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NODAL AND LOOP ANALYSIS TECHNIQUES
LEARNING GOALS
Develop systematic techniques to determine all the voltagesand currents in a circuit
NODAL ANALYSISLOOP ANALYSIS
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
k
VI a
62 :SOHM' 0321 III :KCL
3*3 IkVb :SOHM' …OTHER OPTIONS...
4
34
*4
124
12
IkV
II
b
5
345
*3
0
IkV
III
C
:SOHM'
:KCL
k12kk 12||4
k6
kk 6||6
k
VI
12
121
)12(93
3
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
VVV
VVV
1
10
ba
ba
VVV
VVV
3
3 0
______ca
V
QUESTIONDRILL
REFERENCE
SV
aV
bV
cV
1V
KVL
3V
KVL
5V
WHAT IS THE PATTERN???
KVL
05 bc VVV
ONCE THE VOLTAGES ARE KNOWN THE CURRENTS CANBE COMPUTED USING OHM‟SLAW
A GENERAL VIEW
R
v
NmR vvv
5 b cV 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
R
vvi 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
R
vvi mN 'LAW SOHM'
'ii PASSIVE SIGN CONVENTION RULES!
DEFINING THE REFERENCE NODE IS VITAL
SMEANINGLES IS4V VSTATEMENT THE 1
UNTIL THE REFERENCE POINT IS DEFINED
BY 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 ANALYSIS1. 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 IIIV
a
4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES
0369
k
VV
k
V
k
VV baasa AND GET ALGEBRAIC EQUATIONS IN THE NODE VOLTAGES ...
REFERENCE
SV
aV
bV
cV
0:@543 IIIV
b
0:@65 IIV
c
0943
k
VV
k
V
k
VVcbbab
039
k
V
k
VVcbc
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
R
VV
R
VV
R
VVIII cbdbba
0LEAVING CURRENTS
00321
321
R
VV
R
VV
R
VVIII cbdbba
0 NODE INTO CURRENTS
2I
3I1I
a b c
d
aV
bV
cV
dV
1R 3R
2R'2I
'3I'
1I
00321
'3
'2
'1
R
VV
R
VV
R
VVIII bcdbab
0LEAVING CURRENTS
00321
'3
'2
'1
R
VV
R
VV
R
VVIII 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: THE FORMAL MANIPULATION OF EQUATIONS MAY BE SIMPLER IF ONE USES CONDUCTANCES INSTEAD OF RESISTANCES.
@ NODE 1
02
21
1
1
R
vv
R
viA SRESISTANCE USING
0)( 21211 vvGvGiA ESCONDUCTANC WITH
REORDERING TERMS
@ NODE 2
REORDERING TERMS
THE MODEL FOR THE CIRCUIT IS A SYSTEMOF ALGEBRAIC EQUATIONS
THE MANIPULATION OF SYSTEMS OF ALGEBRAICEQUATIONS CAN BE EFFICIENTLY DONEUSING MATRIX ANALYSIS
EXAMPLE
@ NODE 1 WE VISUALIZE THE CURRENTSLEAVING AND WRITE THE KCL EQUATION
REPEAT THE PROCESS AT NODE 2
03
12
4
122
R
vv
R
vvi
OR VISUALIZE CURRENTS GOING INTO NODE
WRITE THE KCL EQUATIONS
mA15
A
B
C
k8 k8k2 k2
SELECT ASREFERENCE
AV
BV
MARK THE NODES (TO INSURE THAT NONE IS MISSING)
ANOTHER EXAMPLE OF WRITING NODE EQUATIONS
WRITE KCL AT EACH NODE IN TERMS OFNODE VOLTAGES 015
82@ mA
k
V
k
VA AA
01528
@ mAk
V
k
VB BB
A MODEL IS SOLVED BY MANIPULATION OFEQUATIONS AND USING MATRIX ANALYSIS
THE NODE EQUATIONS
kRRkR
mAimAi BA
6,12
4,12
321
THE MODEL
REPLACE VALUES AND SWITCH NOTATIONTO UPPER CASE
NUMERICAL MODEL
USE GAUSSIAN ELIMINATION
ALTERNATIVE MANIPULATION
k12/*
k6/*
242
1223
21
21
VV
VV
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
A
AAdjA
FOR THE ADJOINT REPLACEEACH ELEMENT BY ITSCOFACTOR
AND DO THE MATRIX ALGEBRA ...
kkkV
6
104
3
10118
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
k
VV
k
VmAV
1264:@ 211
1
USING KCL
0126
2:@ 1222
k
VV
k
VmAV
BY “INSPECTION”
mAVk
Vkk
412
1
12
1
6
121
mAVkkk
212
1
6
1
12
12
LEARNING EXTENSION
mA6
1I
2I
3I
036
6:@ 222
k
V
k
VmAV 2
12V V
CURRENTS COULD BE COMPUTED DIRECTLYUSING KCL AND CURRENT DIVIDER!!
