Maximum Power Transfer and Thevenin-Norton Equivalents, Slides of Electrical Circuit Analysis

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Engineering 43 MaxPower SuperPosition Docsity.com 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               sL LLs Ls LS Ls S Ls LS Ls S Ls LS LL L Ls LS Ls L S LL L Ls L S L L L L RR RRR RR RV RR V RR RV RR V RR RV RR P RR RV RR R V RR V P RR R VV R V P R P                                               2 2 2 d d d d 0 1 d d 0 3 2 2 2 3 2 2 2 2 2 2 2 2 2 2 Docsity.com Norton’s Equivalence Theorem  Norton Equivalent Circuit for PART A  iN = Norton Equivalent CURRENT Source  RN = Norton Equivalent PARALLEL RESISTANCE LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B a b_ Ov  i LINEAR CIRCUIT PART B a b_ Ov  i NRNi PART A Docsity.com Example: VOC, ISC, RTH = RN • Use Thevénin and Norton for find the OutPut Voltage in the Circuit Below • Recall: VTH = VOC & IN = ISC Docsity.com Maximum Power Transfer • Consider The Amp-Speaker Matching Issue From PreAmp (voltage ) To speakers + - RTH VTH Docsity.com Max Power Xfer cont • Find Max Power Condition Using Differential Calculus  Set The Derivative To Zero To Find MAX or MIN Points • For this Case Set To Zero The NUMERATOR                3 2 2 LTH LLTH TH L L RR RRR V dR dP  Solving for “Best” (Pmax) Load THL RR  *  This is The Maximum Power Transfer Theorem • The load that maximizes the power transfer for a circuit is equal to the Thevenin equivalent resistance of the circuit       02 0 2 3 2 *              LLTH LTH LLTH TH RL L RRR RR RRR V dR dP L Docsity.com Max Power Quantified • By Calculus we Know RL for PL,max THL RR  *  Recall the Power Transfer Eqn   2 2 TH LTH L L V RR R P    Sub RTH for RL   2 2max, TH THTH TH L V RR R P     2 2 2 2max, 42 TH TH TH TH TH TH L V R R V R R P   So Finally TH TH L R V P 2 max, 4 1  Docsity.com Max Pwr Xfer Example • Determine RL for Maximum Power Transfer • Need to Find RTH – Notice This Ckt Contains Only INDEPENDENT Sources  Thus RTH By Source Deactivation kkkkRTH 6634   This is Then the RL For Max Power Transfer  To Find the AMOUNT of Power Transferred Need the Thevenin Voltage  Then use RTH = 6kΩ along with VTH a b Docsity.com Max Pwr Xfer cont • The Controlling Variable  Now Short Ckt Current • The Added Wire Shorts the 2k Resistor  Remember now the partition points   VkIV mAII III II OC X 82 and4 so 426 so 2 12 122 2 '     mAI II SC XX 4 020000 ""    Then RTH  k mA V I V R SC OC TH 2 4 8 V8 c d  The RTH for ckt at a-b = 2kΩ+4kΩ; So kRL 6 *  ][ 3 8 ][ 6*4 82 max mWmWP  a b Docsity.com Thevenin & Norton Summary • 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  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 Docsity.com WhiteBoard Work  Let’s Work this nice Max Power Problem  Find Pmax for Load RL Docsity.com Linearity cont. • Linearity Characteristics – Additivity  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       2121 uTuTuuT   uTuT  )( • Homogeneity – a.k.a. Scaling Docsity.com Linearity cont. • Using Node Analysis For Resistive Circuits Yields Models Of The Form fAv   The Model Can Be Made More Detailed BsAv  • 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 Docsity.com Past Techniques  Case Study • Find Vo  Redraw the Ckt to Reveal Special Cases • After Untangling  Solution Techniques Available?       Docsity.com Use Homogeneity Analysis • Find Vo by Scaling   1V EQR  If Vo is Given Then V1 Can Be Found By The Inverse Voltage Divider OO V R RR VV RR R V 2 21 11 21 2     Now Use VS As a 2 nd Inverse Divider O EQ EQ EQ EQ S V R RR R RR V R RR V 2 214 1 4      Then Solve for Vo  Assume That The Answer Is KNOWN • How to Find The Input In A Very Easy Way ? Docsity.com Homogeneity Analysis cont • The Procedure Can Be Made Entirely Algorithmic 1. Give to Vo Any Arbitrary Value (e.g., V’o = 1V ) 2. Compute The Resulting Source Value and Call It V’s 3. Use linearity 0 ' 0 ''' 0 VkVVkVVV SSS  4. The given value of the source (Vs) corresponds to ' S S V V k  5. Then The Desired Output ' 0' ' 00 V V V kVV S S Docsity.com Homogeneity Comment • 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 Docsity.com Source Superposition • This Technique Is A Direct Application Of Linearity • Normally Useful When The Circuit Has Only A Few Sources Docsity.com Illustration  Src Superposition • Consider a Circuit With Two Independent Sources: VS, IS  Calculated By Setting The CURRENT Source To ZERO (OPEN ckt) And Solving The Circuit circuit + - VS IS + VL _ IL V a V a IL S S 1 2 SVBY ONCONTRIBUTI 1 LV SIBY ONCONTRIBUTI 2 LV 1 LV 2 LV  Calculated By Setting The VOLTAGE Source To ZERO (SHORT ckt) And Solving The Circuit  Now by Linearity Docsity.com Illustration cont 1 LI 1 LV 2 LI 2 LV= +  Circuit With Current Source Set To Zero • OPEN Ckt  Circuit with Voltage Source set to Zero • SHORT Ckt  By Linearity 2121 LLLLLL VVVIII  Docsity.com Numerical Example • Find Vo By SuperPosition  Set to Zero The V-Src • i.e., SHORT it Current division Ohm’s law ][460 ' 0 VkIV    mA 3 2 mA 9 6 9 3 mA 20 I Contribution by Isrc → Docsity.com Numerical Example cont. • Find Vo By SuperPosition  Set to Zero The I-Src • i.e., OPEN it + - V0 "6k 3k 3V ][6240 " 0 ' 00 VV VVV    Yields Voltage Divider (UN- tangle)  Finally, Add by SuperPosition  By V-Divider ][2 V Contribution by Vsrc Docsity.com WhiteBoard Work  Let’s Work this Nice SuperPosition Problem  Find IO Docsity.com Numerical Example cont.2 • Determine – Current I2 By Current Divider – V”o Using Ohm’s Law  When in Doubt REDRAW 2I 2k||4k 2k 6k I2 + V"0 _ 2mA   mAmA kkkk kkk I 7 5 2 )4||2(26 )4||2(2 2     VVV VVVVV VkIV O OOO O 857.6 7 48 7 30 7 18 7 30 6 "' 2 "     The Current Division  Finally The SuperPosition Addition Docsity.com Sample Problem • Determine Io by Source SuperPosition • First Consider Only the Voltage Source • Yields  Then  Second Consider Only the 3 mA I-Source mA k V I 5.1 6 9 01    • Yields Current Divider Docsity.com Sample Prob cont • Determine Io by Source SuperPosition • By IO2 Current Divider  Third Consider 4mA Src  The Current will Return on the Path of LEAST Resistance; Thus  So by Source Superposition mA k k mAI 5.1 6 3 302     003 I mAIIII 30302010  Docsity.com