Hi

can someone explain how to connect 2 voltage 4:1 baluns in series to
achieve 16:1

Thanks
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can someone explain how to connect 2 voltage 4:1 baluns in series to
achieve 16:1
=================================

With say one 4:1 balun having 2 windings Aa and Bb and the second 4:1
balun having 2 windings Cc and Dd with A & a etc the ends of a
particular winding , connect the windings as follows :

a to B , b to C , c to D

Connect grounded unbalanced side to bC connection

Connect the hot unbalanced side either to aB or cD connection

Connect balanced line between A and d


The above applies when the 2 existing baluns being wound with 2 windings
each with the 2 corresponding ends being A and B , a and b for one
balun and C and D , c and d for the other balun.


I should have made an ASCII drawing of the 4 windings in series ,but I
didn't expect it to be transmitted correctly ,so I hope the above
description makes sense .

Frank GM0CSZ / KN6WH

Cheers wrote:
can someone explain how to connect 2 voltage 4:1 baluns in series to
achieve 16:1

Assuming you are trying to match 50 ohms to 800
ohms, the first one would need to be a 50:200 unun
and the second one would need to be a 200:800 balun.

Where does the 800+j0 ohm impedance that you
are trying to match come from?
--
73, Cecil http://www.w5dxp.com

Hi Cecil

In article ,
says...

Assuming you are trying to match 50 ohms to 800
ohms, the first one would need to be a 50:200 unun
and the second one would need to be a 200:800 balun.

Where does the 800+j0 ohm impedance that you
are trying to match come from?

is 900 ohms 3-wire Version of the WBFD

|----------|A|----------|
|---------'\/\/\,-------|
|----------|B|----------|

http://www.cebik.com/wire/wbfd.html



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The information contained in this post is copyright of
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In article ,
says...
can someone explain how to connect 2 voltage 4:1 baluns in series to
achieve 16:1
=================================

With say one 4:1 balun having 2 windings Aa and Bb and the second 4:1
balun having 2 windings Cc and Dd with A & a etc the ends of a
particular winding , connect the windings as follows :

a to B , b to C , c to D

Connect grounded unbalanced side to bC connection

Connect the hot unbalanced side either to aB or cD connection

Connect balanced line between A and d


The above applies when the 2 existing baluns being wound with 2 windings
each with the 2 corresponding ends being A and B , a and b for one
balun and C and D , c and d for the other balun.


I should have made an ASCII drawing of the 4 windings in series ,but I
didn't expect it to be transmitted correctly ,so I hope the above
description makes sense .

Frank GM0CSZ / KN6WH

Hi Frank

Thanks for the info., will have a go at it.

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the poster, and specifically may not be published in,
or used by http://www.jlaforums.com

You need to be aware that there's a good chance that this might not
work, or at least not like you think it will.

An impedance transforming balun includes one or more windings connected
across the two conductors. The impedance of the winding itself appears
across the conductors as a shunt impedance. In order for the winding
itself not to have an appreciable effect on the system, the winding
impedance must be at least several times the impedance of the circuit to
which it's connected. A rule of thumb is that it should be a minimum of
5 times the impedance of the circuit, with 10 times being better. That
means that a 4:1 balun intended to transform 200 to 50 ohms would have a
high-Z winding impedance of around 1000 - 2000 ohms. When a balun is
designed for multiband use, the impedance is likely to be on the low
side at the lower end of the frequency range, since this is where it's
hardest to get high impedance. Also, if the impedance is too high at the
low end, you can run into resonance or other problems at the high end.

The bottom line of all this is that you're likely to be putting a
winding impedance of somewhere around 1000 - 2000 ohms in parallel with
your 800 ohm antenna. If this impedance is pretty purely inductive, it
will have a profound effect on the tuning of the antenna. If it's fairly
resistive, it'll also dissipate a good fraction of the power you apply
to it. So don't be too surprised if you see either or both these effects.

