On Mon, 28 Feb 2005 06:28:24 GMT, "George, W5YR"
wrote:

IF the Thevenin source approach worked, we would have to be content with max
50% efficient amplifiers. We know we can do better than that substantially.

Hi George,

I know you won't appreciate this, but it is a telling, simple test of
a practical situation with a practical Amateur grade transistor model
100W transmitter commonly available for more than 20-30 years now:
1. Presuming CW mode into a "matched load" (any definition will do);
2. Report the DC power consumed before hitting the key;
3. Report the DC power consumed while holding the key.

Concurrently note:
A. Report Heat Sink Temperature for a previously idle/rcv condition;
B. Report Heat Sink Temperature after 10 minute key-down.

For a hypothetical "100W" model (again, a contemporary, common example
for Amateur use) available through standard commercial venues:
2. About 20W - 30W
3. About 200W - 250W
A. About 20 degrees C (or room temperature)
B. Well above 37 degrees C (or skin temperature)

Now, if we are to be any judge of efficiency (Thevenin does not have
to be invoked, condemned, or venerated); then it runs close to 50%
(±10%). Others can invoke their favorite deity to explain.

Now, if we are to be any judge of dissipation (no requirement for
advanced degree); then heat as a loss by virtue of less than 100%
efficiency is quite evident. Others can invoke photons to describe
why.

To forestall any armchair engineers, yes, this efficiency is System
efficiency. However, I would be surprised if a practical common
Amateur grade transistor model transmitter commonly available for more
than 20-30 years now has any configuration that does not apply supply
voltage directly to the final transistors; and instead adds a
significant current path outside of this load (citations to available
schematics would be compelling, but any argument without this would be
speculation). It takes very little effort to subtract out the power
drain of the receive mode (being very representative of the similar
power demand of supporting circuitry for transmit up to the driver
stage). Barring such amazing evidence of a significant power drain
not found in the finals, it follows that a simple computation of
efficiency has its merit and has been met.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
To forestall any armchair engineers, yes, this efficiency is System
efficiency...

.... which is not the definition of "plate (anode) efficiency"
from the IEEE Dictionary: "The ratio of load circuit power
(alternating current) to the *plate* power input (direct current)."
For amplifier efficiency calculations, only the DC power input
to the plate-collector-drain is considered, by definition. The
power dissipated in the surrounding support circuitry is not
included in the definition of "plate efficiency".

Reference: _Electronic_Fundamentals_and_Applications_ by
John D. Ryder, Copyright 1954, by Prentice-Hall, page 348,
Section 10-6. Plate-Circuit Efficiency
--
73, Cecil http://www.qsl.net/w5dxp


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On Mon, 28 Feb 2005 08:23:53 -0600, Cecil Moore
wrote:
Reference:
An obscure deity indeed....

Richard Clark wrote:

Cecil Moore wrote:
Reference: John D. Ryder

An obscure deity indeed....

He authored three of Prentice-Hall's Electrical
Engineering Series books back in the 50's. He
was Michigan State's Dean of Engineering at the
time.
--
73, Cecil http://www.qsl.net/w5dxp


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On Mon, 28 Feb 2005 10:57:24 -0600, Cecil Moore
wrote:
An obscure deity indeed....
He authored three
What an impoverished disciple of an obscure deity. Did he get his
ankles wet measuring supply current - or are we talking water level
above the knees?

Richard Clark, KB7QHC wrote:
"However, I would be surprised if a practical common Amateur grade
transistor model transmitter commonly available for more than 30 years
now has any configuration that does not supply voltage directly to the
final transistors---."

My Kenwood TS-130S manual has specifications on page 2. Transmitter
output impedance is specified as: 50 OHMS!

Om page 30 is a level diagram. Output from the transmitter`s low-pass
filter is measured as 70.7 volts at 14.25 MHz, which is 100 watts into
50 ohms (square root of PR).

D-C power is fed to the center-tap of a push-pull output transformer to
the final transistors.

From the specifications page also, the power reguirement is TX: 18A
13.8V DC. It`s a linear amplifier. Only 40% efficiency. The designer
probably was more interested in low harmonics than efficiency. The final
by itself only takes part of the 18A ao its efficiency is more than 40%.

