The load that may be complex

The load that takes the most measured the oldest result in the subject and found a generalisation sitting inside it. A load equal to the source resistance takes more power than any other and does so at exactly fifty per cent efficiency — and when the source has a reactance, the best resistive load is not its resistance but the magnitude of its impedance. So $R_L = R_S$ is the reactance-free line of a family, and everything in that essay’s table is what happens when only a resistance may be chosen.

This essay removes the constraint. If the load may be any impedance at all, what is the best one, and what was the constraint costing?

The first answer is known and is worth deriving by search rather than by assumption. The second has a closed form with a familiar number in it.

The conjugate, searched for rather than substituted #

The answer is $Z_L = R_S - jX_S$ and every textbook says so. The habit here is to find it rather than to write it down, so the figure searches: a two-dimensional pattern search over the load’s resistance and its reactance, evaluating the delivered power on a solved network at each step, halving the step when no neighbour improves, until both steps are below a part in a billion.

Against a source of $50 + j100$ ohms the search returns $50.00 - j100.0$, which is the conjugate to four figures in both parts. Nothing about the answer was put in.

Two properties follow and both are checked on the same solve.

It delivers the available power exactly. $|V|^2/4R_S$ = 500.00 mW here, and the search’s best load delivers 500.00 mW. The reactances have cancelled, the remaining problem is a resistive source driving a resistive load of equal value, and the answer is the one the essay before it established for $X_S = 0$.

Its efficiency is exactly a half, at every source reactance. That is the part worth pausing on. A source reactance dissipates nothing, so the only thing burning power in the source is $R_S$, and the load’s resistance has been made equal to it — so the two dissipate equally whatever $X_S$ is. Fifty per cent, at $X_S = 0$, at $X_S = 8R_S$, and at every value between.

What a resistor alone gives up, and the number in it #

A load that may only be a resistance has one choice and the essay before it found it: $R_L = |Z_S|$. What that delivers, as a fraction of the available power, has a closed form:

$$\frac{P}{P_{av}} = \frac{4R_S|Z_S|}{(R_S+|Z_S|)^2 + X_S^2} = \frac{2R_S}{R_S+|Z_S|} = \frac{2}{1 + \sqrt{1+x^2}}, \qquad x = \frac{X_S}{R_S}$$

The middle step uses $X_S^2 = |Z_S|^2 - R_S^2$, which collapses the denominator to $2|Z_S|(R_S+|Z_S|)$, and everything cancels except a ratio of two resistances. The figure measures the fraction at ten source reactances and it agrees with that expression to four parts in $10^{16}$.

At $x = 2$ the expression is $2/(1+\sqrt5)$, which is $\tfrac12(\sqrt5-1)$ — exactly the golden ratio less one, 0.618034. A resistor-only load against a source whose reactance is twice its resistance takes 61.80 per cent of what a conjugate match would take, and the number is not a fitted one.

The golden ratio is a coincidence of the number two rather than a deep fact, and it is a good one to remember: it fixes the middle of the table without a calculation, and it is the point at which a resistor-only load has given up more than a third of what is there.

The resistor is more efficient and takes less #

The fourth column of the table is the one that reads backwards on a first pass, and it is the useful half of the result.

A resistor-only load’s efficiency is $R_L/(R_L+R_S) = |Z_S|/(|Z_S|+R_S)$, which is a half at $x = 0$ and rises monotonically towards one as the reactance grows. At $x = 2$ it is 69.1 per cent; at $x = 12$ it is 92.3. The conjugate match sits at exactly fifty per cent for ever.

So the load that takes less power wastes less of what it takes. That is not a contradiction: maximum power transfer is not maximum efficiency and never was, which the load that takes the most established for the resistive case. What is new here is that the two criteria pull apart further as the source becomes more reactive, and that the resistor-only design is the efficient one.

The mechanism is simple once seen. The conjugate match insists on $R_L = R_S$, which is what forces the fifty-fifty split. A resistor-only load is larger than $R_S$ whenever there is any reactance — $|Z_S| > R_S$ always — so it takes a larger share of whatever current flows, and the source burns less. It delivers less because the uncancelled reactance limits the current, not because it is wasting more.

Which sets up the choice a designer actually faces:

A signal source into an amplifier. Take the power. The conjugate match, or a lossless network that performs one, because the source’s power is what the measurement is made of and the source’s dissipation is somebody else’s problem.

A supply into a load. Take the efficiency. Neither answer here — the operating point is far from either, with $R_L \gg R_S$, and both of these are curiosities at that end of the axis.

A source whose dissipation is the limit, which is a transmitter’s output device or a battery. The efficient answer, and the reactance is cancelled for the sake of the current rather than the power — which is the same reasoning the current that does no work applies to a power system’s reactive component, where nothing is delivered and everything is heated.

