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Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

Dec 22, 2015

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Page 1: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

• Meshes and Loops

• Steps of Mesh Analysis

• Supermesh

• Examples

Lecture 6. Mesh Analysis

1

Page 2: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

2

Circuit Analysis – A Systematic Approach

• Mesh Analysis is another general method that is almost as powerful as Nodal 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 analysis results in a system of linear equations which must be solved for unknown currents.

Page 3: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

3

Learning by Examples: 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.

+

-

Vout

1k

1k

1k

V1

+

-V2

+

-

Vout = (V1 + V2)/3

Page 4: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

4

Mesh Analysis: The Recipe

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.

Page 5: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

5

Mesh 2

1k

1k

1k

Step 1: Identifying the Meshes

V1

+

-V2

+

-Mesh 1

Page 6: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

6

Mesh Analysis: The Recipe

1. Identify mesh (loops).

2. Assign a current to each mesh.

3. Apply KVL around each mesh to get an equation in terms of the mesh currents.

4. Solve the resulting system of linear equations.

Page 7: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

7

1k

1k

1k

Step 2: Assigning Mesh Currents

V1

+

-V2

+

-I1 I2

Page 8: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

8

Mesh Analysis: The Recipe

1. Identify mesh (loops).

2. Assign a current to each mesh.

3. Apply KVL around each mesh to get an equation in terms of the mesh currents.

4. Solve the resulting system of linear equations.

Page 9: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

9

Step 3: Voltages from Mesh Currents

R

I1

+ -VR

VR = I1 R

R

I1

+ -VRI2

VR = (I1 - I2 ) R

Page 10: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

10

1k

1k

1k

V1

+

-V2

+

-I1 I2

KVL Around Mesh 1

-V1 + I1 1k + (I1 - I2) 1k = 0

I1 1k + (I1 - I2) 1k = V1

Page 11: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

11

1k

1k

1k

V1

+

-V2

+

-I1 I2

KVL Around Mesh 2

(I2 - I1) 1k + I2 1k + V2 = 0

(I2 - I1) 1k + I2 1k = -V2

Page 12: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

12

Mesh Analysis: The Recipe

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.

Page 13: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

13

Step 4: Solve the Equations

• The two equations can be combined into a single matrix/vector equation.

2

1

2

1

k1k1k1

k1k1k1

V

V

I

I

I1 + (I1 - I2) 1k = V1

(I2 - I1) 1k + I2 1k = -V2

• Re-organize the equations:

- 1kI1 + (1k + 1k ) I2 = -V2

(1k + 1k )I1 - 1k I2 = V1

Page 14: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

14

Using MATLAB

>> A = [1e3+1e3 -1e3;

-1e3 1e3+1e3];

>> v = [7; -4];

>> i = inv(A)*v

i =

0.00333

-0.00033 I1 = 3.33mAI2 = -0.33mA

Vout = (I1 - I2) 1k = 3.66V

Page 15: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

15

Another Example

1k

2k

2k

12V+

-4mA

2mA

I0

Page 16: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

16

Mesh 2

Mesh 3

Mesh 1

Identify Mesh’s

1k

2k

2k

12V+

-4mA

2mA

I0

Page 17: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

17

Assign Mesh Currents

1k

2k

2k

12V+

-4mA

2mA

I0

I1 I2

I3

Page 18: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

18

How to Deal with 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?

• The 4mA current source sets I2:

I2 = -4mA

• The 2mA current source sets a constraint on I1 and I3:

I1 - I3 = 2mA

• We have two equations and three unknowns. Where is the third equation?

Page 19: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

19

1k

2k

2k

12V+

-4mA

2mA

I0

I1 I2

I3

The Supermesh surrounds this source!

The Supermesh

does not include this

source!

SuperMesh

Page 20: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

20

KVL Around the Supermesh

-12V + I3 2k + (I3 - I2)1k + (I1 - I2)2k = 0

I3 2k + (I3 - I2)1k + (I1 - I2)2k = 12V

Page 21: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

21

Solve the Equations

• The three equations can be combined into a single matrix/vector equation.

V12

mA2

mA4

1k2k2k1k2k

101

010

3

2

1

I

I

I

Page 22: Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

22

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

-0.0008I1 = 1.2mAI2 = -4mAI3 = -0.8mA

I0 = I1 - I2 = 5.2mA