Systematic Circuit Analysis (T&R Chap 3)
Node-voltage analysis
Using the voltages of the each node relative to a ground
node, write down a set of consistent linear equations for
these voltages
Solve this set of equations using, say, Cramer’s Rule
Mesh current analysis
Using the loop currents in the circuit, write down a set of
consistent linear equations for these variables. Solve.
This introduces us to procedures for systematically
describing circuit variables and solving for them
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Nodal Analysis
Node voltages
Pick one node as the ground node
Label all other nodes and assign voltages vA, vB, …, vN
B vB
and currents with each branch i1, …, iM
+
Recognize that the voltage across a branch
v1
v5
+
is the difference between the end node
+
v3
A vA
voltages
+
+
Thus v3=vB-vC with the direction
v4
v2
as indicated
C vC
Write down the KCL relations at each node
Write down the branch i-v relations to express branch
currents in terms of node voltages
Accommodate current sources
Obtain a set of linear equations for the node voltages
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Nodal Analysis – Example 3-1 p.72
vA
Apply KCL
R1
i0
Node A: i0-i1-i2=0
Node B:i1-i3+i5=0
Node C: i2-i4-i5=0
iS1
vB
i3=G3vB
i4=G4vC
i5=iS2
R2
vC
i4
i3
R3
Write the element/branch eqns
i0=iS0
i1=G1(vA-vB)
i2=G2(vA-vC)
i5
i1
iS2
i2
R4
Reference node
Substitute to get node voltage equations
Node A: (G1+G2)vA-G1vB-G2vC=iS1
Node B:
-G1vA+(G1+G3)vB=is2
Node C:
-G2vA+(G2+G4)vC=-iS2
& G1 +G2
$ 'G1
$
% 'G2
'G1
'G2 #& v A #
G1 +G3
0 !$ v B ! =
!$ !
0
G2 +G4 "% vC "
& iS 1 #
$ iS 2 !
$ 'i !
% S2 "
Solve for vA, vB, vC then i0, i1, i2, i3, i4, i5
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Systematic Nodal Analysis
& G1 +G2 'G1 'G2 #& v A #
$ 'G1 G1 +G3
0 !$ v B ! =
$
!$ !
'
G
0
G
+
G
% 2
2 4 "% vC "
& iS 1 #
$ iS 2 !
$ 'i !
% S2 "
R1
i0
iS1
vA
i5
vB
i1
iS2
i2
R2
vC
i4
i3
R3
R4
Writing node equations by inspection
Reference node
Note that the matrix equation looks
just like Gv=i for matrix G and vector v and i
G is symmetric (and non-negative definite)
Diagonal (i,i) elements: sum of all conductances connected
to node i
Off-diagonal (i,j) elements: -conductance between nodes i
and j
Right-hand side: current sources entering node i
There is no equation for the ground node – the column sums
give the conductance to ground
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43
Nodal Analysis Example 3-2 p.74
R/2
Node A:
Conductances
G/2B+2GC=2.5G
Source currents entering= iS
vA
2R
iS
vB
R/2
2R
vC
R
Node B:
Conductances
G/2A+G/2C+2Gground=3G
Source currents entering= 0
Node C:
Conductances
2GA+G/2B+Gground=3.5G
Source currents entering= 0
& 2.5G ' 0.5G ' 2G #& v A # & iS #
$
!$ ! $ !
3G
' 0.5G !$ vB ! = $ 0 !
$ ' 0.5G
$ ' 2G ' 0.5G 3.5G !$ v ! $ 0 !
%
"% C " % "
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Nodal Analysis – some points to watch
1. The formulation given is based on KCL with the
sum of currents leaving the node
0=itotal=GAtoB(vA-vB) +GAtoC(vA-vC)+…+ GAtoGroundvA+ileavingA
This yields
0=(GAtoB+…+GAtoGround)vA-GAtoBvB-GAtoCvC…-ienteringA
(GAtoB+…+GAtoGround)vA-GAtoBvB-GAtoCvC…=ienteringA
2. If in doubt about the sign of the current source,
go back to this basic KCL formulation
3. This formulation works for independent current
source
For dependent current sources (introduced later) use your
wits
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Nodal Analysis Exercise 3-2 p. 72
2KΩ
vA
vB
+
iS1
1KΩ
iS2 500Ω
vO
_
& 1.5 (10 '3
$
$ ' 0.5 (10 '3
%
' 0.5 (10 '3 #!& v A # & iS1 #
!
