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EE 215 Fundamentals of Electrical Engineering
Lecture Notes
Nodal Analysis
7/31/01
Rich Christie
Nodal Analysis Overview:
Systematic method of circuit analysis.
Works on all circuits - very powerful method.
No creative thinking needed.
Makes big sets of equations with many unknowns - computationally time consuming to
solve unless you are a computer.
Hence basis of almost all computerized circuit solution methods.
Process:
Write KCL at every node in terms of the node voltages.
Special handling for sources.
What's a node voltage?
+
Voltages are measured
between two nodes.
a
We have been using the ends
of circuit elements as the
nodes, and getting the voltage
across the element.
But what if we chose one
node and measured all other
voltages with respect to that
node?
vab
-
5!
b
+
6A
10 !
+
va
-
vb
15 !
-
We call the one node we choose the reference node. Usually we choose the lowest
voltage node in the system, or the negative terminal of the largest voltage source, or then
node at the bottom of the circuit drawing. In theory we could choose any node, but it
makes computation and understanding easier to choose a reference node that we can
easily think of as having zero voltage. (Higher up the page => higher voltage.)
We mark the reference node with a special symbol, also called the ground symbol,
because in practice we actually connect the reference node in many circuits to actual dirt
(ground), and claim it has zero voltage.
215A
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The voltage between any node and the reference node is then a node voltage.
Note that the node voltage of the reference node is always zero!
Note that you can find the voltage across any circuit element (branch) by taking the
difference between two node voltages.
Notation:
We'll leave the reference node unnumbered and letter the other nodes a, b, c, etc. (We
could number the nodes, but this tends to cause confusion with numbered branch
elements.)
A single subscript is the node
voltage of the node designated
by the subscript. Thus va is the
node voltage of node a, the
voltage measured between
node a and the reference node.
Two subscripts is a branch
voltage, the voltage between
two nodes. Thus, vab is the
voltage between nodes a and
b, with node a as the positive
terminal (as before).
+
a
vbc
-
b
c
+
+
vb
-
-
vc
d
Then from KVL
v ab = v a ! v b
Note that there does not have to be a loop of circuit elements for KVL to work - just a
loop of voltages. e.g.
v b ! v bc ! v c = 0
v bc = v b ! v c
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You'll notice that once all the node voltages are known, it is easy to calculate the branch
voltages, and then the current through all the branches. When we know the values of all
the node voltages, we say we have a complete solution, and know the system state.
Even better, we already know one of the node voltages - the reference node is always
zero.
Thus, for a circuit with N nodes, we need only write N-1 equations in N-1 unknowns.
In nodal analysis we do this by writing KCL at each node except the reference node, in
terms of the node voltages.
Let's start at node a. There are
two currents leaving node a,
iar going to the reference node
and iab going to node b.
a
iar
By KCL
iar + iab = 0
iab
R1
is
+
vab
b
R3
c
R2
R4
(At this point the critical
student will wonder why we
are using currents leaving
node a instead of entering
node a, as used in the canonical statement of KCL. In fact, using currents entering the
node works fine, and I will occasionally use this, but using currents leaving the node
makes writing the equations directly less error prone. Check out the signs of va in the
equation we will develop for node a.)
We can see that for the current source,
iar = !i s
From Ohm's Law,
i ab =
v ab v a ! v b
=
R1
R1
so, subbing in for the currents,
! is +
va ! vb
=0
R1
is KCL for node a in terms of node voltages.
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We'd actually like to be able to write KCL in terms of node voltages directly, without
writing the currents, and without writing the branch voltages. Let's try it for node c:
vc ! vb vc
+
=0
R3
R4
It's actually vc - vref over R4, but vref is zero so it just drops out of the equation.
Right now there are two equations in three unknowns, va, vb and vc. We just need one
more, at node b
vb ! va vb vb ! vc
+
+
=0
R1
R2
R3
Notice how each term has the positive value of the node voltage for the node where we
are writing KCL. In this case, vb is positive in each term. This is a good consistency
check to be sure you got the polarities of the voltages correct. If you have +vb and -vb on
the same side of the node b equation, you probably made an error.
