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

Engineering 43. MaxPower SuperPosition. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. OutLine : MaxPwr & SuperPose. Work On WhtBd Student Suggest HomeWork Problem Thevénin & Norton Review Example Problem ( WhtBd )

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

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  1. Engineering 43 MaxPowerSuperPosition Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. OutLine: MaxPwr & SuperPose • Work On WhtBd Student Suggest HomeWork Problem • Thevénin& Norton Review • Example Problem (WhtBd) • Maximum Power Transfer Theorem Derivation • MaxPwr Application Examples • Thevénin & Norton Summary

  3. OutLine: MaxPwr & SuperPose • Linearity & Homogeneity • Guess Solution, Work BackWards, Scale Guess • Comparative Case Study • SuperPosition→ Activate & DeActivate • Example Problem (WhtBd)

  4. Thevénin’s Equivalence Theorem • vTH = Thévenin Equivalent VOLTAGE Source • RTH = Thévenin Equivalent SERIES RESISTANCE • Thevenin Equivalent Circuit for PART A

  5. Norton’s Equivalence Theorem • iN = Norton Equivalent CURRENT Source • RN = Norton Equivalent PARALLEL RESISTANCE • Norton Equivalent Circuit for PART A

  6. Example: VOC, ISC, RTH = RN • Use Thevéninand Norton for find the OutPutVoltage in the Circuit Below • Recall: VTH = VOC & IN = ISC

  7. Now Isc

  8. Consider The Amp-Speaker Matching Issue From PreAmp (voltage ) To speakers Maximum Power Transfer

  9. The Simplest Model for a Speaker is to Consider it as a RESISTOR only BASIC MODEL FOR THE ANALYSIS OF POWER TRANSFER Maximum Power Xfer Cont • Since the “Load” Does the “Work” We Would like to Transfer the Maximum Amount of Power from the “Driving Ckt” to the Load • Anything Less Results in Lost Energy in the Driving Ckt in the form of Heat

  10. Consider Thevenin Equivalent Ckt with Load RL Find Load Pwr by V-Divider Maximum Power Transfer • Consider PL as a FUNCTION of RL and find the maximum of such a function  have at left! • i.e., Take 1st Derivative and Set to Zero • For every choice of RL we have a different power. • How to find the MAXIMUM Power value?

  11. Find Max Power Condition Using Differential Calculus Max Power Xfer cont • Set The Derivative To Zero To Find MAX or MIN Points • For this Case Set To Zero The NUMERATOR • Solving for “Best” (Pmax) Load • This is The Maximum Power Transfer Theorem • The load that maximizes the power transfer for a circuit is equal to the Thevenin equivalent resistanceof the circuit

  12. By Calculus we Know RL for PL,max Max Power Quantified • Sub RTH for RL • Recall the Power Transfer Eqn • So Finally

  13. Determine RL for Maximum Power Transfer Need to Find RTH Notice This Ckt Contains Only INDEPENDENT Sources a b Max Pwr Xfer Example • Thus RTH BySource Deactivation • To Find the AMOUNT of Power Transferred Need the Thevenin Voltage • Then use RTH = 6kΩalong with VTH • This is Then the RL For Max Power Transfer

  14. To Find VTH Use Meshes The Eqns for Loops 1 & 2 Max Pwr Xfer Example cont • Solving for I2 • Recall • At Max: PL = PMX, RL = RTH • Now Apply KVL for VOC

  15. Determine RL and Max Power Transferred Find Thevenin Equiv.At This Terminal-Set a c b d Max Pwr Xfer • Use Loop Analysis • Recall for Max Pwr Xfer • This is a MIXED Source Circuit • Analysis Proceeds More Quickly if We start at c-d and Adjust for the 4kΩ at the end • Eqns for Loops 1 & 2

  16. The Controlling Variable c a d b Max Pwr Xfer cont • Remember now the partition points • Now Short Ckt Current • The Added Wire Shorts the 2k Resistor • The RTH for ckt at a-b = 2kΩ+4kΩ; So • Then RTH

  17. Independent Sources Only RTH = RN by Source Deactivation VTH = VOC or = RN·ISC IN = ISC or = VOC/RTH Mixed INdep and Dep Srcs Must Keep Indep & dep Srcs Together in Driving Ckt VTH = VOC IN = ISC RTH = RN= VOC/ ISC Thevenin & Norton Summary • DEPENDENT Sources Only • Must Apply V or I PROBE • Pick One, say IP = 1.00 mA, then Calculate the other, say VP • VTH = IN = 0 • RTH = RN= VP/ IP

  18. WhiteBoard Work • Let’s Work this nice Max Power Problem • Find Pmax for Load RL

  19. [Independent Srcs] Vsrc’s in Series Isrc’s in Parallel Series & Parallel Resistors Previous Equivalent Circuits • The Complementary Configs are Inconsistent with Source Definitions

  20. Models Used So Far Are All LINEAR Mathematically This Implies That They satisfy the principle of SUPERPOSITION The Model T(u) is Linear IF AND ONLY IF Linearity • For All Possible • Input Pairs: u1 & u2 • Scalars α1 & α2 • AN Alternative, And Equivalent, Linearity & Superposition Definition • The Model T(u) is Linear IF AND ONLY IF It Exhibits • ADDITIVITY • HOMOGENEITY

