Nodal Analysis 3 1 Dr Holbert February 22

Nodal Analysis (3. 1) Dr. Holbert February 22, 2006 ECE 201 Lect-9 1

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. ECE 201 Lect-9 2

Summing Circuit 500 W + I 1 500 W V 1 k. W 500 W I 2 – Solution: V = 167 I 1 + 167 I 2 ECE 201 Lect-9 3

Can you analyze this circuit using the techniques of Chapter 2? ECE 201 Lect-9 4

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 ECE 201 Lect-9 5

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. ECE 201 Lect-9 6

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. ECE 201 Lect-9 7

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 node; express currents in terms of node voltages. 4. Solve the resulting system of linear equations. ECE 201 Lect-9 8

Reference Node 500 W + I 1 500 W V 1 k. W 500 W I 2 – The reference node is called the ground node. ECE 201 Lect-9 9

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 node; express currents in terms of node voltages. 4. Solve the resulting system of linear equations. ECE 201 Lect-9 10

Node Voltages V 1 500 W V 500 W 2 1 I 1 2 500 W V 3 3 1 k. W 500 W I 2 V 1, V 2, and V 3 are unknowns for which we solve using KCL. ECE 201 Lect-9 11

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 node; express currents in terms of node voltages. 4. Solve the resulting system of linear equations. ECE 201 Lect-9 12

Currents and Node Voltages V 1 500 W V 2 V 1 500 W ECE 201 Lect-9 13

KCL at Node 1 V 1 I 1 500 W V 2 500 W ECE 201 Lect-9 14

KCL at Node 2 V 1 500 W V 2 500 W V 3 1 k. W ECE 201 Lect-9 15

KCL at Node 3 V 2 500 W V 3 500 W I 2 ECE 201 Lect-9 16

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 node; express currents in terms of node voltages. 4. Solve the resulting system of linear equations. ECE 201 Lect-9 17

System of Equations • Node 1: • Node 2: ECE 201 Lect-9 18

System of Equations • Node 3: ECE 201 Lect-9 19

Equations • These equations can be written by inspection. • The left side of the equation: – 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. ECE 201 Lect-9 20

Equations • The right side of the equation: – The right side of the equation is the sum of currents from sources entering the node. ECE 201 Lect-9 21

Matrix Notation • The three equations can be combined into a single matrix/vector equation. ECE 201 Lect-9 22

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 ECE 201 Lect-9 23

Solving the Equation with MATLAB I 1 = 3 m. A, I 2 = 4 m. A >> A = [1/500+1/500 -1/500 0; -1/500+1/1000+1/500 -1/500; 0 -1/500+1/500]; >> i = [3 e-3; 0; 4 e-3]; ECE 201 Lect-9 24

Solving the Equation >> v = inv(A)*i v = 1. 3333 1. 1667 1. 5833 V 1 = 1. 33 V, V 2=1. 17 V, V 3=1. 58 V ECE 201 Lect-9 25

Matrix Refresher • Given the 2 x 2 matrix A • The inverse of A is ECE 201 Lect-9 26

Class Examples • Learning Extension E 3. 1 • Learning Extension E 3. 3 • Learning Extension E 3. 5 ECE 201 Lect-9 27