# Coax Collinear Antenna

Discussion in 'General Electronics Discussion' started by flippineck, Mar 15, 2015.

1. ### flippineck

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Sep 8, 2013

There seem to be many variations of designs, many plans on the net for this type of vertical omnidirectional radio antenna.

They all seem to share the common features, most notably a chain of elements each one half wavelength long.

Loads of plans on the net but very little explanation of how it works.

There seems to be big gulf between "just follow the plans, don't concern yourself with theory" on one hand and the odd site that does go into the theory, but soo deeply and mathematically that it's impenetrable to the average guy (especially the average guy who is number-blind and failed maths, like me)

Can anyone on here explain the workings in terms of voltages, currents, wave interactions, standing waves etc without going too deeply mathematical? Or know of any net pages that do a good job of it?

One thing I struggle with is when discussions of impedance start creeping in.

I want to try building one for 1800MHz and one for 800MHz.

I already built this variation - http://martybugs.net/wireless/collinear.cgi - adapted for 1800MHz and it does seem to work well, I believe the full coax version might be even better though?

2. ### duke37

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Jan 9, 2011
To get an antenna gain all the parts of the antenna should pass current in phase so that when looked at from a distance, the radiation from the whole length adds up.

Stacking a half wavelength on top of another gives a null since the currents in each part will be in antiphase.

The coax antenna screens the opposite current and the outer does the radiating in phase.
The loop connected elements uses a delay line to alter the phase. I would think this would need to be made very accurately and two or three sections would be as much as would be possible without too many cumulative errors.

You may be able to find an antenna modelling program but things are complicated since every section will interact with every other section.

3. ### flippineck

320
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Sep 8, 2013
it's a case of going from push - pull - push - pull in a plain wire, to swapping round every half cycle so the continuous length presented to the aether goes push - push - push - push?

is this the main part of the mechanics of the antenna? I can't help wondering what the ends do. sites often talk about a quarter-wave whip on the top, and a quarter-wave sleeve on the bottom. I don't understand where these fit in.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Yeah, pretty much.

The "things on the end" are probably attempts to ensure a correct impedance.

Perhaps @davenn can comment in far more detail since he works at frequencies so high you can almost see the radiation pattern with your unaided eyes! :-D

davenn likes this.
5. ### davennModerator

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Sep 5, 2009
haha, Steve

The one I have build and successfully for the 1300MHz ( 1240 - 1300) ham band the top end was shorted rather than a terminating resistor

here is my one on my www site .....
http://www.sydneystormcity.com/microwave.htm

scroll 2/3 way down the page ( or have a good read as you mosey down )

Dave

Last edited: Mar 16, 2015
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6. ### flippineck

320
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Sep 8, 2013
Thanks chaps. Dave, see the copper tube balun bit, for matching the feedline - how do you calculate the dimensions for that bit?

Nice yagis on your page btw.. I tried some yagis here on my application (mobile network voice / data) but found them unsuitable because the transmitters I'm trying to talk to are at all different points of the compass & the equipment keeps changing from one to another.

7. ### davennModerator

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Sep 5, 2009
its a 1/4 wavelength for this one it is 58mm

300/1296 = 0.23 ( 23cm) / 4 = 0.0578 ( 5.8cm) ( 58mm)

8. ### flippineck

320
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Sep 8, 2013
Slightly off-topic relative to the current thread, but I had something of an epiphany in my understanding of the workings of antennas and radio waves last night, whilst watching a 1940's Canadian Air Force training video.

I could never quite 'get' how a radio wave could travel through anything including a perfect dark vacuum. Mentally for years I have been searching for some kind of 'aether' to 'carry' the wave, even if it isn't matter of some flavour.

Then this guy comes along and says, well, the radio wave is two things - a fluctuating magnetic field and a fluctuating electric field, both readily realisable/feelable in the common world in the form of bar magnets and static electricity etc.

Yes, I know that.. I thought.. virtually all explanations include that.. then he says...

The important thing being, a fluctuating magnetic field will cause a fluctuating electric field, and vice versa.. and no actual conductor need be present!

And that's how the wave travels through empty space.. magnetic causes electric causes magnetic causes electric causes magnetic... and so on

Wow, blinding flash of revelation. I get it now. It's only taken me half a century.

(*steve*), davenn and Supercap2F like this.
9. ### davennModerator

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Sep 5, 2009
Woohoooo Those moments are awesome huh .... I remember a looooong time ago when I finally realised how to use a transistor as a switch
it also was a stunning revelation and like your one, for me took some 15 years hahaha
No-one had ever told me, I hadn't had any transistor theory tuition, google didn't exist and hell, the internet was only just taking shape

yeah, I saw that comment the other day and was going to correct it but let it fly and kept to the topic in hand

The next revelation that may come for you is ( if it hasn't already) when you have an audio signal in a cable etc. That audio signal is also emitting an RF signal from the wire at the same frequency

When an AC signal ( any AC signal) is applied to the circuit, the electrons in the wire etc oscillate back and forward at that frequency thus producing an oscillating electric field at that freq, thus an oscillating magnetic field etc

The EM field, as well as being radiated outwards, also moves along the outside of a conductor. The conductor is acting as a waveguide. And that EM field is carrying a lot of the energy in the circuit. It travels at near the speed of light, c*, around 95 - 98% (0.95 - 0.98) for a bare conductor. For an insulated conductor the EM propagation speed drops considerably down to around 60% (0.6) to 90% depending mainly on the type of insulation and its thickness
This is why you see transmission lines ... coax cable etc have a stated Velocity Factor

* c (lower case) denotes speed of light in a vacuum

Now going out on a limb here, you do realise light from IR through visible to UV and then beyond to X-rays etc is also EM radiation ?
If you knew that, then you need to ask yourself .... If light is EM radiation and it can travel through the vacuum of space without a medium,
then radio waves which are also EM must be able to do the same thing

cheers
Dave

10. ### davennModerator

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Sep 5, 2009
This is very important and needs to be taken into consideration when designing RF striplines on a PCB, tuned lengths of transmission lines for filters and BALUNS etc. Because you cannot use the freespace speed of light, c to work out tuned lengths of conductors

you take what would be the freespace length and multiply it by the velocity factor Vf eg.....

lets work out a 1/4 wavelength at 100MHz
( when all calc's are done in MHz, we can use 300 as c and all measurement results will be in metres) .....

therefore
300 / 100 = 3 metres wavelength
3 / 4 = 0.75 m ( 75 cm or 750 mm)

therefore
0.75 is the free space 1/4 wavelength for 100MHz
now to get the electrical 1/4 wavelength .... lets assume the coax has a 0.66 Vf ( that's typical for a 1/4 inch diameter, 50 Ω coax, say RG58)

therefore
0.75 x 0.66 = 0.495 m ( 49.5cm or 495mm) is the electrical 1/4 wavelength at 100 MHz

cheers
Dave

Last edited: Mar 18, 2015
11. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Back to the main subject, I think I have an article at home which deals with the theory behind this antenna type. I'll try to dig it up.

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13. ### davennModerator

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Sep 5, 2009
I assume you have a subscription to SC, Steve ?

only the first 2 pages are visible without a subscription

14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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That's correct, I have a hard copy subscription.