  21. Linearity Characteristics Additivity Linearity cont. • NOTE • Technically, Linearity Can Never Be Verified Empirically on a System • But It Could Be Disproved by a SINGLE Counter Example. • It Can Be Verified Mathematically For The Models Used • Homogeneity • a.k.a. Scaling

  22. Using Node Analysis For Resistive Circuits Yields Models Of The Form Linearity cont. • The Model Can Be Made More Detailed • Where • A and B are Matrices • s Is A Vector Of All Independent Sources • For Ckt Analysis Use The Linearity Assumption To Develop Special Analysis Methods • Where • v Is A Vector Containing All The Node Voltages • f Is a Vector Containing Only independent Sources

  23. Find Vo Past Techniques  Case Study • Redraw the Ckt to Reveal Special Cases • After Untangling      • Solution Techniques Available? 

  24. - - - - - - - Case Study cont. • Loop Analysis for Vo • Node Analysis • Out → Positive

  25. Case Study cont. • Series-Parallel Resistor-Combinations • In other Words • By VOLTAGE Divider

  26. Find Vo by Scaling Use Homogeneity Analysis • If Vo is Given Then V1 Can Be Found By The Inverse Voltage Divider • Now Use VS As a 2nd Inverse Divider • Assume That The Answer Is KNOWN • How to Find The Input In A Very Easy Way ? • Then Solvefor Vo

  27. Homogeneity Analysis cont • The Procedure Can Be Made Entirely Algorithmic • Give to Vo Any Arbitrary Value (e.g., V’o = 1V ) • Compute The Resulting Source Value and Call It V’s • Use linearity • The given value of the source (Vs) corresponds to • Then The Desired Output

  28. This is a Nice Tool For Special Problems Normally Useful When There Is Only One Source Best Judgment Indicates That Solving The Problem BACKWARDS Is Actually Easier Than the Forward Solution-Path Homogeneity Comment

  29. Solve Using Homogeneity (Scaling) Illustration  Homogeneity • Assume • V’out = V2 = 1volt • Then By Ohm’s Law

  30. Solve Using Homogeneity Illustration  Homogeneity cont • Scaling Factor • Again by Ohm’s Law • Using Homogeneity • Scale from Initial Assumption: • Then

  31. Source Superposition • This Technique Is A Direct Application Of Linearity • Normally Useful When The Circuit Has Only A Few Sources

  32. Consider a Circuit With Two Independent Sources: VS, IS Illustration  Src Superposition • Now by Linearity • Calculated By Setting The CURRENT Source To ZERO (OPEN ckt) And Solving The Circuit • Calculated By Setting The VOLTAGE Source To ZERO (SHORT ckt) And Solving The Circuit

  33. = Illustration cont + • Circuit With Current Source Set To Zero • OPEN Ckt • Circuit with Voltage Source set to Zero • SHORT Ckt • By Linearity

  34. = + Illustration cont. • This approach will be useful if solving the two, 1-Src circuits is simpler, or more convenient, than solving a circuit with two sources • We can have any combination of sources. And we can partition any way we find convenient • The Above Eqns Illustrate SUPERPOSITION

  35. Loop equations Contribution of v1 Contribution of v2 Example  Solve for i1 = + • Alternative for i1(t) By SuperPosition: • Find i1’’ by I-Divider • Once we know the “partial circuits” we need to be able to solve them in an efficient manner

  36. Find Vo By SuperPosition Numerical Example • Set to Zero The V-Src • i.e., SHORT it Current division Contribution by Isrc→ Ohm’s law

  37. Find Vo By SuperPosition Numerical Example cont. • Set to Zero The I-Src • i.e., OPEN it • By V-Divider Contribution by Vsrc • YieldsVoltageDivider(UN-tangle) • Finally, Add by SuperPosition

  38. WhiteBoard Work • Let’s Work this Nice SuperPosition Problem • Find IO

  39. Find Vo Using Source SuperPosition Example  SuperPosition • Set to Zero The I-Src • i.e., OPEN it • Set to Zero The V-Src • i.e., SHORT it

  40. Define V1 on the V-Src ckt Example cont • If V1 is known then V’o is obtained using the 6&2 Voltage-Divider • V1 can be obtained by series parallel reduction and divider

  41. Determine Current I2 By Current Divider V”o Using Ohm’s Law Numerical Example cont.2 • When in Doubt REDRAW • Finally The SuperPosition Addition • The Current Division

  42. Determine Io by Source SuperPosition First Consider Only the Voltage Source Sample Problem • Yields • Second Consider Only the 3 mA I-Source • Yields Current Divider • Then

  43. Determine Io by Source SuperPosition By IO2 Current Divider Sample Prob cont • The Current will Return on the Path of LEAST Resistance; Thus • Third Consider 4mA Src • So by Source Superposition

  44. Use Source Superposition to Determine Io Illustration • Open the Current Source • Next Short the V-Source • By Equivalent Resistance

  45. Looks Odd & Confusing → REDRAW 2 2 2 1 2 2 2 3 1 3 Illustration cont • Now Use I-Divider • Finally By Linearity

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