Hello Walt,

We have just a voltage source of 212Vrms with 100 Ohms in series. This
is all in a black box and called my PA. That black box has a UHF
socket and that mentions "100 W into 50 Ohms".

When you connect 50 Ohms to the SO 239 socket, you get a voltage
divider. 2/3 of the EMF is across the internal 100 Ohms and 1/3 is
across the 50 Ohms load. This results in 70.7V across the load (so
100W).

Does this make sense?

Wim
PA3DJS

On 14 mayo, 01:28, K7ITM wrote:
On May 11, 7:15*am, Wimpie wrote:



On 10 mayo, 23:37, walt wrote:

[deleted]

Now Wim, your math is very challenging: You state 100w delivered by
the source, but at one point you also state 111w is delivered to the
100-load--at another point you state that 100w is delivered to the 100-
ohm load. Which is it?

You also state that voltage out of the source is 212.1v--sum ting wong
here. 212.1v across 50 ohms yields 899.73w, and 212.1v across 100 ohm
yields 449.86w. These power values are nowhere near the values
appearing in you statements.

In my calculations, with 100w the voltage across a 50-ohm line is
70.71v, and across a 100-ohm line the voltage is 100v.

With a 50-ohm line terminated in 100 ohms, the vrc is 0.3333 as you
stated, thus the power-reflection coefficient is 0.1111. This means,
as I've stated continually, that with 100w delivered by the source,
the power reflected at the 2:1 mismatch is 11.111w, which when added
to the 100w supplied by the source, makes the forward power 111.11w.
Now with 11.111w of power reflected at the mismatch, this leaves 100w
delivered to the 100-ohm load, not 111w. Please tell me where the 111w
came from. You also haven't told us why you calculate 130w forward and
15w reflected. Someone else may have made these calculations, but it
was you I asked for an explanation, because you repeated those
calculations.

When using the correct physics and math in this example, how can the
results *be so different?

To close, let me present the procedure I use to calculate the total
forward power--I guess it is related to Ohm's Law:

With the power-reflection coefficient as prc, the forward power PF = 1/
(1 - prc).

So Wim, can you clarify the confusion you appear to have made?

Walt

Hello Walt,

The source (that is the PA in this case) produces an EMF (as mentioned
before) of 212.1Vrms, Source impedance is 100 Ohms (so forget the
"conjugated match" thing for this case).

When terminated according to the numbers above the socket ("100W into
50 Ohms"), this results in 100W *( Vout = 212.1*50/(50+100) = 70.7V,
just a voltage divider consisting of 100 Ohms and 50 Ohms ).

70.7V into 50 Ohms makes 100W.

Now we remove the 50 Ohms load and create a mismatch (referenced to 50
Ohms), by connecting a 100 Ohms load (VSWR = 100/50 = 2)

Now the output voltage will be:

Vout = 212.1*100/(100 +100) = 106V

106V into 100 Ohms makes 112W (forgive me the one Watt difference).

This all without any transmission line theory.

Now the forward and reflected power balance (this is exactly the same
as in my previous posting, except for using 112W instead of 111W):

100 Ohms equals |rc| = 0.3333 (50 Ohms reference).

Hence "reflected power" = 0.333^2*"incident power" * = 0.111*"incident
power".

"net delivered power" = "incident power" – "reflected power" = 112W.

"net delivered power" = (1-0.11111)*"incident power" = 112W.

"incident power"= 126W, "reflected power" = 14 W

As far as I can see, there is nothing new with respect to the previous
calculation, except for the truncation/rounding.

Similar reasoning can be applied to my class-E PA. It is designed for
500W into 4.5 Ohm, but the output impedance is below 1 Ohms. If you
try to achieve conjugate match, you fry the active devices. *Same is
valid for most audio amplifiers, they may mention: *"80W into 8 Ohm",
but that does not mean that Zout = 8 Ohms.

