CIRCUITS 1DEVELOP 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 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

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

BASIC CONCEPTSLEARNING 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

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

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

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?

1 2 3 4 5 610

102030

Charge(pC)

Time(ms)

Here we are given thecharge flow as functionof time.

)/(10100102

10101010 93

1212

sCsCm

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

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

This example illustrates the various waysin which the current notation can be used

b

a

I

A3

ab

cb

IAIAI

42

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

sAmN

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 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

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

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

VVAB 5

V5

WHICH POINTHAS THE HIGHER 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

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

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

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

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

2A

2abI A

4P W ABSORBS POWER

1

2

1

2

AIVV 4,12 1212 AIVV 2,4 1212

WHEN IN DOUBT LABEL THE TERMINALSOF THE COMPONENT

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 ABDORBED 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(

][1 AIO

POWER BALANCE