System Design for Optimum Performance

 

Example 2.3-3: Maximum Power Transfer

in a Speaker-Amplifier System

 

 

1.   Understand the purpose of the problem.

·       To optimize the efficiency of power transfer from an amplifier to a speaker, the electrical resistance of the speaker must be “matched” to the amplifier.

·       We must identify what speaker resistance will receive the maximum power from a given amplifier (so called maximum power transfer).

 

2.   Collect the known information. Realize that some of it might be discovered later to be unnecessary.

·       We know the specifications of the amplifier.  In particular we know the source resistance (Rs) and the source voltage (Vs).

 

3.   Determine what information you must find.

·       The power delivered to the speaker (the resistive load) is related to its resistance and the voltage it across it. The optimum load resistance (the one getting the most power for a given amplifier source voltage and source resistance) is what we need to find. 

 

4.   Simplify the problem only enough to allow the required information to be obtained. State any assumptions you make.

·       We need to assume a relatively simple model for the amplifier. In this case, we’ll model it as a voltage source (or power source) and a simple resistor in series.

·       We will assume that the speaker can be modeled as a simple resistor.

 

5.   Draw a sketch and label any necessary variables.

 

 

6.   Determine which of the fundamental principles are applicable.

·       Kirchhoff’s Voltage Law and Ohm’s Law give rise to what is called the “voltage divider” equation:

                                 

 

·       Power (measured in Watts) dissapated in a resistive load is equal the voltage squared divided by the resistance:

               

 

·       We have no control over Vs or Rs.  We wish to maximize pL(RL) with respect to RL for a fixed value of Vs and Rs.

·       We can do this by calculating pL for a range of values of RL assuming that Rs and Vs are known.

 

7.   Think about your proposed solution approach in general and consider other approaches before proceeding with the details.

·       We could use a system simulation package (PSPICE) and try to search for the optimal solution to the problem.

·       Since we have a mathematical model for the system based on our initial assumptions, we could mathematically optimize using calculus.

·       We could solve the equations on a spreadsheet and search for the optimum value.

 

8.   Label each step of the solution process.

·       Set up an array of possible load resistances (RL)

·       Fix Rs and Vs at the appropriate value for your amplifier

·       Compute the value pL for all the RL values

·       Search for where pL is maximum, this will correspond to the best RL.

 

9.   Solve the problem. If you are solving the problem with a program:

·       State the problem concisely.

Maximize 

as a function of RL assuming Vs and RS are known and fixed.

·       Specify the data to be used by the program. This is the “input”.

 

Vs=20 Volts

Rs=10 W

 

·       Specify the information to be generated by the program. This is the “output”.

 

A plot of Power vs. RL

The optimum RL for a given Vs and Rs

 

·       Work through the solution steps by hand or with a calculator; use a simpler data set if necessary.

 

Here is a partial solution we can use to reality check the program output

 

Vs=20 Volts, Rs=10 W, RL=5 W à pL=8.88 Watts

 

·       Write and run the program.

Let’s do it by writing an m-file!

 

%

% Solution to Example 2.3-3

%

% Optimization of the output power of a

% amplifier-speaker system

%

% File: ex233.m

%

 

clear all;close all;clc

 

% Specify a range on values for RL (Step 1)

RL=linspace(5,50,100);

% Could have also used RL=[5:.5:50];

 

% Define input parameters (Step 2)

Rs=10;

Vs=20;

 

% Calculate power for each value of RL (Step 3)

pL=(RL./(Rs+RL).^2)*Vs^2;

 

% And find the optimal value of RL (Step 4)

[maxvalue,indexofmax]=max(pL);

disp('The optimum load resistance is');

RL(indexofmax)

 

% Output the results graphically

figure

plot(RL,pL)

xlabel('RL (Ohms)')

ylabel('pL (Watts)')

          title('Example 2.3-3: Power as a function of RL (Vs=20 V, Rs=10 Ohms)')

grid

 

·       Check the output of the program with your hand solution.

 

>>        pL(1)

 

ans =

 

         8.8889

 

Hand calculated value checks out!

 

·       Run the program with your input data (if different from your simple data set) and perform a reality check on the output.

 

Output range seems reasonable and matches our expectations for the case where RL=5 W.

 

·       If the program will be used as a general tool in the future, test it by running it for a range of reasonable data values; perform a reality check on the results.

 

We can try our program for different values of Rs.

 

Append these lines to your m-file…

 

% Generate results for each value or Rs

 

Rs=10;

p1=(RL./(Rs+RL).^2)*Vs^2;

Rs=15;

p2=(RL./(Rs+RL).^2)*Vs^2;

Rs=20;

p3=(RL./(Rs+RL).^2)*Vs^2;

Rs=25;

p4=(RL./(Rs+RL).^2)*Vs^2;

 

% Plot the results

 

figure

plot(RL,p1, 'b-',RL,p2, 'r--',RL,p3, 'g-.',RL,p4, 'm: ');

legend('Rs=10','Rs=15','Rs=20','Rs=25')

title('Example 2.3-3: Power as a function of RL (Vs=20 V)')

xlabel('RL (Ohms)');

ylabel('pL (Watts)')

grid

 

 

10.                   Present your results.

 

 

Based on the figures above, It appears that the optimum value of RL is when RL=RS. This seems a rather nice and simple conclusion.

 

Do not state the answer with any greater precision than is justified by any of the following:

·       The precision of the given information.

·       The simplifying assumptions.

·       The requirements of the problem.

 

These are not applicable here. The graphical solution “hides” the precision of the calculation and the main conclusion stated in a way that avoids precision issues.

 

 

 

This document was originally created by Dr. Malcolm Daniels of the Department of Electrical and Computer Engineering and was modified by Dr. Russell Hardie.