1/4 vs 1/2 wavelength antenna

Hi I am building an rf transmitter for a short range data link at 433MHZ and am almost done, but I would like to understand better exactly what I am seeing with regard to antenna performance.

All technical notes I have read recommend a 1/4 wave whip over ground plane as offering the best performance, statements like: "Best range is achieved with either a straight piece of wire, rod or PCB track @

1/4 wavelength over a ground plane", I understand many factors effect performance however I have found that a "bent" 1/2 wavelength length of wire offers better performance.

If I use a 1/4 wavelength I need (due to case requirements) to have two 90 degree bends in it (feed -> up, across, up). If I use a 1/2 wavelength I need to run it once around the (plastic) case (feed -> up, around the case, up).

I hope this makes some sense, anyway I have found the 1/2 wave is less effected by polarisation and offers generally better performance. However while more ground plane may help a 1/4 wave it seems to hinder the 1/2 wave, I guess because it shields the loop around the case?

Regards

Reply to
Nug
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A 1/4 wavelength antenna really needs to be straight and at right angles to the ground plane. That is probably why the 1/2 wavelength antenna works better in your case.

Leon

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Leon Heller, G1HSM
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Reply to
Leon Heller

[.. 1/4 wave and 1/2 wave ...]

An antenna looks like an LC tuned circuit loaded by the radiation resistance. Your output stage has some impedance that correctly matches to it (there are exceptions we will ignore) and it is this impedance you want the antenna system to have. When the correct matching is done, the antenna works as an impedance mathcing network that matches the output stages impedance to the radiation resistance.

The normal (90 degrees to) 1/4 wave whip over a ground plane is one half of a dipole that is 1/2 wave length. The ground plane operates like a mirror. The electrostatic lines of force follow the same path with the mirroring as they would if the other 1/2 of the dipole was there. This lets you use a smaller (1/4 wave) antenna to get the same effect as the

1/2 wave.

In your case, you are not using a whip antenna. If I've read what you wrote correctly, the antenna spends more of its length parallel to the surface of the PCB than it does running 90 degrees away from it. You have some circuit with a ground plane and a limitted sized box to work with, so the mechanical shape is constained by the box and not the ideal electronics.

Since the box is small: If you have the equipment to do so, I suggest you measure (estimate) the impedance of the longest single loop of wire that will fit within the case. ie: connect to both ends. You have to have the electronics PCB in the case when you do this. If you are very lucky, its impedance will not be too hard to match to the output stage.

--
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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

I would make the grandiose statement that since you are bending the wire it no longer exhibits the performance of a "standard" 1/4 or 1/2 wave antenna. I would suggest that if you indeed made a 1/4 wave GP that protruded from the box surface (with a suitable counterpoise) it would outperform the 1/2 wave bent one..

Assuming you have to put the antenna inside the box or wrapped around it I suggest you look into tuning it with some C and/or L. In that case you would construct the antenna to fit your case parameters and adjust the matching for best radiation. Keep in mind that the C/L tuning components could be lengths of coax and open feeder/wire. (because of the high operating freq)

Cheers Bob VK2YQA (Sydney Australia)

Nug wrote:

Reply to
Bob Bob

Sorry OM,

This was all nonsense.

73's Richard Clark, KB7QHC
Reply to
Richard Clark

RF transmitters are not impedance matched to antennae in the sense of maximum transfer of power. "Maximum transfer of power" is a small signal (ideal linear parameters) issue, not a large signal issue. That is, the antenna/load are not conjugately matched. What is said, is that a TX'er will deliver some given power into, for example, 50 ohms. This says nothing about the output impedance of the PA.

Power amplifiers are concerned with DC input power to RF output power efficiency, thus they are load-line "matched," not impedance matched. The concept of "output impedance" breaks down for large signal devices. For example, what is the output impedance of a class C or D amp taken when the transistor is on or off? I suppose one could consider the time-averaged impedance, but I'm not sure of the utility (to be fair, the time-averaged reactive output component is tuned out as best possible). The vague output impedance is a problem even for large signal class A devices. Again, RF PA's should be load-line matched. Output-Z is irrelevent.

Reply to
gwhite

Hi OM,

You seem to be shy of facts and long on claims. Got any experience at the bench, or is this all arm-chair philosophy?

