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Loop (Mesh) Analysis

Loop (Mesh) Analysis. Dr. Holbert January 30, 2008. 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

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Loop (Mesh) Analysis

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  1. Loop (Mesh) Analysis Dr. Holbert January 30, 2008 EEE 202

  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 EEE 202

  3. 1kW 1kW + + – + – V1 Vout 1kW V2 – Another Summing Circuit • The output voltage V of this circuit is proportional to the sum of the two input voltages V1 and V2 EEE 202

  4. 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. EEE 202

  5. 1. Identifying the Meshes 1kW 1kW + – Mesh 1 Mesh 2 + – V1 V2 1kW EEE 202

  6. 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. EEE 202

  7. 2. Assigning Mesh Currents 1kW 1kW 1kW + – + – V1 V2 I1 I2 EEE 202

  8. 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. EEE 202

  9. VR + – Voltages from Mesh Currents VR + – I2 R R I1 I1 VR = I1R VR = (I1 –I2 ) R EEE 202

  10. 1kW 1kW 1kW + – + – V1 V2 I1 I2 3. KVL Around Mesh 1 –V1 + I1 1kW + (I1–I2) 1kW = 0 I1 1kW + (I1–I2) 1kW = V1 EEE 202

  11. 1kW 1kW 1kW + – + – V1 V2 I1 I2 3. KVL Around Mesh 2 (I2–I1) 1kW + I2 1kW + V2 = 0 (I2–I1) 1kW + I2 1kW = –V2 EEE 202

  12. 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. EEE 202

  13. Matrix Notation • The two equations can be combined into a single matrix/vector equation EEE 202

  14. 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 EEE 202

  15. Another Example 2kW 2mA 1kW + – 12V 2kW 4mA I0 EEE 202

  16. 1. Identify Meshes 2kW Mesh 3 2mA 1kW + – Mesh 1 2kW Mesh 2 12V 4mA I0 EEE 202

  17. I3 I1 I2 2. Assign Mesh Currents 2kW 2mA 1kW + – 2kW 12V 4mA I0 EEE 202

  18. 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? EEE 202

  19. 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? EEE 202

  20. The Supermesh! The Supermesh does not include this source! 2kW The Supermesh surrounds this source! 2mA I3 1kW + – 2kW 12V 4mA I1 I2 I0 EEE 202

  21. 3. KVL Around the Supermesh -12V + I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 0 I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V EEE 202

  22. Matrix Notation • The three equations can be combined into a single matrix/vector equation EEE 202

  23. 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 EEE 202

  24. Solution I1 = 1.2 mA I2 = – 4 mA I3 = – 0.8 mA I0 = I1–I2 = 5.2 mA EEE 202

  25. 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 EEE 202

  26. 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 EEE 202

  27. 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 EEE 202

  28. Class Examples • Drill Problems P2-12, P2-14, P2-15 EEE 202

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