Matching the helix to 50 ohms

One technique used for matching a helical antenna (typically 120-140 ohms) to a feed line (50 or 75 ohms) is to use the bottom of the first turn as a tapered transmission line from the feed at the periphery. One could spend some time analyzing this in horrid detail (none of which would be replicated with usual construction tolerances), or you can do it empirically (which is what most folks do). Some analysis is useful to get a feel for what to expect though. There is probably a shortest possible length for the tapered line, as well as an optimal shape (exponential spacing variation, as it happens), but given the usual construction tolerances, we just bend it until it works. If someone has a pointer to a more complete theoretical treatment, I'd love to reference it. (There are some papers in IEEE Transactions on this).

One way to think of it is as a single wire above a ground plane. Two formulae are commonly applied here (one being an approximation of the other):

 Z = 138 * log10(4h/d) where: d is the diameter h is the height of the conductor above the ground plane.

This is essentially a restatement of the standard two wire transmission line equation, where the ground plane provides an image of the conductor:

 Z =120 * cosh-1(D/d) = 276 log10(2D/d)   where: D is distance between conductors d is the diameter of the conductors

For the "strip conductor" case (i.e. using copper foil tape or shim stock), modeling as a coplanar transmission line would be a good bet. It gets even more complicated when one considers that the conductor (in the strip case, especially) is often mounted on some sort of dielectric.

### Some calculations for typical sizes of interest

Assuming we want to match 140 ohms to 50, the important thing is the h/d ratio.

• 140 ohms: h/d = 2.5
• 50 ohms: h/d = 0.575

note that in the 50 ohm case, the actual clearance between the conductor and the ground plane is pretty small:
clearance = h-d/2
h = 0.575 * d
so clearance = 0.575*d - 0.5*d = 0.075 * d

For an antenna using AWG #10 wire for the helix conductor (approximately 0.1" in diameter), the clearance is .0075 inches (around the thickness of a couple sheets of paper!).

I didn't find a handy reference for using a flat conductor perpendicular to the ground plane, but a reasonable assumption might be to consider it a round conductor where the diameter is half the width of the sheet.

### Just how good a match do you need, anyway

In the case where the receiver or transmitter is close to the antenna (i.e. not much transmission line), one might not need a very good match, since there is little power being lost in line resistance. The input (for receivers) and output (for transmitters) impedances of many radios isn't particularly well characterized either. All you can really count on is that the numbers in the spec sheet were obtained with 50 ohm test equipment. For all you know, changing the match might actually make it work better. Of course, it might also make a harmonic problem worse (i.e. the manufacturer was depending on a good match at the desired, and a bad match (or no worse) at the harmonics).

The basic approach I use is to use a matching technique that is fairly broadband, verify that the impedance is in the right general area and doesn't have any huge bumps or dips (indicative of some other problems), and just try it out.

radio/helixmatch.htm - 3 December 2001 - Jim Lux
900 MHz helical antennas