Maximizing Power Transfer in an Amplifier-Speaker Circuit, Slides of Calculus for Engineers

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Amplifier Driving Speaker

• Consider an Amplifier Circuit

connected to a Speaker

Driving

Circuit

a.k.a. the

“SOURCE”

Speaker

a.k.a. the

“LOAD”

Circuit Simplification

• Thévenin’s Equivalent Circuit Theorem (c.f. ENGR43) Allows

Tremendous Simplification of the Amp Ckt

Thevenin +



RS

VS

The Final Ckt Model

Source Load

 Driving Circuit  The Speaker

Electrical Power Physics

• For ANY Electrical

Device with a:

• Potential, V , across it
• A current, I , thru it

+ V −

I

 Then the Power

Used by the Device:

 Now OHM’s Law

Relates the Voltage-

across and Current-

Thru a resistor

P = VI

R

−

v

i

V = RI

Summary to This Point

• 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

Find R L such the Driving

Ckt Operates at the

Highest Efficiency; i.e.,

we seek R L that will

MAXIMIZE Driver→Load

Power Transfer

Analytical Game Plan

• Goal Find R L to Maximize P L (RL )

• From the Physics we Know

S L

L

L S

L L L

L L L

R R

R

V V

V R I

P V I

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.

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.

The MATLAB Code

% Bruce Mayer, PE % ENGR22 * 20Jan07 * Rev. 13Sep % 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 .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. 0.0240 0.0167 0.0122 0. 0.0222 0.0163 0.0125 0. 0.0204 0.0156 0.0123 0. 0.0187 0.0148 0.0120 0.

max_ratio =

0.0250 0.0167 0.0125 0.

row =

1 2 3 4

RS = 10 RS = 15 RS = 20 RS = 25 RL = 10 RL = 15 RL = 20 RL = 25 RL = 30