$ !=$
'3 ! v
2.5 (10 "% B " % ' iS 2 "
Solve this using standard linear equation solvers
Cramer’s rule
Gaussian elimination
Matlab
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Nodal Analysis with Voltage Sources
Current through voltage source is not computable
from voltage across it. We need some tricks!
They actually help us simplify things
Method 1 – source transformation
vA
RS
vA
Rest
of the
circuit
+
_ vS
vB
!
vS
RS
Rest
of the
circuit
RS
vB
Then use standard nodal analysis – one less node!
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Nodal Analysis with Voltage Sources 2
Method 2 – grounding one node
vA
+
_ vS
vB
Rest
of the
circuit
This removes the vB variable – simpler analysis
But can be done once per circuit
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Nodal Analysis with Voltage Sources 3
supernode
Method 3
Create a supernode
Act as if A and B were one node
KCL still works for this node
Sum of currents entering
supernode box is 0
Write KCL at all N-3 other nodes
N-2 nodes less Ground node
using vA and vB as usual
Write one supernode KCL
Add the constraint vA-vB=vS
vA
+
_ vS
vB
Rest
of the
circuit
These three methods allow us to deal with all cases
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Nodal Analysis Example 3-4 p. 76
R1
+_ v
S1
vA
R2
+
v0
_
R3
vS2
+_
!
vS 1
R1
R1
vS 2
R2
+
R2
R3 v0
_
This is method 1 – transform the voltage sources
Applicable since voltage sources appear in series with Rs
Now use nodal analysis with one node, A
(G1 + G2 + G3 )v A = G1vS1 + G2vS 2
G1vS1 + G2vS 2
vA =
G1 + G2 + G3
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Nodal Analysis Example 3-5 p. 77
iin
R/2
Rin
vA
2R
_
vS +
vB
R/2
2R
vC
What is the circuit input
resistance viewed
through vS?
R
v A = vS
!0.5Gv A +3GvB !0.5GvC =0
! 2Gv A ! 0.5Gv B + 3.5vC = 0
Rewrite in terms of vS, vB, vC
This is method 2
2.75vS
6.25vS
, vC =
10.25
10.25
v !v v !v 11.75vS
iin = S B + S C =
2R
R / 2 10.25 R
10.25 R
Rin =
= 0.872 R
11.75
3Gv B − 0.5GvC = 0.5Gv S
− 0.5Gv + 3.5Gv = 2Gv
B
C
S
vB =
Solve
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Nodal Analysis Example 3-6 p. 78
vA i2
i1
R2
R1
supernode
+
vB
_
vS1
i3 v
C
i4
R3
+
+
vO
_ vS2
R4
reference
Method 3 – supernodes
KCL for supernode: i1 + i2 + i3 + i4 = 0
Or, using element equations
G1v A + G2 (v A ! v B ) + G3 (vC ! v B ) + G4 vC = 0
Now use v B = vS 2
(G1 + G3 )v A + (G3 + G4 )vC = (G2 + G3 )v S 2
Other constituent relation
MAE140 Linear Circuits
v A ! vC = v S1
Two equations in two unknowns
52
Summary of Nodal Analysis
1. Simplify the cct by combining elements in series or parallel
2. Select as reference node the one with most voltage sources
connected
3. Label node voltages and supernode voltages – do not label
the nodes directly connected to the reference
4. Use KCL to write node equations. Express element currents
in terms of node voltages and ICSs
5. Write expressions relating node voltages and IVSs
6. Substitute from Step 5 into equations from Step 4
Write the equations in standard form
7. Solve using Cramer, gaussian elimination or matlab
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Solving sets of linear equations
& 5 ' 2 ' 3 #& x1 # & 4 #
$
!$ ! $
!
$ ' 5 7 ' 2 !$ x2 ! = $ ' 10 !
$ ' 3 ' 3 8 !$ x ! $ 6 !
