Richard Harrison wrote:
1/2 or more of the power received by a receiving antenna is re-radiated.

Nearly all of the power received by a transmitting antenna is
transmitted.

Expanding a bit to make the receiving and transmitting systems symmetrical
with respect to power: If the transmitter is linear (like the antenna
is linear), i.e. Class-A, 1/2 or more of the generated power will be lost
in the source. In a linear resonant system, about 1/2 of the power sourced
reaches the antenna and about 1/2 of the received power makes it to the
receiver. It's the old maximum power transfer theorem at work.

A receiving antenna must be resonant to enable full acceptance of
available energy, and it must be matched to avoid re-radiation of more
than 50% of the energy it is able to grab.
If off-resonance, the receiving antenna has too-high impedance for
significant induced current. Of course, we have such good receivers we
can do without good efficiency.

A properly tuned antenna tuner ensures that the *antenna system* is resonant
for both transmit and receive (assuming the receiver's input impedance is the
same as the transmitter's output impedance). Note that an off-resonant antenna
*wire* is integrated into a resonant antenna *system* through the use of an
antenna tuner. Chapter 7 in _Reflections_II_ explains how even though it might
better have been titled, "My Transmatch Really Does Tune My Antenna" *SYSTEM*.
--
73, Cecil http://www.qsl.net/w5dxp



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On Sun, 22 Feb 2004 13:07:46 -0600 (CST),
(Richard Harrison) wrote:

A receiving antenna must be resonant to enable full acceptance of
available energy,

Where did you come up with that one?

I suggest you revisit capture area.

Danny, K6MHE

Cecil, W5DXP wrote:
"If the transmitter is linear (like the antenna is linear), i.e.
Class-A, 1/2 or more of the generated power weill be lost in the
source."

True, that would be an equalizer between reception and transmitting
system efficiencies of antennas, but Class A isn`t the only way to get
linear amplification, Hi-Fi nuts to the contrary not withstanding. Class
B is often used to combine efficiency with high undistorted output
capability. Class B amplifiers are biased to cut-off so they draw no
current when there is no signal input. A class B amplifier may have 60%
efficiency at full power output, for example. Such an amplifier will
have only about 30% efficiency at 1/2 of its maximum power output.

Turman writes on page 354 of his 1955 edition:

"With the largest signal that the (Class-B) amplifier can be expected to
handle satisfactorily, Emin/Eb will be small, and the actual efficiency
at full power is commonly of the order of 60%."

The receiving antenna can never be more than 50% efficient due to
re-radiation which I don`t seem to be able to explain. Sorry.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
The receiving antenna can never be more than 50% efficient due to
re-radiation which I don`t seem to be able to explain. Sorry.

It's because receiving antennas are linear devices which I don't
seem to be able to explain. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Don, K6MHE wrote:
"Where did you come up with that one?"
(A response to my statement that a receiving antenna must be resonant to
enable full acceptance of available energy)

I`ve tweaked antenna trimmers which dramatically boosted the signal when
reasonance was reached. I`ve seen grounded 1/4-wave structures near a
broadcast station detuned, thus eliminating the distortion they had
caused in the station`s radiation pattern. If they`re not resonant, they
don`t accept enough energy to make any difference in the station`s
pattern.

!/2-wave wires in free-space are resonant. Resonance is defined as unity
power factor, that is, XL=XC. At resonance, reactance is balanced out
and only resistance is left to oppose current in a wire. Usually the
wire has a radiation resistance which is large as compared with its loss
resistance in practical antennas.

At frequencies below first resonance, the ungrounded wire is less than a
1/2-wavelencth. It has a low radiation resistance and a high capacitive
reactance. We can add inductance to tune the wire to resonance.

At frequencies above first resonance, the ungrounded wire is more than a
1/2-wavelength, and if it is not much longer, the wiire has an inductive
reactance. The phase flip-flop at resonance is abrupt and the reactance
is an impediment to the current on either side of resonance. The correct
series capacitor can be placed in series with the roo-long wire to tune
out its excess inductive reactance.

A mechanical analog is the vibrating-reed frequency meter used at power
frequencies. All the reeds are in the power-frequency field. Only the
reed of resonant length has so little opposition to the excitation that
it vibrates freely.

A versatile antenna tuner can insert either inductive or capacitive
reactance in series with an antenna to correct its power factor (tune it
to resonance) so it can accept maximum excitation.

Best regards, Richard Harrison, KB5WZI

On Mon, 23 Feb 2004 08:54:50 -0600 (CST),
(Richard Harrison) wrote:

A versatile antenna tuner can insert either inductive or capacitive
reactance in series with an antenna to correct its power factor (tune it
to resonance) so it can accept maximum excitation.