If you have an antenna analyzer you can terminate the high Z end with
around 800 ohms of resistance and measure the input impedance. I'm sure
you won't see anything that much resembles 50 + j0, at least not over
much of a frequency range. If your antenna isn't resonant so it doesn't
present close to 800 + j0 ohms to the balun, you'll likely see
additional effects.

Roy Lewallen, W7EL

In article , says...
You need to be aware that there's a good chance that this might not
work, or at least not like you think it will.

An impedance transforming balun includes one or more windings connected
across the two conductors. The impedance of the winding itself appears
across the conductors as a shunt impedance. In order for the winding
itself not to have an appreciable effect on the system, the winding
impedance must be at least several times the impedance of the circuit to
which it's connected. A rule of thumb is that it should be a minimum of
5 times the impedance of the circuit, with 10 times being better. That
means that a 4:1 balun intended to transform 200 to 50 ohms would have a
high-Z winding impedance of around 1000 - 2000 ohms. When a balun is
designed for multiband use, the impedance is likely to be on the low
side at the lower end of the frequency range, since this is where it's
hardest to get high impedance. Also, if the impedance is too high at the
low end, you can run into resonance or other problems at the high end.

The bottom line of all this is that you're likely to be putting a
winding impedance of somewhere around 1000 - 2000 ohms in parallel with
your 800 ohm antenna. If this impedance is pretty purely inductive, it
will have a profound effect on the tuning of the antenna. If it's fairly
resistive, it'll also dissipate a good fraction of the power you apply
to it. So don't be too surprised if you see either or both these effects.

If you have an antenna analyzer you can terminate the high Z end with
around 800 ohms of resistance and measure the input impedance. I'm sure
you won't see anything that much resembles 50 + j0, at least not over
much of a frequency range. If your antenna isn't resonant so it doesn't
present close to 800 + j0 ohms to the balun, you'll likely see
additional effects.

Roy Lewallen, W7EL

Hi Roy

to my understanding the load is purely resistive according to this
article http://www.cebik.com/wire/wbfd.html
anyway have a read and tell me what you think I am quiet interested in
others opinion on this one.
finding the right non inductive resistor has proven to be a challenge so
far.
no one told me home brewing antenna is a dark art :-)



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Cheers wrote:
finding the right non inductive resistor has proven to be a challenge so
far.

I have a bunch of 600 ohm 50 watt non-inductive resistors.
My email address is
--
73, Cecil http://www.w5dxp.com

On Feb 28, 7:30 pm, Cheers wrote:
In article , says...

You need to be aware that there's a good chance that this might not
work, or at least not like you think it will.

An impedance transforming balun includes one or more windings connected
across the two conductors. The impedance of the winding itself appears
across the conductors as a shunt impedance. In order for the winding
itself not to have an appreciable effect on the system, the winding
impedance must be at least several times the impedance of the circuit to
which it's connected. A rule of thumb is that it should be a minimum of
5 times the impedance of the circuit, with 10 times being better. That
means that a 4:1 balun intended to transform 200 to 50 ohms would have a
high-Z winding impedance of around 1000 - 2000 ohms. When a balun is
designed for multiband use, the impedance is likely to be on the low
side at the lower end of the frequency range, since this is where it's
hardest to get high impedance. Also, if the impedance is too high at the
low end, you can run into resonance or other problems at the high end.

The bottom line of all this is that you're likely to be putting a
winding impedance of somewhere around 1000 - 2000 ohms in parallel with
your 800 ohm antenna. If this impedance is pretty purely inductive, it
will have a profound effect on the tuning of the antenna. If it's fairly
resistive, it'll also dissipate a good fraction of the power you apply
to it. So don't be too surprised if you see either or both these effects.

