TOTAL



In‐Class Activity 2: For the following circuit, use the mesh current method to find the mesh currents and voltages across the 8Ω and 6Ω resistors. Compare your solution to the result of Lecture 4‐2 In‐Class Activity 2.



In‐Class Activity 3: For the following circuit, find the power developed in the dependent voltage source.



Lecture 5‐1 Mesh‐current method (special cases)

Reading: 4.6 – 4.7

Objectives:

• Be able to apply the mesh‐current method to circuits with current sources to solve for voltage, current, and power

**************************************

Recall that the mesh‐current method uses KVL to determine the unknown mesh currents in a circuit. The steps to implement the mesh‐current method include labeling the meshes, write the KVL equation for each mesh, and solve the simultaneous set of equations. FE Preview: What is the current, IB, through the 40 V battery? a) 1.3 A b) 2.9 A c) 6.7 A d) 13 A Just like there was a special case of the node‐voltage method when it involves voltage sources, there are also special cases of the mesh‐current method when a current source is shared between meshes or on an outer branch of a circuit. The following 2 circuits give illustrations of how current sources affect mesh analysis.

(a) current source only in one mesh (b) two meshes have the current source in common

CASE I CURRENT SOURCE IN ONLY ONE MESH When a current source exists only in one mesh, the mesh current is equal to the value of the source. The sign is dependent upon the direction of the source current. For the circuit in (a) i2 = ‐5A.



CASE II TWO MESHES HAVE THE CURRENT SOURCE IN COMMON When the current source exists between two meshes, create a supermesh by excluding the current source and any elements connected in series with it. If a circuit has two or more supermeshes that intersect, they should be combined for form a larger supermesh. A supermesh results when two meshes have a (dependent or independent) current source in common. Properties of a supermesh: 1. The current source in the supermesh provides the constraint, (i.e., i2 – i1 = 6A) 2. A supermesh has no current of its own 3. A supermesh requires the application of both KVL and KCL (i.e., ‐20 + v6Ω + v10Ω + v4Ω = 0) Node‐Voltage Method versus Mesh‐Current Method Networks that contain many series‐connected elements, voltage sources, or supermeshes are more suitable for mesh analysis. Networks with parallel‐connected elements, current sources, or supernodes are more suitable for nodal analysis.

• A circuit with fewer nodes than meshes is better analyzed using nodal analysis. • A circuit with fewer meshes than nodes is better analyzed using mesh analysis. • The key is to select the method that results in the smaller number of equations. • Also, consider the information that is required. • If node voltages are required, it may be expedient to apply node analysis. • If branch or mesh currents are required, it may be more expedient to use mesh analysis.

Concept Question: For the following circuit, would you select the node‐voltage or mesh‐current method to solve for the voltage across each passive circuit element? Why?



In‐Class Activity 1: For the above circuit, use the mesh‐current method to find the mesh currents. In‐Class Activity 2: For the above circuit, use the mesh‐current method to confirm that the circuit satisfies the law of conservation of energy.


Power (W) absorbed

20 V 6 A 2 Ω 4 Ω 6 Ω 10 Ω TOTAL



In‐Class Activity 3: For the following circuit, use the mesh‐current method to find all of the mesh currents. Find the voltage across the two current sources. Compare the result the lecture 4‐1 In‐Class Activity 2.



Lecture 5‐2 Source Transformations

Reading: 4.9

Objectives:

• Be able to briefly and clearly explain the process of applying source transformations to simplify a circuit

• Be able to apply source transformations to simplify a resistive circuit • Be able to use source transformations to solve for voltage, current, and power in a

resistive circuit

**************************************In‐Class Activity 1: For the following circuit, use the mesh‐current or node‐voltage method to find v1, v2, v3, vo and Io.



One method to simplify passive elements in a circuit is series‐parallel reductions. One technique to simplify active elements in a circuit is source transformations. A source transformation allows a voltage source in series with a resistor to be replaced by a current course in parallel with the same resistor or vice versa.

NOTE: The arrow of the current is directed toward the positive terminal of the voltage source. When performing the transformation from a current source to a voltage source, the value is found by using Ohm’s law, vs = isR When performing the transformation from a voltage source to a current source, the value is found by using Ohm’s law, is = vs/R Special Cases: When a resistance, Rp is in parallel with the voltage source or a resistance, Rs is in series with the current source, there is no difference in the transformation circuit.

Important note: Based upon the concept of KVL, voltage sources in series add algebraically



Based upon the concept of KCL, current Sources in parallel add algebraically

In‐Class Activity 1: For the following circuit, use source transformations to simplify the circuit to one source and two resistors to find vo. Find the power developed by the 12 V and 3A sources.



