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 ) 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 NODAL ANALYSIS OF CIRCUITS WITHOUT VOLTAGE SOURCES 2.1 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:
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 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 relative to the voltage of the reference node 0. We call this node voltage v ; its positive reference (the plus sign) is assumed to be at node ; 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 R c is called a grounded element because one of its leads is connected to the reference node, or ground. The other resistor, R b, 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 R b and R c in more detail: 2
The element voltage of the grounded resistor v Rc is the same as the node voltage 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,. Consider now the floating resistor with element voltage v Rb : This element voltage is the difference of two node voltages:. The only alternative would be to reverse the polarities defining the element voltage, in which case the element voltage would be equal to. 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 v i 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:
i resistors away from node Topic 4A Nodal Analysis = i current sources towards node v i v j R a + v i v m R b = i s1 i s2 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 h i, h j and h m are analogous to the node voltages, v i, v j and v m. 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,, and v 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
The KCL equation is: i resistors = i current sources away from node towards node R a + R b = i s1 i s2 Note that the voltage on the far terminal of current source i s2 does not affect the current at node 1 since the connection from node to node 1 is through an independent current source of defined current. On the other hand that voltage on the far terminal of R b, 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: + + v = i s4 R c R b R d At this node, only one current source is attached so there is only one current term. The partial circuit for node is as follows: The KCL equation for node is: v R e + v R d = i s2 + i s We can see that the current in the floating, or non-grounded, resistor R d appears in the equations for each of the two nodes to which it is connected (nodes 2 and ); 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. Tidying-up the KCL Equations We can re-write our nodal equations as a linear function of unknown voltages: 5
1 + 1 R a R b + v 2 1 R b + v 0 ( ) = i s1 i s2 1 R b + v 1 2 + 1 + 1 R b R c R d + v 1 R d = i s4 ( ) + 1 R d 0 + v 1 + 1 R d R e = i s2 + i s Consider the general set of equations in the same three variables: y 11 + y 12 + y 1 v = i 1 y 21 + y 22 + y 2 v = i 2 y 1 + y 2 + y v = i This looks a bit of a mess. Much of the problem is because the unknown variables, and v are repeated on each line; by putting them as a heading above each column in a tabular format, we need only write them once: v const y 11 y 12 y 1 = i 1 y 21 y 22 y 2 = i 2 y 1 y 2 y = i The set of unknown voltages can be rotated and placed by the side of the set of coefficient terms: y 11 y 12 y 1 i 1 y 21 y 22 y 2 v 2 = i 2 y 1 y 2 y v i It is understood that is the variable for the first column of coefficients,, 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 1 0 R a R b R b 1 1 + 1 + 1 1 v 1 i s1 i s2 v R b R b R c R d R 2 = i s4 d 0 1 1 + 1 v i s2 + i s R d R d R e 6
Finally, we can substitute the component values: R a = 4 Ω, R b = 2 Ω, R c = 2 Ω, R d = Ω, R e = 2 Ω. i s1 = 11 A, i s2 = 4 A, i s = A, i s4 = 4 A. 1 4 + 1 1 2 2 1 1 2 2 + 1 2 + 1 0 1 0 1 1 + 1 2 v 1 4 2 = 1 4 2 0 1 0 1 5 6 v 7 = 4 7 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: 2 0 28 8 2 v 2 = 24 0 2 5 v 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: = 12 V, = 4 V, and v = 10 V. Notice that the last operation of scaling can destroy the symmetry property of the matrix. A NODAL ANALYSIS EXAMPLE Find the current i for the following circuit using nodal analysis: Solution 7
There are N = 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 and are defined. The partial circuit pertaining to node 1 is as follows: The KCL equation for node 1 is: 4 + = 1 The partial circuit pertaining to node 2 and its KCL equation is: The matrix form for our example circuit is: We carry out the additions: 1 4 + 1 1 7 12 1 6 + 1 1 6 + 1 1 1 2 = 8 = 1 8 = 1 8 And scale the two rows by common denominators 12 and 6, respectively: 7 4 2 v 2 = 12 48 We now consider solving the equations produced by nodal analysis. 8