Awaiting your comment,

Wim
PA3DJSwww.tetech.nl

Of course, with pretty much any decently designed power amplifier, no
matter the frequency, included protection circuits will limit the
dissipation and voltage in the critical areas (e.g., the output active
devices). *Loads become disconnected or shorted sometimes; winds blow
stray wires or tree limbs across antennas. *Through a piece of
transmission line, such a change can reflect back to become any
possible phase angle, and any of a very wide range of magnitudes, at
the amplifier's output.

Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. *Off
topic: *the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. *That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. *

Hello Tom,

I agree with your statement. If you have a CLC LPF section that
introduces about 90 degrees phase delay (see it as a quarter wave
section) and it looks into an open load (so zero power is delivered
to this open load), this will result in a low impedance at the input
of the CLC network. If the plate is connected at this CLC network,
the plate's dissipation will greatly increase.

Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. *Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

I also fully agree on this one.


Cheers,
Tom

With kind regards,

Wim
PA3DJS
www.tetech.nl

On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote:

Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. Off
topic: the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

I'm going to slip a mickey into this by a careful, editorial change of
focus BACK to the subject line.

Though it's a red herring typical of Ham-speak, most modern retail
100W RF transmitters for amateur service have a very low output
impedance, a fraction of an ohm [ editorial: until, of course, it goes
to the Z transformer that precedes the bandpass filter].

It must be a rare condition for which there is a transition frequency
below which audio amps source rated at sub Ohm Z feeding loads up to
10 times the sub-Ohm design work - and above which retail Amateur
transmitters at loads up to 10 times the sub-Ohm design do not work
[editorial: without that Z transformer]. Curious logic.

I wonder if that works (sic) backwards? Would the RF deck's finals
feed a speaker with audio with equal power performance of the Audio
amp? Both are sub-Ohm sources, and the RF deck certainly has the
bandwidth.

I await the specification of this transition frequency and the
analysis of how the 100W RF final deck would fail to supply 100W audio
(however crummy it may sound).

Any problems that might arise for the retail Ham transmitter are
already handled by protection circuitry so that's a wash.

73's
Richard Clark, KB7QHC

On May 13, 8:57*pm, Richard Clark wrote:
On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote:
Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. *Off
topic: *the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. *That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. *Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. *Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

I'm going to slip a mickey into this by a careful, editorial change of
focus BACK to the subject line.

Though it's a red herring typical of Ham-speak, most modern retail
100W RF transmitters for amateur service *have a very low output
impedance, a fraction of an ohm [ editorial: until, of course, it goes
to the Z transformer that precedes the bandpass filter].

It must be a rare condition for which there is a transition frequency
below which audio amps source rated at sub Ohm Z feeding loads up to
10 times the sub-Ohm design work - and above which retail Amateur
transmitters at loads up to 10 times the sub-Ohm design do not work
[editorial: without that Z transformer]. *Curious logic.

I wonder if that works (sic) backwards? *Would the RF deck's finals
feed a speaker with audio with equal power performance of the Audio
amp? *Both are sub-Ohm sources, and the RF deck certainly has the
bandwidth. *

I await the specification of this transition frequency and the
analysis of how the 100W RF final deck would fail to supply 100W audio
(however crummy it may sound).

Any problems that might arise for the retail Ham transmitter are
already handled by protection circuitry so that's a wash.

73's
Richard Clark, KB7QHC

Yes Wim, I understand your black box, and the voltage-divider action
that occurs when terminated into 50 ohms. But you didn't answer my
question, as you normally don't do. So I'll ask it again, as I did in
my previous post:

"( Vout = 212.1*50/(50+100) = 70.7V," I asked for the mathematical
basis for this equation. I agree that it's mathematically correct, but
212.1 is voltage, or E, but where does 212.1*50 come from? The
equation appears to be of the form E/R = E, which is absurd. In what
equational form is this equation? I'm trying to learn here.
Walt
"

On 5/13/2011 8:08 PM, walt wrote:
On May 13, 8:57 pm, Richard wrote:
On Fri, 13 May 2011 16:28:52 -0700 (PDT), wrote:
Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. Off
topic: the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

I'm going to slip a mickey into this by a careful, editorial change of
focus BACK to the subject line.