In‐Class Activity 2: For the following circuit, use source transformations to find ix.(hint: simplify to one voltage source and 2 resistors)



Lecture 5‐3 Thevenin and Norton Equivalents

Reading: 4.10 – 4.11

Objectives:

• Be able to briefly and clearly explain the concept of Thevenin and Norton equivalent circuits

• Be able to construct a Thevenin or Norton equivalent circuit

**************************************Thevenin and Norton equivalents are circuit simplification techniques that focus on the voltage and current characteristics at the terminal of a circuit. Recall from source transformations that two circuits are said to be equivalent if they have the same voltage‐current relationship at their terminals. When a circuit can be simplified and described at its terminals in terms of a voltage source in series with a resistance, this is the Thevenin equivalent circuit.

The Thevenin Voltage, Vth is the open circuit voltage across terminals a and b. When a circuit can be simplified at its terminals and is described in terms of a current source in parallel with a resistance, this is the Norton equivalent circuit. The Norton current, IN, is the short circuit current between terminals a and b. The relationships between the Thevenin Voltage and Norton current are based upon Ohm’s law.

R = Voc/Isc → Rth = Vth/IN Note that RTh = RN



There are several techniques that can be used to find the Thevenin equivalent of a circuit. When there are only independent sources, source transformations can be used if the configuration allows.

In‐Class Activity For the above circuit, find the Thevenin equivalent to the left of terminals a and b. In‐Class Activity For the above circuit, find the Norton equivalent to the left of terminals a and b. In‐Class Activity For the above circuit, if there is a 5Ω load placed across the terminals of the circuit, what is the power delivered to the load?



When you cannot use source transformations to find the Thevenin or Norton equivalent of a circuit use one of the following techniques. For independent sources only,

• Find the equivalent or input resistance at a‐b when all independent sources are turned off, Rab = RTh

For dependent sources only,

• Apply a test voltage or current across terminals a and b, the relationship between the voltage across and current through the test source is RTh = vtest/itest

• If Rth produces a negative value in your calculations, this implies that the circuit is actually absorbing power if it was assumed to be delivering power.

For any combination of independent and dependent sources,

• Find the open circuit voltage across terminals a‐b, voc = VTh

• Find the short circuit current through terminals a‐b, isc = IN

• Find the open circuit voltage and short circuit current between the terminals and Rth = voc/isc.

The question is, “how do you turn off a voltage and/or current source in circuit analysis?” To turn off a voltage source means that it becomes 0 volts. To turn off the voltage source, replace the voltage source with a short circuit which has a net resistance of 0Ω and a net voltage of 0V. To turn of a current source means that it becomes 0 amperes. To turn off the current source, replace the current source with an open circuit which has an infinitely large resistance and a net current of 0A.


For the following circuit, find the Thevenin equivalent resistance, Rth by turning off all independent sources.



In‐Class Activity For the following circuit, find the Thevenin equivalent voltage, Vth by finding the open circuit voltage between terminals a and b (Vab).

In‐Class Activity For the above circuit, find the Norton equivalent current, IN by finding the short circuit current between terminals a and b (Iab). In‐Class Activity For the above circuit, confirm that you can also find Rth by using the formula Vth/IN.



Lecture 6‐1 Thevenin and Norton Equivalents

Reading: 4.10 – 4.11

Objectives:

• Be able to briefly and clearly explain the concept of Thevenin and Norton equivalent circuits

• Be able to construct a Thevenin or Norton equivalent circuit

**************************************FE Preview 1: What are the Thevenin equivalent resistance and voltage between terminals A and B? a) Rth = 3 Ω, Vth = 45 V b) Rth = 7.5 Ω, Vth = 7.5 V c) Rth = 7.5 Ω, Vth = 60 V d) Rth = 12 Ω, Vth = 5 V FE Preview 2: What are the Norton equivalent source and resistance values for the circuit shown? a) IN =5 A; RN =5 Ω b) IN = 10 A; RN =20 Ω c) IN = 1 A; RN =5 Ω d) IN = 1 A; RN =10 Ω



In‐Class Activity 1: For the following circuit, find the Thevenin and Norton equivalent circuit across terminals a and b.



In‐Class Activity 2: For the following circuit, find the Thevenin and Norton equivalent circuit across terminals a and b.



In‐Class Activity 3: For the following circuit, find the Thevenin and Norton equivalent circuit across terminals a and b. Use the equivalent circuit to find ix.



Lecture 6‐2 Maximum Power Transfer

Reading: 4.12

Objectives:

• Be able to calculate the resistance of a load for maximum power transfer • Be able to calculate the power deliver to the load resistance of a Thevenin or Norton

equivalent circuit

**************************************The primary purpose for examining the terminal characteristics of a circuit is to identify the effect that the circuit will have on a load. There are 2 types of power transfer based upon efficiency of the transfer and the amount of power transferred. Maximum power transfer involves determining the value of a load resistor, RL that permits maximum power delivery. Based upon the derivation on page 127, the maximum power is transferred to the load when RL = RTH.

To find the power delivered to the load, use KVL, ‐VTh + iRTh + iRL = 0 solving for i yields, i = VTh/(RTh + RL) using the power formula, p = VTh

2 RL /(RTh + RL)

2 if RL = RTh, then the maximum power is pmax = VTh

2 /4RL



In‐Class Activity 1: For the above circuit, what is the value of a load resistor placed across terminals a and b for maximum power transfer. What is the value of the power across the resistor selected?



