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Chapter 3 – Nodal Analysis Read pages 65 - 80 Nodal Analysis: Nodal analysis is a systematic application of KCL that generates a system of equations which can be solved to find voltage at each node in a circuit. (We sum currents at each node to find the node voltages.) Homework: •online HW, Nodal #1 and Nodal #2 •3FE-1 and 3FE-3 •Due 9/24/01 Fall 2001 ENGR201 Nodal Analysis 1 Nodal Analysis Steps: 1. Label all nodes in the circuit, 2. Select one node as the reference node (also called common). The voltage at every other other node in the circuit is measured with respect to the reference node. 3. Write a KCL equation (i = 0) at each node. 4. Solve the resulting set of equations for the node voltages. Branches connected to a node will have one of three types of elements: current sources (independent or dependent) resistors voltage sources (independent or dependent) Fall 2001 ENGR201 Nodal Analysis 2 Nodal Analysis – Branches With Curent Sources Since we are applying KCL, current sources (either independent or dependent) connected to a node provide terms for our KCL equation that we can write down by inspection. IS = IR1 + IR2 + IR3 The next step is to write each resistive current in terms of the node voltages. If a current source is dependent, we must also write the dependent current in terms of the node voltages. Fall 2001 ENGR201 Nodal Analysis 3 Nodal Analysis – Resistive Branches Consider a single resistor connected between two arbitrary nodes: A + VAB R B IAB 0V By KVL, the voltage drop from node-A to node-B is the difference between the voltage at node-A (VA0 = VA) and the voltage at node-B (VB0 = VB) . VAB VA VB R R The current leaving node-A going toward node-B, IAB, is: I AB The current leaving node-B going toward node-A is: I BA Fall 2001 ENGR201 Nodal Analysis VBA VB VA R R 4 Example 1 If we apply the previous techniques to the resistors connected to node-X in the following circuit and apply KCL at node-X, we get the following equation. Note that the equation should have five terms since there are five branches connected to node-X and each branch will have a corresponding current B A I2 X R1 R2 0 C R3 I1 VX VA VX VC VX VE I 2 I1 0 R1 R2 R3 E VX VA VX VC VX VE I1 I 2 0 R1 R2 R3 D currents leaving node-X resistive branches Fall 2001 ENGR201 Nodal Analysis currents entering node-X current Sources 5 Example 2 Use nodal analysis to find Ix. 12 k Ix 4 mA 6 k Step 1, Label nodes: 4 mA Fall 2001 V1 2 mA 6 k 12 k Ix 6 k ENGR201 Nodal Analysis V2 6 k 2 mA 6 Example 2 - continued Use nodal analysis to find Ix. Step 2: Write KCL equations at each node (except reference node): V1 4 mA Ix 6 k 6 k V1 V2 V1 0 4mA 12k 6k V2 V1 V2 0 2mA 12k 6k Fall 2001 V2 12 k 2 mA 1 1 1 V1 V2 4V 12 6 12 1 1 1 V1 V2 2V 12 12 6 ENGR201 Nodal Analysis 7 Example 2 - continued V1 12 k Ix 4 mA Use nodal analysis to find Ix. 6 k In matrix form: V2 6 k 2 mA 1 1 1 12 6 12 V 4 1 V 1 1 1 V2 2 12 12 6 Solving these equations (shown on the following slide) yields: V1 = -15 V and V2 = 3 V. Fall 2001 ENGR201 Nodal Analysis 8 Example 2 - continued Use nodal analysis to find Ix. V1 = -15 V 12 k V2 = 3 V 4 mA Ix 6 k 6 k 2 mA In terms of the node voltages: Ix = (V1 - V2)/12k = (-15 – 3)/12 k = -18v/ 12k Ix = -1.5mA Fall 2001 ENGR201 Nodal Analysis 9 TI-86 Solution 1 1 1 V1 V2 4V 12 6 12 1 1 1 V1 V2 2V 12 12 6 3 1 2 4 Fall 2001 ENGR201 Nodal Analysis 10 TI-86 Solution 1 1 1 V1 V2 4V 12 6 12 1 1 1 V1 V2 2V 12 12 6 5 7 6 Fall 2001 ENGR201 Nodal Analysis 11 Dependent Sources • Circuits containing dependent sources generally introduce another unknown - the parameter (voltage or current) that controls the dependent source. • This requires that the additional unknown be eliminated by writing an equation that expresses the controlling parameter in terms of the node voltages. • The resulting equations, with the additional unknown eliminated, are solved in a conventional manner. • The following example illustrates. Fall 2001 ENGR201 Nodal Analysis 12 Dependent Source Example V1 Io V2 The nodal equations are: R2 R1 Node #1: R3 Io IS V1 V1 V2 Io 0 R1 R2 V2 V2 V1 IS Node #2: R3 R2 • There are three unknowns in the equations, V1, V2 and Io. • Another equation is needed that relates Io to V1 and/or V2. The additional equation can be formed by noting the Io is the current through R3, and by Ohm’s law Io = V2/R3. This relation can be used to form a system of three- equations or to eliminate Io from the first equation, leaving a two-by-two system to solve. Fall 2001 ENGR201 Nodal Analysis 13 Example 3 Use nodal analysis to find node voltages V1 and V2. V1 V2 2Io Io The node equations are: V1 V V 1 2 4ma 10k 10k V 2 V2 V1 2Io 0 10k 10k The “extra” unknown, Io, can be expressed as: V1 Io 10k The equations become: V1 V V2 1 4ma 10k 10k V2 V V1 V 2 2 1 0 2V V 40 volts 1 2 10k 10k 10k V1 16v and V2 8v V1 2V2 0 Fall 2001 ENGR201 Nodal Analysis 14