ECE201 Lect-91 Nodal Analysis (3.1) Dr. Holbert February 22, 2006

ECE201 Lect-92 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.

ECE201 Lect-93 + – V 500  1k  500  I1I1 I2I2 Summing Circuit Solution: V = 167I I 2

ECE201 Lect-94 Can you analyze this circuit using the techniques of Chapter 2?

ECE201 Lect-95 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

ECE201 Lect-96 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.

ECE201 Lect-97 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.

ECE201 Lect-98 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.

ECE201 Lect-99 Reference Node The reference node is called the ground node. + – V 500  1k  500  I1I1 I2I2

ECE201 Lect-910 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.

ECE201 Lect-911 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

ECE201 Lect-912 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.

ECE201 Lect-913 Currents and Node Voltages 500  V1V1 V1V1 V2V2

ECE201 Lect-914 KCL at Node  I1I1 V1V1 V2V2

ECE201 Lect-915 KCL at Node  1k  500  V2V2 V3V3 V1V1

ECE201 Lect-916 KCL at Node  I2I2 V2V2 V3V3

ECE201 Lect-917 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.

ECE201 Lect-918 System of Equations Node 1: Node 2:

ECE201 Lect-919 System of Equations Node 3:

ECE201 Lect-920 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.

ECE201 Lect-921 Equations The right side of the equation: –The right side of the equation is the sum of currents from sources entering the node.

ECE201 Lect-922 Matrix Notation The three equations can be combined into a single matrix/vector equation.

ECE201 Lect-923 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

ECE201 Lect-924 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];

ECE201 Lect-925 Solving the Equation >> v = inv(A)*i v = V 1 = 1.33V, V 2 =1.17V, V 3 =1.58V

ECE201 Lect-926 Matrix Refresher Given the 2x2 matrix A The inverse of A is

ECE201 Lect-927 Class Examples Learning Extension E3.1 Learning Extension E3.3 Learning Extension E3.5