In‐Class Activity 2: For the above circuit, outline the steps to find the value of a load resistor for maximum power transfer. Be detailed as possible.




a) Calculate the value of resistor R for maximum power transfer b) Determine the maximum power absorbed by the resistor, R



Lecture 6‐3 Superposition Reading: 4.13

Objectives:

• Be able to clearly and concisely describe the principle of superposition in your own words

• Be able to apply the principle of linearity (superposition) to find unknown, voltages and current in a circuit

• Be able to explain why superposition cannot be used to find power in a circuit element • Be able to redraw a circuit with ideal sources are “turned off”

**************************************

A linear system obeys the principle of superposition which states that when a linear electric circuit is excited, or driven, by more than one independent source of energy, the total response is the sum of the individual response due to each source. Note that superposition can only be applied to find voltage and current responses in a circuit, not power. Superposition cannot be used on power because it does not have a linear relationship. For example, if, p1 = i1

2R and p2 = i22R, total power in the resistor,

p ≠ p1 + p2 Steps to apply superposition to an electric circuit:

1. Redraw the circuit with each independent source turned on one at a time 2. For each circuit, apply a circuit analysis technique to calculate the relevant voltage or

current response (v′ or i′) 3. Repeat step 2 for each independent source in the circuit 4. Use the principle of linearity or superposition to calculate the total response (i.e. v =

v′ + v′′ + … or v = i′ + i′′ + … ) The question is, “how do you turn off a voltage and/or current source in circuit analysis?”

To turn off a voltage source means that it becomes 0 volts. To turn off the voltage source, replace the voltage source with a short circuit which has a net resistance of 0Ω and a net voltage of 0V.

To turn of a current source means that it becomes 0 amperes. To turn off the current source, replace the current source with an open circuit which has an infinitely large resistance and a net current of 0A.



Concept Question: In the following circuit, all of the light bulbs are identical. When the switch is closed the voltage across bulb C becomes

a) V1/2 b) V1 c) 1.5V1 d) 0.67V1

In‐Class Activity 1: For the following circuit, redraw the circuit with each of the independent sources turned on one at a time.

In‐Class Activity 2: For the above circuit, use superposition to find vo and io.



In‐Class Activity 3: For the following circuit, use superposition to find ix.

s



In‐Class Activity 4: For the following circuit, use superposition to find the voltage and current for the 5Ω resistor. Use the total voltage and current for the 5Ω resistor to find the power delivered.

Operational AmplifiersInverting, Non-inverting,

Summing, DifferenceCascaded

More realistic op ampsOp Amp ApplicationsTerminal Equations

Two-Port parametersTerminated two-port circuits



Lecture 7‐1 Introduction to Operational Amplifiers

Reading: 5.1 – 5.2

Objectives:

• Be able to name the five op amp terminals and describe the voltage and current constraints for an ideal model

• Be able to analyze simple circuits containing ideal op amps using KCL • Be able to recognize the following types of op amps: noninverting, inverting, summing,

and difference

**************************************

The operational amplifier (opamp) is an electronic unit that behaves like a voltage‐controlled voltage source. It is a high‐gain amplifier that has high input impedance and low outpu resistance. An operational amplifier (opamp) is an active circuit element designed to perform mathematical operations of addition, subtraction, multiplication, division, differentiation, and integration.



The following figure shows an example of an operational amplifier integrated circuit as well as the symbol used to represent an op amp in circuit analysis.

The five terminals of interest on the operational amplifier are the • Inverting input (Vn) • Non inverting input (Vp) • output • positive power supply (V+) • negative power supply (V‐)

The voltage characteristic of an op amp has a linear and nonlinear region as shown by the following figure. When the op amp is operating in the linear region, the output voltage is equal to the difference in its input voltages times the multiplying constant or gain.

Vo = AVd = A (Vp – Vn) A is called the open‐loop voltage gain because it is the gain of the op amp without any external feedback from output to input. When the output of the operational amplifier exceeds the positive or negative power supply it is in saturation. When the op amp is in saturation, the outputs are equal to the positive or negative power supply.

An ideal op amp has infinite input resistance and zero output resistance and infinite gain. There are 2 constraints or assumptions to make for an ideal operational op‐amp

1. the virtual short condition states that the voltage at the positive terminal equals the voltage at the negative terminal (vp = vn)

2. the infinite input impedance condition states that the current into the positive and negative terminals of the opamp is negligible (ip = in = 0A)

It is possible to analyze ideal operational amplifiers by applying the 2 assumptions and Kirchhoff’s current laws at the input and output terminals.




For the following circuit, use the ideal assumptions and KCL to determine the gain, vo/vs.




For the following circuit, use the ideal assumptions and KCL to determine the gains for each of

the inputs and the output, vo.