1
2
3
8
3(6 ) 2
3 6
6(6 ) 4
3 6
I mA
kI mA mA
k k
kI mA mA
k k
LEARNING EXTENSION
IN MOST CASES THEREARE SEVERAL DIFFERENTWAYS OF SOLVING APROBLEM
NODE EQS. BY INSPECTION
mAVVk
6202
121
mAVkk
V 63
1
6
10 21
116V V
k
VI
k
VI
k
VI
362
23
22
11
Once node voltages are known
1
1@ : 2 6 0
2
VV mA mA
k
Node analysis
CIRCUITS WITH DEPENDENT SOURCES CANNOTBE MODELED BY INSPECTION. THE SYMMETRYIS LOST.
A PROCEDURE FOR MODELING•WRITE THE NODE EQUATIONS USING DEPENDENTSOURCES AS REGULAR SOURCES.•FOR EACH DEPENDENT SOURCE WE ADD ONE EQUATION EXPRESSING THE CONTROLLINGVARIABLE IN TERMS OF THE NODE VOLTAGES
02
21
1
1
R
vv
R
vio
02
12
3
2
R
vv
R
viA
MODEL FOR CONTROLLING VARIABLE
3
2
R
vio
NUMERICAL EXAMPLE
mAvkk
vk
vkk
vkk
23
1
12
1
6
1
06
1
3
2
6
1
12
1
21
21
0111
2
23
1
21
v
RRv
RR
REPLACE AND REARRANGE
AivRR
vR
2
32
1
2
111
k4/*
k6/*
][123
02
21
21
VVV
VV
ADDING THE EQUATIONS ][125 2 VV
VV5
241
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
VA
mAimAikR
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 row
0, -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
k
VVmA
k
VV
NODE EQUATIONS
010
210
:@ 2122
k
VI
k
VVV O
CONTROLLING VARIABLE (IN TERMS ON NODEVOLTAGES)
k
VIO
10
1
REPLACE
01010
210
010
410
2112
211
k
V
k
V
k
VV
k
VVmA
k
V
REARRANGE AND MULTIPLY BY 10k
02
][402
21
21
VV
VVV eqs. add and 2/*
VVVV 16805 11
VVV
V 82
21
2
O V VOLTAGETHE FIND
NODE EQUATIONS
063
2 k
V
k
VmA xx
NOTICE REPLACEMENT OF DEPENDENT SOURCE IN TERMS OF NODE VOLTAGE
012126
k
V
k
V
k
V OOx
k6/*
k12/*
][4022
][4][123
VVVV
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
k
VV
k
VV
k
V
][5.1][64
0)()(2
22
12322
VVVV
VVVVV
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
VV
VV
THESE ARE THE REMAININGTWO NODE EQUATIONS
+
-
2SI
3SI 1SV
1SI1R
2R
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
k
V
k
VVmA
R
V
R
VVIV S
021
12
1][4
0:@
42212
2
42
3
32
1
1232
k
VV
k
V
k
VVmA
R
VV
R
VV
R
VVIV S
02
][4][2
0:@
24
2
24214
k
VVmAmA
R
VVIIV 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
k
V
k
VVmA
021
12
1][4 42212
k
VV
k
V
k
VVmA
02
][4][2 24
k
VVmAmA
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
4
32
4
110
152
023
3
2
1
V
V
V
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 SIk
VmA
@V_2 012
4 2 k
VmAIS
2 eqs, 3 unknowns...Panic!! The current through the source is notrelated to the voltage of the source
Math solution: add one equation
][621 VVV
Efficient solution: enclose the source, and all elements in parallel, inside a surface.
Apply KCL to the surface!!!
04126
6 21 mAk
V
k
VmA
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
k12/*
][6
046126
21
21
VVV
mAmAk
V
k
V
(2)
(1)
Equations The
ALGEBRAIC DETAILS
...by Multiply Eq(1). in rsdenominato Eliminate 1.
