Kevin, I suspected that the formula may present a problem (it shows up ok on our in-house network).
The formula is: Rho = [1/(2*Pi*sigma*delta)]*[(1/rhob)+(1/rhoa)] in Ohms/meter Where, Rho is attenuation constant Sigma is conductivity Delta is skin depth Rhoa is the radius if the inner shield Rhob is the radius of the primary conductor Speaking for Ted, since he is not on this list server, he is saying that you should look at the ratio of (1/rhoa) and (1/rhob) to get a feel for the relative importance of each term. (1/rhoa) will be a much larger term, indicating that most loss is from the primary conductor, not the shield return. Hence, changing the shield shouldn't have great impact on the loss. Jim ------------------------------------------------------------------------ --------------------------- Dr. Jim Knighten NCR 17095 Via del Campo San Diego, CA 92127 Telephone: 619-485-2537 Fax: 619-485-3788 e-mail: [email protected] ---------- From: Kevin Harris [SMTP:[email protected]] Sent: Friday, July 31, 1998 12:40 PM To: 'Knighten, Jim' Subject: RE: Twisted Shielded Pair Hi Jim, It maybe just me but the "simple formula" appears as blank space on my computer Regards, Kevin Harris > -----Original Message----- > From: Knighten, Jim [SMTP:[email protected]] > Sent: Friday, July 31, 1998 12:27 PM > To: 'emc-pstc list server' > Cc: DiBene, Ted > Subject: FW: Twisted Shielded Pair > > Jeff, > > I pass this along in hopes that it helps you with your question. > > Jim Knighten > ---------------------------------------------------------------------- > -- > --------------------------- > Dr. Jim Knighten > NCR > 17095 Via del Campo > San Diego, CA 92127 > Telephone: 619-485-2537 > Fax: 619-485-3788 > e-mail: [email protected] > > > ---------- > From: DiBene, Ted > Sent: Friday, July 31, 1998 9:11 AM > To: Knighten, Jim > Subject: RE: Twisted Shielded Pair > > Jim; I have an answer but I am not sure it is the answer he is > looking > for. I believe he is really talking about the reduction in overall > attenuation from increasing the braid coverage to reduce return loss. > In going from 85% to 90% shield coverage the change in return loss is > not significant. Neither is it the dominant source of conductive > losses. Even if this was a single-ended communication link with low > impedance (say 50 ohms) the return loss would still be only a fraction > of the overall conduction loss. Since we are now talking about a > twisted pair and a differential communication path, if the signals are > well balanced then the signal and return paths are intrinisically the > primaries. However, if there is strong even mode coupling in > structure > then there will be currents on the shield that will contribute to the > overall conduction losses. This, of course is very much dependent > upon > this modal impedance which is construction dependent. If the signals > are imbalanced then, as you know, the common-mode current return path > is > the shield and the overall losses may be increased slightly due to > signal energy loss. > > Basically, this is a very difficult question to answer accurately > since > it involves so many variables relating to construction and signal > propagation. However, all the theory aside, there is a simple formula > he can use that is really for approximations to coaxial line return > loss. The differential mode will be much different and thus the loss > contribution will be different however if he is looking for a swag, > here > it is: > > Assuming that your friend knows how to determine the attenuation > constant as function of resistive losses one can use the simple > formula > below; > > > > where rb is the radius of the inner shield and ra is the radius of the > primary conductor. If he takes the first part of the formula and > ratios > this with second he can get a feel for the percentage of loss > differences. Again, this is assuming the frequencies are high enough > to > consider the dominant losses to be due to skin effect. The DC > resistance calculation is much more difficult with a twisted wire and > braided shield construction. The formula above is a very crude > estimate > and will almost always be worst case but I think it is a good start. > > Ted. > > ---------- > From: Knighten, Jim > Sent: Thursday, July 30, 1998 5:33 PM > To: DiBene, Ted > Subject: FW: Twisted Shielded Pair > Importance: High > > Ted, > > You may have an answer to this guy's question. If so, send it > to me and I will post it. > > Jim K. > > ---------------------------------------------------------------------- > -- > --------------------------- > Dr. Jim Knighten > NCR > 17095 Via del Campo > San Diego, CA 92127 > Telephone: 619-485-2537 > Fax: 619-485-3788 > e-mail: [email protected] > > > ---------- > From: [email protected] [SMTP:[email protected]] > Sent: Thursday, July 30, 1998 3:13 PM > To: [email protected] > Subject: Twisted Shielded Pair > > I am looking for a good rule of thumb regarding attenuation > versus shield coverage for a twisted shielded pair cable. > Specifically > what level of attenuation change will be expected in going from a 85% > coverage to 90% coverage. > Thanks in advance > Jeff Duvall
