Who makes

Bifwynee/Bruce, good reply. Atleast I appreciate that you have taken the time to get a deeper understanding of the amp-speaker electrical interface. It's a complicated affair there - more complicated than meets the eye once one delves into the details. I'm sure that it's made you a better audio consumer & you can better make the trade-offs when you go out next to purchase audio gear. Like the SYMS clothing company once said in their TV ads "an educated consumer is a better consumer".

To Cdc:
before answering your question, let's take a step back & find out what is really meant by a "voltage paradigm". In a voltage paradigm amplifier, the output power doubles each time the speaker impedance halves (let us assume for the sake of simplicity that the power amplifier output has infinite current source/sink capacity. Such an amplifier is not available practically but it makes the discussion easier). The output power doubles because the output current capacity doubles; not because the output voltage capacity doubles.
For example:
300W into 8 Ohms. P=V^/R implies V = 49V approx.
600W into 4 Ohms. P=V^2/R implies V = 49V again!
1200W into 2 Ohms. P=V^2/R implies V = 49V once again!
So, as you can see, as impedance halves, output power doubles, the output voltage remains the same.
OTOH,
300W into 8 Ohms. P=I^2*R implies I = 6.12Amps
600W into 4 Ohms. P=I^2*R implies I = 12.24Amps. Output current doubled...
1200W into 2 Ohms. P=I^2*R implies I=24.49Amps. Output current doubled yet again!
So, output power is doubling as impedance is halving because output current is doubling.

Now that you understand this, a tube amp cannot double its output current as speaker impedance halves because tubes are output voltage devices & have large output impedance. The output transformer makes things much better to drive a speaker load but the ratio of amp output impedance to speaker impedance is quite high making it hard for the tube amp to output large amounts of current into a low speaker impedance.

Solid-state amps have output impedances that are sub-1Ohm (because the output stage of a s.s. amp is almost always the BJT emitter or a MOSFET source of a JFET source, which has very low output impedance & several semiconductor devices are paralleled to make the net output impedance even lower). So, even if the speaker impedance drops 8 --> 4 --> 2 & even 1 Ohms, the s.s. amp output impedance is still an order of magnitude (ie. 10X) lower than the speaker impedance. By the physics of this, a s.s. amp is able to output progressively more output current into a lower speaker impedance *while maintaining its output voltage* (as shown above).

A "power paradigm" amplifier cannot maintain its output voltage constant as speaker impedance varies over the audio band. This is the key difference between the voltage paradigm & power paradigm.

One would have to make a very high output impedance s.s. amp such that this s.s. amp cannot deliver large output current into a lower speaker impedance. As Bruce has pointed out "your questions are counter-intuitive and contradictory" because the very nature of s.s. amplifier design ensures very low output impedance: paralleling output devices decreases output impedance, negative feedback (local or global) reduces output impedance, increase in temperature reduces output impedance, etc. You have to go the extra mile to make the output impedance high in a s.s amp. IOW, s.s. are naturally voltage paradigm amplifiers.

Now I am stating & asking the forum at large: could McIntosh's philosophy of using output auto-formers be one technique to increase the output impedance of a s.s. amp (I believe that McIntosh makes several s.s. amps w/ autoformers, no?) such that these s.s. amps show some/a lot of the characteristics of tube amplifiers?

Another devislish way to make a s.s. amplifier behave in a power paradigm manner would be to use it with a very low impedance speaker such as using a s.s. amp to drive an Apogee Full Range speaker where the midrange is driven directly by the amp & has a speaker impedance of 0.14 Ohms (I did not make a mistake here!!). Now, the s.s. amp will see a speaker load that is in the same ball-park as its own output impedance & current will be limited (just like a tube amplifier seeing a 4-Ohm speaker load when its own output impedance is in the 4 Ohm region). Such an amplifier's life could be very short-lived if not chosen correctly. Devilish, as I wrote! ;-) Also, how many such speakers exist today? Almost zero. so, this scenario is not realistic.

I don't know of any other s.s. manuf that makes their s.s. amps power paradigm but memory & experience could be failing me.....
It would be great to hear from other far more knowledgeable members. Thanks.

Is the current output more important than the voltage as that is what gives the speaker its dynamic range? What is voltage good for? It seems like a cheap way make the amp appear to be powerful when reading the specs.

voltage & current are duals of each other - where you will find voltage, you'll find current flow. The converse is also true - Where you'll find current, you'll find voltage. Voltage & current cannot exist without each other.
Think of voltage as electric pressure (old engineering texts had voltage as E) similar to water pressure. Current flows from higher electric pressure to lower electric pressure.
When an amp is delivering power into a speaker it's a combination of voltage & current. The voltage impressed (by the amp) at the speaker terminals causes current to flow in the passive x-over components. Depending on the various values of R, C, L in the x-over components, these respective components develop a voltage across them proportional to the music signal. This gets xferred to the speaker drivers. It's the voltage signal that causes pistonic movement in the speaker drivers & that produces sound. No voltage, no sound. So, voltage is good for a whole lot of things. How much current can be output from the power amp depends on how robust its power supply is. So, as Mapman suggested, not all watts are created equal in the sense that 2 100W amps might have vastly different power supplies. in such a case, the power amp with a more robust power supply will be able to drive a more difficult speaker load because it will be able to output more current into the difficult load to create a proportional-to-music-signal voltage for the speaker drivers.
Amps with a weaker power supply cannot drive difficult loads (yet it will still be a 100W/ch amp).

