The load that takes the most

Maximum power transfer is the one result in this subject that is remembered as a rule and applied as a goal, and the two are not the same thing. It is a genuine maximum, it is exact, and the efficiency at it is fifty per cent — which means that following it as a design goal means deliberately throwing away half the energy in the source. Whether that is right depends entirely on what is scarce.

The maximum, measured #

A ten-volt source with fifty ohms inside it, driving a resistive load, solved at seventy-one load values from a twentieth of the source resistance to thirty times it. The power into the load is read as the current squared times the resistance, from the solved branch current, and the efficiency is that over the total the source delivered.

The peak is at a load-to-source ratio of 1.000000 and the efficiency there is 0.500000. Neither is assumed: the ratio is found by golden-section search on the continuous curve rather than by taking the largest of the seventy-one samples, for the reason the filters field records — a maximum located on a drawing grid reports the grid’s resolution and calls it a measurement.

The curve either side has a shape worth naming.

A load half the source resistance delivers exactly as much power as one twice it — 0.8889 in both rows — and the same is true of 0.8 and 1.25, and of any pair whose ratios are reciprocals. The power is $4x/(1+x)^2$, which is unchanged by replacing $x$ with $1/x$, so the curve is symmetric on a logarithmic axis. That is not a coincidence of these values; it is the reason the peak looks as flat as it does, and it is why the curve is drawn against the ratio on a logarithmic axis rather than against the resistance.

The flatness is the practically useful part: the load may be anywhere from 0.82 to 1.22 times the source resistance and still deliver 99% of the available power. A twenty per cent error in a matching network costs one per cent of the power, which is why matching is a forgiving business and why it is usually specified as a return loss rather than as a resistance.

The efficiency is the load’s share, and nothing else #

The right-hand column has a form that is worth stating because it makes the whole trade obvious.

The efficiency is $x/(1+x)$: the load’s share of the total resistance in the loop. That is all it is. It has no maximum, it rises monotonically from zero to one as the load grows, and at the matched point it is a half because the two resistances are equal.

So the two curves are not two views of one design variable being optimised. They are a quantity with a maximum and a quantity with no maximum, plotted together, and the design question is which of them is the constraint.

Where the source is fixed and the energy is free, take the maximum. A receiving antenna gets what it gets from the sky and there is nothing to conserve; the whole art is in not throwing away any of it, so a match is the right target and fifty per cent is not a loss in any sense that matters.

Where the source is a supply and the energy is paid for, do not go anywhere near it. A power converter running matched would be fifty per cent efficient and would dissipate its full output rating internally.

And where the source is a small signal and the noise is the constraint, the answer is neither — which is the noise field’s, and the optimum source resistance for noise is a different number from either. It is the ratio of an amplifier’s voltage noise to its current noise, which for an ordinary part is a few kilohms, and it has no connection whatever to the source’s own resistance. An input stage matched for power from a fifty-ohm source and one matched for noise from the same source are two different circuits, and the second is usually a transformer away from the first.

What ninety per cent costs #

The most useful number here is the price of efficiency, and it is not small.

Ninety per cent efficiency needs $x/(1+x) = 0.9$, so a load nine times the source resistance. The power delivered there is $4 \times 9/100 = 0.36$ of the maximum. Ninety per cent efficiency delivers 36% of the available power. Ninety-five per cent delivers 19%.

Put the other way: going from matched to ninety per cent efficient costs 64% of the power and saves the source from dissipating 64% of what it was. Both statements are true, and which one a designer cares about is the whole question.

This is the arithmetic behind two familiar facts that look unrelated. A power amplifier’s output stage is not matched to its loudspeaker — a modern one has an output impedance of milliohms into eight ohms, a ratio of thousands, and is running at the far right of this curve on purpose. And a transmitter’s output stage is matched to its antenna, because the scarce thing there is the antenna rather than the supply.

When the source has reactance in it #

Put reactance in the source and the familiar rule stops being the right one.

