44-1

Chapter 4 Systematic analysis methods4.1 Node analysis

node voltage, matrix node equation, floating voltage source, supernode

4.2 Mesh analysismesh current, matrix mesh equation, interior current source, supermesh

4.3 Systematic analysis with controlled sourcesmesh analysis, node analysis

4.4 Applications of systematic analysisequivalent resistance, Thevenin parameters

4.5-4.6 (optional)node analysis with ideal op-amp, -Y transformation

4

4.1 Node analysis Basics1. Node voltage: potential at nonreferenced node to a specified

reference node Suppressing sources to identify node numbers, the number of

unknown node voltages is one less than the node numbers. Select the node with its connection to the maximum voltage

sources as the reference node. All branch voltages can be determined from the values of node

voltages connected.

4-2

1 3 2

4

2. An example

+=

+++

++

=++=++=++

=

s

sas

fee

eedcc

ccba

sfe

edc

scbsa

KCL

i

vGi

vvv

GGGGGGGG

GGGG

ivGvvGvvGvGvvG

ivvGvGvvG

i

00

0

0)(:3 node0)()(:2 node

0)()(:1 node

0out flow analysis node

3

2

1

323

32212

2111

node

symmetricmatrix

into node

4-3

44-4

3. Floating voltage source Floating voltage source with series resistance convert to current

source voltagenodet independenan not is xnsx vvvv +=

3 nodes

x m s n m s n mx

s s s s

v v v v v v v viR R R R +

= = = +

xi

4

Floating ideal voltage source use supernode

3 nodes

(1) v2 or v3 is not an independent unknown, and is difficult to write node equation on iy

(2) Enclose both terminals of the floating ideal source within asupernode easy to write node equations with 2 unknowns

4-5

4

4. Matrix node equationsStep 1: identify reference node, and identify known node voltages by fixed voltage sourcesStep 2: perform floating voltage source conversionStep 3: apply KCL to write the matrix node equation

4-6

[ ][ ] [ ]

[ ]

11 12 1 1 1

21 22 2 2 2

1 12

:

: conductance matrix,

: sum of conductances connected to node i: sum of conductan

s

N s

N s

N NN N sN

ij ji

ii

ij

KCL G v i

G G G v iG G G v i

G G G v i

G G G

GG

=

=

=

ces directly connected to nodes i and j

: source-current vector sum of current sources into node isii

4

1. Ex.4.1

1

1

1 1 1 18 60one-node equation: ( )6 12 4 6 4

node voltage 2418 ( 24) 7

624 24 ( 60)2, 9

12 4

a

b c

v

v

i

i i

+ + =

=

= =

= = = =

Discussion

4-7

4

2. Ex.4.2 3 nodes2 node voltages

=

=

=

=

==

=

+

+=

+++

++++

25

4050

06

3030

4030

96

103

101

101

31

55016

301

601

121

51)

601

121(

)601

121(

601

121

151

61

2

1

2

1

2

1

b

a

i

i

vv

vv

vv

4-8

4

3. Ex.4.3

==

=

+

=

+

+

++

2010

20

31

32

30

10001

201

51

510

51

101

51

101

0101

101

41

21

10001

3

2

1

3

2

1

vvv

vvv

Note: Inserting a R in series with 3mA current source does not change the results.

