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

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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On Sat, 14 May 2005 09:21:57 -0500, Cecil Moore
wrote:

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.


Thanks again for the examples Cecil, I missed the typo but your
formulas appear to be accurate and that's what I was after.

73 Gary