Balanced linked tuner turns ratio

[AMLOVER]

Hi to all,

I am preparing a linked balanced tuner for a 1/8 wave length 600 ohm ladder line feeding a resonant low band monobander dipole.
I have calculated the resistive part of the impedance on the edge of my ladder line to around 118 ohm and the reactive part to about +615jx.
As far my tuner space is limited and the antenna is monobander I am wondering if the turns ratio of the inductors is transforming the resistive part and the capacitor cares for the reactive part.
If so I'll only need few turns for the big inductor. Let say 118 ohm:50 ohm =2.4 and sqr 2.4=1.55 which will be the turns ratio factor for multiplying the link's turns. Per example 5 turns for the link inductor and 1.55X5=8 turns for the big inductor. For safety I'll make 20-24 turns which will save me lot of space in comparison with the 45 turns for a multiband tuner.
Is this makes any sense or somewhere I miss something and I'll anyway need a 45 turns inductor?

Greetings
Stefano

[DMOD]

See Cebicks article:

http://www.antentop.org/w4rnl.001/link.html

or see 4. Link Coupling

https://www.dj0ip.de/antenna-matchboxes/symmetrical-matchboxes/

[W7TFO]

Taking into account some of us build with balanced output right from the finals, a balun isn't necessary in the circuit.

I usually copy the output section of the Raytheon AM BC transmitters. Yes, they were able to match 50Z all the way up to 600Z.

73DG

[ka1bwo]

Stefano, I hope this will help you. Image 1, circuit diagram.  L is a modeled 75 meter dipole at 50 feet with an overall length of 125 feet, T1 is 1/8 wavelength of 600 ohm line. Image 2, impedance at the end of the 600 ohm line vs frequency. A is the link, C1 is the tuning capacitor to match the antenna from 3.5 to 4Mhz and G is the transmitter model. Image 3, SWR vs frequency vs C1, C1 was swept from 66p to 141p. Maximum SWR is 1.53:1 at 4 mhz, C1=66pf.      

[W4AMV]

Wrote a short script in MATLAB to solve
for the link coupled transformer where
the load is complex and the source is
real. Like your case where the reactance
is not tuned out. There is an exact solution
but the values are not practical. If you
resonate out the j615 ohm at at say
3.75 MHz, then the link coupled match is
nice. The series C is 69 pF and the
balanced side L is 6 uH and the 50
ohm side L is 1.35 uH. The coupling
k value is reasonable and easy to set
and measure. An exact match on the
50 ohms side requires a shunt C.
The Q values are low and the response
pretty broadband.

The response is shown below,
with schematic and S parameter response. 

[W4AMV]

This is the equation being solved in the script.

I use T.J. Maresca paper in QST Oct 1959 to seed
the MATLAB script with initial values. See his case
equations for Parallel-Resonant Primary Untuned
Secondary on page 30 of QST.
 

[AMLOVER]

DMOD,
Very usefull info at the provided links, I spend time to learn more in deep.

W7TFO,
This is a succesfull technic but as far I use a narrow band (50-100 Khz) I try to find out the right turns ratio in order to keep the inductor, if possible, as small as possible.

Ka1bwo,
Your calculation is almost as mine but I use a cap (C2) in parallel with the balance inductor and a link coil with reactance 50 ohm (per example 2.1225uH for 3750Khz) so there is no need for C1. With C2 in parallel with balanced inductor I would expect no reactance such as the X=450 Ohm in your scheme. Redrawing your scheme is how is my set up and follows in the attachment with some modifications.
The perfect scenario is to cancell the +588jx or +615jx in my case with C2 and the 118 Ohm resistance to 50 Ohm with the ratio of the balanced inductor to the link coil. Could you please recalculate with the C2 in this place?
In this case the balanced inductor will be not that big and the tuner will be of course usefull only for a narrow band.

W4AMV,
This is a very nice set up. I have never seen the parallel to the link coil 1.7nf cap.
All schematics have it in series with the link in order to cancel if needed some of its reactance.
Could you please explain me if the 69pf cap which cancells the +615jx at 3750Khz is in parallel with the load as it is drawn in the 2nd attachment (TUNER) as C2? or it should be in series with one of the ladder line legs?

