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ECE 3144 Lecture 11
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Reminder from Lecture 10
•Nodal analysis
–A circuit contains n (n > 2) nodes
–Circuit variables: node voltage
–Any one node from these n nodes can be selected as the reference node.
–The node voltages are defined with respect to the reference node.
–With the selection of the reference node, the node voltages for other nodes, say a, b, c,
etc. can be represented as va, vb, vc, etc. So voltage at node a is va, which is really the
voltage at node a with respect to the ground node.
•How to relate the node voltages with voltage across resistors and current
through resistors
•For a N-node network, since one node is selected as the ground node (v= 0),
the number of linearly independent KCL equations is (N-1).
•Nodal analysis case 1: only independent current sources and resistors
–Ohm’s law in matrix form: GV = I
–If G-1 exists, V = G-1I
2
Example 1: nodal analysis with only
v
vb independent current sources
a
a
i2
iA
R1
b
R2
i
i
3
B
3
Given iA, iB, and resistors R1, R2, R3, find
the node voltages va, vb, vc
R
i1
c
•There are 3 nodes, thus only 2 independent equations are needed.
•Node c is selected as the reference node => vc = 0.
•At node a applying KCL: -iA + i1 + i2 = 0 =>
-iA + G1(va – vc) + G2(va-vb) = 0
(1)
•At node b applying KCL: -i2 + iB + i3 = 0 =>
-G2(va-vb) + iB + G3(vb – vc) = 0
(2)
•(1) + (2) => -iA + G\(va-vc) + iB + G3(vb – vc) = 0.
(3)
• Equation (3) is the KCL equation at reference node c.
• Any two of the three equations can be used to derive the third one. Thus there are
only 2 linearly independent equations for this 3-node circuit.
•In general, for a circuit with N nodes, only (N-1) linearly independent equations are
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needed to solve the (N-1) unknowns va and vb.
Example 1: cont’d
•
In the matrix format: GV = I
G1  G2
G
  G2
•
•
•
 G2 
G2  G3 
v a 
V  
 vb 
 iA 
I  
 iB 
If G-1 exists, V = G-1I. The matrix operations we have
learned from Lecture 9 can be used to solve the problem.
It is not difficult to find that the formula shows here is the
Ohm’s law in the matrix format. Later on, we will explain a
systematic way on how to get the Ohm’s law in matrix
format.
One more example will be shown in order to get the general
procedures for determining the formats for matrices G, V
and I.
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Example 2: nodal analysis with only
R3
v1 i2
1
i3
R2
v R5
2
2
iA
i1
•In node 1 applying KCL:
i5 v
33
i4
i1-iA+i2-i3 = 0
or
G1v1 – iA + G2(v1-v2) –G3(v3-v1) = 0 or
iB
R4
R1
independent current sources
(G1+G2+G3)v1 –G2v2-G3v3 = 0
(1)
•In node 2 applying KCL:
-i2 + i4 –i5 = 0
or
-G2(v1-v2) + G4v2-G5(v3-v2) = 0 or
-G2v1 + (G2+G4+G5)v2 – G5v3 = 0
(2)
•In node 3 applying KCL:
i3 + i5 +iB = 0
or
G3(v3-v1) + G5(v3-v2) + iB= 0
-G3v1 – G5v2 + (G3+G5)v3 = 0
or
(3)
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Example 2: cont’d
•Three equations for three unknowns. The equations can be transformed into
the matrix format:
GV = I
 G2
 G3 
G1  G2  G3
G    G2
G2  G4  G5  G5 
  G3
 G5
G3  G5 
 iA 
I   0 
 i B 
 v1 
V  v2 
v3 
•The G matrix is in symmetrical format.
•The symmetry is not accidental. Actually for all the circuits containing only
independent current sources and resistors, the G matrix is in this symmetrical
form.
•We can take advantage of this fact and learn to write the equations by
inspection.
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R3
Nodal analysis with only
independent current sources
 iA 
 G2
 G3 
G1  G2  G3
G  
 G2
G2  G4  G5
 G5  I   0  R1
 


