"Richard Harrison" bravely wrote to "All" (01 Mar 05 08:29:27)
--- on the heady topic of " Say what you mean."

RH From: (Richard Harrison)
RH Xref: aeinews rec.radio.amateur.antenna:26189

RH The Class A amplifier gets all its power from the d-c supply and it is
RH constant, signal or no signal. With signal power output, some of the
RH power in exits to the load.

Let's look at it from the dynamic point of view. When there is no
signal the circumstances are as you describe. When the signal swing
turns the device nearly OFF there is a maximum voltage across the
device but the current is a minimum or close to zero.

Conversely when the signal swings in the opposite polarity the device
turns ON hard but the voltage is at a minimum or near zero too. In
either case the "ideal" device dissipates no extra power because the
two products are always zero. In fact the power difference is
always zero, so there is no reason for the device to cool.

However, no device is ideal, it has losses, and generates harmonics.
If the rectification is such that the average current decreases then
the device will cool. There is a gotcha however, because the harmonics
rob power from the desired output.


RH I`ll use Cecil, W5DXP`s argument. Energy must be conserved. Energy in
RH equals energy out. If some goes to a load it does not stay within the
RH amplifier to make feat.

As you said, if the amplifier dissipates 100W when idle and it can
drive 100W into a load then the power source will still supply 100W
when so doing, of course being an ideal device and neglecting all
other losses. It is easy to test by paralleling a substantial
capacitance and measuring the supply current before the cap. The
current should remain the same or rise only very slightly.


RH Asimov also wrote:
RH "---linear source not operating Class A?"

RH I`ll give an example. The Class B amplifier is biased near current
RH cut-off. Current is near zero when the signal is. Yet, output can
RH favorably vie with that from a Class A amplifier for purity. I learned
RH that nearly 60 years ago when I built my first 6N7 phonograph
RH amplifier.

Class B is hardly linear if it only amplifies 1/2 of the sinewave.
Time for our periodic 10 year review. g

A*s*i*m*o*v

.... That's a cute trick.

Asimov wrote:
In fact the power difference is
always zero, so there is no reason for the device to cool.

Here's the reason the device cools under load.

Plate loss = Eb*Ib - Ip^2*RL

With no signal, the plate loss is Eb*Ib. When
the signal is increased from zero, any power
delivered to the load is subtracted from Eb*Ib,
thus causing the device to cool.
--
73, Cecil http://www.qsl.net/w5dxp


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Asimov wrote:
"Let`s look at it from the dynamic point of view (loss in a Class A
amplifier)."

The no-signal loss of a Class A amplifier is 100%. It equals volts x
amps and appears in the amplifier. Now feed a signal to the ideal
amplifier set just below the clipping level. Average d-c power is
unchanged from the unloaded and no-signal conditions.

Connect a matched load resistor to the amplifier output. If physically
small, the resistor may become warm with heat that were it not for the
load would be otherwise dissipated in the amplifier. Input power to the
Class A amplifier is unchanging.

Finding the internal resistance theoretically is simple. It is simply
the open-circuit output voltage divided by the short-circuit current.

Open-circuit voltage at full output and short-circuit current may be
severe. Internal resistance can be found under less stressful
conditions. Internal resistance will drop the voltage to any load
reasistance. Use the voltage-divider formula to calculate the internal
resistance.

With pure resistances, half the open-circuit volts are dropped by the
internal resistance when the load is a match.

A power amplifier`s internal impedance can be determined.

Power output from a Class A amplifier cools it.

Best regards, Richard Harrison, KB5WZI

Richard, your method of measuring internal impedance of an amplifier sounds
interesting. (o/c volts divided by s/c current)

At present I have no facilities to make such measurements.

Perhaps a few curious readers, who do have facilities, might make crude
measurements on their HF rigs and report approximate impedance values on one
or two bands on this newsgroup. Some sort of average could be obtained.
----
Reg, G4FGQ
===================================

"Richard Harrison" wrote in message
...
Asimov wrote:
"Let`s look at it from the dynamic point of view (loss in a Class A
amplifier)."

The no-signal loss of a Class A amplifier is 100%. It equals volts x
amps and appears in the amplifier. Now feed a signal to the ideal
amplifier set just below the clipping level. Average d-c power is
unchanged from the unloaded and no-signal conditions.

Connect a matched load resistor to the amplifier output. If physically
small, the resistor may become warm with heat that were it not for the
load would be otherwise dissipated in the amplifier. Input power to the
Class A amplifier is unchanging.

