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NODAL AND LOOP ANALYSIS TECHNIQUES. LEARNING GOALS. NODAL ANALYSIS LOOP ANALYSIS. Develop systematic techniques to determine all the voltages and currents in a circuit. CIRCUITS WITH OPERATIONAL AMPLIFIERS. Op-amps are very important devices, widely available, - PowerPoint PPT Presentation
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NODAL AND LOOP ANALYSIS TECHNIQUESLEARNING GOALS
CIRCUITS WITH OPERATIONAL AMPLIFIERS
Develop systematic techniques to determine all the voltagesand currents in a circuit
NODAL ANALYSISLOOP ANALYSIS
Op-amps are very important devices, widely available,that permit the design of very useful circuit…
and they can be modeled by circuits with dependent sources
NODE ANALYSIS
• One of the systematic ways to determine every voltage and current in a circuit
The variables used to describe the circuit will be “Node Voltages” -- The voltages of each node with respect to a pre-selected reference node
IT IS INSTRUCTIVE TO START THE PRESENTATION WITHA RECAP OF A PROBLEM SOLVED BEFORE USING SERIES/PARALLEL RESISTOR COMBINATIONS
COMPUTE ALL THE VOLTAGES AND CURRENTS IN THIS CIRCUIT
SECOND: “BACKTRACK” USING KVL, KCL OHM’S
kVI a62 :SOHM' 13 2KCL: I I I
4
34
*4124
12
IkV
II
b
5
345
*30IkV
III
C
:SOHM' :KCL
k12kk 12||4
k6
kk 6||6
kVI
1212
1 )12(
933
aV
3I
FIRST REDUCE TO A SINGLE LOOP CIRCUIT
THE NODE ANALYSIS PERSPECTIVETHERE ARE FIVE NODES. IF ONE NODE IS SELECTED AS REFERENCE THEN THERE AREFOUR VOLTAGES WITH RESPECTTO THE REFERENCE NODE
THEOREM: IF ALL NODE VOLTAGES WITHRESPECT TO A COMMON REFERENCE NODEARE KNOWN THEN ONE CAN DETERMINE ANY OTHER ELECTRICAL VARIABLE FORTHE CIRCUIT
aS
aS
VVVVVV
1
1 0
ba
ba
VVVVVV
3
3 0
______caVQUESTIONDRILL
REFERENCE
aV bV cV
KVL KVL
WHAT IS THE PATTERN???
KVL
05 bc VVV
ONCE THE VOLTAGES ARE KNOWN THE CURRENTS CANBE COMPUTED USING OHM’SLAW
A GENERAL VIEW
Rv
NmR vvv
5 cbV V V
THE REFERENCE DIRECTION FOR CURRENTS IS IRRELEVANT
'i
Rv
USING THE LEFT-RIGHT REFERENCE DIRECTIONTHE VOLTAGE DROP ACROSS THE RESISTOR MUSTHAVE THE POLARITY SHOWN
Rvvi Nm
LAW SOHM'IF THE CURRENT REFERENCE DIRECTION IS REVERSED ...
'Rv
THE PASSIVE SIGN CONVENTION WILL ASSIGNTHE REVERSE REFERENCE POLARITY TO THE VOLTAGE ACROSS THE RESISTOR
Rvvi mN
'LAW SOHM''ii PASSIVE SIGN CONVENTION RULES!
DEFINING THE REFERENCE NODE IS VITAL
SMEANINGLES IS4V VSTATEMENT THE 1 UNTIL THE REFERENCE POINT IS DEFINEDBY CONVENTION THE GROUND SYMBOLSPECIFIES THE REFERENCE POINT.
V4
ALL NODE VOLTAGES ARE MEASURED WITHRESPECT TO THAT REFERENCE POINT
V2
_____?12 V
12V
THE STRATEGY FOR NODE ANALYSIS 1. IDENTIFY ALL NODES AND SELECT A REFERENCE NODE
2. IDENTIFY KNOWN NODE VOLTAGES
3. AT EACH NODE WITH UNKNOWN VOLTAGE WRITE A KCL EQUATION(e.g.,SUM OF CURRENT LEAVING =0)
0:@ 321 IIIVa
4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES
0369
kVV
kV
kVV baasa AND GET ALGEBRAIC EQUATIONS IN
THE NODE VOLTAGES ...
REFERENCE
SV aV bV cV
0:@ 543 IIIVb
0:@ 65 IIVc
0943
kVV
kV
kVV cbbab
039
kV
kVV cbc
SHORTCUT:SHORTCUT: SKIP WRITING THESE EQUATIONS...
