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Author Topic: Antenna tuners?  (Read 3540 times)
KG4NEL
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« Reply #15 on: May 21, 2012, 09:52:27 AM »

Quote
The tuners always lose power from around 5% and up. A 10% loss is very common and even higher loss on unusual loads is common. Add this to the loss you already have on the feedline before it gets to the tuner.

Not to say that the voltages can't get significant with even a 5% loss in the tuner, but isn't that inaudible on the receiving end?

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W5DQ
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« Reply #16 on: May 21, 2012, 11:37:00 AM »

In regards to W5FYI's comment about 1.4 is only about 2.78 percent loss.  Is there a chart to that or how did you come up with that percent?


~Kirk, KA3VEZ

Just remember this one point and you'll be fine. Almost every modern transceiver (solid state - no tubes) will easily handle a 2 to 1 SWR and with the SWR @ 2:1, that is only an 11% loss of output power.  Keep your SWR below 2:1 and you should not have any probelms in most cases.

In regards to an antenna tuner, like has been stated, the tuner doesn't actually tune anything. It matches the impedance of the transceiver to that seen on the TX end of the feedline. That brings a whole set of other problems to mitigate in many situations like high SWR on coax feedline, high power mismatches causing arcing and burns inside the tuner, the impedance matching range and power level handing capabilities of the tuner itself, bands of coverage vs multiband antennas just to name a few.

Many QRO amplifiers will run ok at 2:1 but some of the processor controlled Alphas will shutdown (self protect) much lower than 2:1, many at 1.5:1 or less.

Years ago, many of old farts using transceivers with tube finals didn't even know what our SWR was so long as the power out was what it was supposed to be (or close to it) and at that operating level, the measured plate / grid currents were within the acceptable range. If those numbers were ok, we'd go for it and transmit. When I first started out, I used to tune up in to a 100W light bulb and I'm sure the SWR wasn't even close to 50 ohms there Smiley

Gene W5DQ
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Gene W5DQ
Ridgecrest, CA - DM15dp
www.radioroom.org
WB6BYU
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« Reply #17 on: May 21, 2012, 01:35:52 PM »

Quote from: W5DQ

...and with the SWR @ 2:1, that is only an 11% loss of output power...




No, do NOT remember that.  It is false.

Just as I said upthread, this statement is confusing reflected power with lost power.
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W5DQ
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« Reply #18 on: May 21, 2012, 01:55:38 PM »

Quote from: W5DQ

...and with the SWR @ 2:1, that is only an 11% loss of output power...




No, do NOT remember that.  It is false.

Just as I said upthread, this statement is confusing reflected power with lost power.

Strange you disagree as all nomographs I have ever seen and used say EXACTLY what I stated, SWR of 2:1 is appx 11% of lost OUTPUT power. You can call it REFLECTED POWER if you want but what is a fact is regardless whether it is lost in feedline losses or reflected back into the final turning into disappated power (i.e. heat), it IS NOT radiated and is LOST from output signal.

Gene W5DQ
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Gene W5DQ
Ridgecrest, CA - DM15dp
www.radioroom.org
WB6BYU
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« Reply #19 on: May 21, 2012, 02:24:48 PM »

That's because such nomographs are written for those who don't know any better.

Those who do understand the losses in such a system know that you can't specify
the output losses strictly as a function of SWR, so wouldn't write such a nomograph
in the first place, at least not labeling it as "lost power".  You have to consider the
coax losses and the load impedance (not just the SWR).

If you see my earlier post, running a length of coax at a 10 : 1 SWR actually resulted in marginally
lower feedline loss than at 1 : 1 SWR.  If your nomograph doesn't allow for that
possibility, then it isn't a realistic reflection of reality.

You can't even say for sure what happens to the reflected power:  in many case it is
actually reflected from the rig end and does get radiated.


Here's a problem - let's compare the calculated results using the two methods:

We'll feed a 50 ohm antenna on 20m with 63 feet of Wireman type 553 window twinlead.  The
characteristic impedance of this line is about 390 ohms, so the SWR on it is 7.8 : 1.
With 100 watts out from the transmitter, how much power gets radiated and how much
gets lost at this high of an SWR?

