Notes for course EE1.1 Circuit Analysis 2004-05
TOPIC 4 – NODAL ANALYSIS
OBJECTIVES
1) To develop Nodal Analysis of Circuits without Voltage Sources
2) To develop Nodal Analysis of Circuits with Voltage Sources
3) To develop Nodal Analysis by Inspection
1
INTRODUCTION
The nodal analysis methods we have introduced so far rely on intuition; we need to recognise series
circuits, parallel circuits and subcircuits that can be transformed in order to simplify the circuit such
that a an unknown voltage or current can be determined.
Nodal analysis is quite different in that it is a systematic method that can be applied in a systematic
way to any circuit containing certain types of elements; it is the method employed by computer
circuit analysis programs such as SPICE.
2
2.1
NODAL ANALYSIS OF CIRCUITS WITHOUT VOLTAGE SOURCES
Nodes, Node Voltages and Element Voltages
Consider a circuit having a reasonable level of complexity:
It is complicated enough that we need an algorithmic method of analysis.
We will use it as motivation for our explanation of the nodal method.
Though it has only resistors and current sources, the algorithm we develop for solving it will be
applicable to circuits with voltage sources as well
We have labelled all the elements of our circuit with both literal and numeric values because we
want to show how each element enters into the analysis.
As you might expect from the name, nodal analysis concentrates on the circuit nodes.
Remember that the nodes are the connected islands of conductor (including element leads) that
remain when we erase the bodies of all the circuit elements.
The nodes of our example circuit are shown below:
Topic 4A – Nodal Analysis
Identifying and counting the nodes is an important part of nodal analysis.
There are N = 4 nodes in our circuit.
We have labelled them 0 through 3 in the figure (rather than 1 through 4).
The reference node is usually assigned the index 0 (this convention applies to SPICE).
Now look at the following:
We show the circuit nodes, but we have also added a voltmeter measuring the voltage at node 3
relative to the voltage of the reference node 0.
We call this node voltage v3; its positive reference (the plus sign) is assumed to be at node 3; its
negative reference (the minus sign) is assumed to be at the reference node, node 0.
We will not show these reference signs explicitly in the future, but will assume tacitly that each of
the non-reference nodes carries a plus reference sign.
We now show the complete set of node voltages relative to reference node 0:
Note that we show the ground reference as a ground symbol on the circuit diagram indicating that
the node to which it is connected is the reference for all other node voltages.
To see why the node voltages are so important, consider the following circuit:
We show the nodes and only two of the resistor elements.
The resistor labelled Rc is called a grounded element because one of its leads is connected to the
reference node, or ground.
The other resistor, Rb, does not have this property. Both of its leads are connected to non-reference
nodes; hence, it is called a floating element, or non-grounded element.
Let us focus on the resistors Rb and Rc in more detail:
2
Topic 4A – Nodal Analysis
The element voltage of the grounded resistor vRc is the same as the node voltage v2 at the nonreference node to which it is connected.
The only other way we could define this element voltage would be to reverse its polarity – in which
case it would be the negative of the non-reference node voltage, –v2.
Consider now the floating resistor with element voltage vRb:
This element voltage is the difference of two node voltages: v1 – v2.
The only alternative would be to reverse the polarities defining the element voltage, in which case
the element voltage would be equal to v2 – v1.
From this discussion, we can draw the following important conclusion: each and every element
voltage in the circuit is completely determined by the node voltages; in fact, each is either a node
voltage, the negative of a node voltage, or the difference of two node voltages.
Thus, if we are able to find the node voltages, we will then be able to compute all element voltages.
2.2
Setting-up nodal analysis
It should be clear that if there are N nodes in a circuit there will be N – 1 node voltages: ie those at
the N - 1 non-reference nodes with respect to the voltage at the reference node.
We therefore need N – 1 independent equations in order to compute them.
KVL only allows us to determine the element voltages from the node voltages.
So we must turn to KCL for these equations. It seems logical to write one KCL equation at each of
the N – 1 non-reference nodes.
Consider an arbitrary node whose voltage is vi and to which are connected some resistors and
current sources:
The reference direction for the resistor currents is in principle an arbitrary choice.
However, choosing these references to be pointed away from node i, as shown, results in an
equation that is in a compact and attractive form for solution.
We can express these resistor currents in terms of element voltages using Ohm's law and then
express the element voltages in terms of the node voltages.
This allows us to write a KCL equation in the following form:
3
Topic 4A – Nodal Analysis
∑
away from node
∑
iresistors =
icurrent sources
towards node
vi − v j vi − vm
+
= is1 − is2
Ra
Rb
If we do this for each of the N – 1 non-reference nodes, we will obtain N – 1 equations whose
unknown variables are the N – 1 node voltages.
We can develop a hydraulic analogy for the situation described by the above KCL equation:
The two constant flow rate water pumps are analogous to the i-sources and the pipes carrying water
away from the central hill (analogous to node i) to the other hilltops (analogous to nodes j and m)
are analogous to the resistors. The heights hi, hj and hm are analogous to the node voltages, vi, vj
and vm.
