I'd like to build the antenna from the following page;

http://www.frars.org.uk/cgi-bin/rend...62&goback=1064

But there are a few details I need to sort out before I try.



1) It uses UT141, Could I sub 9913?

2) Or, should I use the copper tube method shown here,

http://martybugs.net/wireless/biquad/

3) The author states:

"Next, cut four pieces of UT141 coax 80mm long, and strip 1.5mm of outer
from each end, leaving a total outside length of 77mm. These will form part
of the phasing assembly. Follow the diagram as to the connection of these
pieces of coax. Connection of two double quads in parallel will result in an
impedance of 25 ohms."



Please view 4double quad back jpg at bottom of page as reference.



I assume he's describing the 77mm pieces that run at a 45* angles, however
the picture does not show

UT141 coax as he describes.



4) Again the author states:

" Next for the 25ohm to 100 ohm matching sections, cut two pieces of 61mm
UT141, again strip the outer back 1.5mm from each end to leave a outside
length of 58mm. Of course 2 100 ohm coax's in parallel will result in an
impedance of 50 ohms. Connect each piece to the joined 77mm lengths as per
the diagram. Under the connector, both centre conductors from the 58mm
lengths are connected to the centre pin of the SMA socket. Solder the outer
of these lengths to the SMA socket body to ensure a good earth connection."



Again he says use UT141 but I don't see it in the picture.



I'm looking forward to your input!

Thanks, Mike

"amdx" wrote in message
...
I'd like to build the antenna from the following page;

http://www.frars.org.uk/cgi-bin/rend...62&goback=1064

But there are a few details I need to sort out before I try.



1) It uses UT141, Could I sub 9913?

2) Or, should I use the copper tube method shown here,

http://martybugs.net/wireless/biquad/

3) The author states:

"Next, cut four pieces of UT141 coax 80mm long, and strip 1.5mm of
outer from each end, leaving a total outside length of 77mm. These will
form part of the phasing assembly. Follow the diagram as to the connection
of these pieces of coax. Connection of two double quads in parallel will
result in an impedance of 25 ohms."



Please view 4double quad back jpg at bottom of page as reference.



I assume he's describing the 77mm pieces that run at a 45* angles, however
the picture does not show

UT141 coax as he describes.



4) Again the author states:

" Next for the 25ohm to 100 ohm matching sections, cut two pieces of 61mm
UT141, again strip the outer back 1.5mm from each end to leave a outside
length of 58mm. Of course 2 100 ohm coax's in parallel will result in an
impedance of 50 ohms. Connect each piece to the joined 77mm lengths as per
the diagram. Under the connector, both centre conductors from the 58mm
lengths are connected to the centre pin of the SMA socket. Solder the
outer of these lengths to the SMA socket body to ensure a good earth
connection."



Again he says use UT141 but I don't see it in the picture.



I'm looking forward to your input!

Thanks, Mike

Hi Mike

You have given some great reference sites. Thanks.
If *I* was intending to feed two 25 ohm loads to a 50 ohm line, I'd
consider a series connection rather than transforming with a balanced line
transformer.
Besides, I didnt follow how the transforming of 25 ohms on a 100 ohm line
produced an impedance that is appropriate for matching to 50 ohms. But, I
havent studied the electrical equivalent lengths of the coax lengths.

Jerry

"Jerry Martes" wrote in message
news:yvJsi.35$zg3.14@trnddc04...

"amdx" wrote in message
...
I'd like to build the antenna from the following page;

http://www.frars.org.uk/cgi-bin/rend...62&goback=1064

But there are a few details I need to sort out before I try.



1) It uses UT141, Could I sub 9913?

2) Or, should I use the copper tube method shown here,

http://martybugs.net/wireless/biquad/

3) The author states:

"Next, cut four pieces of UT141 coax 80mm long, and strip 1.5mm of
outer from each end, leaving a total outside length of 77mm. These will
form part of the phasing assembly. Follow the diagram as to the
connection of these pieces of coax. Connection of two double quads in
parallel will result in an impedance of 25 ohms."



Please view 4double quad back jpg at bottom of page as reference.



I assume he's describing the 77mm pieces that run at a 45* angles,
however the picture does not show

UT141 coax as he describes.



4) Again the author states:

" Next for the 25ohm to 100 ohm matching sections, cut two pieces of 61mm
UT141, again strip the outer back 1.5mm from each end to leave a outside
length of 58mm. Of course 2 100 ohm coax's in parallel will result in an
impedance of 50 ohms. Connect each piece to the joined 77mm lengths as
per the diagram. Under the connector, both centre conductors from the
58mm lengths are connected to the centre pin of the SMA socket. Solder
the outer of these lengths to the SMA socket body to ensure a good earth
connection."



