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Nodal Analysis (3.1)
Dr. Holbert
September 27, 2001
ECE201 Lect-10
1
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.
ECE201 Lect-10
2
Summing Circuit
500W
500W
+
I1
500W
V
1kW
500W
I2
–
Solution: V = 167I1 + 167I2
ECE201 Lect-10
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Can you analyze this circuit using
the techniques of Chapter 2?
ECE201 Lect-10
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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
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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-10
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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-10
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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-10
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Reference Node
500W
500W
+
I1
500W
V
1kW
500W
I2
–
The reference node is called the ground node.
ECE201 Lect-10
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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-10
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Node Voltages
V1
500W V 500W
2
1
I1
2
500W
V3
3
1kW
500W
I2
V1, V2, and V3 are unknowns for which we
solve using KCL.
ECE201 Lect-10
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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-10
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Currents and Node Voltages
V1
500W
V2
V1
V1
500W
500W
V1  V2
500W
ECE201 Lect-10
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KCL at Node 1
V1
I1
500W
V2
V1  V2
V1
I1 

500W 500W
500W
ECE201 Lect-10
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KCL at Node 2
V1
500W
V2 500W
V3
1kW
V2  V1 V2 V2  V3


0
500W 1kW 500W
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KCL at Node 3
V2 500W
V3
500W
I2
ECE201 Lect-10
V3  V2
V3

 I2
500W 500W
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.
ECE201 Lect-10
17
System of Equations
• Node 1:
1 
V2
 1
V1 

 I1

 500W 500W  500W
• Node 2:
V3
V1
1
1 
 1

 V2 


0

500W
 500W 1kW 500W  500W
ECE201 Lect-10
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System of Equations
• Node 3:
V2
1 
 1

 V3 

  I2
500W
 500W 500W 
ECE201 Lect-10
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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.
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Equations
• The right side of the equation:
– The right side of the equation is the sum
of currents from sources entering the
node.
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Matrix Notation
• The three equations can be combined into a
single matrix/vector equation.
1
 1
 500W  500W

1
 
500W


0

1

500W
1
1
1


500W 1kW 500W
1

500W
ECE201 Lect-10

 V
 1   I1 

1
 V2    0 

500W     
1
1  V3   I 2 

500W 500W 
0
22
Matrix Notation
• The equation can be written in matrixvector form as
Av = i
• The solution to the equation can be written
as
v = A-1 i
ECE201 Lect-10
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Solving the Equation with
MATLAB
I1 = 3mA, I2 = 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];
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Solving the Equation
>> v = inv(A)*i
v =
1.3333
1.1667
1.5833
V1 = 1.33V, V2=1.17V, V3=1.58V
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Matrix Refresher
• Given the 2x2 matrix A
a b 
A

c d 
• The inverse of A is
 d  b
1
1
A 
a d  b c  c a 
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Class Examples
• Learning Extension E3.1
• Learning Extension E3.3
• Learning Extension E3.5
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