We now have enough mathematical background to solve the circuit.
We choose the ground reference as before to be at the bottom node and assign unknown node
voltages v1, v2, and v3 to the other nodes:
The circuit "prepared for nodal analysis" is as follows:
Remember that the six ground symbols all refer to the single node chosen as reference, ie node 0.
The portion of the circuit pertaining to the equation at node 1 is as follows:
4
Topic 4A – Nodal Analysis
The KCL equation is:
∑
∑
iresistors =
away from node
icurrent sources
towards node
v1 v1 − v2
+
= is1 − is2
Ra
Rb
Note that the voltage on the far terminal of current source is2 does not affect the current at node 1
since the connection from node 3 to node 1 is through an independent current source of defined
current. On the other hand that voltage on the far terminal of Rb, node 2, does affect the current at
node 1, as we can see from the KCL equation.
The partial circuit concerning node 2 is as follows:
The KCL equation for node 2 is:
v2 v2 − v1 v2 − v3
+
+
= −is4
Rc
Rb
Rd
At this node, only one current source is attached so there is only one current term.
The partial circuit for node 3 is as follows:
The KCL equation for node 3 is:
v3 v3 − v2
+
= is2 + is3
Re
Rd
We can see that the current in the floating, or non-grounded, resistor Rd appears in the equations for
each of the two nodes to which it is connected (nodes 2 and 3); these currents are equal in
magnitude but of opposite sign. Currents in the grounded resistors appear just once, in the KCL
equation for the node to which they are connected.
2.3
Tidying-up the KCL Equations
We can re-write our nodal equations as a linear function of unknown voltages:
5
Topic 4A – Nodal Analysis
⎛ 1
⎛ 1 ⎞
1 ⎞
v1 ⎜
+
+ v2 ⎜ − ⎟ + v3 ( 0 ) = is1 − is2
⎟
⎝ Ra Rb ⎠
⎝ Rb ⎠
⎛ 1 ⎞
⎛ 1
⎛ 1 ⎞
1
1 ⎞
v1 ⎜ − ⎟ + v2 ⎜
+
+
+
v
3
⎜⎝ − R ⎟⎠ = −is4
⎝ Rb ⎠
⎝ Rb Rc Rd ⎟⎠
d
⎛ 1 ⎞
⎛ 1
1⎞
v1 ( 0 ) + v2 ⎜ − ⎟ + v3 ⎜
+ ⎟ = is2 + is3
⎝ Rd ⎠
⎝ Rd Re ⎠
Consider the general set of 3 equations in the same three variables:
y11v1 + y12 v2 + y13v3 = i1
y21v1 + y22 v2 + y23v3 = i2
y31v1 + y32 v2 + y33v3 = i3
This looks a bit of a mess. Much of the problem is because the unknown variables v1, v2 and v3 are
repeated on each line; by putting them as a heading above each column in a tabular format, we need
only write them once:
×v1 ×v2
y11 y12
y21 y22
y31 y32
×v3
const
y13 =
i1
y23 =
i2
y33 =
i3
The set of unknown voltages can be rotated and placed by the side of the set of coefficient terms:
⎡ y11
⎢y
⎢ 21
⎢⎣ y31
y12
y22
y32
y13 ⎤ ⎡ v1 ⎤ ⎡ i1 ⎤
y23 ⎥⎥ ⎢⎢ v2 ⎥⎥ = ⎢⎢i2 ⎥⎥
y33 ⎥⎦ ⎢⎣ v3 ⎥⎦ ⎢⎣ i3 ⎥⎦
It is understood that v1 is the variable for the first column of coefficients, v2, for the second column,
and so on.
The unknown voltages, the source currents and the coefficients depending on the resistor values
have each been put in square brackets to show that they belong together.
The notation we have used is called matrix form. The matrix is a powerful concept which underlies
many computational methods and computer algorithms where large matrices may be treated as
single variables.
Matrix algebra defines the rules for operations on matrices, such as addition and subtraction.
We will not use matrix algebra at this point, as it has not been covered in Mathematics; we just use
the matrix as a convenient way of expressing the equations in tidy systematic form.
The set of equations for our example circuit may now be expressed in matrix form:
⎡ 1
1
+
⎢
⎢ Ra Rb
⎢
1
⎢ −
Rb
⎢
⎢
0
⎢
⎣
⎤
⎥
⎥ ⎡ v1 ⎤ ⎡ is1 − is2 ⎤
1 ⎥⎢ ⎥ ⎢
⎥
−
⎥ ⎢ v2 ⎥ = ⎢ −is4 ⎥
Rd ⎥
⎢⎣ v3 ⎥⎦ ⎢⎣is2 + is3 ⎥⎦
1
1 ⎥
+
⎥
Rd Re ⎦
1
Rb
1
1
1
+
+
Rb Rc Rd
1
−
Rd
−
0
6
Topic 4A – Nodal Analysis
Finally, we can substitute the component values:
Ra = 4 Ω, Rb = 2 Ω, Rc = 2 Ω, Rd = 3 Ω, Re = 2 Ω. is1 = 11 A, is2 = 4 A, is3 = 3 A, is4 = 4 A.
