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

206420164

84912

321

321

321

=+−−=++−=−−

VVVVVV

VVV

LATER THE MODELS WILL BE DIFFERENTIAL EQUATIONS OF THE FORM

fdtdfy

dtdy

dtyd

fydtdy

4384

3

2

2+=++

=+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

1 1 2 2 1 1 2 2

Model

Principle of Superposition( ) ( ) ( )

y Tu

T u u T u T uα α α α

=

+ = +

a b2 TERMINALS COMPONENT

characterized by thecurrent through it andthe voltage differencebetween terminals

NODE

NODE

ELECTRIC CIRCUIT IS AN INTERCONNECTION OF ELECTRICAL COMPONENTS

LOW DISTORTION POWER AMPLIFIER

+-

L

C

1R

2RSv −

+

Ov

TYPICAL LINEAR CIRCUIT

The concept of node is extremely important. We must learn to identify a node in any shape or form

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

Information at the foundation ofmodern science and technologyfrom the Physics Laboratory of NIST

Detailed contents

Values of the constants and related informationSearchable bibliography on the constants

In-depth information on the SI, the modernmetric system

Guidelines for the expressionof uncertainty in measurement

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 CHARGE GAINS 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 FORCE TO DRIVE THROUGH ONE AMPERE OF CURRENT

AV

=Ω

IT IS REQUIRED ONE WATT OF POWER TO DRIVE ONE AMPER OF CURRENT AGAINST AN ELECTROMOTIVE 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 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

+

++

)(tq

+

EXAMPLE

])[120sin(104)( 3 Cttq π−×=

=)(ti )120cos(120104 3 tππ×× − ][A

][)120cos(480.0)( mAtti ππ=

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 the charge flow as function of 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 TO UNITS

CONVENTION FOR CURRENTS

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 b

a

a

ab

b

b

A3

A3− A3

A3−

THE DOUBLE INDEX NOTATION

IF THE INITIAL AND TERMINAL NODE ARE LABELED ONE CAN INDICATE THEM AS SUBINDICES FOR THE CURRENT NAME

a bA5 AIab 5=

AIab 3=

AIba 3−=

AIab 3−=

AIba 3=

POSITIVE CHARGES FLOW LEFT-RIGHT

POSITIVE CHARGES FLOW RIGHT-LEFT

baab II −=

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

This example illustrates the various ways in 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 OF ONE 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 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

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 REFERENCE POLARITY

V 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

VVAB 2=

VVAB 5−= VVBA 5=BAAB VV −=

ENERGY VOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…

CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB OR RELEASE ENERGY – THEY MAY TRANSFER ENERGY FROM ONE POINT TO ANOTHER

BASIC FLASHLIGHT Converts energy stored in battery to thermal energy in lamp filament which turns incandescent and glow

The battery supplies energy to charges. Lamp absorbs energy from charges. The net effect is an energy transfer

EQUIVALENT CIRCUIT

Charges gain energy here

Charges supply Energy here

WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROM POINT B TO POINT A IN THE CIRCUIT?

VVAB 2=

JVQWQWV 240==⇒=

THE CHARGES MOVE TO A POINT WITH HIGHER VOLTAGE -THEY GAINED (OR ABSORBED) ENERGY THE CIRCUIT SUPPLIED ENERGY TO THE CHARGES

ENERGY VOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE… CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB OR RELEASE ENERGY

VVAB 5=

−

+V5

WHICH POINT HAS THE HIGHER VOLTAGE?

THE VOLTAGE DIFFERENCE IS 5V

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

][1072.9

13600102700

3

3

C

HrHrs

SCQ

×=

××

×= −

TOTAL ENERGY STORED THE CHARGES ARE MOVED THROUGH A 7.2V VOLTAGE DIFFERENTIAL

][10998.6

][2.71072.9][

4

3

J

JCJVCQW

×=

××=

×=

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]

HOW DO WE RECOGNIZE IF AN ELEMENT SUPPLIES OR RECEIVES POWER?

VIP =IN GENERAL ∫=

2

1

)(),( 12

t

tdxxpttw

PASSIVE SIGN CONVENTION

POWER RECEIVED IS POSITIVE WHILE POWER SUPPLIED 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 CURRENT ARE BOTH POSITIVE THE CHARGES MOVE FROM HIGH TO LOW VOLTAGE AND THE COMPONENT RECEIVES ENERGY --IT IS A PASSIVE ELEMENT

a b

−+ abVGIVEN THE REFERENCE POLARITY

a babI

IF THE REFERENCE DIRECTION FOR CURRENT IS 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 ON PASSIVE 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

S2 S1

We must examine the voltage across the component and 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 REVERSED IN ONE OF THE BATTERIES?

DETERMINE WHETHER THE ELEMENTS ARE SUPPLYING OR RECEIVING POWER AND HOW MUCH

a

b

a

b

WHEN IN DOUBT LABEL THE TERMINALS OF 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 TERMINALS OF THE COMPONENT

SELECT VOLTAGE REFERENCE POLARITY BASED ON CURRENT REFERENCE DIRECTION

−

+

)5(][20 AVW AB ×=−

][4 VVAB −= −

+

IVW ×−= )5(][40

][8 AI −=

SELECT HERE THE CURRENT REFERENCE DIRECTION BASED 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 = 12W P2 = 36W P3 = -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 DEPENDENT SOURCES

CURRENT DEPENDENT SOURCES

?,,, βµ rg FOR UNITS

EXERCISES WITH DEPENDENT SOURCES

OVFIND ][40 VVO = OIFIND mAIO 50=

DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES

][40 V

][80])[2])([40( WAVP −=−=

TAKE VOLTAGE POLARITY REFERENCE TAKE CURRENT REFERENCE DIRECTION

][160])[44])([10( WAVP −=×−=

POWER ABSORBED OR SUPPLIED BY EACH ELEMENT

][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