Transcript No Slide Title
CIRCUITS 1
DEVELOP TOOLS FOR THE ANALYSIS AND DESIGN OF BASIC LINEAR ELECTRIC CIRCUITS
A FEW WORDS ABOUT ANALYSIS USING MATHEMATICAL MODELS
BASIC STRATEGY USED IN ANALYSIS
MATHEMATICAL ANALYSIS DEVELOP A SET OF MATHEMATICAL EQUATIONS 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 TECHNIQUES TO DEVELOP MATHEMATICAL MODELS FOR ELECTRIC CIRCUITS THE MATHEMATICS CLASSES - LINEAR ALGEBRA, DIFFERENTIAL EQUATIONS- PROVIDE THE TOOLS TO SOLVE THE MATHEMATICAL MODELS FOR THE FIRST PART WE WILL BE EXPECTED TO SOLVE SYSTEMS OF ALGEBRAIC EQUATIONS 12
V
1 4
V
1 2
V
1 9
V
16 4 2
V V
2 2 4
V
3
V
3 6
V
3 8 0 20 THE MODELS THAT WILL BE DEVELOPED HAVE NICE MATHEMATICAL PROPERTIES.
IN PARTICULAR THEY WILL BE LINEAR WHICH MEANS THAT THEY SATISFY THE PRINCIPLE OF SUPERPOSITION
T
(
)
1
T u
1
Principle of Superposition
y
Tu
Model
2 2
)
LATER THE MODELS WILL BE DIFFERENTIAL EQUATIONS OF THE FORM 3
dy dt
y
f d
2
y dt
2 4
dy dt
8
y
3
df dt
4
f
ELECTRIC CIRCUIT IS AN INTERCONNECTION OF ELECTRICAL COMPONENTS 2 TERMINALS COMPONENT
a b
characterized by the current through it and the voltage difference betweeb terminals NODE NODE
v S
+ -
L R
1
R
2
v O
TYPICAL LINEAR CIRCUIT
C
LOW DISTORTION POWER AMPLIFIER
LEARNING GOALS
BASIC CONCEPTS
•
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
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SI DERIVED BASIC ELECTRICAL UNITS
ONE AMPERE OF CURRENT CARRIES ONE COULOMB OF CHARGE EVERY SECOND.
A
C
s
1 COULOMB 6.28
10 18 (e) (e) IS THE CHARGE OF ONE ELECTRON VOLT IS A MEASURE OF ENERGY PER CHARGE. TWO POINTS HAVE A VOLTAGE DIFFERENCE OF ONE VOLT IF ONE COULOMB OF CHARGE GAINS ONE JOULE OF CHARGE WHEN IT IS MOVED FROM ONE POINT TO THE OTHER.
V
J C
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 FORCE TO DRIVE THROUGH ONE AMPERE OF CURRENT
V A
IT IS REQUIRED ONE WATT OF POWER TO DRIVE ONE AMPER OF CURRENT AGAINST AN ELECTROMOTIVE DIFFERENCE OF ONE VOLTS
W
V
A
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.
q
(
t
) What is the meaning of a negative value for
q(t)?
PROBLEM SOLVING TIP IF THE CHARGE IS GIVEN DETERMINE THE CURRENT BY DIFFERENTIATION IF THE CURRENT IS KNOWN DETERMINE THE CHARGE BY INTEGRATION A PHYSICAL ANALOGY THAT HELPS VISUALIZE ELECTRIC CURRENTS IS THAT OF WATER FLOW. CHARGES ARE VISUALIZED AS WATER PARTICLES
EXAMPLE
q
(
t
)
q
(
t
) 4 10 3 sin( 120
t
)[
C
]
i
(
t
) 4 10 3 120 cos( 120
t
) [
A
]
i
(
t
) 0 .
