1. Fall 2001ENGR201 Nodal Analysis1 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 Chapter 3 – Nodal Analysis

2. Fall 2001ENGR201 Nodal Analysis2 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. Nodal Analysis Steps: Branches connected to a node will have one of three types of elements: current sources (independent or dependent) resistors voltage sources (independent or dependent)

3. Fall 2001ENGR201 Nodal Analysis3 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. 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. Nodal Analysis – Branches With Curent Sources I S = I R1 + I R2 + I R3

4. Fall 2001ENGR201 Nodal Analysis4 Consider a single resistor connected between two arbitrary nodes: AB R + V AB - By KVL, the voltage drop from node-A to node-B is the difference between the voltage at node-A (V A0 = V A ) and the voltage at node-B (V B0 = V B ). +VA-+VA- +VB-+VB- The current leaving node-A going toward node-B, I AB, is: I AB The current leaving node-B going toward node-A is: Nodal Analysis – Resistive Branches 0 V

5. Fall 2001ENGR201 Nodal Analysis5 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 A B C D 0 E X I1I1 I2I2 R1 R2 R3 Example 1 currents leaving node-X resistive branches currents entering node-X current Sources

6. Fall 2001ENGR201 Nodal Analysis6 12 k  2 mA 4 mA IxIx 6 k  Use nodal analysis to find I x. IxIx Step 1, Label nodes: Example 2 4 mA 6 k  12 k  6 k  2 mA V1V1 V2V2

7. Fall 2001ENGR201 Nodal Analysis7 Use nodal analysis to find I x. Example 2 - continued 12 k  2 mA 4 mA IxIx 6 k  V1V1 V2V2 Step 2: Write KCL equations at each node (except reference node): 

8. Fall 2001ENGR201 Nodal Analysis8 Use nodal analysis to find I x. Example 2 - continued 12 k  2 mA 4 mA IxIx 6 k  V1V1 V2V2 In matrix form: Solving these equations (shown on the following slide) yields: V 1 = -15 V and V 2 = 3 V.

9. Fall 2001ENGR201 Nodal Analysis9 Use nodal analysis to find I x. Example 2 - continued 12 k  2 mA 4 mA IxIx 6 k  V 1 = -15 VV 2 = 3 V In terms of the node voltages: I x = (V 1 - V 2 )/12k  = (-15 – 3)/12 k  = -18v/ 12k  I x = -1.5mA

10. Fall 2001ENGR201 Nodal Analysis10 1 2 3 4 TI-86 Solution

11. Fall 2001ENGR201 Nodal Analysis11 TI-86 Solution 6 5 7

12. Fall 2001ENGR201 Nodal Analysis12 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. Dependent Sources

13. Fall 2001ENGR201 Nodal Analysis13 IoIo IoIo R1R1 R2R2 R3R3 ISIS V1V1 V2V2 The nodal equations are: Node #1: Node #2: There are three unknowns in the equations, V 1, V 2 and I o. Another equation is needed that relates I o to V 1 and/or V 2. The additional equation can be formed by noting the I o is the current through R 3, and by Ohm’s law I o = V 2 /R 3. This relation can be used to form a system of three- equations or to eliminate I o from the first equation, leaving a two-by-two system to solve. Dependent Source Example

14. Fall 2001ENGR201 Nodal Analysis14 Use nodal analysis to find node voltages V 1 and V 2. 2I o IoIo V1V1 V2V2 The node equations are: The “extra” unknown, I o, can be expressed as: The equations become:  Example 3