Yes, you are correct, Bob. Here we are talking about power amps & power amps driving speakers so in both cases there is an electrical circuit completed around the voltage source (the amp) which will cause current flow.
I assumed this due to nature of the question from the OP.

If you play a song at 90 dB with 110dB musical peaks like rim shots, don't you need the current to give that dynamic range for the 10 milliseconds?

yes, you do. this sudden burst of current comes from the power supply capacitors. like you wrote, the rim shots (your example) are extremely quick & fleeting events. By the time the bridge rectifier reacts to this quick event, the event itself has passed. The power supply itself cannot react fast enough to quick events & that is by design - it's supposed to be a DC power supply that remains steady no matter what (given the amp is being driven within its limits). SO, it's the capacitor bank of the power supply that reacts to these quick events. That's why many amp manuf boast about how much power supply cap they have + you'll find many other amp manuf to have bypass caps in parallel w/ the power supply main cap. These bypass caps are much smaller (10,000uF) with very low ESR such that they can react very quickly to rim shot events.

Ralph,

Now if the speaker is only going down to 4 ohms, the fact that the amp can't double power into that impedance does not mean that it is not a voltage source. This is due to the fact that the feedback of the amplifier will make it act like a voltage source independently of the amp's ability to double power.

ok. feedback keeps the output impedance low such that the lower impedance of the speaker is still much higher than the amp output impedance & the amp acts like a voltage source.

Now its understood that adding negative feedback to an amplifier reduces its output impedance, right? But right here we see that this really is not the case at all.

i'm having a lot of trouble accepting this. There's a closed form equation that clearly shows that negative feedback reduces output impedance. output impedance is reduced by a factor of gain*feedback factor. Now, if gain of the amp falls off, then you can keep adding negative feedback & it will not reduce the output impedance much at all. Most power amps are AC-coupled amps so their response at the low end is a high-pass. is that why amp gain is rolling off at low freq & negative feedback is not having the desired effect.

If a circuit really has a lower output impedance, it can therefore drive lower impedance loads without loss of performance. So if negative feedback really did reduce output impedance, you could make any amplifier drive 2 ohms without losses just by adding more feedback!

I don't think so. ability to drive a lower speaker impedance will depend on the output stage (more output current needs to be shared by more output devices), how much current the power transformer can supply, heatsinking ability (all these points you've mentioned in your next sentence). You can keep adding negative feedback but If the amp is incapable of supplying the current, additional negative feedback does nothing.

From this we can see that the term 'output impedance' as used by the Voltage Paradigm does not in fact refer to the actual output impedance of the amplifier at all! Instead, it refers to the how the amplifier *reacts* to its load impedance with its voltage response. That is something quite different.

I have no idea what you've written here!

So in our example of the inexpensive solid state amp that cannot quite double its power into 4 ohms, it is still a voltage source as its feedback causes it to *limit* its output power into lower impedances, based on what it can linearly do into higher impedances.

i'm not sure that this making any sense. your statement seems to imply that this example power amp has intelligence in that it can figure out how much power it can output linearly into a higher impedance & store that in its memory & then restrict its output to that same power level when it encounters lower impedances. Nah, I don't think that happens. I believe that your example power amp will simply run out of ability to drive a lower impedance when it draws all the current it can based on its power output stage & its power transformer.

Perhaps Ralph or Bombaywalla will have some additional thoughts on your question, but that's the best I can do on it.

Thanks, Al.
like Ralph & you, I was trying to understand what the strength of the ARC Ref150 power supply is when Bruce wrote 1040J. Tell me if I'm wrong:
150W/ch into 8 ohms - I calculate that the secondary is at 35VAC.
Energy = Voltage * Current * time
Then, 1040J/35VAC gives me a Current * time product = 29.71.
So, *supposing* the Ref150 power supply can provide 10Amps, then 29.71/10Amps = 2.971 secs. That's a long time to supply that much current.
How I read this is that with 1040J of storage energy, the Ref150 can supply 10Amps for 2.971 secs while maintaining 35VAC on the secondary.
With music program material no transient is going to last that long meaning that the Ref150 power is pretty darn robust....

12-06-13: Almarg
Ralph, thanks for the thorough explanations. I too was having some difficulty understanding some of this, but after reading your two posts on the subject a couple of times I think I follow what you are saying.

Thank you, Al, for this post. Good to know that I was not the only one having trouble interpreting what Ralph wrote. I know that if YOU are having trouble interpreting Ralph then I'm well within the right to be confused as well. ;-)

Ralph, thanks for your 2nd post - you have made things much clearer than what you wrote in your orig post on this matter. I think that I now follow what you are saying.

I like Al's question to you tho' as I was thinking along those lines as well. Please clarify further. Thank you.