AND PRACTICE WRITING
THESE DIRECTLY
a b c
d
aV
bV
cV
dV
1R 3R
2R
WHEN WRITING A NODE EQUATION...AT EACH NODE ONE CAN CHOSE ARBITRARYDIRECTIONS FOR THE CURRENTS
AND SELECT ANY FORM OF KCL.WHEN THE CURRENTS ARE REPLACED IN TERMSOF THE NODE VOLTAGES THE NODE EQUATIONSTHAT RESULT ARE THE SAME OR EQUIVALENT
00321
321
RVV
RVV
RVVIII cbdbba
0LEAVING CURRENTS
00321
321
RVV
RVV
RVVIII cbdbba
0 NODE INTO CURRENTS
2I3I1I
a b c
d
aV
bV
cV
dV
1R 3R
2R '2I
'3I'
1I
00321
'3
'2
'1
RVV
RVV
RVVIII bcdbab
0LEAVING CURRENTS
00321
'3
'2
'1
RVV
RVV
RVVIII bcdbab
0 NODE INTO CURRENTS
WHEN WRITING THE NODE EQUATIONSWRITE THE EQUATION DIRECTLY IN TERMSOF THE NODE VOLTAGES.BY DEFAULT USE KCL IN THE FORMSUM-OF-CURRENTS-LEAVING = 0
THE REFERENCE DIRECTION FOR THECURRENTS DOES NOT AFFECT THE NODEEQUATION
CIRCUITS WITH ONLY INDEPENDENT SOURCES
HINT:HINT: THE FORMAL MANIPULATION OF EQUATIONS MAY BE SIMPLER IF ONE USES CONDUCTANCES INSTEAD OF RESISTANCES.
@ NODE 1
02
21
1
1 Rvv
RviA SRESISTANCE USING
0)( 21211 vvGvGiA ESCONDUCTANC WITH
REORDERING TERMS
@ NODE 2REORDERING TERMS
THE MODEL FOR THE CIRCUIT IS A SYSTEMOF ALGEBRAIC EQUATIONS
THE MANIPULATION OF SYSTEMS OF ALGEBRAICEQUATIONS CAN BE EFFICIENTLY DONEUSING MATRIX ANALYSIS
LEARNING BY DOING
@ NODE 1 WE VISUALIZE THE CURRENTSLEAVING AND WRITE THE KCL EQUATION
REPEAT THE PROCESS AT NODE 2
03
12
4
122
Rvv
Rvvi
OR VISUALIZE CURRENTS GOING INTO NODE
WRITE THE KCL EQUATIONS
mA15
A
B
C
k8 k8k2 k2
SELECT ASREFERENCE
AV
BVMARK THE NODES (TO INSURE THAT NONE IS MISSING)
ANOTHER EXAMPLE OF WRITING KCL
WRITE KCL AT EACH NODE IN TERMS OFNODE VOLTAGES 015
82@ mA
kV
kVA AA
01528
@ mAkV
kVB BB
A MODEL IS SOLVED BY MANIPULATION OFEQUATIONS AND USING MATRIX ANALYSIS
THE NODE EQUATIONS
kRRkRmAimAi BA
6,124,12
321
THE MODEL
REPLACE VALUES AND SWITCH NOTATIONTO UPPER CASE
NUMERICAL MODEL
USE GAUSSIAN ELIMINATION
ALTERNATIVE MANIPULATION
k12/*
k6/*
2421223
21
21
VVVV
RIGHT HANDSIDE IS VOLTS.COEFFS ARENUMBERS
][122 1 VV ADD EQS
equations) add (and3/*][604 2 VV
LEARNING EXAMPLE
SOLUTION USING MATRIX ALGEBRA
PLACE IN MATRIX FORM
USE MATRIX ANALYSIS TO SHOW SOLUTION
PERFORM THE MATRIX MANIPULATIONS
||)(1
AAAdjA
FOR THE ADJOINT REPLACEEACH ELEMENT BY ITSCOFACTOR
AND DO THE MATRIX ALGEBRA ...
kkkV
6104
310118
332
1
SAMPLE
AN EXAMPLE OF NODE ANALYSIS
1@v
2@ v
3@ v
Rearranging terms ...
