FreeLists / si-list / [SI-LIST] Re: Metallic Transmission (UTP modelling)

["Chris Chalmers" <cchalmers@xxxxxxxxxxx>]

John and others,

Thanks for replying,

In Dr Johnson's book he has some worst case figures
for cat 5E wire.  Its stated that

Ro = 1.253 ohms/m and zo = 85ohms ( you rightly stated)

From equation 3.115 this gives,

Attenuation = 20.4 db/100m

You said that you would need to include the DC losses.

These are

  DC attenuation = 4.34  *(Rdc / Z0)

worst case Rdc = 0.1876 ohm/m from Dr J

  DC attenuation = 0.958 ohm/100m

This is the attenuation in the RC region.  Can you just
 add the two figures together?  the skin effect resistance
 formula does not have this included?

This would give  21.36db/100m  + dielectric losses

Chris







Chris,

Figure 8.2 page 464 is based on the EIA568B specification, and are the
worst case values
for a CAT5E cable, this corresponds to the data from Addison cables :-

http://www.addison-tech.com/Addison%20Product%20Specification%20Download/Eng
lish%20Version/Lan%20Cable/UTP%20CAT5E%20CABLE.pdf

Cat 5E cable has Zo(min) = 85 Ohms (100Ohms +/- 15) so  3.115 yeilds

Loss =  19.2dB/100m

add in the DC loss (0.17Ohms/m) and dielectric losses and you'll be in
the right ball park

Regards

John




Chris Chalmers wrote:

>Si List,
>    I have been trying to calculate the attenuation in the skin effect
> loss region of a piece of CAT 5E cable operating at 100MHz however
>I am not happy with the answer I am getting.  I am working from Dr
Johnson's
>High
> speed Signal Propagation Book.  I am not sure whether I am
> applying the formula correctly.
>
>     I have tried to take the example in the HSSP book to ,
>   calulate the attenuation ( in the skin effect region) at 100MHz for Cat
>5E cable.
>
>      taking Ro = 1.189ohms/m  24 AWG    from formula on page 461
>      and wo  = 10MHz
>
>  attenuation  in skin effect region  from page 192
>
>  attenuation db/m = 4.34 (Ro/Zo) * SQRT(w/wo)    (equation Equation 3.115)
>
>                           = 4.34 (1.189/100) * SQRT ( (2x PI x 100000000)/
>2 x PI 10000000)
>                           = 4.34 (0.01189) *  3.1623
>                           = 0.163  db/m
>
>  From the graph on page 464 for cat 5E, the attenuation looks like
>22db/100m
>  not  16.3 db/100m like I have calculated.
>
>   The example in the book works out the values Ro and Wo for Cat 3 cable
>but I
>   can't see how the terms in it could be any different for Cat 5E cable
>since the
>   terms in it are mostly to do with the AWG of the wire.
>
>  I am possibly being very silly here but any help you can provide me with
>   would help loads.
>
>  For those of you who don't have this excellent book, a sample chapter of
>this
>   book is available at http://www.phptr.com/articles/article.asp?p=101149.
>   Well worth buying!
>
>  Best Regards
>
>  Chris
>



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