# Re: 2.4ghz helix

• Subject: Re: [amsat-bb] 2.4ghz helix
• From: "John P. Toscano" <tosca005@xxxxxxxxxx>
• Date: Fri, 24 Nov 2000 21:37:58 -0600

```Bob W7LRD wrote:

> With a Drake converter at the antenna with 60 feet of LMR-400 to the
> shack, what would be the  helix size want to be for  3D.  I will be
> building one soon.

I'd recommend buying a copy of the ARRL Antenna Book.  It's an
excellent publication, and has a chapter on helical antennas.

In general:

C_lambda = Circumference of the helical coil = 0.75 to 1.33
wavelengths, optimal is close to 1.00 wavelength.

D_lambda = Diameter of the helical coil = C_lambda / Pi

S_lambda = Axial length of each turn of the coil = 0.2126 to 0.2867
times C_lambda.  If the coil circumference is one wavelength,
then the coil length per turn should be 0.2126 to 0.2867 wavelengths.

Pitch angle = arctan(S_lambda / C_lambda) = 12 to 16 degrees.  If you
have followed the guidelines above for S_lambda and C_lambda, the
pitch angle should come out okay.

G = Diameter of the ground plane behind the helix if circular, or side
length if square = 0.8 to 1.1 wavelengths.

g = Height of first turn above the groundplane = 0.12 to 0.13
wavelengths.

Absolute minimum number of turns to guarantee axial mode radiation
pattern = 3.  Number of turns actually present = n.

As the antenna is lengthened (as n increases), the change in gain (in
db) is approximately proportional to the change in log(n).  The formula
listed in the ARRL antenna book is an old one, which has been
subsequently shown to be an overly optimistic one.  It predicts, for
example, the following relationship between n and gain, for an antenna
with C_lambda = 1 wavelength and S_lambda = 0.25 wavelengths.  To make
it a bit more concrete, I have also added the helix length in cm and
inches for 2400 MHz:

n        gain            helix length = n x S_lambda
---    ----------   ---------------------------------------
4      11.8 dbi    1 wavelength   =  12.5 cm =  4.9 inches
8      14.8 dbi    2 wavelengths  =  25.0 cm =  9.8 inches
12      19.6 dbi    3 wavelengths  =  37.5 cm = 14.8 inches
16      17.8 dbi    4 wavelengths  =  50.0 cm = 19.7 inches
24      19.6 dbi    6 wavelengths  =  75.0 cm = 29.5 inches

According to work by D. T. Emerson at the National Radio Astronomy
Observatory, a more realistic gain estimate would be as follows:

> An empirical expression for the maximum possible gain Gmax of the
> helical antenna as a function of its length L in wavelengths is:
>
> Gmax(dB) = 10.25 + 1.22 L - 0.0726 L^2
>
> This expression is only valid for lengths L between 2 and 7
> wavelengths.

Reference:  http://www.tuc.nrao/edu/~demerson/helixgain/helix.htm

Re-doing the calculations for the table above, we get:

n        gain           helix length = n x S_lambda
---    ---------    ----------------------------------------
4      11.4 dbi    1 wavelength   =  12.5 cm =  4.9 inches
8      12.4 dbi    2 wavelengths  =  25.0 cm =  9.8 inches
12      13.3 dbi    3 wavelengths  =  37.5 cm = 14.8 inches
16      14.0 dbi    4 wavelengths  =  50.0 cm = 19.7 inches
24      15.0 dbi    6 wavelengths  =  75.0 cm = 29.5 inches

The ARRL antenna book also talks about a better way to increase the
gain of the helix, if you have some room to spare.  This is to make
an array of four helices in a square on a common ground plane that
is 2.5 wavelengths on a side, with the helices spaced 1.5 wavelengths
apart.  This would give 6 db of additional gain over whatever the
single-helix version gives.  It also brings the impedance of the
antenna down from its initial 140 ohms to 35 ohms, which is easier
to match to a 50 ohm feed.  There is a bit of a trick to getting the
matching correct, which is best read from the ARRL book directly.

Happy building!

John (KBØZEV)

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