electric circuit is an interconnection of electrical components low distortion power amplifier...
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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
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
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
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
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
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