Though it's a red herring typical of Ham-speak, most modern retail
100W RF transmitters for amateur service have a very low output
impedance, a fraction of an ohm [ editorial: until, of course, it goes
to the Z transformer that precedes the bandpass filter].

It must be a rare condition for which there is a transition frequency
below which audio amps source rated at sub Ohm Z feeding loads up to
10 times the sub-Ohm design work - and above which retail Amateur
transmitters at loads up to 10 times the sub-Ohm design do not work
[editorial: without that Z transformer]. Curious logic.

I wonder if that works (sic) backwards? Would the RF deck's finals
feed a speaker with audio with equal power performance of the Audio
amp? Both are sub-Ohm sources, and the RF deck certainly has the
bandwidth.

I await the specification of this transition frequency and the
analysis of how the 100W RF final deck would fail to supply 100W audio
(however crummy it may sound).

Any problems that might arise for the retail Ham transmitter are
already handled by protection circuitry so that's a wash.

73's
Richard Clark, KB7QHC

Yes Wim, I understand your black box, and the voltage-divider action
that occurs when terminated into 50 ohms. But you didn't answer my
question, as you normally don't do. So I'll ask it again, as I did in
my previous post:

"( Vout = 212.1*50/(50+100) = 70.7V," I asked for the mathematical
basis for this equation. I agree that it's mathematically correct, but
212.1 is voltage, or E, but where does 212.1*50 come from? The
equation appears to be of the form E/R = E, which is absurd. In what
equational form is this equation? I'm trying to learn here.
Walt
"

I apologize for stepping in, Walt, but, maybe this will help:

Vout = 212.1 * (50/(50+100))

In other words, Vout is equal to Vsource times the divider ratio.

I know you know this. I think you just misread it.

Cheers,
John

On 14 mayo, 03:08, walt wrote:
On May 13, 8:57*pm, Richard Clark wrote:



On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote:
Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. *Off
topic: *the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. *That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. *Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. *Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

I'm going to slip a mickey into this by a careful, editorial change of
focus BACK to the subject line.

Though it's a red herring typical of Ham-speak, most modern retail
100W RF transmitters for amateur service *have a very low output
impedance, a fraction of an ohm [ editorial: until, of course, it goes
to the Z transformer that precedes the bandpass filter].

It must be a rare condition for which there is a transition frequency
below which audio amps source rated at sub Ohm Z feeding loads up to
10 times the sub-Ohm design work - and above which retail Amateur
transmitters at loads up to 10 times the sub-Ohm design do not work
[editorial: without that Z transformer]. *Curious logic.

I wonder if that works (sic) backwards? *Would the RF deck's finals
feed a speaker with audio with equal power performance of the Audio
amp? *Both are sub-Ohm sources, and the RF deck certainly has the
bandwidth. *

I await the specification of this transition frequency and the
analysis of how the 100W RF final deck would fail to supply 100W audio
(however crummy it may sound).

Any problems that might arise for the retail Ham transmitter are
already handled by protection circuitry so that's a wash.

73's
Richard Clark, KB7QHC

Yes Wim, I understand your black box, and the voltage-divider action
that occurs when terminated into 50 ohms. But you didn't answer my
question, as you normally don't do. So I'll ask it again, as I did in
my previous post:

"( Vout = 212.1*50/(50+100) = 70.7V," *I asked for the mathematical
basis for this equation. I agree that it's mathematically correct, but
212.1 is voltage, or E, but where does 212.1*50 come from? The
equation appears to be of the form E/R = E, which is absurd. In what
equational form is this equation? I'm trying to learn here.
Walt
"

Walt,

First you should change language, because if you don't understand
something, it is not always the other party who is wrong.

