What is the most powerful audio output tube?

boomer#6877250@none.com wrote:

> What is the most powerful audio output tube, as far as RMS wattage
> output?
> 
4CX25000A would be good!   Hmmm, I think the filament runa about 10 V at 170 
A, you can run at least 10 KV on the anode at several amps.  The output 
transformer would be about the size of a small car.

I was at a Greatful Dead concert in 1969 at the Fox Theater in St. Louis, 
and they used the modulator out of an AM broadcast transmitter as an audio 
amp.  The tubes were about a foot in diameter.

Jon

>"(55V/1.4)^2/4 is about 400W. Allow some clipping... "

It's 378. Number one, how much clipping do I allow. Ten percent ? That is hard to take even for regular people let alone anything even close to an audiophile. Also, that 55 volts was measured with no load. The units have basic iron, not a tightly regulated SMPS. No way in hell it maintains that. 

>"From the above, you must think they use 10mohm speakers.  ;-) "

Watts is volts times amps. The speaker impedance does not matter. I remember when Fostgate had not been bought by Rockford and they rated into eight ohms and used higher rail voltages. After the buyout they started going to the lower impedance load for their power ratings. 

You do realize that higher impedance dynamic speakers are more efficient right ? All that lower impedances do is to not make you kick up the voltage as much, but per watt, with all else equal, lower impedance is less efficient per watt. NOT per volt. 

In fact even house stereo speaker manufacturers have gotten away from 1 watt/ 1 meter sensitivity ratings because to get the good sound they have to use that choke in the crossove that makes the load more reactive, and plus like in the case of my Boston Acoustics which sounded great, a 3.1 ohm woofer. And it had inductance in front of it to bring out that bottom octave. 

So now many of them use @ 2.83 VRMS/1 meter for that spec. And in the meantime most consumer amps cannot handle that load. I have had a few pairs of speakers with good bass that kicked in the current limiting on some amps. 

So now you are telling me they don't even use a sine wave ? how the hell do they measure the distortion then ? To have to old rating system it had to be a sine wave because the distortion meter requires a sine wave. 

Unless they use the null method but then it has to be phase compensated to be fair, and that is pretty much a bunch of shit. 

Generally I take the easy way out. I just use integrated program material, rock usually and crank it into a real load until it clips, and I see the rails dropping and the ripple in it in the envelope on the scope. I take that peak value and just say "Your amp clips at ____ watts". Older amps do better because they didn't have quite the linearity at the upper ranges of their power, and that is what gave us 3dB clipping headroom, because like Pioneer, all of them, they did not want to claim that power rating at 8 % distortion. And it cost them in bigger heatsinks ad all that, but their customers didn't mind the extra watts. Or the extra weight. 

<"You have to look at the definitions.  They're not always what you
think they should be but they are what they are. "

Then apply the same standards and accept the "RMS watt". I will still be measuring it the same way.

boomer#...
>
> 
> 
> It would be nice to know why (RMS watts is a misnomer),
>

** People who see a term being used in trade or commerce etc and think they know what it means from a literal interpretation of the words are called idiots.

A *term* means what the folk USING it intend it to mean.

A bottle of "Steak Sauce" contains no steak, but from the term it ought to be made from steak. 

The term "watts RMS" is defined and used to mean the average power output measured with a sine wave signal and specified load. 

Only morons reverse the word order and imply that "RMS watts" are a special kind.


....  Phil 
 

On Sat, 10 Dec 2016 19:47:28 -0600, boomer#6877250@none.com wrote:

>On Sat, 10 Dec 2016 17:51:34 -0500, krw <krw@somewhere.com> wrote:
>
>>>It would be nice to know why (RMS watts is a misnomer), without having
>>>to watch a frikkin video, which in my case is not possible. Some of us,
>>>myself included do not have access to high speed internet. Videos can be
>>>fun, but why make a video to state something that can be stated in a
>>>paragraph?
>>
>>"RMS" is short for "root means square", or the square-root of the
>>"average" of the values squared.  Power is voltage-squared times
>>resistance.  The power produced by a voltage source is the effective
>>voltage times the resistance.  For DC, the effective voltage is the
>>same as the voltage.  In the case of a sine wave, the "effective"
>>voltage is the peak to peak voltage divided by SQRT(2).  For other
>>wave shapes the difference between the effective (or RMS) voltage and
>>the peak voltage is different but the RMS voltage is the effective
>>voltage.  
>>
>>Now go back to power, (RMS)Voltage * (RMS) voltage / resistance =
>>power.  It makes no sense to say RMS * RMS = RMS.  It's not, it's just
>>"power".  RMS has no physical meaning for power.
>> 
>
>Thanks for the useful info. Is there any method to determine the ACTUAL
>power output from an amplifier? 

