Chapter 4 Systematic analysis methods -...
Transcript of Chapter 4 Systematic analysis methods -...
-
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