Transcript Nodal Analysis
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.
2.
3.
Select one of the
n
reference node.
nodes as a reference node (that we define to be zero voltage, or ground). Assign voltages
v
1 ,
v
2 , …
v n
-1 to the remaining
n
-1 nodes. These voltages are referenced with respect to the 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.
Solve the resulting simultaneous equations to obtain the node voltages
v
1 ,
v
2 , …
v n
-1 .
3
2A
v
1
R
1
R
2
Example
v
2
R
4
R
3
v
3
i s R
5 Select a reference node as ground. Assign voltages
v
1 ,
v
2 , and
v
3 to the remaining
3
nodes.
4
2A
i
1
v
1
R
1
i
2
Example
i
4
v
2
R
2
R
3
i
3
R
4
v
3
i s i
5
R
5 Apply KCL to each of the
3
non-reference nodes (sum of currents
leaving
node is zero).
Node 1: 2 1
i
2 0 Node 2: 0 Node 3: 2 4
i i s
3
i
4
i
5 0 5
2A
i
1
v
1
R
1
i
2
Example
i
4
v
2
R
2
R
3
i
3
R
4
v
3
i s i
5
R
5
i
1 Now express
i
1 ,
i
2 , …
i
5 in terms of
v
1 ,
v
2 ,
v
3 (the node voltages). Note that current flows from a higher to a lower potential.
v
1
R
1 0
i
2
v
1
v
2
R
2
i
3
v
2
R
0 4
i
4
v
2
v
3
R
3
i
5
v
3 0
R
5 6
Node 1: Node 2: Node 3: 2 4 2
i
3
i
4
i
1
s i i
2 5 0
v
1
R
1
v
1
R
2
v
2
R
2 0
v
1
R
2
v
2
R
2
v
2
R
4
v
2
R
3
v
3
R
3 0
v
2
R
3
v R
3 3
s v
3
R
5
i
1
v
1 0
R
1
i
2
v
1
v
2
R
2
i
3
v
2 0
R
4
i
4
v
2
v
3
R
3
i
5
v
3 0
R
5 7
Node 1: 0
G
1
v
1
R
1
v
1
R
2
v
2
R
2 2 1
G v
2 2 0
v
3 2 Node 2: 0
v
1
R
2
v
2
R
2
v
2
R
4
G v
2 1
G
2
G
3
v
2
R
3
v
3
R
3 4 2
G v
3 3 0 Node 3: 0
v
2
R
3
v
3
R
3
s v R
5 3 0
v
1
G v
3 2
G
3 5 3
i s
8
G
1 2 1
G v
2 2 0
v
3 2
G v
2 1
G
2
G
3 0
v
1
G v
3 2
G
3 4 2
G v
3 3 0 5 3
i s
These three equations can be written in matrix form as
G
1
G
2
G
2 0
G
2
G
2
G
3
G
3
G
4 0
G
3
G
3
G
5
v
v
2
v
3 2
i
0
s
9
G
1
G
2 0
G
2
G
2
G
2
G
3
G
3
G
4 0
G
3
G
3
G
5
v
1
v
2
v
3 2
i
0
s
i Gv
i G
is an (
n
–1) x (
n
–1)
symmetric
conductance matrix
v
is a 1 x (
n
-1) vector of node voltages is a vector of currents representing “known” currents 10
2A
Writing the Nodal Equations by Inspection
v
1
v
2
v
3
R
2
R
3
i
1
R
1
i
3
R
4
i s i
5
R
5
G
1
G
2
G
2 0
G
2
G
2
G
3
G
3
G
4 •The matrix
G
is symmetric,
g
kj are negative or zero.
=
g
jk 0
G
3
G
3
G
5
v
2 2 0
i s
and all of the off-diagonal terms The
g
kk terms are the sum of all conductances connected to node
k
.
The
g
kj terms are the negative sum of the conductances connected to BOTH node
k
and node
j
.
The
i
k (the
k
th 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
G
1
G
2 0
G
2
G
2
G
2
G
3
G
3
G
4 0
G
3
G
3
G
5
v
v
2
v
3 2
i
0
s
Gv
i v
1
G i
12
Test with numbers
2A
v
1 1 2 1/2 S 1 S
v
2 3 1/3 S 4 1/4 S
v
3 1A 1/2 S 2 1 0 1 1 3 0 1 3
v
2 1.5
1 0 1 1.583
0.333
0 0.333
0.833
v
2 2 0 1 2 0 1 13
MATLAB Run
2A 1.5
1 0 1 1.583
0.333
0 0.333
0.833
v
2 2 0 1
v
1 2 1
v
2 4 3
v
3 1A 2 14
MATLAB:
PSpice Simulation
15
What happens if we have dependent current sources in the circuit?