Let's assemble and rearrange the equations
& 1 #
& 1 #
v a $$ !! + v b $$ ' !! = i s
% R1 "
% R1 "
& 1
& 1 #
& 1 #
1
1 #
!! + v c $$ '
!! = 0
v a $$ ' !! + v b $$ +
+
% R1 "
% R1 R2 R3 "
% R3 "
& 1 #
& 1
1 #
!! + v c $$
!! = 0
v b $$ '
+
% R3 "
% R3 R 4 "
There are lots of ways to solve these equations. One way is to solve the first equation for
va in terms of vb, substitute in to the second equation, solving for vb in terms of vc, and
substitute in to the third equation to find vc.
We can also write the equations in matrix form (computers like this).
& 1
$
$ R1
$' 1
$ R1
$
$ 0
$%
1
R1
1
1
1
+
+
R1 R2 R3
1
'
R3
'
#
!
! &v a # &i s #
1 !$ ! $ !
'
vb = 0
R3 ! $ ! $ !
1
1 ! $% v c !" $% 0 !"
!
+
R3 R4 !"
0
5
And the matrix equations can be solved for the voltages using Cramer's Rule, or by
inverting the matrix and premultiplying the right hand side. (See math notes or book for
Cramer's Rule and determinants.)
Mind you, I would never solve the example circuit by nodal analysis unless someone
made me do it. Instead, I would use opportunistic circuit theory, where I would apply a
sequence of simple circuit manipulations, involving one or two circuit elements at a time.
Specifically, I would :
1. Do a series combination of R3 and R4
2. Do a parallel combination of the result of step 1 with R2
3. Apply Ohm's Law to the result of step 2 to find vb
4. Apply Ohm's Law again on R1 and add to vb to get va (Okay, this is two steps)
5. Apply voltage divider from vb to get vc
This would be a lot faster for me to do with paper, pencil and calculator. But I'm not a
computer.
Let’s try an example where we have to apply nodal analysis:
The objective here is to
find the value of i.
This configuration of
resistors is called a
Wheatstone Bridge, or
bridge for short. It is used
when measuring small
resistances, e.g
thermocouples.
i
5!
10!
10!
2A
3!
2!
1A
The hopover shows that
these two wires are NOT
connected where they
cross.
(There are two conventions for showing connections. Either you put a visible dot at every
connection point where more than two wires meet - a dot means a connection, no dot
means no connection - OR, you don't put dots but do use hops. Cross with no hop means
a connection. Cross with hop means no connection. Sometimes people use hop and dot.
Sometimes people use neither one and you just have to guess (most common in back of
the envelope circuits, not technical drawing).
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Hmm, resistors are not in parallel - not in series - can't combine! Can't do voltage divider,
can't do current divider - stuck - get systematic!
1. Assign reference node choose bottom node in resistor
connection
2. Find and label the other
nodes.
b
i
5!
10!
10!
2A
a
3. Write KCL at each node, in
terms of node voltages:
c
3!
2!
1A
va va ! vc va ! vb
+
+
+ 1 = 0 note there are as many terms as connections
3
10
10
to a, all the va's have the same (positive) sign, and the current flowing out of a is positive.
Do this consistency check on each equation you write.
v ! vc vb ! va
at (b) ! 2 + b
+
=0
5
10
v
v ! va vc ! vb
at (c) ! 1 + c + c
+
=0
2
10
5
at (a)
form matrices
1
&1 1
$ 3 + 10 + 10
$
1
$ '
10
$
$ ' 1
$%
10
1
10
1 1
+
5 10
1
'
5
'
1
#
10 ! &v a # & ' 1#
!$ ! $ !
1
'
! vb = 2
5
! $$ v !! $$ 1 !!
1 1 1!% c " % "
+ +
2 10 5 !"
