The reactance that is not free

The load that takes the most found the familiar result by search rather than by derivation: a load equal to the source’s resistance takes the most power, and takes it at exactly fifty per cent efficiency. The load that may be complex freed the load’s reactance and found the rest of the family. Against a source with a reactance, the best load is the source’s conjugate, which cancels the reactance and takes all the available power at half efficiency whatever the reactance. A load that may only be a resistance takes $2/(1 + \sqrt{1 + x^2})$ of that power, with $x$ the source’s reactance over its resistance — at $x = 2$ exactly the golden ratio less one, 0.618.

The conjugate match’s advantage rests on one assumption that the earlier essay named and left: that the cancelling reactance is free. A real capacitor or inductor has losses, summarised by its quality factor, and a lossy cancelling element adds a resistance in series with the source’s. The earlier essay asked at what quality factor the conjugate match stops beating the resistor-only one. It has an exact answer. It also turns out to be the wrong question, because the best use of a lossy component is not to cancel all of the reactance with it.

The two loads, for reference #

The source here has 50 Ω of resistance and 100 Ω of reactance at the frequency of interest, $x = 2$. Its available power, the most any load can take, is 500 mW for a 10 V source. A conjugate load takes all of it. A resistor alone, the best one being 111.8 Ω, the magnitude of the source’s impedance, takes 0.618 of it: 309 mW.

A capacitor with a quality factor #

The conjugate load is two parts: a resistance equal to the source’s, and a capacitive reactance equal and opposite to the source’s inductive one. The capacitor is the part that is not free. A capacitor’s losses are summarised by its quality factor $Q$, the ratio of its reactance to its equivalent series resistance, so a capacitor of reactance $X_c$ carries a resistance $X_c/Q$ that dissipates power and delivers none to the load.

The quality factors that matter in practice span the whole range this page draws. A good film or ceramic capacitor at audio frequencies has a Q of hundreds or thousands; an electrolytic, a few to a few tens; a wound inductor at a kilohertz, perhaps ten or twenty; a small inductor at radio frequency, tens. So the thresholds below are not academic limits at the edge of what components can do. A cancelling inductor with a Q of three sits squarely in the region where the answer depends on how it is used.

Put that capacitor in series with the source and cancel the reactance completely. The source now looks like a pure resistance $R_S + X_S/Q$, and the best load resistor matches it: the load takes $V^2/(4(R_S + X_S/Q))$, which is

$$\frac{P}{P_{\text{avail}}} = \frac{1}{1 + x/Q}$$

of the power a lossless conjugate would have taken. A capacitor with a Q of ten, cancelling a reactance of twice the source’s resistance, costs a sixth of the available power. And the efficiency is exactly a half again, at every Q — the load takes as much as the source and the capacitor dissipate between them. The earlier essay expected the efficiency to move off a half once the cancelling element was lossy. It does not, because the load resistor is re-matched to the new total resistance and the division of the power is then the resistive problem once more.

The first threshold #

The full cancellation’s curve rises from nothing towards one as Q grows, and crosses the resistor-alone line at one quality factor. Setting $1/(1 + x/Q)$ equal to $2/(1 + \sqrt{1 + x^2})$ gives it:

$$Q_{\text{full}} = \frac{2\left(\sqrt{1 + x^2} + 1\right)}{x},$$

3.2361 at $x = 2$ — which is twice the golden ratio, $1 + \sqrt 5$, exactly. Below that quality factor a designer who cancels the whole reactance delivers less power than one who fits the best resistor and cancels nothing. A capacitor with a Q of three, which is a poor capacitor but a plausible inductor at some frequencies, costs more than it saves.

That answers the question the earlier essay posed. It is not the end of it, because full cancellation and no cancellation are not the only two choices.

The threshold depends on how much reactance there is to cancel, and it moves the opposite way from what one might guess.

For a source of 50 Ω + j25 Ω, whose reactance is half its resistance, a resistor alone already takes 0.9443 of the available power, and there is little left to win. Full cancellation must reach a quality factor of 8.47 before it helps, and any cancellation at all needs 4.24. The smaller the reactance, the better the capacitor must be, because the mismatch it removes shrinks faster than its own loss does. So the cancelling capacitor is worth fitting for a badly mismatched source and not for a mildly mismatched one, which is the reverse of the instinct that a small correction is always cheap. At a quality factor of three, where partial cancellation still beat a resistor alone for the j100 Ω source, the best design for the j25 Ω source cancels nothing.

Cancelling part of the reactance #

The capacitor’s value is free to choose. A smaller reactance cancels less of the source’s and carries less resistance of its own, since its loss is its reactance over Q. So there is a trade inside the conjugate match itself: each ohm of reactance cancelled removes an ohm of mismatch and adds $1/Q$ of an ohm of loss, and once the remaining reactance is small the mismatch it leaves costs less than the loss that would remove it.