I say the answer is about 88% radiated and 12% lost in the line.  That's about the same
as the 11% loss you are claiming for an SWR of 2 : 1 without regard to the type or length
of the transmission line, and the SWR is much higher in this case.

[edited to add:]

And does the length of the feedline make any difference in your model?  Because I plan to
assemble this system and measure the actual power delivered to the 50 ohm load.  If the
power lost is dependent only on the 7.8 : 1 SWR and not the length of the feedline, then
I may use a shorter length for convenience.
« Last Edit: May 21, 2012, 03:12:24 PM by WB6BYU » Logged
STAYVERTICAL
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« Reply #20 on: May 21, 2012, 04:22:13 PM »

Lets keep it simple:

A dipole sufficiently high above the ground has a theoretical radiation resistance of 73 ohms.
I am assuming you are using a 50 ohm transmission line.
The best swr possible with that setup is 73/50 which equals 1.46 to 1
If you use a 70 ohm transmission line you could get 1.04 to 1

Now, back to assuming you are using 50 ohm transmission line and 1.4 to 1.
The SWR losses will depend entirely on the quality of your transmission line, that is, its loss characteristics.
The lower loss the line, the less effect SWR has on your transmission line losses.

Remember, the reason loss due to SWR occurs in a transmission line is that the reflected signal is bouncing back and forth between antenna and transmitter.
Every transit it pays a percentage toll, equivalent to the loss value of the transmission line.
With a higher SWR, more of your signal bounces, and because a percentage is taken, the value of the toll taken is higher.
If your transmission line had no loss( the toll was zero), all of the reflected power would eventually be radiated, and SWR would not cause any loss.
This is why open wire line can be used with high SWR, it is a very low loss transmission line.

The purpose of a tuner at the transmitter is only to provide a nice impedance for the transmitter, nothing to do with reducing transmission line loss.
So with or without a tuner at the transmitter end, your SWR based feedline loss is the same.
However, the tuner will have some loss itself, so you will end up with lower power, all other things being equal.

However, if your transmitter folds back power under high SWR, the tuner will provide the transmitter with its optimum load,
and so avoid a protective power reducing operation.

Now if you put a tuner at the antenna/feedline junction (remote tuner), then that is a different thing entirely.
In this situation, your feedline can have a SWR of 1:1, and everyone is happy, and losses are minimal.
The tuner will however have some loss, how much depends on its design.


Thats it, simple isn't it?

73 - Rob


« Last Edit: May 21, 2012, 05:00:43 PM by STAYVERTICAL » Logged
W5DXP
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« Reply #21 on: May 22, 2012, 05:40:00 AM »

The purpose of a tuner at the transmitter is only to provide a nice impedance for the transmitter, ...

Simply put, what the tuner does is to prohibit the reflected energy wave from reaching the source by forcing that reflected energy to take a different path. Since in a transmission line, there are only two possible directions of energy flow, the conservation of energy principle tells us that the reflected energy will be redistributed back toward the antenna as part of the forward wave.

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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
STAYVERTICAL
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Posts: 854




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« Reply #22 on: May 22, 2012, 02:55:41 PM »

The purpose of a tuner at the transmitter is only to provide a nice impedance for the transmitter, ...

Simply put, what the tuner does is to prohibit the reflected energy wave from reaching the source by forcing that reflected energy to take a different path. Since in a transmission line, there are only two possible directions of energy flow, the conservation of energy principle tells us that the reflected energy will be redistributed back toward the antenna as part of the forward wave.



Well put.
It's all a matter of adjusting phases and magnitudes and the tuner does that with its reactive elements (inductors and capacitors).
Sometimes we overcomplicate answers (I am as guilty as anyone).

73 - Rob
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KA3VEZ
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« Reply #23 on: May 23, 2012, 05:00:05 PM »

Wow!  Thank you all for all the ideas.  I guess I bit more off than I can chew.  Let me ask this while everyone is here.  Would air chokes reduce SWR issues or is that to reduce RF from back down the line?


~Kirk, KA3VEZ
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WB6BYU
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Posts: 13113




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« Reply #24 on: May 23, 2012, 05:59:16 PM »

Quote from: KA3VEZ

...Would air chokes reduce SWR issues or is that to reduce RF from back down the line?



