1. Lecture 261 Nodal Analysis

2. Lecture 262 Example: A Summing Circuit The output voltage V of this circuit is proportional to the sum of the two input currents I 1 and I 2. This circuit could be useful in audio applications or in instrumentation. The output of this circuit would probably be connected to an amplifier.

3. Lecture 263 + - V 500  1k  500  I1I1 I2I2 Summing Circuit V = 167I 1 + 167I 2

4. Lecture 264 Can you analyze this circuit using the techniques of Chapter 2?

5. Lecture 265 Not This One! There are no series or parallel resistors to combine. We do not have a single loop or a double node circuit. We need a more powerful analysis technique: Nodal Analysis

6. Lecture 266 Why Nodal or Loop Analysis? The analysis techniques in Chapter 2 (voltage divider, equivalent resistance, etc.) provide an intuitive approach to analyzing circuits. They cannot analyze all circuits They cannot be easily automated by a computer.

7. Lecture 267 Node and Loop Analysis Node analysis and loop analysis are both circuit analysis methods which are systematic and apply to most circuits. Analysis of circuits using node or loop analysis requires solutions of systems of linear equations. These equations can usually be written by inspection of the circuit.

8. Lecture 268 Steps of Nodal Analysis 1.Choose a reference node. 2.Assign node voltages to the other nodes. 3.Apply KCL to each node other than the reference-express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

9. Lecture 269 Reference Node The reference node is called the ground node. + - V 500  1k  500  I1I1 I2I2

10. Lecture 2610 Steps of Nodal Analysis 1.Choose a reference node. 2.Assign node voltages to the other nodes. 3.Apply KCL to each node other than the reference-express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

11. Lecture 2611 Node Voltages V 1, V 2, and V 3 are unknowns for which we solve using KCL. 500  1k  500  I1I1 I2I2 123 V1V1 V2V2 V3V3

12. Lecture 2612 Steps of Nodal Analysis 1.Choose a reference node. 2.Assign node voltages to the other nodes. 3.Apply KCL to each node other than the reference-express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

13. Lecture 2613 Currents and Node Voltages 500  V1V1 V1V1 V2V2

14. Lecture 2614 KCL at Node 1 500  I1I1 V1V1 V2V2

15. Lecture 2615 KCL at Node 2 500  1k  500  V2V2 V3V3 V1V1

16. Lecture 2616 KCL at Node 3 500  I2I2 V2V2 V3V3

17. Lecture 2617 Steps of Nodal Analysis 1.Choose a reference node. 2.Assign node voltages to the other nodes. 3.Apply KCL to each node other than the reference-express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

18. Lecture 2618 System of Equations Node 1: Node 2:

19. Lecture 2619 System of Equations Node 3:

20. Lecture 2620 Equations These equations can be written by inspection-the left side: –The node voltage is multiplied by the sum of conductances of all resistors connected to the node. –Other node voltages are multiplied by the conductance of the resistor(s) connecting to the node and subtracted.

21. Lecture 2621 Equations The right side of the equation: –The right side of the equation is the sum of currents from sources entering the node.

22. Lecture 2622 Matrix Notation The three equations can be combined into a single matrix/vector equation.

23. Lecture 2623 Matrix Notation The equation can be written in matrix- vector form as Av = i The solution to the equation can be written as v = A -1 i

24. Lecture 2624 Solving the Equation with MATLAB I 1 = 3mA, I 2 = 4mA >> A = [1/500+1/500 -1/500 0; -1/500 1/500+1/1000+1/500 -1/500; 0 -1/500 1/500+1/500]; >> i = [3e-3; 0; 4e-3];

25. Lecture 2625 Solving the Equation >> v = inv(A)*i v = 1.3333 1.1667 1.5833 V 1 = 1.33V, V 2 =1.17V, V 3 =1.58V