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Two coax as substitute for open line
Let's suppose that we use to identical length of coax, and on each
one, we connect the shield to the inner conductor at each of its ends. If we use these two independant (?) sections as a substitute for an open-wire line, could we use the usual formula Z=276*log(2S/D) to compute the impedance of that line? In that case, the characteristic impedance Zo of the coax could be neglected. And what about the matched losses in dB/100' of the resultant line? Twice the one of the coax? 73 de Pierre VE2PID |
Two coax as substitute for open line
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Two coax as substitute for open line
Pierre,
The matched loss of the resulting open-wire line is not related to the matched loss of the coax cablebecause those cables are not individually operating in "Transmission Line" mode. For the same reason, the velocity factor of the open-wire line will not be the same as the coax. Here's a write-up of an experiment I did to produce low Zo balanced line. I didn't measure the loss because my application was for a very short run: http://www.karinya.net/g3txq/twin_feed/ 73, Steve G3TXQ |
Two coax as substitute for open line
steveeh131047 wrote in news:f6f2c446-5128-4851-aab2-
: Here's a write-up of an experiment I did to produce low Zo balanced line. I didn't measure the loss because my application was for a very short run: Steve, you haven't let on much detail about the application of your low Zo line. I think I suggested elsewhere that you could fabricate such a thing with parallel coaxes with all of their shields at each end bonded. I explored using square tubes as a rigid transmission line. This finds application in a range of areas, eg the boom of a LP. See http://www.vk1od.net/calc/tstl.htm . A 50 ohm line (if that was your objective) requires an air spaced line of 1.19 D/d. The article contains a calculator to solve the curve fit. Owen |
Two coax as substitute for open line
On Feb 24, 11:28*am, Owen Duffy wrote:
ve2pid wrote in news:cd290c79-c0c7-4308-af41- : Let's suppose that we use to identical length of coax, and on each one, we connect the shield to the inner conductor at each of its ends. If we use these two independant (?) sections as a substitute for an open-wire line, could we use the usual formula Z=276*log(2S/D) to compute the impedance of that line? In that case, the characteristic impedance Zo of the coax could be neglected. And what about the matched losses in dB/100' of the resultant line? Twice the one of the coax? Pierre, The formula you give is an approximation which is not good for low Zo, and it ignores the effect of the dielectric (coax jacket). I have seen discussion of this recently in another place. Yes, you can fabricate a two wire open line like this. You need to find a way to bind the two cables for consistent physical spacing, if the jacket is PVC you are using a lossy dielectric, the braided conductor is lossier at RF than an equivalent solid copper tube. So an expensive low Zo line, not a low as you might think, with poor performance. Why would you do this? Owen I certainly agree with Owen that it's not a particularly good way to get a low-loss two-wire balanced line. To me, the attraction of a two- wire balanced line is the low loss when it's properly implemented at relatively higher impedance. That said, however, one can buy "Siamese twin" cables with two pieces of coax in the same jacket, arranged to make it easy to separate them. The web of jacket between the two lines is relatively thin. Such line is common in CATV installations in homes. At low frequencies (HF, especially lower HF frequencies), the dielectric loss may not be too bad, especially for the low impedance involved. It may be a cheap way to get an approximately 100 ohm balanced line. Even though the braided conductors will have higher loss than equivalent diameter smooth conductors, the copper loss for typical RG-6-type line used that way (about 5mm OD outer conductor) will almost certainly be less than that using, say, 2mm diameter solid conductors. You may be able to find surplus (e.g. reel-ends) of this sort of line at very attractive prices. Even new, it's not too bad--a lot cheaper than equivalent diameter solid copper! Cheers, Tom |
Two coax as substitute for open line
On Feb 24, 8:16*pm, Owen Duffy wrote:
steveeh131047 wrote in news:f6f2c446-5128-4851-aab2- : Here's a write-up of an experiment I did to produce low Zo balanced line. I didn't measure the loss because my application was for a very short run: Steve, you haven't let on much detail about the application of your low Zo line. I think I suggested elsewhere that you could fabricate such a thing with parallel coaxes with all of their shields at each end bonded. I explored using square tubes as a rigid transmission line. This finds application in a range of areas, eg the boom of a LP. Seehttp://www.vk1od..net/calc/tstl.htm. A 50 ohm line (if that was your objective) requires an air spaced line of 1.19 D/d. The article contains a calculator to solve the curve fit. Owen Owen. The application is no secret - the first paragraph in my linked page explains it. I was looking for line with a Zo in the range 50ohms-70ohms that could be used for the band interconnects on the hexbeam. Arguably, balanced line would be easier to handle, and might have some common-mode advantages. Total length is about 4ft so loss is not an issue. It would need to be very easily replicated, using materials readily available to constructors world-wide. At the end of the day it became more of a "self learning" exercise than a serious quest ! 73, Steve G3TXQ |
Two coax as substitute for open line
In the ARRL's Antenna Book 21st ed page 24-21, we see that if we
connect the two shields of the coax cables together, we obtain 'Shielded parallel Lines' . In that case, the resultant impedance is simply the sum of the characteristic impedances of each coax. So, there is quite a difference between the two independent coax I mentioned in my first message (we connect the shield to the inner conductor at each of its ends) (A) and the 'Shielded Parallel Lines' case (B). I am trying to understand why and it is the reason I posted my first message... In (A), the Z=276*log(2S/D) applies, so the Zo of each coax does not matter.. but in (B), Z=Zo1+Zo2, so the value of each Zo matters. Am I right? And how to compute matched line loss in case (A) and in case (B) ? |
Two coax as substitute for open line
steveeh131047 wrote in
: The application is no secret - the first paragraph in my linked page explains it. I missed that! I was looking for line with a Zo in the range 50ohms-70ohms that could be used for the band interconnects on the hexbeam. Arguably, balanced line would be easier to handle, and might have some common-mode advantages. Total length is about 4ft so loss is not an issue. It I am not so sure there would be much difference. would need to be very easily replicated, using materials readily available to constructors world-wide. If you tried my calculator for the square tubes, you would have found that you get lower Zo for the same D/d ration as for round tubes. For example, a pair of 12mm tubes clamped together with 2mm spacing gives Zo of around 50 ohms. If such an element could be used to replace one of the spreaders, then you have neat solution... but it would take some custom moulding to support such a spreader on the hub. At the end of the day it became more of a "self learning" exercise than a serious quest ! Well, that is a serious quest. Cheers Owen 73, Steve G3TXQ |
Two coax as substitute for open line
On Feb 24, 5:21*pm, Owen Duffy wrote:
... mainly because the effective RF resistance of the braid is not easy to estimate. Is there no data for the RF resistance of the center conductor vs the RF resistance for the braid? One would think it could be ascertained by comparing known coax losses to known parallel line losses when the wires are the same size. -- 73, Cecil, w5dxp.com |
Two coax as substitute for open line
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Two coax as substitute for open line
ve2pid wrote:
In the ARRL's Antenna Book 21st ed page 24-21, we see that if we connect the two shields of the coax cables together, we obtain 'Shielded parallel Lines' . In that case, the resultant impedance is simply the sum of the characteristic impedances of each coax. So, there is quite a difference between the two independent coax I mentioned in my first message (we connect the shield to the inner conductor at each of its ends) (A) and the 'Shielded Parallel Lines' case (B). I am trying to understand why and it is the reason I posted my first message... In (A), the Z=276*log(2S/D) applies, so the Zo of each coax does not matter.. but in (B), Z=Zo1+Zo2, so the value of each Zo matters. I am sure that in case B it is as for resistors in parallel, ie 2 50ohm cables in parallel give you 25ohms Zo. Jeff |