⎡1 1
⎢4 + 2
⎢
⎢ −1
⎢ 2
⎢
⎢ 0
⎢⎣
1
2
1 1 1
+ +
2 2 3
1
−
3
−
⎤
⎡ 3
0 ⎥
⎢ 4
⎥ ⎡ v1 ⎤ ⎢
1
1
− ⎥ ⎢⎢ v2 ⎥⎥ = ⎢ −
⎢ 2
3 ⎥
⎢⎣ v3 ⎥⎦ ⎢
⎥
1 1⎥
⎢ 0
+
⎢⎣
3 2 ⎥⎦
1
⎤
0 ⎥
2
⎥ ⎡ v1 ⎤ ⎡ 7 ⎤
4
1⎥⎢ ⎥ ⎢ ⎥
− ⎢ v2 ⎥ = ⎢ −4 ⎥
3
3⎥
⎢v ⎥ ⎢ 7 ⎥
1 5 ⎥⎥ ⎣ 3 ⎦ ⎣ ⎦
−
3 6 ⎥⎦
−
Notice that each diagonal term in position (i, i) is the sum of the conductances connected to node i.
Notice also that each off-diagonal term in position (i, j) is the negative of the conductance
connecting nodes i and j; the conductance connecting nodes j and i is precisely the same, which
means that the coefficient matrix is symmetric.
These observations make it possible to write the matrix (ie equations) straight down without
performing the KCL analysis; we will look at this later.
For any equation that is true, we can multiply both sides by the same factor and it is still true; let us
multiply the three rows by common denominators 4, 6 and 6, respectively:
⎡ 3 −2 0 ⎤ ⎡ v1 ⎤ ⎡ 28 ⎤
⎢ −3 8 −2 ⎥ ⎢ v ⎥ = ⎢ −24 ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥
⎢⎣ 0 −2 5 ⎥⎦ ⎢⎣ v3 ⎥⎦ ⎢⎣ 42 ⎥⎦
We now have three tidied-up equations in three unknowns.
The equations may be solved by a variety of methods, including use of a calculator.
We will consider the method of Gaussian elimination later.
The solution is: v1 = 12 V, v2 = 4 V, and v3 = 10 V.
Notice that the last operation of scaling can destroy the symmetry property of the matrix.
3
A NODAL ANALYSIS EXAMPLE
Find the current i for the following circuit using nodal analysis:
Solution
7
Topic 4A – Nodal Analysis
There are N = 3 nodes; thus, we expect to have to write N – 1 = 2 KCL nodal equations.
We select the reference to be the bottom node and assign voltages to the other nodes:
Now let's write one KCL equation at the nodes where v1 and v2 are defined.
The partial circuit pertaining to node 1 is as follows:
The KCL equation for node 1 is:
v1 v1 − v2
+
= −1
4
3
The partial circuit pertaining to node 2 and its KCL equation is:
v2 v2 − v1
+
=8
6
3
The matrix form for our example circuit is:
⎡1 1
⎢4 + 3
⎢
⎢ −1
⎢⎣ 3
1 ⎤
3 ⎥ ⎡ v1 ⎤ ⎡ −1⎤
=
⎥
1 1 ⎥ ⎢⎣ v2 ⎥⎦ ⎢⎣ 8 ⎥⎦
+
6 3 ⎥⎦
−
We carry out the additions:
1⎤
⎡ 7
⎢ 12 − 3 ⎥ ⎡ v1 ⎤ ⎡ −1⎤
⎢
⎥⎢ ⎥ = ⎢ ⎥
⎢ − 1 1 ⎥ ⎣ v2 ⎦ ⎣ 8 ⎦
⎢⎣ 3 2 ⎥⎦
And scale the two rows by common denominators 12 and 6, respectively:
⎡ 7 −4 ⎤ ⎡ v1 ⎤ ⎡ −12 ⎤
⎢ −2 3 ⎥ ⎢ v ⎥ = ⎢ 48 ⎥
⎣
⎦⎣ 2⎦ ⎣
⎦
We now consider solving the equations produced by nodal analysis.
8
Topic 4A – Nodal Analysis
4
GAUSSIAN ELIMINATION
The equation in matrix from for the second example may be written:
⎡ y11
⎢y
⎣ 21
y12 ⎤ ⎡ v1 ⎤ ⎡ i1 ⎤
=
y22 ⎥⎦ ⎢⎣ v2 ⎥⎦ ⎢⎣i2 ⎥⎦
Which corresponds to the equations
y11v1 + y12 v2 = i1
y21v1 + y22 v2 = i2
Let us multiply the second equation by y12/y22, and adopt our tabular representation:
v1
y11
y21
v2
y12
y22
const
i1
i2
v1
y11
y21y12
y22
→
v2
y12
y12
const
i1
i2 y12
y22
This has made the coefficients of v2 equal. If we form the difference of the two equations, v2 will
disappear:
v1
y y
y11 − 21 12
y22
v2
0
const
i y
i1 − 2 12
y22
We can now determine v1:
v1 =
i1y22 − i2 y12
y11y22 − y21y12
Having determined v1, we can use either of the original equations to determine v2.
