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EEL 3041 Sp 2012 Principles of Electrical Engineering. Instructor: Dr. Harold Klee Office:Harris Center 206 Office Hours:tbd Telephone:(407) 823-2270 Email:[email protected] Class Hours:M,W 12:00-13:15 Classroom:HEC 104 - PowerPoint PPT Presentation
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EEL 3041 Sp 2012Principles of Electrical
Engineering• Instructor: Dr. Harold Klee• Office: Harris Center 206• Office Hours: tbd• Telephone: (407) 823-2270• Email: [email protected]• Class Hours: M,W 12:00-13:15• Classroom: HEC 104• Textbook: A Brief Introduction to Circuit Analysis, J.
D. Irwin, Wiley, 2003
• Course Website http://www.eecs.ucf.edu/courses/eel3041/spr2012
Learning Objectives
• Fundamental laws of electrical circuits and circuit analysis, fundamentals of electronics and power systems:– 1 Course– 2 Basic Variables: voltage and current– 3 Fundamental Laws: Ohm’s, KCL, KVL
Topics/Chapters
• Basic concepts of electrical engineering
• Resistive circuit analysis
• Transient circuit analysis
• AC steady-state analysis
• Steady state power analysis
• Variable frequency network
• Electronics fundamentals
• Unannounced quizzes will be given throughout the semester and will add to your exam scores and therefore affect your final grade. NO individual makeup opportunity will be given for any missing quiz.
• Homework will be assigned on a weekly basis but will not be collected or graded. It is important that you do the homework. Some of the quiz questions will be similar or even identical to homework problems. A list of all home work problems is included in the syllabus.
Exam 1 30% Exam 2 30% Final Exam (comprehensive) 40% Quizzes
Final average will be determined fromAve = 0.25×(Exam1+Exam2+Sumof Quiz scores) + 0.4×Final
Grade System
• A 90-100
• B 80-89
• C 70-79
• D 60-69
• F 0-59
CIRCUITS 1
DEVELOP TOOLS FOR THE ANALYSIS AND DESIGN OF BASIC LINEAR ELECTRIC CIRCUITS
A FEW WORDS ABOUT ANALYSISUSING MATHEMATICAL MODELS
BASIC STRATEGY USED IN ANALYSIS
MATHEMATICAL ANALYSIS
DEVELOP A SET OF MATHEMATICALEQUATIONS THAT REPRESENT THE CIRCUIT - A MATHEMATICAL MODEL -
LEARN HOW TO SOLVE THE MODEL TO DETERMINE HOW THE CIRCUIT WILL BEHAVE IN A GIVEN SITUATION
THIS COURSE TEACHES THE BASIC TECHNIQUESTO DEVELOP MATHEMATICAL MODELS FORELECTRIC CIRCUITS
THE MATHEMATICS CLASSES - LINEAR ALGEBRA,DIFFERENTIAL EQUATIONS- PROVIDE THE TOOLSTO SOLVE THE MATHEMATICAL MODELS
THE MODELS THAT WILL BE DEVELOPED HAVENICE MATHEMATICAL PROPERTIES.IN PARTICULAR THEY WILL BE LINEAR WHICHMEANS THAT THEY SATISFY THE PRINCIPLE OFSUPERPOSITION
FOR THE FIRST PART WE WILL BE EXPECTEDTO SOLVE SYSTEMS OF ALGEBRAIC EQUATIONS
20642
0164
84912
321
321
321
VVV
VVV
VVV
LATER THE MODELS WILL BE DIFFERENTIALEQUATIONS OF THE FORM
fdt
dfy
dt
dy
dt
yd
fydt
dy
4384
3
2
2
LOW DISTORTION POWER AMPLIFIER
a b2 TERM INALS C O M PO NENT
characterized by thecurrent through it andthe vo ltage diff erencebetweeb term inals
NO DE
NO DE
ELECTRIC CIRCUIT IS AN INTERCONNECTION OF ELECTRICAL COMPONENTS
+-
L
C
1R
2R
Sv
Ov
TYPICAL LINEARCIRCUIT
BASIC CONCEPTS
LEARNING GOALS
•System of Units: The SI standard system; prefixes
•Basic Quantities: Charge, current, voltage, power and energy
•Circuit Elements: Active and Passive
http://physics.nist.gov/cuu/index.html
In fo rm a tio n a t th e fo u n d a tio n o fm o d e rn s c ie n c e a n d te c h n o lo g yfro m th e P h y s ic s L a b o ra to ry o f N IS T
D e ta ile d c o n te n ts
V a lu e s o f th e c o n s ta n ts a n d re la te d in fo rm a tio nS e a rc h a b le b ib lio g ra p h y o n th e c o n s ta n ts
In -d e p th in fo rm a tio n o n th e S I, th e m o d e rnm e tric s y s te m
G u id e lin e s fo r th e e x p re s s io no f u n c e rta in ty in m e a s u re m e n t
A b o u t th is referen ce. F eed b ack.
