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VK2ZRG > TECH 10.08.04 12:57l 59 Lines 2944 Bytes #999 (0) @ WW
BID : 813_VK2ZRG
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Subj: TX lines
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Sent: 040810/0724Z @:VK2WI.#SYD.NSW.AUS.OC #:45104 [SYDNEY] FBB7 $:813_VK2ZRG
From: VK2ZRG@VK2WI.#SYD.NSW.AUS.OC
To : TECH@WW
VK2ZRG/TPK 1.83d Msg #:813 Date:10-08-04 Time:8:15Z
Hello all techies,
I've been writing a small programme to calculate transmission
lines in the last few weeks. I've added four lines of code to calculate loss
at a nominated frequency. The calculated loss is the sum of copper loss and
dielectric loss.
In the past I had wondered why balanced transmission lines had very much
lower loss than coaxial lines. It now seems clear that the reason why this
is true has nothing to do with balanced lines being balanced and little to
do with a balanced line being air spaced. The real reason is simply that
balanced lines are high impedance and coaxial lines are low impedance.
If you compare a 200 ohm balanced line with a 200 ohm air spaced coaxial
line having a centre conductor the same diameter as wires in the balanced
line, the loss of the coaxial line will be close to half (around 52%) that
of the balanced line!
Also the reason why foam coaxial cables have lower losses than their
equivalent solid PE cable, has little to do with losses in the dielectric
for frequencies of less than 2 GHz. The real reason is that foam cables have a
larger centre conductor which means less copper loss.
Am I right? All comments appreciated.
I've also been wondering about what is the correct way to calculate the
Zo of a square coaxial cable. i.e A round inner conductor in a square tube.
There is a formula on page 5-36 of the ARRL UHF / Microwave Handbook that
simply multiplies the ratio of outer to inner dimensions by 1.08 for the
standard formula for a round cable. This is the formula.
Zo = 60 / ûEr * LN(1.08*(Outside /Inner))
This formula is probably adequate for most practical cases but obviously
is wrong for very low impedances. Multiplying the ratio by 1.08 is the same
as adding 4.62 ohms to the result from the standard formula for round cables.
So for the case where inner diameter equals outer diameter, the Zo for a
round cable is 0 ohms and 4.62 ohms for a square cable by this formula.
This cannot be true, both must be zero ohms. The solution to the problem
lies in the calculation of the capacitance of a round conductor in a square
tube and then finding an equivalent outer diameter for a round cable with
the same size inner conductor.
Does anyone have a formula for doing this? I have a formula for round
cables. This is it.
Capacity = 55.555 * Er / LN(OutSideDia/InnerDia) pF per metre.
I have a programme called APPCAD that came as freeware from Agilent (AKA
Hewlett Packard in times gone by). You may be able to find it on their web
site. This has a section that can calculate a number of different
transmission lines including square coaxial types. It would appear that
APPCAD uses the 1.08 factor for square coaxial line too.
73s from Ralph VK2ZRG@VK2W1.#SYD.NSW.AUS.OC
/ack
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