4-9

1 1 2 1

3 2 3

-30 -node 1: 3 02 10 4-node 3: 3 05 20

v v v v

v v v

+ + =

+ + =

4

4. Ex. 4.4 find ia, ib

3 nodes

no need for source conversionfor a non-floating voltage source

==

=

=

==

+

=

++

++

mAi

mAi

vv

vv

b

a

02000

0

33000

615

60

624

315

648

424

10001

61

31

11

11

11

41

21

11

10001

2

1

2

1

4-10

4

5. Ex. 4.5 find v1, v2

supernode

1 2 1 2 11 2

2 1 2 2 11 2

( 30) 50 1 1 1 1 1 30 50supernode 1: 1 0 ( ) ( ) 12 10 5 10 2 5 10 2 2 5

( 30) 1 1 1 1 1 30( ) ( ) 7node 2 7 010 2 1 10 2 22 1 10

v v v v v v v

v v v v v v v

+ + = + + + = + +

+ + + = + + =

1 21 1

2 21 2

1 1 1 1 1 4 330 50( ) 261 4010 2 5 10 2 5 52 5101 1 1 1 1 3 830( ) 87

10 2 1 10 2 5 52

v vv vv vv v

+ + + =+ + = = = + + + + =

4-11

4

2 1 1 2 11 2

2 1 2 2 11 2

30 30 50 1 1 1 1 1 30 50 30supernode 1: 1 0 ( ) ( ) 12 10 5 10 2 5 10 2 10 5

( 30) 1 1 1 1 1 30( ) ( ) 7node 2 7 010 2 1 10 2 102 1 10

v v v v v v v

v v v v v v v

+ + + + = + + + = + +

+

+ + + = + + + =

(alternate)

1 21 1

2 21 2

1 1 1 1 1 30 50 30 4 3( ) 1 2 1010 2 5 10 2 10 5 5 5101 1 1 1 1 30 3 8( ) 7 10

10 2 1 10 2 10 5 5

v vv vv vv v

+ + + + + = = = = + + + + + =

4-11

4

4.2 Mesh analysis Basics1. Dual relation

2. mesh: a closed current path containing no closed paths within it,a special loop

node voltage mesh currentmatrix node equation matrix mesh equation (for planar

circuit only)KCL KVLfloating voltage source conversion interior current source conversion

4-12non-planar circuit

planar circuit

44-13

3. The minimum number of meshes is determined by suppressing all sources.

4. An example

mesh

1 1 1 2

2 1 2 2 3

3 2 3

1

2

3

mesh analysis voltage drop in mesh 0

mesh 1: ( ) ( ) 0mesh 2 : ( ) ( ) 0node 3 : ( ) 0

0

0

KVL

a s b c s

c d e

e f s

a b c c

c c d e e

e e f

i

R i i R i R i i vR i i R i R i iR i i R i v

R R R R iR R R R R i

R R R i

=

+ + = + + = + + =

+ + + + +

0s a s

s

v R i

v

+ =

same directionof mesh current

+_a s

R i

4

5. Interior current source Interior current sources with parallel resistance convert to

voltage source

is not an independent mesh currents x n xx n s x x s n s s s

i i i ii i i v i R i R i R=

= + = = +

+xv

4-14

4

Interior ideal current source use supermesh

2 2 1

2 1 2

2 2 1 2 1 2

( ) 0( ) ( ) 0

( ) ( ) ( ) 0

d b y

y c y e y

d b c y e y

R i R i i vv R i i i R i i

R i R i i R i i i R i i

+ = + + + + = + + + + + =

2 2 1 2 1 2( ) ( ) ( ) 0d b c y e yR i R i i R i i i R i i+ + + + + =

(1) Two meshes exist with source compressed.(2) i2 or i3 is not an independent unknown.(3) Use supermesh to enclose both meshes of the interior ideal source write mesh equation with unknown mesh currents

4-15

4

6. Matrix mesh equationsStep 1: identify known mesh currents fixed by current sourceStep 2: label the remaining unknown mesh currentsStep 3: apply KVL to write the matrix mesh equation

4-16

[ ][ ] [ ]

[ ]

11 12 1 1 1

21 22 2 2 2

1 12

:

: resistance matrix,

: sum of resistances connected to mesh : sum of resistances

s

N s

N s

N NN N sN

ij ji

ii

ij

KVL R i v

R R R i vR R R i v

R R R i v

R R R

R iR

=

=

=

shared by meshes and

: source-voltage vector sum of voltage sources in mesh to give current in the same direction of mesh current

si

i j

vi

44-17-10W12)]-(1[(60//12):3 source

0W10)(50:2 source0,30:1 source2

012//60150

30512//601515

156150505)1)(12//60()15(:2mesh

0306)15(:1mesh

0analysismesh

21

2

1

2

1

2212

121

===

==

=

++

+

=++++=+

=

iiii

ii

iiiiiii

vmesh

KVL

Discussion1. Ex.4.6 (Ex.4.2) find i1, i2

same direction as mesh

+

_1 60//12

4

2. Ex.4.7 find i1, i2, i3

===

+=

++

++

mAimAimAi

iii

213

2012

41020

16606633

033810

3

2

1

3

2

1

Note: Inserting a R in parallel with 20V voltage source does notchange the results.

