Transcript Loop (Mesh) Analysis - Arizona State University

Loop (Mesh) Analysis
Dr. Holbert
January 30, 2008
Lect5
EEE 202
1
Loop Analysis
• Nodal analysis was developed by applying
KCL at each non-reference node
• Loop analysis is developed by applying
KVL around loops in the circuit
• Loop (mesh) analysis results in a system
of linear equations which must be solved
for unknown currents
Lect5
EEE 202
2
Another Summing Circuit
• The output voltage V of this circuit is
proportional to the sum of the two input
voltages V1 and V2
1kW
1kW
+
V1
+
–
Vout
1kW
+
–
V2
–
Lect5
EEE 202
3
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations for the mesh/loop currents.
Lect5
EEE 202
4
1. Identifying the Meshes
1kW
V1
+
–
1kW
Mesh 1
Mesh 2
+
–
V2
1kW
Lect5
EEE 202
5
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations for the mesh/loop currents.
Lect5
EEE 202
6
2. Assigning Mesh Currents
1kW
V1
Lect5
+
–
1kW
1kW
I1
I2
EEE 202
+
–
V2
7
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations for the mesh/loop currents.
Lect5
EEE 202
8
Voltages from Mesh Currents
+ VR
R
+
–
–
R
I1
I1
VR = (I1 – I2 ) R
VR = I1 R
Lect5
VR
I2
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9
3. KVL Around Mesh 1
1kW
V1
+
–
1kW
1kW
I1
I2
+
–
V2
–V1 + I1 1kW + (I1 – I2) 1kW = 0
I1 1kW + (I1 – I2) 1kW = V1
Lect5
EEE 202
10
3. KVL Around Mesh 2
1kW
V1
+
–
1kW
1kW
I1
I2
+
–
V2
(I2 – I1) 1kW + I2 1kW + V2 = 0
(I2 – I1) 1kW + I2 1kW = –V2
Lect5
EEE 202
11
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations for the mesh/loop currents.
Lect5
EEE 202
12
Matrix Notation
• The two equations can be combined into a
single matrix/vector equation
 1kW   I 1   V1 
1kW  1kW




  1kW

1kW  1kW  I 2   V2 

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EEE 202
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4. Solving the Equations
Let:
V1 = 7V and V2 = 4V
Results:
I1 = 3.33 mA
I2 = –0.33 mA
Finally
Vout = (I1 – I2) 1kW = 3.66V
Lect5
EEE 202
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Another Example
2kW
2mA
1kW
12V
Lect5
+
–
2kW
I0
EEE 202
4mA
15
1. Identify Meshes
2kW
2mA Mesh 3
1kW
12V
Lect5
+
–
Mesh 1
2kW
Mesh 2
I0
EEE 202
4mA
16
2. Assign Mesh Currents
2kW
2mA
12V
+
–
I3
1kW
2kW
I1
I2
4mA
I0
Lect5
EEE 202
17
Current Sources
• The current sources in this circuit will have
whatever voltage is necessary to make the
current correct
• We can’t use KVL around any mesh
because we don’t know the voltage for the
current sources
• What to do?
Lect5
EEE 202
18
Current Sources
• The 4mA current source sets I2:
I2 = –4 mA
• The 2mA current source sets a constraint
on I1 and I3:
I1 – I3 = 2 mA
• We have two equations and three
unknowns. Where is the third equation?
Lect5
EEE 202
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2kW
The
Supermesh
surrounds
this source!
12V
2mA
+
–
I3
2kW
I2
I1
1kW
The
Supermesh
does not
include this
source!
4mA
I0
Lect5
EEE 202
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3. KVL Around the Supermesh
-12V + I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 0
I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V
Lect5
EEE 202
21
Matrix Notation
• The three equations can be combined into
a single matrix/vector equation
1
0
 0
  I 1   4mA
 1
  I    2mA 
0

1

 2  

2kW  1kW  2kW 2kW  1kW  I 3   12V 
Lect5
EEE 202
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4. Solve Using MATLAB
>> A = [0 1 0; 1 0 -1;
2e3 -1e3-2e3 2e3+1e3];
>> v = [-4e-3; 2e-3; 12];
>> i = inv(A)*v
i = 0.0012
-0.0040
-0.0008
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EEE 202
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Solution
I1 = 1.2 mA
I2 = – 4 mA
I3 = – 0.8 mA
I0 = I1 – I2 = 5.2 mA
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EEE 202
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Advantages of Nodal Analysis
• Solves directly for node voltages
• Current sources are easy
• Voltage sources are either very easy or
somewhat difficult
• Works best for circuits with few nodes
• Works for any circuit
Lect5
EEE 202
25
Advantages of Loop Analysis
• Solves directly for some currents
• Voltage sources are easy
• Current sources are either very easy or
somewhat difficult
• Works best for circuits with few loops
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EEE 202
26
Disadvantages of Loop Analysis
• Some currents must be computed from
loop currents
• Does not work with non-planar circuits
• Choosing the supermesh may be difficult.
• FYI: Spice uses a nodal analysis approach
Lect5
EEE 202
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Class Examples
• Drill Problems P2-12, P2-14, P2-15
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EEE 202
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