Kenwood says its transmitter presents an impedance of 50 ohms at its
"OUT" terminals in the diagram on page 30. 100 watts at 70.7 volts.

Best regards, Richard Harrison, KB5WZI

On Mon, 28 Feb 2005 09:55:34 -0600, (Richard
Harrison) wrote:

From the specifications page also, the power reguirement is TX: 18A
13.8V DC. It`s a linear amplifier. Only 40% efficiency. The designer
probably was more interested in low harmonics than efficiency. The final
by itself only takes part of the 18A ao its efficiency is more than 40%.

Hi Richard,

My point exactly. However, how much more than 40% and how much of
that 18A (248W) is not devoted to the finals? Your "typical" rig
would most like be composed of a three stage design with 5W/20W/100W
kind of progression - very common.

The manufacturer of your finals transistors would be quick to spec
them with a ballpark of 45% efficiency (Class AB, in a CW application)
in the best of circumstances for full power.

Perhaps someone else has better Pout/Pin figures for a commonly
available, retail Amateur 100W multi-mode HF rig? Say like 13.8V @
10A for 100W CW?

Again, appeals to deities are always amusing (say what you pray), but
we are missing the photon (sic) explanation (see what you mean) to
completely round this out.

73's
Richard Clark, KB7QHC

"Richard Clark" bravely wrote to "All" (27 Feb 05 23:49:44)
--- on the heady topic of " Say what you mean."

RC From: Richard Clark
RC Xref: aeinews rec.radio.amateur.antenna:26138

RC On Mon, 28 Feb 2005 06:28:24 GMT, "George, W5YR"
RC wrote:

IF the Thevenin source approach worked, we would have to be content with max
50% efficient amplifiers. We know we can do better than that substantially.

RC Hi George,

RC I know you won't appreciate this, but it is a telling, simple test of
RC a practical situation with a practical Amateur grade transistor model
RC 100W transmitter commonly available for more than 20-30 years now:
RC 1. Presuming CW mode into a "matched load" (any definition will do);
RC 2. Report the DC power consumed before hitting the key;
RC 3. Report the DC power consumed while holding the key.

RC Concurrently note:
RC A. Report Heat Sink Temperature for a previously idle/rcv condition;
RC B. Report Heat Sink Temperature after 10 minute key-down.

RC For a hypothetical "100W" model (again, a contemporary, common example
RC for Amateur use) available through standard commercial venues:
RC 2. About 20W - 30W
RC 3. About 200W - 250W
RC A. About 20 degrees C (or room temperature)
RC B. Well above 37 degrees C (or skin temperature)

RC Now, if we are to be any judge of efficiency (Thevenin does not have
RC to be invoked, condemned, or venerated); then it runs close to 50%
RC (110%). Others can invoke their favorite deity to explain.


The maximum power transfer 50% efficiency Po/Pin figure is only valid
for a "linear" amplifier termed Class A (current flows through the
whole cycle). Other amplifier classes were invented which improved on
that 50%. i.e. Class B (push-pull) can approach just shy of 70%
(current flows through 50% of cycle), and Class C close to 90%
(current flow less than 50% of cycle). Did I misunderstand the
problem?

A*s*i*m*o*v

.... "Beware of all enterprises that require new clothes." -- THOREAU

Asimov wrote:
.. The maximum power transfer 50% efficiency Po/Pin figure is only valid
for a "linear" amplifier termed Class A (current flows through the
whole cycle). Other amplifier classes were invented which improved on
that 50%. i.e. Class B (push-pull) can approach just shy of 70%
(current flows through 50% of cycle), and Class C close to 90%
(current flow less than 50% of cycle). Did I misunderstand the
problem?

What you missed is Richard C. trying to redefine the
efficiency of a final amp. :-)
--
73, Cecil http://www.qsl.net/w5dxp


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On Monday, 28 Feb 2005 09:21:38 -500, "Asimov"
wrote:

Did I misunderstand the problem?

Hi Isaac,

Probably. It only asked two things (both power readings).

73's
Richard Clark, KB7QHC