2&1 BETWEEN ESCONDUCTANC 3&1 BETWEEN ESCONDUCTANC 3&2 BETWEEN ESCONDUCTANC
NODE TO CONNECTED ESCONDUCTANC
COULD WRITE EQUATIONS BY INSPECTION
WRITING EQUATIONS “BY INSPECTION”FOR CIRCUITS WITH ONLY INDEPENDENTSOURCES THE MATRIX IS ALWAYS SYMMETRIC
THE DIAGONAL ELEMENTS ARE POSITIVE
THE OFF-DIAGONAL ELEMENTS ARE NEGATIVE
Conductances connected to node 1
Conductances between 1 and 2
Conductances between 1 and 3
Conductances between 2 and 3
VALID ONLY FOR CIRCUITSWITHOUT DEPENDENTSOURCES
kVV
kVmAV
1264:@ 211
1
USING KCL
0126
2:@ 1222
kVV
kVmAV
BY “INSPECTION”
mAVk
Vkk
412
112
161
21
mAVkkk
212
161
121
2
LEARNING EXTENSION
mA6
1I
062:@ 11 mAmAIVKCL
2I
3I
036
6:@ 222 k
VkVmAV
mAImAIVV
4212
3
2
2
CURRENTS COULD BE COMPUTED DIRECTLYUSING CURRENT DIVIDER!!
mAmAkk
kI
mAmAkk
kI
4)6(63
6
2)6(63
3
3
2
LEARNING EXTENSION
IN MOST CASES THEREARE SEVERAL DIFFERENTWAYS OF SOLVING APROBLEM
NODE EQS. BY INSPECTION
mAVVk
62021
21
mAVkk
V 631
610 21
kVI
kVI
kVI
3622
32
21
1
CIRCUITS WITH DEPENDENT SOURCES CANNOTBE MODELED BY INSPECTION. THE SYMMETRYIS LOST.
A PROCEDURE FOR MODELINGA PROCEDURE FOR MODELING•WRITE THE NODE EQUATIONS USING DEPENDENT SOURCES AS REGULAR SOURCES.•FOR EACH DEPENDENT SOURCE WE ADD ONE EQUATION EXPRESSING THE CONTROLLING VARIABLE IN TERMS OF THE NODE VOLTAGES
02
21
1
1 Rvv
Rvio
02
12
3
2
Rvv
RviA
MODEL FOR CONTROLLING VARIABLE
3
2
Rvio
NUMERICAL EXAMPLE
mAvkk
vk
vkk
vkk
231
121
61
061
32
61
121
21
21
01112
231
21
v
RRv
RR
REPLACE AND REARRANGE
AivRR
vR
2
321
2
111
k4/*
k6/*
][12302
21
21
VVVVV
ADDING THE EQUATIONS ][125 2 VV
VV524
1
CIRCUITS WITH DEPENDENT SOURCES LEARNING EXAMPLE
LEARNING EXAMPLE: CIRCUIT WITH VOLTAGE-CONTROLLED CURRENT
WRITE NODE EQUATIONS. TREAT DEPENDENTSOURCE AS REGULAR SOURCE
EXPRESS CONTROLLING VARIABLE IN TERMS OFNODE VOLTAGES
FOUR EQUATIONS IN OUR UNKNOWNS. SOLVEUSING FAVORITE TECHNIQUE
OR USE MATRIX ALGEBRA
REPLACE AND REARRANGE
CONTINUE WITH GAUSSIAN ELIMINATION...
USING MATLAB TO SOLVE THE NODE EQUATIONS
]/[2,4,2,4
,2,1
4
321
VAmAimAikR
kRRkR
BA
» R1=1000;R2=2000;R3=2000;R4=4000; %resistances in Ohm» iA=0.002;iB=0.004; %sources in Amps» alpha=2; %gain of dependent source
DEFINE THE COMPONENTS OF THE CIRCUIT
DEFINE THE MATRIX G » G=[(1/R1+1/R2), -1/R1, 0; %first row of the matrix-1/R1, (1/R1+alpha+1/R2), -(alpha+1/R2); %second row0, -1/R2, (1/R2+1/R4)], %third row. End in comma to have the echo
G =
0.0015 -0.0010 0 -0.0010 2.0015 -2.0005 0 -0.0005 0.0008
Entries in a row areseparated by commas (or plain spaces).Rows are separated bysemi colon
DEFINE RIGHT HAND SIDE VECTOR » I=[iA;-iA;iB]; %end in ";" to skip echo
SOLVE LINEAR EQUATION» V=G\I % end with carriage return and get the echo
V = 11.9940 15.9910 15.9940
LEARNING EXTENSION: FIND NODE VOLTAGES
010
410
:@ 2111
kVVmA
kVV
NODE EQUATIONS
010
210
:@ 2122
kVI
kVVV O
CONTROLLING VARIABLE (IN TERMS ON NODEVOLTAGES)
kVIO 10
1
REPLACE
01010
210
010
410
2112
211
kV
kV
kVV
kVVmA
kV
REARRANGE AND MULTIPLY BY 10k
02][402
21
21
VVVVV eqs. add and 2/*
VVVV 16805 11
VVVV 82 21
2
O V VOLTAGETHE FIND
NODE EQUATIONS
063
2 kV
kVmA xx
NOTICE REPLACEMENT OF DEPENDENT SOURCE IN TERMS OF NODE VOLTAGE
012126 k
Vk
VkV OOx
k6/*
k12/*
][4022][4][123VVVV
VVVV
OxO
xx
LEARNING EXTENSION
3 nodes plus the reference. In principle one needs 3 equations...
