Signals and Networks: Maximum Power Transfer Theorem

The maximum power transfer theorem states that, to obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.

fixed source resistance:

R S

load resistance:

R L

I = \frac{V}{R_\mathrm{S} + R_\mathrm{L}}.

P_\mathrm{L} = I^2 R_\mathrm{L} = \left(\frac{V}{R_\mathrm{S} + R_\mathrm{L}}\right)^2 R_\mathrm{L} = \frac{V^2}{R_\mathrm{S}^2 / R_\mathrm{L} + 2 R_\mathrm{S} + R_\mathrm{L}}.

L

will be maximum, when the denominator

R_\mathrm{S}^2 / R_\mathrm{L} + 2 R_\mathrm{S} + R_\mathrm{L}

will be minimum.

\frac{d}{d R_\mathrm{L}} \left(R_\mathrm{S}^2 / R_\mathrm{L} + 2R_\mathrm{S} + R_\mathrm{L}\right) = -R_\mathrm{S}^2 / R_\mathrm{L}^2 + 1.

For a maximum or minimum, the first derivative is zero, so

R_\mathrm{S}^2 / R_\mathrm{L}^2 = 1

R_\mathrm{S} = R_\mathrm{L}. \,\!

The theorem results in maximum power transfer, and not maximum efficiency. If the resistance of the load is made larger than the resistance of the source, then efficiency is higher, since a higher percentage of the source power is transferred to the load, but the magnitude of the load power is lower since the total circuit resistance goes up.

We define the efficiency

η

as the ratio of power dissipated by the load to power developed by the source.

\eta = \frac{R_\mathrm{load}}{R_\mathrm{load} + R_\mathrm{source}} = \frac{1}{1 + R_\mathrm{source} / R_\mathrm{load}}.

Consider three particular cases:

If $R_\mathrm{load} = R_\mathrm{source}$ , then $\eta = 0.5,$
If $R_\mathrm{load} = \infty$ or $R_\mathrm{source} = 0,$ then $\eta = 1,$
If $R_\mathrm{load} = 0$ , then $\eta = 0.$

In Reactive circuits:

|I| = { |V_\mathrm{S}| \over |Z_\mathrm{S} + Z_\mathrm{L}| }.

\begin{align} P_\mathrm{L} & = I_\mathrm{rms}^2 R_\mathrm{L} = {1 \over 2} |I|^2 R_\mathrm{L} = {1 \over 2} \left( {|V_\mathrm{S}| \over |Z_\mathrm{S} + Z_\mathrm{L}|} \right)^2 R_\mathrm{L} \\ & = {1 \over 2}{ |V_\mathrm{S}|^2 R_\mathrm{L} \over (R_\mathrm{S} + R_\mathrm{L})^2 + (X_\mathrm{S} + X_\mathrm{L})^2}, \end{align}

where the resistance

R_\mathrm{S}

and reactance

X_\mathrm{S}

are the real and imaginary parts of

Z_\mathrm{S}

, and

X_\mathrm{L}

is the imaginary part of

Z_\mathrm{L}

To determine the values of

R_\mathrm{L}

and

X_\mathrm{L}

(since

V_\mathrm{S}

R_\mathrm{S}

, and

X_\mathrm{S}

are fixed) for which this expression is a maximum, we first find, for each fixed positive value of

R_\mathrm{L}

, the value of the reactive term

X_\mathrm{L}

for which the denominator

(R_\mathrm{S} + R_\mathrm{L})^2 + (X_\mathrm{S} + X_\mathrm{L})^2 \,

is a minimum. Since reactances can be negative, this denominator is easily minimized by making

X_\mathrm{L} = -X_\mathrm{S}.\,

The power equation is now reduced to:

P_\mathrm{L} = {1 \over 2}{{|V_\mathrm{S}|^2 R_\mathrm{L}}\over{(R_\mathrm{S} + R_\mathrm{L})^2}}\,\!

and it remains to find the value of

R_\mathrm{L}

which maximizes this expression. However, this maximization problem has exactly the same form as in the purely resistive case, and the maximizing condition

R_\mathrm{L} = R_\mathrm{S}

can be found in the same way.

The combination of conditions

$R_\mathrm{L} = R_\mathrm{S}\,\!$
$X_\mathrm{L} = -X_\mathrm{S}\,\!$