4-18

4

3. Ex.4.8 (interior current source) to find va, ib

interior mesh

=+=+===

==

++

=

++

++

63,7

8)7(8

46

76187820

102622824

2

1

2

1

2

1

b

xxb

a

iiiii

iv

ii

ii

3 node eqs, 2 mesh eqsmesh analysis easier4-19

4

4. Ex4.9 (interior ideal current source) to find i1

2 nodes, 1 mesh supermesh

interior current source

220)4(310)5(6 1111 ==+++ iiii

4-20

4

4.3 Systematic analysis with controlled sourcesBasics1. An example on mesh analysis

1 1

1

1

( - ) 0mesh equation ( )

( - ) : constraint equation

a s b a

a b s a a

a a s

KVL R i i R i vR R i i R v

v R i i

+ + =

+ =

=

1 1

1 1

1

11 11 1 11 11

( ) ( - )( )( ) (1 )

( ) , , , (1 )

a b s a a s a a s

a b a s a a s

a b a a s

s a b a s a s

R R i i R v i R R i iR R i R i i R R iR R R i R i

R R i v R R R R R v R i

+ = =

+ = + =

= = + = =

1

11 1( )

1 1

11 1 11 1 11

11 11 1

reformulate as the usual mesh eq.

( ) (1 )

, (1 ) ,

( )

a a s

sv R i i

s s a a s a a s a s a

s s a s a

s

R i v

v i R v i R R i i R i R i

R i v R i v R i R R

R R i v

=

=

= = = +

= + = =

=

4-21

44-22

2. Mesh analysis

[ ]

[ ] [ ] [ ][ ] [ ][ ][ ] [ ] [ ][ ] [ ] ~~R :4 step

termssourcet independen:~ ,~~

equations constraintth vector wivoltage-source write:3 step equation constraintsource controlledeach for :2 step

matrix resistancecurrentsmesh unknown identify :1 step

ss

sss

viRRvi

viRvv

R

==

+=

3. Node analysis[ ]

[ ] [ ] [ ][ ] [ ][ ][ ] [ ] [ ][ ] [ ]

~~G :4 step

termssourcet independen:~ ,~~

equations constraintctor with current ve-source write:3 step equation constraintsource controlledeach for :2 step

matrix econductanc voltagesnodeunknown identify :1 step

ss

sss

ivGGiv

ivGii

G

==

+=

4

Discussion1. Ex.4.10 find i1, i2

1 1 2 1 2

2 1 2 2 1 2

2 1 2 2 1 2 1 2

11 2

2

mesh 1 6 10 4( ) 14 4 6mesh 2 4( ) 7 3( 8 ) 0,

4 4 7 3 24 24 0 20 10 014 4 6

1, 220 10 0

a a

i i i i ii i i i i i i i

i i i i i i i ii

i ii

= + = + + + = =

+ + + = =

= = =

4-23

4

2. Ex.4.10 find i1, i2

vc=24ia

3

[ ]

[ ] [ ] [ ][ ]

[ ] [ ][ ] [ ] 2,106

1020414

,~~1020414~

~~242400

06

24246

836

vector voltage-source

equation constraint,144

4143744

4410

212

1

2

1

21

21

==

=

=

=

+=

+

=

+

=

=

=

=

++

+=

iiii

viRRRR

iRvii

iiiv

iiiR

s

sa

s

a

4-24

4

3. Ex.4.11 find the current gain using mesh analysis

[ ]

[ ] [ ] [ ][ ]