The classical method of measuring arbitrary waveform power up into the
RF, microwave and light frequencies is to feed the unknown waveform to
a resistor.

An other identical resistor is heated by a variable DC source, which
is adjusted until the resistors are at the same temperature. Then jus
measure the DC source voltage and current and multiply them together.

Of course, this is a slow method, so this is a good way for long time
output power measurements.

A more modern method of measuring how much power is actually delivered
into a speaker is to put a small (less than 0.1 ohm) in series with
the speaker. Using a two channel (stereo) ADC, measure the amplifier
output voltage directly from the amplifier terminals on one channel
(L) and the voltage drop (current) across the resistor (R). 

Put the resistor in the speaker return wire to avoid ADC common mode
issues.

Assuming the ADC runs at 48 kHz, for each sample first scale both
samples so that they represents Volts resp. Amperes. Then multiple the
L and R channel scaled samples with each other (representing the
instant ious energy for that sample) and add the product to an
accumulator. After 1 second divide the accumulated value by 48000 to
get directly watts. If you want 1 minute average, after one minute,
divide by  2,880,000. Then reset accumulator for next period.
      

Phil Allison <pallison49@gmail.com> wrote:

> boomer#...@none.com wrote:
> > 
> >
> > Thanks for the useful info. Is there any method to determine the ACTUAL
> > power output from an amplifier?
> >
> 
> ** Maker's specs are normally quite correct for any AC powered amplifier.

That may be true in the industrial amplifier market, but in the consumer
market I don't think it is even vaguely true.   Such meaningless
expressions as "peak audio power" are common.




> 
> > Yes, I know that with a sine wave, there
> > will be bursts of power, with most being from the bass or low end.
> 
> ** That has no comprehensible meaning. 
> 
> 
> > But
> > there has to be a way to state the actual Maximum power any amp can
> > produce.
> >
> 
> **  See the maker's specs. 
> 
> > Not only to know the abilities of an amp, but also to choose a
> > speaker(s) that can handle the maximum power. 
> 
> ** There is no easy way to do that and a lot depends on what the purpose
> of the amplifier and speaker is - plus who is going to operate the system.
> 
> Unlike amplifiers, most speakers have no fixed limit on max power input -
> only  a limit on the average power they can handle over a period of time.
> The problem is that speaker maker's never tell you what that is.
> 
> 
> ...   Phil 


-- 

Roger Hayter

<boomer#6877250@none.com> wrote:

> On Sat, 10 Dec 2016 17:51:34 -0500, krw <krw@somewhere.com> wrote:
> 
> >>It would be nice to know why (RMS watts is a misnomer), without having
> >>to watch a frikkin video, which in my case is not possible. Some of us,
> >>myself included do not have access to high speed internet. Videos can be
> >>fun, but why make a video to state something that can be stated in a
> >>paragraph?
> >
> >"RMS" is short for "root means square", or the square-root of the
> >"average" of the values squared.  Power is voltage-squared times
> >resistance.  The power produced by a voltage source is the effective
> >voltage times the resistance.  For DC, the effective voltage is the
> >same as the voltage.  In the case of a sine wave, the "effective"
> >voltage is the peak to peak voltage divided by SQRT(2).  For other
> >wave shapes the difference between the effective (or RMS) voltage and
> >the peak voltage is different but the RMS voltage is the effective
> >voltage.  
> >
> >Now go back to power, (RMS)Voltage * (RMS) voltage / resistance =
> >power.  It makes no sense to say RMS * RMS = RMS.  It's not, it's just
> >"power".  RMS has no physical meaning for power.
> > 
> 
> Thanks for the useful info. Is there any method to determine the ACTUAL
> power output from an amplifier? Yes, I know that with a sine wave, there
> will be bursts of power, with most being from the bass or low end. But
> there has to be a way to state the actual Maximum power any amp can
> produce. Not only to know the abilities of an amp, but also to choose a
> speaker(s) that can handle the maximum power. 
> 
> If this is all based on mathematics, I am a not very good with math...
> (I'm being honest about that).

Maths may be needed to understand the measurements, but very little is
needed to do the actual measurement.  The power output of an audio
amplifier is defined as the power produced into a defined load (often
four or eight ohms) as an audio sine wave at a given frequency, or range
of frequencies, with a given degree of distortion.   The power in an
audio sine wave can be at most about a quarter of maximum voltage peak
the amplifier can achieve into a suitable load resistance sqared and
divided by that resistance.   Because by defiinition we are talking
about power in an audio sine wave.  Any power above that would be very
distorted.  In theory an amplifier could produce about four times as
much power as a square wave as it could as a sine wave.    So to measure
it you need a standard resistive load, a voltmeter and a distortion
measuring  meter of some kind.  Oh, and a signal generator to drive the
amplifier with an audio signal of the desired frequency. 