If you have an antenna analyzer you can terminate the high Z end with
around 800 ohms of resistance and measure the input impedance. I'm sure
you won't see anything that much resembles 50 + j0, at least not over
much of a frequency range. If your antenna isn't resonant so it doesn't
present close to 800 + j0 ohms to the balun, you'll likely see
additional effects.

Roy Lewallen, W7EL

Hi Roy

to my understanding the load is purely resistive according to this
articlehttp://www.cebik.com/wire/wbfd.html
anyway have a read and tell me what you think I am quiet interested in
others opinion on this one.
finding the right non inductive resistor has proven to be a challenge so
far.
no one told me home brewing antenna is a dark art :-)

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The information contained in this post is copyright of
the poster, and specifically may not be published in,
or used by http://www.jlaforums.com

You may find it better to just wind your own transformer. 16:1
impedance ratio is 4:1 turns ratio. You can twist four wires together
to make a quadrifilar cable that you use to put a few turns on a
ferrite toroid. Then connect the windings all in series (say
1--2--3--4). The center tap, between 2 and 3, would connect to the
coax feedline outer conductor, and the tap between 1 and 2 to the coax
center. The outer leads--the start and stop of the whole thing--
connect to the antenna. As for impedances of the windings and losses,
this is one of those "your mileage may vary" things. I'd claim you
don't need the impedance at the lowest frequency of interest to be too
much higher than the 800 ohms resistive, since reactance is in
quadrature to resistance and it's not the same as two resistances in
parallel. For example, if you use a core that gives you, say, 0.3uH/
turn, and you put 4 turns of the quadrifilar cable on it, the 16 turns
when you put the four wires in series gives you about 256*0.3 or
75uH. At 3.5MHz (looks like the practical low end for that antenna),
that's 1.65k ohms inductive, and that in parallel with 800 ohm
resistive is about 650+j300 ohms. Through 16:1 (if the 16:1 is
perfect, which of course it won't be, but should be decent), that's
about 40+j20, which is about 1.6:1 SWR, which shouldn't upset the
apple cart. Though coils wound on ferrites get resistive at higher
frequencies, if you pick the right ferrite, by the time the balun's
winding impedance is significantly resistive, it should also be quite
high, so loss is not necessarily a big deal. Also, you can put about
2300pF in series between the balun and the feedline center conductor,
and it will significantly correct the reactance at 3.5MHz -- and
become progressively more like a short at higher frequencies, taking
itself out of the picture. If the winding reactance behaves like a
linear inductor versus frequency, the series capacitor will work very
well. Assuming the idealized case--800 ohm resistive load, 4:1 turns
ratio transformer that looks like a pure 75uH across the full winding--
the worst SWR is 1.24 at 3.5MHz, 1.18 at 4.0MHz, and falls
monotonically to practically 1:1 at 30MHz. (And I can get to even
better match across the whole frequency range, even with only 50uH
transformer inductance, by adding a cap in series between the
transformer output and the antenna...) My advice: stay with a
relatively low impedance, relatively few turns on the 4:1 turns-ratio
transformer, and correct the low end with a bit of series
capacitance. That will help you avoid problems with parasitic
capacitances (etc) at high frequency. Use an appropriate core
material that doesn't "die" and become resistive at 10MHz. Though my
model is idealized, I expect you should be able to come acceptably
close to it in practice.

Though my suggested design does not agree with Roy's suggested
reactance 5 times the resistance it's across (my suggestion being only
a couple times, or even less), I want to be clear that I'm in full
agreement about the need to be careful about the _effects_ of low
reactance at low frequencies, resonances and capacitive reactance
parasitics at high frequencies, and dissipative ferrite at high
frequencies. You're bound to learn a lot in the effort to get it
working really right.

Cheers,
Tom

In article ,
says...
Cheers wrote:
finding the right non inductive resistor has proven to be a challenge so
far.

I have a bunch of 600 ohm 50 watt non-inductive resistors.
My email address is

Hi

Sent you an email to the above address few days ago, have you received
it?


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