Lecture 7‐3 Inverting and Summing Operational Amplifiers

Reading: 5.3 – 5.4

Objectives:



and difference

**************************************

An inverting amplifier produces an output voltage that is an inverted, scaled value of the input voltage (the gain can be less than 1) A summing amplifier is an inverting amplifier with multiple inputs and produces an output voltage that is a scaled sum of the input voltages (some or all of the gains can be less than one) The following table summarizes the input‐output relationships, gain, and circuit configurations for the inverting and summing amplifiers. These relationships can be derived using KCL and the 2 basic assumptions for ideal operational amplifiers. Type Vo Gain Circuit

Inverting

i1

f vRR−

1

f

i

o

RR

vv

−=

Summing 3

3

f2

2

f1

1

f vRRv

RRv

RR

−−−

v1: 1

f

1

o

RR

vv

−=

v2: 2

f

2

o

RR

vv

−=

v3: 3

f

3

o

RR

vv

−=



In‐Class Activity 1: For the following circuit, what range of input voltages will avoid op amp saturation?



In‐Class Activity 2: For the following circuit, determine the values of ia, va, io, and vo.



In‐Class Activity 3: For the following circuit, what type of operational amplifier is it? If va = 1.2 V, vb = 1.5 V and vc = ‐2 V, what are the values for vo and io?



Lecture 8‐1 Noninverting, Cascaded, Difference Operational Amplifiers

Reading: 5.5 – 5.6

Objectives:



and difference • Be able to determine the gain and output for cascaded operational amplifier circuits

**************************************

A noninverting amplifier produces an output voltage that is a scaled value of the input voltage (the gain must be greater than one) A difference amplifier produces an output voltage that is a scaled value of the input voltage difference (the gain can also be less than 1) The following table summarizes the input‐output relationships, gain, and circuit configurations for the noninverting and difference amplifiers. These relationships can be derived using KCL and the 2 basic assumptions for ideal operational amplifiers.

Type Vo Gain Circuit Non inverting

i1

f vRR1 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+

1

f

i

o

RR

1vv

+=

Difference (general)

( )( ) 1

1

22

431

214 vRRv

RRRRRR

−++

1

2

1

o

RR

vv

−=

=2

o

vv ( )

( )431

214

RRRRRR

++

Difference (simplified)

⎟⎟⎠

⎞⎜⎜⎝

⎛=

4

3

2

1

RR

RR

( )121

2 vvRR

−

1

2

1

o

RR

vv

−=

1

2

2

o

RR

vv

=



Recall that operational amplifiers are designed using transistors and diodes and are used to perform mathematical operations. There are 2 basic assumptions or conditions used to analyze ideal operational amplifiers: virtual short circuit, infinite input impedance. Using these assumptions and KCL, it is possible to find the output of any operational amplifier circuit. In‐Class Activity 1: Analyze the following circuit to determine the input and output relationship. What type of op amp configuration is it? What is the output if the input is 5 V?

In‐Class Activity 2: For the following circuit, assuming the op amp is operating in its linear region, what is vo?




For the following circuit, use superposition to find the output voltage, Vo.

In‐Class Activity 4: For the following circuit, find the input‐out relationship. If va = 2 V, what range of values for vb will result in linear operation?



Op amp circuits may be cascaded without changing their input‐output relationships. A cascaded operational amplifier connection is a head‐to‐tail arrangement of two or more op amp circuits such that the output of one is the input of the next. When op amp circuits are cascaded, each circuit is called a stage. The overall gain of a cascaded connection is the product of the gains of the individual op amp circuits, or

A = A1A2A3A4A5…An_ In‐Class Activity 5: Assuming the following op amps are operating in a linear region, what is the output, vo? What is the numerical value of vo if v1 = 10 mV and v2 = 15 mV?



Lecture 8‐2 More realistic op amps

Reading: 5.7

Objectives:

• Be able to state the properties of a more realistic model of an operational amplifier • Be able to able to analyze a more realistic model of an operational amplifier for the gain

and input‐output characteristics **************************************

A more realistic model of an operational amplifier does not have the 3 simplifying assumptions

The following table shows the more typical range for the op amp parameters

Parameter Ideal Values Typical Range

Open‐loop gain, A ∞ 105 to 108

Input resistance, Ri ∞ Ω 105 to 1013 Ω

Output resistance, Ro 0 Ω 10 to 100 Ω

Supply Voltage, VCC 5 to 24 V




The following inverting amplifier has an input resistance of 500 kΩ, an output resistance of 5 kΩ, and an open‐loop gain of 300, 000. Assume that the operational amplifier is operating in its linear region.

a) Calculate the voltage gain (vo/vg) b) Calculate the value of vn in microvolts when vg = 1V c) Calculate the resistance seen by the signal source (vg) d) Compare the answers to (a) – (c) using the ideal model for the op amp




A 741 op amp has an open‐loop voltage gain of 2 x 105, input resistance of 2 MΩ, and output resistance of 50 Ω. The op amp is used in the circuit of the following circuit. Find the closed‐loop gain vo/vs. Determine current i when vs = 2V. Compare this current to the value for the ideal model.