73's Richard Clark, KB7QHC
Reply to
Richard Clark

Nice articulation. I don't know who OM is, but RF transmitter power amps are not "impedance matched." Neither are audio power amps for that matter.

Reply to
gwhite

"OM" is an amateur radio term. It is short for "Old Man". It is a respectful term for all other males that is quick to transmit via Morse code. Richard Clark appears to be an amateur radio operator or the like.

RF transmitter power amps are certainly "impedance matched" to the intended load. Take a look in the ARRL "The radio amateur's handbook". If you have the 1944 addition, you will need to start reading at page 96 in the lower right column. If you don't have that, try Motorola's AN-721.

As for audio amp, you are 1 for 3 my friend.

--
--
kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

Dear gwhite [no call, no location]: Notwithstanding the clear limitations on making conclusions about what happens inside of a circuit that has been modeled using Thevenin's theorem, it is part of Religion that the least important theorem in circuit theory is applicable. Debates about Faith are a waste of energy. Avoid the tar-baby.

73 Mac N8TT

-- J. Mc Laughlin; Michigan U.S.A. Home: snipped-for-privacy@Power-Net.Net

"gwhite"

Reply to
J. Mc Laughlin

A CMOS Class-E amp is in full saturation (0.5v at 2a) for 10% of a cycle and off (12v at 0a) for the other

90% of a cycle. The tank circuit changes the digital energy to analog energy by filtering out everything except the fundamental frequency component. How in the world does one determine the steady-state impedance of the CMOS source? Isn't the best one can do with a digital switch is to keep it within specified parameters? The CMOS device dissipates 2 watts for 10% of the time - therefore 0.2 watts steady-state.

--

73, Cecil
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Reply to
Cecil Moore

Sorry, should have been: The CMOS device dissipates one watt for 10% of the time - therefore 0.1 watts.

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Reply to
Cecil Moore

Now, now, Cecil! Don't sully the thread with facts !-)

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

It appears that I may have canceled an earlier posting by accident so will repeat it.

A certain Class-E CMOS amp is in full saturation for

10% of a cycle, 0.5v at 2a. For the rest of the time it is off. The supply voltage is 12v. What is the steady-state impedance of the source at the fundamental frequency?

--

73, Cecil
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Reply to
Cecil Moore

My stereo amp has a spec on output impedance. As I recall, it was around

0.16 Ohms. Intended load is 4 - 16 Ohms.

Tam

Reply to
Tam/WB2TT

Efficiency

maximize

Sure are. A 30 year ago 500 level course I took called Non-Linear Transistor Design said so. The way you handle class C etc. is by handling each harmonic separately in your analysis of the transistor plus tank circuit. You match to the harmonic you want. It may be the fundamental, or the third for a tripler, etc.

tom K0TAR

Reply to
Tom Ring

Correct - in the case of an audio amplifier damping factor is critical to accurate reproduction of transients. In an RF amp, it is normally not required or desired.

tom K0TAR

Reply to
Tom Ring

"Reg Edwards" wrote in message news:cvovbf$6m2$ snipped-for-privacy@sparta.btinternet.com...

I suppose the same could be said of any block that is susceptible to having some feedback put around it. Therefore the term "output impedance" should never be used at all. And of course, any term that could, or has ever been known to lead to an argument, with any uninformed person that might come along, should be eliminated from our vocabulary.

Uuugh. Mmmmph. Me drag woman to cave by hair.

Nonsense.

If I wanted to speak of an impedance inside of some circuit, I might loosely speak of it as "internal", but in any useful discussion, it would be spoken of as either an output impedance or an input impedance, and, with most people I have such discussions with, there would be no need to add that some unknown additional feed- back not part of the present discussion could alter the observable impedance.

I hope your post was a troll.

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--Larry Brasfield
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Reply to
Larry Brasfield

So what impedance does the reflected wave encounter?

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73, Cecil  http://www.qsl.net/w5dxp


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Reply to
Cecil Moore

Let's assume the designer is an amateur who didn't provide any protection for his tube's output. The lower the resistive load, the more current the output device draws until it fails. What is the output impedance of the device?

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73, Cecil  http://www.qsl.net/w5dxp


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Reply to
Cecil Moore

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