Part 7 - Designing Vertical Aerials

Before we begin - there is an Excel spreadsheet you can download that has the formulas built in but it's hidden at the bottom of the page, please feel free to download it and play around and look at how the efficiency changes, now onto the boring stuff.....

Why use one?

A half-wave dipole can be quite large to fit into a small garden. It needs to be mounted high-up to be effective and may not be radiating in the places you need at the time you need it - rotating it is not always an option.

If we use a vertical aerial then it will radiate in all directions equally; has a small footprint; can be mounted on the ground (or high up) and it has been shown to have a low angle of radiation - just what we need. So why don't we all use them? Well, to be really effective they have to be a 1/4 of a wavelength long and on the lower bands that can be a challenge.

I'm sure you have seen wonderous adverts about vertical aerials and how they can be used on all the bands with low SWR. Well this page is going to show you how to make one of those wonderous verticals with a low SWR on all bands and all for £10 - It will be 5.2m tall and require almost no radials!

Sound too good to be true - read on, battle through the maths (I've tried to make it as simple but as full as possible) and see if you want to make it with a low SWR on all bands? Maybe you would prefer my higher SWR version for the same price - not tempted? Read on and you might be!

What is a vertical aerial?

What Let's start out with the idea that the vertical Antenna is a dipole turned on its side.

Then let's remove the bottom wire and replace it with the ground. Now we have 'a vertical' , it is important to remember (and you may have picked up on it) that replacing the dipole's other wire with ground is not the same as getting rid of the other half of the dipole!

We still have a dipole, it is just that one half is the wire and the other half is the ground. Sound a bit silly? Agreed, ground doesn’t conduct like a wire but we are in a little bit of luck here as it does conduct, just not fantastically well but we can just assume for the moment that it does conduct.

It's Capacitance

Any conductors in close proximity to each other will have a capacitance between them (if you're unsure about capacitance go here and read the advice, then come back). We have two conductors, the wire and the ground therefore we have a capacitance between them. The value is dependant on how the ground and surrounding objects are made up - so it is dependant on your particular location. All that follows is based on the assumption that the capacitance is about 3.8pF for every foot of aerial wire (for a base loaded vertical – more options later).

So if we know the length of the aerial we can calculate its total capacitance. I'm making an aerial that is 5.2m long (that is the longest spare length of wire I have at the moment!). 5.2 m is 17.1 feet (use Google and type convert 5.2m to feet to get the answer).

17.1 feet x 3.8 pF/foot = 64 pF

Its Reactance

Where do we go from here? Well, If we have a capacitor in an RF circuit it gives us a Capacitive Reactance (X_C) which we can work out using the formula.

Putting our numbers in gives us:

Which give us the answer L=8.1uH.

WOW! I’ve we put an 8.1uH inductor in the aerial then its XL will cancel the XC of the aerial and it will be resonant, I can’t see any drawbacks at the moment so let’s do it!

Uuummm – How do I make an 8.1uH inductor?

Building the inductor - what size and how many turns?

Well there are several formulas for this – and it can get very messy so let’s choose a good one for approximating it. This is for a single layer coil with an air core

Where: T = the number of turns, L = inductance, r = radius in inches, l = length in inches. OK there is a lot in there so we need to look at what we have got and see if we can make a stab at the rest!

We have L, nothing else! So we must take a stab at r and l. Lots of practical coils are wound on a piece of piping and I have some spare PVC drainage pipe that has a radius of 25 mm (0.98 inches) and I reckon a length of 150 mm (5.9 inches) might do the trick (I’ll provide and Excel Spreadsheet so you can experiment later).

Notice I’ve converted back to mm and the answer will be in mm. Use inches and the answer will be in inches.

The answer is 4,714 mm, so 5m of wire will do me fine. Now I can make my coil – we’ll get to that a bit later, let’s see what we have achieved so far:

1. We got a 5.2 m piece of wire we wanted to use on 7 MHz.

2. The wire has a capacitance per foot, we have a guideline for so we calculated the total capacitance.

3. We know capacitance has a reactance so we worked out what that would be at 7MHz.

4. To make the aerial resonant we need to cancel the XC and we can do that by adding in an inductive reactance XL.

5. We worked out the inductance required to give XC = XL.

6. Once we had done this we calculated the number of turns on a size of plastic pipe.

7. We also worked out the length of wire required.

All this is based on the assumption that the capacitance is 3.8pF per foot so if we change that we change everything. This is where a little experimentation comes in because it

will not be 3.8 pF/ foot!

What else can we work out about our aerial?

We could walk away and think that this is now a resonant aerial so it will work perfectly – not the case! We can now work out the aerial’s efficiency and it’s not going to be 100%.

Efficiency is given to us by the equation:

The Total System Resistance is the Radiation Resistance + Ground Resistance + Coil Resistance. So we can re-write this as:

So if we want a high efficiency we need a high Radiation Resistance and a low Total system resistance.

To find the answer we need to find the Radiation Resistance; Ground Resistance and Coil resistance.

Coil Resistance

The Coil Resistance is easy, I’ll measure it with my meter – it’s 1 Ω.

Radiation Resistance

The radiation resistance is given by the formula (thank you to AD5X - Phil Silas for the help on this one, by the way Phil does not not endorse this page, just this equation!).

Where l is the length of the wire and λ is the wavelength we want to operate on, so we had better work that out first.

We should all know that c is the speed of light and F is the Frequency. Putting those in we get…

= 42.85 m

Revisiting the equation for Radiation resistance, we can put the 42.8m in:

= 5.8 Ω

Ground Resistance

Ground Resistance is the only part of the formula left and we can’t measure it (well apparently you can) so we are going to rely on some lovely people who have done much more experimentation and maths than we have and let is know it is usually between 7 -15 Ω. My ground is average (whatever that is) so let’s go for 10 Ω. [This is the only assumption I've made - I would like more information on it].