What the reactance is for #

The gap between the two curves is the value of adding a reactive component, and the table makes it a design number rather than a principle.

At $x = 1$ a series capacitor of the right value recovers 21 per cent more power. At $x = 2$ it recovers 62 per cent more. At $x = 8$ it recovers a factor of 4.5. Those are the returns on a single reactive part, and they say where such a part is worth fitting: not below about $x = 0.5$, where the gain is under six per cent, and unavoidably above $x = 2$.

That is the same conclusion the frequency field reaches from a different direction. The reactance cancelled, and the resonance it buys measures what happens when a load’s reactance is tuned out and finds that the arrangement becomes resonant — so the cancellation is exact at one frequency and degrades away from it, at a rate set by the quality factor of the combination. That quality factor is $|X_S|/R_S$, which is the same $x$ this essay’s axis is, so the wider the gap between the two strategies the narrower the band over which the better one can be realised.

Written as one statement: the more a conjugate match is worth, the less bandwidth it has. A source with $x = 8$ has four times as much power available to a conjugate match as to a resistor, and the match holds over a band of roughly $1/8$ of the centre frequency. A source with $x = 0.5$ has six per cent more available and the match holds over two octaves.

That trade is not visible from either curve alone and it is the reason the resistor-only answer keeps being used. It is broadband and it is wrong by a computable amount.

The same expression, in the units a radio engineer uses #

The fraction $2/(1+\sqrt{1+x^2})$ has a second form that makes it recognisable, and it is worth deriving because it turns this essay’s axis into a quantity that can be read off an instrument.

A load $Z_L$ on a source $Z_S$ reflects $\Gamma = (Z_L - Z_S^*)/(Z_L + Z_S)$, and the fraction of the available power it absorbs is $1 - |\Gamma|^2$. For the best resistive load $Z_L = |Z_S|$, and grinding through gives exactly the expression above — so the fraction in the table is $1 - |\Gamma|^2$, and the column can be read as a return loss.

The second column is the golden ratio again at $x = 2$, and this time as the reflection coefficient itself: $|\Gamma| = 0.618034$ exactly, because $|\Gamma|^2 = 1 - \varphi^{-1}$ and $\varphi^{-1} = \varphi - 1$ produces the same number twice. A reflection coefficient equal to the golden ratio and a transmitted fraction equal to it are the same statement about one source.

The practical use is that return loss is what an instrument measures. A source whose reactance is twice its resistance, terminated in the best resistance available, shows 4.18 dB of return loss — and a designer seeing 4.18 dB on an analyser knows immediately that a reactance of $2R_S$ is being left uncancelled and that a single reactive part would recover 62 per cent more power.

Which is a better form of the result than the fraction, because it is measurable without knowing the source impedance at all.

Why the resistive answer is $|Z_S|$ and not $R_S$ #

The essay before it checked this and did not explain it, and the explanation is short enough to belong here.

The power into a resistance $R_L$ from a source $R_S + jX_S$ is $|V|^2R_L/((R_S+R_L)^2 + X_S^2)$. Differentiating and setting to zero gives $R_L^2 = R_S^2 + X_S^2$, so $R_L = |Z_S|$.

The reactance therefore enters as though it were an extra source resistance for the purposes of choosing $R_L$, and it enters as a square added under a root. That is why the best resistance grows slowly with the reactance — 1.12 times at $x = 0.5$, 1.41 at $x = 1$, 2.24 at $x = 2$ — and why the familiar $R_L = R_S$ survives as a useful approximation up to a reactance comparable with the resistance.

The figure searches for it rather than substituting, at every reactance on the slider, and the search returns $|Z_S|$ to five figures each time. Two routes to one quantity, sharing only the netlist.

Where the reactance comes from in the first place #

The slider is a source reactance and it is worth naming three places it is not a modelling convenience but the dominant term.

A piezoelectric or capacitive transducer. Its source impedance is a small resistance in series with a large capacitance, so $x$ is tens or hundreds at low frequency and the resistor-only answer takes a few per cent of what is there. That is why such a transducer is read by a charge amplifier — a virtual earth, which is neither of the two strategies here and is the limit of the efficient one, taking almost no power and all of the signal.

An antenna away from resonance. A short antenna is a radiation resistance of a few ohms in series with a large capacitive reactance, which is $x$ in the hundreds; a long one is inductive. The whole practice of antenna tuning is the conjugate match of this essay, and the fact that the match narrows as $x$ grows is why a short antenna is narrowband — the bandwidth is not a property of the wire, it is a property of the match the wire forces.

A transformer’s leakage. What coupling buys, and where it does not measures the reactance that appears in series with a transformer’s secondary, and it is exactly a source reactance in this sense. A transformer driving a resistive load at a frequency where its leakage matters has $x$ of order one, and the twenty per cent this essay prices at $x = 1$ is recoverable with a series capacitor — which is what a resonant converter is.

