CIRCUITS 1
description
Transcript of CIRCUITS 1
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CIRCUITS 1DEVELOP TOOLS FOR THE ANALYSIS AND DESIGN OF BASIC LINEAR ELECTRIC CIRCUITS
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A FEW WORDS ABOUT ANALYSISUSING MATHEMATICAL MODELS
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BASIC STRATEGY USED IN ANALYSIS
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MATHEMATICAL ANALYSIS
DEVELOP A SET OF MATHEMATICALEQUATIONS THAT REPRESENT THE CIRCUIT - A MATEMATICAL 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
FOR THE FIRST PART WE WILL BE EXPECTEDTO SOLVE SYSTEMS OF ALGEBRAIC EQUATIONS
206420164
84912
321
321
321
VVVVVVVVV
LATER THE MODELS WILL BE DIFFERENTIALEQUATIONS OF THE FORM
fdtdfy
dtdy
dtyd
fydtdy
4384
3
2
2
THE MODELS THAT WILL BE DEVELOPED HAVENICE MATHEMATICAL PROPERTIES.IN PARTICULAR THEY WILL BE LINEAR WHICHMEANS THAT THEY SATISFY THE PRINCIPLE OFSUPERPOSITION
1 1 2 2 1 1 2 2
Model
Principle of Superposition( ) ( ) ( )
y Tu
T u u T u T u
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a b2 T ERM INALS C O M PO NENT
characterized by thecurrent through it andthe vo ltage diff erencebetween term inals
NO DE
NO DE
ELECTRIC CIRCUIT IS AN INTERCONNECTION OF ELECTRICAL COMPONENTS
LOW DISTORTION POWER AMPLIFIER
+-
L
C
1R
2R
Sv
Ov
TYPICAL LINEARCIRCUIT
The concept of node is extremelyimportant. We must learn to identify a node in any shape or form
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BASIC CONCEPTSLEARNING GOALS
• System of Units: The SI standard system; prefixes
• Basic Quantities: Charge, current, voltage, power and energy
• Circuit Elements: Active and Passive
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http://physics.nist.gov/cuu/index.html
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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
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SI DERIVED BASIC ELECTRICAL UNITS
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ONE AMPERE OF CURRENT CARRIES ONE COULOMB OF CHARGE EVERY SECOND.
ELECTRON ONE OF CHARGE THE IS (e)(e) 106.28 COULOMB 1 18
sCA
VOLT IS A MEASURE OF ENERGY PER CHARGE. TWO POINTS HAVE A VOLTAGE DIFFERENCE OF ONE VOLT IF ONE COULOMB OF CHARGEGAINS ONE JOULE OF CHARGE WHEN IT IS MOVED FROM ONE POINT TO THE OTHER.
CJV
OHM IS A MEASURE OF THE RESISTANCE TO THE FLOW OF CHARGE.THERE IS ONE OHM OF RESISTENCE IF IT IS REQUIRED ONE VOLT OF ELECTROMOTIVE FORCETO DRIVE THROUGH ONE AMPERE OF CURRENT
AV
IT IS REQUIRED ONE WATT OF POWER TO DRIVE ONE AMPER OF CURRENT AGAINST ANELECTROMOTIVE DIFFERENCE OF ONE VOLTS
AVW
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CURRENT AND VOLTAGE RANGES
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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 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
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)(tq
EXAMPLE
])[120sin(104)( 3 Cttq
)(ti )120cos(120104 3 t ][A
][)120cos(480.0)( mAtti
EXAMPLE
000
)( 2 tmAet
ti t
FIND THE CHARGE THAT PASSES DURING IN THE INTERVAL 0<t<1
1
0
2 dxeq x )21(
21
21 02
1
0
2 eee x
FIND THE CHARGE AS A FUNCTION OF TIME
t t
xdxedxxitq 2)()(
0)(0 tqt
t
tx edxetqt0
22 )1(21)(0
And the units for the charge?...
