In article ,
Jeff Liebermann wrote:
Oh-oh.
*grin*
Ground #1: the ferrite beads would only block non-balanced current
flow back down the outside of the coax. They will have no effect at
all on power which is reflected back down the inside of the coax (the
center conductor and the inside of the shield) from any impedance
mismatch where the coax meets the folded dipole.
Agreed, if there were a balun present. Without a balun, there's going
to be some radiation from the coax cable, presumably from the shield.
Disclaimer: I don't fully understand how baluns work, and certainly
don't understand what the piece of coax with ferrite beads on the
MFJ-1800 is suppose to do.
A piece of coax, with several ferrite beads or cores around it, *IS* a
balun. It's a 1:1 "current balun".
It has the effect of ensuring that current flow into the radiator (the
folded dipole, in this case) is balanced - that is, the current
flowing into each of the FD's two feedpoint ends is equal.
In effect, the presence of the beads (if they're choking the coax
properly) actually ensures that the transmitter *does* see the true
effect of any impedance mismatch. The transmitter is just as likely
to see a higher SWR than a lower one, when the beads are added.
You lost me here. Just what are the beads suppose to do if not block
the reflected signal? Presumably, they do serve some useful purpose.
Now, you're suggesting that they could make the VSWR worse?
Well, let's see. From your questions, and from your comment that you
don't really understand how baluns work, I suspect that there are
several aspects of what's going on here which are being "mushed
together". Let's tease them apart, with an example, and see if this
clarifies thing.
We're dealing with the following inter-related things: characteristic
impedance, VSWR, reflections, balance, RF flowing where you don't want
it, and feedline radiation.
Let's start with a simple example: a folded dipole (characteristic Z
of 300 ohms), being fed by a twin-wire feedline (characteristic Z also
300 ohms), from a balanced RF source. RF is going to flow up the
feedline... and the current at a point on each conductor will be
equal-and-opposite that of the point immediately opposite it on the
other conductor. The feedline is "balanced".
At each point on the feedline, the RF voltage and the RF current will
be in a specific relationship, as defined by Ohm's Law (E=IR).
Rewrite this as R=E/I, divide the voltage you measure by the current
you measure, and you'll get 300 ohms as the result... this ratio is
the characteristic impedance of the line.
When the RF reaches the folded dipole, it will "see" the same
characteristic impedance and will flow into the dipole without any
sort of disruption. Because the dipole is physically symmetrical, and
(just as important) because there's nowhere else for the current
flowing up either side of the feedline to go, the antenna is also
balanced, and equal-and-opposite currents will flow into the two sides
of the antenna.
This is a nice setup, electrically - everything behaves as if we would
like.
Now, let's change the FD to a loop, with a feedlpoint impedance of
around 100 ohms. That is to say, when current flows into the
feedpoint, it must do so where the ratio of voltage-over-current is
lower.
When RF flowing up the (balanced, 300-ohm) feedline hits the loop's
feedpoint... well, we have a mismatch. The voltage-to-current ratios
in the feedline and the loop will differ. The physical effect of this
is that only part of the RF power successfully enters the loop. The
rest is reflected back down the feedline, as a wave travelling in the
opposite of the original direction. [The same thing happens when a
radar wave in space hits an object - the impedance difference between
free space and the material of the object reflects much of the power
back towards its origin.]
The forward (original) and reverse (reflection-generated) waves will
reinforce (constructively) and interfere (destructively) as they
travel along the lines. At points where their voltages are in phase,
the voltages will add and the voltage across the line will be higher.
At points where they interfere (phases are opposed) the voltages will
be lower. The ratio between these two (higher-voltage and
lower-voltage) points defines the VSWR, and the transmitter will "see"
this.
In this case, we have reflected power... and we have a VSWR of higher
than 1:1. However, the whole system is still balanced - it's
physically and electrically symmetrical.
Now, let's set that example aside for a moment, and look at another
one.
Let's consider a simple dipole (center-fed, non-folded), with an
impedance of 72 ohms. This is a physically symmetrical (balanced)
antenna.
Let's feed this with coax, having a characteristic of 72 ohms.
Is the coax balanced? Yes and no. In one sense, it's balanced... if
you look at the current flow on the inside of the coax, you'll find
that the flows on the center conductor, and the inside of the shield,
are equal-and-opposite. Also, the EM fields created by the RF flow
are entirely within the coax (if the shield has full coverage, as in a
hardline), which means that RF doesn't "leak out" and inteference has
a hard time "leaking in". This is why you can run coax through
conduit or next to metal objects without causing problems.
In another sense, though, the coax is not balanced, because it
consists of *three* conductive surfaces... the center conductor, the
inside of the shield, and the outside of the shield. When current
flowing up the coax reaches the end (e.g. the antenna feedpoint), the
symmetry can be broken:
- Current flowing up the center conductor has only one place it can
go - to one side of the antenna feedpoint.
- Current flowing up the inside of the shield has *two* paths to
which it can flow. It can flow into the other side of the antenna
feedpoint, *and* it can "wrap around" the end of the shield and
flow back down the outside of the coax. It will tend to do both,
in a complex ratio which depends on the impedances that it "sees"
for each of these two paths.