Greetings
Stefano

[W4AMV]

To clarify. I took your stated measured or calculated impedance at the end of the 600 ohm line to be 118 ohms series with j 615 ohms. That series Z is the R2 + jX2 value across the secondary link L2. The reflected Z is calculated  via the equation I posted using the MATLAB script and the resulting reflected Z is in addition to the primary inductance on the low Z side, namely the 50 ohm side. I convert the series Z on the primary side to its parallel equivalent and hence the parallel C that you spoke of results. The result is the primary side is REAL and matched to 50 ohms.

[W4AMV]

Oh, the Z load you stated is a series Z. I assume that is the case. Correct me if wrong. So yes, the 69 pF is in series tuning out the j615 ohms at 3.75 MHz.  Now the problem is just taking the 118 ohm and transferring it to 50 ohms at the desired frequency. If I desire to use no secondary tuning C and reflect the complex load to the primary, that is possible. However, the resulting component values are not nice. This is partially drive by the fact that the Q of your load is higher than desired.

[ka1bwo]

Ka1bwo,
Your calculation is almost as mine but I use a cap (C2) in parallel with the balance inductor and a link coil with reactance 50 ohm (per example 2.1225uH for 3750Khz) so there is no need for C1. With C2 in parallel with balanced inductor I would expect no reactance such as the X=450 Ohm in your scheme. Redrawing your scheme is how is my set up and follows in the attachment with some modifications.
The perfect scenario is to cancell the +588jx or +615jx in my case with C2 and the 118 Ohm resistance to 50 Ohm with the ratio of the balanced inductor to the link coil. Could you please recalculate with the C2 in this place?
In this case the balanced inductor will be not that big and the tuner will be of course usefull only for a narrow band.



Stefano, here is the rev schematic and SWR chart. Your scheme will be more efficient due to less series resistance
 in the inductors. Link turns ratio = 7.538

Joe

[W4AMV]

You can convert your series antenna Z to its parallel equivalent. This provides yet another solution. The basic configuration follows the link coupled topology from the QST paper discussed as resonant primary non resonate secondary, see figure. However, the author of this paper did not address the complex load. If you convert series to parallel then the 69 pF cap is in shunt with the balanced line and secondary winding and a new set of primary and secondary inductance values emerge. As well, the k factor can change as desired as you alter the matching bandwidth or Q of the source and load side of the link coupled network.

[ka1bwo]

You can convert your series antenna Z to its parallel equivalent. This provides yet another solution. The basic configuration follows the link coupled topology from the QST paper discussed as resonant primary non resonate secondary, see figure. However, the author of this paper did not address the complex load. If you convert series to parallel then the 69 pF cap is in shunt with the balanced line and secondary winding and a new set of primary and secondary inductance values emerge. As well, the k factor can change as desired as you alter the matching bandwidth or Q of the source and load side of the link coupled network.

W4AMV MATCHING SOLUTION:

[AMLOVER]

You can convert your series antenna Z to its parallel equivalent. This provides yet another solution. The basic configuration follows the link coupled topology from the QST paper discussed as resonant primary non resonate secondary, see figure. However, the author of this paper did not address the complex load. If you convert series to parallel then the 69 pF cap is in shunt with the balanced line and secondary winding and a new set of primary and secondary inductance values emerge. As well, the k factor can change as desired as you alter the matching bandwidth or Q of the source and load side of the link coupled network.

W4AMV MATCHING SOLUTION:

Joe,

Could you please try once more with the attached scheme which is the usual 'balanced link coupled tuner'?

Greetings
Stefano

[W4AMV]

This follows the QST paper guideline, parallel resonant primary,  series resonate secondary. The solution is not unique. You can vary the Q on wither side of the network despite the give antenna complex Z. So you can vary the coupling coefficient, k, of the link coupled coils. The only mod from the paper is tuning out the reactance of the antenna. Another solution is obtained by converting the antenna Z from its series to parallel equivalent. Then proceed again with the link coupled equations. I encourage you to read the paper.

[W4AMV]

Here is the network for the parallel equivalent Z. The coupling is light. Easier to implement. Increased coupling is possible, just  decrease the Q of primary/secondary sides. The antenna equivalent Z highlighted in box
and the shunt C to resonate is absorbed into the shunt C that results from solving the link coupling equations. Plotted the return loss as REAL RETURN LOSS as a positive number. 

[ka1bwo]

You can convert your series antenna Z to its parallel equivalent. This provides yet another solution. The basic configuration follows the link coupled topology from the QST paper discussed as resonant primary non resonate secondary, see figure. However, the author of this paper did not address the complex load. If you convert series to parallel then the 69 pF cap is in shunt with the balanced line and secondary winding and a new set of primary and secondary inductance values emerge. As well, the k factor can change as desired as you alter the matching bandwidth or Q of the source and load side of the link coupled network.

