Nodal Analysis Objective: To analyze circuits using a systematic technique: the nodal analysis.

Transcription

1 Circuits (MTE 20) (Spring 200) Nodal Analysis Objective: To analyze circuits using a systematic technique: the nodal analysis. University of Waterloo, Electrical and Computer Engineering Dep. Node Voltages: In the figure below, V means the voltage of the node a with respect to node r. If we consider the node r as a reference node, then V is simply written as V, because the voltage of the reference node is zero. V means the voltage of the node b with respect to node r. If we consider the node r as a reference node, then V is simply written as V, because the voltage of the reference bode is zero. Node voltages V ar RΩ V br V a RΩ V b Vr=0, we can redraw the circuit as follows r 0 V Remark: A current through a resistor can be calculated by using node-voltages as follows: Node votlages Node votlages i 2 V R V 2 V i 2 R V 2 I 2 = V -V 2 I 2 = R V 2 -V R

2 Circuits (MTE 20) (Spring 200) A voltage across a resistor can be calculated by using node-voltages as follows: Nodal Analysis The nodal analysis is a systematic way of applying KCL at each essential node of a circuit and represents the branch current in terms of the node voltages. This will give us a set of equations that we solve together to find the node voltages. Once we find the node voltages we can use them to calculate any other currents or voltages of interest. Case #: Circuits with Independent Current Sources. Example #: Find the node voltages of the following circuit. Solution Step # Identify all of the essential nodes and choose one of them as a reference node. Step #2 Assign voltages variable to all nodes except the reference node. Step # 2

3 Circuits (MTE 20) (Spring 200) Apply the KCL at each node except the reference node.in this step for each node we assume the branch current is leaving from the node, and then we describe the branch current in term of node voltages. Apply KCL at Node # Node# has three branches, so we apply the KCL to obtain an equation with three terms as follows: V 2 V V 6 V 2V 2 Apply KCL at Node # 2 Node#2 has three branches, so we apply the KCL to obtain an equation with three terms as follows: V V V 6 6 V 2V 2 2 Step # Solving question and 2 for the unknown node voltages (V and V 2 ): V 6V V V

4 Circuits (MTE 20) (Spring 200) Example #2: Use the nodal analysis to find the i x, i y, i z, i n, i m, V a, V b, V c, V d. Solution: First we apply the nodal analysis technique in order to find the node voltages, then we use the node voltages to calculate: i x, i y, i z, i n, i m, V a, V b, V c, V d. Step # Identify all of the essential nodes and choose one of them as a reference node. Step #2 Assign voltages variable to all nodes except the reference node. Step # Apply KCL at each node except the reference node.in this step for each node we assume the branch current is leaving from the node, and then we describe the branch current in term of node voltages. Apply KCL at Node # Node# has three branches, so we apply the KCL to obtain an equation with three terms as follows:

5 Circuits (MTE 20) (Spring 200) V 0 V V V V V.2 V.0 V Apply KCL at Node # 2 Node#2 has three branches, so we apply KCL to obtain an equation with three terms as follows: 0 V V V V V. V. V 2 Apply KCL at Node # 2 Node#has three branches, so we apply KCL to obtain an equation with three terms as follows: V V V V V V. V. V Step # Solving the questions,2 and for the unknown node voltages( V,V 2 and V ): V. V, V 72.7 V, V V After we find the node voltages, we can use the node voltages to calculate any other unknowns: Now, let us use the node voltages to find the unknown currents i x, i y, i z, i n, i m : i V V A i V V 20 i V V A 2.27 A

6 Circuits (MTE 20) (Spring 200) i V i V A. A Now, let us use the node voltages to calculate the unknown voltages:, V a, V b, V c and V d. V V V V V.V V V V V V V V V Case #2: Circuits with Independent Voltage Sources (a) If the voltage sources are connected to the reference node Example #: Find i 0 by using the nodal analysis: Solution: 6

7 Circuits (MTE 20) (Spring 200) We do not need to apply the KCL at Node, because the node voltage at this node is known, V =2 v. Also, we do not need to apply the KCL at Node, because the node voltage at this node is known V =-6 v. Notice that the V is negative because its polarity is opposite to the polarity of the voltage source. Notice that, for node voltages the negative sign of the polarity is at the reference node. Now, we have only one unknown node voltage (V 2 ). Apply KCL at node #2: V 6 k V V 2 k V V 2 k Substitute the values of V and V : V 6 k V 2 2 k V 6 2 k Solve for V 2 V 2 v (b) If the Voltage Source Are Connected Between Two Unreferenced Nodes Example #: Find the node voltages for the following circuit. Solution: 7