Solution
][6
][242
21
21
VVV
VVV
][10][303 11 VVVV
2 Veliminate to equations Add2.
][4][612 VVVV
2 Vcompute to Eq(2) Use 3.
1V
2V
1sI
2sI
1R2R
3R
SV
][6],[10],[20
4,10
21
321
mAImAIVV
kRkRR
ssS
FIND THE NODE VOLTAGESAND THE POWER SUPPLIEDBY THE VOLTAGE SOURCE
2012VV
0101010
21 mAk
V
k
V ][10010/*21
VVVk
][2021
VVV
][40100
][60
21
2
VVV
VV
:adding
TO COMPUTE THE POWER SUPPLIED BY VOLTAGE SOURCEWE MUST KNOW THE CURRENT THROUGH IT
VI
k
VVmA
k
VI
V
106
10
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 FIND
SUPERNODE
1231 VVCONSTRAINT SUPERNODE
123V
KCL @ SUPERNODE
VVVV 12,642 KNOWN NODE VOLTAGES
LEARNING EXAMPLE
VV
VV
4
6
4
1
SOURCES CONNECTED TO THEREFERENCE
SUPERNODE
CONSTRAINT EQUATION VVV 1223
KCL @ SUPERNODE
02
)4(
212
63322
k
V
k
V
k
V
k
Vk2/*
VVV 22332
VVV 1232 add and 3/*
VV 385 3
mAk
VI
O8.3
2
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
R
V
R
VV
R
VV
KCL @SUPERNODE (Careful not to omit any current)
04
32
5
34
6
4
3
5
2
15
1
12
R
VV
R
VV
R
V
R
V
R
VV
R
VV
CONSTRAINTS DUE TO VOLTAGE SOURCES
11 SVV
252 SVVV
345 SVVV
5 EQUATIONS IN FIVE UNKNOWNS.
supernode
1V
2V 3
V
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 REFERENCE
VV 31
0263
: 212
xI
k
V
k
VV VKCL@
2
CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES k
VI
x
6
2
REPLACE
06
263
2212
k
V
k
V
k
VVk6/*
VVVV 602212
mAk
VVI
O1
3
21
OI FIND
LEARNING EXAMPLE
SUPER NODE WITH DEPENDENT SOURCE
VOLTAGE SOURCE CONNECTED TO REFERENCE
VV 63
SUPERNODE CONSTRAINT xVVV 2
21
KCL AT SUPERNODE
k12/*
062)6(22211
VVVV
CONTROLLING VARIABLE IN TERMS OF NODES
2VV
x
213VV
183321 VV 184
1 V
CURRENT CONTROLLED VOLTAGE SOURCE
CONSTRAINT DUE TO SOURCE xkIVV 2
12
KCL AT SUPERNODE0
22
24 21
k
VmA
k
VmA
CONTROLLING VARIABLE IN TERMS OF NODES
k
VI
x
2
1121 22 VVkIV x
][421
VVV
0221 VV
add and 2/*
][832
VV
mAk
VI
O
3
4
2
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
k
VI X
X2
„a‟ has units of [Volt/Amp]
22
XaVV
What happens when a=8?
1V X
V 2V
X
S
aIV
VV
10002
1
0322
2
k
vV
k
V
k
VV XXSX
KCL @ Vx
k
aVkV X
2
*12
REPLACE Ix IN V2
SX
XXXSX
VVa
aVVVVV
3)8(
0)2/(23)(3
REPLACE V2 IN KCL
IDENTIFY AND LABEL NODES
LEARNING EXAMPLE FIND THE VOLTAGE Vo
AT SUPER NODE
1 22
XV V V
2 3 1 32 14
2 01 1 1 1
V V V VV V VmA
k k k k
3 2 3 1
3@ : 2 0
1 1
V V V VV mA
k k
4 4@ : 4V V V
CONTROLLING VARIABLE2X
V V
1k
1k
SOLVE EQUATIONS NOW
1
1 3
1 3
3
2 2 6
2 2
X
X
X
V V
V V V V
V V V V
VARIABLE OF INTEREST1 3O
V V V
LEARNING EXAMPLE Find the current Io
FIND NODES – AND SUPER NODES
2 2@ : 12V V V
3 3@ : 2
XV V V
4 1
1 3 4 3 4 51 2 4
@super node:
6 (constraint eq.)
2 01 1 1 1 1
X
V V V
V V V V V VV V VI
k k k k k
5 4 5
5@ : 2 0
1 1X
V V VV I
k k
1 2
4
CONTROLLING VARIABLES
1
X
X
V V V
VI
k
7 eqs in 7 variables
VARIABLE OF INTEREST 5
1O
VI
k
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