1/4 vs 1/2 wavelength antenna

Nug · Feb 21, 2005

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

Richard Clark · Feb 23, 2005

On Wed, 23 Feb 2005 19:08:20 GMT, gwhite <gwhite@deadend.com> wrote:

RF transmitters are not ....

Sorry OM,

This was all nonsense.

73's
Richard Clark, KB7QHC

gwhite · Feb 25, 2005

Richard Clark wrote:

On Wed, 23 Feb 2005 19:08:20 GMT, gwhite <gwhite@deadend.com> wrote:

RF transmitters are not ....

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.

Richard Clark · Feb 25, 2005

On Fri, 25 Feb 2005 03:17:12 GMT, gwhite <gwhite@deadend.com> wrote:

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 · Feb 25, 2005

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: JCM@Power-Net.Net

"gwhite" <gwhite@deadend.com>

Cecil Moore · Feb 25, 2005

Ken Smith wrote:

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

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Cecil Moore · Feb 25, 2005

Ken Smith wrote:

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

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Jim Thompson · Feb 25, 2005

On Fri, 25 Feb 2005 09:47:16 -0600, Cecil Moore <w5dxp@hotmail.com>
wrote:

Ken Smith wrote:
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?

480)460-2350 | |
| E-mail Address at Website Fax

gwhite · Feb 25, 2005

Ken Smith wrote:

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/

Ken Smith · Feb 26, 2005

In article <balu11ho51ui947sboak8fbsg86dp281rq@4ax.com>,
Jim Thompson <thegreatone@example.com> wrote:

On Fri, 25 Feb 2005 09:47:16 -0600, Cecil Moore <w5dxp@hotmail.com
wrote:
[...]
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 !-)

This needs a 12V transistor that switches in about 25pS. Would that be
likely on a current chip?

--
--
kensmith@rahul.net forging knowledge

Ken Smith · Feb 26, 2005

In article <421FBB39.6BB0153C@deadend.com>, gwhite <gwhite@deadend.com> wrote:

Ken Smith wrote:

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.

I still say they are. Motorola AN-721 takes on the theory. AN-758 does a
practical example matching 12.5 Ohms into 50 Ohms

--
--
kensmith@rahul.net forging knowledge

Tom Ring · Feb 26, 2005

Ken Smith wrote:

In article <421FBB39.6BB0153C@deadend.com>, gwhite <gwhite@deadend.com> wrote:

Ken Smith wrote:

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.

I still say they are. Motorola AN-721 takes on the theory. AN-758 does a
practical example matching 12.5 Ohms into 50 Ohms

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

Rich Grise · Feb 26, 2005

On Sat, 26 Feb 2005 00:40:09 +0000, Ken Smith wrote:

In article <421FBB39.6BB0153C@deadend.com>, gwhite <gwhite@deadend.com> wrote:
Ken Smith wrote:

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.

I still say they are. Motorola AN-721 takes on the theory. AN-758 does a
practical example matching 12.5 Ohms into 50 Ohms

Evidently, the guy's never tuned up a 40 meter pi-net output transmitter. ;-)

If that's not impedance matching, I don't know what it is! (Oh, "Load line"
matching? What are the two parameters of the load line? Voltage and Current,
right? What's the slope of the load line? Impedance!)

Cheers!
Rich

Rich Grise · Feb 26, 2005

On Fri, 25 Feb 2005 17:59:56 -0500, Tam/WB2TT wrote:

"gwhite" <gwhite@deadend.com> wrote in message
news:421E98CC.6E3BDC78@deadend.com...
Richard Clark wrote:

On Wed, 23 Feb 2005 19:08:20 GMT, gwhite <gwhite@deadend.com> wrote:

RF transmitters are not ....

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.

My stereo amp has a spec on output impedance. As I recall, it was around
0.16 Ohms. Intended load is 4 - 16 Ohms.

That works because the transmission line is less than 0.01 wavelength.
So impedance matching becomes moot. If the speaker line were 1/4 wavelength
long, there would be almost no signal transferred at all.

Cheers!
Rich

Ken Smith · Feb 26, 2005

In article <pan.2005.02.26.02.11.30.772360@example.net>,
Rich Grise <richgrise@example.net> wrote:
[... audio matching ..]

That works because the transmission line is less than 0.01 wavelength.
So impedance matching becomes moot. If the speaker line were 1/4 wavelength
long, there would be almost no signal transferred at all.

( Unless you use the specially tapered euphonic cable! )

Actually, if you used a thick enough wire, you could go 1/4 wavelength at
audio frequencies. If you connect a speaker straight onto the mains,
chances are you are going further than that from the generator.

--
--
kensmith@rahul.net forging knowledge

Cecil Moore · Feb 26, 2005

gwhite wrote:

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.

So what impedance does the reflected wave encounter?
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore · Feb 26, 2005

Ken Smith wrote:

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.

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

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Cecil Moore · Feb 26, 2005

Rich Grise wrote:

Evidently, the guy's never tuned up a 40 meter pi-net output transmitter. ;-)

If that's not impedance matching, I don't know what it is! (Oh, "Load line"
matching? What are the two parameters of the load line? Voltage and Current,
right? What's the slope of the load line? Impedance!)

And there's the catch. If the load line is the source
impedance, the load (not the designer) effects the source
impedance.
--
73, Cecil http://www.qsl.net/w5dxp

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Reg Edwards · Feb 26, 2005

The phrase "output impedance" in connection with amplifiers is ambiguous and
likely to result in arguments.

The correct description is "internal impedance" or "internal resistance" and
should always be used.
----
Reg.

Cecil Moore · Feb 26, 2005

Ken Smith wrote:

If you then
put in the output device protection they didn't include, you end up with
the matching as I explained elsewhere.

SWR foldback is part of impedance matching?
--
73, Cecil http://www.qsl.net/w5dxp

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1/4 vs 1/2 wavelength antenna

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