8 Circuits (MTE 20) (Spring 200) Since we can not express the current through the voltage source in term of node voltages, we combine the two nodes 2 and as one supernode (as shown in the figure below) and relate one of the node voltages (V 2 or V ) in term of the other one as follows: V V 22 Then V V 22 Now, apply the KCL at the supernode: We have six branches connected to the supernode (as indicated in the above figure): 8

9 Circuits (MTE 20) (Spring 200) V V V Apply the KCL at node # V 22 V 7 V V 70 We have four branches connected to node #: 8 V V V V 22 2 V 22 7 V 7 V 77 2 Solving equations and 2 for V and V 2 : V. V V. V Then V V V Example #: In the pervious example, find the current through the 22-voltage source? Solution: We assume the direction of the current as show below (you can assume the opposite direction), then we apply the KCL at node 2 or to find the current i : Apply the KCL at node #2: 9

10 Circuits (MTE 20) (Spring 200) i V V V V V V i. Also, we can find i by applying the KCL at node # V 22 V i V 22 V V 22 2 V 22 2 i. 22. A. A Example #6: (an extra problem) Write the node-voltage equations for the following circuits. Solution 0

11 Circuits (MTE 20) (Spring 200) / Ω 2 v / Ω /6 Ω V 2 V V A Ω / Ω -2 A In order to find the current in the (/6 Ω) resistor, we assume a node voltage V x between the voltage source and the (/6 Ω) resistor (as show in the figure below) and relate V x to the node voltage V and the value of the voltage source (2 v) as follows: V V 2 Then, V V 2 Now, we can easily find the current through the voltage source and (/6 Ω) resistor in term of node voltages ((V 2 and V. / Ω V / Ω V x =V -2 2 v V 2 /6 Ω V A Ω / Ω -2 A Apply the KCL at node # 2 V V V V 2 V 6 V 6 V V 7

12 Circuits (MTE 20) (Spring 200) Apply the KCL at node #2 V V V V V 2 6 V V 6 V 2 Apply the KCL at node # 8 V V V V 2 7 V V V 8 Case # Nodal Analysis for Circuits with Dependent Sources For circuits that include dependent sources, first we ignore the fact that the dependent source is a dependent source and we write the node-voltage equations as we would for a circuit with independent sources. The node-voltage equations will have extra unknown variables for the dependent sources beside to the normal unknown node voltages. All the extra unknown variable of the dependent sources must be eliminated by writing an expression for each one in term of the node voltages. There MUST be a relation between the unknown variable of the dependent and the node voltages, because node voltages can be used to describe any current or voltage of any branch in the circuit. The following examples will show you how to apply the nodal analysis for circuits with dependent sources. (a) Nodal Analysis for Circuits with Dependent Current Sources Example #7: Find the node voltages for the following: Solution 2

13 Circuits (MTE 20) (Spring 200) Apply the KCL at node # : 2 i V 2 k V V 6 k a The above equation has an extra unknown variable i. We should relate the extra unknown variable of the dependent source to the node voltages. From the above figure it is clear that: Substitute i in equation a: Apply the KCL at node #2: Solving for the node voltages i V k 2 x V k V 2 k V V 6 k V 2 V b 2 ma V k V V 6 k V V 2 V 2 V V 2 V

14 Circuits (MTE 20) (Spring 200) (b) Nodal Analysis for Circuits with Dependent Voltage Sources Example #8: Find the current I 0 by using the nodal analysis. 6 Ω 2 V x kω 2 Ω I 0 6 Ω + V + x v Solution Since there is a voltage source between two unreferenced nodes, we combine the nodes in one supernode. Then, we relate one of the node voltages (V ) to the other one (V 2 ) using the value of the voltage source between them. From the above figure: V V 2V Then, V V 2V

15 Circuits (MTE 20) (Spring 200) Apply the KCL at the supernode: The supernode has four branches so the node equation should have four terms: V 2 V 2 k V 2 V V 6 k V 6 k V V a 2 k From the figure it is clear that V 6, so we do have to write the node-voltage equation for node #. Substitute the value of V, in equation a : V 2 V V 2 V 6 V 2 k 6 k 6 k V 6 b 2 k The above equation has an extra unknown variable. We should relate the extra unknown variable of the dependent source to the node voltages. From the above figure it is clear that: V V Substitute V in equation b: V 2 V V 2 V 6 V 2 k 6 k 6 k V 6 2 k V 2 V 2 V 2 V 6 2 V V 6 Solving for V 2 V. V and V V. Then, V V 2V V. 2.. V I V..7 A 2 2