1 / 25

LOOP ANALYSIS

LOOP ANALYSIS. VSR. 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.

alyssa-walls
Download Presentation

LOOP ANALYSIS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. LOOP ANALYSIS VSR VSR E&E Dept

  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. VSR E&E Dept

  3. 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. VSR E&E Dept

  4. Summing Circuit Solution: Vout = (V1 + V2)/3 1kW 1kW + + – + – V1 Vout 1kW V2 – VSR E&E Dept

  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. VSR E&E Dept

  6. Identifying the Meshes 1kW 1kW 1kW + – Mesh 1 Mesh 2 + – V1 V2 VSR E&E Dept

  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. VSR E&E Dept

  8. Assigning Mesh Currents 1kW 1kW 1kW + – + – V1 V2 I1 I2 VSR E&E Dept

  9. 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. VSR E&E Dept

  10. VR + – Voltages from Mesh Currents VR + – I2 R R I1 I1 VR = I1R VR = (I1 -I2 ) R VSR E&E Dept

  11. 1kW 1kW 1kW + – + – V1 V2 I1 I2 KVL Around Mesh 1 -V1 + I1 1kW + (I1 - I2) 1kW = 0 I1 1kW + (I1 - I2) 1kW = V1 VSR E&E Dept

  12. 1kW 1kW 1kW + – + – V1 V2 I1 I2 KVL Around Mesh 2 (I2 - I1) 1kW + I2 1kW + V2 = 0 (I2 - I1) 1kW + I2 1kW = -V2 VSR E&E Dept

  13. 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. VSR E&E Dept

  14. Matrix Notation • The two equations can be combined into a single matrix/vector equation. VSR E&E Dept

  15. 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 VSR E&E Dept

  16. 2kW 2mA 1kW + – 12V 2kW 4mA I0 Another Example VSR E&E Dept

  17. 1. Identify Meshes 2kW Mesh 3 2mA 1kW + – Mesh 1 2kW Mesh 2 12V 4mA I0 VSR E&E Dept

  18. I3 I1 I2 2. Assign Mesh Currents 2kW 2mA 1kW + – 2kW 12V 4mA I0 VSR E&E Dept

  19. 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? VSR E&E Dept

  20. 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? VSR E&E Dept

  21. The Supermesh! The Supermesh does not include this source! 2kW The Supermesh surrounds this source! 2mA I3 1kW + – 2kW 12V 4mA I1 I2 I0 VSR E&E Dept

  22. KVL Around the Supermesh -12V + I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 0 I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V VSR E&E Dept

  23. Matrix Notation • The three equations can be combined into a single matrix/vector equation. VSR E&E Dept

  24. Solution I1 = 1.2 mA I2 = -4 mA I3 = -0.8 mA I0 = I1 - I2 = 5.2 mA VSR E&E Dept

  25. THANK YOU VSR E&E Dept

More Related