In message , Brian Howie
writes
In message , John S
writes
On 10/1/2015 3:29 AM, Brian Howie wrote:
In message , Wayne
writes


"Ian Jackson" wrote in message
...

In message , rickman
writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff
writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the fact
that the antenna feedpoint impedance was not purely resistive, but
was actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both for 20
ohms resistive, and for 20-j130)?

A quick model.

A vertical antenna about 4.2m long with a wire radius of 0.5mm
approximates to what you have . The devil is in the j130 If you use a
1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a
bad match over the band

Expanding on the original question.... Antenna feedpoint
approximately 20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that
leave the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

No the line SWR is still about 36:1. If you run a bit of poke, you might
melt the coax.

If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+
j87) and use a series C ~130pf to take out the the +j87. You have to
do this at the antenna end.

It goes without saying you need a good ground.

Brian

What do you think of this while leaving his antenna unchanged?

2.3uH
___
'-----o-----UUU---------------
.-. | ^
| | | |
20 | | C| '
'-' C| 4uH
| C| 50 ohms
| |
--- | .
-130--- | |
| | v
'-----o------------------------
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


Yes that works too. Lots of ways to do it.

I'm taking the opportunity to refresh my admittedly rudimentary skills
with the Smith Chart - and in particular, single- and double-stub
matching (little used for over 50 years!!).

However, in Wayne's situation, the length of the coax is only 15'. On
14MHz, that's just over a quarterwave (taking the velocity factor into
account). But even with a horrendous SWR, how much loss does this length
of 213 coax have? It might be a lot more convenient to do all the
matching in the warmth and comfort of the shack.

I have done exactly this with a 130' inverted-L Marconi-type antenna,
fed at the far end directly with around 100' of old (early 1960s)
semi-airspaced TV trunk cable (with a good ground there). It worked fine
on 160-80-40m (the bands I was interested in working), but it loaded up
fine up to 10m - and as it seemed lively enough on receive, I'm sure it
would have put out a reasonable signal. Although I eventually treated
myself to a remote automatic ATU, I'm not convinced the system works any
better than it did with the direct coax connection.
--
Ian

On 10/1/2015 2:37 PM, John S wrote:
On 10/1/2015 12:27 PM, rickman wrote:
On 10/1/2015 1:21 PM, John S wrote:
On 10/1/2015 12:18 PM, rickman wrote:
On 10/1/2015 1:09 PM, John S wrote:
On 10/1/2015 11:56 AM, Brian Howie wrote:
In message , John S
writes
On 10/1/2015 3:29 AM, Brian Howie wrote:
In message , Wayne
writes


"Ian Jackson" wrote in message
...

In message , rickman

writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff

writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the
fact
that the antenna feedpoint impedance was not purely resistive,
but
was actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both
for 20
ohms resistive, and for 20-j130)?

A quick model.

A vertical antenna about 4.2m long with a wire radius of 0.5mm
approximates to what you have . The devil is in the j130 If you
use a
1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and
not a
bad match over the band

Expanding on the original question.... Antenna feedpoint
approximately 20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that
leave the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

No the line SWR is still about 36:1. If you run a bit of poke, you
might
melt the coax.

If you want a single 50 ohm feed, extend the antenna to 5.7m (
~50+
j87) and use a series C ~130pf to take out the the +j87. You
have to
do this at the antenna end.

It goes without saying you need a good ground.

Brian

What do you think of this while leaving his antenna unchanged?

2.3uH
___
'-----o-----UUU---------------
.-. | ^
| | | |
20 | | C| '
'-' C| 4uH
| C| 50 ohms
| |
--- | .
-130--- | |
| | v
'-----o------------------------
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


Yes that works too. Lots of ways to do it.

Brian

Oh, yes of course. I try to avoid capacitors as much as I can because
one end can float with static voltage while an inductor does not. I've
had issues with nearby static lightning discharges.

Just my paranoia. To each his own.

But it is connected by a 20 ohm resistor. How bad can that be?


