lecture 8 1

Loop (Mesh) Analysis (3.2)

Prof. Phillips

February 17, 2003

lecture 8 2

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.

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Summing Circuit

Solution: Vout = (V1 + V2)/3

+

–

Vout

1k

1k

1k

V1 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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Mesh 2

1k

1k

1k

Identifying the Meshes

V1 V2Mesh 1+–

+–

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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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1k

1k

1k

Assigning Mesh Currents

V1 V2I1 I2

+–

+–

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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

R

I1

+ –VR

VR = I1 R

R

I1

+ –VR

I2

VR = (I1 - I2 ) R

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KVL Around Mesh 1

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

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

1k

1k

1k

V1 V2I1 I2

+–

+–

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KVL Around Mesh 2

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

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

1k

1k

1k

V1 V2I1 I2

+–

+–

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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.

2

1

2

1

k1k1k1

k1k1k1

V

V

I

I

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Solving the Equations

Let: V1 = 7V and V2 = 4V

Results:

I1 = 3.33 mA

I2 = -0.33 mA

Finally

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

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Another Example

1k

2k

2k

12V 4mA

2mA

I0

+–

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Mesh 2

Mesh 3

Mesh 1

1. Identify Meshes

1k

2k

2k

12V 4mA

2mA

I0

+–

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2. Assign Mesh Currents

I1 I2

I31k

2k

2k

12V 4mA

2mA

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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1k

2k

2k

12V 4mA

2mA

I0

I1 I2

I3

The Supermesh surrounds this source!

The Supermesh

does not include this

source!

+–

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KVL Around the Supermesh

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

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

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Matrix Notation

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

V12

mA2

mA4

1k2k2k1k2k

101

010

3

2

1

I

I

I

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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 Examples