Again he says use UT141 but I don't see it in the picture.



I'm looking forward to your input!

Thanks, Mike

Hi Mike
You have given some great reference sites. Thanks.

Google is a friend to all!

If *I* was intending to feed two 25 ohm loads to a 50 ohm line, I'd
consider a series connection rather than transforming with a balanced
line transformer.

I don't really have a good understanding, but, looks like he has
two 50 ohm antennas connected thru two 77mm wires and then a repeat of
two 50 ohm antennas connected thru two 77mm wires. Then both sides are
connected
thru two 58mm wires to the center conductor of the connector.
(that all seems like 12.5ohms to me)
But I all ready know there is to much info missing here.


Besides, I didnt follow how the transforming of 25 ohms on a 100 ohm line
produced an impedance that is appropriate for matching to 50 ohms. But,
I havent studied the electrical equivalent lengths of the coax lengths.

I know he talks about 100 ohm line but I don't see any.
I hope one of the gurus helps out on this.
Mike

In article ,
amdx wrote:

I don't really have a good understanding, but, looks like he has
two 50 ohm antennas connected thru two 77mm wires and then a repeat of
two 50 ohm antennas connected thru two 77mm wires. Then both sides are
connected
thru two 58mm wires to the center conductor of the connector.
(that all seems like 12.5ohms to me)
But I all ready know there is to much info missing here.

Besides, I didnt follow how the transforming of 25 ohms on a 100 ohm line
produced an impedance that is appropriate for matching to 50 ohms. But,
I havent studied the electrical equivalent lengths of the coax lengths.

I know he talks about 100 ohm line but I don't see any.
I hope one of the gurus helps out on this.

Let's start out by assuming that the biquads are each 50 ohms (which
is something you can ensure by placing them at the correct distance
from the reflecting ground-plane).

Each 50-ohm biquad is fed through a suitable length of 50-ohm
semi-rigid coaxial line. Since the antenna's impedance matches that of
the coax, there will be a 1:1 SWR on the coax, and the same 50-ohm
impedance will appear at the far end of the coax. In this particular
case, the semi-rigid coax will consist of the part which goes through
the reflector (supporting the biquad) plus the trimmed 80mm section
(which ends up being 77mm after stripping and soldering).

There are four such 77mm lengths, which come together as two pairs.
At this point, each pair is soldered together, placing two 50-ohm
loads in parallel. This creates a 25-ohm load.

Now, each 25-ohm connection is soldered to a further length of 50-ohm
cable, 58mm long plus a bit for making the connection. The two 58mm
sections come together, at the SMA connector.

These 58mm sections act as impedance transformers. One of the useful
characteristics of a coax which is 1/4 or 3/4 or 5/4 electrical
wavelength long, is that it transforms a resistive impedance at one
end, to a different resistive impedance at the other... and the
geometric mean of these two impedances is equal to the characteristic
impedance of the feedline.

In this case, if you have a 25-ohm impedance at one end, and the cable
has a 50-ohm impedance, then the impedance at the other end of the
cable will be 100 ohms.

This places two 100-ohm impedances in parallel where they're connected
to the SMA jack, resulting in a 50-ohm impedance.

In the case of this particular design, the impedance-matching sections
(58mm) aren't actually the 3/4 electrical wavelength that one might
expect. UT141 has a velocity factor of 0.7, and my slide-rule
calculation suggests that they'd be around 65mm long for a true 3/4
electrical wavelength at 2.4 GHz.

I would guess that the biquad antennas aren't exactly 50 ohms
resistive, and that this impedance is being transformed somewhat by
the 77mm sections, and that the length of the final impedance-matching
section has been hand-tweaked to 58mm to give the best practical match
to 50 ohms (and thus SWR) for the antenna and phasing system.

That being the case, I'd probably recommend trying to reproduce the
antenna design as-published. If you switch to a cable having a
different velocity factor, you may need to do some cutting-and-trying
to get the right impedance match.

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

"amdx" wrote in message
...

"Jerry Martes" wrote in message
news:yvJsi.35$zg3.14@trnddc04...

"amdx" wrote in message
...
I'd like to build the antenna from the following page;

http://www.frars.org.uk/cgi-bin/rend...62&goback=1064

But there are a few details I need to sort out before I try.



1) It uses UT141, Could I sub 9913?

2) Or, should I use the copper tube method shown here,

http://martybugs.net/wireless/biquad/

3) The author states:

"Next, cut four pieces of UT141 coax 80mm long, and strip 1.5mm of
outer from each end, leaving a total outside length of 77mm. These will
form part of the phasing assembly. Follow the diagram as to the
connection of these pieces of coax. Connection of two double quads in
parallel will result in an impedance of 25 ohms."