%
"% 3 " %
"
Cramer’s Rule Thomas & Rosa Appendix B pp. A-2 to A-11
5 ! 2 !3
7 !2
! 2 !3
! 2 !3
# = !5 7 ! 2 = 5
! ( !5)
+ ( !3)
!3 8
!3 8
7 !2
!3 !3 8
=5( 7"8!( !3)"( !2) )+5( ( !2)"8!( !3)"( !3) )!3( ( !2)"( !2) !7"( !3) )
= 250 ! 125 ! 75 = 50
4 ! 2 !3
7 !2
! 2 !3
! 2 !3
#1 = !10 7 ! 2 = 4
! ( !10)
+6
!3 8
!3 8
7 !2
6 !3 8
!1 100
x1 = =
=2
! 50
= 4( 7"8!( !3)"( !2) )+10( ( !2)"8!( !3)"( !3) )+ 6( ( !2)"( !2) !7"( !3) )
= 200 ! 250 + 150 = 100
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Solving sets of linear equations (contd)
5
4 !3
!10 ! 2
4 !3
4 !3
" 2 = !5 !10 ! 2 = 5
! ( !5)
+ ( !3)
6
8
6 8
!10 ! 2
!3 6
8
= !340 + 250 +114 = 24
5
!2
! 24
x2 = 2 = =0.48
! 50
4
7 !10
!2 4
!2 4
" 3 = !5 7 !10 = 5
! ( !5)
+ ( !3)
!3 6
!3 6
7 !10
!3 !3 6
=60 !0 + 24 =84
!3 84
x3 =
=
= 1.68
! 50
Notes:
This Cramer is not as much fun as Cosmo Kramer in Seinfeld
I do not know of any tricks for symmetric matrices
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Solving Linear Equations – gaussian elimination
& 5 '2 '3 #& x1 # & 4 #
$ '5 7 '2 !$ x2 !=$ '10 !
$ '3
%
'3
8
!$ x ! $
"% 3 " %
row2+row1
row3+row1×3/5
&
#
$ 5 ' 2 '3 4 !
$ 0 5 '5 ' 6 !
$
' 21 31 42 !
$0
!
%
5
5 5"
row3÷10
&
#
$ 5 ' 2 '3 4 !
$ 0 5 '5 ' 6 !
$
84 !
$0 0 1
!
%
50 "
MAE140 Linear Circuits
6
Augment
matrix
!
"
& 5 ' 2 '3 4 #
$
!
$ '5 7 ' 2 '10 !
$ '3 '3 8
6 !"
%
row3×5
& 5 ' 2 '3 4 #
$
!
$ 0 5 '5 ' 6 !
$ 0 ' 21 31 42 !
%
"
Row
operations
only
row3+row2×21/5
&
#
$ 5 ' 2 '3 4 !
$ 0 5 '5 ' 6 !
$
84 !
$ 0 0 10
!
%
5"
(row2+row3×5) ÷5
(row1+row2 ×2
+ row3×3) ÷5
&
#
4 !
$ 5 ' 2 '3
$ 0 1 0 24 !
50 !
$
$ 0 0 1 84 !
%
50 "
&
100 #
$1 0 0
50 !
$ 0 1 0 24 !
50 !
$
$ 0 0 1 84 !
%
50 "
56
Solving Linear Equations - matlab
A=[5 –2 –3; -5 7 –2; -3 –3 8]
A =
5 –2 –3
-5 7 –2
-3 –3 8
inv(A)*B
ans = 2.0000
0.4800
1.6800
B=[4;-10;6]
A\B
B =
4
-10
6
ans = 2.0000
0.4800
1.6800
inv(A)
ans = 1.0000
0.9200
0.7200
0.5000
0.6200
0.4200
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0.5000
0.5000
0.5000
57
Mesh Current Analysis
Dual of Nodal Voltage Analysis with KCL
Mesh Current Analysis with KVL
Mesh = loop enclosing no elements
Restricted to Planar Ccts – no crossovers (unless you are really
clever)
Key Idea: If element K is contained in both mesh i and mesh j then
its current is ik=ii-ij where we have taken the reference directions
as appropriate
Same old tricks you already know
R1
+ v1 +
+
v3
v0 + vS1
- iA -
R2
+ v2
R3
iB
-
+
v4 + vS2
- -
Mesh A: v0-v1-v3=0
Mesh B: v3-v2-v4=0
(R1+R3)iA-R3iB=vS1
-R3iA+(R2+R3)iB=-vS2
MAE140 Linear Circuits
v1=R1iA
v2=R2iB
v3=R3(iA-iB)
v0=vS1
v4=vS2
& R1 + R3 ' R3 #& i A # = & vS1 #
$ 'R
!$ i ! $ 'v !