4 GAUSSIAN ELIMINATION The equation in matrix from for the second example may be written: Which corresponds to the equations y 11 y 12 y 21 y 22 v 2 = i 1 i 2 y 11 + y 12 = i 1 y 21 + y 22 = i 2 Let us multiply the second equation by y 12 /y 22, and adopt our tabular representation: const y 11 y 12 i 1 y 21 y 22 i 2 9 const y 11 y 12 i 1 y 21 y 12 y 22 y 12 i 2 y 12 y 22 This has made the coefficients of equal. If we form the difference of the two equations, v2 will disappear: We can now determine : const y 11 y 21y 12 y 22 0 i 1 i 2y 12 y 22 = i 1y 22 i 2 y 12 y 11 y 22 y 21 y 12 Having determined, we can use either of the original equations to determine. 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: const y 11 y 12 i 1 y 21 y 22 i 2 const y 11 y 21y 12 y 22 0 i 1 i 2y 12 y 22 Note that we are eliminating the second row and column. The element on the main diagonal of the row and column to be eliminated, y 22 in this case is called the pivot. Other elements that will disappear in the elimination are those in the second column, y 12, and those in the second row, y 21 and i 2. 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 y 12 and y 21. We then form the product of these elements and divide it by the pivot, eg y 12 y 21 /y 22. The target element is the intersection of the row of y 12 and the column of y 21, ie y 11. We then subtract y 12 y 21 /y 22 from y 11. The same rule explains the modification of the i 1 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.
The basic operation in the element-by-element form of Gaussian elimination may be represented as follows: Element d is the pivot. a c b d a b c 0 d 0 0 In the example above, the target element is to the left of the pivot; it may also be to the right: a c b d 0 b a d c 0 0 We can now apply Gaussian elimination to the 2 2 example above: Hence: const 7 4 12 2 48 7 4 const ( ) ( 2) const 1 0 156 = 156 1 = 156 1 = 12 V We can substitute into either of the original equations: ( ) = 1 ( 48 + 24 ) = 72 = 24 V = 1 48 + 2V 1 We were asked for branch current i in the original circuit: 0 12 4 ( ) 48 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. i = = 12 24 = 4 A We can now solve the nodal equations we obtained in the first example: 2 0 28 8 2 v 2 = 24 0 2 5 v 42 We first eliminate row and column ; the element 5 is the pivot; since there are zeros in column and in row, we only have two products to consider, namely ( 2) ( 2) and ( 2) (42). We use the tabular layout: 10
v const 2 0 28 Topic 4A Nodal Analysis v const 2 0 28 8 2 24 8 4 0 24 + 84 0 2 5 42 5 5 Finally, we have scaled row 2 by 5 to tidy it up. const 2 28 6 5 6 5 const F 2 28 15 6 6 We now eliminate row and column 2; the element 6 is the pivot; there are two products to consider, namely ( 2) ( 15) and ( 2) ( 6): Hence: const 2 28 15 6 6 const 0 6 0 28 72 6 = 26 78 6 = 6 = 12 V We can then use one of the equations after the first elimination to obtain : = 1 2 V 1 28 const 78 6 ( ) = 1 ( 12 28 ) 2 = 8 2 = 4 V Finally, we can use any of the original equations to obtain v : v = 1 ( 5 42 + 2V 2 ) = 1 ( 5 42 + 2 4) = 50 5 = 10 V 5 INDEPENDENCE AND SOLVABILITY 0 26 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
In the de-activated circuit, nodes 2 and do not have a path to the reference node. This circuit can still be described by KCL equations for its 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 R b 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 NODAL ANALYSIS OF CIRCUITS WITH VOLTAGE SOURCES 6.1 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 = 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
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 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. Within the larger Kirchhoff boundary, we arbitrarily denoted one of the node voltages as v x ; due to constraints imposed by the voltage sources, the other node voltages within this boundary are related to v x and are v x + 8 and v x + 6. The smaller Kirchhoff boundary has only a single node voltage which we denote as v y. 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. 1
We may choose as ground reference node either of the nodes within the generalised node (nodes 1, 2 or ) or node 4. To begin, let us choose the reference to be at node 2; this forces v x to be 0 V; node 1 now has a known voltage of 8 V and node 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: v y 8 + v y 6 1 The solution is v y = 2 V. We now know all of the node voltages and can compute any voltage or current of interest. = 6 Consider what happens what happens if we choose the ground reference to be at node 4. This forces v y 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