)1(21 2 eq Units?
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1 2 3 4 5 610
102030
Charge(pC)
Time(ms)
Here we are given thecharge flow as functionof time.
)/(10100102
10101010 93
1212sC
sCm
1 2 3 4 5 610
102030
Time(ms)
) Current(nA40
20
DETERMINE THE CURRENT
To determine current we must take derivatives.PAY ATTENTION TOUNITS
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CONVENTION FOR CURRENTSIT 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 b
a
a
ab
b
b
A3
A3 A3
A3
THE DOUBLE INDEX NOTATIONIF 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
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
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This example illustrates the various waysin which the current notation can be used
b
a
I
A3
ab
cb
IAIAI
42
A2
c
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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
sAmN
COULOMBJOULEVOLT
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
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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
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THE TWO-INDEX NOTATION FOR VOLTAGESINSTEAD OF SHOWING THE REFERENCE POLARITY WE AGREE THAT THE FIRST SUBINDEX DENOTESTHE POINT WITH POSITIVE REFERENCE POLARITY
VVAB 2
VVAB 5 VVBA 5BAAB VV
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ENERGYVOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…
CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB ORRELEASE ENERGY – THEY MAY TRANSFER ENERGY FROM ONE POINT TO ANOTHER
BASIC FLASHLIGHT Converts energy stored in batteryto thermal energy in lamp filamentwhich turns incandescent and glows
The battery supplies energy to charges.Lamp absorbs energy from charges.The net effect is an energy transfer
EQUIVALENT CIRCUIT
Charges gainenergy here
Charges supplyEnergy here
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WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROMPOINT B TO POINT A IN THE CIRCUIT?
VVAB 2
JVQWQWV 240
THE CHARGES MOVE TO A POINT WITH HIGHERVOLTAGE -THEY GAINED (OR ABSORBED) ENERGYTHE CIRCUIT SUPPLIED ENERGY TO THE CHARGES
ENERGYVOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB ORRELEASE ENERGY
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VVAB 5
V5
WHICH POINTHAS THE HIGHER VOLTAGE?
THE VOLTAGEDIFFERENCE IS 5V
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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
HrHrs
SCQ
TOTAL ENERGY STOREDTHE CHARGES ARE MOVED THROUGH A 7.2VVOLTAGE DIFFERENTIAL
][10998.6
][2.71072.9][
4
3
J
JCJVCQW
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
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PASSIVE SIGN CONVENTIONPOWER RECEIVED IS POSITIVE WHILE POWERSUPPLIED IS CONSIDERED NEGATIVE
A CONSEQUENCE OF THIS CONVENTION IS THATTHE REFERENCE DIRECTIONS FOR CURRENT ANDVOLTAGE 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
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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
IVPIVP
UNDERSTANDING PASSIVE SIGN CONVENTION
0,0 '''' BABA IVS2 ON
V
I
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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?
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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
2A
2abI A
4P W ABSORBS POWER
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1
2
1
2
AIVV 4,12 1212 AIVV 2,4 1212
WHEN IN DOUBT LABEL THE TERMINALSOF THE COMPONENT
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SELECT VOLTAGE REFERENCE POLARITYBASED ON CURRENT REFERENCE DIRECTION
)5(][20 AVW AB
][4VVAB
IVW )5(][40
][8 AI
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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
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+-
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 ABDORBED OR SUPPLIED BY EACH ELEMENT
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CIRCUIT ELEMENTS
PASSIVE ELEMENTS
INDEPENDENT SOURCES
VOLTAGE DEPENDENTSOURCES
CURRENTDEPENDENTSOURCES
?,,, rg FOR UNITS
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EXERCISES WITH DEPENDENT SOURCES
OVFIND ][40VVO OIFIND mAIO 50
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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
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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
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USE POWER BALANCE TO COMPUTE Io
W12
))(6( OI)9)(12(
)3)(10( )8)(4( )11)(28(
][1 AIO
POWER BALANCE