I don't understand. Do you mean the antenna's feed point resistance of
20 ohms? My understanding of the installation is that the antenna is not
directly connected to ground. Am I off track here?

My bad. I didn't realize that was the antenna. But the capacitor could
be bypassed with a large value resistor if static charge is your
concern. A kohm should do the job without impacting the circuit
significantly. But wait! Isn't the -130 cap also the antenna then?


Yes. I guess I should have enclosed the combination in a box to
represent the antenna. Sorry.

So there is no cap, right? No cap, no worry. Oh, wait again. I see
there are two circuits being discussed. So *that* cap can be bypassed
with a 1 kohm resistor and not impact the circuit, right?

--

Rick

On 10/1/2015 2:41 PM, rickman wrote:
On 10/1/2015 2:37 PM, John S wrote:
On 10/1/2015 12:27 PM, rickman wrote:
On 10/1/2015 1:21 PM, John S wrote:
On 10/1/2015 12:18 PM, rickman wrote:
On 10/1/2015 1:09 PM, John S wrote:
On 10/1/2015 11:56 AM, Brian Howie wrote:
In message , John S

writes
On 10/1/2015 3:29 AM, Brian Howie wrote:
In message , Wayne
writes


"Ian Jackson" wrote in message
...

In message , rickman

writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff

writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the
fact
that the antenna feedpoint impedance was not purely resistive,
but
was actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both
for 20
ohms resistive, and for 20-j130)?

A quick model.

A vertical antenna about 4.2m long with a wire radius of 0.5mm
approximates to what you have . The devil is in the j130 If you
use a
1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and
not a
bad match over the band

Expanding on the original question.... Antenna feedpoint
approximately 20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that
leave the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

No the line SWR is still about 36:1. If you run a bit of poke, you
might
melt the coax.

If you want a single 50 ohm feed, extend the antenna to 5.7m (
~50+
j87) and use a series C ~130pf to take out the the +j87. You
have to
do this at the antenna end.

It goes without saying you need a good ground.

Brian

What do you think of this while leaving his antenna unchanged?

2.3uH
___
'-----o-----UUU---------------
.-. | ^
| | | |
20 | | C| '
'-' C| 4uH
| C| 50 ohms
| |
--- | .
-130--- | |
| | v
'-----o------------------------
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


Yes that works too. Lots of ways to do it.

Brian

Oh, yes of course. I try to avoid capacitors as much as I can because
one end can float with static voltage while an inductor does not.
I've
had issues with nearby static lightning discharges.

Just my paranoia. To each his own.

But it is connected by a 20 ohm resistor. How bad can that be?


I don't understand. Do you mean the antenna's feed point resistance of
20 ohms? My understanding of the installation is that the antenna is
not
directly connected to ground. Am I off track here?

My bad. I didn't realize that was the antenna. But the capacitor could
be bypassed with a large value resistor if static charge is your
concern. A kohm should do the job without impacting the circuit
significantly. But wait! Isn't the -130 cap also the antenna then?


Yes. I guess I should have enclosed the combination in a box to
represent the antenna. Sorry.

So there is no cap, right? No cap, no worry. Oh, wait again. I see
there are two circuits being discussed. So *that* cap can be bypassed
with a 1 kohm resistor and not impact the circuit, right?


No, there is only one cap. The 20 ohm resistor in series with the -130
ohm cap represents the antenna impedance of 20 - j130 ohms.

On 10/1/2015 1:56 PM, Ian Jackson wrote:
In message , Brian Howie
writes
In message , John S
writes
On 10/1/2015 3:29 AM, Brian Howie wrote:
In message , Wayne
writes


"Ian Jackson" wrote in message
...

In message , rickman
writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff
writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the fact
that the antenna feedpoint impedance was not purely resistive, but
was actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both for 20
ohms resistive, and for 20-j130)?

A quick model.

