Nodal analysis is more commonly used than mesh or loop analysis for analysing networks. It can be used to determine the unknown node voltages of both planar and non-planar circuits. Nodal equations are usually formed by applying Kirchoff’s Current Law to the nodes with unknown voltages, whereas equations based on Kirchoff’s Voltage Law are used to form the mesh equations. In order to apply nodal analysis to a circuit, the first step is to select a reference node or datum node and then assign a voltage at each of the other nodes with respect to the reference node. In a circuit with dc sources, the node that has the lowest voltage is usually selected as the reference node and then the other node voltages would be positive. Often the node that has the maximum elements connected to it in a circuit tends to be the reference node. Many electronic circuits have a metallic chassis and the reference node, usually the negative terminal of the dc source present in the circuit, happens to be connected to the chassis. It is common practice to connect the chassis to earth terminal of the utility supply. Then the reference node is at earth or ground potential and hence the reference terminal is referred to as the ground terminal, even if it is not earthed. In a power system, the casing of power appliances is usually earthed and the neutral of the utility supply remains connected to earth at the source. The reference node in such a power system is then at zero potential.
In nodal analysis, the voltage at the reference node is assumed to be zero. The voltages at other nodes are expressed with reference to the datum node. Since the unknown node voltages are determined by nodal analysis, it is logical to write the KCL equations at nodes. The procedure can be summarized as follows.
● Select a reference node and treat it to be at zero or ground potential.
● Label the nodes with unknown voltages.
● At each of these nodes, mark currents in the elements as flowing away from the node.
● Form KCL equations and solve the set of simultaneous equations for the unknown voltages.
The voltage across an element can be expressed as the difference of voltages at nodes to which it is connected. Use the passive sign convention and mark the currents to be flowing away from the node at which the KCL is applied. Then the mutual conductance term would have a negative sign whereas the self-conductance at a node would be a positive value. This aspect should become clear after a few examples. It is worth stressing that labelling the nodes and assigning current direction properly are really important. The practice of this technique would enable one to write the KCL equations and the matrix equations by inspection.
Nodal analysis is the ideal technique for analysis when all the sources in a circuit are independent current sources and the first example shows how to analyse such a circuit.
Worked Example 1
Obtain the unknown node voltages V1 and V2 of the circuit in Fig. 1.
Solution:
The circuit shown in Fig. 1 is presented again in Fig. 2 to show that there are only three nodes in this circuit and that it is appropriate to select the reference node as shown. It is seen that the node that has the highest number of elements connected to it is the reference or datum node. It turns out that the potential at the datum node is lowest. Let it be assumed that the voltage at the datum node is zero volt.
The datum node has been marked and the unknown node voltages have also been labelled. The next task is to form the KCL equations at these nodes.
For the node with the voltage V1, the circuit in Fig. 3 is used. The two nodes with unknown voltages are marked A and B in the circuit shown in Fig. 3.
At node A, the sum of currents entering it should equal the sum of currents leaving it. Then
It is seen from the circuit in Fig. 3 that
Substituting for Ia and Ib from equation (2) into equation (1), we get that
On grouping terms, equation (3) becomes:
That is,
It is seen that the coefficient of V1 is G11, the sum of conductances connected to node A. This term, G11, reflects the self-conductance connected to node A, whereas the term, G12, is the mutual conductance linking nodes A and B. Note that it has a negative sign.
The KCL equation at node B is obtained from the circuit in Fig. 4.
It is seen from the circuit in Fig. 4 that
Substituting for Ic and Id from equation (7) into equation (6), we get that
On grouping terms, equation (8) becomes:
That is,
The solution can be verified easily. For the node voltages defined above, the current through the 6 W resistor is 2 A, the current through the 4 W resistor is 1 A, and the current through the 2 W resistor is 4 A. It is seen that the KCL equations at nodes A and B are satisfied.
Worked Example 2
Obtain the unknown node voltages V1 and V2 of the circuit in Fig. 5.
Solution:
The KCL equation at node A is:
The KCL equation at node B is:
These two equations, (12) and (13), can be expressed in the form of a matrix equation.
On solving, we obtain that V1 = 12 Volts and V2 = 8 Volts.
Worked Example 3
Find the current through the 8 W resistor in the circuit shown in Fig. 8.
Solution:
The only unknown node voltage to be solved for is the voltage at node a. At this node,
Then Va = 64 V.
It is not appropriate to use the mesh analysis for this problem, because there are four unknowns and the solution would not be as simple as that presented now.
Find Va, Vb and Vc marked in the circuit shown in Fig. 9.
Solution:
Since three unknown node voltages have to be determined, three independent equations are required to be formed. By applying KCL to those nodes, three equations can be formed.
KCL equation at node a:
KCL equation at node b:
KCL equation at node c:
From equations (15), (16) and (17), we get that
The set of simultaneous equations can also be solved by elimination. This technique is illustrated below. Equations (15), (16) and (17) can be represented as follows:
Then
Find V1, V2 and V3 marked in the circuit shown in Fig. 10.
Solution:
Since three unknown node voltages have to be determined, three independent equations are required to be formed. By applying KCL to node 2, one equation can be formed. Since the voltage source is connected to two nodes, the second equation can be formed. The current through the voltage source has been marked as I. Two expressions for I can be formed, one from the KCL applied to node 1 and the other from node 3. By equating these two expressions, the third equation can be formed.