Finding the internal resistance theoretically is simple. It is simply
the open-circuit output voltage divided by the short-circuit current.

Open-circuit voltage at full output and short-circuit current may be
severe. Internal resistance can be found under less stressful
conditions. Internal resistance will drop the voltage to any load
reasistance. Use the voltage-divider formula to calculate the internal
resistance.

With pure resistances, half the open-circuit volts are dropped by the
internal resistance when the load is a match.

A power amplifier`s internal impedance can be determined.

Power output from a Class A amplifier cools it.

Best regards, Richard Harrison, KB5WZI

Reg Edwards wrote:
Richard, your method of measuring internal impedance of an amplifier sounds
interesting. (o/c volts divided by s/c current)

Reg, I tried that one time in college. I got as far as torching
two ARC-5 1625's driving an open circuit so I don't know what
would have happened with a short circuit.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Reg Edwards wrote:

Richard, your method of measuring internal impedance of an amplifier
sounds
interesting. (o/c volts divided by s/c current)


Reg, I tried that one time in college. I got as far as torching
two ARC-5 1625's driving an open circuit so I don't know what
would have happened with a short circuit.

Some of Motorola's devices might actually be able to stand
Reg's little test. The MRF150, for instance is advertised as
being able to withstand a 30:1 VSWR at all phase angles. I
think Reg should offer to pay for any damage sustained as the
result of this test.
73,
Tom Donaly, KA6RUH

Reg Edwards wrote:
Richard, your method of measuring internal impedance of an amplifier
sounds
interesting. (o/c volts divided by s/c current)

Reg, I tried that one time in college. I got as far as torching
two ARC-5 1625's driving an open circuit so I don't know what
would have happened with a short circuit.
--
73, Cecil

==============================
Cecil,
Presumably you hadn't heard of Ohms Law.

The internal resistance of a generator is independent of its internal
voltage. Just reduce the drive level to some small, no particular value,
and stop making excuses.

( Somebody will say it IS dependent. But we are interested only in
ball-park accuracy. And in any case the operating point will be within the
normal range of operation between no-drive and full drive, or between no
modulation and full modulation.)
----
Reg, G4FGQ

Reg Edwards wrote:
The internal resistance of a generator is independent of its internal
voltage. Just reduce the drive level to some small, no particular value,
and stop making excuses.

( Somebody will say it IS dependent. But we are interested only in
ball-park accuracy. And in any case the operating point will be within the
normal range of operation between no-drive and full drive, or between no
modulation and full modulation.)

That might work better for Class-AB than Class-C. If one
reduces drive level on a Class-C, one is reducing the 'on'
time of the device, thus changing the internal impedance
probably somewhat inversely proportional to the 'on' time.

If I remember correctly, my IC-706 would not fold back
under any load condition when run at minimum power.
--
73, Cecil http://www.qsl.net/w5dxp

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"Richard Harrison" bravely wrote to "All" (02 Mar 05 10:35:29)
--- on the heady topic of " Say what you mean."

RH From: (Richard Harrison)
RH Xref: aeinews rec.radio.amateur.antenna:26249

RH Asimov wrote:
RH "Let`s look at it from the dynamic point of view (loss in a Class A
RH amplifier)."

RH The no-signal loss of a Class A amplifier is 100%. It equals volts x
RH amps and appears in the amplifier. Now feed a signal to the ideal
RH amplifier set just below the clipping level. Average d-c power is
RH unchanged from the unloaded and no-signal conditions.

RH Connect a matched load resistor to the amplifier output. If physically
RH small, the resistor may become warm with heat that were it not for the
RH load would be otherwise dissipated in the amplifier. Input power to
RH the Class A amplifier is unchanging.
[,,,]
RH Power output from a Class A amplifier cools it.

Okay, it cools. Enough said. I was confusing it with Class B which
does warm up the heatsink. ;-)

A*s*i*m*o*v

.... May you find the light and walk the mountain tops.

Asimov wrote:
"I was confusing it with Class B which does warm up the heatsink"

True. Class B amplifiers can be single-ended for R-F if used with the
right tank circuit to supply the missing 1/2-cycle. More often they are
push-pull to reduce harmonics.

SSB final amplifiers are biased to near current cut-off. During
modulation gaps, current falls to nearly zero. When current is zero, so
is the dissipation.

Reg mentioned the blistering heat of the 6N7. Thai`s my recollection too
when getting nearly 10 watts of audio from the small twin-triode
octal-based metal tube. Reduce the audio volume and the heat is reduced
simultaneously.

Best regards, Richard Harrison, KB5WZI