'
&0.533 ' 0.1 ' 0.1# &v a # & ' 1#
$ ' 0.1 0.3 ' 0.2! $ v ! = $ 2 !
$
!$ b ! $ !
$% ' 0.1 ' 0.2 0.8 !" $% v c !" $% 1 !"
Solve by inverting with Matlab
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&v a # &0.5459#
$ v ! = $9.2729!
$ b! $
!
$% v c !" $%3.6365!"
Solve by Cramer's Rule:
1. Need determinant of matrix
0.533 ! 0.1 ! 0.1
# = ! 0.1
! 0.1
0.3 ! 0.2
! 0.2 0.8
= 0.533 " 0.3 " 0.8 + (!0.1)(! 0.2 )(! 0.1)+ (! 0.1)(! 0.1)(! 0.2 )
! [(! 0.1)(0.3)(! 0.1)+ (! 0.2 )(! 0.2 )(0.533)+ (0.8)(! 0.1)(! 0.1)]
= 0.0916
2. For vb, stick RHS column in second column, and find determinant
0.533 ! 1 ! 0.1
! 0.1 2 ! 0.2 = 0.8494
! 0.1 1
0.8
3.
Cramer's Rule: v b =
0.533 " 1 " 0.1
" 0.1 2 " 0.2
" 0.1 1
0.8
!
=
0.8494
= 9.2729
0.0916
Similarly,
0.533 " 0.1 " 1
" 0.1 0.3
2
vc =
" 0.1
" 0.2
!
1
= 3.6365
And let's remember what we were looking for (forgetting the question is a common
problem in exams):
i=
v b ! v c 9.2729 ! 3.6365
=
= 1.127 A
5
5
8
So it's ugly math the computer loves to do (try matlab - it makes matrix solutions trivial
for YOU) and we get a solution where opportunistic methods fail.
(OK, so there exists a Y-Δ resistance transformation that could be applied with paper and
pencil to solve the circuit. But very few people remember how to do a Y-Δ transform,
and the computations are complicated enough that nodal analysis is probably an
efficiency win.)
Circuits with Voltage Sources:
So far we have done nodal analysis on circuits with current sources. What about voltage
sources?
There are two cases. Case 1, the source is attached to the reference node. Since we often
get to choose a reference node, this is a good reason why the reference node is often at
the negative terminals of voltage sources!
In this case, it should be pretty obvious that the node voltage at the other end of the
voltage source is just the source voltage:
va = v s
a
R1
R3
b
c
Then at node b
vb ! v s vb vb ! vc
+
+
=0
R1
R2
R3
vs
+
-
R2
is
and at node c
! is +
vc ! vb
=0
R3
Note that knowing va has reduced the problem from a three unknown problem to a two
unknown problem.
In matrix form
1
1
&1
$R + R + R
2
3
$ 1
1
$
'
$%
R3
or
1#
R3 ! &v b #
=
!
1 ! $%v c !"
R3 !"
'
& vs #
$R !
$ 1!
% is "
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&
$
$ 1
$' 1
$ R1
$
$ 0
$%
0
1
1
1
+
+
R1 R2 R3
1
'
R3
#
!
0 ! &v a # & v s #
1
' ! $v b ! = $ 0 !
R3 ! $ ! $ !
1 ! $% v c !" $% ' i s !"
!
R3 !"
(This is how the computer does it. The top line is called an "auxiliary equation" because
it is not KCL.)
Case 2: The voltage source is not connected to the reference node. This could be because
there are two or more voltage sources in the circuit, or because someone has cleverly put
the reference node elsewhere. The computer has to handle all possibilities!
In this case, the nodes on either side of the voltage source are combined into a supernode,
or generalized node. KCL can still be written covering all currents crossing the boundary
of the supernode. (You can have a supernode without a voltage source, around a resistor,
say, but its really only useful where there is a voltage source.)
