[VHFcontesting] Power Divider?

Paul Decker kg7hf at comcast.net
Fri Jul 9 06:23:06 PDT 2010





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 at contesting.com 
[ mailto : vhfcontesting -bounces at contesting.com] On Behalf Of N9TZL 
Sent: Wednesday, July 07, 2010 10:18 AM 
To: vhfcontesting @contesting.com; vhf at 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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