5. Nodal analysis
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Transcript 5. Nodal analysis
Lecture 5. Nodal Analysis
• Nodes and reference nodes
• Steps of Nodal Analysis
• Supernodes
• Examples
1
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.
2
Learning by Examples: A Summing Circuit
500W
I1s
+
500W V
-
500W
1kW
500W
I2s
• 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?
3
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.
4
Step 1. Reference Node
500W
500W
+
I1s
500W
V
1kW
500W
I2s
–
The reference node is often called the ground node
5
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.
6
Step 2. Node Voltages
V1
500W V 500W
2
1
I1s
2
500W
V3
3
1kW
500W
I2s
V1, V2, and V3 are unknowns for which we solve using KCL.
7
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.
8
Step 3. Currents and Node Voltages
V1
500W
V1 V2
500W
V2
V1
V1
500W
500W
9
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.
10
Step 4. KCL at Node 1
V1
I1s
500W
500W
V2
V1 V2
V1
I1s
0
500W 500W
11
Step 4. KCL at Node 2
V1
500W
V2 500W
V3
1kW
V2 V1 V2 V2 V3
0
500W 1kW 500W
12
Step 4. KCL at Node 3
V2 500W
V3
500W
I2s
V3 V2
V3
I 2s 0
500W 500W
13
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.
14
Step 5. Solving the Equations
• Re-organize the Equations
1
1
1
V
V2 0 V3 I1s
1
500W
500W 500W
V1 V2
V1
I1s
0
500W 500W
V V
V V
V
2 1 2 2 3 0
500W 1kW 500W
V3 V2
V
3 I 2s 0
500W 500W
1
1
1
1
1
V1
V3 0
V2
500W
500
W
1
k
W
500
W
500
W
0 V1
1
1
1
V2
V3 I 2 s
500W
500W 500W
• 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.
15
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
V
1 I1
1
V2 0
500W
1
1 V3 I 2
500W 500W
0
• 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
16
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
17
A General Solution
500W
500W
+
I1s
500W
V
1kW
500W
I2s
–
Solution: V = 167I1 + 167I2
• Can you prove this?
18
Another Example: A Linear Large Signal
Equivalent to a Transistor
0.7V
Ib
5V
+
–
1kW
+ –
50W
100Ib
2kW
+
Vo
–
19
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.
20
Another Example: A Linear Large Signal
Equivalent to a Transistor
0.7V
1
5V
Ib V2
V1
+
–
1kW
2
+ –
V3
V4
3 50W
4
+
Vo
100Ib
2kW
–
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.
22
KCL @ Node 4
Node 1: V1
5
0.7V
1
5V
Ib V2
V1
+
–
1kW
2
+ –
V3
V4
3 50W
4
+
Vo
100Ib
2kW
–
V3 V4
V4
100 I b
0
Node 4:
50W
2kW 23
How to Treat the Dependent Source
• We must express Ib in terms of the node voltages:
Ib
V1 V2
1kW
• Equation from Node 4 becomes
V3 V4
V V
V
100 1 2 4 0
50W
1 kW 2kW
24
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!
25
Supernode
0.7V
1
V2
V1
+
–
1kW
Ib
+ –
V3
50W
100Ib
V4
+
4
Vo
2kW
V2 V1 V3 V4
0
1kW
50W
And don’t forget
–
V2 V3 0.7
• If a voltage source is not connected to the reference node, then it is
supernode!
26
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.
27
Step 5. Solving the Equations
V1 5
Great, one variable is already known!
V2 V3 0.7
V2 V1 V3 V4
0
1kW
50W
V2 5 V3 V4
0
1kW
50W
V3 V4
V1 V2
V4
100
0
50W
1 kW 2kW
V3 V4
5 V2
V4
100
0
50W
1 kW 2kW
• Write the equations for V2, V3 and V4:
V2 V3 0V4 0.7
V
V2
V
5
3 4
1kW 50W 50W 1kW
100
1
1
1
500
V2
V3 (
)V4
1 kW
50W
50 2kW
1 kW
28
Class Examples
• Drill Problems P2-8 and P2-10
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