With the load resistor always chosen as the magnitude of whatever impedance remains — the resistor-only rule applied to the partly cancelled source — the delivered power is $V^2/(2(R + |Z|))$, with $R$ the source’s resistance plus the capacitor’s and $|Z|$ the magnitude of what is left. Minimising $R + |Z|$ over the capacitor’s reactance gives the best partial cancellation, and at every quality factor it is at least as good as either extreme.

At $Q = 3$ it takes 0.6650 of the available power, where full cancellation takes 0.5971 and a resistor alone 0.6180: better than both, with a capacitor that full cancellation had made not worth fitting. At $Q = 100$ it takes 0.9807 against 0.9806 — the two converge, because a good enough capacitor should cancel nearly everything.

The best fraction to cancel is zero below a threshold quality factor, and rises smoothly above it towards one. At a Q of ten the best design cancels 88 per cent of a reactance twice the source’s resistance; at a hundred, 99. Full cancellation is the limit the fraction approaches as the capacitor becomes lossless. It is not a design choice anyone should make with a real capacitor, since at every finite Q cancelling slightly less does slightly better.

In numbers, at a Q of ten against this source: full cancellation delivers 1/(1 + 2/10) of the available 500 mW, 417 mW, and burns 83 mW in the capacitor. The best partial cancellation cancels 88 per cent of the reactance, leaves 12 Ω of it in place, matches the load to what remains, and delivers 421 mW. Four milliwatts is not much; the point is the direction, and that it costs nothing. The capacitor is smaller, which is usually cheaper, and the design is more tolerant of the capacitor’s own value, since it sits on a flat part of the curve rather than at its edge.

The second threshold, and why it is the golden ratio #

The threshold below which cancelling anything hurts comes from asking whether a first small ohm of cancellation helps. At zero cancellation, adding a reactance $X_c$ reduces the remaining mismatch at a rate $X_S/|Z_S|$ and adds loss at a rate $1/Q$ to both the resistance and, through it, the magnitude. It helps when

$$Q > \frac{\sqrt{1 + x^2} + 1}{x} = \frac{|Z_S| + R_S}{X_S},$$

which is half of the full-cancellation threshold. At $x = 2$ it is $(\sqrt 5 + 1)/2$, the golden ratio, 1.6180, and the search over the capacitor’s value finds exactly that: below it the best capacitor has no reactance at all, and above it the best one has some.

The golden ratio appears for the same reason as in the earlier essay. There the resistor-only load took $2/(1 + \sqrt{1 + x^2})$, whose denominator at $x = 2$ is $1 + \sqrt 5$; here the threshold is that same $1 + \sqrt{1 + x^2}$ divided by $x$, and $x = 2$ halves it. The number is not a coincidence of the chosen reactance so much as a property of $\sqrt 5$, which is what $|Z_S|/R_S$ is when the reactance is twice the resistance.

The search lands on exactly zero below the threshold, not on a small value, and that is worth a sentence because it is unusual for an optimum. The quantity being minimised, $R + |Z|$, has a slope with respect to the cancelling reactance that is constant in sign across the whole range when Q is below the threshold: every ohm cancelled costs more than it gains, the first ohm included, and the best design is at the boundary of what is allowed. Above the threshold the slope at zero is negative and the optimum moves into the interior, continuously, starting from nothing. So the fraction cancelled does not jump as Q crosses the threshold; it grows from zero, which is what the fraction figure shows as the curves leave the axis one by one.

Drawn against the source’s reactance, both thresholds fall as the reactance grows, towards one and two respectively. The direction is the useful part. A small reactance is expensive to cancel: at a tenth of the source’s resistance, a capacitor needs a Q of 20 before cancelling any of it helps, because there is little mismatch to remove and every ohm of cancelling reactance brings its full loss. A large reactance is cheap to cancel: at five times the resistance a Q of 1.22 is enough, since almost any reduction of a large mismatch is worth its loss.

So the rule of thumb that a matching network is worth fitting only if its components are good has the dependence the wrong way round for the cases where it matters most. A badly mismatched source rewards even a poor cancelling component; a nearly matched one needs a good component or none.

Where the loss is the whole story #

At the other extreme, where the source’s reactance dwarfs its resistance, the thresholds stop mattering and the loss formula takes over. An electrically small antenna or a small piezoelectric transducer can present a reactance twenty times its radiation or motional resistance. At $x = 20$ the partial threshold is a Q of barely more than one, so any real cancelling component is worth fitting — a resistor alone would take only 9.5 per cent of the available power. But a cancelling inductor with a Q of 50 takes $1/(1 + 20/50)$, 71 per cent, and the best partial cancellation cancels 99.7 per cent of the reactance and takes 71.5. The mismatch is nearly all removed and the component’s own loss is the limit: the inductor’s resistance raises the source’s effective resistance by forty per cent, and the 29 per cent of the available power that the load does not receive is that increase, whatever the design.

