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• Nodes and reference nodes
• Steps of Nodal Analysis
• Supernodes
• Examples
Lecture 5. Nodal Analysis
1
2
Circuit Analysis – A Systematic Approach
• We’ve learned several tricks to perform circuit analysis: Single loop circuits, equivalent resistor, superposition etc.
• Nodal Analysis is a rather general method that allows you to analyze virtually all the linear circuits via a well defined recipe.
3
Learning by Examples: A Summing Circuit
• The output voltage V of this circuit is proportional to the sum of the two input currents I1s and I2s.
• This circuit could be useful in audio applications or in instrumentation.
• The output of this circuit would probably be connected to an amplifier.
• Can you solve this problem using superposition method?
+
-
V 500
500
1k
500
500I1s I2s
4
Nodal Analysis: The Recipe
1. Choose a reference node and assign 0 voltage to it.
2. Assign node voltages to the other nodes.
3. Express currents in terms of node voltages.
4. Apply KCL to each node other than the reference node.
5. Solve the resulting system of linear equations.
5
Step 1. Reference Node
The reference node is often called the ground node
+
–
V 500
500
1k
500
500I1s I2s
6
Nodal Analysis: The Recipe
1. Choose a reference node and assign 0 voltage to it.
2. Assign node voltages to the other nodes.
3. Express currents in terms of node voltages.
4. Apply KCL to each node other than the reference node.
5. Solve the resulting system of linear equations.
7
Step 2. Node Voltages
V1, V2, and V3 are unknowns for which we solve using KCL.
500
500
1k
500
500I1s I2s
1 2 3
V1 V2 V3
8
Nodal Analysis: The Recipe
1. Choose a reference node and assign 0 voltage to it.
2. Assign node voltages to the other nodes.
3. Express currents in terms of node voltages.
4. Apply KCL to each node other than the reference node.
5. Solve the resulting system of linear equations.
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Step 3. Currents and Node Voltages
500
V1500V1 V2
50021 VV
5001V
10
Nodal Analysis: The Recipe
1. Choose a reference node and assign 0 voltage to it.
2. Assign node voltages to the other nodes.
3. Express currents in terms of node voltages.
4. Apply KCL to each node other than the reference node.
5. Solve the resulting system of linear equations.
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Step 4. KCL at Node 1
500
500I1s
V1 V2
0500500
1211
VVV
I s
12
Step 4. KCL at Node 2
500
1k
500 V2 V3V1
0500k1500
32212
VVVVV
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Step 4. KCL at Node 3
0500500 2
323
sI
VVV
500
500
I2s
V2 V3
14
Nodal Analysis: The Recipe
1. Choose a reference node and assign 0 voltage to it.
2. Assign node voltages to the other nodes.
3. Express currents in terms of node voltages.
4. Apply KCL to each node other than the reference node.
5. Solve the resulting system of linear equations.
15
Step 5. Solving the Equations
• The left side of the equation is a sum of a linear combination of node voltages (variables to be determined).
• The right side of the equation is a sum of currents from sources entering the node.
• Re-organize the Equations
sIVVV 1321 0500
1
500
1
500
1
0500
1
500
1
k1
1
500
1
500
1321
VVV
sIVVV 2321 500
1
500
1
500
10
0500500
1211
VVV
I s
0500k1500
32212
VVVVV
0500500 2
323
sI
VVV
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Matrix Notation• The three equations can be combined into a single matrix/vector
equation.
2
1
3
2
1
0
500
1
500
1
500
10
500
1
500
1
k1
1
500
1
500
1
0500
1
500
1
500
1
I
I
V
V
V
• 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
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Solving the Equation with MATLAB
I1s = 3mA, I2s = 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];
>> v = inv(A)*i v = 1.3333 1.1667 1.5833
18
+
–
V 500
500
1k
500
500I1s I2s
A General Solution
Solution: V = 167I1 + 167I2
• Can you prove this?
19
Another Example: A Linear Large Signal Equivalent to a Transistor
5V100Ib
+
–
Vo
50
Ib
2k1k+–
+ –
0.7V
20
Nodal Analysis: The Recipe
1. Choose a reference node and assign 0 voltage to it.
2. Assign node voltages to the other nodes.
3. Express currents in terms of node voltages.
4. Apply KCL to each node other than the reference node.
5. Solve the resulting system of linear equations.
5V 100Ib
+
–
Vo
50
Ib
2k
1k
0.7V
12 3 4
V1V2 V3 V4
+–
+ –
Another Example: A Linear Large Signal Equivalent to a Transistor
22
Nodal Analysis: The Recipe
1. Choose a reference node and assign 0 voltage to it.
2. Assign node voltages to the other nodes.
3. Express currents in terms of node voltages.
4. Apply KCL to each node other than the reference node.
5. Solve the resulting system of linear equations.
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KCL @ Node 4
0k2
10050
443
VI
VVb
5V 100Ib
+
–
Vo
50
Ib
2k
1k
0.7V
12 3 4
V1V2 V3 V4
+–
+ –
Node 1:
Node 4:
51 V
24
How to Treat the Dependent Source
• We must express Ib in terms of the node voltages:
• Equation from Node 4 becomes
k1
21 VVIb
0k2k1
10050
42143
VVVVV
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How to Deal With Nodes 2 and 3?
• The 0.7-V voltage supply makes it impossible to apply KCL to nodes 2 and 3, since we don’t know what current is passing through the supply.
• We do know that
V2 – V3 = 0.7 V
• We need another equation!
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100Ib
+
–
Vo
50Ib
2k
1k
0.7V
14
V1 V2 V3 V4
+–
+ –
050k1
4312
VVVV
Supernode
0.732 VVAnd don’t forget
• If a voltage source is not connected to the reference node, then it is supernode!
27
Nodal Analysis: The Recipe
1. Choose a reference node and assign 0 voltage to it.
2. Assign node voltages to the other nodes.
3. Express currents in terms of node voltages.
4. Apply KCL to each node other than the reference node.
5. Solve the resulting system of linear equations.
28
Step 5. Solving the Equations
51 V
050k1
4312
VVVV
0.732 VV
0k2k1
10050
42143
VVVVV
Great, one variable is already known!
050k1
5 432
VVV
0k2k1
5100
504243
VVVV
• Write the equations for V2, V3 and V4:
k1
5
5050k1432 VVV
0.70 432 VVV
k1
500)
k2
1
50
1(
50
1
k1
100432 VVV
29
Class Examples
• Drill Problems P2-8 and P2-10
