8 Lesson Node Voltage Analysis (Section 3-1)(CLO 3-1) One of the basic understandings that will help students write node voltage equations is to recognize that this technique is nothing more than applying KCL. The first thing to ask the students to do is to count and label the nodes, and then to select one of the nodes as the reference node or ground. Although any node can be selected as a ground, if they choose wisely they can simplify the analysis. Ask them to look for voltage sources. If one or more share a common node, that node makes a good reference candidate. Encourage them to select the negative terminal of a voltage source as their reference. That will automatically make the positive node equal to the voltage of the source that is one less node they have to worry about. The node labels are ground symbol (0V) and VA, VBVN. Some of the node voltages may be known since they are connected to a source whose other end is connected to ground e.g. VA = VS. Writing the node voltage equations follows from KCL. Here the trick is to always assume that the currents leaving the node are positive (i.e. the sum of currents leaving a node is zero). VA is connected to ground via a voltage source VS. Hence, VA is known and we write equations at each unknown node (VB and VC)
VA IX VS
I
R1
VB R3 IS
VC + VI _
VA = VS VB VS VB 0 VB VC + + =0 R1 R2 R3 VC VB + IS = 0 R3
R2
Tell the students that they should collect all the unknown terms on the left side of the equation and all the sources on the right. This will enable the students to reduce the equations to a set of linear equations that are then ready to be solved using one of several analytical techniques and or computer tools. Draw the currents this one time to help them see that we are indeed summing currents at a node, but then erase the currents because they can easily get confused if they draw them at each node. Stress to the students that they can and should write node equations without drawing in the currents.
VB (
1 1 1 1 V + + ) VC ( ) = S R1 R2 R3 R3 R1 VV B + C = IS R3 R3
These equations are simple enough to solve by hand using substitution or simultaneously. If the parameters numeric had values, they could be solved using MATLAB or their hand calculators. If they are in literal form, they will need to use the symbolic processor in MATLAB. For up to a 3 3, they could solve it by hand using Cramers Rule. Anyway, they should already know how to solve a set of linear equations. It is also important for the students to know how to find a desired current or a voltage across a particular circuit element. Once they have found the node voltages the desired current can be found by appealing to the desired element equation. For example the current identified in the circuit above is I = (VS VB)/R1. Point out to the students that had the arrow gone the other way the current would be found as I = (VB VS)/R1. VI, the voltage across the current source, is simply VC0 or VC.
Draw a circuit that does not have the ground node at the bottom of the circuit. That is, place the voltage source on top of the page. Students at times get used to the ground at the bottom and place it there even if that is not the best place to put it. Another issue is that students like to write node equations even at known nodes VA for example. This can be done but is not normally necessary; moreover, it often adds undesired complexity to the solution. If one really wants to write a node equation at a known node then one must identify a current through the voltage source and include it in the equation, IX, in the example circuit. The equation at node VA then would be:
VA VB + IX = 0 R1
The last thing to cover is how to handle a circuit with a voltage source connecting two nonground nodes. We will use the Supernode technique. Consider the circuit below. Regardless of where one picks a ground one source will always be connected between two non ground nodes. To use a super node you draw a node around the non-grounded source i.e. VS2. Then treat the currents exiting that supernode as if they were exiting from one node. Supernode
V
S2
V
A
R
1
V
B
V R
3
C
VA 0 VA VS1 VC VS1 + + + IS = 0 R2 R1 R3
Then we need the equation relating nodes VA and VC namely, VA VC=VS2. Since VB is known this will allow the node voltages to be solved.
R
2
V
S1
IS