Maximum power theorem: Wikis

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Encyclopedia

In electrical engineering, 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 be made the same as that of the source. It is claimed that Moritz von Jacobi was first to discover the maximum power (transfer) theorem which is referred to as "Jacobi's law".

The theorem applies to maximum power, and not maximum efficiency. If the resistance of the load is made larger than the resistance of the source, then efficiency is higher, since most of the power is generated in the load, but the overall power is lower since the total circuit resistance goes up.

If the internal impedance is made larger than the load then most of the power ends up being dissipated in the source, and although the total power dissipated is higher, due to a lower circuit resistance, it turns out that the amount dissipated in the load is reduced.

Maximizing power transfer versus power efficiency

The theorem was originally misunderstood (notably by Joule) to imply that a system consisting of an electric motor driven by a battery could not be more than 50% efficient since, when the impedances were matched, the power lost as heat in the battery would always be equal to the power delivered to the motor. In 1880 this assumption was shown to be false by either Edison or his colleague Francis Robbins Upton, who realized that maximum efficiency was not the same as maximum power transfer. To achieve maximum efficiency, the resistance of the source (whether a battery or a dynamo) could be made close to zero. Using this new understanding, they obtained an efficiency of about 90%, and proved that the electric motor was a practical alternative to the heat engine.

The condition of maximum power transfer does not result in maximum efficiency. If we define the efficiency η as the ratio of power dissipated by the load to power developed by the source, then it is straightforward to calculate from the above circuit diagram that

$\eta = {R_\mathrm{load} \over {R_\mathrm{load} + R_\mathrm{source} } } = { 1 \over { 1 + { R_\mathrm{source} \over R_\mathrm{load} } } } \,\!$

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 $\eta=0\,\!$.

The efficiency is only 50% when maximum power transfer is achieved, but approaches 100% as the load resistance approaches infinity (or the source resistance approaches zero), though the total power level tends towards zero. When the load resistance is zero, all the power is consumed inside the source (the power dissipated in a short circuit is zero) so the efficiency is zero.

Note that for a time varying voltage and current, the above applies only to the resistive component of the power, not the reactive component. As shown below, a proper pure reactive match can theoretically achieve 100% power transfer AND 100% efficiency simultaneously in a perfect resistance free reactive circuit. In an imperfect reactive circuit (i.e. one with both reactance and resistance), the efficiency will be the power dissipated in the load divided by the power provided by the source. When the resistance is a small fraction of the reactance, the efficiency will then be very high even at the maximum power transfer point. This is a key reason why power grids use AC versus DC power, and why radio frequency (RF) circuits can operate with relatively high efficiency.

Impedance matching

A related concept is reflectionless impedance matching. In radio, transmission lines, and other electronics, there is often a requirement to match the source impedance (such as a transmitter) to the load impedance (such as an antenna) to avoid reflections in the transmission line.

Calculus-based proof for purely resistive circuits

(See Cartwright[1] for a non-calculus-based proof)

In the diagram opposite, power is being transferred from the source, with voltage $V \,\!$ and fixed source resistance $R_\mathrm{S} \,\!$, to a load with resistance $R_\mathrm{L} \,\!$, resulting in a current $I \,\!$. By Ohm's law, $I \,\!$ is simply the source voltage divided by the total circuit resistance:

$I = {V \over R_\mathrm{S} + R_\mathrm{L}} \,\!$

The power $P_\mathrm{L} \,\!$ dissipated in the load is the square of the current multiplied by the resistance:

$P_\mathrm{L} = I^2 R_\mathrm{L} = {{ \left( {V \over {R_\mathrm{S} + R_\mathrm{L}}} \right) }^2} R_\mathrm{L} = {{V^2} \over {R_\mathrm{S}^2 / R_\mathrm{L} + 2R_\mathrm{S} + R_\mathrm{L}}} \,\!$

We could calculate the value of $R_\mathrm{L} \,\!$ for which this expression is a maximum, but it is easier to calculate the value of $R_\mathrm{L} \,\!$ for which the denominator