P riv a c y S ta te m e n t / S e c u rity N o tic e - N IS T D is c la im e r
SI DERIVED BASIC ELECTRICAL UNITS
ONE AMPERE OF CURRENT CARRIES ONE COULOMB OF CHARGE EVERY SECOND.
ELECTRON ONE OF CHARGE THE IS (e)
(e) 106.28 COULOMB 1 18
VOLT IS A MEASURE OF ENERGY PER CHARGE. TWO POINTS HAVE A VOLTAGE DIFFERENCE OF ONE VOLT IF ONE COULOMB OF CHARGEGAINS (OR LOSES) ONE JOULE OF ENERGY WHEN IT IS MOVED FROM ONE POINT TO THE OTHER.
1 ( ) 1 ( ) per ( )Volt V Joule J Coulomb C
OHM IS A MEASURE OF THE RESISTANCE TO THE FLOW OF CHARGE.THERE IS ONE OHM OF RESISTANCE BETWEEN TWO POINTS IF ONE VOLT OF ELECTROMOTIVE FORCE IS REQUIRED TO MOVE ONE AMPERE OF CURRENT BETWEEN THE TWO POINTS
1 ( ) 1 ( ) per ( )Ohm Volt V Ampere A
IT REQUIRES ONE WATT OF POWER TO MOVE ONE AMPERE OF CURRENT BETWEENTWO POINTS WITH A VOLTAGE DIFFERENCE OF ONE VOLT
1 (W) 1 ( ) 1 ( )Watt Volt V Ampere A
1 A 1 coulomb per sec
CURRENT AND VOLTAGE RANGES
Strictly speaking current is a basic quantity and charge is derived. However, physically the electric current is created by a movement of charged particles.
)(tq
What is the meaning of a negative value for q(t)?
PROBLEM SOLVING TIPPROBLEM SOLVING TIP IF THE CHARGE IS GIVEN DETERMINE THE CURRENT BYDIFFERENTIATIONIF THE CURRENT IS KNOWN DETERMINE THE CHARGE BYINTEGRATION
A PHYSICAL ANALOGY THAT HELPS VISUALIZE ELECTRICCURRENTS IS THAT OF WATER FLOW. CHARGES ARE VISUALIZED AS WATER PARTICLES
)(tq
EXAMPLE
])[120sin(104)( 3 Cttq
)(ti )120cos(120104 3 t ][A
][)120cos(480.0)( mAtti
EXAMPLE
0
00)( 2 tmAe
tti t
FIND THE CHARGE THAT PASSES DURING IN THE INTERVAL 0<t<1
1
0
2 dxeq x)
2
1(
2
1
2
1 021
0
2 eee x
FIND THE CHARGE AS A FUNCTION OF TIME
t t
xdxedxxitq 2)()(
0)(0 tqt
t
tx edxetqt0
22 )1(2
1)(0
And the units for the charge?...
)1(2
1 2 eq Units?
1 2 3 4 5 610
102030
Charge(pC)
Time(ms)
Here we are given thecharge flow as functionof time.
)/(10100102
10101010 93
1212
sCs
Cm
1 2 3 4 5 610
102030
Time(ms)
) Current(nA
40
20
DETERMINE THE CURRENT
To determine current we must take derivatives.PAY ATTENTION TOUNITS
CONVENTION FOR CURRENTS
IT IS ABSOLUTELY NECESSARY TO INDICATETHE DIRECTION OF MOVEMENT OF CHARGED PARTICLES.