Two coax as substitute for open line
Jeff wrote in news:hm5dj4$4ls$1
@speranza.aioe.org: I am sure that in case B it is as for resistors in parallel, ie 2 50ohm cables in parallel give you 25ohms Zo. Jeff, you you offer more explanation that just that your are "sure". If you can't explain it, it speaks of whether you are sure. Owen |
Two coax as substitute for open line
Owen Duffy wrote in
: steveeh131047 wrote in : The application is no secret - the first paragraph in my linked page explains it. Steve, Another thought comes to mind, and that is a pair of rectangular conductors, preformed to the parabolic like shape and screwed togeter with insulated fixings. You could achieve the Zo you seek with say 20mm wide flat with manageable spacing... however it would be a water trap to some extent (just like LP feeds using twin booms). Owen |
Two coax as substitute for open line
Owen Duffy wrote:
Jeff wrote in news:hm5dj4$4ls$1 @speranza.aioe.org: I am sure that in case B it is as for resistors in parallel, ie 2 50ohm cables in parallel give you 25ohms Zo. Jeff, you you offer more explanation that just that your are "sure". If you can't explain it, it speaks of whether you are sure. Owen Well having just tried it for real on a network analyser, and simulated it on Ansoft designer I am now convinced rather than being sure!! Jeff |
Two coax as substitute for open line
On Feb 25, 4:51*pm, Owen Duffy wrote:
Owen Duffy wrote : steveeh131047 wrote in : The application is no secret - the first paragraph in my linked page explains it. Steve, Another thought comes to mind, and that is a pair of rectangular conductors, preformed to the parabolic like shape and screwed togeter with insulated fixings. You could achieve the Zo you seek with say 20mm wide flat with manageable spacing... however it would be a water trap to some extent (just like LP feeds using twin booms). Owen Owen, Thanks for the suggestion - water shouldn't be a problem because this "band interconnect" TL can be installed within the hexbeam centre post where it would be protected. One chap has already fabricated a "square coaxial" version - second photo, he http://lema.epfl.ch/images/stories/L...HB9MCZ_wb.html but that will be beyond the capability of many constructors! It's not a major issue - most constructors simply use RG213; but some find the physical terminations "fiddly", especially if you try to assemble it inside the centre-post for neatness and weather protection. I'm pretty sure that twin line using the right gauge ecw with teflon tubing on one of the wires would be about the correct Zo and easy to fabricate. It's just another thing on my "to do" list - a list which seems to get longer rather than shorter :) 73, Steve G3TXQ |
Two coax as substitute for open line
steveeh131047 wrote in
: .... Thanks for the suggestion - water shouldn't be a problem because this "band interconnect" TL can be installed within the hexbeam centre post where it would be protected. One chap has already fabricated a "square coaxial" version - second photo, he http://lema.epfl.ch/images/stories/L...exbeam/hexbeam _HB9MCZ_wb.html Seeing the construction, I think a pair of rectangular tubes would be easier to construct, and raise less water retention issues. I didn't realise that the line section is straight, I had visualised it taking the shape of one of the spreaders if you understand what I mean. Anyway, my recommendation would be like in the pic, but using two side by side rectangular tubes, say 20mm with 4mm spacing. Owen |
Two coax as substitute for open line
Jeff wrote in
: Owen Duffy wrote: Jeff wrote in news:hm5dj4$4ls$1 @speranza.aioe.org: I am sure that in case B it is as for resistors in parallel, ie 2 50ohm cables in parallel give you 25ohms Zo. Jeff, you you offer more explanation that just that your are "sure". If you can't explain it, it speaks of whether you are sure. Owen Well having just tried it for real on a network analyser, and simulated it on Ansoft designer I am now convinced rather than being sure!! That is not an explanation at all. Your confirmation might just be confirmation of a wrong interpretation of the B configuration. Owen |
Two coax as substitute for open line
Owen Duffy wrote in
: .... Anyway, my recommendation would be like in the pic, but using two side by side rectangular tubes, say 20mm with 4mm spacing. I should have added that I would attach the coax at the bottom, and use a ferrite cored broadband balun. Owen |