I believe you are confusing matching with antenna operation (gain and
efficiency). Using a simple dipole for example, you can calculate both
gain and efficiency using NEC. Doing that I have modeled a dipole at
3/8, 1/2 and 5/8-wavelengths (80-meter band). The antenna was modeled
using #14 copper wire in free space. Here is what NEC reported:

Wavelength Gain(DBi) Efficiency

3/8 1.8 96.36%
1/2 (resonate) 2.02 97.22%
5/8 2.29 97.72%

Clearly you should be able to see that only advantage is that the
longer antenna has greater the gain and efficiency. The fact antenna
is resonance or not is not the determining factor.

73
Danny, K6MHE

"Richard Harrison" wrote in message
...
teve Nosko wrote:
"BTW--what is your line, Richard?"

I apologize for a critical tone in my response to Steve`s posting. An
ancient previous discussion of dissipationless resistance in this
newsgroup leaves me primed to comment when it appears unappreciated.

I only found it mildly critical. Your tone was not interpreted as
hostile in any way, just driven to add.

Dissipationless resistance is the stuff which allows a Class C amplifier
exceed 50% efficiency.

See comment on this later...

I won`t say I`ve been teaching X years, as I`ve never had that role.
Long ago, I found my patience and temperament unsuited to tutoring. I am
a long retired electrical engineer and find entertainment in the

Ahhh! I say, with an air of new found understanding.


newsgroups.

Best regards, Richard Harrison, KB5WZI

Danny,
You can't ever discard the factor Q in any discusion with respect to antenna
efficiency or any calculation for that matter.
Q is intrinsic in any calculation that determines efficiency especialy when
considering what the object of an antenna is.
Cecil's aproach to what it is the 'object' is to provide a total system ,
not a system that is led around by its nose by a predetermined antenna
structure is an example. The idea of designing a house around a workable
door that is pre-supplied is what we do today with respect to communication,
and is why I use a different antenna to the norm. When I am confident that
personal attacks come to a halt per Antennex statement I will be happy to
explain more in depth.
If you are content with what you have then that is understandable as humans
always resist change, including myself. You being an antenna guru I
understand even more the resistance to accept the possibility of advancement
from one who is less educated in the field than oneself.
Regards
Art




"Dan Richardson @mendolink.com" ChangeThisToCallSign wrote in message
...
On Sun, 22 Feb 2004 13:07:46 -0600 (CST),
(Richard Harrison) wrote:

A receiving antenna must be resonant to enable full acceptance of
available energy,

Where did you come up with that one?

I suggest you revisit capture area.

Danny, K6MHE

"Richard Harrison" wrote in message
...
Reg, G4FGQ wrote:
"What allows a class-C amplifier to exceed 50% efficiency is a small
operating angle."

While this is too vague, Richard tries to add detail, but mis-steps just
a bit... and Steve goes into an extended "You ain't quite correct blurb..."


Exactly, and during the majority of the degrees it`s switched completely
off. It draws no current and suffers no "IsquaredR loss" during the
amplifier off-time. Impedance is approximately E/I, but I is the average
I, which is much less than the bursts of I during the conduction angle.

We must be careful with the word "average" here.

First, my "class C" model is a follows:
I liken it to digital or "switched modes". While I have never scoped the
plate to observe this... When the tube is cut off for a large part of the
cycle, there is a high voltage on the tube (I believe it swings higher than
the supply dou to the "ringing" of the plate tuned circuit), but no current.
Hence, ExI=0.
When the tube is on, it is slammed hard on by the "high" grid signal and
there is a high plate current, but the plate voltage is very low (anybody
know how low and if I am all wet? ... tubes aren't quite like transistors
in the digital mode)--therefore ExI=somthing, but since the E is low, it is
lower than in class A during that part of a cycle. There may also be some
effect due to the fact that the plate tank is swinging low allowing the
plate voltage to be even lower.

Did you know that in class A, the plate power dissipated goes DOWN by
the amount that is delivered to the load??? Cool! huh? Isn't physics neat!

Second, it is the RMS current through the tube which will waste power,
so it is what we must be concerned with. Yes, if the tube is off the
current is zero at that time, but the RMS must be considered and it does not
go down as fast you might think. As an example, for the same current
pulses, but spaced out to half the duty cycle, the average drops to half,
but the RMS only drops to .707. There is a square root in there.

[[Anybody see the "AC Watt meter article in QST]] It is an OOPS! Most
power supplies don't draw sine wave current. It is pulses. I have been in
contact with both Bob Shrader (the author) and Stu Cohen (Tech editor) and I
just finished an analysis and am going to make more measurements to verify,
but the numbers Bob published can be as much as 1/3 the true power values
(depending upon the DVM he used and the current waveform of the supplies he
measured.

--
Steve N, K,9;d, c. i My email has no u's.

wa-da ya blokes think.



The switched-off time makes the I in the denominator of E/I very small
indeed and the solution to Ohm`s law is a high impedance without the
dissipation of a resistance that remains in place continuously while
agitating the atoms of a poor conductor to limit current.

