I posted an article along a similair vein to this in rec.radio.cb recently
(just hold that spit in your mouth for a moment!), but response has been
typically pathetic so...
I am no expert on the subject, so I may appear somewhat ignorant here (excuse
me!), but I'm intrigued by a glass-mounted antenna I've seen on sale here
in the UK - an antenna which claims to be the only one of its kind in the
world - designed to operate on 11m (CB) and, it is claimed, 10m (hence the
posting here).
It works on the same principle as some cellular 'phone antennae - an externally
mounted section stuck to the glass, coupled to a (for want of a
better description) black box mounted on the inside which is attatched
to the co-ax downlead.
I'm not a regular reader of this news group, so I don't know if this sort of
thing has been discussed in the past (probably ad-nauseum, knowing my luck).
I know that this type of antenna is available for other bands, but I'm sceptical
about the performance of something like this on 10-11m. Has anyone had
experience of this antenna (its called the Moonraker Astro Fantom - snappy
name, huh!), or anything similair and can comment on its effectiveness or
otherwise.
I'm basically facing the usual dilemma of not wanting to bore extra holes
in my car bodywork (secondhand buyers in Scotland are particularly
sensitive about non-standard apertures on cars on account of all the rain we
get here!)
Any advice would be much appreciated before I fork out the necessary 16.95
sterling.
Enrico V Vanni
------------------------------
Date: 12 Apr 91 14:50:42 GMT
From: sdd.hp.com!hp-col!hpctdlb!drn@ucsd.edu
Subject: 50 to 75 ohm transformer?
To: info-hams@ucsd.edu
One of the problems I see in using the 1/2 wave principle for matching the
75 ohm cable to the 50 source is attaching the connector. In his article
in _73 Amateur Radio_ July 1989, Edward Krome, KA9LNV, solves both the
impedance matching and the connector attachment problem by using some real
easy-to-get materials. For most of us, some type of connector attachment
arrangement would have to be fabricated to attach a type N connector to 3/4
inch hardline. Why not incorporate the matching transformers (only 8 inches
long at 432) at the same time?
I would be happy to forward a copy of the article if you can not find it in
your local library.
73's
Dave Novotny, WA6IFI
drn@hpdctdlb.col.hp.com
719-531-4494
------------------------------
Date: 12 Apr 91 14:39:41 GMT
From: sdd.hp.com!hp-col!hpctdlb!drn@ucsd.edu
Subject: 50 to 75 ohm transformer???
To: info-hams@ucsd.edu
There is a good article about constructing 75 ohm to 50 ohm impedance convertersin the July 1989 edition of 73 Amateur Radio. The construction techniques use
easy to get materials from your local parts store (N-flange connectors),
hardware store (copper pipe), and hobby store (copper tubing). Dimensions are
given for 144, 432, 902,and 1296 mHz bands. If you need a copy of the article,
E-mail me your address.
73's
Dave Novotny, WA6IFI
drn@hpctdlb.col.hp.com
HP Colorado Telecommunications Division.
------------------------------
Date: 11 Apr 91 16:01:28 GMT
From: hpfcso!hpfcdc!perry@hplabs.hpl.hp.com
Subject: Antenna Matching Gedanken Experiment
To: info-hams@ucsd.edu
It seems we have two camps here:
#1 sez that power hits the bad antenna, bounces back, and fries the
finals.
#2 sez that the power doesn't go to the antenna, stays in the
transmitter, and fries the finals.
I can't figure out the difference. Either way, you replace the finals.
:-)
In case anyone is counting votes, I support the theory #2. A (good)
feedline doesn't consume power, neither does a bad antenna. The power
has nowhere to go, and stays bottled up in the finals. Well-designed
finals sense the heat rise and turn down the DC input power, reducing
In article <1991Apr9.145118.24707@ux1.cso.uiuc.edu> sc80@ux1.cso.uiuc.edu (sc Student) writes:
>gary@ke4zv.UUCP (Gary Coffman) writes:
>
>>A transmitter is not a load! There seems to be a general misconception about
>
>Sure it's a load. Try pumping power into the feed line at the antenna end.
>Then measure what
>happens at the transmitter final. That's just what a high VSWR does. That is
>real power being reflected, and it will raise the voltage on the plate or
>collector. That's why transistors are PROTECTED with shut down circuits
>when high VSWR is present.