Putting the numbers back into the Efficiency formula:

or 32%

For every 100 Watts in I’ll be radiating 32 Watts. The rest of the power is lost in the coil and ground.

So out of 100 Watts, 32 Watts is ‘used’ in the Radiation Resistance. This is really what the Radiation Resistance is - the equivelent resistance that would didssipate the power transmitted - don't quote it as it is more complicated!

So we really want to get the radiation resistance as high as possible. We can only do that by making the length of the aerial longer. Look at the formula for radiation resistance

it’s value is dependant on the bit and that can only go larger if the top number increases. So my 5.2m aerial would be much better if it was 10 m long. In fact it would be about 65% efficient – not bad! We could also reduce the Ground Resistance and Coil Resistance to increase the efficiency. Well my coil is made from copper so that is pretty good and not going to improve it much.

That leaves us with the Ground Resistance. Can we reduce it? Well we could give it a helping hand! This statement is so obvious but I’ve got to make it - we could make it much more conductive by adding a conductor. There I said it – but what conductor can we add to the ground. Water works – it really does. The other option would be to add some metal, copper is the obvious choice.

So now if we add lots of copper to the soil (and connect it to the ground connection of our aerial) it will Reduce our Ground Resistance and the efficiency increases. Yes it does and it’s a great idea to do it, we know them as radials (not to be confused with a raised ground plane - this page from the University of Hawaii Ham club explains more) .

But wait – aren’t you just adding back in the other half of the dipole you took away when we created this vertical? Not really, we are assisting the bigger half of the aerial which is now the ground. There are loads of resources and theories about what are the best radials, how many, how long, how deep to bury them, if it’s better higher-up etc. and although loads of people have the right answer they are mostly different!

There is a common theme – more is better, longer is better. I would go as far as to say that just below the ground's surface is also better and that they should not be insulated. It seems to be often stated that lengths over 1/3 of a wavelength are a waste - much better to use the copper for more shorter radials. '

Below is a graph of the Antenna Efficiency for different Ground Resistances, you can't argue it - the better the lower the Ground the better the efficiency.

Even more things we can do...

So now we have a resonant antenna and we know that it is 30% efficient - what is this in our old favourite dB. Well the formula is:

That certainly isn’t 1:1 and an ATU is required to match the feedline to the transmitter, or an impedence transformer at the base of the aerial.

How does it perform on all the bands?

For my 5.2m wire the following table shows me the Inductance, Efficiency and SWR for all he bands I might use it for:

In theory I could operate this aerial on 14-28 MHz with an SWR of less than 3:1 by changing the tap on the inductor and if we pushed it we could include 40 m as well.

Now I’ve created all these calculations in a handy spreadsheet that you can freely download at the bottom of this page. More to do though....

Making it Sell...

But I want to sell this aerial with great claims about the SWR (remember it only costs £10).

What happens if we try to reduce the SWR to 1:1?

Well we would need to increase the 17.8 Ω closer to 50Ω. The 17.8Ω is the total system resistance which equals the Ground, Coil and Radiation resistances added up.

If we increase any of these then we also affect the efficiency formula (they are all in that formula) and the larger we make any of them the less efficient the aerial becomes. Let’s try and see:

We’ll increase the Ground Resistance to make the Total System Resistance = 50 Ω and see what the efficiency is now the SWR is 1:1

or 12%.

Is that a surprise? Making the SWR 1:1 decreases the efficiency and in this example quite drastically (it was 32%).

A low SWR does not mean an efficient aerial! Just a very efficient protector for your PA.

So now I am going to replace my lovely copper coil with a metal ‘slinky’ spring which has a resistance of 10Ω and I’m going to remove the radials so I have a higher ground resistance of 20Ω. Let’s look at the same table with those little changes...

Now I can rightly claim to have less than 2:1 SWR across all bands from 80-10m! But look at the efficiency drop – staggeringly worse!

So it is possible to produce a base loaded, tapped inductor, all-bander with an SWR of less than 2:1 on all bands. And you can sell it with the amazing strap-line that "No radials are required" - damn right - they would make it more efficient and screw up the SWR, think of the returned aerials beacause they are too efficient! You can also add the strap-line "perfect for internal ATUs". It just gets better. [note: for all those who don't recognise irony this paragraph is an example]

If you read any reviews of all-banders and see the “great contacts” and “quality of my audio” - don’t be fooled, we know that they were really only putting out about 43% of whatever they were using.

Before I get loads of mails – You could put the coil half way up the aerial to increase efficiency and there is a calculator for that too in the spreadsheet. Changing bands will be more difficult, as is construction but the increases in efficiency are worth it if you can do it.

Parts

5 m "Twin and Earth" mains cable with 1.5 mm2- £6. (opened up mine was 5.2m - free copper!). The Earth makes the coil and one of the other conductors is the aerial. The other will become one of the many a radials.

6m fishing pole - £4 (was 10m but it cracked!)

Pipe – found it.

Total - £10

By the way the formulas work for all frequencies – try 144Mhz and a 0.4m aerial - OK at VHF the ground does not play a part but humour me!

Also – this is not the end of the story, this is just a very simple vertical aerial with a base loading coil. As with all things aerial related – there are hundreds of different designs that will do lots of different things like band changing using traps or capacitor/inductor combinations, stubs and a whole host of other things. I wish I could afford some of them and the XYL wouldn’t mind me putting them up, but in the end the longer the better and if it says radials aren't required, don't buy and you know why!