http ://home. teleport .com/~ oldaker / power_dividers . htm
Slightly off topic, but I remember going through this page a few months ago and
I thought I found a typo in one of the examples.
The drawing for a 4 port power divider shows an 11/32 inner conductor. The
example, when worked shows to use a 17/32 which works to 25 Ohms, great for 4
50 Ohm antennas.
When I built mine, I cut my center conductor in half and center drilled all the
ends, then tapped them with 4-40 machine screws. I put a piece of small bolt
stock to connect the two halves. On the coax connectors instead of extensions,
I soldered wire lugs. Then, you can place the connectors into the outer shell
and screw the entire Assembly together. There is no soldering required between
the center rod and the connectors, and no extra access holes on the back side.
You can also cut your stock slightly longer and heat it up to fold over the
ends to seal it shut.
Paul, kg7hf.
----- Original Message -----
From: "Greg Zenger [N2GZ]" <n2gz@ gregzenger .com>
To: VHFcontesting @contesting.com
Cc: "N9TZL" <n9tzl@ owc .net>
Sent: Wednesday, July 7, 2010 9:31:14 PM GMT -05:00 US/Canada Eastern
Subject: Re: [ VHFcontesting ] Power Divider?
It will work fine. It will result in slightly larger center conductor, so
the power handling will increase and insertion loss will decrease. I have
not built transformers with .063 wall tubing, but I think you will do just
fine if you can source appropriately sized center conductor.
As usual, I got excited and started to play with numbers:
Z0=59.9584916* ln (1.0787*S/d) (Ref A)
Z0=138*log(1.08*S/d) (Ref B)
d=outside diameter of inner conductor
S=inside dimension of square conductor
Solving for d we get
d=S*1.0787*e^(-Z0/59.9584916)
d=S*1.08*10^(-Z0/138)
Common transmission line impedances used in 1/4 wave transformers are 25,
35.35, 61.24 and 70.71 ohms. { Z= sqrt (Z1*Z2) } Doing the math it looks like
the following sizes would be ballpark ideal for .063 wall tube:
5/8 24.6 ohm
15.5mm 26.0
16mm 24.2
17/32 34.4
13mm 36.62
13.5mm 34.4
11/32 60.5
8.5mm 62.1
9/32 72.5
7mm 73.7
7.5mm 69.6
The impedances are not exactly ideal, but assuming the loads are 50 ohms,
theoretical VSWR mismatch will be less than 1.05:1
Keep in mind that your antennas are probably not exactly 50 ohms, so their
combined and transformed impedance won't be exactly 50 ohms either. If you
do know if the impedances of your antennas are greater, or lower than 50
ohms, you can select a power divider topology and impedance that can
compensate if you wish to nit-pick every little bit of performance out of
the system. Just keep in mind the tolerances that you are able to construct
to (or test to) before you get too carried away.
-Greg Zenger, N2GZ/1, FN31xi
(Ref A) http :// fermi .la. asu . edu /w9cf/articles/square/index. html
(Ref B) http ://home. teleport .com/~ oldaker / power_dividers . htm
-----Original Message-----
From: vhfcontesting -bounces@contesting.com
[ mailto : vhfcontesting -bounces@contesting.com] On Behalf Of N9TZL
Sent: Wednesday, July 07, 2010 10:18 AM
To: vhfcontesting @contesting.com; vhf@w6yx. stanford . edu
Subject: [ VHFcontesting ] Power Divider?
Hello:
Has anyone built power dividers with the .063 wall thickness, 1" square Al
tubing instead of using the .125 wall thickness, 1" square Al tubing?
Asuming you use the correct inside rod diameter do the ones with .063 wall
work as well as those with .125 wall thickness?
Reason for the question is I already have some of the .063 wall tubing.
Thanks for any and all info,
73, Dennis N9TZL
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