Feeders, SWR and Matching

The feeder is the least glamorous part of a station and the part that quietly ruins the most contacts. It is also where the examiner puts his arithmetic: the SWR questions are the only aerial questions that need a calculation, and they are easy marks once you know which of two sums to do. Beyond the exam, this lesson explains the three things that go wrong on real Indian rooftops — cheap thin coax on 70 cm, an ATU used as though it were a cure, and a dipole fed without a balun.

What a feeder has to do

A feeder carries radio-frequency energy between transmitter and aerial, and it has three duties: deliver as much of the power as possible; radiate nothing itself, since anything it radiates leaves in an uncontrolled direction; and pick up nothing on the way in, because a feeder acting as a receiving aerial brings the household's electrical noise straight to your receiver.

The three types you will meet

Coaxial cable is an inner conductor, a plastic dielectric, a braided outer conductor and a jacket. The field is entirely contained between inner and braid, so it is unbalanced and can be clipped to a wall or run through a window frame without changing its behaviour. Transmitting cable is 50 Ω; television and receiving cable is 75 Ω.

Open-wire or ladder line is two parallel wires held apart by spacers with air as the dielectric: balanced, very low loss, usually about 600 Ω, and — because its field extends outside the conductors — needing stand-off insulators and clearance from walls and metal. Twin lead is the flat 300 Ω moulded version: balanced, cheap, the natural feeder for the folded dipole, badly degraded when wet or dirty.

The trade is straightforward: open wire wins on loss and loses on practicality, coax the other way round. On a terrace with a parapet, a water tank, a washing line and a monsoon, most people choose coax and accept the cost — defensible on HF, expensive at UHF.

Characteristic impedance

A feeder has inductance along its conductors and capacitance between them, both spread evenly along its length. The ratio of the two decides how much voltage accompanies a given current in a wave travelling along the line, and that ratio is the characteristic impedance:

L and C are fixed by the physical construction alone — the diameter of the conductors, the spacing between them and the dielectric between them. So Zo does not depend on length (cut a 50 Ω cable in half and you have two 50 Ω cables), it does not depend on frequency (a 50 Ω cable is 50 Ω on 80 m and on 70 cm), and it depends neither on the power going through it nor on what is connected at the far end. The exam tests all three.

Velocity factor

A wave travels more slowly inside a cable than in free space, because the dielectric holds part of the field. That ratio is the velocity factor: about 0.66 for solid polythene coax, 0.8 for foam, 0.97 for open wire. It does not matter when the cable is simply carrying power, and it matters enormously the moment you cut a piece to a given electrical length, because an electrical quarter wave of cable is physically shorter than a quarter wave in air.

Worked example 1 — a quarter-wave section for 20 m

You need a quarter-wave matching section at 14.175 MHz, made from ordinary solid-dielectric coax with a velocity factor of 0.66.

75 ÷ 14.175 = 5.29 m in free space.

5.29 × 0.66 = 3.49 m of actual cable.

Cut it at 5.29 m, ignoring the velocity factor, and it is half as long again as it should be and will not match anything.

Loss, and why it rises with frequency

Two mechanisms eat power in a feeder and both worsen with frequency: skin effect crowds the current into an ever thinner surface layer, raising the conductors' effective resistance, and the dielectric absorbs more as the field reverses more often. Loss is quoted in decibels per hundred metres, and 3 dB means half your power is gone before it reaches the aerial.

This is what makes thin cable a false economy at VHF and UHF. A 30 metre run of RG-58 costs a little over a decibel on 40 m, barely noticeable; the same run at 435 MHz costs of the order of ten decibels. Ten watts leaving the transceiver arrives at the aerial as about one watt, and received signals are cut by the same amount coming back. On 70 cm use thick, low-loss cable and keep the run short.

Matched and mismatched lines

Connect a feeder to a load whose impedance equals its Zo — 50 Ω coax into a 50 Ω aerial — and the load absorbs everything that arrives. The line is matched: one wave travelling one way, the same voltage measured anywhere along it, nothing coming back.

Connect it to anything else and the load cannot absorb it all. The remainder has nowhere to go but back up the line towards the transmitter. That returning wave meets the outgoing wave, and where the two are in step they add and where they are out of step they subtract. Because the pattern is fixed in place by the position of the load, it does not travel: it is a standing wave.

SWR, two ways

Standing wave ratio — strictly VSWR, voltage standing wave ratio — is defined from that picture:

It is always one or greater, and exactly one when the line is matched, because then the maximum and the minimum are the same number. Nobody measures those two voltages directly; there are two practical routes to the same figure.

From the impedances

For a purely resistive load, divide the larger of the two impedances by the smaller, so that the answer comes out greater than one:

Worked example 2 — from two impedances

A 50 Ω transmission line feeds a 300 Ω load.

300 ÷ 50 = 6, so the SWR is 6:1.

Worked example 3 — the mismatch the other way

The same 50 Ω line feeds a 25 Ω load. Now the line impedance is the larger one:

50 ÷ 25 = 2, so the SWR is 2:1. A 100 Ω load on the same line also gives 2:1 — SWR alone cannot tell you which way the mismatch lies.