…but two nodes are connected tothe reference through voltage sources. Hence those node voltages are known!!!
…Only one KCL is necessary
012126
12322
kVV
kVV
kV
][5.1][640)()(2
22
12322
VVVVVVVVV
EQUATIONS THE SOLVING
Hint: Each voltage sourceconnected to the referencenode saves one node equation
One more example ….
CIRCUITS WITH INDEPENDENT VOLTAGE SOURCES
][6][12
3
1
VVVV
THESE ARE THE REMAINING
TWO NODE EQUATIONS
+-
2SI
3SI 1SV
1SI 1R2R
3R
4R
Problem 3.67 (6th Ed) Find V_0 R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2kIs1 =2mA, Is2 = 4mA, Is3 = 4mA,Vs = 12 V
IDENTIFY AND LABEL ALL NODES
WRITE THE NODE EQUATIONS
][12:@ 33 VVVV VS VOLTAGENODE KNOWN
021
][2
0:@
121
4
1
1
2111
kV
kVVmA
RV
RVVIV S
021
121
][4
0:@
42212
2
42
3
32
1
1232
kVV
kV
kVVmA
RVV
RVV
RVVIV S
02
][4][2
0:@
24
2
24214
kVVmAmA
RVVIIV SS
NOW WE LOOK WHAT IS BEING ASKED TO DECIDE THE SOLUTIONSTRATEGY.
210 VVV
1V2V 3V
4V
OV
FOR NEEDED AREONLY OVVV 21,
021
][2 121
kV
kVVmA
021
121
][4 42212
kVV
kV
kVVmA
02
][4][2 24
kVVmAmA
TO SOLVE BY HAND ELIMINATE DENOMINATORS
][423 21 VVV */2k
*/2k ][3252 421 VVVV
*/2k ][442 VVV
Add 2+3 ][3642 21 VVV
][10][404 11 VVVV
][14][564 22 VVVV
FINALLY!! ][4210 VVVV
4324
110152
023
3
2
1
VVV
ALTERNATIVE: USE LINEAR ALGEBRA
(1)
(2)
(3)
So. What happens when sources are connected between two non reference nodes?
][423 21 VVV add and 2/*
We will use this example to introduce the concept of a SUPERNODE
Conventional node analysis requires all currents at a node
@V_1 06
6 1 SIkVmA
@V_2 012
4 2 kVmAIS
2 eqs, 3 unknowns...Panic!! The current through the source is notrelated to the voltage of the sourceMath solution: add one equation
][621 VVV
Efficient solutionEfficient solution: enclose the source, and all elements in parallel, inside a surface.Apply KCL to the surface!!!
04126
6 21 mAkV
kVmA
The source current is interiorto the surface and is not required
We STILL need one more equation
][621 VVV Only 2 eqs in two unknowns!!!
SUPERNODE
THE SUPERNODE TECHNIQUE
SI
k 12 / *
][6
04612621
21
VVV
mAmAkV
kV
(2) (1)
Equations TheALGEBRAIC DETAILS
...by Multiply Eq(1). in rsdenominato Eliminate 1.Solution
][6][242
21
21
VVVVVV
][10][303 11 VVVV 2 Veliminate to equations Add2.
][4][612 VVVV 2 Vcompute to Eq(2) Use 3.