Joe,

Could you please try once more with the attached scheme which is the usual 'balanced link coupled tuner'?

Greetings
Stefano

Transmission line end impedance parallel equivalent

[W4AMV]

sim smith provides a solution with k = 1. This simplifies the prior posted equation and I suppose you can choose a Q for the primary and secondary loop within some constraints. Need to visit those details. Anyway, if you get a k of 1, great. A k of 0.8 or greater is needed otherwise the match deteriorates as shown in this plot. I assume one is able to enter a desired k into sim smith. Nicely done.

 

[aa5wg]

Hi Stefono,

What band is this mono-band antenna for?

And, you might consider using 300 ohm instead of 600 ohms for your feedline impedance.  This lowers the high voltages on the feedline and thus a lower voltage stress on your link antenna output components (inductors and capacitors - potential arcing of capacitors).  

Let us start with above and move on from there.  What do you think?

Chuck

[ka1bwo]

sim smith provides a solution with k = 1. This simplifies the prior posted equation and I suppose you can choose a Q for the primary and secondary loop within some constraints. Need to visit those details. Anyway, if you get a k of 1, great. A k of 0.8 or greater is needed otherwise the match deteriorates as shown in this plot. I assume one is able to enter a desired k into sim smith. Nicely done.

 

Yes, you are correct a k of 1 was used for the analysis you can enter the desired k. Here is a plot of the ideal power delivered to the 600 ohm transmission line vs k, with a transmitter output of 100 watts.

[AMLOVER]

W4AMV,
I read the QST article you reccommended me and saw that it is for linked coupling of the tube output to 50 Ohm. It is advised to keep K less than 0.4 and when did that the Q should be 12 or more. Anyway in that article both parts of coupler are considering as resistive. In order to use those equations I have to cancell the reactive part of the ladder line. This is not however the mainstream technic. I see now that things become easier with k=1 but still the series cap is a must. This cap is not involved in any balanced linked couple tunner consideration.

Joe,
I have studied very carefully your program's results but I can't understand why the LI which is on the ladder side facing 118 Ohm is smaller than LR which is on the link side facing 50 Ohm. Normally should be opossed.

Chuck,
It is for the low band as I have mentioned in my first post. 300 Ohm line seems much better in calculations but it is not very safe because the wires (12 gauge) have to be less then 2'' closed together, this distance needs more spacers per foot and is unsafe to winds, rain and snow.    

In conclusion I think that the parallel cap must be on the edge of a sufficient for the frequency inductor to keep the Q and the circulating currents low and the line should touch the inductor where the resistance is correcttly transformed. I wonder if I can manage to use a not so big inductor without sacrificing much gain. My tuner should look like the one in the photo  http://amfone.net/Amforum/index.php?topic=36685.0 where the antenna set up including the OWL length is like mine. I only wish not to have so many turns in the main inductor.

[W4AMV]

Perhaps you can clarify with a schematic the exact circuit topology you desire. The limited view of the hardware requires quite a bit of extrapolation!

[W4AMV]

After a careful re read of your post, is this the topology desired?

No tuning on the secondary implies the secondary reflected Z will also be complex and reactance cancellation on the primary should permit the primary port Z to be 50 + j0.

[aa5wg]

Stefono,

When you say low band do mean 160 meters?

Thank you.

Chuck

[W4AMV]

Hi aa5wg...

per prior posts, looks like... 80 meters phone.

(per example 2.1225uH for 3750Khz)

[W4AMV]

If the topology is the one I mentioned earlier... Where the link is required to absorb the antenna reactance (inductive) and there is NO tuning on the balanced transmission line side, then this results in the schematic attached. Note, counter intuitive, the link inductance is larger than the balanced side inductor. There is only the 50 ohm side of the tuner with a series cap to tune out the reflected impedance. The result is the consequence of absorbing the reactance of the termination. Obviously different results will occur dependent on the 600 ohm side of the balanced line Z and the selection of circuit Q values. Finding a set of component values is easy using the Matlab script and the full linked coupled equation from the prior post. You might consider the balanced tapped inductance with link as I think showed in your photo. K1JJ tuner. I have not analyzed this from design to component value. Perhaps someone has already.