A 741 op amp has an open‐loop voltage gain of 2 x 105, input resistance of 2 MΩ, and output resistance of 50 Ω. The op amp is used in the circuit of the following circuit. Calculate the closed‐loop gain vo/vs and find io when vs =1V. Compare this current to the value for the ideal model.



Lecture 8‐3 Operational Amplifier Applications

Objectives:

• Be able to design operational amplifiers to meet a given design specification • Be able to list applications of operational amplifiers and analyze operational amplifier

circuits for these applications

************************************** Operational amplifiers are used frequently in electronics for signal conditioning,

instrumentation amplifiers, digital‐to‐analog converters, filters, comparators, gyrators,

oscillators, rectifiers, and regulators.


Design an operational amplifier circuit to find the average of 3 input voltages: va, vb, vc.


Design an operational amplifier that provides a relationship between output voltage vo and

input voltage vs such that vo = 12vs – 10. Two op‐amps, a 6‐V battery, and several resistors are

available.




The following operational amplifier is used to measure changes in the shape of elastic solids by

using a strain gage. If a pair of strain gages is mounted on a bar to measure the strain and used

in a Wheatstone bridge, one pair of gages measures as the bar is lengthened and represented

by R + ΔR. As the bar is shortened, the other pair of strain gage resistance is represented by R ‐

ΔR. Suppose the strain gage has a value of 120 Ω ± 1%, Analyze the following circuit to

determine the relationship between the output voltage, vo and the change in resistance, ΔR.

Determine the output vo if ΔR = 0.1 Ω.




The following circuit is a voltage‐to‐current converter where iL = Avi, if R1 = R3 = 1 kΩ and R2 = R4

= 2.2 kΩ, determine the constant term A. Select the value of RL such that it absorbs 10 mW of

power.




The digital‐to‐analog converter (DAC) transforms digital signals to analog form. This can be

implemented using an op amp as a binary weighted ladder. The bits are weighted according to

the magnitude of their place value by descending values of Rf/Rn, so that each lesser bit has half

the weight of the next higher. For the following three‐bit DAC

a) Determine |Vo| for [V1V2V3] = [010]

b) Find |Vo| if [V1V2V3] = [110]

c) If |Vo| = 1.25 V is desired, what should be the values of [V1V2V3]?

d) If |Vo| = 1.75 V is desired, what should be the values of [V1V2V3]?



Lecture 9‐1 Terminal Equations, Two‐Port Parameters

Reading: 18.1 – 18.2

Objectives:

• Be able to identify two‐port circuits and explain the difference between two‐port and one‐port networks

• Be able to define terminal equations and two‐port parameters • Be able to define the terms symmetrical and reciprocal • Be able to define the terms immittance, transmission, and hybrid parameters • Be able to calculate two‐port parameters for a simple resistive circuit

**************************************

Terminal pairs on a circuit where signals are either fed in or extracted are referred to as the ports of the system. A two‐port network is an electrical network with two separate ports for input and output.

Restrictions: i. there can be no energy stored within the circuit ii. there can be no independent sources within the circuit iii. the current into the port must equal the current out of the port iv. all external connections must be made to either the input port or the output port Similar to the fundamental principle of analysis for Thevenin and Norton equivalence and operational amplifiers only the terminal variables (I1, I2, V1, V2) are of interest not the currents and voltages inside the circuit. The terms that relate the voltages and currents of a two‐port network are the parameters.



The open‐circuit impedance parameters or z parameters have the units of ohms. These are found by open‐circuiting the input or output port.

z11 = 0I1

1

2

IV

=

z12 = 0I2

1

1

IV

=

z21 = 0I1

2

2

IV

=

z22 = 0I2

2

1

IV

=

z11 = open‐circuit input impedance z12 = open‐circuit transfer impedance from port 1 to port 2 z21 = open‐circuit transfer impedance from port 2 to port 1 z22 = open‐circuit output impedance When z11 = z22 the two‐port network is said to be symmetric. When the two‐port network is linear and has no dependent sources, the transfer impedances are equal (z12 = z21) and the two‐port network is said to be reciprocal. Any two‐port network made entirely of resistors, capacitors, and inductors must be reciprocal. The impedance parameters can also be described using the following matrix relationship.

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

2

1

2221

1211

2

1

II

zzzz

VV

In‐class Activity 1: Find the z parameters for the following circuit.



In‐class Activity 2: Find the z‐parameters for the following circuit.



In‐class Activity 3: Find I1 and I2 in the following circuit.



Lecture 9‐3 Two‐Port Parameters

Reading: 18.2

Objectives:

• Be able to calculate two‐port parameters for a simple resistive circuit

************************************** Admittance parameters or y parameters have the units of Siemens. The values are determined by setting V1 = 0 or V2 = 0 (input and output short‐circuited). These parameters are also called the short‐circuit admittance parameters.