For a solid state amplifier the supply voltage to the output stage and
for a valve amplifier the designed anode dissipation of the output
valves can give a very good idea of the maximum possible power an audio
amplifier could produce.  It is an interesting exercise to compare that
with the claimed power output of an amplifier.  At the very least,
comparing this with the claimed power output can give one an idea of
what class of people the makers are trying to sell the amplifier to.

-- 

Roger Hayter

Roger Hayter wrote:


> > >
> > > Thanks for the useful info. Is there any method to determine the ACTUAL
> > > power output from an amplifier?
> > >
> > 
> > ** Maker's specs are normally quite correct for any AC powered amplifier.
> 
> That may be true in the industrial amplifier market,
>

** No, it is generally true of audio amplifiers for hi-fi and pro audio. 


> but in the consumer
> market I don't think it is even vaguely true. 

 ** Think again. 

> Such meaningless
> expressions as "peak audio power" are common.
> 
>

** In car audio and portables. 

 Not "AC powered ".


....  Phil 

Roger Hayter wrote:

>
> 
> Maths may be needed to understand the measurements, but very little is
> needed to do the actual measurement.  The power output of an audio
> amplifier is defined as the power produced into a defined load (often
> four or eight ohms) as an audio sine wave at a given frequency, or range
> of frequencies, with a given degree of distortion.
>

** Correct.

> The power in an
> audio sine wave can be at most about a quarter of maximum voltage peak
> the amplifier can achieve into a suitable load resistance sqared and
> divided by that resistance. 

** Wrong - and how very dumb.


> n theory an amplifier could produce about four times as
> much power as a square wave as it could as a sine wave. 

** Wrong again, same basic error. 

   The power ratio is 2:1.


> For a solid state amplifier the supply voltage to the output stage and
> for a valve amplifier the designed anode dissipation of the output
> valves can give a very good idea of the maximum possible power an audio
> amplifier could produce. 
>

** Wrong about valve amplifiers. 

A pair of output valves can be run in class A, class AB, class B and class B split rail modes.

Efficiency varies from 30% to 75% over that range  - so the power output varies by a large factor. 

....  Phil 

Phil Allison <pallison49@gmail.com> wrote:

> Roger Hayter wrote:
> 
> >
> > 
> > Maths may be needed to understand the measurements, but very little is
> > needed to do the actual measurement.  The power output of an audio
> > amplifier is defined as the power produced into a defined load (often
> > four or eight ohms) as an audio sine wave at a given frequency, or range
> > of frequencies, with a given degree of distortion.
> >
> 
> ** Correct.
> 
> > The power in an
> > audio sine wave can be at most about a quarter of maximum voltage peak
> > the amplifier can achieve into a suitable load resistance sqared and
> > divided by that resistance. 
> 
> ** Wrong - and how very dumb.
> 
Noted.  


> 
> > n theory an amplifier could produce about four times as
> > much power as a square wave as it could as a sine wave. 
> 
> ** Wrong again, same basic error. 
> 
>    The power ratio is 2:1.

You are of course correct.  I was confused by the rail to rail voltage,
as indeed some advertisers seem to be!

> 
> 
> > For a solid state amplifier the supply voltage to the output stage and
> > for a valve amplifier the designed anode dissipation of the output
> > valves can give a very good idea of the maximum possible power an audio
> > amplifier could produce. 
> >
> 
> ** Wrong about valve amplifiers.
> 
> A pair of output valves can be run in class A, class AB, class B and class
> B split rail modes.
> 
> Efficiency varies from 30% to 75% over that range  - so the power output
> varies by a large factor.
> 
> ....  Phil

I was saying that the maximum possible can be predicted - if the
designer decides to use class A then even this won't be achieved.   Just
a rule of thumb to exclude ridiculous claims about "peak music power" or
whatever.


-- 

Roger Hayter

On Sat, 10 Dec 2016 21:02:34 -0800 (PST), Phil Allison
<pallison49@gmail.com> wrote:

>
>Unlike amplifiers, most speakers have no fixed limit on max power input - only  a limit on the average power they can handle over a period of time. The problem is that speaker maker's never tell you what that is.  

The speaker coil resistance increases with temperature (and hence of
previous power input). The power limit can be calculated by knowing
the coil maximum temperature at a specific power (actually voltage
output for a typical amplifier).

The power limits for a closed speaker is more predictable, but for
bass reflex enclosures, you can easily blow the cone out at 20 Hz (not
to mention 3 Hz, which is typically the frequency, at which the final
amplifier drops to 1, due to the feedback loop) amplifier.