THE UNIVERSALLY ACCEPTED CONVENTION INELECTRICAL ENGINEERING IS THAT CURRENT ISFLOW OF POSITIVE CHARGES.AND WE INDICATE THE DIRECTION OF FLOWFOR POSITIVE CHARGES -THE REFERENCE DIRECTION-
A POSITIVE VALUE FORTHE CURRENT INDICATESFLOW IN THE DIRECTIONOF THE ARROW (THEREFERENCE DIRECTION)
A NEGATIVE VALUE FORTHE CURRENT INDICATESFLOW IN THE OPPOSITEDIRECTION THAN THE REFERENCE DIRECTION
a b
a
a
ab
b
b
A3
A3 A3
A3
THE DOUBLE INDEX NOTATION
IF THE INITIAL AND TERMINAL NODE ARELABELED ONE CAN INDICATE THEM AS SUBINDICES FOR THE CURRENT NAME
a bA5 AIab 5
AIab 3
AIba 3
AIab 3
AIba 3
POSITIVE CHARGESFLOW LEFT-RIGHT
POSITIVE CHARGESFLOW RIGHT-LEFT
baab II
This example illustrates the various waysin which the current notation can be used
b
a
I
A3
ab
cb
I
AI
AI
4
2
A2
c
CONVENTIONS FOR VOLTAGES
ONE DEFINITION FOR VOLT TWO POINTS HAVE A VOLTAGE DIFFERENTIAL OFONE VOLT IF ONE COULOMB OF CHARGE GAINS (OR LOSES) ONE JOULE OF ENERGY WHEN IT MOVES FROM ONE POINT TO THE OTHER
a
b
C1
IF THE CHARGE GAINSENERGY MOVING FROMa TO b THEN b HAS HIGHERVOLTAGE THAN a.IF IT LOSES ENERGY THENb HAS LOWER VOLTAGETHAN a
DIMENSIONALLY VOLT IS A DERIVED UNIT
sA
mN
COULOMB
JOULEVOLT
VOLTAGE IS ALWAYS MEASURED IN A RELATIVE FORM AS THE VOLTAGE DIFFERENCE BETWEEN TWO POINTS
IT IS ESSENTIAL THAT OUR NOTATION ALLOWS US TO DETERMINE WHICH POINT HAS THE HIGHER VOLTAGE
THE + AND - SIGNS DEFINE THE REFERENCEPOLARITY
V IF THE NUMBER V IS POSITIVE POINT A HAS VVOLTS MORE THAN POINT B.IF THE NUMBER V IS NEGATIVE POINT A HAS|V| LESS THAN POINT B
POINT A HAS 2V MORETHAN POINT B
POINT A HAS 5V LESSTHAN POINT B
THE TWO-INDEX NOTATION FOR VOLTAGES
INSTEAD OF SHOWING THE REFERENCE POLARITY WE AGREE THAT THE FIRST SUBINDEX DENOTESTHE POINT WITH POSITIVE REFERENCE POLARITY
VVAB 2
VVAB 5 VVBA 5BAAB VV
WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROMPOINT B TO POINT A IN THE CIRCUIT?
2 120 240WV W V Q V C JQ
THE CHARGES MOVE TO A POINT WITH HIGHERVOLTAGE -THEY GAINED (OR ABSORBED) ENERGY
VVAB 5
V5
WHICH POINT HAS THEHIGHER VOLTAGE?
THE VOLTAGEDIFFERENCE IS 5V
EXAMPLEA CAMCODER BATTERY PLATE CLAIMS THATTHE UNIT STORES 2700mAHr AT 7.2V.WHAT IS THE TOTAL CHARGE AND ENERGYSTORED?
CHARGETHE NOTATION 2700mAHr INDICATES THATTHE UNIT CAN DELIVER 2700mA FOR ONE FULL HOUR
][1072.9
13600102700
3
3
C
HrHr
s
S
CQ
TOTAL ENERGY STOREDTHE CHARGES ARE MOVED THROUGH A 7.2VVOLTAGE DIFFERENTIAL
][10998.6
][2.71072.9][
4
3
J
JC
JVCQW
ENERGY AND POWER
2[C/s] PASSTHROUGH THE ELEMENT
EACH COULOMB OF CHARGE LOSES 3[J]OR SUPPLIES 3[J] OF ENERGY TO THE ELEMENT
THE ELEMENT RECEIVES ENERGY AT A RATE OF 6[J/s]
THE ELECTRIC POWER RECEIVED BY THEELEMENT IS 6[W]
HOW DO WE RECOGNIZE IF AN ELEMENTSUPPLIES OR RECEIVES POWER?