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

2

1

2221

1211

2

1

VV

yyyy

II

y11 = 0V1

1

2

VI

=

y12 = 0V2

1

1

VI

=

y21 = 0V1

2

2

VI

=

y22 = 0V2

2

1

VI

=

y11 = short‐circuit input admittance y12 = short‐circuit transfer admittance from port 2 to port 1 y21 = short‐circuit transfer admittance from port 1 to port 2 y22 = short‐circuit output admittance The impedance and admittance parameters are collectively referred to as the immittance parameters. The convert between the immittance parameters, use the following y to z parameters z to y parameters

z11 = yy22

Δ y11 = z

z22

Δ

z12 = yy12

Δ−

y12 = zz12

Δ−

z21 = yy21

Δ−

y21 = zz21

Δ−

z22 = yy11

Δ y22 = z

z11

Δ

Δy = y11y22 – y12y21 Δz = z11z22 – z12z21



In‐class activity 1: Obtain the y parameters for the following circuit.

In‐class activity 2: Obtain the y parameters for the following circuit.



The a and b parameters are called the transmission parameters because they describe the voltage and current at one end of the two‐port network in terms of the voltage and current at the other end. V1 = a11V2 ‐ a12I2 I1 = a21V2 ‐ a22I2

V2 = b11V1 ‐ b12I1 I2 = b21V1 ‐ b22I1

a11 = 0I2

1

2

VV

=

open‐circuit voltage ratio b11 = 0I1

2

1

VV

=

open‐circuit voltage gain

a12 = 0V2

1

2

IV

=

− negative short‐circuit transfer

impedance (Ω)

b12 = 0V1

2

1

IV

=

− negative short‐circuit transfer

impedance (Ω)

a21 = 0I2

1

2

VI

=

open‐circuit transfer admittance

(S)

b21 = 0I1

2

1

VI

=

open‐circuit transfer admittance

(S)

a22 = 0V2

1

2

II

=

− negative short‐circuit current

ratio

b22 = 0V1

2

1

II

=

− negative short‐circuit current

gain In‐class activity 3: The following measurements pertain to a two‐port circuit operating in the sinusoidal steady state. With port 2 open, a voltage equal to 150 cos 4000t V is applied to port 1. The current into port 1 is 25 cos(4000t ‐ 45°) A, and the port 2 voltage is 100 cos (4000t + 15°) V. With port 2 short‐circuited, a voltage equal to 30 cos 4000t V is applied to port 1. The current into port 1 is 1.5 cos(4000t + 30°) A, and the current into port 2 is 0.25 cos (4000t + 150°) A. Find the a parameters that can describe the sinusoidal steady‐state behavior of the circuit.



The h and g parameters relate cross variables, that is an input voltage and output current to an output voltage and input current. The h and g parameters are called hybrid parameters. The g parameters are also called inverse hybrid parameters. V1 = h11I1 + h12V2 I2 = h21I1 + h22V2

I1 = g11V1 + g12I2 V2 = g21V1 + g22I2

h11 = 0V1

1

2

IV

=

short‐circuit input impedance (Ω) g11 = 0I1

1

2

VI

=

open‐circuit input admittance (S)

h12 = 0I2

1

1

VV

=

open‐circuit reverse voltage gain g12 = 0V2

1

1

II

=

short‐circuit reverse current gain

h21 = 0V1

2

2

II

=

short‐circuit forward current gain g21 = 0I1

2

2

VV

=

open‐circuit forward voltage gain

h22 = 0I2

2

1

VI

=

open‐circuit output admittance (S) g22 = 0V2

2

1

IV

=

short‐circuit output impedance (Ω)

In‐class activity 4: Determine the Thevenin equivalent at the output port of the following circuit.



Lecture 10‐1 Terminated two‐port circuits

Reading: 18.3

Objectives:

• Be able to calculate voltage, current, and power for terminated two‐port circuits for a simple resistive circuit

• Be able to define the terms symmetrical and reciprocal

************************************** In summary, two‐port parameters can be used to describe electronic devices such as transistors and there are special relationships between the different parameters. Table 18.1 in the textbook describes all of the relationships between the two‐port parameters.

If a two‐port circuit is reciprocal then the following relationships must exist.

i. z12 = z21

ii. y12 = y21

iii. a11a22 – a12a21 = Δa = 1

iv. b11b22 – b12b21 = Δb = 1

v. h12 = ‐h21

vi. g12 = ‐g21

A two‐port circuit is reciprocal if the interchange of an ideal voltage source at one port with an

ideal ammeter at the other port produces the same ammeter reading. A two‐port circuit is also

reciprocal if the interchange of an ideal current source at one port with an ideal voltmeter at

the other port produces the same voltmeter reading.

