Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Engineering 43 Super Node/MeshThevenin/Norton Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Need Only ONE KCL Eqn ReCall Node (KCL) Analysis • The Remaining Eqns From the Indep Srcs • 3 Nodes Plus the Reference. In Principle Need 3 Equations... • But two nodes are connected to GND through voltage sources. Hence those node voltages are KNOWN • Solving The Eqns

3. Consider ThisExample Conventional Node Analysis Requires All Currents at a Node SuperNode Technique SUPERNODE • But Have Ckt V-Src Reln • More Efficient solution: • Enclose The Source, and All Elements In parallel, Inside a Surface. • Call That a SuperNode • 2 eqns, 3 unknowns...Not Good (IS unknown) • Recall: The Current thru the Vsrc is NOT related to the Potential Across it

4. Apply KCL to the Surface Supernode cont. SUPERNODE • The Source Current Is interiorto the Surface and is NOT Required • Still Need 1 More Equation – Look INSIDE the Surface to Relate V1 & V2 • Now Have 2 Equations in 2 Unknowns • Then The Ckt Solution Using LCD Technique • See Next Slide

5. The Equations Now Apply Gaussian Elim • Use The V-Source Rln Eqn to Find V2 SUPERNODE • Mult Eqn-1 by LCD (12 kΩ) • Add Eqns to Elim V2

6. Find the node voltages And the power supplied By the voltage source To compute the power supplied by the voltage source We must know the current through it: @ node-1 BASED ON PASSIVE SIGN CONVENTION THE POWER IS ABSORBED BY THE SOURCE!!

7. Write the Node Equations KCL At v1 Illustration using Conductances  • At The SuperNodeHave V-Constraint • v2−v3= vA • KCL Leaving Supernode   • Now Have 3 Eqnsin 3 Unknowns • Solve Using Normal Techniques

8. Find Io Known Node Voltages SUPERNODE Example • The SuperNodeV-Constraint • Now use KCL at SuperNode to Find V3 • Mult by 2 kΩLCD, collectTerms to Find:

9. Find Io Using Nodal Analysis Known Voltages for Sources Connected to GND SUPERNODE Numerical Example • Now Notice That V2 is NOT Needed to Find Io • 2 Eqns in 2 Unknowns • The Constraint Eqn • Now KCL at SuperNode • By Ohm’s Law

10. Dependent Sources • Circuits With Dependent Sources Present No Significant Additional Complexity • The Dependent Sources Are Treated As Regular Sources • As With Dependent CURRENT Sources Must Add One Equation For Each Controlling Variable

11. Find Io by Nodal Analysis Notice V-Source Connected to the Reference Node Numerical Example – Dep Isrc • Sub Ix into KCL Eqn • KCL At Node-2 • Mult By 6 kΩ LCD • Controlling Variable In Terms of Node Potential • Then Io

12. Find Io Supernode Constraint Current Controlled V-Source • Controlling Variable in Terms of Node Voltage • Multiply by LCD of 2 kΩ • Recall • Then • KCL at SuperNode • So Finally

13. Use Mesh Analysis ReCall MESH Analysis • Sub for I1 to Find I2 • So Vo

14. Create Mesh Currents SUPERMESH SuperMeshes Write Constraint Equation Due To Mesh Currents SHARING Current Sources Write Equations For Remaining Meshes Define A Supermesh By (Mentally) Removing The Shared Current

15. Write KVL For The SuperMesh as we do NOT know the Voltage Across the 4 mA Current-Source SUPERMESH SuperMeshes cont. • We Now have 3 Eqns in 3 Unknowns and the Math Model is Complete • Solve for I1, I2, I3 using standard techniques KVL

16. Use superNODE to AVOID a V-source current, IVs, in KCL Eqns SuperNODE vs SuperMESH • Use superMESH to AVOID an I-source ΔVIs in KVL Eqns

17. Strategy Define Loop Currents That Do NOT Share Current Sources Even If It Means ABANDONING Meshes For Convenience, Begin by Using Mesh Currents Until Reaching Shared CURRENT Source as V-across an I-source is NOT Known At That Point Define a NEW Loop Shared Isrc – General Loop Approach • To Guarantee That The New Loop Gives An Independent Equation, Must Ensure That It Includes Components That Are NOT Part Of Previously Defined Loops