For a voltage divider:

Vout = Vin*R1/(R1+R2).

R1 is the lower resistor , here 50 Ohms, R2 is the upper resistor,
here 100 Ohms (inside the black box). I can't make it easier. Vin =
212.1 Vrms.

Wim

On May 13, 9:17*pm, Wimpie wrote:
On 14 mayo, 03:08, walt wrote:









On May 13, 8:57*pm, Richard Clark wrote:

On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote:
Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. *Off
topic: *the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. *That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. *Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. *Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

I'm going to slip a mickey into this by a careful, editorial change of
focus BACK to the subject line.

Though it's a red herring typical of Ham-speak, most modern retail
100W RF transmitters for amateur service *have a very low output
impedance, a fraction of an ohm [ editorial: until, of course, it goes
to the Z transformer that precedes the bandpass filter].

It must be a rare condition for which there is a transition frequency
below which audio amps source rated at sub Ohm Z feeding loads up to
10 times the sub-Ohm design work - and above which retail Amateur
transmitters at loads up to 10 times the sub-Ohm design do not work
[editorial: without that Z transformer]. *Curious logic.

I wonder if that works (sic) backwards? *Would the RF deck's finals
feed a speaker with audio with equal power performance of the Audio
amp? *Both are sub-Ohm sources, and the RF deck certainly has the
bandwidth. *

I await the specification of this transition frequency and the
analysis of how the 100W RF final deck would fail to supply 100W audio
(however crummy it may sound).

Any problems that might arise for the retail Ham transmitter are
already handled by protection circuitry so that's a wash.

73's
Richard Clark, KB7QHC

Yes Wim, I understand your black box, and the voltage-divider action
that occurs when terminated into 50 ohms. But you didn't answer my
question, as you normally don't do. So I'll ask it again, as I did in
my previous post:

"( Vout = 212.1*50/(50+100) = 70.7V," *I asked for the mathematical
basis for this equation. I agree that it's mathematically correct, but
212.1 is voltage, or E, but where does 212.1*50 come from? The
equation appears to be of the form E/R = E, which is absurd. In what
equational form is this equation? I'm trying to learn here.
Walt
"

Walt,

First you should change language, because if you don't understand
something, it is not always the other party who is wrong.

For a voltage divider:

Vout = Vin*R1/(R1+R2).

R1 is the lower resistor , here 50 Ohms, R2 is the upper resistor,
here 100 Ohms (inside the black box). *I can't make it easier. Vin =
212.1 Vrms.

Wim


Thank you for the clarification, Wim. I was unaware of that equation
relative to voltage dividers. I'm going to have to do some thinking on
that equation before I'll understand the relevance of the numerator,
or how it relates to the calculation.

Walt

On May 13, 9:36*pm, walt wrote:
On May 13, 9:17*pm, Wimpie wrote:









On 14 mayo, 03:08, walt wrote:

On May 13, 8:57*pm, Richard Clark wrote:

On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote:
Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. *Off
topic: *the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. *That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. *Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. *Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

I'm going to slip a mickey into this by a careful, editorial change of
focus BACK to the subject line.

Though it's a red herring typical of Ham-speak, most modern retail
100W RF transmitters for amateur service *have a very low output
impedance, a fraction of an ohm [ editorial: until, of course, it goes
to the Z transformer that precedes the bandpass filter].

It must be a rare condition for which there is a transition frequency
below which audio amps source rated at sub Ohm Z feeding loads up to
10 times the sub-Ohm design work - and above which retail Amateur
transmitters at loads up to 10 times the sub-Ohm design do not work
[editorial: without that Z transformer]. *Curious logic.

I wonder if that works (sic) backwards? *Would the RF deck's finals
feed a speaker with audio with equal power performance of the Audio
amp? *Both are sub-Ohm sources, and the RF deck certainly has the
bandwidth. *

I await the specification of this transition frequency and the
analysis of how the 100W RF final deck would fail to supply 100W audio
(however crummy it may sound).