v x + 8 + v x ( v x + 8) 4 Topic 4A Nodal Analysis ( + vx + 8) v x 4 + v x + 6 1 + v x ( v x + 6) ( + vx + 6) v x 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: = 6 Application of KCL to this surface results in the much simpler (but equivalent) equation v x + 8 + v x + 6 1 The solution is v x = 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 Ω resistor and the 4 Ω resistor are both, in fact, redundant. 6. 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.. 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. = 6 5. Solve the resulting equations for the unknown node voltages. 6. Solve for any element voltages or currents. 15
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 N v v-sources, the number of nodes is reduced by N v. 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
We see that, apart from the reference node, there are two nonessential nodes and one supernode. Known voltages (0 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: The solution is v = 8 V. v 0 2 ( ) + v 4 + v 20 + v + 15 1 = 10 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 = 1 6.6 General Form for the Matrix Nodal Equation Consider the following circuit example: 17
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 v s2 and g b and the one between v s and g f ) and N v = v-sources; hence, N KCL = 7 1 =. We identify the ordinary and generalized nodes by erasing all element bodies except those of the v- sources: 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 nodes each provide a KCL equation: Supernode: Middle essential node: Right essential node: g a ( v s1 ) + g b ( v s2 ) + g c ( ) = i s1 i s2 g c ( ) + g d + g e ( v ) = i s2 18
g e ( v ) + g f ( v v s ) = i s These equations may be written compactly in matrix form: g a + g b + g c g c 0 g c g c + g d + g e g e 0 g e g e + g f v i s1 i s2 + g a v s1 + g b v s2 = i s2 i s + g f v s 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 i- sources 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 g x 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
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. 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): g a + g b + g c + g f g c g f g c g f g c + g d + g e + g f v 2 = i s + g a v s1 + g f v s i s + g e v s2 g f v s 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: G 11 G 12 G 21 G 22 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 G ii is the sum of the conductances connected to essential node i or supernode i. 20
2) Each off-diagonal entry G ij is the negative of the sum of the conductances connected between node or supernode i and node or supernode j. ) The nodal conductance matrix G is symmetric, that is G ji = G ij. For example, the sum of the conductances connected to supernode 2 is g c + g d + g e + g f, 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 g c + g f, 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: g a + g b + g c + g f g c g f g c g f g c + g d + g e + g f v 2 = i s + g a v s1 + g f v s i s + g e v s2 g f v s We can move the terms on the right hand side to the left hand side: ( g a + g b + g c + g f ) + ( g c g f ) i s + g a v s1 + g f v s ( g c g f ) + ( g c + g d + g e + g f ) i s + g e v s2 g f v s We can now apply superposition. ( ) = 0 ( ) = 0 The currents due to the original left hand side can be obtained by setting the independent source values i s and v s1, v s2, v s to zero; then the currents due to the independent sources can be obtained by setting the nodal voltages and 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: i source1 = i s + g a v s1 + g f v s i source2 = i s + g e v s2 g f v s 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 i s and v s1, v s2, v s to zero; this corresponds to de-activation all voltage and currents sources in the circuit: 21
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: g a + g b + g c + g f g c g f g c g f g c + g d + g e + g f v 2 = i s + g a v s1 + g f v s i s + g e v s2 g f v s 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 and : 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
1 8 + 1 8 + 1 4 + 1 2 1 4 + 1 2 1 4 + 1 2 1 8 + 1 8 + 1 4 + 1 2 1 = 4 1 4 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 8 + 16 2 + 5 8 8 16 = 16 2 5 12 Assembling these two parts gives the complete matrix nodal equation: The solution is easily computed to be: 7.2 Example 4.6 1 4 1 4 = 16 V = 16 12 = 0 V Write the nodal equations for the following circuit by inspection and solve them: The circuit prepared for nodal analysis is as follows: 2
Two KCL equations will be needed. We next draw the deactivated network: The nodal conductance matrix, by inspection, is 1 4 + 1 2 1 4 The forced network is as follows: 1 4 1+ 1 4 + 1 2 = 4 1 4 1 4 7 4 Summing the currents directed toward the two essential nodes gives 2 + 4 2 12 + 4 = 4 8 1 Assembling the component parts gives the complete matrix nodal equation: The solution is easily computed to be: 4 1 4 = 4 V 1 4 7 4 = 4 8 = 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