"Owen Duffy" wrote in message
...
"Jerry" wrote in
:

...
Hi Owen

Richard Clark once told me how to combine 4 antennas in an array.
He got
me to feed 4 antennas, 50 ohms each with 50 ohm coax with no dividers.
I just fed each antenna with 50 ohm coax. At the point where the 4
coaxes get combined, I connected two coaxes in series and the other

Can you explain in more detail what you mean by "I connected two coaxes in
series"?

two also in series. Then parallel them to get back to 50 ohms. The
result is two 50 ohm loads in series to make 100 ohms and with the
other 100 ohms in parallel, the combination is a good 50 ohm load.
You can see a sketch in the Feb 2008 QST. It works pretty slick when
the antennas are 50 or 70 ohms where it is easy to get the right coax
impedance.

Owen

Hi Owen

I dont know how to include pictures in this text.

I would draw two touching circles to represent the outer conductors. The
generator is fed between to the inner conductors. Hence, two 50 ohm loads
on the coaxes will look like a 100 ohm load to the generator.

Jerry

"Jerry" wrote in
:


"Owen Duffy" wrote in message
...
"Jerry" wrote in
:

...
Hi Owen

Richard Clark once told me how to combine 4 antennas in an array.
He got
me to feed 4 antennas, 50 ohms each with 50 ohm coax with no
dividers.
I just fed each antenna with 50 ohm coax. At the point where the
4
coaxes get combined, I connected two coaxes in series and the other

Can you explain in more detail what you mean by "I connected two
coaxes in series"?

two also in series. Then parallel them to get back to 50 ohms.
The result is two 50 ohm loads in series to make 100 ohms and with
the other 100 ohms in parallel, the combination is a good 50 ohm
load. You can see a sketch in the Feb 2008 QST. It works pretty
slick when the antennas are 50 or 70 ohms where it is easy to get
the right coax impedance.

Owen

Hi Owen

I dont know how to include pictures in this text.

I would draw two touching circles to represent the outer conductors.
The
generator is fed between to the inner conductors. Hence, two 50 ohm
loads on the coaxes will look like a 100 ohm load to the generator.

Jerry

Ok, I understand, you the two inner conductors with a 100 ohm load
between them.

You have another pair like that from the other two arrays.

How do you connect them to the main feedline.


Owen

"Owen Duffy" wrote in message
...
"Jerry" wrote in
:


"Owen Duffy" wrote in message
...
"Jerry" wrote in
:

...
Hi Owen

Richard Clark once told me how to combine 4 antennas in an array.
He got
me to feed 4 antennas, 50 ohms each with 50 ohm coax with no
dividers.
I just fed each antenna with 50 ohm coax. At the point where the
4
coaxes get combined, I connected two coaxes in series and the other

Can you explain in more detail what you mean by "I connected two
coaxes in series"?

two also in series. Then parallel them to get back to 50 ohms.
The result is two 50 ohm loads in series to make 100 ohms and with
the other 100 ohms in parallel, the combination is a good 50 ohm
load. You can see a sketch in the Feb 2008 QST. It works pretty
slick when the antennas are 50 or 70 ohms where it is easy to get
the right coax impedance.

Owen

Hi Owen

I dont know how to include pictures in this text.

I would draw two touching circles to represent the outer conductors.
The
generator is fed between to the inner conductors. Hence, two 50 ohm
loads on the coaxes will look like a 100 ohm load to the generator.

Jerry

Ok, I understand, you the two inner conductors with a 100 ohm load
between them.

You have another pair like that from the other two arrays.

How do you connect them to the main feedline.


Owen

Hi Owen

The two 100 ohm loads in parallel give a 50 ohm load to the 50 ohm coax
main feed line. I used a ferrite "balun" where the unbalanced 50 ohm
*main feed line* connects to the center conductors that are connected to be
a 50 ohm load. .

Jerry KD6JDJ

"Jerry" wrote in
:

....
The two 100 ohm loads in parallel give a 50 ohm load to the 50 ohm
coax
main feed line. I used a ferrite "balun" where the unbalanced 50
ohm *main feed line* connects to the center conductors that are
connected to be a 50 ohm load. .

You didn't mention the balun in your fist posting. Without an effective
balun, the system would be quite poor.

In each pair of antenna side coax lines where the shields are tied together
and the inners are used for a 100 ohm connection point, you drive one coax
in opposite phase to the other. Your description did not note that there is
particular phasing requirement for the coax lines to the antennas, and
polarity of connection of the DEs.

BTW, you have constructed a type of power divider. There are a lot of
different ways to do it.

Owen

"Owen Duffy" wrote in message
...
"Jerry" wrote in
:

...
The two 100 ohm loads in parallel give a 50 ohm load to the 50 ohm
coax
main feed line. I used a ferrite "balun" where the unbalanced 50
ohm *main feed line* connects to the center conductors that are
connected to be a 50 ohm load. .

You didn't mention the balun in your fist posting. Without an effective
balun, the system would be quite poor.

