Yes I should know this.. However an issue has come up in my place of
work I am trying to get my head around. On a 10GHz microwave TX there is
a stripline parallel to the output track that is suppose to detect
forward power. Nothing strange about that. What is weird is that it
appears to be roughly a half wave length (or more).

The question is this. If one hangs a 1/2 wave dipole in free space I
assume it receives such that current maximums are at the centre and
voltage maxiums at the ends. Is this the case whether a feedline is
connected or not?

If I then take a stripline cct terminated at one end with a 50r resistor
and a detect diode at the other and is a half wave long, what is the
current/voltage distribution in this configuration.

What I wonder is if it is a halfwave there may be no voltage at the
detect diode input. (It could be 3/4 wavelength when one factors in pcb
dielectric and end loading)

Thoughts?

Cheers Bob VK2YQA

Bob Bob wrote:
If I then take a stripline cct terminated at one end with a 50r resistor
and a detect diode at the other and is a half wave long, what is the
current/voltage distribution in this configuration.

Since the ends of the 1/2WL conductor are open circuits, standing
waves on the conductor are the result. The standing wave(s) consists
of the superposition of the forward wave(s) and the reflected wave(s).
This is exactly why the forward and reflected waves cannot be ignored
(as some gurus advise) if understanding is the goal.

For a 1/2WL conductor, the standing wave has a cosine envelope with
a phase near zero all along the line because the forward and reflected
phasors are rotating in opposite directions. The current is obviously
zero at the ends and maximum in the center.

The resistor termination at one end dissipates either the forward
wave or reflected wave depending upon which end it is located. The
dipole at the other end rectifies the other wave. If you are measuring
the forward power, the forward wave is rectified by the diode and the
reflected wave is dissipated/attenuated by the resistor. I have an
old Heathkit SWR meter that operates the same way. There's two pickup
conductors, one for forward power and one for reflected power.

In short, the resistor dissipates one of the component traveling waves
thus preventing reflections while the diode rectifies the other traveling
wave, thus providing a voltage proportional to that other traveling wave.
Which wave is dissipated and which wave is rectified depends upon which
end of the conductor the resistor and diode are located.
--
73, Cecil http://www.qsl.net/w5dxp

With 50 Ohms at the end, I suppose the line is producing progressive waves
and not standing waves, and no matter its length

73

André F5AD
http://f5ad.free.fr

Clearly in a center-fed half-wave dipole, if there is no load at the
center, the current there is quite low, and depends on the capacitance
from wire end to wire end. If you short across the feedpoint, it will
be resonant and the center will be a current node.

In the case of coupled microstrip hybrids, maximum coupling occurs at
odd multiples of 1/4 wavelength, and falls to zero at even multiples of
1/4 wavelength -- that is, at multiples of 1/2 wavelength. If you
DON'T properly terminate one end of the coupled line, you'll loose
directivity because of reflections at that end.

I suspect that if you include the propagation velocity of the
microstrip, you'll find that it's not a half-wave long.

Cheers,
Tom

K7ITM wrote:
Clearly in a center-fed half-wave dipole, if there is no load at the
center, the current there is quite low, and depends on the capacitance
from wire end to wire end. If you short across the feedpoint, it will
be resonant and the center will be a current node.

Maybe you would like to rethink that, Tom. Consider a parasitic 1/2WL
element on a Yagi. The currents at the ends are obviously zero. The
current 1/4WL away from the zero points are the maximum current
points. If you short across the feedpoint of a 1/2WL dipole, it will
be resonant but the center will be a current loop (antinode), i.e.
a maximum current point. The center of a *one wavelength* dipole will
be a current node.
--
73, Cecil http://www.qsl.net/w5dxp

On Tue, 28 Feb 2006 14:25:32 -0600, Bob Bob wrote:
The question is this. If one hangs a 1/2 wave dipole in free space I
assume it receives such that current maximums are at the centre and
voltage maxiums at the ends. Is this the case whether a feedline is
connected or not?

Yes.

If I then take a stripline cct terminated at one end with a 50r resistor
and a detect diode at the other and is a half wave long, what is the
current/voltage distribution in this configuration.

The point of your need at work is to insure that stripline presents a
50 Ohm characteristic as an untuned line, not a resonant line. Of
course, you can perform the same using a tuned line, but this seems
unlikely barring new details offered by you.