This is clumsy with variables but much easier with numbers which we will consider a little later.
We now try to develop a simple algorithm for applying Gaussian elimination. Rather than consider
operation on a whole row, we consider applying it element-by-element.
Let's compare the original and reduced forms of the equations:
v1
y11
y21
v2
y12
y22
const
i1
i2
→
v1
y y
y11 − 21 12
y22
v2
0
const
i y
i1 − 2 12
y22
Note that we are eliminating the second row and column. The element on the main diagonal of the
row and column to be eliminated, y22 in this case is called the pivot. Other elements that will
disappear in the elimination are those in the second column, y12, and those in the second row, y21
and i2.
The process can be explained as follows: We take an element in the column to be eliminated and an
element in the row to be eliminated, eg y12 and y21. We then form the product of these elements and
divide it by the pivot, eg y12y21/y22. The target element is the intersection of the row of y12 and the
column of y21, ie y11. We then subtract y12y21/y22 from y11. The same rule explains the modification
of the i1 term. In this case there are only these two products of terms in the row and column to be
eliminated (apart from the pivot); in general we must make sure we do this for all such products. If
one term in such a product is zero, no modification occurs to the target element.
9
Topic 4A – Nodal Analysis
The basic operation in the element-by-element form of Gaussian elimination may be represented as
follows:
a
b
c
d
a−
→
b×c
d
0
0
0
Element d is the pivot.
In the example above, the target element is to the left of the pivot; it may also be to the right:
a
b
c
d
a×d
c
0
0 b−
→
0
We can now apply Gaussian elimination to the 2 × 2 example above:
v1
7
−2
v2
−4
3
const
−12
48
v1
→
→
7−
v1
13
3
( −4 ) × ( −2 )
3
v2
0
v2
0
const
−12 −
( −4 ) × 48
3
const
156
3
Hence:
v1 =
156 3 156
=
= 12 V
13 3
13
We can substitute v1 into either of the original equations:
v2 =
1
1
72
( 48 + 2V1 ) = ( 48 + 24 ) = = 24 V
3
3
3
We were asked for branch current i in the original circuit:
In order to obtain a branch current, we need to express the branch voltage in terms of the node
voltages and then use Ohm's law.
v −v
12 − 24
i= 1 2 =
= −4 A
3
3
We can now solve the 3 nodal equations we obtained in the first example:
⎡ 3 −2 0 ⎤ ⎡ v1 ⎤ ⎡ 28 ⎤
⎢ −3 8 −2 ⎥ ⎢ v ⎥ = ⎢ −24 ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥
⎢⎣ 0 −2 5 ⎥⎦ ⎢⎣ v3 ⎥⎦ ⎢⎣ 42 ⎥⎦
We first eliminate row and column 3; the element 5 is the pivot; since there are zeros in column 3
and in row 3, we only have two products to consider, namely (–2) × (–2) and (–2) × (42). We use
the tabular layout:
10
Topic 4A – Nodal Analysis
v1 v2 v3 const
v1
v2
v3
const
v1 v2
3 −2 0
28
3
−2
0
28
3 −2
−3 8 −2 −24 →
4
84 →
36
−3 8 −
0 −24 +
−3
5
5
5
0 −2 5
42
Finally, we have scaled row 2 by 5 to tidy it up.
const
v1 v2 const F
28 → 3 −2
28
36
−15 36 −36
−
5
We now eliminate row and column 2; the element 36 is the pivot; there are two products to
consider, namely (–2) × (–15) and (–2) × (–36):
v1 v2 const
3 −2
28
−15 36 −36
→
v1
30
3−
36
v2
0
const
72
28 −
36
→
v1
78
36
v2
const
0
26
Hence:
v1 =
26
36
=
= 12 V
78 36
3
We can then use one of the equations after the first elimination to obtain v2:
v2 =
1
1
8
( 3V1 − 28 ) = ( 3 × 12 − 28 ) = = 4 V
2
2
2
Finally, we can use any of the original equations to obtain v3:
v3 =
5
1
1
50
( 42 + 2V2 ) = ( 42 + 2 × 4 ) = = 10 V
5
5
5
INDEPENDENCE AND SOLVABILITY
In general the process we have described leads to a set of equations which can be solved to yield the
nodal equations. We might ask under what conditions can it be guaranteed that the equations are
soluble. The answer is that the equations are soluble provided that when the network is deactivated, by replacing every current source by an open-circuit, then every node should have a path
to the reference node through one or more resistors. Let us consider a couple of examples:
Consider our first example circuit:
We show the de-activated circuit alongside the circuit itself. It can be seen that the de-activated
circuit satisfies our criterion since every node has a resistive path to the reference node. Hence the
equations will have a valid solution for any element values.