 G3
 G5
G3  G5 
 iB 
v1 i2
i 1
i3
R2
v2
2
R5
i5 v3
3
A
iB
R4
i1
i4
•At node 1, the coefficient of v1 is the sum of all the conductances connected to node 1; the coefficient
of v2 is the negative of the conductances connected between node 1 and node 2; the coefficient of v3 is
the negative of the conductances connected between node 1 and node 3; the right side of the equation is
the sum of currents entering the node 1 through current sources. This is KCL at node 1.
•At node 2, the coefficient of v2 is the sum of all the conductances connected to node 2; the coefficient
of v1 is the negative of the conductances connected between node 2 and node 1; the coefficient of v3 is
the negative of the conductances connected between node 2 and node 3; the right side of the equation is
the sum of currents entering the node 2 through current sources. This is KCL at node 2.
•At node 3, the coefficient of v3 is the sum of all the conductances connected to node 3; the coefficient of
v1 is the negative of the conductances connected between node 3 and node 1; the coefficient of v2 is the
negative of the conductances connected between node 3 and node 2; the right side of the equation is the
sum of currents entering the node 3 through current sources. This is KCL at node 3
•In general, KCL is applied to node j with node voltage vj, the coefficient of vj, which is the element
gjj of G matrix, is the sum of all the conductances connected to node j. The coefficient of any other
node voltage, say i (ij), is the negative of the sum of the conductances connected directly between
node j and node i. The right hand side of the equation is equal to the sum of the currents entering the
node via independent current resources.
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I3
G2
a
I1
G3
b
G1
I2
d
 G2
G1  G2  G1


G    G1 G1  G3
 G3

  G2
 G3 G2  G3  G4 
c
G4
Given the circuit network,
using the nodal analysis
techniques to get the Ohm’s
law equations in the matrix
format
v a 
V  vb 
vc 
 I1  I 3 
I    I 2 
 I 2 
GV = I => V = G-1I
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Nodal analysis case 2: circuits with dependent current sources
•The presence of the dependent current sources may destroy the symmetrical form
of the nodal analysis
•Consider the following circuit
•In node 1, applying KCL:
v
v
1
i0
2
i0 
R2
R1
R3
iA
i0
v1
v  v2
 1
 0 (1)
R1
R2
•In node 2, applying KCL:

v1  v2
 i0  i A  0 (2)
R2
•Replace i0 = v2/R3 into both equations (1) and (2)
(G1  G2 )v1  (G2  G3) v2  0
 G2 v1  (G2  G3 )v2  i A
•The matrix form is:
G1  G2
G
  G2
 (G2  G3 )
(G2  G3 ) 
0
I  
i A 
 v1 
V  
v 2 
GV = I =>V = G-1I
•Notice that the G matrix is not symmetric anymore
9
Summary for nodal analysis containing current sources
• For circuits only containing independent current
sources, construct the G matrix and I matrix
directly by taking the advantage that the G matrix
is symmetric.
• For circuits containing both independent and
dependent current sources, the symmetric format
of G matrix may not exist anymore.
– Treat the dependent sources as independent sources
first.
– Derive the KCL equations at each node.
– Replace the dependent sources by the control formula
given.
– Solve the equations for nodal voltages.
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Nodal analysis case 3: circuits with
independent voltage source
• There are two scenarios for circuits containing
independent voltage sources
– Simpler one: The independent voltage source is
connected to the reference node.
– More complicated one: the independent voltage
source is connected between two nonreference
nodes.
• We will illustrate both scenarios.
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Scenario 1: independent voltage source connected to
the reference node
•Any time an independent voltage source is connected between an reference node and a
nonreference node, the voltage for the nonreference node is known.
•So the presence of voltages sources in this case will make the voltage equations simpler.
Say the given network has N nodes. If k independent voltages sources are connected
between a reference node and a nonreference node, then the number of linearly independent
equations needed to solve the node voltages is reduced from (N-1) to (N-1-k).
9k
-
12k
V3
6k
+
+ 12 V
V2
-
12k
V1
6V
•For the circuit given, immediately we know
V1 = 12 V and V2 = -6V.
•The only unknown voltage variable is V3.
• Applying KCL at node 2
V1  V2 V3  V2 V2


0
12k
12 K
6K
12  V2  6  V2 V2


0
12k
12 K
6K
V2 =1.5 V
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Scenario 2: independent voltage source connected
between the nonreference nodes
V1
6V
+
-
V2
12k
6 mA
6k
•If we follow the nodal analysis brute force
manner, we will have a problem: the current
4 mA on the branch where independent voltage
source is can be not calculated, which means
KCL equations at node 1 and node 2 can not
be established.
•A method called supernode technique is introduced.
First circle the voltage source and the two connecting nonreference nodes to form
a supernode.
Write the equations that defines the voltage relationships between the two nonreference
nodes as a result of the presence of the voltage source.
(1)
V1 – V2 = 6
Write the KCL equation for the super node.
V
V
(2)
 6mA  1  2  4mA  0
6k 12k
Then we have two equations (1)& (2) and two unknowns.
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Homework for lecture 11
• Problem 3.9, 3.10, 3.15, 3.22
• Due Feb 4
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