Instead, we have a low-resistanc in high conducton for short spurts.
On-time is limited, instead of conduction, to produce a certain
effective resistance.

Another way of saying just wjat I did above, but "effective resistance"
is one way of thinking about it and this resistance must be calculated using
the RMS values.

An automobile Kettering ignition system may use a dwell-meter to
indicate how much of the time the points are closed. An ohmmeter
indicates the resistance between its test prods.

I'd be willing to place a bet (knowing how an analog ohm meter works,
that the *diflection* of the two meter pointers is the same (see below).
Both meters respond to the averacge current through them and both will show
full scale when the points are open (I think thta is the correct polarity).
Here's the "below":
There is, however, the confusion added by the coil/cap waveform for which
the ohm meter is not equiped to limit - whereas, I believe the dwell meter,
if well designed, will have something to limit so as to remofe it as a
complicatin.

--
Steve N, K,9;d, c. i My email has no u's..

The two test circuits
are almost the same although limitation of the deflection of the
dwell-meter is different from limitation of the deflection of the
ohmmeter due to the difference between limited conduction angle ignition
points, and the continuous conduction through a current-limiting
resistor. There`s an analogy between Class C and Class A amplifiers in
there somewhere.

Best regards, Richard Harrison, KB5WZI

On Mon, 23 Feb 2004 16:51:43 GMT, "aunwin"
wrote:

You can't ever discard the factor Q in any discusion with respect to antenna
efficiency or any calculation for that matter.
Q is intrinsic in any calculation that determines efficiency especialy when
considering what the object of an antenna is.

Hi Art,

Q is NOT the arbiter of all that is efficient. In fact high Q can
lead to very poor communication links.

The most efficient and simplest antenna, the dipole, exhibits a very
low Q for the very obvious reason: it is built to lose power by
design. The loss to radiation resistance, Rr, is indistinguishable to
Ohmic loss when computing Q. This in itself directly states that
maximizing Q is inimical to transmitting power if you do not separate
out the two losses.

Terman treats this inferentially in his discussion of Power Amplifiers
and their Plate Tank's Q. To select one that exhibits too high a
value is to risk very poor operation. He suggests that a Q of 8 to 15
is a reasonable value. This confounds many who seek to peak their
designs and fail to come to terms with unloaded and loaded Q
valuations.

It is the Q's relation to power loss to heat that makes the
difference, not to the power curve in isolation of this loss. Small
antennas suffer from high Q for this very reason - too little thought
is given to the radiation resistance's correlation to the Ohmic loss
of the system. As Richard has pointed out, one Ohm loss within the
structure is hardly a loss leader for an antenna with 73 Ohms Rr. To
achieve 50% efficiency requires your antenna to exhibit less than this
same value of Ohmic loss (however, let's be generous in comparisons to
1/1000th that value). The rage of "High Q" antennas is in various
loops of small diameters.

Let's look at small loops' Rr for various sizes in tabulated form:
Fo 1M diameter Efficiency with 1 mOhm loss
160M 29 µOhms 2.8%
80M 500 µOhms 33%
60M 1.5 mOhms 60%
40M 7.5 mOhms 88%
30M 24 mOhms 96%
20M 120 mOhms 99%

Let's examine the validity of that generous assignment of 1 mOhm loss
and see if it is reasonably warranted. Skin effect is the single
largest contributor to this loss as a source (aside from poor
construction techniques). Using the 1M diameter loop as being a
practically sized construction, and if were using 2.54cM diameter
copper wire/tubing we find:
Fo skin effect loss
160M 13.8 mOhms
80M 20 mOhms
60M 23 mOhms
40M 28 mOhms
30M 33 mOhms
20M 39 mOhms

Well, 1 mOhm was too generous and if we look at those loops' Rr once
again against a robust, thick loop element:
Fo 1M diameter Efficiency with skin effect loss
160M 29 µOhms 0.2%
80M 500 µOhms 2.4%
60M 1.5 mOhms 6%
40M 7.5 mOhms 21%
30M 24 mOhms 42%
20M 120 mOhms 75%

These "High Q" loops are NOT efficient, they are convenient. The two
terms are not the same at all and yet in common discussion they are
confused to mean the same thing.

I would point out further, that commercial vendors do not use 1 inch
tubing (as the numbers above would force to even bigger conductors).
Instead they use much larger tubing; but even here, one vendor uses a
flat strap which is a very poor substitute as the skin resistance is
defined at the edges and the face of the flat strap is far less
conductive. The physics of conduction forces current to seek the
smallest radius (the edge) to the exclusion of the broad surface (this
is why we use tubular conductors and not flat ones).

I will leave it to the student to reverse-engineer the required
conductor size to obtain the same 1 mOhm results of the first table
above. Even then, it will be seen that the common dipole still reigns
supreme in efficiency.

73's
Richard Clark, KB7QHC