Consider now the following circuit and its de-activated version:
11
Topic 4A – Nodal Analysis
In the de-activated circuit, nodes 2 and 3 do not have a path to the reference node. This circuit can
still be described by KCL equations for its 3 nodes; however, the equations will not be soluble
because they are not linearly independent.
But our knowledge of circuit analysis reveals immediately that there are problems with this circuit.
Firstly, we have two independent current sources in series, which is an invalid sub-circuit.
Secondly, since the voltage of a current source is arbitrary, it is obvious that the sub-circuit
comprising Rb and its two nodes has voltages that cannot be defined with respect to the reference
node. Hence, this is an unsatisfactory circuit that no student of Imperial College would design!
For sensible circuits the criterion for a valid solution is satisfied and node voltages can be
determined.
6
6.1
NODAL ANALYSIS OF CIRCUITS WITH VOLTAGE SOURCES
Generalized Nodes
Consider the following example circuit:
We have labelled the four nodes, which can be identified by erasing the body of each of the
elements but leaving the leads.
If the two v-sources were replaced by i-sources, we would have to write 4 – 1 = 3 nodal equations
in order to solve for three independent non-reference node voltages.
With the v-sources in the circuit, however, the node voltages are not independent.
Clearly, to solve the circuit, we should know how many independent node voltages there are.
To do this, we deactivate the voltage and current sources and determine the number of nodes in the
circuit obtained:
12
Topic 4A – Nodal Analysis
We see that now there are just two nodes in this circuit; thus, we would anticipate only 2 – 1 = 1
nodal equation.
Notice that nodes 1, 2, and 3 merge together when the voltage sources are deactivated. This reflects
the fact that their node voltages are not independent, but are constrained by the v-sources.
We have drawn a closed Kirchhoff surface around these three nodes and one around node 4 in the
deactivated network..
Now let's see how this dependency is reflected in the original circuit (before we de-activated the
sources):
The shaded areas each define a Kirchhoff current law boundary; one comprises node 4 and the other
comprises the combination of nodes 1, 2 and 3.
Within the larger Kirchhoff boundary, we arbitrarily denoted one of the node voltages as vx; due to
constraints imposed by the voltage sources, the other node voltages within this boundary are related
to vx and are vx + 8 and vx + 6. The smaller Kirchhoff boundary has only a single node voltage
which we denote as vy.
We call any set of nodes connected by a path of v-sources a generalized node.
We see that there is only one independent node voltage for each generalized node.
We may refer to nodes to which no v-source is connected, such as node 4, simply as nodes (or
"ordinary" nodes).
6.2
Node Classification
Only one nodal equation is required to solve our example circuit because it is essentially a 2-node
circuit.
It therefore seems logical to anticipate writing one KCL equation for either the node 4 or for the
generalized node.
13
Topic 4A – Nodal Analysis
We may choose as ground reference node either of the 3 nodes within the generalised node (nodes
1, 2 or 3) or node 4.
To begin, let us choose the reference to be at node 2; this forces vx to be 0 V; node 1 now has a
known voltage of 8 V and node 3 has a known voltage of 6 V.
When one node in a generalised node is chosen as the reference node and the voltages at the other
nodes within the generalised node become known, these other nodes are known as non-essential
nodes.
Non-essential refers to the fact that KCL does not need to be applied at these nodes because their
node voltages are known.
We now call the ordinary non-reference node (node 4) an essential node to distinguish it from the
others; the voltage at an essential node is unknown.
Thus the voltages at non-essential nodes are known values and the voltages at essential nodes are
unknown.
We may now label the circuit as follows:
We now write one KCL equation at the essential node:
vy − 8
3
+
vy − 6
1
= −6
The solution is vy = 2 V. We now know all of the node voltages and can compute any voltage or
current of interest.
Consider what happens what happens if we choose the ground reference to be at node 4.
This forces vy to be 0 V.
The shaded Kirchhoff boundary is a generalized node which does not contain the reference node.
We refer to such a boundary as a supernode.
A supernode is a generalised node which does not contain the reference node.
KCL for the entire boundary surrounding the supernode is:
14
Topic 4A – Nodal Analysis
vx + 8 vx − ( vx + 8 ) ( vx + 8 ) − vx vx + 6 vx − ( vx + 6 ) ( vx + 6 ) − vx
+
+
+
+
+
=6
3
4
4
1
3
3
Before we solve this equation, let's make a simple observation: there are two pairs of terms that are
equal but opposite in sign (the second and third and the fifth and sixth); they correspond to the
currents through the two resistors connected directly between two of the nodes making up the
supernode. Each resistor current appears once leaving the Kirchhoff surface around the supernode
and once-with opposite sign re-entering.