The purpose of a choke balun is to reduce the common mode current flowing on the
outside of the coax rather than matching the SWR.

Now, if you have common mode currents, that means the outside of the coax is part
of your antenna.  And that means you are adding the chokes to your antenna, so you
may see a change in SWR, but it could go either up or down.
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K4FMX
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« Reply #25 on: May 23, 2012, 06:56:54 PM »

Quote from: W5DQ

...and with the SWR @ 2:1, that is only an 11% loss of output power...




No, do NOT remember that.  It is false.

Just as I said upthread, this statement is confusing reflected power with lost power.

Strange you disagree as all nomographs I have ever seen and used say EXACTLY what I stated, SWR of 2:1 is appx 11% of lost OUTPUT power. You can call it REFLECTED POWER if you want but what is a fact is regardless whether it is lost in feedline losses or reflected back into the final turning into disappated power (i.e. heat), it IS NOT radiated and is LOST from output signal.

Gene W5DQ

You are correct that a 2:1 SWR will give an 11% loss of power in a system where the generator source is a 50 ohm resistor such as a typical lab signal generator. In this case most all reflected power will be absorbed in the source resistance of the generator.

This is where "mismatch loss" calculations come from. With a 2:1 SWR in this situation you truly do lose 11% of the power because it is reflected back to the source and absorbed by it.

However a typical solid state transmitter, even though specified as having a 50 ohm output, is nowhere near a 50 ohm source. It is usually a very low impedance and as such it will absorb very little reflected power seen coming back to it.

The transmitter is comonly called a 50 ohm output but what is really meant is that it will deliver maximum power out when it sees a 50 ohm load.

So if the reflected power is not absorbed by the transmitter then there is only one place for it to go and that is to be re-reflected back to the load and dissipated there. If the load is an antenna it will be ultimately radiated.

An easy proof of where reflected power goes can be observed with a wattmeter such as a Bird thru line. you can observe the reflected power in the reverse direction on the meter. You will also notice that when reflected power is present that the forward power reading on the meter will be higher than normal.
Looking at the Bird manual it will tell you that to find the true power delivered to the load you must subtract the reflected power indication from the forward power indication.

As an example with 100 watts out of the transmitter and a 3:1 SWR which is 25% reflected power you will see the meter show 25 watts on the reflected scale and 125 watts on the forward scale. Subtracting the 25 watts reflected from the forward 125 will yield 100 watts of power that will be delivered to the load.

There is really 125 watts of forward power in this situation as the re-reflected 25 watts power combines with the original 100 watts forward power.

Now for simplicity of explanation we are considering that the transmitter is not reducing power due to any reflected power and that we have no feed line loss, which is possible if the load is right near our test setup.

Some small amount of reflected power may be absorbed by the transmitter rather than being re-reflected but that will usually be small.
If a tuner is used at the transmitter before the wattmeter then a near perfect conjugate match can be had and no amount of reflected power will reach the transmitter. It will all be re-reflected. This will also insure that the transmitter does not fold back and reduce its output because of high SWR. With the tuner in place the transmitter will never see the high SWR.

It is interesting to install another wattmeter between the transmitter and the tuner to observe that you really do have only 100 watts output when the second wattmeter between the tuner and 3:1 load indicates 125 watts forward.
Note that the 125 watt reading may be slightly less due to any loss in the tuner.

Reading the Bird wattmeter manual can be very enlightening on the subject.

73
Gary  K4FMX
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W5DXP
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« Reply #26 on: May 24, 2012, 06:11:16 AM »

When a tuner is properly adjusted, there is zero reflected energy on the piece of 50 ohm coax between the transmitter output and the tuner input. That's all one needs to know to realize that zero reflected power is being dissipated in the transmitter because zero reflected energy is being allowed to reach the transmitter. In a low-loss system with a properly adjusted tuner, most of the reflected power is redistributed back toward the antenna as forward power.