1V 2V
1sI
2sI
1R 2R
3R
SV
][6],[10],[204,10
21
321
mAImAIVVkRkRR
ssS
FIND THE NODE VOLTAGESAND THE POWER SUPPLIEDBY THE VOLTAGE SOURCE
2012 VV
0101010
21 mAkV
kV ][10010/* 21 VVVk
][2021 VVV
][40100][60
21
2
VVVVV
:adding
TO COMPUTE THE POWER SUPPLIED BY VOLTAGE SOURCEWE MUST KNOW THE CURRENT THROUGH IT
VI
kVVmA
kVIV 10
610
211 mA8
BASED ON PASSIVE SIGN CONVENTION THEPOWER IS RECEIVED BY THE SOURCE!!
mWmAVP 160][8][20
WRITE THE NODE EQUATIONS
1@v
:supernode) (leavingKCL :CONSTRAINT
SUPERNODE @Avvv 32
THREE EQUATIONS IN THREE UNKNOWNS
LEARNING EXAMPLE
OI FINDSUPERNODE
1231 VVCONSTRAINT SUPERNODE
123 V
KCL @ SUPERNODE
VVVV 12,6 42 KNOWN NODE VOLTAGES
LEARNING EXAMPLE
VVVV4
6
4
1
SOURCES CONNECTED TO THE
REFERENCE
SUPERNODE
CONSTRAINT EQUATION VVV 1223
KCL @ SUPERNODE
02
)4(212
6 3322
kV
kV
kV
kV
k2/*
VVV 223 32 VVV 1232 add and 3/*VV 385 3
mAkVIO 8.32
3 LAW SOHM'
OIV FOR NEEDED NOT IS 2
LEARNING EXTENSION
+-
+ -
+-1R
2R
3R
4R
5R
6R
7R
Supernodes can be more complex
Identify all nodes, select areference and label nodes
Nodes connected to reference througha voltage source
Voltage sources in between nodes and possible supernodes
EQUATION BOOKKEEPING:KCL@ V_3, KCL@ supernode, 2 constraints equationsand one known node
KCL@V_3 07
3
5
43
4
23
RV
RVV
RVV
KCL @SUPERNODE (Careful not to omit any current)
04
32
5
34
6
4
3
5
2
15
1
12
RVV
RVV
RV
RV
RVV
RVV
CONSTRAINTS DUE TO VOLTAGE SOURCES
11 SVV
252 SVVV
345 SVVV
5 EQUATIONS IN FIVE UNKNOWNS.
supernode
1V
2V 3V
4V
5V
WRITE THE NODE EQUATIONS
CIRCUITS WITH DEPENDENT SOURCESPRESENT NO SIGNIFICANT ADDITIONAL COMPLEXITY. THE DEPENDENT SOURCESARE TREATED AS REGULAR SOURCES
WE MUST ADD ONE EQUATION FOR EACHCONTROLLING VARIABLE
VOLTAGE SOURCE CONNECTED TO REFERENCEVV 31
0263
: 212
xIkV
kVV VKCL@ 2
CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES k
VI x 62
REPLACE
06
263
2212
kV
kV
kVV k6/*
VVVV 602 212
mAkVVIO 1
321
OI FIND
LEARNING EXAMPLE
SUPER NODE WITH DEPENDENT SOURCE
VOLTAGE SOURCE CONNECTED TO REFERENCEVV 63
SUPERNODE CONSTRAINT xVVV 221
KCL AT SUPERNODE
k12/*
062)6(2 2211 VVVV
CONTROLLING VARIABLE IN TERMS OF NODES
2VVx 21 3VV
1833 21 VV 184 1 V
CURRENT CONTROLLED VOLTAGE SOURCE
CONSTRAINT DUE TO SOURCE xkIVV 212
KCL AT SUPERNODE 02
22
4 21 kVmA
kVmA
CONTROLLING VARIABLE IN TERMS OF NODES
kVI x 2
1121 22 VVkIV x
][421 VVV 02 21 VV
add and 2/*
][83 2 VV
mAkVIO 3
42
2
k2 k3
k6k2xaI1000
xISV
An example with dependent sources
2 nodes are connected to the reference through voltage sources
EXPRESS CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES
kVI X
X 2
‘a’ has units of [Volt/Amp]
22XaVV
What happens when a=8?
1V XV 2V
X
S
aIVVV10002
1
0322
2
kvV
kV
kVV XXSX
KCL @ Vx
kaVkV X
2*1
2
REPLACE Ix IN V2
SX
XXXSX
VVaaVVVVV
3)8(0)2/(23)(3
REPLACE V2 IN KCL
IDENTIFY AND LABEL NODES