We can include any such element within the Kirchhoff surface itself:
Application of KCL to this surface results in the much simpler (but equivalent) equation
vx + 8 vx + 6
+
=6
3
1
The solution is vx = –2 V. This is consistent with our earlier result that the voltage of node 4 was
+2 V relative to node 2.
Since a resistor in parallel with a voltage source is equivalent to the voltage source alone, the upper
3 Ω resistor and the 4 Ω resistor are both, in fact, redundant.
6.3
The Nodal Analysis Algorithm
The following algorithm summarizes the nodal method as a procedure:
1. Deactivate the circuit and determine all of the ordinary nodes and generalized nodes.
The latter are those that merge together when the v-sources are deactivated.
2. Choose the ground reference. If it is chosen within a generalized node, all of the other nodes
within that generalized node are called nonessential nodes, for they carry known node voltages. If it
is chosen at an ordinary node, each generalized node is referred to as a supernode. Ordinary nonreference nodes are now referred to as essential nodes.
3. Assign node voltages – one unknown node voltage to each essential node and one unknown node
voltage to an arbitrarily chosen node in each supernode. Each of the other node voltages within each
supernode is expressed in terms of the arbitrarily chosen one and the v-source values. Any
nonessential nodes are labelled with their known values of node voltage.
4. Write one KCL equation for each essential node and one for each supernode in terms of the
unknown node voltages assigned in step 3.
5. Solve the resulting equations for the unknown node voltages.
6. Solve for any element voltages or currents.
15
Topic 4A – Nodal Analysis
6.4
Number of nodal equations required
To answer the question of how many KCL equations we need to analyse a given circuit, consider
the deactivated version of our circuit:
The dark dots represent the terminals of the original sources, where each v-source has been replaced
by a short circuit and the current source by an open circuit.
Notice that when each v-source was deactivated it merged two nodes, thus reducing the number of
nodes by one; hence, if there are Nv v-sources, the number of nodes is reduced by Nv.
Therefore, the number of unknown node voltages (and also the number of KCL equations) is:
N KCL = N − 1 − N v
Each v-source reduces the complexity of nodal analysis by one equation.
Notice, however, that deactivation of the i-source did not affect the number of nodes.
6.5
Example 4.2
Find the current i in the following circuit:
Solution
First, we identify nodes and generalized nodes:
Note that in this case there are no ordinary nodes, only two generalized nodes.
Next, we choose our ground reference node at the bottom left node within generalized node 1.
16
Topic 4A – Nodal Analysis
We see that, apart from the reference node, there are two nonessential nodes and one supernode.
Known voltages (30 V and –20 V) are assigned to the two nonessential nodes.
The voltages of the nodes within the supernode are defined by denoting one of its node voltages as
unknown; we denote the top node voltage as v.
The voltage at the other node in the supernode is expressed in terms of v.
We can now write one KCL equation at the supernode.
Before that, let's the units we might use. Ohm's law can be written in any convenient set of units
desired. For instance, we can have:
V = mA × kΩ
If we express all resistances in kΩ and all currents in mA, then the voltages will still be in volts.
Working in these units, the equation for the supernode is:
v − 30 v v − ( −20 ) v + 15
+ +
+
= −10
2
4
3
1
The solution is v = –8 V.
It is only once that we have determined the node voltages, that we consider the current i that we
were asked to find.
We can label all nodes with their now-known voltages.
We can use the differences of the node voltages (that is, KVL) and Ohm's law to find the resistor
currents and KCL to find currents in elements, resulting in the values shown.
By applying KCL at the ground reference node, we see that i = 10 + 4 + 7 – 2 = 19 mA.
This checks with the current in the 2 kΩ resistor.
We could have determined the number of KCL equations before we started:
N KCL = N − 1 − N v = 5 − 1 − 3 = 1
6.6
General Form for the Matrix Nodal Equation
Consider the following circuit example:
17
Topic 4A – Nodal Analysis
Note that each resistor has been labelled with its conductance.
Let's determine the number of KCL equations needed; we count N = 7 nodes (node between vs2 and
gb and the one between vs3 and gf) and Nv = 3 v-sources; hence, NKCL = 7 – 1 – 3 = 3.
We identify the ordinary and generalized nodes by erasing all element bodies except those of the vsources:
We see that there are two ordinary nodes and two generalized nodes.
Next, selecting the ground reference as the bottom node, we redraw the original circuit prepared for
nodal analysis, with the nodes further classified into essential, nonessential, and supernodes:
We will not need a KCL equation for the non-essential node because its voltage is known.
