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*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) ---- Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA. To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org

**References**:**2.4ghz helix***From:*Bob W7LRD

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