A vertical antenna about 4.2m long with a wire radius of 0.5mm
approximates to what you have . The devil is in the j130 If you use a
1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and
not a
bad match over the band

Expanding on the original question.... Antenna feedpoint
approximately 20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that
leave the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

No the line SWR is still about 36:1. If you run a bit of poke, you
might
melt the coax.

If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+
j87) and use a series C ~130pf to take out the the +j87. You have to
do this at the antenna end.

It goes without saying you need a good ground.

Brian

What do you think of this while leaving his antenna unchanged?

2.3uH
___
'-----o-----UUU---------------
.-. | ^
| | | |
20 | | C| '
'-' C| 4uH
| C| 50 ohms
| |
--- | .
-130--- | |
| | v
'-----o------------------------
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


Yes that works too. Lots of ways to do it.

I'm taking the opportunity to refresh my admittedly rudimentary skills
with the Smith Chart - and in particular, single- and double-stub
matching (little used for over 50 years!!).

However, in Wayne's situation, the length of the coax is only 15'. On
14MHz, that's just over a quarterwave (taking the velocity factor into
account). But even with a horrendous SWR, how much loss does this length
of 213 coax have? It might be a lot more convenient to do all the
matching in the warmth and comfort of the shack.

He is already doing that with his ATU and is pretty happy with it. And
he has indicated that he doesn't want to add matching components up at
the feed point. Brian and I were just bouncing ideas back and forth.

I have done exactly this with a 130' inverted-L Marconi-type antenna,
fed at the far end directly with around 100' of old (early 1960s)
semi-airspaced TV trunk cable (with a good ground there). It worked fine
on 160-80-40m (the bands I was interested in working), but it loaded up
fine up to 10m - and as it seemed lively enough on receive, I'm sure it
would have put out a reasonable signal. Although I eventually treated
myself to a remote automatic ATU, I'm not convinced the system works any
better than it did with the direct coax connection.

On 10/1/2015 9:16 PM, John S wrote:
On 10/1/2015 2:41 PM, rickman wrote:
On 10/1/2015 2:37 PM, John S wrote:
On 10/1/2015 12:27 PM, rickman wrote:
On 10/1/2015 1:21 PM, John S wrote:
On 10/1/2015 12:18 PM, rickman wrote:
On 10/1/2015 1:09 PM, John S wrote:
On 10/1/2015 11:56 AM, Brian Howie wrote:
In message , John S

writes
On 10/1/2015 3:29 AM, Brian Howie wrote:
In message , Wayne
writes


"Ian Jackson" wrote in message
...

In message , rickman

writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff

writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the
fact
that the antenna feedpoint impedance was not purely resistive,
but
was actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both
for 20
ohms resistive, and for 20-j130)?

A quick model.

A vertical antenna about 4.2m long with a wire radius of 0.5mm
approximates to what you have . The devil is in the j130 If you
use a
1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1
and
not a
bad match over the band

Expanding on the original question.... Antenna feedpoint
approximately 20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does
that
leave the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

No the line SWR is still about 36:1. If you run a bit of poke,
you
might
melt the coax.

If you want a single 50 ohm feed, extend the antenna to 5.7m (
~50+
j87) and use a series C ~130pf to take out the the +j87. You
have to
do this at the antenna end.

It goes without saying you need a good ground.

Brian

What do you think of this while leaving his antenna unchanged?

2.3uH
___
'-----o-----UUU---------------
.-. | ^
| | | |
20 | | C| '
'-' C| 4uH
| C| 50 ohms
| |
--- | .
-130--- | |
| | v
'-----o------------------------
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


Yes that works too. Lots of ways to do it.

Brian

Oh, yes of course. I try to avoid capacitors as much as I can
because
one end can float with static voltage while an inductor does not.
I've
had issues with nearby static lightning discharges.

Just my paranoia. To each his own.

But it is connected by a 20 ohm resistor. How bad can that be?


I don't understand. Do you mean the antenna's feed point resistance of
20 ohms? My understanding of the installation is that the antenna is
not
directly connected to ground. Am I off track here?