RF flowing back down the outside of the coax will generate radiation,
which will add to / interfere with the RF radiatiating from the
dipole. This will tend to alter the antenna's pattern somewhat.
In this example (72-ohm coax connected to 72-ohm dipole), we won't
necessarily see an even current flow into the dipole, with a nice
clean 1:1 VSWR on the line. We can still see some reflected power.
Why? Because of the "third wire" - the outside of the coax. Its
impedance will appear in parallel with that of one side of the
dipole... it will 'accept' some RF current (in effect, diverting it
from its side of the dipole) and this will alter the feedpoint
impedance of the antenna system. The imbalance of the coax (the
presence of the "third wire") has disrupted our nice pretty
theoretical picture. We've got more current flowing into the
center-conductor side of the antenna (where there's only one path)
than we have flowing into the other side (since part of the current
has been diverted to flow back down the outside of the coax), and the
antenna is no longer behaving in a balanced fashion.
How do we prevent this from happening? By using a balun... a device
which has an unbalanced (usually coaxial) feed on one side, and which
*forces* an electrical balance on the other side. Some baluns force
equal-and-opposite voltages on the two terminals on the balanced side,
while others force equal-and-opposite currents on either side.
Some baluns are designed to present the same impedance on each side
(1:1 baluns) while others are designed to have different impedances
connected to either side (the 4:1 balun is the best-known).
A series of ferrites, clamped around a coax, are a simple (and often
very effective) form of current balun. There's no impedance
transformation built into this design - it's a 1:1 type.
How does this "choke balun" create balance? Quite simply: it
blocks current flow on the "third wire" (the outside of the coax) by
placing a high RF impedance in series with any current flow on this
path.
As an example: consider the 72-ohm coax-and-dipole example. Let's
assume, for the sake of argument, that the RF impedance seen along the
outside of the coax (the "third wire") is on the order of 100 ohms or
so (not unreasonable if it's a couple of wavelengths long). Without a
balun of some sort, there's going to be a fair amount of current flow
back along this "third wire", because the impedance isn't all that
much higher than the 36 ohms (that of one-half of the dipole) in which
it is in parallel with.
Now, let's clamp a few nice ferrites onto the coax. These add *lots*
of inductance to the "third wire"... for the sake of argument let's
say it's on the order of 1000j ohms.
All of a sudden, the third wire presents a *much* higher impedance at
the feedpoint than it used to... it's 30 times as high as that of the
dipole segment to which it's connected. Only a tiny amount of current
can flow into the "third wire".
Since the current flowing up the inside of the feedline is balanced,
and since the current reaching the feedpoint now has only one place it
can go effectively (into the antenna), the current flow into the
antenna is symmetrical - balance has been restored. Because the
feedpoint impedance is no longer being significantly affected by the
"third wire", we see almost exactly a 1:1 VSWR.
If we've added enough inductance (or enough RF resistance) to the
"third wire", and choked off the current flow to essentially nothing,
then the power loss in the choke is negligible... because there isn't
enough current flowing to generate power to dissipate.
This is beginning to sound like the discussion about whether it's
acceptable to use a high VSWR antenna. The standard answer is that if
the transmitter can take it without going into protection or
oscillation, then it's acceptable.
I think that's oversimplifed a bit. I think there are really three
factors to consider:
- Will the transmitter be stable and happy (not self-protect, not
oscillate, not burn up)?
- Will the transmitter actually deliver full rated power (or close to
it) into the impedance that it "sees" as a result of the higher
SWR? This impedance may be lower or higher than its rated load,
and may have significant reactance in it.
- Will the additional ("excess") losses in the transmission line, as
a result of the higher current at some points, be a problem?
I don't have a problem with that,
although I've never seen a Yagi-Uda antenna with a 50 ohm coax hung
directly onto a folded dipole because there are usually easy ways to
do the matching and balanced to unbalanced conversion.
The "Cheap Yagi" design is the closest to that... a half-folded
dipole, with the directors and reflectors positioned so that the HFD
has a 50-ohm feedpoint impedance. Direct coax connection (ferrite
bead is optional).
Digging a deeper hole, I've been assuming that if the ferrite beads
were not there, the coax cable will radiate. After all, that's one
purpose of a balun, to prevent coax radiation from mangling the
pattern.
It could radiate, depending on the length of the coax.
That's still a dubious proposition due to the large length
of exposed center conductor at both ends of the coax piece, which
certainly will radiate some. I can add that to the model, but I don't
know how to model the ferrite beads.
I believe you'd want to model the coax balun as a combination of two
elements: a pure transmission line, and a "third wire" connected at
the antenna feedpoint and running back parallel to the transmission
line (this being the effect of the outside of the coax).
To model the ferrite beads, add a lumped impedance in series with the
third wire, at the point at which it connects to the antenna. As a
wild-and-almost-certainly-wrong SWAG, try adding 100+100j (to mimic
the purely inductive, and the lossy-resistive aspects of the ferrite).
--
Dave Platt AE6EO
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