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