Nodal Analysis
Discussion D2.3
Chapter 2
Section 2-7
1
Nodal Analysis
• Interested in finding the NODE VOLTAGES,
which are taken as the variables to be
determined
• For simplicity we start with circuits containing
only current sources
2
Nodal Analysis Steps
1. Select one of the n nodes as a reference node (that
we define to be zero voltage, or ground). Assign
voltages v1, v2, … vn-1 to the remaining n-1 nodes.
These voltages are referenced with respect to the
reference node.
2. Apply KCL to each of the n-1 non-reference
nodes. Use Ohm’s law to express the branch
currents in terms of the node voltages.
3. Solve the resulting simultaneous equations to
obtain the node voltages v1, v2, … vn-1.
3
Example
v1
R2
2A
v3
v2
R1
R3
R4
is
R5
Select a reference node as ground. Assign voltages v1,
v2, and v3 to the remaining 3 nodes.
4
Example
i2
v1
2A
i1
v2
R2
R1
i4
i3
R3
R4
v3
is
i5
R5
Apply KCL to each of the 3 non-reference nodes
(sum of currents leaving node is zero).
Node 1:
2  i1  i2  0
Node 2:
i2  i3  i4  0
Node 3:
i4  is  i5  0
5
Example
i2
v1
2A
i1
R2
R1
i4
v2
i3
R3
R4
v3
is
i5
R5
Now express i1, i2, …i5 in terms of v1, v2, v3 (the node voltages).
Note that current flows from a higher to a lower potential.
v1  0
i1 
R1
v2  0
v1  v2
i3 
i2 
R4
R2
v2  v3
i4 
R3
v3  0
i5 
R5
6
Node 1:
v1 v1 v2
2  i1  i2  0  2   
R1 R2 R2
v1 v2 v2 v2 v3
Node 2: i2  i3  i4  0  
   
R2 R2 R4 R3 R3
v3
v2 v3
  is 
Node 3: i4  is  i5  0  
R3 R3
R5
v1  0
i1 
R1
v2  0
v1  v2
i3 
i2 
R4
R2
v2  v3
i4 
R3
v3  0
i5 
R5
7
Node 1:
v1 v1 v2
0  2 

R1 R2 R2
G1  G2  v1  G2v2  0v3  2
v1 v2 v2 v2 v3
Node 2: 0  


 
R2 R2 R4 R3 R3
G2v1  G2  G3  G4  v2  G3v3  0
Node 3:
v3
v2 v3
0     is 
R3 R3
R5
0v1  G3v2  G3  G5  v3  is
8
G1  G2  v1  G2v2  0v3  2
G2v1  G2  G3  G4  v2  G3v3  0
0v1  G3v2  G3  G5  v3  is
These three equations can be written in matrix form as
 G1  G2

 G2
 0

G2
G2  G3  G4
G3
  v1   2 
   
G3   v2    0 
G3  G5   v3   is 
0
9
 G1  G2

 G2
 0

G2
G2  G3  G4
G3
  v1   2 
   
G3   v2    0 
G3  G5   v3   is 
0
Gv  i
G is an (n –1) x (n –1) symmetric conductance matrix
v
is a 1 x (n-1) vector of node voltages
i
is a vector of currents representing “known” currents
10
Writing the Nodal Equations by Inspection
v1
2A
i1
v3
v2
R2
R1
i3
 G1  G2

 G2
 0

R3
is
R4
G2
G2  G3  G4
G3
i5
R5
  v1   2 
   
G3   v2    0 
G3  G5   v3   is 
0
•The matrix G is symmetric, gkj = gjk and all of the off-diagonal terms
are negative or zero.
The gkk terms are the sum of all conductances connected to node k.
The gkj terms are the negative sum of the conductances connected to
BOTH node k and node j.
The ik (the kth component of the vector i) = the algebraic sum of the
independent currents connected to node k, with currents entering the
11
node taken as positive.
MATLAB Solution of Nodal Equations
 G1  G2

 G2
 0

G2
G2  G3  G4
G3
  v1   2 
   
G3   v2    0 
G3  G5   v3   is 
0
Gv  i
1
vG i
12
Test with numbers
v1
1
1S
2A
2
1/2 S
v2
3
1/3 S
4
1/4 S
v3
1A
1/2 S
2
1
0  v1   2 
1  1 2

   

1
1

1
4

1
3

1
3

 v2    0 
 0
 v   1 

1
3
1
3

1
2

 3   
1
0   v1   2 
1.5

   

1
1.583

0.333

  v2    0 
 0 0.333 0.833   v   1 

 3   
13
MATLAB Run
1
0   v1   2 
1.5

   