KCL at node a: i1 + i2 + i3 = 0
i1
i2
KCL at node b: i4 + i5 + i6 = 0
a
Note: i4 = !i2
i3
i5
i4
b
i6
Add: i1 + i3 + i5 + i6 = 0
Here a supernode (the dotted line) is
drawn around nodes a and b. i2 and i4 are
inside the supernode and do not cross the
supernode boundary. What is the sum of
currents entering the supernode?
i1
i5
a
b
i6
i3
i1 + i3 + i5 + i6 = 0
So KCL works for supernodes.
Supernode
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Let's try nodal analysis with a supernode.
There's no point moving the
reference node in this case!
vs1
a
+
We have
R1
vc = vs2
-
R3
b
R2
c
+
-
We have
v a ! v b = v s1
(This is the auxiliary equation for the supernode we are about to draw.)
We draw a supernode around the voltage source between nodes a and b and write KCL
for it (with current leaving the supernode to keep the signs of voltage straight).
va vb vb ! vc
+
+
=0
R1 R2
R3
Now we can sub vc = vs2 and vb = va - vs1 from the two source equations, (or for va)
leaving one equation in one unknown (the human way), or we could form matrices (the
computer way). Let’s compute.
&
$0
$1
$1
$
% R1
0
'1
1
1
+
R 2 R3
#
1 ! &v a # &v s 2 #
0 ! $ v b ! = $ v s1 !
1 !$ ! $ !
' ! $% v c !" $% 0 !"
R3 "
Let's try with numbers. Find the node voltages:
10!
8V
a
3A
+
b
40!
vs2
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Voltage Source:
v b ! v a = 8V
KCL at Supernode:
va
v
!3+ b = 0
10
40
vb = 8 + va
4v a ! 120 + v b = 0
Solve:
4v a ! 120 + 8 + v a = 0
5v a = 112
v a = 22.5V
v b = v a + 8 = 30.5V
Check: Write node voltages at nodes, compute currents, look for KCL
10!
22.5 V
a
8V
-
+
30.5 V
b
2.25
40!
3A
0.75
3
At the supernode, compute currents, do they add to zero? Yes!
Nodal Analysis with Dependent Sources:
The key thing here is to express the dependent source value in terms of the node voltages.
a
4V
+
-
2!
4!
b
Right away note that
va = 4V
c
i
4i
Also, b is the same node as
reference (attached by a wire).
But let's keep them separate for a
second.
The dependent source has a term
in i. Let's find i from node
voltages by writing KCL at node
b.
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vb ! va
v ! vc
+i+ b
=0
2
4
And let's apply va = 4, vb = 0
va ! vb vc ! vb
i=
+
2
4
i=
v
4 vc
+
= 2+ c
2 4
4
Now write KCL at node c, plugging in the node voltage version of i:
vc
! 4i = 0
4
vc
v
'
! 4% 2 + c
4
4
&
3
! vc = 8
4
v c = !10.67V
$
"=0
#
And we can check. At node b, iab = 2 A, ibc = 10.67/4=2.666A, i must be .666, 4i is 2.67A
check!
Summary:
Full blown process for nodal analysis is
1. First decide you are using nodal analysis.
2. Write dependent source equations in terms of node voltages.
3. Next look for the voltage sources. Place a reference node at the low end of a voltage
source to minimize the number of supernodes.
4. For voltage sources attached to the reference node, write the auxiliary equation.
(Basically, a node voltage becomes the source voltage or its negative.)
5. For supernodes, write the auxiliary equation and KCL for the supernode.
6. Write KCL for all other nodes.
7. Check your equations for consistency: Are the signs of the node voltage for the node
equation all the same? Are there as many terms as there are connections? Are the current
signs consistent with leaving the node?
8. Solve, either by algebra or matrix mathematics (linear algebra).
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9. Check your answer by using the node voltages to compute branch currents and
checking that KCL is satisfied.
10. Find the value you were looking for in the first place, which may or may not be a
node voltage, by applying KCL, KVL or Ohm's Law as needed.