Please view 4double quad back jpg at bottom of page as reference.



I assume he's describing the 77mm pieces that run at a 45* angles,
however the picture does not show

UT141 coax as he describes.



4) Again the author states:

" Next for the 25ohm to 100 ohm matching sections, cut two pieces of
61mm UT141, again strip the outer back 1.5mm from each end to leave a
outside length of 58mm. Of course 2 100 ohm coax's in parallel will
result in an impedance of 50 ohms. Connect each piece to the joined 77mm
lengths as per the diagram. Under the connector, both centre conductors
from the 58mm lengths are connected to the centre pin of the SMA socket.
Solder the outer of these lengths to the SMA socket body to ensure a
good earth connection."



Again he says use UT141 but I don't see it in the picture.



I'm looking forward to your input!

Thanks, Mike

Hi Mike
You have given some great reference sites. Thanks.

Google is a friend to all!

If *I* was intending to feed two 25 ohm loads to a 50 ohm line, I'd
consider a series connection rather than transforming with a balanced
line transformer.

I don't really have a good understanding, but, looks like he has
two 50 ohm antennas connected thru two 77mm wires and then a repeat of
two 50 ohm antennas connected thru two 77mm wires. Then both sides are
connected
thru two 58mm wires to the center conductor of the connector.
(that all seems like 12.5ohms to me)
But I all ready know there is to much info missing here.


Besides, I didnt follow how the transforming of 25 ohms on a 100 ohm
line produced an impedance that is appropriate for matching to 50 ohms.
But, I havent studied the electrical equivalent lengths of the coax
lengths.

I know he talks about 100 ohm line but I don't see any.
I hope one of the gurus helps out on this.
Mike

I got some help offline (thanks Dave) the wires I thought were solid are
actually
a coax (UT141). It make a little more sense now.
Mike

Thank you for that clear explanation.
And regarding the distance from ground-plane, I've seen two different
specs one 30.5mm and the other 18mm.
Thanks again, Mike

Let's start out by assuming that the biquads are each 50 ohms (which
is something you can ensure by placing them at the correct distance
from the reflecting ground-plane).

Each 50-ohm biquad is fed through a suitable length of 50-ohm
semi-rigid coaxial line. Since the antenna's impedance matches that of
the coax, there will be a 1:1 SWR on the coax, and the same 50-ohm
impedance will appear at the far end of the coax. In this particular
case, the semi-rigid coax will consist of the part which goes through
the reflector (supporting the biquad) plus the trimmed 80mm section
(which ends up being 77mm after stripping and soldering).

There are four such 77mm lengths, which come together as two pairs.
At this point, each pair is soldered together, placing two 50-ohm
loads in parallel. This creates a 25-ohm load.

Now, each 25-ohm connection is soldered to a further length of 50-ohm
cable, 58mm long plus a bit for making the connection. The two 58mm
sections come together, at the SMA connector.

These 58mm sections act as impedance transformers. One of the useful
characteristics of a coax which is 1/4 or 3/4 or 5/4 electrical
wavelength long, is that it transforms a resistive impedance at one
end, to a different resistive impedance at the other... and the
geometric mean of these two impedances is equal to the characteristic
impedance of the feedline.

In this case, if you have a 25-ohm impedance at one end, and the cable
has a 50-ohm impedance, then the impedance at the other end of the
cable will be 100 ohms.

This places two 100-ohm impedances in parallel where they're connected
to the SMA jack, resulting in a 50-ohm impedance.

In the case of this particular design, the impedance-matching sections
(58mm) aren't actually the 3/4 electrical wavelength that one might
expect. UT141 has a velocity factor of 0.7, and my slide-rule
calculation suggests that they'd be around 65mm long for a true 3/4
electrical wavelength at 2.4 GHz.

I would guess that the biquad antennas aren't exactly 50 ohms
resistive, and that this impedance is being transformed somewhat by
the 77mm sections, and that the length of the final impedance-matching
section has been hand-tweaked to 58mm to give the best practical match
to 50 ohms (and thus SWR) for the antenna and phasing system.

That being the case, I'd probably recommend trying to reproduce the
antenna design as-published. If you switch to a cable having a
different velocity factor, you may need to do some cutting-and-trying
to get the right impedance match.

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

In article ,
amdx wrote:

Thank you for that clear explanation.

Quite welcome!

And regarding the distance from ground-plane, I've seen two different
specs one 30.5mm and the other 18mm.

I believe that those two distances are optimizing for different things.

If I recall correctly, the 30.5 mm spacing is optimizing the antenna
for gain (the sharpest forward-looking lobe, likely).

The narrower spacing seems to optimize the design for the best
impedance match (the biquad is very close to 50 ohms in this version).

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!