From forward and reflected power

A directional wattmeter in the line reads forward power and reflected power. Power goes as the square of voltage, so take the square root of the power ratio first to get the reflection coefficient, Γ:

Worked example 4 — from a wattmeter reading

The meter reads 100 W forward and 4 W reflected.

Γ = √(4 ÷ 100) = √0.04 = 0.2

SWR = (1 + 0.2) ÷ (1 − 0.2) = 1.2 ÷ 0.8 = 1.5:1

Four per cent of the power came back, which is a fifth of the voltage. Dividing 100 by 4 and answering 25:1 is the mistake the paper is fishing for.

Reflection coefficient and return loss

Γ is the fraction of the voltage reflected, running from 0 for a perfect match to 1 for total reflection at an open or short circuit. The same information in decibels is the return loss — how far below the forward power the reflected power is:

For the example above, 20 log₁₀ (1 ÷ 0.2) = 20 × 0.699 = 14 dB. A big return loss is good news — very little came back. An SWR of 1.5:1, a Γ of 0.2 and a return loss of 14 dB are three ways of saying the same thing.

Work both routes yourself

The lab below takes either pair of numbers — two impedances, or a wattmeter's forward and reflected readings — and shows the working, the reflection coefficient, the return loss and the standing wave that results. Do worked examples 2 to 4 on paper first, then enter them here and watch the trace flatten as the mismatch improves.

What a high SWR actually costs

Two real penalties, and one myth.

• Extra loss in the feeder. Reflected power does not vanish. It runs back up the line, is partly re-reflected at the transmitter end and travels the cable again, paying the cable's normal loss on every trip. On low-loss cable at HF the extra is trivial; on thin cable at UHF, where you were already losing most of it, a high SWR turns a bad situation into a hopeless one.
• Transmitter foldback. A solid-state PA sees higher voltage and current peaks on a mismatched line, so the rig contains a protection circuit that senses reflected power and reduces drive. Above about 2:1 your output starts dropping. The rig is not faulty; it is defending itself.

Matching devices

The ATU, and what it really does

An aerial tuning unit is a network of inductance and capacitance at the transmitter end of the feeder. Adjust it and the transmitter looks into a comfortable 50 Ω and delivers full power without folding back. What it has not done matters just as much:

• It has not retuned the aerial, which is still off resonance and still the wrong impedance.
• It has not removed the standing wave between itself and the aerial. Everything beyond the ATU is exactly as it was, extra feeder loss included, and an SWR meter between rig and ATU now reads a comforting 1:1 that is telling you about the ATU, not about the roof.

An ATU is a genuinely useful device — it is what lets one wire work on five bands — but it is a matching network, not a repair. Matching at the aerial itself always beats matching in the shack.

Gamma and delta matches

Both fix the mismatch where it happens, by tapping the element where its impedance is the value you want. A gamma match takes the coax inner through a series capacitor to a point a little out from the centre, braid on the centre of the element: it suits unbalanced coax feeding a metal-tube element such as a Yagi's, and needs no balun and no break in the element. A delta match fans the two sides of a balanced feeder out to two points either side of centre.

The quarter-wave transformer

A quarter-wave length of line of the right impedance, inserted between two different impedances, matches them. Its Zo must be the geometric mean of the two:

Worked example 5 — the matching section

A 300 Ω line is to feed a 75 Ω aerial.

Zo = √(75 × 300) = √22500 = 150 Ω

So a quarter wavelength of 150 Ω line joins them, cut to physical length using its own velocity factor. Note the trap: 187.5 Ω is the average of 75 and 300 and is the wrong answer — it must be the geometric mean.

The balun

A balanced aerial or feeder is symmetrical about earth, with equal and opposite currents on its two conductors. A dipole is balanced: two identical legs either side of the feedpoint. Coax is unbalanced — an inner conductor and an outer screen that is grounded, and the two are not interchangeable.

Join them directly and the current returning from one leg has a choice: back up the inside of the braid, as intended, or down the outside of the braid, which is a separate conductor as far as RF is concerned. Some of it takes the outside path, and then:

• the feeder carries current and therefore radiates, distorting the dipole's pattern and filling in its nulls;
• the dipole is fed unequally, so it is no longer the aerial you cut;
• RF comes back down the outside of the coax into the shack, where it produces the familiar symptoms — a hot microphone, a bite off the case, an SWR reading that changes when you touch the rig, and interference to the neighbours' television.

A balun — balanced to unbalanced — prevents that. The simplest useful kind is a choke or current balun: several turns of the coax itself wound into a coil, or threaded through ferrite rings, at the feedpoint. It presents a high impedance to any current trying to flow on the outside of the braid while doing nothing to the wanted current inside. A dipole needs a 1:1; a 4:1 is used where the impedance must be stepped too, as with a folded dipole.

Keep the feeder at right angles

A feeder running back close to and parallel with a dipole's own wire is coupled to it: it picks up current, becomes part of the aerial and unbalances it — the same fault as having no balun, by a different route. Bring the coax away from the feedpoint at right angles to the wire, and keep it that way for at least a quarter wavelength; on an inverted V that means straight down the mast. It costs nothing and it is the difference between the pattern you designed and the one you happen to get.