18. A Possible Approach Create a Loop by Avoiding The Current Source General Loop Approach cont. • The Eqns for Current Source Loops • The Eqns for 3rd Loop (3 Eqns & 3 Unknowns) • The Loop Currents Obtained With This Method Are Different From Those Obtained With A SuperMesh • A SuperMesh used previously Defined Mesh Currents

19. For Loop Analysis Note Three Independent Current Sources Four Meshes One Current Source Shared By Two Meshes Example  Find V Across R’s • Careful Choice Of Loop Currents Should Make Only One Loop Equation Necessary • Three Loop Currents Can Be Chosen Using Meshes And Not Sharing Any Source • Mesh Equations For Loops With I-Sources

20. KVL for I4 Loop Example  Find V Across R’s cont. • Solve For The Current I4 Using The ISj • Now Use Ohm’s Law To Calc Required Voltages • Note that Loop-4 does NOT pass thru ANY CURRENT-Sources • This AVOIDS the UNknown potentials across the I-sources

21. General Approach Treat The Dep. Source As Though It Were Independent Add One EquationFor The Controlling Variable Example at Rt.: Mesh Currents Defined by Sources Dependent Sources • Mesh-3 by KVL • Mesh-4 by KVL

22. The Controlling Variable Eqns Dependent Sources cont. • Combine Eqns,Then Divide by 1kΩ • In Matrix Form • Solve by Elimination or Linear-Algebra

23. Consider the Ckt Loop & Node Compared • Find Vo by NODE Analysis • ID Nodes • Make a SuperNode • Vsrc to GND • SuperNode Constraint

24. KCL at SuperNode Loop & Node Compared (2) • Mult. By 1kΩ LCD • The Node 3 KCL • Mult. By 1kΩ LCD • The SuperNode Eqn

25. The Controlling Var. Loop & Node Compared (3) • Thus 3 Eqns in Unknowns V1, V2, V3 • Recall the GOAL • In SuperNodeEqn

26. Now by Loops Loop & Node Compared (4) • The Mesh/Loop Eqns • Loop-1: • Loop-3: • Loop-2: • Note I3 = –2 mA • Loop-4: • Start with 3 Meshes • Add a General Loop to avoid the Isrc • Note I1 = 2 mA

27. The Controling Var Loop & Node Compared (5) • By Net Current & Ohm’s Law • SubOut Vx in Loop-2 & Loop-4 Eqns: • As Before Vx = V2 • And V2 is related to the net current From Node-2 to GND • After Subbing Find:

28. Summarize Loops Loop & Node Compared (6) • Recall the GOAL • General Comments • Nodes (KCL) are generally easier if we have VOLTAGE Sources • Loops (kVL) are generally easier if we have CURRENT Sources • Using The Loop/Mesh Eqns Find

29. Thevenin’s & Norton’sTheorems • These Are Some Of The Most Powerful Circuit analysis Methods • They Permit “Hiding” Information That Is Not Relevant And Allow Concentration On What Is Important To The Analysis

30. From PreAmp (voltage ) To speakers Low Distortion Power Amp • to Match Speakers And Amplifier One Should Analyze The Amp Ckt

31. To Even STAND A CHANCE to Match the Speakers & Amp We Need to Simplify the Ckt Consider a Reduced CIRCUIT EQUIVALENT Replace the OpAmp+BJT Amplifier Cktwith a MUCH Simpler (Linear) Equivalent Low Dist Pwr Amp cont • The Equivalent Ckt in RED “Looks” The Same to the Speakers As Does the Complicated Circuit

32. Thevenin’s Equivalence Theorem • vTH = Thevenin Equivalent VOLTAGE Source • RTH = Thevenin Equivalent Series (Source) RESISTANCE • Thevenin Equivalent Circuit for PART A

33. Norton’s Equivalence Theorem • iN = Norton Equivalent CURRENT Source • RN = Norton Equivalent Parallel (Source) RESISTANCE • Norton Equivalent Circuit for PART A