480 cos( 120
t
) [
mA
] EXAMPLE
i
(
t
) 0
e
2
t mA t
0
t
0 FIND THE CHARGE THAT PASSES DURING IN THE INTERVAL 0 q 0 1 e 2 x dx 1 2 e 2 x 1 0 1 2 e 2 ( 1 2 e 0 ) q 1 2 ( 1 e 2 ) Units? FIND THE CHARGE AS A FUNCTION OF TIME q ( t ) t i ( x ) dx t e 2 x dx t 0 q ( t ) 0 t 0 q ( t ) 0 t e 2 x dx And the units for the charge?... 1 2 ( 1 e 2 t ) DETERMINE THE CURRENT Here we are given the charge flow as function of time. m 10 10 12 2 10 3 10 0 10 12 C s 10 10 9 ( C / s ) To determine current we must take derivatives. PAY ATTENTION TO UNITS IT IS ABSOLUTELY NECESSARY TO INDICATE THE DIRECTION OF MOVEMENT OF CHARGED PARTICLES. THE UNIVERSALLY ACCEPTED CONVENTION IN ELECTRICAL ENGINEERING IS THAT CURRENT IS FLOW OF POSITIVE CHARGES. AND WE INDICATE THE DIRECTION OF FLOW FOR POSITIVE CHARGES -THE REFERENCE DIRECTION A POSITIVE VALUE FOR THE CURRENT INDICATES FLOW IN THE DIRECTION OF THE ARROW (THE REFERENCE DIRECTION) A NEGATIVE VALUE FOR THE CURRENT INDICATES FLOW IN THE OPPOSITE DIRECTION THAN THE REFERENCE DIRECTION a THE DOUBLE INDEX NOTATION IF THE INITIAL AND TERMINAL NODE ARE LABELED ONE CAN INDICATE THEM AS SUBINDICES FOR THE CURRENT NAME 5 A b I ab A a 3 A I ab 3 A b a 3 A I ab 3 A b a 3 A b a 3 A b I ba 3 A I ba 3 A POSITIVE CHARGES FLOW LEFT-RIGHT POSITIVE CHARGES FLOW RIGHT-LEFT I ab I ba a I 3 A 2 A b I 2 A I cb I ab 4 A c This example illustrates the various ways in which the current notation can be used ONE DEFINITION FOR VOLT TWO POINTS HAVE A VOLTAGE DIFFERENTIAL OF ONE VOLT IF ONE COULOMB OF CHARGE GAINS (OR LOSES) ONE JOULE OF ENERGY WHEN IT MOVES FROM ONE POINT TO THE OTHER 1 C a b IF THE CHARGE GAINS ENERGY MOVING FROM a TO b THEN b HAS HIGHER VOLTAGE THAN a. IF IT LOSES ENERGY THEN b HAS LOWER VOLTAGE THAN a DIMENSIONALLY VOLT IS A DERIVED UNIT VOLT JOULE COULOMB N m A s 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 V THE + AND - SIGNS DEFINE THE REFERENCE POLARITY IF THE NUMBER V IS POSITIVE POINT A HAS V VOLTS MORE THAN POINT B. IF THE NUMBER V IS NEGATIVE POINT A HAS |V| LESS THAN POINT B POINT A HAS 2V MORE THAN POINT B POINT A HAS 5V LESS THAN POINT B THE TWO-INDEX NOTATION FOR VOLTAGES INSTEAD OF SHOWING THE REFERENCE POLARITY WE AGREE THAT THE FIRST SUBINDEX DENOTES THE POINT WITH POSITIVE REFERENCE POLARITY V AB 2 V WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROM POINT B TO POINT A IN THE CIRCUIT? V W Q W VQ 240 J THE CHARGES MOVE TO A POINT WITH HIGHER VOLTAGE -THEY GAINED (OR ABSORBED) ENERGY V AB 5 V V AB V BA V BA 5 V 5 V THE VOLTAGE DIFFERENCE IS 5V WHICH POINT HAS THE HIGHER VOLTAGE? V AB 5 V EXAMPLE A CAMCODER BATTERY PLATE CLAIMS THAT THE UNIT STORES 2700mAHr AT 7.2V. WHAT IS THE TOTAL CHARGE AND ENERGY STORED? CHARGE THE NOTATION 2700mAHr INDICATES THAT THE UNIT CAN DELIVER 2700mA FOR ONE FULL HOUR Q 2700 10 3 C S 3600 s Hr 1 Hr 9 . 72 10 3 [ C ] TOTAL ENERGY STORED THE CHARGES ARE MOVED THROUGH A 7.2V VOLTAGE DIFFERENTIAL W Q [ C ] V J C 9 . 72 10 3 7 . 2 [ J ] 6 . 