A reciprocal two‐port circuit is symmetric if its ports can be interchanged without disturbing

the values of the terminal currents and voltages. Figure 18.6 in the textbook shows four

examples of symmetric two‐port circuits. The following relationships exist among the port

parameters.



i. z11 = z22

ii. y11 = y22

iii. a11 = a22

iv. b11 = b22

v. h11h22 – h12h21 = Δh = 1

vi. g11g22 – g12g21 = Δg = 1

18.3 Analysis of the Terminated Two‐Port Circuit In a typical application of a two‐port network, the circuit is driven at port 1 and loaded at port 2. The following figure shows an example of a terminated two‐port model.

Terminated Two‐Ports

The six characteristics of the terminated two‐port circuit define its terminal behavior:

• the input impedance Zin = V1/I1 or the admittance Yin = I1/V1

• the output current I2

• the Thevenin voltage and impedance with respect to port 2

• the current gain I2/I1

• the voltage gain V2/V1

• the voltage gain V2/Vg

Table 18.2 in the textbook includes all of the terminated two‐port equations. The z and y

parameters for a terminated two‐port model are given below on the table.



z parameters y parameters input impedance/ admittance

Zin/Yin

L22

211211 Zz

zzz+

− L22

L211211 Zy1

Zyyy+

−

output current I2

( )( ) 2112L22g11

g21

zzZzZzVz

−++−

Lgg11L22

g21

ZyZZyZy1Vy

Δ+++

Thevenin voltage Vth ( ) g

g11

21 VZz

z+

( ) gg22

21 VyZy

yΔ+

−

Thevenin impedance

Zth

g11

211222 Zz

zzz+

− g22

g11

yZyZy1

Δ++

current gain I2/I1

L22

21

Zzz+

−

L11

21

yZyyΔ+

voltage gain V2/V1 zZz

Zz

L11

L21

Δ+

L22

L21

Zy1Zy

+−

voltage gain V2/Vg ( )( ) 2112L22g11

L21

zzZzZzZz

−++ ( )( )L22g11Lg2112

L21

Zy1Zy1ZZyyZy

++−

determinant of parameter matrix

Δz = z11z22 – z12z21

Δy = y11y22 – y12y21

In‐class activity 1: For the following circuit, a. determine the value of a load resistor placed across the output port for maximum power

transfer b. For the selected resistor, what is the value of the power delivered to the load?

Conversion from h to y parameters Δh = h11h22 – h12h21 ⎥

⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ −

=⎥⎦

⎤⎢⎣

⎡

1111

21

11

12

11

2221

1211

hh

hh

hh

h1

yyyy

Δ



In‐class activity 2: The following two‐port circuit is described in terms of the following y parameters. y11 = 0.15 S y21 = ‐0.25 S y12 = ‐0.05 S y22 = 0.25 S a) Find the phasor voltage, V2 b) Find the average power deliver to the 5kΩ load. c) Find the average power delivered to the input port.

In‐class activity 3: The a parameters and source and load characteristics of a terminated two port network are: a11 = 0.5m a12 = 10Ω a21 = 1μS a22 = ‐30m Vg = 50∠0° mV Zg = 100 Ω ZL = 5kΩ a) Calculate the average power delivered to the load resistor b) Calculate the load resistance for maximum average power c) Calculate the maximum average power delivered to the resistor in b) Conversions from a to z parameters (Δa = a11a22 – a12a21)

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎦

⎤⎢⎣

⎡

21

22

21

2121

11

2221

1211

aa

a1

aa

aa

zzzz

Δ



Lecture 10‐2 Cascaded two‐port networks

Reading: 18.4

Objectives:

• Be able to calculate voltage, current, and power for cascaded two‐port networks for a simple resistive circuit

************************************** Synthesizing a large, complex system is usually made easier by first designing subsections of the system. If the subsections are modeled by two‐port circuits, synthesis involves the analysis of interconnected two‐port circuits. Two‐port circuits may be interconnected five basic configurations: a) cascade (port 2 feeds into port 1 of the second network) b) series‐series (port1a is in series with port 1b, port 2a is in series with port 2b) c) parallel‐parallel (port1a is in parallel with port 1b, port 2a is in parallel with port 2b) d) series‐parallel (port 1a is in series with port 1b, port 2a is in parallel with port 2b) e) parallel‐series (port 1a is in parallel with port 1b, port 2a is in series with port 2b) These five basic configurations are shown in the following figure.



Using the corresponding two‐port matrix representations, complicated networks can be analyzed by connecting elementary two ports.

a) cascaded connection: use a‐parameters: a = [a1][a2] b) series‐series connection: use z‐parameters: z = [z1] + [z2] c) parallel‐parallel connection: use y‐parameters: [y] = [y1] + [y2] d) series‐parallel connection: use h‐parameters: [h] = [h1] + [h2] e) parallel‐series connection: use g‐parameters: [g] = [g1] + [g2]