The load here is a pure resistance — which is the practical case, since a resistive load is what most things are — and the best resistive load is no longer the source’s resistance. It is the magnitude of the source impedance:

The best load is $\sqrt{R^2 + X^2}$ to six figures at every row, which is checked rather than assumed. And the efficiency at the optimum is no longer a half — it rises to 80% with four times the resistance in reactance, because the reactance carries current without dissipating anything, so the source’s own loss is a smaller share of what it is handling.

The power available falls at the same time, and that is the cost the reactance imposes: 39% of what a resistive source of the same resistance could give. Which is why the answer, when it is available, is to cancel the reactance rather than to live with it — a conjugate match returns the full power and the fifty per cent efficiency, and everything in the table is what happens when only the resistance can be chosen.

Reading it as a return loss #

The flatness of the peak is why radio engineering never quotes a match as a resistance. It quotes a reflection coefficient, and the two are worth converting between once.

A load $x$ times the source resistance reflects $(x-1)/(x+1)$ of the incident wave, and the power delivered is $1 - |\Gamma|^2$ of the available. So the 99% band above — $x$ from 0.82 to 1.22 — is $|\Gamma|$ from −0.099 to +0.099, which is a return loss of 20 dB. That is the number a specification actually carries, and it means: twenty decibels of return loss is one per cent of the power.

The two load ratios in each row are reciprocals, which is the symmetry above stated once more — the reflection’s magnitude is the same for a load and its reciprocal, and only the sign changes. That is why a mismatch specification is a single number rather than a range, and why a two-to-one standing wave ratio, which sounds severe, costs eleven per cent of the power.

It also explains why the far right of the efficiency curve is not reached by matching networks. A load nine times the source is a return loss of 1.9 dB — a reflection of 0.8 — which in a transmission-line system is a catastrophe and in a power-supply system is simply a well-designed regulator. The same number, read against two different scarcities.

The theorem’s own conditions #

Two of them, and both are worth checking before the rule is applied.

The source has to be linear, which is what makes an internal resistance a well-defined thing to have. A solar cell, a battery near its limit and a current-limited supply are all sources whose terminal curve is not a straight line, and for those the load that takes the most power is found by maximising the product of the terminal voltage and current on the actual curve — which is what a maximum-power-point tracker does, and it is somewhere else entirely.

And the source resistance has to be fixed. The theorem says: given this source, which load takes the most. It does not say: choose both. If the source resistance can be reduced, reducing it always increases the power and the efficiency, and the matched condition is a worse design than an unmatched one with a better source. That is the reading of the theorem that produces genuinely bad engineering, and it is common enough to be worth naming.

The arithmetic makes the second point sharply. A ten-volt source with fifty ohms inside it can deliver at best half a watt, into fifty ohms, at fifty per cent efficiency. The same ten volts behind five ohms delivers five watts into five ohms — ten times as much — and driving the original fifty-ohm load it delivers 1.65 W at 91% efficiency, which is three times the power and twice the efficiency of the matched case it replaced. Every quantity in the problem improved, and the improvement came from the half of the circuit the theorem treats as given. A design conversation that starts at “what load should this drive” has already conceded the larger of the two factors available.

The other fifty per cent, which is a different theorem #

Two results in this collection produce exactly one half, and they are not the same result.

This one: a matched load takes the most power available, and the source dissipates as much as the load receives. It is about a maximum, in the steady state, and the half falls out of the two resistances being equal at the optimum.

The other, in the transients field: charging a capacitor from a step dissipates exactly as much as the capacitor stores, whatever it is dissipated in. That is about a total over a transient, and the resistance does not appear at all.

Neither implies the other and the mechanisms have nothing in common — one is a maximum of a smooth function and the other is an exact conservation statement. It is worth having both in view because the coincidence of the number invites reading them as one theorem, and a designer who does that concludes that a slower charge would be fifty per cent efficient, which it is not.