Current I through the voltage source flows away from node 1 to node 3, as shown in Figs. 11 and 12. Then
That is,
KCL at node 2:
Since the voltage source is connected to nodes 1 and 3,
From equations (27), (28) and (29), we get that
This example is used again next to illustrate what a supernode is and how a circuit can be analysed using a supernode.
By forming a supernode, it may be possible to reduce the number of unknowns by one and the solution thus can become simpler. The circuit diagram used in worked example 5 is re-presented in Fig. 14 with the box enclosed by broken lines forming a supernode. As can be seen, the use of a supernode is appropriate when the circuit contains a floating voltage source. A floating voltage source has neither of its terminals to the datum node. We can apply Kirchoff’s current law to supernode. Then
That is,
The above equation can be simplified as shown next.
KCL at node 2:
From equations (30) and (31), we obtain that
To form a supernode, we enclose a region and that region becomes a supernode. The algebraic sum of currents incident at the supernode is also to be zero. The use of supernode is useful for circuits containing floating voltage sources . If the floating voltage source is a dependent source, there may be no reduction in the number of equations required to be solved for, but nonetheless the formation of one of the equations may be slightly easier.
EXAMPLES WITH DEPENDENT SOURCES
Worked Example 7
Obtain voltages V2 and V3 of the circuit in Fig. 15.
Solution:
Two unknown node voltages have to be determined. One KCL equation can be formed by equating the two expressions for current marked in Fig. 15. The second equation is obtained from the expression for the dependent source.
The KCL equations obtained at nodes 2 and 3 are presented below.
KCL Equation at node 2:
KCL Equation at node 3:
From equations (32) and (33), we obtain that
From the circuit in Fig. 15,
On solving equations (34) and (35), we obtain that
Worked Example 8
Find V1, V2 and V3 marked in the circuit shown in Fig. 16.
Solution:
Since three unknown node voltages have to be determined, three independent equations are required to be formed. By applying KCL to node 2, one equation can be formed. Using dependency of the floating voltage source connected to nodes 1 and 3, the second equation can be formed. The current through the voltage source has been marked as Iy in Fig. 16. Two expressions for Iy can be formed, one from the KCL applied to node 1 and the other from node 3. By equating these two expressions, the third equation can be formed.
KCL at node 2:
Since the voltage source connected to nodes 1 and 3 is dependent on Ia,
Current Iy through the voltage source flows away from node 1 to node 3. Then
That is,
From equations (36), (37) and (38), we get that
It is possible to solve this problem using a supernode. All that is necessary is to obtain equation (8) using a supernode. For this purpose, the circuit in Fig. 16 is re-presented below.
Worked Example 9
Find V1, V2 and V3 marked in the circuit shown in Fig. 17.
Solution:
Three unknown node voltages have to be determined and hence three equations have to be formed.
KCL Equation at node 1:
KCL Equation at node 2:
On simplifying, we get that
KCL equation at node 3:
After substituting appropriate expressions for Va and Vb and simplifying, we get that
From equations (39), (40) and (41), we can obtain the values of V1, V2 and V3.
Worked Example 10
The task is to find the ratio of output voltage to input voltage, given that the op
amp is not ideal. It has a finite gain, a finite input resistance and a finite output
resistance.
Solution:
The first step is to draw the equivalent circuit, using the voltage amplifier model,
as shown in Fig. 19.
As shown in Fig. 19, the input resistance is the internal resistance of the op amp between its two input terminals. The voltage across the input resistance is amplified by the op amp with a gain of -A, and the output resistance is in series with the dependent voltage source. In order to solve using nodal analysis, it is necessary to mark the nodes. The voltages at these nodes are marked as VA, VB, and VC.
We need to form KCL equations at the nodes at which the voltages are not known. The KCL equation at node with voltage VA is shown below.
The voltage output VC of the dependent source can be expressed in terms of the node voltage VA and the source voltage VA, as shown below.
The KCL equation at the output node can be obtained, and it is presented below. Since there can be a load connected to the output terminals, it has been included in the circuit in Fig. 21. When the op amp has a non-zero output resistance, the output voltage is a function that has the output resistance, and the load resistance as two of its parameters. . When the op amp has zero output resistance, the output voltage is independent of load resistance.
Equations (43), (44) and (45) can be combined into one matrix equation, and it is expressed below.
Given numerical values, it is not difficult to get the ratio of output voltage to source voltage. Since it turns out to be a long expression, it is not derived here.
Next it is shown how we can simplify the ratio of output voltage to source voltage progressively, as we drop the non-ideal parameters of the op amp one by one.
Let the output resistance be zero. Then the matrix simplifies to 2 by 2 matrix, as the output of the dependent source is the output voltage of the circuit, as shown in Fig. 22.
In addition, if the input resistance is infinite, the equivalent circuit can be drawn,
as shown by the circuit in Fig. 23. In this case, the matrix equation, gets
simplified, as shown below.
As input resistance tends to infinity and output resistance tends to zero, we get that
Now we can get the ratio of output voltage to source voltage, as shown below.
Given that the gain of the op amp is a finite value, the gain of the non–inverting amplifier circuit can be expressed by the expression shown above. The same expression was derived in the chapter on devices. This example shows the usefulness of nodal analysis. If this technique is not used, it is much more difficult to obtain the gain of the circuit, if the op amp has a finite gain, a finite input resistance and a finite output resistance.
This chapter has shown nodal analysis can be applied to simple circuits and to circuits with dependent sources. The next chapter describes the sources or signals used for exciting an electrical circuit and their properties.