vhf rf amp to antenna... how to interface ?
Now I need to feed this to an efficient/ effective antenna.
Ideally I would like to use a vertical whip. But end-fed wires are - supposedly- simple. I could go with a dipole, or even a folded dipole.
My questions is - for the best possible efficiency/ ease of implementation,
(1) which should I select
(2) how do I interface the rf amp output to this antenna. Meaning what network/ circuit/ baluns/ etc etc should I design and use for best results.
I have tried researching this and implementing a few quickie options, but clearly this is a 'black-art', and i need help & guidance !
thanks guys
Depends on the purpose of your FM transmitter system.
If you need to broadcast and cover a territory around, then use a simple quarter-wave monopole.
To improve coverage, use a vertical array of dipoles.
To transmit a "beam" in a selected direction, use a Yagi .
You can buy commercial antennas for FM reception, they can work with your 1.4 Watt input.
Find more details in ARRL Antenna Book, any edition.
Purpose is to broadcast an FM signal over as large an area as possible. So as per your advise I will use a 1/4 monopole. Though I've read somewhere that 5/8th monopoles are 'better' ?
Next is the more vexing question - how do I interface/ connect this with the output of the amp ?
Again, follow my advice to read ARRL Antenna Book. You will find there also suitable matching devices to interface your 50 Ohm output to antenna input.
To add one advice: transmiting a high power FM requires a license in civilized nations. Without it you may face a legal punishment with quite bad result.
Ah yes... the legality response. Would you say that 1.4w is "high power" ?
In many countries the unlicensed limit is 1mW and even then only within certain frequency ranges. If you are thinking of 1.4W in the broadcast VHF band, you need a license. Don't despair, if you apply for one there is a good chance it will be granted but the authorities will tell you the maximum ERP and the frequency you are permited to use.
The difference between vertical antenna lengths, provided they have a suitable ground plane, is the matching impedance and radiation angle. Impedance matching is straightforward, use an appropriate LC network. The radiation angle is the measure of the angle between horizontal and the line of maximum radiation, to cover the largest area you probably want it to be as low as possible so it targets most power outwards rather than upwards. You have to think in 3 dimensions, what you need for best coverage is a radiation pattern like a flat horizontal disk so nothing gets wasted aiming at the sky or the ground. It's tricky to achieve but the Antenna book will give clues about the different patterns you get from each antenna configuration.
Brian.
Oh great. Now setting aside the legal issues please ?
What I read is that the impedance presented by a monopole is 36.5 + j21.25 over a perfect ground plane.
I have no perfect ground plane, and will probably mount the 1/4wave on a pole with a reasonable counterpoise. Connect this to the shack with a 50ohm coax.
Now, what I read is that the antenna impedance will appear down in the shack as a completely different complex impedance which is dependent on the coax length, amongst other things. Not a simple problem.
This further leads me to think that I would need some sort of fancy ATU along with an swr-meter to actually realise a 'good' transfer of power. i.e. swr < 2:1
But surely that's too complicated ! Isn't there a simpler solution ? Or am I missing something ?
And - please - no more recommendations to read ARRL handbooks.
If someone complains, you will be for the high jump my friend.
Not simple at all. Too many variables and unknowns.
Yes indeed.
It only matters if you actually want to radiate some of that 1.4 watts.
The standard solution for all antenna matching problems is a variable CLC pi matching network ("matchbox") combined with a VSWR meter.
I understand you already know that this stuff is discussed in detail in the ARRL handbook and other amateur radio publications.
Yes of course. Thanks FvM ...as alway.
But here's the thing - my RF amp -and many homebrews like it- are designed to deliver a power level based on the impedance "presented to it". From the well publicised equation P = (Vc- Ve)^2 / 2.R
The concept of "output impedance" is hard to understand or fit into this context.
Next issue - a basic SWR meter is designed for a nominal source & load of 50 ohm. There are several elements in a typical swr meter which are based on this. HOWEVER, if my source loading (NOT source impedance!) is based on the above equation, and my final load (antenna + coax) is a totally unknown impedance which we have agreed to above, then what is this swr meter measuring ?!? Will the swr meter give me a valid useful reading despite this ?
And I have found the ARRL Antenna books as well as the ARRL Handbooks as confusing today as they were 35 years ago, so please do not refer me to these again. I do not subscribe to any pubs. I need YOUR advice and experience for my specific questions.
thanks folks !
Yeah thanks Tony, not helpful at all.
Regarding high jumps, i live elsewhere.
cheers !
This would refer to an ideal voltage source with zero ohms output impedance, hard to believe for a RF amplifier. You can make amplifier with relative low output impedance (a few ohms) up to several 100 MHz, but the effort is much higher than a for simple output stage with real impedance.
It's also required for a transmitter output stage to include a harmonic filter to comply at least halfways with harmonic emission requirements. The filter can be only designed based on a specific output impedance (might be zero, at least theretically).
A transmitter output impedance different from 50 ohm (or whatever your cable impedance is) will be transformed by the transmission line into a quite different value at the other end, depending on cable length and frequency. Another strong reason to have a defined real transmitter output impedance.
A matching network can nevertheless connect sources and loads of arbitrary impedances, and a SWR meter can measure forward and reflected power at a specific cable point. In case of your imagined zero ohm source, there would be no reflected power because the transmitter is bouncing all energy back to the load. But as said, this doesn't sound real.
Nevertheless, maximum power transfer into a load will always occur when the source and load impedances are conjugate and impedance matched.
Yes it will.
If source and load are both 50 ohms there will be zero indication of any reflected power. The ideal and almost impossible to obtain outcome.
As you say, where the situation is far from perfect "some" reflected energy will be indicated.
All a reflected power indication is telling you is that there is a problem, and a somewhat vague indication of the magnitude of the problem.
So you use the reflected power indication to adjust experimentally by trial and error for a minimum reflected power indication.
Your best bet will be to obtain an swr meter and experiment with both the antenna itself and an adjustable tuning unit between transmitter and feedline.
As there are so many variables it can be a very frustrating business.
But unless you have some very sophisticated measuring equipment available, its really the only way.
I think the classic method is two matching networks - the first between your transmitter output and the feeder cable, so you have a known impedance to match to and can use the feeder efficiently and a second network between the feeder and the antenna. Use an SWR meter at the antenna end to match for highest forwards/backwards ratio, if the cable is matched you know the input to the SWR bridge is 50 Ohms or whatever so you can adjust the match to the unknown antenna with the LC network.
If you want to do it experimentally, terminate the feeder with a resistor (50 Ohm non-inductive resistor for 50 Ohm cable etc.) and use the SWR bridge to match the transmitter to the cable then remove the resistor and fit the SWR bridge to the antenna end instead.
Brian.