A transmitter is not a load. Let's try to explain by analogy. If you have
an open fire hydrant gushing water represent a transmitter and a guy with
a garden hose trying to squirt water into the open hydrant as your reflected
power, you can see that the water is totally reflected by the stream coming
out of the hydrant. Now the impedance against the water flowing out of the
hydrant increases, so without rematching, the flow from the hydrant
decreases by exactly the amount that is reflected from the garden hose
because the effective pressure required to deliver a given quanity of
water has changed. Therefore the total amount of water delivered out of the
hydrant, the original gush and the re-reflected squirt from the hose remains
the same. So your statement that the voltage will rise on the plate is
correct, *in the absence of retuning*, but this is simply because the
impedance presented at the transmitter output has changed. It is not because
reflected power is being absorbed by the "load" of the transmitter. By simply
retuning the transmitter's output network, or a matchbox, to match the new
impedance presented by the line, the voltage on the plate will return to normal.
>>this floating around. Let's see if we can clear it up. In classical
>>transmission line theory textbooks it is common to see a source represented
>>as a generator with a series resistor Rs. This Rs is referred to as the
>>"equivalent source resistance". It is stated that maximum power transfer
>>occurs when the load resistance Rl is equal to Rs. The system is said to
>>be matched under this condition. Now with two resistors of equal value
>>in series with a generator, half the power of the generator is dissipated
>>across each resistor. Therefore if this were a true description of our
>>transmitters, we could never exceed 50% efficiency in delivering power
>>to the load. Half the power would have to be dissipated in the transmitter
>>source resistance Rs. This is totally false, real transmitters have
>>efficiencies well over 50%.
>
>The increased efficiency is usually gained by shortening the ON cycle, such as in class C. The Q of the circuit then supplies the completion of the the
>cycle. In this "pulse" mode, the tube is ON less than one half the cycle,
>thereby giving higher efficiency. Compare the efficiency of class C, B, and
>A amplifiers.
Right answer to the wrong question. Restating the question, if in a matched
transmitter where Rs=Rl, and they are in series with the generator, the load,
Rl, will receive only half the power produced by the generator while Rs, the
purported real impedance of the final, dissipates the other half. This means
that the efficiency can never exceed 50% regardless of the duty cycle. A
low duty cycle only means that a *given* device can supply more peak power
than it's steady state rating. That a matched transmitter cannot deliver
more than 50% of it's power to the load is a false statement in a real
transmitter because Rs is not a real impedance. It is a fictitious value
derived from the instantaneous value of Ep/Ip and represents the operating
point of the amplifier, nothing more.
>>So what's the deal? The key here is the word "equivalent". What Rs represents
>>in a real transmitter is the *load-line* of the active device as transformed
>>by the output matching network. This is the *operating point* as defined
>>by the instanteous E/I of the output device. This is *not* a resistor. It
>>can't dissipate *any* power. To a signal being forced *into* the output
>>of an operating transmitter, it looks like an *open* circuit. Hence we
>>get *total* re-reflection.
Slight correction, looking into the transmitter one doesn't see an exact
open circuit, though conjugate match theory says it should be. Instead one
sees an extreme mismatch that for all practical purposes behaves just like
an open circuit. Current and voltage in the forward wave are in phase while
current and voltage in the reflected wave are 180 degrees out of phase. A
perfect conjugate match would behave like a quarterwave shorted stub causing
a perfect phase reversal of the reflected current in the re-reflected
wave. Nothing is perfect in the real world.
>True, the equivelent resistance is somewhat fictitious, but the power is
>very real, ie. plate voltage times plate current equals DISSIPATION. Try
False statement! Plate voltage times plate current = DC INPUT POWER.
Dissipation = DC input power - RF output power. It is a measure of the
inefficiency of a circuit in converting DC to RF.
>tuning your tube rig to maximum output on CW into a perfectly matched load
>at the antenna. Then mismatch the load to above 3:1 and watch those little
>babies turn red and melt down. Thats REAL power being absorbed by a real
>resistor (plate).
That's real *inefficiency* causing increased dissipation because you didn't
*retune* to match the new impedance at the transmitter output caused by
the 3:1 SWR. Remember that the definition of a transmission line transformer
is a length of line with a SWR other than 1:1, so what you have done is
create a transformer that is presenting an impedance to the transmitter
that it is not tuned for. The correct answer is to retune the transmitter.
>You don't use a matching device (presumably you meant at the transmitter,
>not the antenna) to lower the VSWR on the coax. You have to match the
>transmission line impedance at the antenna to significantly lower the
>VSWR on the line, although back matching at the generator will have a small
>effect - Roger K9ALD
Sorry, I misstated. (Fell into my own trap so to speak) What I was trying
to say was that you can use a matching device *at the transmitter* to
match the impedance presented by the line with other than 1:1 SWR *and*
that any additional coax losses suffered by operating this way will only be
*double* the coax losses of a flat line. At HF this is totally insignificant,
on the order of .46 db for 100 feet of RG-8. The reason that this is true
is because the *line* is acting as a transmission line transformer in
*both* directions. Therefore the re-reflected wave will be transformed
exactly the right amount on it's *second* trip up the line to match the
