1/4 vs 1/2 wavelength antenna

[Nug]

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

 

[Leon Heller]

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?

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

 

[Bob Bob]

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)

 

[Rich Grise]

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?

A 1/4 wave antenna will match to a low impedance, unbalanced. A 1/2 wave
dipole will match to a low impedance, balanced. A 1/2 wave piece of wire
fed at the end will match to a high impedance.

What kind of circuit are you using for your output?

Thanks,
Rich

 

[Ken Smith]

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.

[.. 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.

 

[gwhite]

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.

[.. 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.

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.

 

[Richard Clark]

RF transmitters are not ....

Sorry OM,

This was all nonsense.

73's
Richard Clark, KB7QHC

 

[gwhite]

Sorry OM,

This was all nonsense.

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.

 

[Ken Smith]

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.

"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.

 

[Richard Clark]

RF transmitter power amps are
not "impedance matched." Neither are audio power amps for that matter.

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

 

[J. Mc Laughlin]

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

 

[Cecil Moore]

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.

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.

 

[Cecil Moore]

The CMOS device dissipates 2
watts for 10% of the time - therefore 0.2 watts
steady-state.

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

 

[Cecil Moore]

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.

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?

 

[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?

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

...Jim Thompson

 

[Tam/WB2TT]

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.

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

 

[gwhite]

RF transmitter power amps are certainly "impedance matched" to the
intended load.

I'm sorry, but they are not. Nor are any power amps that I know of. Efficiency
(and thus necessarily output swing) is what matters for power amps. To maximize
swing requires load line matching, not impedance matching.

If you want to study RF PA's I suppose Cripps is one of the best I know of for a
modern text:

http://www.amazon.com/exec/obidos/tg/detail/-/0890069891/

 

[gwhite]

What is the steady-state impedance of the
source at the fundamental frequency?

Exactly Cecil. That's exactly why I said the concept of impedance is dubious
for large signal devices and thus provides a second argument, albeit weaker,
against the "impedance matching" idea for power amps.

"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?"

The strongest argument for dropping the impedance matching concept is PA
efficiency, and therefore maximum signal swing. Obtaining maximum swing is a
load line issue.

 

[Ken Smith]

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.

For what you say here really to be true the transistors must switch very
fast. About 25pS switching speed is needed at about 400KHz. If we take
that to be the case however, I think you will see why matching still
applies.

Lets take the reactive component first. If there is a reactive component
to the loading, the current in the switch will have a higher RMS value
without that increase in RMS increasing the radiated power of the system.
So the reactive component of the matching is fairly obvious.

Imagine that you have a well designed Class-E circuit loaded with the load
the designer optimized it for.

Now imagine that you slightly increase the resistance slightly. When you
do so, the current into the load will decrease but the voltage will not
increase enough to compensate for this.

Now lets assume that you slightly decrease the resistance. Since we are
assuming that this is a well designed case, we can assume that the
designer took steps to ensure that the output devices would be protected
from excess currents. This could be done by reducing the operating
voltage of the output section, for example. In any case, the voltage on
the load will decrease by a larger factor than the current will increase.

So it is obvious that the reactive part is matched and the resistive part
is matched just as it would be in a non-class-E output section.

 

[Ken Smith]

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

I answered your other copy of the posting.