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#1
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Hi all.
Throwing together a 20m single quad loop (triangular) antenna. In the past I have fed these with a 75r Q section. Nothing strange here.. I use to coil up most of it in a 5" coil to help reduce the line radiation problem - if any... No tuner in use.. This time however I am going to use two pieces of 50 ohm coax in parallel giving a 100r balanced Q section feeder. This more closely matches the loop Z. The shields are joined together at each end of the section and the coax centre conductors become the two antenna connection points. Question; Should I be worried about inbalance in the 50 ohm cable from the radio to the Q section. ie does the Q section reduce any of that problem? If that isnt the case I can make up the usual coil of coax before it meets the Q section.. or can I coil up the Q section instead. I also wonder about Q section shield grounding as I hadnt intended grounding either end. Will this be an issue? Cheers Bob W5/VK2YQA |
#2
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On Mon, 12 Sep 2005 15:39:23 -0500, Bob Bob wrote:
Hi all. Throwing together a 20m single quad loop (triangular) antenna. In the past I have fed these with a 75r Q section. Nothing strange here.. I use to coil up most of it in a 5" coil to help reduce the line radiation problem - if any... No tuner in use.. This time however I am going to use two pieces of 50 ohm coax in parallel giving a 100r balanced Q section feeder. This more closely matches the loop Z. The shields are joined together at each end of the section and the coax centre conductors become the two antenna connection points. I have a question, Why? The loop Z is going to be in the neighborhood of 120 ohm. A 75 ohm Q section is a better deal than 100 ohm. Question; Should I be worried about inbalance in the 50 ohm cable from the radio to the Q section. ie does the Q section reduce any of that problem? If that isnt the case I can make up the usual coil of coax before it meets the Q section.. or can I coil up the Q section instead. I also wonder about Q section shield grounding as I hadnt intended grounding either end. Will this be an issue? Cheers Bob W5/VK2YQA |
#3
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The modelled Z is around 140 ohms. The "top" element of the loop is
about 20% longer than the other two. It is starting to approach a folded dipole. Bob Wes Stewart wrote: I have a question, Why? The loop Z is going to be in the neighborhood of 120 ohm. A 75 ohm Q section is a better deal than 100 ohm. |
#4
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Hi Roger
Okay bad choice of words. Try series... With the centre conductor of both coaxes joined at each end you do indeed get 25 ohms.. "Parallel" was more a case of describing the physical rather than electrical layout. When you connect just the shields you effectively have a 100 ohm balanced but shielded feeder with the two centre conductors being the "real" transmission line. Also apparently good for placing inside a metal pipe without it affecting the transmission system - but I am not doing that. And where the 50 ohm coax from the radio joins the bottom of the Q section its shield goes to one wire and the centre conductor to the other. Okay? Roger Leone wrote: My math says that 2 sections of 50 ohm coax in parallel makes a 25 ohm feeder, not a 100 ohm feeder. Also, if your intention is to use the 2 coax center conductors to connect to the loop feedpoint, where will the differential mode currents which are necessary at the feedpoint come from? And if you don't connect the coax shields to something at the feedpoint, where does the current flowing in the shield go? |
#5
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![]() "Bob Bob" wrote in message news ![]() Okay? Bob: Okay, I see what you are planning to do. Seems reasonable, but there is still a strangeness to the coax shields of the Q section being connected only to each other. Let us know how it works. Roger |
#6
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On Mon, 12 Sep 2005 18:38:34 -0500, Bob Bob wrote:
The modelled Z is around 140 ohms. The "top" element of the loop is about 20% longer than the other two. It is starting to approach a folded dipole. Okay. 75 ohm is -still- better than 100. |
#7
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Roger Leone wrote:
My math says that 2 sections of 50 ohm coax in parallel makes a 25 ohm feeder, not a 100 ohm feeder. The two sections are not electrically paralleled but instead are used as balanced feeders. They are physically parallel but electrically in series, thus Z0 = 100 ohms. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#8
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Hi Roger
Might get around to it today! (Being unemployed is sometimes fun..) Also looking at the possibility of harmonic operation w/out adjustment. Not useful for a 20M quad as the next usable freq is about 38Mhz. I found though that if I (modelled) cut for 10.1Mhz it is also usable 28.4-29.0. This is also a good reason for using the high side impedence transformation (ie 50r to 100r = 122r vs 50r to 75r = 90r) as the Z on 10M is higher. Radiation pattern isnt much good for long DX though.. Cheers Bob Roger Leone wrote: Okay, I see what you are planning to do. Seems reasonable, but there is still a strangeness to the coax shields of the Q section being connected only to each other. Let us know how it works. Roger |
#9
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Err why?
50r to 100r = 122r 50r to 75r = 90r Do I have the calcs right? Wes Stewart wrote: Okay. 75 ohm is -still- better than 100. |
#10
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Bob Bob wrote:
Err why? 50r to 100r = 122r SQRT(50*200) = 100 50r to 75r = 90r SQRT(50*100) = 70.7 Do I have the calcs right? Wes Stewart wrote: Okay. 75 ohm is -still- better than 100. SQRT(50*140) = 84 ohms. Z0=75 ohms is closer to 84 ohms than is Z0=100 ohms. If you had a 200 ohm feedpoint, Z0=100 ohms would be perfect. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
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