Two coax as substitute for open line
Owen Duffy wrote:
Jeff wrote in : Owen Duffy wrote: Jeff wrote in news:hm5dj4$4ls$1 @speranza.aioe.org: I am sure that in case B it is as for resistors in parallel, ie 2 50ohm cables in parallel give you 25ohms Zo. Jeff, you you offer more explanation that just that your are "sure". If you can't explain it, it speaks of whether you are sure. Owen Well having just tried it for real on a network analyser, and simulated it on Ansoft designer I am now convinced rather than being sure!! That is not an explanation at all. Your confirmation might just be confirmation of a wrong interpretation of the B configuration. Owen and you just might be trolling. Jeff |
Two coax as substitute for open line
Jeff wrote in
: Owen Duffy wrote: Jeff wrote in : Owen Duffy wrote: Jeff wrote in news:hm5dj4$4ls$1 @speranza.aioe.org: I am sure that in case B it is as for resistors in parallel, ie 2 50ohm cables in parallel give you 25ohms Zo. Jeff, you you offer more explanation that just that your are "sure". If you can't explain it, it speaks of whether you are sure. Owen Well having just tried it for real on a network analyser, and simulated it on Ansoft designer I am now convinced rather than being sure!! That is not an explanation at all. Your confirmation might just be confirmation of a wrong interpretation of the B configuration. Owen and you just might be trolling. And you just might be wrong on both counts. I will offer you an explanation of why the ARRL position is IMHO correct. The configuration is a pair of identical coaxial cables, shields tied together at each end, and the inner conductors used as a two wire transmission line (http://www.vk1od.net/transmissionline/stcm/Fig02.png). Question is what is Zo of the combination? for each of the coaxial lines, its intinsic Zo defines the ratio of V/I for a travelling wave at any point. If the two inner conductors are driven by a differential signal, then I in one conductor equals -I in the other (defined by differential mode), and V conductor to conductor is twice V conductor to shield, so V'=2*(I*Zo), and Zo'=V'/I=2*(I*Zo)/I=2*Zo. Owen |
Two coax as substitute for open line
Owen Duffy wrote in
: .... .... in the other (defined by differential mode), and V conductor to conductor is twice V conductor to shield, so V'=2*(I*Zo), and Zo'=V'/I=2*(I*Zo)/I=2*Zo. I should have expected that this is not obvious to you Jeff, and expand that to: .... in the other (defined by differential mode), and V conductor to conductor is (by definition of differential mode) V'=(I*Zo-(-I*Zo))=2*I*Zo, and Zo'=V'/I=2*(I*Zo)/I=2*Zo. Owen |
Two coax as substitute for open line
On Feb 24, 3:36*pm, Cecil Moore wrote:
On Feb 24, 5:21*pm, Owen Duffy wrote: ... mainly because the effective RF resistance of the braid is not easy to estimate. Is there no data for the RF resistance of the center conductor vs the RF resistance for the braid? One would think it could be ascertained by comparing known coax losses to known parallel line losses when the wires are the same size. -- 73, Cecil, w5dxp.com There was an article published maybe 15 years ago (RF Design magazine?? Electronics Design?? EDN??), authored by a fellow from Andrew as I recall, about some of the finer points of coaxial cable. I thought it was quite a good article. He had "rules of thumb" for loss in stranded center conductors versus solid that I remember for sure, and perhaps for braid as well--I don't recall that for certain. It wasn't huge, just a few percent, for the stranded center. Of course in coax, since there's a lot more surface area to the outer conductor than the inner, the braid would have to be considerably worse than a solid conductor to significantly add to the total series RF resistance, so it wouldn't be trivial to resolve by measuring coax with a solid outer versus a braided outer. You'd have to go to considerable effort to keep the rest of the construction identical to nail down the contribution of the braid versus solid outer. In any event, it seems a reasonable way to get a large diameter RF conductor that remains flexible...I'd be very surprised if a 5mm diameter coax braid was a worse two-wire line conductor than 2.5mm solid, smooth copper. I would want to keep the jacket on it, sealed against weather at the ends, to keep the copper clean. Cheers, Tom |
Two coax as substitute for open line
On Feb 24, 3:12*pm, ve2pid wrote:
In the ARRL's *Antenna Book 21st ed page 24-21, we see that if we connect the two shields of the coax cables together, we obtain 'Shielded parallel Lines' . In that case, the resultant impedance is simply the sum of the characteristic impedances of each coax. So, there is quite a difference between the two independent coax I mentioned in my first message (we connect the shield to the inner conductor at each of its ends) (A) and the 'Shielded Parallel Lines' case (B). *I am trying to understand why and it is the reason I posted my first message... In (A), the Z=276*log(2S/D) *applies, so the Zo of each coax does not matter.. but in (B), Z=Zo1+Zo2, so the value of each Zo matters. Am I right? And how to compute matched line loss in case (A) and in case (B) ? I suppose what follows has already been covered, but perhaps different words to say it will help... In the case of the two pieces of coax, used as coax, the fields are entirely inside the coaxial lines. The center conductors carry current, and as they have relatively little surface area compared with the outer conductors, they contribute most of the loss (assuming good dielectric with negligible loss). Balance depends on the load and how the lines are fed; there is nothing inherently balanced about the line itself. In fact, you could make such a line out of two pieces of coax with different constructions: same outer conductor diameter, but solid polyethylene dielectric for one and foamed polyethylene for the other. The result would be different attenuations in the two, and different propagation velocities, so that for the same physical length for each, fed 180 degrees out of phase, the signals at the other end would not be "balanced." Fed as a two-wire line, it would be balanced (assuming it's kept away from un-balancing structures). The effective impedance for two identical lines (back to the same dielectric, etc.) fed that way is just the twice the impedance of a single line. In the case of the two pieces arranged with constant spacing, fed as a balanced line, all the fields are external to the lines. There is essentially no field inside the outer conductor of each line. The center conductors carry no current, so there is no particular need to connect them electrically to the outer conductors. The loss is determined by the RF resistance to the current carried on the outside of the outer conductor, and by loss in the dielectric between the wires (or more properly, dielectric in the area around the wires where the electric field is significant). Since the fields are external to the coax (not used as coax), the impedance depends only on the physical configuration outside the coax -- not on the coax inner conductor diameter, and not on the dielectric inside the coax. Those are invisible to the fields carrying energy on the two-wire balanced line. Note that it's possible to transmit three independent TEM signals on a couple pieces of coax configured as a two-wire line: one on each coax and one on the balanced line. There's a useful formula for estimating the loss of coax or balanced lines. It includes a term for loss in the resistance of the conductors, and another term for loss in the dielectric. For coax using good dielectric, up to a few hundred MHz at least, the second term can almost always be ignored. In English units (conversion to metric left as an exercise...), it's: A100 = 4.34*Rt/Zo + 2.78*f*Fp*sqrt(epsilon) where A100 = matched line loss in dB/100 feet Rt = total conductor RF resistance (see below), ohms Zo = characteristic impedance of the line f = operating frequency in MHz Fp = power factor of dielectric at frequency f epsilon = permittivity of the dielectric, relative to air (Note: Fp is essentially the same as the dielectric power factor for any dielectric you'd care to use in an RF transmission line.) A good estimate of Rt for copper conductors thicker than a couple of skin-depths is: Rt = 0.1*(1/d + 1/D)*sqrt(f) where d = diameter of inner coax conductor, or the wire diameter for two- wire balanced line D = diameter of the outer coax conductor, or = d for two-wire balanced line This formula is from the "Transmission Lines" chapter of Sams' "Reference Data for Engineers." You can replace 1/d and 1/D in the Rt formula with Rrf/d and Rrf/D for non-copper conductors, where Rrf is the RF resistance of the conductor relative to copper. Normally, that would be sqrt(Rdc(material)/Rdc(copper)), for non-magnetic conductors. So for aluminum, use 1.23 in the numerator for that conductor instead of 1.00. It is _probably_ reasonable to use 1.07 instead of 1.00 for stranded copper, and maybe 1.10 for braid, but "your mileage may vary." Cheers, Tom |