What I wonder is if it is a halfwave there may be no voltage at the
detect diode input. (It could be 3/4 wavelength when one factors in pcb
dielectric and end loading)

And this would be reason why to NOT have it be a resonant line. I
presume there's a ground in the vicinity for both this sniffer section
and the main feed (which also needs to exhibit a characteristic Z).
Such devices are generally useful over an octave range.

Thoughts?

This, of course, does require attention to factors such as pcb
dielectric as you have already taken care of end loading with the
resistor and the diode.

73's
Richard Clark, KB7QHC

I earlier wrote, "If you short across the feedpoint, it will
be resonant and the center will be a current node."

Belay that. Change it to, "If you short across the feedpoint, it will
be resonant and the center will be a current maximum."

Richard, Cecil and Tom

I should explain the "problem" somewhat.

The transmitter designs I believe are not very good. Perhaps they were
designed to work on a different or narrower frequency and were hastily
adapted by beancounters instead of designers. grin What is happening
is that at a certain critical band of freqs (around 11.2GHz) the output
from the detect diode is very close to zero and hence the power control
(via ALC) and calibration table dont work too well. Thinking it was a
design issue I actually moved the ref diode to the limit of its pads and
the problem resolved itself. The effect was of course to make the sensor
length electrically shorter by doing this. I was trying to establish
whether I had a 1/2 or 3/4 wavelength sensor section to maybe take it up
with engineering. (No doubt for another frequency I would move it in the
opposite direction)

I suspect there is also another issue whereby the resonant sensor
section is actually absorbing more power than it should and dissipating
it in the resistor. The effect of this being a loss of available output
power at the SMA connector.

Keep in mind that I am a production line tech rather than a designer. I
am not suppose to modify things, only test and align! (Am actually an IT
professional but had trouble finding that line of work!)

Many thanks for your input. As usual with you gents I am going to have
to work through it all slowly!

To answer you specifically Richard, the PCB is a multilayer fibreglass
thing with a largish ground being under both the output and sensor
striplines. I was surprised that the sensor line was so long and so
close when I first saw it thinking some major overcoupling might be
occurring. The tracks are maybe 3mm wide and about 1.5mm (edges) apart.
I am thinking also that the 50 ohm output is not being preserved as it
goes past the sensor stripline. I think the freq coverage for this model
is about 10.5 to 11.5GHz. Keep in mind that this really is the frist
time I have seen microwave TX's up close so my gut feelings about track
sizes/spacing may be way off.

Cheers Bob VK2YQA

Bob, VK2YQA wrote:
"The question is this. If one hangs a 1/2-wave dipole in free space I
assume it receives such that current maximums are at the centre and
voltage maximums at the ends. Is this the case whether a feedline is
connected or not?"

Yes, but you must have continuity between both halves of the dipole. If
you disconnect the feedline leaving an open circuit gap in the dipole at
its centre, you no longer have a 1/2-wave dipole of the same frequency.
You have (2) lengths of wire and each has its first resonance at about
twice the frequency of first resonance of your original dipole.

Best rergards, Richard Harrison, KB5WZI

On Tue, 28 Feb 2006 20:35:01 -0600, Bob Bob wrote:
I suspect there is also another issue whereby the resonant sensor
section is actually absorbing more power than it should and dissipating
it in the resistor.

Hi Bob,

That is why it is there, among other reasons.

The effect of this being a loss of available output
power at the SMA connector.

This is a symptom, and should not be a cause.

To answer you specifically Richard, the PCB is a multilayer fibreglass
thing with a largish ground being under both the output and sensor
striplines.

That is as it should be.

I was surprised that the sensor line was so long and so

On the order of 1.5 cM?

close when I first saw it thinking some major overcoupling might be
occurring. The tracks are maybe 3mm wide and about 1.5mm (edges) apart.

Sounds like a boilerplate design - which is to say right out of some
book or App. Note.

I am thinking also that the 50 ohm output is not being preserved as it
goes past the sensor stripline.

What does that mean?

I think the freq coverage for this model
is about 10.5 to 11.5GHz. Keep in mind that this really is the frist
time I have seen microwave TX's up close so my gut feelings about track
sizes/spacing may be way off.

Check the resistor.

73's
Richard Clark, KB7QHC