VIP IN GENERAL
2
1
)(),( 12
t
t
dxxpttw
PASSIVE SIGN CONVENTION
POWER RECEIVED IS POSITIVE WHILE POWERSUPPLIED IS CONSIDERED NEGATIVE
A CONSEQUENCE OF THIS CONVENTION IS THAT
THE REFERENCE DIRECTIONS FOR CURRENT AND
VOLTAGE ARE NOT INDEPENDENT -- IF WE
ASSUME PASSIVE ELEMENTS
a b
abV
abI
ababIVP
IF VOLTAGE AND CURRENTARE BOTH POSITIVE THECHARGES MOVE FROM HIGH TO LOW VOLTAGE AND THE COMPONENTRECEIVES ENERGY --IT ISA PASSIVE ELEMENT
a b
abVGIVEN THE REFERENCE POLARITY
a babI
IF THE REFERENCE DIRECTION FOR CURRENTIS GIVEN
THIS IS THE REFERENCE FOR POLARITY
REFERENCE DIRECTION FOR CURRENT
a b
abV
VVab 10
EXAMPLE
THE ELEMENT RECEIVES 20W OF POWER.WHAT IS THE CURRENT?
abI
SELECT REFERENCE DIRECTION BASED ONPASSIVE SIGN CONVENTION
ababab IVIVW )10(][20
][2 AIab
A2
Voltage(V) Current A - A' S1 S2positive positive supplies receivespositive negative receives suppliesnegative positive receives suppliesnegative negative supplies receives
S2S1
We must examine the voltage across the componentand the current through it
0,0 ABAB IV1S ON
0,0 '''' BABA IV2S ON
A A’
B B’
''''2
1
BABAS
ABABS
IVP
IVP
UNDERSTANDING PASSIVE SIGN CONVENTION
0,0 '''' BABA IV
S2 ON
V
I
CHARGES RECEIVE ENERGY.THIS BATTERY SUPPLIES ENERGY
CHARGES LOSE ENERGY.THIS BATTERY RECEIVES THE ENERGY
WHAT WOULD HAPPEN IF THE CONNECTIONS ARE REVERSEDIN ONE OF THE BATTERIES?
DETERMINE WHETHER THE ELEMENTS ARE SUPPLYING OR RECEIVING POWERAND HOW MUCH
a
b
a
b
WHEN IN DOUBT LABEL THE TERMINALSOF THE COMPONENT
AIab 4
VVab 2
WP 8 SUPPLIES POWER
VVab 2
A2
AIab 2
WP 4 RECEIVES POWER
1
2
1
2
AIVV 4,12 1212 AIVV 2,4 1212
SELECT VOLTAGE REFERENCE POLARITYBASED ON CURRENT REFERENCE DIRECTION
)5(][20 AVW AB
][4VVAB
IVW )5(][40
][8 AI
SELECT HERE THE CURRENT REFERENCE DIRECTIONBASED ON VOLTAGE REFERENCE POLARITY
A2
)2(][40 1 AVW
][201 VV
IVW ])[10(][50
][5 AI
WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION
+-
V24
V6
V18
A2
A2
123
P1 = 12WP2 = 36WP3 = -48W
)2)(6(1 AVP
)2)(18(2 AVP
)2)(24()2)(24(3 AVAVP
IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT
COMPUTE POWER ABS ORBED OR SUPPLIED BY EACH ELEMENT
CIRCUIT ELEMENTS
PASSIVE ELEMENTS
INDEPENDENT SOURCES
VOLTAGE DEPENDENTSOURCES
CURRENTDEPENDENTSOURCES
?,,, rg FOR UNITS
EXERCISES WITH DEPENDENT SOURCES
OVFIND ][40VVO OIFIND mAIO 50
DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES
][40V
][80])[2])([40( WAVP
TAKE VOLTAGE POLARITY REFERENCE TAKE CURRENT REFERENCE DIRECTION
][160])[44])([10( WAVP
POWER ABSORBED OR SUPPLIED BY EACHELEMENT
][48)4)(12(1 WAVP
][48)2)(24(2 WAVP
][56)2)(28(3 WAVP
][8)2)(4()2)(1( WAVAIP xDS
][144)4)(36(36 WAVP V
NOTICE THE POWER BALANCE
USE POWER BALANCE TO COMPUTE Io
W12
))(6( OI)9)(12(
)3)(10(
)8)(4( )11)(28(
POWER BALANCE
][1 AIO