The remaining 3 nodes each provide a KCL equation:
Supernode:
ga ( v1 − vs1 ) + gb ( v1 − vs2 ) + gc ( v1 − v2 ) = is1 − is2
Middle essential node:
gc ( v2 − v1 ) + gd v2 + ge ( v2 − v3 ) = is2
Right essential node:
18
Topic 4A – Nodal Analysis
ge ( v3 − v2 ) + g f ( v3 − vs3 ) = −is3
These equations may be written compactly in matrix form:
⎡ ga + gb + gc
⎢
−gc
⎢
⎢
0
⎣
−gc
gc + gd + ge
−ge
0 ⎤ ⎡ v1 ⎤ ⎡is1 − is2 + ga vs1 + gb vs2 ⎤
⎥
⎢
⎥
−ge ⎥ ⎢⎢ v2 ⎥⎥ = ⎢
is2
⎥
⎥
ge + g f ⎥⎦ ⎢⎣ v3 ⎥⎦ ⎢⎣
−is3 + g f vs3
⎦
All of the terms involving the independent v-sources are now on the right-hand side of the equation.
The coefficient matrix on the left of the equation consists only of conductances.
The right-hand side contains the currents entering each node due to the independent sources (both isources and v-sources).
The terms on the right are linear combinations of the voltages and currents of the independent
sources, voltage and current.
Hence, the solution will give the nodal voltages as linear combinations of the independent source
values.
It follows that all voltages and currents in the circuit are linear combinations of the source values.
This leads to a general principle of superposition which was the basis for deriving Thevenin and
Norton equivalent circuits.
Notice that in the above circuit, resistor gx correctly disappears from the equations as it is in parallel
with a voltage source.
6.7
Quiz
Q4.2-1. For each of the generalized nodes shown, compute the voltages at the nodes not labelled
with the unknown voltage. All voltage source values are 6 V.
Q4.2-2. Consider the circuit shown below. In order to classify the nodes into regular nodes and
generalized nodes erase the bodies of all elements except the v-sources:
How many nodes are there?
How many are ordinary and how many generalised.
19
Topic 4A – Nodal Analysis
Q4.2-4. Choose the ground reference at node 0.
Identify essential nodes and non-essential nodes and supernodes.
Express non-essential node voltages in terms of unknown voltages, eg v1.
Check the formula for the number of KCL equations.
7
NODAL ANALYSIS BY INSPECTION
Consider the following circuit:
The circuit is already prepared for nodal analysis. KCL at the essential node and the supernode
leads to (working not shown):
⎡ ga + gb + gc + g f
⎢
−gc − g f
⎣
−gc − g f
⎤ ⎡ v1 ⎤ ⎡ is + ga vs1 + g f vs3 ⎤
=
gc + gd + ge + g f ⎥⎦ ⎢⎣ v2 ⎥⎦ ⎢⎣ −is + gevs2 − g f vs3 ⎥⎦
The 2 × 2 matrix on the left is called the nodal conductance matrix; the 2 × 1 matrix on the right is
called the vector of nodal currents due to independent sources (voltage and current sources).
The nodal conductance matrix has the general form:
⎡ G11 G12 ⎤
⎢G
⎥
⎣ 21 G22 ⎦
where the top first is obtained from the KCL equation for node 1 and the second row is obtained
from the KCL equation for node 2.
We now make some observations about the nodal conductance matrix:
1) Each diagonal entry Gii is the sum of the conductances connected to essential node i or
supernode i.
20
Topic 4A – Nodal Analysis
2) Each off-diagonal entry Gij is the negative of the sum of the conductances connected
between node or supernode i and node or supernode j.
3) The nodal conductance matrix G is symmetric, that is Gji = Gij.
For example, the sum of the conductances connected to supernode 2 is gc + gd + ge + gf, and that is
the entry in the (2, 2) position of the G matrix.
The sum of the conductances between supernode 2 and essential node 1 is gc + gf, and the negative
of this quantity appears in both the (1, 2) and the (2, 1) positions.
These special characteristics allow us to write the nodal conductance matrix by inspection.
We now consider the right-hand side of the matrix equation obtained by KCL.
Let's go back to our KCL equations:
⎡ ga + gb + gc + g f
⎢
−gc − g f
⎣
−gc − g f
⎤ ⎡ v1 ⎤ ⎡ is + ga vs1 + g f vs3 ⎤
=
gc + gd + ge + g f ⎥⎦ ⎢⎣ v2 ⎥⎦ ⎢⎣ −is + gevs2 − g f vs3 ⎥⎦
We can move the terms on the right hand side to the left hand side:
( ga + gb + gc + g f ) v1 + ( −gc − g f ) v2 − (is + gavs1 + g f vs3 ) = 0
( −gc − g f ) v1 + ( gc + gd + ge + g f ) v2 − ( −is + gevs2 − g f vs3 ) = 0
We can now apply superposition.
The currents due to the original left hand side can be obtained by setting the independent source
values is and vs1, vs2, vs3 to zero; then the currents due to the independent sources can be obtained by
setting the nodal voltages v1 and v2 to zero.
We can verify that this is the case by applying short-circuits to the reference node from node 1 and
node 2 of our circuit:
The currents which flow to ground in the short circuits are:
isource1 = is + ga vs1 + g f vs3
isource2 = −is + gevs2 − g f vs3
These are indeed the currents on the right hand side of the KCL equation set.
