23, 13, WLAN Helix antenna design

How to design and build helix antennas is described here. Also some of the theory behind it. There are tree radiation modes of helical antennas. The most used mode is the axial mode, it provides maximum radiation along the helix axis. For axial mode, the circumference of one helix-turn is in the order of one wavelength. In normal mode, radiation goes broadside to the helix axis. It occurs when helix diameter is small in respect to the wavelength. If wavelength is small compared to the helix diameter, in conical mode, radiation will go into multiple lobes.

normal mode

axial mode

conical mode

Opening angle and gain (over dipole) versus number of turns. It is ideal gain, in real life it`s less.

calculation of helix antenna dimensions is pretty easy. It can be done with an Excel spreadsheet, you can download it here. All neccessary formulas are in the screenshot of this spreadsheet above.

Feedpoint impedance of air wound helix antennas in axial mode is 137 Ohm (almost purely resistive). Matching is required if 50 Ohm coax cable is to be connected. There are two common ways to do this matching. Either with a coax section or a stripline above a ground plane, both a quarter wavelenght long and of proper resistance. The "proper resistance" is calculated as follows:

As calculated above, a quaterwave long matching "device" having an impedance of 83 Ohm is required.

To obtain the needed impedance of 83 Ohm, for round coaxial sections the ratio of outher/inner conductor must be about 4. If the outher conductor is of square shape, the factor lowers to 3.4.

More easy to construct mecanically is a quarterwave long matching stripline section over a groundplane. In practise, the first quater turn (from the feed point) of the helix is a copper strip of a certain width and distance from the groundplane (the helix reflector). See the picture above for details.

I have found different formulas to calculate impedances of striplines over a ground plane. In the picture above, the tree most "close" are shown. I finally used formula number 3. A quarter wavelength long (58 mm) piece of copper sheet, 0.5 mm thick, 12.5 mm wide and at 5 mm distance from the reflector will do the job for a 1296 MHz helix. For 13 cm, the piece should be about 3.2 mm.

Practical Examples

 VSWR 23 cm helix

1296 MHz version. The SWR is below 1:2 over a bandwidth of 700 MHz.

 VSWR 13 cm helix

2350 MHz version. The SWR is below 1:2 over a bandwidth of greater then 1000 MHz.