I've read for years ( and never asked why ) that when you're operating
into a high SWR that a high impedance feedline ( say 450 Ohm ladder
line VS 52 Ohm coax ) provides much less loss. I think I recall
someone in this group saying that its mostly current losses. Does the
high impedance line have higher voltage points across its length and
therefore less current flow for a give power level ( say 100 watts )
than the 52 Ohm coax ?

I guess an analogy if the above is true could be made about the 120Kv
+ power lines on tall steel towers that are about 500 feet behind my
shack. ( Lucky me ! ) They have much less loss than trying to run say
120 volts and all the current flow that would entail for the same
wattage delivered to homes, business etc ? I can imagine the size of
the conductors required to deliver the same amount of wattage at 120V
VS 120 Kv +/-.

Thanks .... Gary

Gary wrote:
I've read for years ( and never asked why ) that when you're operating
into a high SWR that a high impedance feedline ( say 450 Ohm ladder
line VS 52 Ohm coax ) provides much less loss. I think I recall
someone in this group saying that its mostly current losses. Does the
high impedance line have higher voltage points across its length and
therefore less current flow for a give power level ( say 100 watts )
than the 52 Ohm coax ?

The Z0 of a feedline forces the ratio of forward voltage to forward
current to be Z0. It also forces the ratio of reflected voltage to
reflected voltage to be Z0. Let's say we have 100 watts forward and
50 watts reflected on both 450 ohm feedline and 52 ohm coax. The
forward voltage on the 450 ohm feedline is SQRT(100*450). The forward
current on the 450 ohm feedline is SQRT(100/450). The forward voltage
on the 52 ohm coax is SQRT(100*52). The forward current on the 52 ohm
coax is SQRT(100/52). The same pattern holds for reflected signals.
The effect of Z0 on voltage and current is easy to see.
--
73, Cecil http://www.qsl.net/w5dxp


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Gary:

This question is out of my league, suspect Roy will be able to answer with
no problem--probably a lot of the others too...

However, in my expermenting, I have adusted the match at an antenna fed with
300 twin lead (causing a high SWR) while watching Field Strength on a meter
which was positioned so it could only "see" a good section of the
feedline... you could watch radiation from the feedline go up with SWR--I
cringe when they say coax has even more "loss"--or perhaps this is not the
"loss" you mean...

Warmest regards,
John
--
Marbles can be used in models with excellent results! However, if forced
to keep using all of mine up... I may end up at a disadvantage... I seem
to have misplaced some!!!


"Gary" wrote in message
...
| I've read for years ( and never asked why ) that when you're operating
| into a high SWR that a high impedance feedline ( say 450 Ohm ladder
| line VS 52 Ohm coax ) provides much less loss. I think I recall
| someone in this group saying that its mostly current losses. Does the
| high impedance line have higher voltage points across its length and
| therefore less current flow for a give power level ( say 100 watts )
| than the 52 Ohm coax ?
|
| I guess an analogy if the above is true could be made about the 120Kv
| + power lines on tall steel towers that are about 500 feet behind my
| shack. ( Lucky me ! ) They have much less loss than trying to run say
| 120 volts and all the current flow that would entail for the same
| wattage delivered to homes, business etc ? I can imagine the size of
| the conductors required to deliver the same amount of wattage at 120V
| VS 120 Kv +/-.
|
| Thanks .... Gary

On Fri, 13 May 2005 23:33:40 -0500, Cecil Moore
wrote:

Gary wrote:
I've read for years ( and never asked why ) that when you're operating
into a high SWR that a high impedance feedline ( say 450 Ohm ladder
line VS 52 Ohm coax ) provides much less loss. I think I recall
someone in this group saying that its mostly current losses. Does the
high impedance line have higher voltage points across its length and
therefore less current flow for a give power level ( say 100 watts )
than the 52 Ohm coax ?

The Z0 of a feedline forces the ratio of forward voltage to forward
current to be Z0. It also forces the ratio of reflected voltage to
reflected voltage to be Z0. Let's say we have 100 watts forward and
50 watts reflected on both 450 ohm feedline and 52 ohm coax. The
forward voltage on the 450 ohm feedline is SQRT(100*450). The forward
current on the 450 ohm feedline is SQRT(100/450). The forward voltage
on the 52 ohm coax is SQRT(100*52). The forward current on the 52 ohm
coax is SQRT(100/52). The same pattern holds for reflected signals.
The effect of Z0 on voltage and current is easy to see.

Thanks Cecil ! In your example it appears that the coax is carrying
about 3 times the current of the 450 Ohm ladder line. That explains a
lot.

73 Gary

Thanks for the reply and example John. I was referring to the loss in
power when operating into a high SWR with coax VS a high impedance
line like twinlead / ladder line / open wire. Someone in a post here
mentioned it was mainly current losses and that piqued my interest.
Cecil answered my question and gave me the formula. In his example the
coax was carrying about 3 times the current of the 450 Ohm ladder
line, which explains it.

