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Nodal Analysis (3.1)

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Example: A Summing Circuit ... circuit is proportional to the sum of the two input currents ... Summing Circuit. Solution: V = 167I1 167I2. ECE201 Lect-6. 4 ... – PowerPoint PPT presentation

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Title: Nodal Analysis (3.1)


1
Nodal Analysis (3.1)
  • Prof. Phillips
  • Feb 5, 2003

2
Example A Summing Circuit
  • The output voltage V of this circuit is
    proportional to the sum of the two input currents
    I1 and I2.
  • This circuit could be useful in audio
    applications or in instrumentation.
  • The output of this circuit would probably be
    connected to an amplifier.

3
Summing Circuit
500W
500W

I1
V
500W
1kW
I2
500W
  • Solution V 167I1 167I2

4
Can you analyze this circuit using the techniques
of Chapter 2?
5
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
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
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
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.

9
Reference Node
500W
500W

I1
V
500W
1kW
I2
500W
  • The reference node is called the ground node.

10
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.

11
Node Voltages
500W
500W
V1
V2
V3
1
2
3
I1
500W
1kW
I2
500W
  • V1, V2, and V3 are unknowns for which we solve
    using KCL.

12
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.

13
Currents and Node Voltages
500W
V1
V2
V1
500W
14
KCL at Node 1
500W
V1
V2
I1
500W
15
KCL at Node 2
500W
500W
V2
V3
V1
1kW
16
KCL at Node 3
500W
V2
V3
500W
I2
17
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.

18
System of Equations
  • Node 1
  • Node 2

19
System of Equations
  • Node 3

20
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.

21
Equations
  • The right side of the equation
  • The right side of the equation is the sum of
    currents from sources entering the node.

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

23
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
Solving the Equation with MATLAB
  • I1 3mA, I2 4mA
  • gtgt A 1/5001/500 -1/500 0
  • -1/500 1/5001/10001/500 -1/500
  • 0 -1/500 1/5001/500
  • gtgt i 3e-3 0 4e-3

25
Solving the Equation
  • gtgt v inv(A)i
  • v
  • 1.3333
  • 1.1667
  • 1.5833
  • V1 1.33V, V21.17V, V31.58V

26
Matrix Refresher
  • Given the 2x2 matrix A
  • The inverse of A is

27
Class Examples
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