04-05-12: 2chnlben
How then would you suggest I utilize my Kimber wire. I have plenty of it to work with. The 4tc does not provide a lot of copper (gauge). I want to increase the amount of copper (gauge) and thus my desire to use four sets per speaker.
Would I need to take two runs of the 4tc and twist them together...then at the ends, separate and combine the negative wires and the positive wires from the two runs. Repeating this procedure for both the top and bottom sets of the speaker's binding posts...? Sounds like a (good) plan! Best regards, -- Al |
Thanks Al.
How then would you suggest I utilize my Kimber wire. I have plenty of it to work with. The 4tc does not provide a lot of copper (gauge). I want to increase the amount of copper (gauge) and thus my desire to use four sets per speaker.
Would I need to take two runs of the 4tc and twist them together...then at the ends, separate and combine the negative wires and the positive wires from the two runs. Repeating this procedure for both the top and bottom sets of the speaker's binding posts...? |
04-05-12: 2chnlben
I want to use each "braided pair" (one complete run of the 4tc) per speaker terminal (2 runs - one to each of the top/bottom positives, and 2 runs - one to each of the top/bottom negatives). Hi Ben, Thanks very much for the nice words. What you are describing is not what I was interpreting your original post to mean. And I wouldn't recommend doing what you are now describing. If you use all of the conduction paths within a single cable to carry just one polarity of the signal (+ or -), you are essentially redesigning the cable to have very different characteristics than what it was designed to provide. Inductance will increase considerably. Also, the effect that Kijanki described about noise being picked up and entering the feedback loop of the amplifier, if it has one, becomes a significantly greater possibility. Capacitance will be reduced, but capacitance is relatively unimportant for a speaker cable as long as it is not extremely high. Twisting with another cable carrying the opposite polarity will reduce the magnitude of the increase in inductance, and the likelihood of noise-related issues, but just somewhat. Best regards, -- Al |
04-04-12: Jbuenech
I have Kimber 3033 and tried to protect them from dust by wrapping them up in a thin plastic sheet, those used to keep food in the fridge. Result was a huge change in sound for the worst. I suppose the reason for that might have been static electricity on the plastic sheet, which is a possible issue that I hadn't thought of when I composed my previous responses. Not sure how or if it may be possible to quantify or predict that risk. Ben, Digikey has a selection of tapes having anti-static properties, which it would probably be a good idea to look through. Under "tape type" select one of the three anti-static entries, and then click "apply filters." Best regards, -- Al |
Jbuenech - that's good input and certainly makes me rethink my original strategy.
Al and Kijanki - You guys are both so knowledgeable and I always look forward to your intelligent responses.
The Kimber 4tc, as you likely know, is comprised of multiple twisted ("braided") wires including both the positive and negative runs. I want to use each "braided pair" (one complete run of the 4tc) per speaker terminal (2 runs - one to each of the top/bottom positives, and 2 runs - one to each of the top/bottom negatives). My thought was to run the four runs parallel and evenly spaced. Perhaps I should twist the runs.
Please help me determine the best approach to twisting:
1) twist the top negative and top positive separately and twist the bottom negative and bottom positive separately.
2) Do step (1) - above - and then twist those two "twisted pairs" together to form one large twisted run (with all four cables).
3) Twist the two positive runs together; twist the two negatives together; and then twist those two twisted pairs together...
4) Don't twist - run parallel/evenly spaced.
5) Other...
Thanks guys.
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Al, that's perhaps why some manufacturers use foam Teflon (including my Acoustic Zen Absolute ICs) - not only to lower dielectric constant but to reduce absorption (introducing air). As for driving 1.5nF - I'll take your word, having limited experience. |
So what else is there (other than resistance, inductance or dielectric) to consider? Nothing else that I can think of, beyond what we've already said. I don't think that dielectric absorption will be an issue. The electric field between adjacent paralleled cables will be minimal, partly because of the one-inch distance, and partly because of the cancellation resulting from the twisted and braided construction of the two conductors in each of the four paralleled cables. Also, depending on which of the two physical configurations I asked about is used, the dielectric between paralleled cables may be mostly air, which as I understand it has essentially no susceptibility to dielectric absorption. Total capacitance (between + and -) for a 2.5 meter run (about 8.2 feet) would be 1448 pf, somewhat on the high side but I suspect still low enough to not have audible consequences with the great majority of amplifiers. Best regards, -- Al |
Al, I didn't notice the link you provided. I agree that 22nH per foot would be extremely low and we can take it out of the equation. So what else is there (other than resistance, inductance or dielectric) to consider?
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Hi Kijanki,
I suspect that you are basing those calculations on typical numbers for two conductors that are not twisted. See the link in my previous post, though, which shows a photo and the specs of the specific cable. And note that the inductance number is for 2.5 meters.
So due to the braided and twisted construction of the cable, it appears that the inductance is only about 87 nH per foot-run, for a single cable. A double-shotgun configuration (four two-conductor cables in parallel for each speaker) would reduce that to about 22 nH per foot-run, which of course is extremely low.
Best regards,
-- Al |
Al, Yes part of the noise would be cancelled. Twisting just offers better cancellation.
Reduction of inductance might be more important than reduction in resistance since xover inductor in series with the woofer has most likely around 0.08ohm limiting DF to 100. Assuming he can cut inductance by 2 it will be 200nH per ft less, making two runs of 10' equal 4uH equivalent to 0.5ohm reactance reduction at 20kHz. |
Hi Ben, I'm not sure if you are saying that the four cables would be laid out between two parallel strips of wide tape, so that most of what is between adjacent cables is air, or if the tape would be very thickly wound around each cable, so that most of what is between them is tape. In either case, though, I don't think that the dielectric properties of the tape would have much if any relevance. Although of course I would be more certain of that in the first of those two situations. Given that each of the cables has its + and - conductors closely and extensively intertwined, and considering the one inch spacing between cables, I would expect any interaction between adjacent cables to be insignificant compared to what occurs between the intertwined conductors, and the dielectric effects of their individual insulation. So I would expect the net effects of what you are proposing to simply be a factor of 4 reduction of the resistance and inductance that is specified for an individual cable, and an increase in the specified capacitance by a factor of 4. None of those changes strike me as unreasonable in any way, with this particular cable, but your guess is as good as mine as to what the resulting sonic effects will be. P.S: Just saw Kijanki's response. Although the four cables would pick up noise unequally, wouldn't the noise picked up by any one of them be cancelled just as well as if it were the only cable that was present? Best regards, -- Al |
2chnlben , What is your objective? Dielectric properties of the tape (you could use Teflon tape) are secondary IMHO to fact that you defeat shielding of the cable achieved by twisting. Twisting two wires exposes them equaly to fields (noise) and since induced identical currents cancel it becomes very effective shielding. Output of your amp is most likely also an input especially at high frequencies. |
Hi Ben, here is my experience. I have Kimber 3033 and tried to protect them from dust by wrapping them up in a thin plastic sheet, those used to keep food in the fridge.
Result was a huge change in sound for the worst. Harsh, unmusical,..., forcing me to remove it. Once done, good sound returned.
I admit I covered the whole lenght of the cables with the plastic sheet. Maybe using a short lenght of tape might not have the same effect.
Good luck.
Chema |