R
+
R
%
3
2 3 "% B " % S 2 "
58
Mesh Analysis by inspection Ri = vS
Matrix of Resistances R
Diagonal ii elements: sum of resistances around loop
Off-diagonal ij elements: - resistance shared by loops i and j
Vector of currents i
As defined by you on your mesh diagram
Voltage source vector vS
Sum of voltage sources assisting the current in your mesh
If this is hard to fathom, go back to the basic KVL to sort
these directions out
vS1
+ iC
R2
R3
+ vS2
-
R1
iA
iB
MAE140 Linear Circuits
+
R4
-
& R1 + R2
$
$ 0
v0
$ 'R
%
2
0
R3 + R4
' R3
' R2 #& i A # & ' vS 2 #
!$ ! $
!
' R3 !$ iB ! = $ vS 2 !
R2 + R3 !"$% iC !" $% ' vS1 !"
59
Mesh Equations with Current Sources
Duals of tricks for nodal analysis with voltage sources
1. Source transformation to equivalent
T&R Example 3-8 p. 91
3KΩ
3KΩ
5KΩ
iO
2KΩ
5V +
-
4KΩ
2mA
1KΩ
& 6000 ' 2000 #& i A # & 5 #
$
!$ ! = $ !
% ' 2000 11000 "% iB " % ' 8 "
MAE140 Linear Circuits
+
5V -
5KΩ
iO
2KΩ
iA
iB
4KΩ
+
8V -
1KΩ
iA=0.6290 mA
iB=-0.6129 mA
iO=iA-iB=1.2419 mA
60
Mesh Analysis with ICSs – method 2
Current source belongs to a single mesh
5KΩ
3KΩ
iO
5V +
-
2KΩ
iA
4KΩ
iC
iB
2mA
Same example
1KΩ
6000i A ! 2000iB = 5
! 2000i A + 11000iB ! 4000iC = 0
iC = !2 mA
Same equations!
Same solution
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Mesh Analysis with ICSs – Method 3
Supermeshes – easier than supernodes
Current source in more than one mesh and/or not in parallel
with a resistance
1. Create a supermesh by eliminating the whole branch
involved
2. Resolve the individual currents last
Supermesh
R1
R2
iB
iS2
iS1
iA R
3
iC
Excluded branch
+
vO
R4
MAE140 Linear Circuits
-
R1 (iB !i A ) + R2iB + R4iC + R3 (iC !i A ) =0
i A = iS1
iB !iC =iS 2
62
Summary of Mesh Analysis
1. Check if cct is planar or transformable to planar
2. Identify meshes, mesh currents & supermeshes
3. Simplify the cct where possible by combining
elements in series or parallel
4. Write KVL for each mesh
5. Include expressions for ICSs
6. Solve for the mesh currents
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Linearity & Superposition
Linear cct – modeled by linear elements and
independent sources
Linear functions
f(Kx)=Kf(x)
Additivity:
f(x+y)=f(x)+f(y)
Superposition –follows from linearity/additivity
Homogeneity:
Linear cct response to multiple sources is the sum of the
responses to each source
1. “Turn off” all independent sources except one and
compute cct variables
2. Repeat for each independent source in turn
3. Total value of all cct variables is the sum of the values
from all the individual sources
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Superposition
+
Turning off sources
vS +
_
Voltage source
Turned off when
i
v=0 for all i
i
v
v
i
+
a short circuit
i
v
-
Current source
Turned off when
v
i
i=0 for all v
an open circuit
i
+
iS
v
v
-
We have already used this in
Thévenin and Norton equiv
i
i
+
v
MAE140 Linear Circuits
-
v
65
Where are we now?
Finished resistive ccts with ICS and IVS
Two analysis techniques – nodal voltage and mesh current
Preference depends on simplicity of the case at hand
The aim has been to develop generalizable techniques for
access to analytical tools like matlab
Where to now?
Active ccts with resistive elements – transistors, op-amps
Life starts to get interesting – design introduced
Capacitance and inductance – dynamic ccts
Frequency response – s-domain analysis
Filters
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