998 10 4 [ J ] ENERGY AND POWER 2[C/s] PASS THROUGH 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 THE ELEMENT IS 6[W] IN GENERAL P VI w ( t 2 , t 1 ) t 1 t 2 p ( x ) dx HOW DO WE RECOGNIZE IF AN ELEMENT SUPPLIES OR RECEIVES POWER? PASSIVE SIGN CONVENTION POWER RECEIVED IS POSITIVE WHILE POWER SUPPLIED IS CONSIDERED NEGATIVE THIS IS THE REFERENCE FOR POLARITY a V ab I ab b P V ab I ab IF VOLTAGE AND CURRENT ARE BOTH POSITIVE THE CHARGES MOVE FROM HIGH TO LOW VOLTAGE AND THE COMPONENT RECEIVES ENERGY --IT IS A PASSIVE ELEMENT A CONSEQUENCE OF THIS CONVENTION IS THAT THE REFERENCE DIRECTIONS FOR CURRENT AND VOLTAGE ARE NOT INDEPENDENT -- IF WE ASSUME PASSIVE ELEMENTS GIVEN THE REFERENCE POLARITY V ab a b REFERENCE DIRECTION FOR CURRENT a b I ab IF THE REFERENCE DIRECTION FOR CURRENT IS GIVEN EXAMPLE V ab 2 A a V ab I ab 10 V b THE ELEMENT RECEIVES 20W OF POWER. WHAT IS THE CURRENT? SELECT REFERENCE DIRECTION BASED ON PASSIVE SIGN CONVENTION 20 [ W ] V ab I ab ( 10 V ) I ab I ab 2 [ A ] UNDERSTANDING PASSIVE SIGN CONVENTION We must examine the voltage across the component and the current through it S1 A B V I A’ B’ S2 P S 1 P S 2 V AB I AB V A ' B ' I A ' B ' ON S 1 V AB 0 , I AB 0 ON S 2 V A ' B ' 0 , I A ' B ' 0 ON S2 V A ' B ' 0 , I A ' B ' 0 CHARGES RECEIVE ENERGY. THIS BATTERY SUPPLIES ENERGY CHARGES LOSE ENERGY. THIS BATTERY RECEIVES THE ENERGY WHAT WOULD HAPPEN IF THE CONNECTIONS ARE REVERSED IN ONE OF THE BATTERIES? DETERMINE WHETHER THE ELEMENTS ARE SUPPLYING OR RECEIVING POWER AND HOW MUCH I ab 4 A a V ab 2 V I ab 2 A 2 A P 8 W SUPPLIES POWER b P 4 W RECEIVES POWER WHEN IN DOUBT LABEL THE TERMINALS OF THE COMPONENT b a V ab 2 V 1 1 2 V 12 12 V , I 12 4 A 2 V 12 4 V , I 12 2 A I 8 [ A ] V AB 4 [ V ] 20 [ W ] V AB ( 5 A ) SELECT VOLTAGE REFERENCE POLARITY BASED ON CURRENT REFERENCE DIRECTION 40 [ W ] ( 5 V ) I WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION V 1 20 [ V ] 2 A I 5 [ A ] 40 [ W ] V 1 ( 2 A ) SELECT HERE THE CURRENT REFERENCE DIRECTION BASED ON VOLTAGE REFERENCE POLARITY 50 [ W ] ( 10 [ V ]) I COMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENT 2 A P 1 ( 6 V )( 2 A ) 6 V 1 P 3 V + 3 2 A ( 24 V )( 2 A ) ( 24 V )( 2 A ) 2 P1 = 12W P2 = 36W P3 = -48W V P 2 ( 18 V )( 2 A ) IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT PASSIVE ELEMENTS INDEPENDENT SOURCES CURRENT DEPENDENT SOURCES VOLTAGE DEPENDENT SOURCES UNITS FOR , g , r , ? EXERCISES WITH DEPENDENT SOURCES FIND V O V O 40 [ V ] FIND I O I O 50 mA DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES 40 [ V ] P ( 40 [ V ])( 2 [ A ]) 80 [ W ] TAKE VOLTAGE POLARITY REFERENCE P ( 10 [ V ])( 4 4 [ A ]) 160 [ W ] TAKE CURRENT REFERENCE DIRECTION POWER ABSORBED OR SUPPLIED BY EACH ELEMENT USE POWER BALANCE TO COMPUTE Io 12 W ( 12 )( 9 ) ( 6 )( I O ) ( 10 )( 3 ) ( 4 )( 8 ) ( 8 2 )( 11 ) P 1 ( 12 V )( 4 A ) 48 [ W ] P 2 ( 24 V )( 2 A ) 48 [ W ] P 3 ( 28 V )( 2 A ) 56 [ W ] P DS P 36 V ( 1 I x )( 2 A ) ( 4 V )( 2 A ) 8 [ W ] ( 36 V )( 4 A ) 144 [ W ] NOTICE THE POWER BALANCE POWER BALANCE I O 1 [ A ]Charge(pC) 30 20 10
10 1 2 3 4 5 6 Time(ms)
40 30 20 ) 10
10 20 1 2 3 4 5 6 Time(ms)
CONVENTION FOR CURRENTS
5
CONVENTIONS FOR VOLTAGES
Voltage(V) Current A - A' positive positive negative negative positive negative positive negative S1 S2 supplies receives
receives supplies
receives supplies
supplies receives
24
18
CIRCUIT ELEMENTS