My bad. I didn't realize that was the antenna. But the capacitor
could
be bypassed with a large value resistor if static charge is your
concern. A kohm should do the job without impacting the circuit
significantly. But wait! Isn't the -130 cap also the antenna then?


Yes. I guess I should have enclosed the combination in a box to
represent the antenna. Sorry.

So there is no cap, right? No cap, no worry. Oh, wait again. I see
there are two circuits being discussed. So *that* cap can be bypassed
with a 1 kohm resistor and not impact the circuit, right?


No, there is only one cap. The 20 ohm resistor in series with the -130
ohm cap represents the antenna impedance of 20 - j130 ohms.

Way back in the thread...

"If you want a single 50 ohm feed, extend the antenna to 5.7m (~50+
j87) and use a series C ~130pf to take out the the +j87. You have to
do this at the antenna end."

Otherwise there is no cap to worry about blowing up.

--

Rick

On 10/2/2015 2:34 AM, Jeff wrote:
___
'-----o-----UUU---------------
.-. | ^
| | | |
20 | | C| '
'-' C| 4uH
| C| 50 ohms
| |
--- | .
-130--- | |
| | v
'-----o------------------------
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


Yes that works too. Lots of ways to do it.

Brian

Oh, yes of course. I try to avoid capacitors as much as I can because
one end can float with static voltage while an inductor does not. I've
had issues with nearby static lightning discharges.

Just my paranoia. To each his own.

But it is connected by a 20 ohm resistor. How bad can that be?


I don't understand. Do you mean the antenna's feed point resistance of
20 ohms? My understanding of the installation is that the antenna is not
directly connected to ground. Am I off track here?

My bad. I didn't realize that was the antenna. But the capacitor could
be bypassed with a large value resistor if static charge is your
concern. A kohm should do the job without impacting the circuit
significantly. But wait! Isn't the -130 cap also the antenna then?


What capacitor???????????

The 20-j130 is the impedance presented by the antenna at the feed point
to gnd, there is no physical capacitor, the 4uH to gnd will provide a dc
path for any static.

Yes, that is what I said. "Isn't the -130 cap also the antenna then?"

I was confused because of poor trimming when someone posted about a
different matching network using a capacitor and a reply saying they
don't like capacitors because they fail. With the context not being
clear I thought they were talking about the capacitor in the diagram
with the inductive matching network, but as you say, this capacitance is
just part of the antenna.

--

Rick

On 10/2/2015 8:14 AM, Jeff wrote:
On 02/10/2015 07:47, Jeff wrote:

However, in Wayne's situation, the length of the coax is only 15'. On
14MHz, that's just over a quarterwave (taking the velocity factor into
account). But even with a horrendous SWR, how much loss does this length
of 213 coax have? It might be a lot more convenient to do all the
matching in the warmth and comfort of the shack.

You still have to bear in mind that the 20-j130 load gives something
like a 20:1 mismatch at the interface between the coax and the load.

This means that only about 17% of your froward power will reach the
antenna on its 1st trip up and down the coax, the other 83% will be
reflected back down the coax towards the ATU and suffer the loos in the
coax again.

This will be repeated again and again, with 83% of what is re-reflected
being re-reflected again, and so on. So you can see that even with a
small loss in the coax if is amplified by the multiple trips up and down
the coax. This higher the VSWR the more power is re-reflected to undergo
multiple losses.

Jeff

Having done a few quick calculations; with a vswr of 20:1 at the
antenna, and a feeder having a loss of 0.5dB, and an ATU matching at the
Tx end, the total power wasted in the cable with a 100W transmitter is
over 55W, with only about 45W actually reaching the antenna.

Even reducing the cable loss to 0.1dB results in about 20W being lost in
the cable.

Reducing the VSWR reduces the loss significantly.

What is the reflection coefficient at the ATU/feedline interface?

--

Rick

On 10/3/2015 1:29 AM, Jeff wrote:

Having done a few quick calculations; with a vswr of 20:1 at the
antenna, and a feeder having a loss of 0.5dB, and an ATU matching at the
Tx end, the total power wasted in the cable with a 100W transmitter is
over 55W, with only about 45W actually reaching the antenna.