73 Gary

On Fri, 13 May 2005 21:47:40 -0700, "John Smith"
wrote:

Gary:

This question is out of my league, suspect Roy will be able to answer with
no problem--probably a lot of the others too...

However, in my expermenting, I have adusted the match at an antenna fed with
300 twin lead (causing a high SWR) while watching Field Strength on a meter
which was positioned so it could only "see" a good section of the
feedline... you could watch radiation from the feedline go up with SWR--I
cringe when they say coax has even more "loss"--or perhaps this is not the
"loss" you mean...

Warmest regards,
John

You will get all sorts of technical reasons for lower loss. But
essentially -

The wires in high impedance balanced pair lines are thicker than the
inner conductor of coaxial lines.

Thicker wires mean lower resistance.

Lower resistance means lower loss.
----
Reg

Gary wrote:

Cecil Moore wrote:
The Z0 of a feedline forces the ratio of forward voltage to forward
current to be Z0. It also forces the ratio of reflected voltage to
reflected voltage to be Z0. Let's say we have 100 watts forward and
^^^^^^^
50 watts reflected on both 450 ohm feedline and 52 ohm coax. The
forward voltage on the 450 ohm feedline is SQRT(100*450). The forward
current on the 450 ohm feedline is SQRT(100/450). The forward voltage
on the 52 ohm coax is SQRT(100*52). The forward current on the 52 ohm
coax is SQRT(100/52). The same pattern holds for reflected signals.
The effect of Z0 on voltage and current is easy to see.

Thanks Cecil ! In your example it appears that the coax is carrying
about 3 times the current of the 450 Ohm ladder line. That explains a
lot.

Yep, just noticed a typo above where "voltage" should have been
"current" above. Hope that was obvious. The ratio of the current
between 50 ohm coax and 450 ohm ladder-line is indeed 3 to 1 *FOR
EQUAL SWRs*.

Taking it to the next level of understanding, what if the SWRs are
not equal? Let's say we have 50 ohm coax and a 50 ohm load. The
system is matched and current flows only one way. Total current
for 100 watts equals SQRT(100/50) = 1.414 amps.

Now let's feed the 50 ohm load with 450 ohm ladder-line. The SWR
will be 9:1. Forward power is 278 watts and reflected power is
178 watts. Forward current is SQRT(278/450) = 0.79 amps. Reflected
current is SQRT(178/450) = 0.63 amps. Both of those currents cause
total I^2*R losses roughly equivalent to their sum. Their sum is
1.414 amps, the same as the forward current in the matched coax.

So the losses in 50 ohm coax and 450 ohm ladder-line are roughly
equivalent using similar size wire and driving a 50 ohm load.

I'll leave it as an exercise as to what happens when 50 ohm coax
vs 450 ohm ladder-line is used to drive a 450 ohm load.
--
73, Cecil http://www.qsl.net/w5dxp

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Gary wrote:
Thanks for the reply and example John. I was referring to the loss in
power when operating into a high SWR with coax VS a high impedance
line like twinlead / ladder line / open wire. Someone in a post here
mentioned it was mainly current losses and that piqued my interest.
Cecil answered my question and gave me the formula. In his example the
coax was carrying about 3 times the current of the 450 Ohm ladder
line, which explains it.

Remember, that is for the *SAME SWR*. When the SWRs are different,
as they will be for a fixed load, that loss ratio figure will vary
away from 3 to 1.
--
73, Cecil http://www.qsl.net/w5dxp

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Reg Edwards wrote:

You will get all sorts of technical reasons for lower loss. But
essentially -

The wires in high impedance balanced pair lines are thicker than the
inner conductor of coaxial lines.

Thicker wires mean lower resistance.

Lower resistance means lower loss.

That is true. But the number one reason that matched line loss
for 450 ohm ladder-line is lower than matched line loss for RG-213
at HF is the effect of (characteristic impedance = load) which is
the same effect as Ohm's law.

Given RG-213 vs 450 ohm ladder-line the losses are *roughly*
equal when:

SWR(coax)/50 = SWR(ladder-line)/450

or, in general, when:

SWR1/Z01 = SWR2/Z02
--
73, Cecil http://www.qsl.net/w5dxp

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The main reason a high impedance feedline equals lower loss is that the
resistance in the feedline is the main loser. Its loss is current
squared times the total resistance. The same power can be conveyed as a
high voltage and a low current (high impedance), or as a low voltage and
a high current (low impedance).

When conveyed as a high current and a low voltage, the power extracted
by resistance is higher for a given power conveyence. Powerline voltages
are often limited by flashover voltage. That`s a reason for d-c
high-voltage power transmission. D-C is at peak value all of the time,
and needs to be insulated for no higher voltage. Maximum voltage means
minimum current for a given power. Ciurrent squared times the resistance
is lowest too.

Loss increases with line length for a particular cross section. A
rule-of-thumb for powerlines is that your transmission voltage should be
1000 volts per mile the energy is to be transported.

Power is volts x amps x cos theta. Cos theta is the power factor which
should be 1 to minimize total current (zero reactive current). Impedance
of the line is volts / amps.

Best regards, Richard Harrison, KB5WZI