Transcript Loop Analysis (3.2)

Loop (Mesh) Analysis (3.2)
Dr. Holbert
February 27, 2006
ECE201 Lect-10
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.
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Example: A Summing Circuit
• The output voltage V of this circuit is
proportional to the sum of the two input
voltages V1 and V2.
• 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
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Summing Circuit
1kW
1kW
+
V1
+
–
Vout
1kW
+
–
V2
–
Solution: Vout = (V1 + V2)/3
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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.
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Identifying the Meshes
1kW
V1
+
–
1kW
Mesh 1
1kW
Mesh 2
ECE201 Lect-10
+
–
V2
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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.
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Assigning Mesh Currents
1kW
V1
+
–
1kW
1kW
I1
I2
ECE201 Lect-10
+
–
V2
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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.
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Voltages from Mesh Currents
+ VR
R
+
–
VR
I2
–
R
I1
I1
VR = I1 R
VR = (I1 - I2 ) R
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KVL Around Mesh 1
1kW
V1
+
–
1kW
1kW
I1
I2
+
–
V2
-V1 + I1 1kW + (I1 - I2) 1kW = 0
I1 1kW + (I1 - I2) 1kW = V1
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KVL Around Mesh 2
1kW
V1
+
–
1kW
1kW
I1
I2
+
–
V2
(I2 - I1) 1kW + I2 1kW + V2 = 0
(I2 - I1) 1kW + I2 1kW = -V2
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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.
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Matrix Notation
• The two equations can be combined into a
single matrix/vector equation.
 1kW   I 1   V1 
1kW  1kW




  1kW

I

V
1
k
W

1
k
W

 2   2 
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Solving the Equations
Let:
Results:
V1 = 7V and V2 = 4V
I1 = 3.33 mA
I2 = -0.33 mA
Finally
Vout = (I1 - I2) 1kW = 3.66V
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Another Example
2kW
2mA
1kW
12V
+
–
2kW
I0
ECE201 Lect-10
4mA
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1. Identify Meshes
2kW
2mA Mesh 3
1kW
12V
+
–
Mesh 1
2kW
Mesh 2
I0
ECE201 Lect-10
4mA
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2. Assign Mesh Currents
2kW
2mA
12V
+
–
I3
1kW
2kW
I1
I2
4mA
I0
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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 the loop because
we don’t know the voltage.
• What to do?
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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?
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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
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KVL Around the Supermesh
-12V + I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 0
I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V
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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 
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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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Solution
I1 = 1.2 mA
I2 = -4 mA
I3 = -0.8 mA
I0 = I1 - I2 = 5.2 mA
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Class Example
• Learning Extension E3.8
• Learning Extension E3.9
• Learning Extension E3.11
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