34. For ANY Part-B Circuit Examine Thevenin Approach • The Thevenin Equiv Ckt for PART-A → • V-Src is Called the THEVENIN EQUIVALENT SOURCE • R is called the THEVENINEQUIVALENT RESISTANCE PART A MUST BEHAVE LIKE THIS CIRCUIT

35. In The Norton Case Norton Examine Norton Approach • The Norton Equiv Ckt for PART-A → • The I-Src is Called The NORTON EQUIVALENT SOURCE

36. Thevenin Norton Interpret Thevenin & Norton • In BOTH Cases • This equivalence can be viewed as a source transformation problem. It shows how to convert a voltage source in series with a resistor into an equivalent current source in parallel with the resistor • SOURCE TRANSFORMATION CAN BE A GOOD TOOL TO REDUCE THE COMPLEXITY OF A CIRCUIT

37. Source transformation is a good tool to reduce complexity in a circuit ...WHEN IT CAN APPLIED “IDEAL sources” are NOT good models for the REAL behavior of sources .e.g., A Battery does NOT Supply huge current When Its Terminals are connected across a tiny Resistance as Would an “Ideal” Source Source Transformations • These Models are Equivalent When • Source X-forms can be used to determine the Thevenin or Norton Equivalent • But There May be More Efficient Methods

38. Example  Solve by Src Xform • In between the terminals we connect a current source and a resistance in parallel • The equivalent current source will have the value 12V/3kΩ • The 3k and the 6k resistors now are in parallel and can be combined • In between the terminals we connect a voltage source in series with the resistor • The equivalent V-source has value 4mA*2kΩ • The new 2k and the 2k resistor become connected in series and can be combined

39. Solve by Src Xform cont. • After the transformation the sources can be combined • The equivalent current source has value 8V/4kΩ = 2mA • The Options at This Point Do another source transformation and get a single loop circuit Use current divider to compute IO and then Calc VO using Ohm’s law

40. EQUIVALENT CIRCUITS Or one more source transformation 3 current sources in parallel and three resistors in parallel PROBLEM Find VO using source transformation Norton Norton

41. These Models are Equivalent Source Xform Summary • Source X-forms can be used to determine the Thevenin or Norton Equivalent • Next Review Several Additional Approaches To Determine Thevenin Or Norton Equivalent Circuits

42. Determine the Thevenin Equiv. • vTH = OPEN CIRCUIT Voltage at A-B if Part-B is Removed and Left UNconnected • iSC = SHORT CIRCUIT Current at A-B if Voltage at A-B is Removed and Replaced with a Wire (a short) • Then by R = V/I

43. One circuit problem 1. Determine the Thevenin equivalent source Remove part B and compute the OPEN CIRCUIT voltage Second circuit problem Remove part B and compute the SHORT CIRCUIT current 2. Determine the SHORT CIRCUIT current Graphically... • Then

44. The Thevenin Equivalent V-Source is computed as the open loop voltage The Thevenin Equivalent Resistance CAN BE COMPUTED by setting to zero all the INDEPENDENT sources and then determining the resistance seen from the terminals where the equivalent will be placed Thevenin w/ Indep. Sources

45. “Part B” “Part B” Thevenin w/ Indep. Sources cont • Since the evaluation of the Thevenin equivalent Resistance for INdependent-Source-Only circuits can be very simple, we can add it to our toolkit for the solution of circuits

46. Find Vo Using Thevenin’s Theorem Identify Part-B (the Load) “PART B” Thevenin Example • Break The Circuit At the Part-B Terminals • DEactivate12V Source to Find TheveninResistance • Produces a SHORT

47. Note That RTH Could be Found using ISC Thevenin Example cont. • Then by I-Divider • By Series-Parallel R’s • Finally RTH • Then Itot • Same As Before

48. Finally the Thevenin Equivalent Circuit Thevenin Example cont.2 • And Vo By V-Divider

49. Let’s do MQ-02e • Find: Vt & Rt

50. Use Thevenin To Find Vo Have a CHOICE on How to Partition the Ckt Make “Part-B” As Simple as Possible “Part B” Thevenin Example • Deactivate the 6V and 2mA Source for RTH