Any problems that might arise for the retail Ham transmitter are
already handled by protection circuitry so that's a wash.

73's
Richard Clark, KB7QHC

Yes Wim, I understand your black box, and the voltage-divider action
that occurs when terminated into 50 ohms. But you didn't answer my
question, as you normally don't do. So I'll ask it again, as I did in
my previous post:

"( Vout = 212.1*50/(50+100) = 70.7V," *I asked for the mathematical
basis for this equation. I agree that it's mathematically correct, but
212.1 is voltage, or E, but where does 212.1*50 come from? The
equation appears to be of the form E/R = E, which is absurd. In what
equational form is this equation? I'm trying to learn here.
Walt
"

Walt,

First you should change language, because if you don't understand
something, it is not always the other party who is wrong.

For a voltage divider:

Vout = Vin*R1/(R1+R2).

R1 is the lower resistor , here 50 Ohms, R2 is the upper resistor,
here 100 Ohms (inside the black box). *I can't make it easier. Vin =
212.1 Vrms.

Wim

Thank you for the clarification, Wim. I was unaware of that equation
relative to voltage dividers. I'm going to have to do some thinking on
that equation before I'll understand the relevance of the numerator,
or how it relates to the calculation.

Walt


Thank you for stepping in, John, as from you and Wim I understand that
that equation represents the voltage-divider circuit. However, I'm
still agonizing over the numerator in the ratio. I just can't fit it
into the voltage-divider concept. Gimme a little time and it will
probably sink in.

Walt

On Fri, 13 May 2011 17:57:27 -0700, Richard Clark
wrote:

I await the specification of this transition frequency and the
analysis of how the 100W RF final deck would fail to supply 100W audio
(however crummy it may sound).


:-)

That was way too hard.

On May 13, 5:57*pm, Richard Clark wrote:
On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote:
Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. *Off
topic: *the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. *That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. *Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. *Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

I'm going to slip a mickey into this by a careful, editorial change of
focus BACK to the subject line.

Though it's a red herring typical of Ham-speak, most modern retail
100W RF transmitters for amateur service *have a very low output
impedance, a fraction of an ohm [ editorial: until, of course, it goes
to the Z transformer that precedes the bandpass filter].

It must be a rare condition for which there is a transition frequency
below which audio amps source rated at sub Ohm Z feeding loads up to
10 times the sub-Ohm design work - and above which retail Amateur
transmitters at loads up to 10 times the sub-Ohm design do not work
[editorial: without that Z transformer]. *Curious logic.

I wonder if that works (sic) backwards? *Would the RF deck's finals
feed a speaker with audio with equal power performance of the Audio
amp? *Both are sub-Ohm sources, and the RF deck certainly has the
bandwidth. *

I await the specification of this transition frequency and the
analysis of how the 100W RF final deck would fail to supply 100W audio
(however crummy it may sound).

Any problems that might arise for the retail Ham transmitter are
already handled by protection circuitry so that's a wash.

73's
Richard Clark, KB7QHC

FWIW, I've built audio amplifiers with output transistors with f-sub-t
around 50MHz. Those transistors certainly would provide decent power
gain at the lower frequency HF ham bands. They were used in the audio
amp to achieve very low distortion across the audio spectrum (a very
few PPM at full rated power output).

One big difference between (typical) audio amplifiers and (typical) RF
power amplifiers is in the use of negative feedback. In an audio
amplifier, voltage-derived negative feedback yields a very low
amplifier output impedance. In almost all RF power amplifiers, little
or no negative feedback is used, so the output impedance is generally
much higher than with an audio amplifier running similar power and
supply voltage to the output devices.

But again, the source impedance of a transmitter is seldom important
in the application of the amplifier. What's important are things like
the optimal load impedance and the rated power output. I'd put things
like distortion specs for a linear amp far above source impedance in
importance. If you (the lurking reader) think source impedance is
important, please explain in detail _why_.

Cheers,
Tom