[ ] [ ][ ] [ ] 10110

105.11

5.0~~

1051000441863

0137~

~~

9696000864000

00

30

)(42126436

30

212436

30

)(46

equation constraint,1940

44110137

3

2

1

1 if

3

2

1

23

1

23

1

=

==

===

=

=

+=

+

=

=

=

==

=

=s

outimAis

s

ss

b

a

s

s

b

a

iiA

mAimAi

mAiviRRRR

iRviiii

iii

i

vv

iv

iiviv

R

s

4-25

44-26

4. Ex.4.12 find the voltage gain using node analysis

[ ]

[ ] [ ] [ ][ ]

[ ] [ ][ ] [ ]

AA

vvA

vvv

AivGG

AGG

vGivv

A

v

Av

v

Av

v

i

vvG

inv

ins

s

inin

d

in

s

d

02.055.105.0

05.0

01.301.0201.0515.0

,~~01.301.0201.0515.0~

~~0200

02

5.0

2

5.0

2ctor current ve-source

equation constraint,01.301.001.0515.0

15.0

1100

1100

1100

12001

1001

21

2

2

1

2

1

1

1

+

==

=

=

=

+=

+

=

=

=

=

=

++

++=

44-27

4.4 Applications of systematic analysisDiscussion1. Examples of determining equivalent resistance or Thevenin

parameters are given to use systematic analysis methods.2. Ex. 4.13 find the equivalent resistance of a resistive bridge

662835

312

243444812

analysismesh

66235

8

12

21

81

41

41

41

31

41

121

analysis node

2121

2

1

2121

2

1

==+===

=

++

++

==+===

=

++

++

eq

eq

Riiiviii

ii

ii

Rvvvivvv

v

v

vv

4 010 //15 6eq

iR

= = =

44-28

3. Ex. 4.14 find the Thevenin parameters of a current amplifier

[ ]

[ ]

[ ]

1

1

1

node analysis with current source: 3 nodesmesh analysis: 2 meshes

21 3,constraint equation

3 7

8 10 2 8 20 20

12 20 00 0

x in

in x in ins

ins

R i i i

i i i i iv

v v

i iv R

v i

= =

+ = = = + = +

[ ]

1

1 3,

3 7

1 3 12 36 183 7 2 2

Norton model

18 , 2 , 36

in inin

sct

sc in t oc sc t in

i

R R

i i v i vi ii v

vi iR

i i R k v i R k i

=

= = = +

=

= = = =

4

4.5-4.6Optional1. Ex. 4.15 find Av of an op-amp circuit

21

21

2122212

11111

122

12

)1(0)11()11(0)(1)1 (1)1(

0, :short virtual

)1...(0)(1) (1:analysis node

KKKK

vvv

KKv

Kvv

RKv

Kv

R

Kvv

RKv

Rvvv

vvRK

vvR

in

outoutinoutinoutin

outx

outxin

outx

+

==++=++

==+=

=+

4-29

44-30

2. Y- (T-) transformation equivalent circuit

( )

mesh analysis:( )

1 1 1

node analysis:1 1 1

b c in c outin

a b b c c aa c c in in

c b c out out c in a c outout

a b b c c a

ca ab ab in

out

ab ca ab

R R v R viR R R R R RR R R i v

R R R i v R v R R viR R R R R R

R R R vv

R R R

+ =

+ ++ = + + =

+ +

+ +

2 2 2

2

( )

( )

, , , ,

ab ca ca bc in ca bc outin

ab bc cain

out ca bc in bc ab ca bc outout

ab bc ca

ab ca bc ab ca bca b c ab bc ca

c a b

ab bc ca a b b c c

R R R R i R R ivR R Ri

i R R i R R R R ivR R R

R R R R R R R R RR R R R R RR R R R R RR R R R R R R R R R R

+ =

+ + = + =

+ +

= = = = = =

+ + + + a

4

3. Ex.4.16 (Ex.4.13)

6)6.14.6//(24

8,32,24

96222

2

=+=

======

=++

eq

bca

abc

cab

accbba

RRRR

RRR

RRR

RRRRRRR

4-31