In‐class activity 1 (cascaded connection): Two identical amplifiers are connected in cascade, as shown in the following Figure. Each amplifier is described in terms of its h parameters. The h parameters are:

h11 = 1 kΩ h12 = 0.15m h21 = 100 h22 = 100μS Find the voltage gain V2/Vg

Step 1. Convert the [h] parameters to the [a] parameters, .a11′ = 21hhΔ

− a12′ = 21

11

hh

− a21′ = 21

22

hh

− a22′ = 21h1

−

Step 2. Convert the [a] parameters to the interconnected cascaded network, [a]=[a′][a′′] a11 = a11′a11′′ + a12′a21′′ a12 = a11′a12′′ + a12′a22′′ a21 = a21′a11′′ + a22′a21′′ a22 = a21′a12′′ + a22′a22′′

Step 3. Find the voltage gain,

g2212Lg2111

L

g

2

ZaaZ)Zaa(Z

VV

+++=



In‐class activity 2 (series‐series connection): In the following figure, let y12 = y21 = 0, y11 = 2 mS, and y22 = 10 mS. Find Vo/Vs.

ans: .09375

Step 1. Convert the [y] parameters to [z] parameters, z11′ = yy22Δ

z12′ = yy12Δ

− z21′ = yy21Δ

− z22′=yy11Δ

Δy = y11y22 – y12y21 Step 2. The 100Ω resistor under the [y] two‐port has z11′′ = z12′′ = z21′′ = z22′′ = 100Ω Step 3. Find the [z] parameters for the overall interconnected series network

[z] = [za] +[zb]

Step 4. Find the voltage gain, ( )( ) 2112L22g11

L21

s

o

zzZzZzZz

VV

−++=



In‐class activity 3 (parallel‐parallel connection): Obtain the h parameters for the following network.

ans: 0.167Ω, 0.5, 0.5, 4.5S Step 1. Obtain the y parameters for the top network [ya] Step 2. Obtain the y parameters for the bottom network [yb] Step 3. Find the y parameters for the overall interconnected parallel network [y] = [ya] +[yb] Step 4. Convert the y parameters to h parameters h11 = 1/y11 h12 = ‐y12/y11 h21 = y21/y11 h22 = Δy/y11



Excerpts from Fundamentals of Engineering Supplied‐Reference Handbook, 8th Edition Page 1 of 2

FORMULAS Ohms Law V = RI Current i(t) = dq/dt Voltage v(t) = dw/dq Resistors in Series Req = R1 + R2 +

… + Rn Resistors in Parallel

1 1 1 1

Special case for 2 parallel resistors

Power Absorbed by a Resistive Element

Kirchhoff’s Laws KVL: ∑ ∑

KCL: ∑ ∑ Thevenin/Norton Equivalent Voc = Vth = IscRth Isc = IN = Voc/Rth ZL = Zth* for maximum power transfer Voltage divider

VR2 =

Current divider

IR2 = ||

Operational Amplifiers

1

Wye‐Delta Equivalents

R1 =

R2 =

R3 =

Ra =

Rb =

Rc =

Two‐Port Relationships

V1 = z11I1 + z12I2 V2 = z21I1 + z22I2 I1 = y11V1 + y12V2 I2 = y21V1 + y22V2

V1 = a11V2 ‐ a12I2 I1 = a21V2 ‐ a22I2 V2 = b11V1 ‐ b12I1 I2 = b21V1 ‐ b22I1 V1 = h11I1 + h12V2 I2 = h21I1 + h22V2

V2 = g21V1 + g22I2 I1 = g11V1 + g12I2



Excerpts from Fundamentals of Engineering Supplied‐Reference Handbook, 8th Edition Page 2 of 2

y to z parameters z to y parameters

z11 = yy22

Δ y11 =

zz22

Δ

z12 = yy12

Δ−

y12 = zz12

Δ−

z21 = yy21

Δ−

y21 = zz21

Δ−

z22 = yy11

Δ y22 =

zz11

Δ

Δy = y11y22 – y12y21 Δz = z11z22 – z12z21 Terminated Two‐Ports

z parameters y parameters input impedance/ admittance

Zin/Yin

L22

211211 Zz

zzz

+−

L22

L211211 Zy1

Zyyy+

−

output current I2

( )( ) 2112L22g11

g21

zzZzZzVz

−++−

Lgg11L22

g21

ZyZZyZy1Vy

Δ+++

Thevenin voltage Vth ( ) g

g11

21 VZz

z+

( ) gg22

21 VyZy

yΔ+

−

Thevenin impedance

Zth

g11

211222 Zz

zzz+

− g22

g11

yZyZy1

Δ++

current gain I2/I1

L22

21

Zzz+

−

L11

21

yZyyΔ+

voltage gain V2/V1 zZz

Zz

L11

L21

Δ+

L22

L21

Zy1Zy

+−

voltage gain V2/Vg ( )( ) 2112L22g11

L21

zzZzZzZz

−++ ( )( )L22g11Lg2112

L21

Zy1Zy1ZZyyZy

++−

determinant of parameter matrix

Δz = z11z22 – z12z21

Δy = y11y22 – y12y21

Documents

Unit 1 Physics Foundation, Circuit Elements, KVL & KC Unit 2