What the theorem assumes #

Maximum power transfer is a statement about a fixed source impedance, and this field measures three things that move it. The far end that rises is the same source read as a voltage, where a leading load can put more voltage on the load than the source has. The current that does no work is the reactive part of the current the theorem’s efficiency figure contains. Exact outside and wrong within is the reduction the theorem is applied to, and it is exact at the terminals and silent about the dissipation inside.

The gate #

The optimum is asserted to be the magnitude of the source impedance, to a part in $10^5$, at every setting of the slider — which is the general statement, of which $R_L = R_S$ is the zero-reactance case.

The efficiency at the optimum is asserted to be exactly a half when the source is resistive, to a part in $10^6$, and asserted to be above a half when it is not — because that is the claim a reader is most likely to carry over from the resistive case without checking.

The optimum is located by search on the continuous curve, not by taking the largest sample. Locating a smooth maximum on a drawing grid measures the grid, which is the fault the ladder essays record and the reason this figure does not repeat it.

Where the theorem is used and where it is a mistake #

Fifty per cent efficiency at the optimum is the sentence that makes this theorem the most misapplied one in the subject, and the collection has cases on both sides of the line.

It is the right criterion wherever the available power is fixed and small and nothing is being wasted that could be recovered — a receiver’s antenna, a sensor, a transducer. The floor a circuit has then adds the qualification that decides the design: the source resistance at which an amplifier is quietest is 6.67 kΩ, the ratio of its two noise generators, and it is not the resistance that transfers maximum power. So even in the case this theorem is for, matching for power and matching for noise are two different matches, and the second is usually the one wanted.

It is the wrong criterion wherever the source is a supply, and the source that is not a source is why: an ideal voltage source’s whole purpose is to hold its voltage at any current, and a load matched to its internal resistance is a load operating at half the voltage it was designed for while the source burns as much as it delivers. A source below a frequency makes the point with a number — a regulator’s output impedance is 0.43 milliohms, so a “matched” load would be a short circuit — which is the reductio that shows the theorem is about a different kind of source entirely.

And exact outside and wrong within is the reason the efficiency half of this essay’s result has to be measured on the real network rather than on an equivalent: a Thévenin reduction is exact about what the terminals deliver and wrong about the heat behind them, by a factor of forty-three in the case measured there. The fifty per cent on this page is a statement about the equivalent’s internal resistance, and the network it stands for may be dissipating something else entirely.

The two-sided shape of the curve #

The last result on this page — that a load half the source resistance delivers exactly as much as one twice it — is the one worth carrying, because it is what makes the theorem usable when the match cannot be exact.

A maximum with a flat top is a maximum a design can miss. Being a factor of two out in either direction costs eleven per cent of the available power, which for most purposes is nothing, and that is why matching networks are built at all: a quarter wave, and the path the current takes back holds its reflection under a tenth over seventeen per cent of a band, and several sections, and the band they buy takes that to 92.6 per cent with four — bands that would be worthless against a sharp optimum and are ample against this one.

The symmetry is also the reason the efficiency half of the result is so often forgotten. The power curve is symmetric in the load ratio and the efficiency curve is monotone in it, so a designer moving away from the optimum in the direction of a larger load loses very little power and gains a great deal of efficiency — ninety per cent at nine times the source, delivering 36 per cent of what was available — while moving the same distance the other way loses the same power and makes the efficiency worse. Two directions that look identical on one axis and are not on the other, which is what a single optimum quoted without its second quantity conceals.

Which is why the ninety-per-cent row is the one worth carrying out of this essay. A load nine times the source resistance delivers 36 per cent of the available power and wastes ten — against a matched load delivering 100 per cent of the available power and wasting as much again. Anywhere the source’s own dissipation is a cost rather than an irrelevance, that is not a trade-off; it is a better design by both measures that matter, and it is what every power circuit in this collection is arranged to do. Maximum power transfer is the right criterion only where the power not delivered is lost anyway — which is true of an antenna and of very little else.