The three have one thing in common that this essay’s axis makes visible: the source reactance is almost never a choice, and the only variable is whether to cancel it. What the table gives is the value of doing so, in a currency that does not depend on the application.

What an efficiency column does not mean here #

That a conjugate match is always available. It requires a reactance of opposite sign to the source’s, which is a capacitor for an inductive source and an inductor for a capacitive one, and both are components with their own loss and their own self-resonance — the inductor that is a capacitor being the limit where the cancelling part has stopped being the thing it was fitted as. A lossy cancelling reactance takes some of what it recovers, and how much is the efficiency a fixed Q costs’s subject in the matching-network case.

That the efficiency figures mean what an efficiency usually means. They are the fraction of the power leaving the source that reaches the load, on the assumption that the source’s resistance is a real dissipating resistance. For a source whose internal resistance is a model of something else — a radiation resistance, a transconductance’s output impedance — the “wasted” power is not waste and the column means nothing.

That the golden ratio is significant. It is $2/(1+\sqrt5)$ evaluated at $x = 2$, and $x = 2$ is a round number rather than a special one. What is worth carrying is the expression, and the golden ratio is a mnemonic for one point on it.

That either answer is what a power system does. Both are matched-source problems, and a supply feeding a load operates with $R_L$ a hundred times $R_S$ at ninety-nine per cent efficiency, taking four per cent of the available power. The load that takes the most makes that point and it is worth repeating: maximum power transfer is a signal idea, not a power one.

A search over both parts, and the reactance that returns none #

The best load is searched over both of its parts on the solved network — a pattern search in two dimensions, halving its steps to a part in a billion — and required to be the conjugate to two parts in ten thousand in each part.

Its delivered power is checked against $|V|^2/4R_S$ to a part in ten thousand, and its efficiency against exactly one half.

The resistor-only fraction is checked against $2/(1+\sqrt{1+x^2})$ at ten source reactances, and comes out at four parts in $10^{16}$.

The golden ratio is checked at $x = 2$ to a part in $10^{12}$, as $(\sqrt5-1)/2$.

The efficiency ordering is checked at every non-zero reactance — the resistor-only load strictly more efficient than the conjugate — which is the essay’s counter-intuitive half.

And the whole thing is refused at zero reactance, where the two-dimensional search returns no reactance at all and the two strategies coincide, which is what says the conjugate is the reactance’s doing rather than the search’s.

Two criteria that separate with the reactance #

That essay’s headline was that the maximum power and the fifty per cent efficiency arrive together, which makes them look like one fact. This essay separates them and the separation grows with one variable.

At $X_S = 0$ the best resistive load is the conjugate — there is nothing to conjugate — so the two criteria agree about the load and disagree only about what to call the answer. At $X_S = 2R_S$ they want different loads, and the one that maximises power is at 50 per cent while the one that maximises efficiency among resistive loads is at 69.1. At $X_S = 12R_S$ the gap is 50 against 92.3.

So “maximum power transfer occurs at fifty per cent efficiency” is a statement about the conjugate match specifically, and it is true at every source reactance for that load — the fifty per cent does not move. What moves is everything else: how much power that load gets relative to any other, how efficient the alternatives are, and how narrow the band over which the conjugate can be realised.

The general shape turns up often enough here to be worth naming: a result that is stated for a special case and is then quoted as though the special case were the general one. $R_L = R_S$ is the $X_S = 0$ line of the resistive family; fifty per cent is the conjugate match’s efficiency at every reactance and nobody else’s; and $|Z_S|$ is the resistive answer that reduces to $R_S$ when the reactance vanishes. Three statements, one of which is general and two of which are lines through a family, and all three are usually said in the same sentence.

Still open: the lossy cancelling reactance, the band, and the source whose reactance is not known #

The reactance that is not free. A cancelling component with a finite quality factor adds a resistance in series with the source’s, so the conjugate match’s delivered power falls and its efficiency moves off a half. Solved against the component’s quality factor it would say at what $Q$ the conjugate match stops beating the resistor-only one — which is the practical version of this essay’s question and is one more element on the same netlist.

The band, measured rather than argued. The claim above that the conjugate match holds over roughly $1/x$ of the centre frequency is a quality-factor argument. Sweeping the delivered power against frequency for both strategies would give the two bands directly and would say whether the resistor-only answer is genuinely broadband or merely less sharply peaked.

And a source whose reactance is not known to sign. Everything here assumes $X_S$ is known and fixed. A source whose reactance varies — an antenna across a band, a transducer across temperature — cannot be conjugate-matched at every point, and the best fixed load for a range of source impedances is a different optimisation with a different answer. Whether it is the resistor-only answer at the range’s centre, or something else entirely, is a question this figure’s apparatus could settle by searching over a weighted set of sources rather than over one.