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

1. Engineering 43 Chp 2 TutorialProblem 2-24Solution Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Amplifier Driving Speaker • Consider an Amplifier Circuit connected to a Speaker Speakera.k.a. the“LOAD” DrivingCircuita.k.a. the“SOURCE”

3. Thévenin’s Equivalent Circuit Theorem (c.f. ENGR43) Allows Tremendous Simplification of the Amp Ckt Circuit Simplification R S Thevenin + V S 

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

5. The Final Ckt Model • Driving Circuit • The Speaker

6. For ANY Electrical Device with a: Potential, V, across it A current, I, thru it Electrical Power Physics • Now OHM’s Law Relates the Voltage-across and Current-Thru a resistor + - V I • Then the Power Used by the Device:

7. Recall the Reduced Ckt Model Voltage Division • Analysis of this “Voltage Divider” Ckt produces a Relationship between VS & VL • This SINGLE LOOP Ckt effectively divides VS across RS and RL

8. What we KNOW By Thévenin Analysis of the Driving Ckt we determined VS & RS Note that VS & RS are FIXED and beyond our Control as Speaker Designers The Speaker Designer CAN, however control the Load Resistance, RL Thus Our Goal Summary to This Point Find RL such the Driving Ckt Operates at the Highest Efficiency; i.e., we seek RL that will MAXIMIZE Driver→Load Power Transfer

9. Analytical Game Plan • Goal Find RL to Maximize PL(RL) • From the Physics we Know Solve on WhiteBoard

10. The MATLAB Problem • And • Define Transfer Ratio, r • RS = 10Ω, 15Ω, 20Ω, 25Ω • Then • So to Maximize PL need to Maximize r • RL = 10Ω, 15Ω, 20Ω, 25Ω, 30Ω

11. MATLAB Game Plan • Concept  Test ALL possible Resistor Combinations then Check for Best • Because we have a small number of allowable values for RS and RL, the most direct way to choose RL is to compute the values of r for each combination of RS and RL. • Since there are four possible values of RS and five values of RL, there are 4(5) = 20 combinations.

12. MATLAB Plan (2) • We can use an array operation in MATLAB to compute r for each of these combinations by defining two 5 × 4 2D-Arrays R_L and R_S. • The five rows of R_L contain the five values of RL, and its four columns are identical. • The four columns of R_S contain the four values of RS, and its five rows are identical.

13. MATLAB Plan (3) • The Arrays we Need • These Arrays MUST have the same size so that we can perform element-by- element operations with them.

14. % Bruce Mayer, PE % ENGR22 * 20Jan07 * Rev. 13Sep08 % Prob 2.24 * file Demo_Prob2_24_0809.m % % Since all COLUMNS in RL are the same, Define one Col and Replicate in Row Vector % Define RL col a = [10;15;20;25;30]; % Make Array R_L by using a in 4-element Row Vector R_L = [a,a,a,a] % % Since all ROWS in RS are the same, Define one Row and Replicate in Col Vector % Define RS row b = [10,15,20,25]; % Make Array R_S by using a in 5-element Col Vector R_S=[b;b;b;b;b] % % Use Element-by-Element Operations to Calc r %% First Sum RS & RL for the 20 combos Rsum = R_S+R_L %% Now sq the 20 sums RsumSq = Rsum.^2 % need "dot" as this is element-by-element %% Finally Divide RL by SQd sums r = R_L./RsumSq % % Use the max(A) command to find the max value in each COL, and the ROW in in Which the max Values Occurs [max_ratio, row] = max(r) The MATLAB Code

15. The .m-File OutPut R_L = 10 10 10 10 15 15 15 15 20 20 20 20 25 25 25 25 30 30 30 30 R_S = 10 15 20 25 10 15 20 25 10 15 20 25 10 15 20 25 10 15 20 25 Rsum = 20 25 30 35 25 30 35 40 30 35 40 45 35 40 45 50 40 45 50 55 r = 0.0250 0.0160 0.0111 0.0082 0.0240 0.0167 0.0122 0.0094 0.0222 0.0163 0.0125 0.0099 0.0204 0.0156 0.0123 0.0100 0.0187 0.0148 0.0120 0.0099 max_ratio = 0.0250 0.0167 0.0125 0.0100 row = 1 2 3 4 RS = 10 RS = 15 RS = 20 RS = 25 RL = 10 RL = 15 RL = 20 RL = 25 RL = 30

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