Even reducing the cable loss to 0.1dB results in about 20W being lost in
the cable.

Reducing the VSWR reduces the loss significantly.

What is the reflection coefficient at the ATU/feedline interface?


Assuming a 'perfect' ATU and a prefect conjugate match, giving 1:1 at
the TX, then it is 1. ie all of the reflected power that reaches the ATU
is re-reflected back up towards the antenna.

Is that good? If the cable length is 1/4 wavelength (as it is in the
info provided for this case) the reflected power is nearly 180 degrees
out of phase with the initial power at the antenna. I think reducing
this through cable losses would not be so bad, or better to dump it in
the ATU?

--

Rick

On 10/4/2015 4:48 AM, Jeff wrote:

What is the reflection coefficient at the ATU/feedline interface?


Assuming a 'perfect' ATU and a prefect conjugate match, giving 1:1 at
the TX, then it is 1. ie all of the reflected power that reaches the ATU
is re-reflected back up towards the antenna.

Is that good? If the cable length is 1/4 wavelength (as it is in the
info provided for this case) the reflected power is nearly 180 degrees
out of phase with the initial power at the antenna. I think reducing
this through cable losses would not be so bad, or better to dump it in
the ATU?

It is good in as much as some of the re-reflected power is radiated (and
some re-re-reflected) since the ATU causes the phase of the re-reflected
wave to be 'in-phase' at the antenna.

It is bad in as much as the reflected power suffers 2 times the cable
loss, and dissipates that in heat, on each return trip, up and down the
coax.

With a high VSWR at the antenna there will be many return trips before
the re-reflected power drops to a negligible level.

No power is 'dumped' in the ATU; although there will be losses, but that
is another story.

If not the ATU, then the transmitter. I'm sure not all of the power is
reflected back from the ATU. Exactly what is the phase of the reflected
power from the ATU? I haven't seen an actual circuit for the ATU in
question. For that matter, what is the phase of the power reflected
from the antenna? I'm pretty confident we are not looking at the return
of the reflected wave in phase with the incident wave.

--

Rick

rickman wrote:

On 10/4/2015 4:48 AM, Jeff wrote:

What is the reflection coefficient at the ATU/feedline interface?


Assuming a 'perfect' ATU and a prefect conjugate match, giving 1:1 at
the TX, then it is 1. ie all of the reflected power that reaches the ATU
is re-reflected back up towards the antenna.

Is that good? If the cable length is 1/4 wavelength (as it is in the
info provided for this case) the reflected power is nearly 180 degrees
out of phase with the initial power at the antenna. I think reducing
this through cable losses would not be so bad, or better to dump it in
the ATU?

It is good in as much as some of the re-reflected power is radiated (and
some re-re-reflected) since the ATU causes the phase of the re-reflected
wave to be 'in-phase' at the antenna.

It is bad in as much as the reflected power suffers 2 times the cable
loss, and dissipates that in heat, on each return trip, up and down the
coax.

With a high VSWR at the antenna there will be many return trips before
the re-reflected power drops to a negligible level.

No power is 'dumped' in the ATU; although there will be losses, but that
is another story.

If not the ATU, then the transmitter. I'm sure not all of the power is
reflected back from the ATU. Exactly what is the phase of the reflected
power from the ATU? I haven't seen an actual circuit for the ATU in
question. For that matter, what is the phase of the power reflected
from the antenna? I'm pretty confident we are not looking at the return
of the reflected wave in phase with the incident wave.

Having specified the transmitter power output, by definiton, no power
can be "lost" in the amplifier, because transmitter power output is
*defined* as the net power it actually manages to get out of its output
socket net of any reflections. The mismatch may make it harder for the
transmitter to achieve that; but, by definition, whatever it does
achieve is its power output. You can't draw a valid defiinition of a
difference between power it never produced at all and power sent back to
it.

--
Roger Hayter