1.
2.
3.
Write the nodal equations in the same way we did for circuits with only independent sources. Temporarily, consider the dependent sources as being independent.
Express the current of each dependent source in terms of the node voltages.
Rewrite the equations with all node voltages on the left hand side of the equality.
16
2A
i
1
v
1
G
1
i
2
Example
i
4
v
2
G
2
G
3
i
3
G
4
v
3
i s
2
i
3
i
5
G
5 Write nodal equations by inspection.
G
1
G
2
G
2 0
G
2
G
2
G
3
G
3
G
4 0
G
3
G
3
G
5
v
v
2 0 2 2
i
3 17
2A
i
1
v
1
G
1
i
2
Example
i
4
v
2
G
2
G
3
i
3
G
4
v
3
i s
2
i
3
i
5
G
5
i
3
G v
4 2
G
1
G
2
G
2 0
G
2
G
2
G
3
G
3
G
4 0
G
3
G
3
G
5
v
v
2 0 2 2
i
3 18
2A
i
1
v
1
G
1
i
2
Example
i
4
v
2
G
2
G
3
i
3
G
4
v
3
i s
2
i
3
i
5
G
5
G
1
G
2
G
2 0
G
2
G
2
G
3
G
3
G
4 0
G
3
G
3
G
5
v v
2 0 2 2
G v
4 2 19
Example
G
1
G
2
G
2 0
G
2
G
2
G
3
G
3
G
4 0
G
3
G
3
G
5
v
2 0 2 2
G v
4 2
G
1
G
2
G
2 0
G
2
G
2
G
3 3 2
G G
4 4 0
G
3
G
3
G
5
v
v
2
v
3 2 0 0 20
2A
i
1
v
1
i
2
G
1
G
2
i
3
i v
2
G
4
G
3 4
v
3
i s
2
i
3
i
5
G
5
G
1
G
2
G
2 0
G
2
G
2
G
3 3 2
G G
4 4 0
G
3
G
3
G
5
v
v
2
v
3 2 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
2A
v
1
i
1
G
1
i
2 DC
Example
v
2
i
4
V
0
G
3
i
3
G
4
v
3
i s i
5
G
5 Assume temporarily that
i
2 is known and write the equations by inspection.
G
1 0 0 0
G
3
G
3
G
4 0
G
3
G
3
G
5
v
1
v
2
v
3
i i
2
s i
2 23
G
0 0 1 0
G
3
G
3
G
4 0
G
3
G
3
G
5
v
v
2
v
3
i i
2
s i
2 There appears to be 4 unknowns (
v
1 ,
v
2 ,
v
3 , and
i
2 ) and only 3 equations. However, from the circuit
V
0 2
v
1 or
v
1 2
V
0 so we can replace
v
1 (we could also replace
v
2 ) and write
G
0 0 1
G
3 0
G
3
G
4 0
G
3
G
3
G
5
v
2
v
2
v
3
V
0
i i
2
s i
2 24
G
1 0 0
G
3 0
G
3
G
4 0
G
3
G
3
G
5
v
2
v
2
v
3
V
0
i i
2
s i
2 Writing the above equation with the unknowns (v2, v3, i2) on the LHS yields
G
3
G
1
G
3
G
4
G
3 0
G
3
G
5 1 1 0
v
v
3
i
2 0
i s G V
1 0 25
Test with numbers
2A
v
1 DC G 1/2 S 2V
i
2
v
2 G 1/3 S G 1/4 S
v
3 1A 1/2 S G 1 2 0 0 0 1 3 0 1 3
v
2 Noting that
v
1 2 2 1 2 0 0 0 1 3 0 1 3
v
2
v
2
v
3 2
i
2
i
2 1
i
2
i
2 1 26
Test with numbers
2A
v
1 DC G 1/2 S 2V
i
2
v
2 G 1/3 S G 1/4 S
v
3 1A 1/2 S G 1 2 0 0 Unknowns: 0 0.5833
1 3 0 1 3 0.8333
v
2
v
2
v
3 2 2 , , ( 3 2
v
1 2 2) 0.5
0 0.5833
0.3333
0.3333 0.8333
0 1
v v
2 3 0 0
i
2
i
2 1 27
MATLAB Run
V V A
v
2
v
3
i
2 2A
v
1 DC G 1/2 S 2V
i
2
v
2 G 1/3 S G 1/4 S
v
3 1A 1/2 S G 0.5
0.5833
0.3333
0 0.3333
0.8333
1 1 0
v
3
v
1 2 2.6316V
28
PSpice Simulation
MATLAB:
v
2
v
3
i
2
v
1 2 2.6316V
29