Two coax as substitute for open line
K7ITM wrote in
: .... In any event, it seems a reasonable way to get a large diameter RF conductor that remains flexible...I'd be very surprised if... Tom, I have attempted to predict the behaviour of bootstrap coax traps, and an important factor is the effective RF resistance of the outside of the coax which forms an inductor. My measurement capacity is quite limited. Trying to reconcile my prediction with W8JI's measurements of such traps makes me think that the effective RF resistance is much poorer than a tube of the same size. Factors causing this would be braiding, proximity effect and dielectric losses in PVC jacket. Owen |
Two coax as substitute for open line
On Feb 25, 12:12*pm, Owen Duffy wrote:
K7ITM wrote : ... In any event, it seems a reasonable way to get a large diameter RF conductor that remains flexible...I'd be very surprised if... Tom, I have attempted to predict the behaviour of bootstrap coax traps, and an important factor is the effective RF resistance of the outside of the coax which forms an inductor. My measurement capacity is quite limited. Trying to reconcile my prediction with W8JI's measurements of such traps makes me think that the effective RF resistance is much poorer than a tube of the same size. Factors causing this would be braiding, proximity effect and dielectric losses in PVC jacket. Owen I think I'll be in good agreement with you, here, Owen, by saying that the right way to find out about something like this is to actually make some measurements on the configuration in question. Maybe I can come up with something... I do have reasonable measurement technology available, though getting it "right" isn't trivial, especially with balanced open-wire line. I do have some ideas and if I can find time will try a couple different ways to see if they agree. Time, right now, will be the problem. Cheers, Tom |
Two coax as substitute for open line
In a line made up of two parallel coax cables with connected shields and
with the shields floating at both ends, let's suppose that one center conductor (cable A) is carrying 1A and the other (cable B) 3A, exactly out of phase with each other -- that is, the line is not balanced. The inside of the cable A shield will have 1A and the inside of the cable shield B will have 3A, again exactly out of phase, and each one out of phase with its corresponding center conductor current. Let's consider electrically short cables for simplicity -- short enough that the currents don't change much along the length. As Tom said, the fields are contained within the coax cables -- on the inside of each one is a canceling pair of currents. But at the ends of the cables, the shield currents can escape to the outside, 1A from cable A, and an out of phase 3A from cable B. The result is a net 2A flowing on the outside of the shield. This current will cause the combined cable to radiate exactly the same amount as if the shield wasn't there at all! All we've done is to physically separate the differential and common mode currents, but we haven't changed their values. So what happens when the conductor currents are balanced, i.e., equal and opposite? Well, then there's no net current on the outside of the shield. Now there's no line radiation, again exactly the same as if the shield wasn't there. The shield likewise has no effect on reception of unwanted signals -- it provides no shielding at all. The only way to make it provide shielding is to block the current path from the inside to the outside of the cables at both ends. In the case of an antenna feedline, however, this would require putting the antenna into a shielded box. The shield has an impact on the differential mode impedance, but it doesn't help line balance or provide shielding at all. Roy Lewallen, W7EL |
Two coax as substitute for open line
Roy Lewallen wrote in
: .... The shield has an impact on the differential mode impedance, but it doesn't help line balance or provide shielding at all. Roy, I quite agree with you. I have received a third hand comment that something in my related article at http://www.vk1od.net/transmissionline/stcm/index.htm is incorrect. I say something because there is a language translation involved in the third hand process. Have you read the article, and have you any issue with it? Thanks Owen |
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