That is the regime in which the quality of the cancelling component is the whole of the matching problem, and it is the reason small antennas are quoted with efficiencies set by their tuning coils rather than by their radiators. In between — a reactance of the order of the resistance — the thresholds on this page are where the design decisions sit.

What this does to a matching network #

Every matching network built of reactances has this problem, and the figures here are its simplest case — one series element, one resistive load. An L-network, a transformer or a tuned coupling has several reactive elements, each with its own Q, and each adding loss in proportion to its reactance. The general lesson survives the extra elements: the best network built of lossy parts does not perform the exact transformation a lossless design calls for. It stops a little short, at the point where the last increment of transformation costs more in loss than it recovers in mismatch.

The Q the components allow measured how a resonant circuit’s quality factor is limited by its components’ own. This page is the same limit acting on power transfer rather than on selectivity: the components’ Q decides not only how sharp a match can be but how much of it is worth making. The efficiency a fixed Q costs measured the same component losses as a loss of efficiency in a network whose transformation is fixed; here the transformation is free, and the loss decides how far to take it.

Efficiency, at a half and above it #

One number does not move with the full cancellation. Its efficiency is exactly a half at every Q, as noted above: the load resistor is re-matched to the source’s resistance plus the capacitor’s, and the power divides equally between the load and everything else, whatever the capacitor wastes.

The partial cancellation is different, and the difference is the earlier essay’s second finding reappearing. It leaves some reactance uncancelled and matches the load resistor to the magnitude of what remains, which is the resistor-only rule applied to a smaller reactance, and the resistor-only rule’s efficiency is $|Z|/(|Z| + R)$ — above a half whenever any reactance is left. So the best lossy design is not only better at delivering power than full cancellation; it is also more efficient, since it leaves the load a little larger than the source’s resistance and refuses a little of the power rather than burning it.

A designer who wants efficiency rather than power still has no reason to fit a lossy cancelling element at all, since the resistor alone is more efficient than any cancellation. One who wants power should fit it only above the golden-ratio threshold and should then cancel a little less than all of the reactance, and gets some efficiency back for doing so. The two goals point in different directions at every quality factor, which is the conclusion the earlier essays reached about maximum power and efficiency, now with the components’ losses in it.

How the numbers were obtained #

Every delivered power is a solve of a netlist: a 10 V source, its 50 Ω resistance and an inductance giving the stated reactance at 1 kHz, a cancelling capacitor with a series resistance of its reactance over Q, and a load resistor, with the power read from the load resistor’s current. The best partial cancellation is a golden-section search over the capacitor’s reactance, with the load resistor at each step set to the magnitude of the remaining impedance, and the power at the optimum is checked against the solved netlist to six figures. The thresholds are compared with the search: just below the partial threshold the search cancels nothing, and just above it cancels something, at every source reactance checked.

What it leaves out #

Loss in the load’s own resistance-setting element. The load here is an ideal resistor of whatever value is best, which a real design reaches through a transformation that is itself lossy; the load’s side of the match has the same problem as the source’s.

A source whose reactance is capacitive, cancelled by an inductor. The algebra is identical with the roles exchanged, but inductors typically have lower Q than capacitors at the same frequency, so the thresholds bind more often on that side.

And frequency. Everything here is at the one frequency the cancellation is designed for; a lossy cancelling element’s reactance and loss both change with frequency, and so does the source’s reactance. How far a partly cancelled match holds across a band is a separate question.

Still open: the band, a source whose reactance is not known, and the lossy L-network #

The band, measured. The conjugate match is exact at one frequency. Sweeping frequency with the cancelling element fixed would say how wide the band is over which the conjugate match beats the resistor-only load, and whether the resistor-only answer is genuinely broadband or merely less sharply peaked.

A source whose reactance is not known exactly. A source whose reactance varies over a range — an antenna across a band, a transducer across temperature — cannot be conjugate-matched at every point. The best fixed load for a range, chosen against the worst case, is a different optimisation, and whether it is the conjugate of the range’s centre, the resistor-only answer, or something between would say what a designer should fit when the source is uncertain.

A lossy two-element network. An L-network transforms a resistance as well as cancelling a reactance, with two elements each carrying its own loss. Whether the best lossy L-network also stops short of the exact transformation, and by how much at a given Q, is the same calculation with one more element.