We said that the currents flowing in the conductances on the left hand side of the KICL equation set
can be obtained by setting the independent source values is and vs1, vs2, vs3 to zero; this corresponds
to de-activation all voltage and currents sources in the circuit:
21
Topic 4A – Nodal Analysis
From this partial circuit, we can determine the conductances connected at each node and the
conductances which are connected between nodes.
Hence the two partial circuits, one with sources de-activated and one with node voltages shortcircuited to ground, allow us to write the nodal equations by inspection:
⎡ ga + gb + gc + g f
⎢
−gc − g f
⎣
−gc − g f
⎤ ⎡ v1 ⎤ ⎡ is + ga vs1 + g f vs3 ⎤
=
gc + gd + ge + g f ⎥⎦ ⎢⎣ v2 ⎥⎦ ⎢⎣ −is + gevs2 − g f vs3 ⎥⎦
This method is the main basis for computer methods of circuit analysis and is used in such packages
as SPICE.
7.1
Example 4.5
Write the nodal equations for the following circuit by inspection and solve for the node voltages v1
and v2:
Solution
The deactivated network is as follows:
From this partial circuit, the left hand side of the KCL equation set, containing the nodal
conductance matrix may be written by inspection:
22
Topic 4A – Nodal Analysis
⎡1 1 1 1
⎢8 + 8 + 4 + 2
⎢
⎢ ⎛ 1 1⎞
⎢ − ⎜⎝ + ⎟⎠
4 2
⎣
⎛ 1 1⎞ ⎤
⎡
−⎜ + ⎟ ⎥
⎝ 4 2 ⎠ ⎡ v1 ⎤ ⎢ 1
⎥⎢ ⎥ = ⎢
1 1 1 1 ⎥ ⎣ v2 ⎦ ⎢ 3
−
+ + + ⎥
⎢⎣ 4
8 8 4 2⎦
3⎤
− ⎥
4 ⎡ v1 ⎤
⎥⎢ ⎥
v
1 ⎥⎣ 2⎦
⎥⎦
The source vector on the right-hand side of the KCL equation set is computed from the circuit with
nodes 1 and 2 short circuited to the reference node (forced network):
Calculation of the currents entering the essential node and the supernode gives:
⎡ 24 16
⎤
⎢ 8 + 2 + 5 ⎥ ⎡ 16 ⎤
⎢
⎥=⎢
⎥
⎢ 8 − 16 − 5 ⎥ ⎣ −12 ⎦
⎢⎣ 8 2
⎥⎦
Assembling these two parts gives the complete matrix nodal equation:
⎡
⎢ 1
⎢
⎢− 3
⎢⎣ 4
3⎤
− ⎥
4 ⎡ v1 ⎤ ⎡ 16 ⎤
⎥⎢ ⎥ = ⎢
v
−12 ⎥⎦
1 ⎥⎣ 2⎦ ⎣
⎥⎦
The solution is easily computed to be:
v1 = 16 V
7.2
v2 = 0 V
Example 4.6
Write the nodal equations for the following circuit by inspection and solve them:
The circuit prepared for nodal analysis is as follows:
23
Topic 4A – Nodal Analysis
Two KCL equations will be needed.
We next draw the deactivated network:
The nodal conductance matrix, by inspection, is
1 ⎤
⎡1 1
⎡ 3
−
⎢4 + 2
⎥
v
⎡ 1⎤ ⎢ 4
4
⎢
⎥⎢ ⎥ = ⎢
⎢ − 1 1 + 1 + 1 ⎥ ⎣ v2 ⎦ ⎢ − 1
⎢⎣ 4
⎢⎣ 4
4 2 ⎥⎦
1⎤
− ⎥
4 ⎡ v1 ⎤
⎥
7 ⎥ ⎢⎣ v2 ⎥⎦
4 ⎥⎦
The forced network is as follows:
Summing the currents directed toward the two essential nodes gives
4 ⎤
⎡
2
+
⎢
2 ⎥ ⎡4⎤
⎢
⎥=⎢ ⎥
⎢ −12 + 4 ⎥ ⎣ −8 ⎦
⎢⎣
1 ⎥⎦
Assembling the component parts gives the complete matrix nodal equation:
⎡ 3
⎢ 4
⎢
⎢− 1
⎢⎣ 4
1⎤
− ⎥
4 ⎡ v1 ⎤ ⎡ 4 ⎤
=
⎥
7 ⎥ ⎢⎣ v2 ⎥⎦ ⎢⎣ −8 ⎥⎦
4 ⎥⎦
The solution is easily computed to be:
v1 = 4 V
v2 = −4 V
Note that for a circuit containing no voltage sources, the currents towards the essential nodes in the
forced network are simply the sums of the currents in the current sources connected to each node.
8
CONCLUSIONS
This completes our present study of nodal analysis. Later we will analyse circuits containing
controlled, as distinct from independent, sources and we will add the further passive elements of the
inductor and the capacitor. We now look at the inductor and the capacitor and consider a different
form of analysis called time domain analysis.
24