Maximum Power Transfer Theorem

The article explains the Maximum Power Transfer Theorem, which states that maximum power output is achieved when the load resistance equals the Thevenin resistance of a circuit. It covers the theorem’s definition, mathematical derivation, efficiency considerations, and practical applications, particularly in signal-processing devices like radio and television receivers.

What is the Maximum Power Transfer Theorem?

Maximum power transfer theorem states that maximum power output is obtained when the load resistance R_L is equal to Thevenin resistance R_th as seen from load Terminals.

Figure 1: Maximum Power Transfer Theorem Circuit Diagram

Any circuit or network may be represented by a Thevenin equivalent circuit.  The Thevenin resistance R_th is comparable to a source internal resistance (R_S), which absorbs some of the power available from the ideal voltage source. In the above figure, a variable load resistance R_L is connected to a Thevenin circuit. The current for any value of load resistance R_L is connected to a Thevenin circuit. The current for any value of load resistance is;

$I_L=\frac{{{V}_{S}}}{{{R}_{S}}+{{R}_{L}}}$

Then by using I²R, the power delivered to the load is,

${{P}_{L}}={{\left( \frac{{{V}_{S}}}{{{R}_{S}}+{{R}_{L}}} \right)}^{2}}{{R}_{L}}~\text{    }\cdots ~~~~~~~\left( 1 \right)$

The load power depends on both R_th (R_S) and R_L; however, R_th (R_S) is considered constant for any particular network. Then, one might get an idea of how PL varies with a change in RL by assuming values for the Thevenin circuit of the above figure and, in turn, calculating PL for different values of RL.

• You May Also Read: Thevenin’s Theorem Explained

Maximum Power Transfer Theorem Derivation 

As we know, the power delivered to the load is,

${{P}_{L}}={{\left( \frac{{{V}_{S}}}{{{R}_{S}}+{{R}_{L}}} \right)}^{2}}{{R}_{L}}$

Taking a derivative on both sides;

$\frac{d{{P}_{L}}}{d{{R}_{L}}}=\frac{V_{S}^{2}{{\left( {{R}_{S}}+{{R}_{L}} \right)}^{2}}-2{{R}_{L}}\left( {{R}_{S}}+{{R}_{L}} \right)}{{{\left( {{R}_{S}}+{{R}_{L}} \right)}^{4}}}$

For P_L to be maximum;

$\frac{d{{P}_{L}}}{d{{R}_{L}}}=0$

So,

$\frac{V_{S}^{2}\left( {{R}_{S}}-{{R}_{L}} \right)}{{{\left( {{R}_{S}}+{{R}_{L}} \right)}^{3}}}=0$

Finally,

${{R}_{S}}={{R}_{L}}$

So maximum power transferred is;

${{P}_{max}}=\frac{V_{S}^{2}}{4{{R}_{S}}}$ (Maximum Power Transfer Formula)

We got the above expression by substituting R_S=R_L into equation (1).

Maximum Power Transfer and Efficiency of Transmission

We observe that power transfer from a real source always produces ohmic heating in the source resistance. Calculations of such internal effects require information about the internal structure and cannot, in general, be based upon Thevenin or Norton equivalent networks. However, the entire load current iL usually passes through the internal resistance of a real source, so we represent the internal conditions by lumped parameters, as shown in Figure 1. The resulting internal power dissipated by R_TH or R_S is then

\[{{P}_{S}}={{R}_{S}}I_{L}^{2}=\frac{{{R}_{S}}}{{{({{R}_{S}}+{{R}_{L}})}^{2}}}*V_{S}^{2}\]

The dashed curve in Figure 2 shows that P_S steadily decreases as R_L increases and that P_S=P_L when R_L/R_S=1.

Figure 2. Maximum Power Transfer and Transmission Efficiency

Since the total power generated by the source is P_L+P_S, the wasted internal power P_S should be small compared to P_L for efficient operation. Formally, we define the power-transfer efficiency as

$Efficiency=\frac{{{P}_{L}}}{{{P}_{L}}+{{P}_{S}}}$

Which is often expressed as a percentage. If the load has been matched for maximum power transfer, then P_S=P_L, and so efficiency,

$Efficiency=\frac{{{P}_{L}}}{2{{P}_{L}}}=50\text{ %}$

Moreover, with R_L=R_S, the terminal voltage drops to V=V_TH/2. Clearly, electrical utilities would not, and should not, strive for maximum power transfer. Instead, they seek higher power-transfer efficiency by making P_S as small as possible.

Maximum Power Transfer Solved Example

Find R_L

Solution

Let’s find V_th first across 150 Ω resistance

${{V}_{th}}={{V}_{S}}=360*\frac{150}{150+30}$

${{V}_{th}}={{V}_{S}}=300~V$

To find R_th or R_S, short circuit the voltage source

${{R}_{th}}={{R}_{S}}=150~||~30=25\Omega $

So, for maximum power transfer, we know that

                    \[{{R}_{L}}={{R}_{th}}=25\text{ }\Omega\]

Now, Find the Maximum power transfer to the load

${{P}_{max}}=\frac{V_{S}^{2}}{4{{R}_{S}}}=900~W$

Maximum Power Transfer Theorem using Matlab Code

Here is the MATLAB code to implement the maximum power transfer theorem in MATLAB.

clear all;close all;clc
%% Circuit Parameters as given in the example (text)
% Matlab Code for Maximum Power Transfer Theorem
V_TH = 300; % Thevenin's Voltage
R_TH = 25; % Thevenin's Equivalent Resistance
R_L = 0:0.5:80; % Load Resistance
%%
%% Load Current & Power Calculation
IL = V_TH./(R_TH + R_L); % Load Current
P_L = IL.^2 .* R_L; % Load Power
%%
% As we know that maximum power transfer occurs when R_TH=R_L
%% Plotting the Results
plot(R_L,P_L,'b')
hold on
title('Maximum Power Transfer using Matlab');
xlabel('Load Resistance R_L');
ylabel('Power to the Load P_L');
gtext('R_TH = R_L = 25 Ohm')
legend('P_L')
grid on

Result

Here is a graph that clearly shows that maximum power transfer occurs when R_th=R_L.

Maximum Power Transfer Theorem using Matlab Simulink

Here, a Matlab Simulink model has been developed for three different cases:

1. When R_L>R_TH
2. When R_L=R_TH
3. When R_L<R_TH

To download and run the model, Click Here.

Maximum Power Transfer Theorem Application

Primarily in applications where voltage and current signals are used to convey information rather than to deliver large amounts of power. For instance, the first stage of a radio or television receiver should get as much power as possible out of the information-bearing signals that arrive via an antenna or cable. Those tiny signals account for only a small fraction of the total power consumed by the receiver, and so power-transfer efficiency is not a significant concern.