In a system with reflections, the fact that SWR=1:1 on the 50 ohm coax between the transmitter output and the tuner input tells us that a 50 ohm Z0-match has been achieved at the tuner input. A Z0-match point guarantees that 100% of the reflected energy is redistributed back toward the antenna at the Z0-match point. Here's a simple lossless example with no tuner to confuse things:

100W XMTR---50 ohm coax---+---1/2WL 600 ohm feedline---50 ohm antenna

The SWR on the 600 ohm line is 12:1. The SWR on the 50 ohm coax is 1:1. That tells us that there is a 50 ohm Z0-match at the '+' point.

The forward power on the 600 ohm feedline is 352 watts and the reflected power is 252 watts. None of the 252 watts makes it past the Z0-match point. All of the 252 watts are redistributed back toward the antenna at the Z0-match point.

The forward power on the 50 ohm coax is 100 watts and the reflected power is 0 watts. All of the 100 watts of forward power is transferred through the Z0-match point. That 100 watts of forward power combines with the 252 watts of reflected power to become the new forward wave of 352 watts.

At the antenna, power delivered to the antenna equals forward power minus reflected power, i.e. 352w-252w=100w. In the above lossless case, that 100w is equal to the transmitter power. In a real world case where overall efficiency is 90%, only 90 watts would reach the antenna but the fact remains that zero reflected energy would be dissipated in the transmitter because of the Z0-match.

The Z0-match established at the input of a tuner does the same thing as the Z0-match in the example above - it prohibits reflected energy from being incident upon the transmitter.
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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
STAYVERTICAL
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« Reply #27 on: May 24, 2012, 03:06:24 PM »

Wow!  Thank you all for all the ideas.  I guess I bit more off than I can chew.  Let me ask this while everyone is here.  Would air chokes reduce SWR issues or is that to reduce RF from back down the line?


~Kirk, KA3VEZ

Hi Kirk,

You can think of the air choke as a balun, since the three wires found in many baluns (there is more than one type),
can be represented by the coax's centre wire, inside surface of the braid, and the outside of the braid.
The braid can be seen as two conductors, since at radio frequencies, energy travels on the surface of the conductors.

These type of chokes are also sometimes called "ugly baluns" and they do work.
They are useful since they do not saturate and are fairly frequency independent, although not entirely.
Construction is based on the range of frequencies you wish it to cover and consists of winding a coil of coax as close to the
antenna feedpoint as possible.
The winding should be neat and like a standard coil, not randomly looped, otherwise interwinding capacitance will affect performance.

These chokes are not a replacement for conventional baluns, but are simple, inexpensive and work.
It is sometimes amazing to see the effect such a choke has on SWR, since many hams don't realise their feedline is unintentionally acting as part of their antenna.

Even vertical antenna's which theoretically don't need a balun, can benefit from this type of choke, since in many cases the radial system is substandard,
and so the feedline also takes part.
If your use of an air choke changes the SWR, it means for whatever reason, your feedline is somehow part of the antenna, not just a pipe for R.F. energy.
This is not necessarily horrible, we live in an imperfect world of compromises, but gives you an indicator for further investigation.

73 - Rob


« Last Edit: May 24, 2012, 03:20:22 PM by STAYVERTICAL » Logged
AA5WG
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« Reply #28 on: May 25, 2012, 05:53:40 AM »

Here is some excellent material to read.

http://w2du.com/Chapter07.pdf

"Chapter 7. My Antennna Tuner Really Does Tune My Antenna"

And,

 http://w2du.com/

Chuck
« Last Edit: May 25, 2012, 06:04:33 AM by AA5WG » Logged
W5DXP
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« Reply #29 on: May 25, 2012, 06:48:07 AM »

"Chapter 7. My Antennna Tuner Really Does Tune My Antenna"

Consider the antenna system, including the tuner, as a tuned circuit. When the tuner is adjusted to 50 ohms with zero reactance at the tuner input, that meets the definition of resonance at that frequency, i.e. the tuner tunes the antenna system to resonance. The tuner has a system-wide resonating effect and increases the voltage, current, and power magnitudes at all points in the system, not just at the tuner. Strangely enough, maximum absolute losses also occur when the antenna system is tuned to resonance.
« Last Edit: May 25, 2012, 06:53:54 AM by W5DXP » Logged

73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
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