1
1.583

0.333

  v2    0 
 0 0.333 0.833   v   1 

 3   
v1
2A
1
2
v2
3
4
v3
2
1A
14
PSpice Simulation
MATLAB:
15
What happens if we have dependent
current sources in the circuit?
1. Write the nodal equations in the same way we
did for circuits with only independent sources.
Temporarily, consider the dependent sources as
being independent.
2. Express the current of each dependent source in
terms of the node voltages.
3. Rewrite the equations with all node voltages on
the left hand side of the equality.
16
Example
i2
v1
2A
i1
G2
G1
i4
v2
i3
G3
G4
v3
is  2i3
i5
G5
Write nodal equations by inspection.
 G1  G2

 G2
 0

G2
G2  G3  G4
G3
  v1   2 
   
G3   v2    0 
G3  G5   v3   2i3 
0
17
Example
i2
v1
2A
i1
G2
G1
i4
v2
i3
G3
G4
v3
is  2i3
i5
G5
i3  G4v2
 G1  G2

 G2
 0

G2
G2  G3  G4
G3
  v1   2 
   
G3   v2    0 
G3  G5   v3   2i3 
0
18
Example
i2
v1
2A
i1
 G1  G2

 G2
 0

G2
G1
i4
v2
i3
G2
G2  G3  G4
G3
G3
G4
v3
is  2i3
i5
G5
  v1   2 
  

G3   v2    0 
G3  G5   v3   2G4v2 
0
19
Example
 G1  G2

 G2
 0

 G1  G2

 G2
 0

G2
G2  G3  G4
G3
G2
G2  G3  G4
G3  2G4
  v1   2 
  

G3   v2    0 
G3  G5   v3   2G4v2 
0
  v1   2 
   
G3   v2    0 
G3  G5   v3   0 
0
20
i2
v1
2A
i1
 G1  G2

 G2
 0

G2
G1
i4
v2
i3
G2
G2  G3  G4
G3  2G4
G3
G4
v3
is  2i3
i5
G5
  v1   2 
   
G3   v2    0 
G3  G5   v3   0 
0
Note: the G matrix will no longer necessarily be symmetric
21
Nodal Analysis for Circuits Containing Voltage Sources
That Can’t be Transformed to Current Sources
• Case 1. If a voltage source is connected between
the reference node and a nonreference node, set
the voltage at the nonreference node equal to the
voltage of the source.
• Case 2. If a voltage source is connected between
two nonreference nodes, assume temporarily that
through the voltage source is known and write the
equations by inspection.
22
Example
i2
v1
2A
i1
v2
DC
V0
G1
i4
v3
G3
i3
G4
is
i5
G5
Assume temporarily that i2 is known and write the equations by inspection.
0
 G1

 0 G3  G4
0
G3

  v1   2  i2 
  

G3   v2    i2 
G3  G5   v3   is 
0
23
0
 G1

 0 G3  G4
0
G3

  v1   2  i2 
  

G3   v2    i2 
G3  G5   v3   is 
0
There appears to be 4 unknowns (v1, v2, v3, and i2) and only 3
equations. However, from the circuit
V0  v2  v1
or
v1  v2  V0
so we can replace v1 (we could also replace v2) and write
0
 G1

 0 G3  G4
0
G3

  v2  V0   2  i2 

 

G3   v2    i2 
G3  G5   v3   is 
0
24
0
 G1

 0 G3  G4
0
G3

  v2  V0   2  i2 

 

G3   v2    i2 
G3  G5   v3   is 
0
Writing the above equation with the unknowns (v2, v3, i2) on
the LHS yields
 G1

 G3  G4
 G
3

0
G3
G3  G5
1   v2   2  G1V0 
  

1  v3   
0


0   i2  
is

25
Test with numbers
v1
2V
2A
v2
DC
i2
G
1/2 S
G
v3
1/3 S
G
1/4 S
1A
1/2 S
G
0
0   v1   2  i2 
1 2

  

0
1
3

1
4

1
3
v

i

 2   2 
 0
 v   1 

1
3
1
3

1
2

 3  

Noting that v1  v2  2
0
0   v2  2   2  i2 
1 2


 

0
1
3

1
4

1
3
v

i

 2   2 
 0
 v   1 

1
3
1
3

1
2

 3  

26
Test with numbers
v1
2V
2A
G
1/2 S
v2
DC
i2
G
v3
1/3 S
G
1/4 S
1A
1/2 S
G
0
0   v2  2   2  i2 
1 2


 

0
0.5833

1
3
v

i

 2   2 
 0
1 3 0.8333   v3   1 

Unknowns: v2 , v3 , i2 (v1  v2  2)
0
0   v2   1
 0.5

   
0.5833
0.3333

1

  v3    0 
 0.3333 0.8333 0   i   1 

 2   
27
MATLAB Run
v1
DC
2V
2A
G
1/2 S
v2
i2
G
v3
1/3 S
G
1/4 S
1A
1/2 S
G
0
1   v2   1
 0.5

   
0.5833

0.3333

1

  v3    0 
 0.3333 0.8333 0   i   1 

 2   
V
V
A
v2
v3
i2
v1  v2  2  2.6316V
28
PSpice Simulation
MATLAB:
v2
v3
i2
v1  v2  2  2.6316V
29