$R_\mathrm{S}^2 / R_\mathrm{L} + 2R_\mathrm{S} + R_\mathrm{L} \,\!$

is a minimum. The result will be the same in either case. Differentiating with respect to $R_\mathrm{L} \,\!$:

${d\over{dR_\mathrm{L}}} \left( {R_\mathrm{S}^2 / R_\mathrm{L} + 2R_\mathrm{S} + R_\mathrm{L}} \right) = -R_\mathrm{S}^2 / R_\mathrm{L}^2+1 \,\!$

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

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

or

$R_\mathrm{L} = \pm R_\mathrm{S} \,\!$

In practical resistive circuits, $R_\mathrm{S} \,\!$ and $R_\mathrm{L} \,\!$ are both positive. To find out whether this solution is a minimum or a maximum, we must differentiate again:

${{d^2} \over {dR_\mathrm{L}^2}} \left( {R_\mathrm{S}^2 / R_\mathrm{L} + 2 R_\mathrm{S} + R_\mathrm{L}} \right) = {2 R_\mathrm{S}^2} / {R_\mathrm{L}^3} \,\!$

This is positive for positive values of $R_\mathrm{S} \,\!$ and $R_\mathrm{L} \,\!$, showing that the denominator is a minimum, and the power is therefore a maximum, when

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

In reactive circuits

The theorem also applies where the source and/or load are not totally resistive. This invokes a refinement of the maximum power theorem which says that any reactive components of source and load should be of equal magnitude but opposite phase. (See below for a derivation.) This means that the source and load impedances should be complex conjugates of each other. In the case of purely resistive circuits, the two concepts are identical. However, physically realizable sources and loads are not usually totally resistive, having some inductive or capacitive components, and so practical applications of this theorem, under the name of complex conjugate impedance matching, do, in fact, exist.

If the source is totally inductive (capacitive), then a totally capacitive (inductive) load, in the absence of resistive losses, would receive 100% of the energy from the source but send it back after a quarter cycle. The resultant circuit is nothing other than a resonant LC circuit in which the energy continues to oscillate to and fro. This is called reactive power. Power factor correction (where an inductive reactance is used to "balance out" a capacitive one), is essentially the same idea as complex conjugate impedance matching although it is done for entirely different reasons.

For a fixed reactive source, the maximum power theorem maximizes the real power (P) delivered to the load by complex conjugate matching the load to the source.

For a fixed reactive load, power factor correction minimizes the apparent power (S) (and unnecessary current) conducted by the transmission lines, while maintaining the same amount of real power transfer. This is done by adding a reactance to the load to balance out the load's own reactance, changing the reactive load impedance into a resistive load impedance.

Proof

In this diagram, AC power is being transferred from the source, with phasor magnitude voltage | VS | (peak voltage) and fixed source impedance ZS, to a load with impedance ZL, resulting in a phasor magnitude current | I | . | I | is simply the source voltage divided by the total circuit impedance:

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

The average power PL dissipated in the load is the square of the current multiplied by the resistive portion (the real part) RL of the load impedance:

\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 RS and reactance XS are the real and imaginary parts of ZS, and XL is the imaginary part of ZL.

In order to determine the values of RL and XL (since VS, RS, and XS are fixed) for which this expression is a maximum, we first find, for each fixed positive value of RL, the value of the reactive term XL 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 RL which maximizes this expression. However, this maximization problem has exactly the same form as in the purely resistive case, and the maximizing condition RL = RS can be found in the same way.

The combination of conditions

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

can be concisely written with a complex conjugate (the *) as:

$Z_\mathrm{S} = Z_\mathrm{L}^*.$

Notes

1. ^ Cartwright, Kenneth V (Spring 2008), "Non-Calculus Derivation of the Maximum Power Transfer Theorem", Technology Interface 8 (2): 19 pages

References

• H.W. Jackson (1959) Introduction to Electronic Circuits, Prentice-Hall.