In each pair of antenna side coax lines where the shields are tied
together
and the inners are used for a 100 ohm connection point, you drive one coax
in opposite phase to the other. Your description did not note that there
is
particular phasing requirement for the coax lines to the antennas, and
polarity of connection of the DEs.

BTW, you have constructed a type of power divider. There are a lot of
different ways to do it.

Owen

the simpler way is to just parallel all 4 of the 50 ohm loads which gives
you a 12 ohm load at the common point, then parallel 2 pieces of 50 ohm coax
1/4 wave long to form a 25 ohm matching section, that will get you back to a
50 ohm common point with no need for a balun.

On Sun, 14 Dec 2008 21:54:45 GMT, "Dave" wrote:

BTW, you have constructed a type of power divider. There are a lot of
different ways to do it.

Owen

the simpler way is to just parallel all 4 of the 50 ohm loads which gives
you a 12 ohm load at the common point, then parallel 2 pieces of 50 ohm coax
1/4 wave long to form a 25 ohm matching section, that will get you back to a
50 ohm common point with no need for a balun.

Unfortunately, this would be a common mode nightmare.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Sun, 14 Dec 2008 21:54:45 GMT, "Dave" wrote:

BTW, you have constructed a type of power divider. There are a lot of
different ways to do it.

Owen

the simpler way is to just parallel all 4 of the 50 ohm loads which gives
you a 12 ohm load at the common point, then parallel 2 pieces of 50 ohm
coax
1/4 wave long to form a 25 ohm matching section, that will get you back to
a
50 ohm common point with no need for a balun.

Unfortunately, this would be a common mode nightmare.

73's
Richard Clark, KB7QHC


.... why? It's a way it's done for broadcasting antennas. There's little
reason why the screen of the feeder can run continuously into the screens of
the transformer sections and on into the screens of the cables that feed the
individual Yagis. If the drive-points of the Yagis are equipped with
adequate baluns then there should be no more common-mode current on the
outside of the continuous screen than for the case of a single Yagi.

Chris

Owen Duffy wrote:
More likely, 2 - 1/4 wave (with velocity factor)50 ohm coax's to a
"Tee" fitting-- Each end also to a "Tee" fitting . ( all 50 ohm coax)
(power devider)

2x50 -----------------

2X Quarter wave | "T" fitting source 50 Ohm

2x50 -----------------

IF this is clear enough-- Jim NN7K

Jerry

Ok, I understand, you the two inner conductors with a 100 ohm load
between them.

You have another pair like that from the other two arrays.

How do you connect them to the main feedline.


Owen

Jim-NN7K . wrote in news:9ch1l.9958$yr3.334
@nlpi068.nbdc.sbc.com:

Owen Duffy wrote:

Firstly, I didn't write the following, Jim did.

More likely, 2 - 1/4 wave (with velocity factor)50 ohm coax's to a
"Tee" fitting-- Each end also to a "Tee" fitting . ( all 50 ohm coax)
(power devider)

2x50 -----------------

2X Quarter wave | "T" fitting source 50 Ohm

2x50 -----------------

IF this is clear enough-- Jim NN7K


Jim, are you introducing another scheme, or were you trying to explain
Jerry's scheme. We sorted Jerry's scheme, he just overlooked some vital
details in his first description. (I haven't said it to date, but I
dislike Jerry's scheme, principally over its use of the balun.)

Yours is another scheme.

There are a lot of ways to do it.

The original question was over an article's diagram that stated that
unequal lines are "WRONG!".

Yours and Jerry's responses have not dealt with the original posting, but
if anything offered alternatives that might be seen to suggest the
original configuration is flawed.

Owen

Keeping in mind simplicity, and , using equal length of 50 0hm
to this power splitter- all antennas should be in phase (all
Left elements , to properly phase), and that it gives 4-50 ohm loads
for one 50 ohm source. Why reinvent the wheel?? Jim NN7K


Owen Duffy wrote:
Jim-NN7K . wrote in news:9ch1l.9958$yr3.334
@nlpi068.nbdc.sbc.com:

Owen Duffy wrote:

Firstly, I didn't write the following, Jim did.

More likely, 2 - 1/4 wave (with velocity factor)50 ohm coax's to a
"Tee" fitting-- Each end also to a "Tee" fitting . ( all 50 ohm coax)
(power devider)

2x50 -----------------

2X Quarter wave | "T" fitting source 50 Ohm

2x50 -----------------

IF this is clear enough-- Jim NN7K


Jim, are you introducing another scheme, or were you trying to explain
Jerry's scheme. We sorted Jerry's scheme, he just overlooked some vital
details in his first description. (I haven't said it to date, but I
dislike Jerry's scheme, principally over its use of the balun.)

Yours is another scheme.

There are a lot of ways to do it.

The original question was over an article's diagram that stated that
unequal lines are "WRONG!".

Yours and Jerry's responses have not dealt with the original posting, but
if anything offered alternatives that might be seen to suggest the
original configuration is flawed.

Owen