Wrong wattage calculation on a review of an audio amp.

[BrianHG]

Wrong wattage calculation on a review of an audio amp.

Help, did this guy calculate the wattages wrong or have I've been doing audio wrong my entire life...
See his table at 10:50 where he lists the scope voltage reading in peak-to-peak and his calculated wattage...



This is what I calculate:
72v p-p into 8 ohm = 36v / 8ohm = 4.5 amp.
4.5 amp X 36v = 162 watts peak.
Or 162 watts peak * 0.707 = 115 watts RMS.
A far cry from what the presenter listed as 81 watts.

It is me that wrong with my 162/115 watts, or the presenter with his half peak power watts of 81?

The rest of his figures all all off by the same amount...

[themadhippy]

It is me that wrong

could be,

72v p-p into 8 ohm = 36v / 8ohm = 4.5 amp.

yer calculating the rms voltage from the peak to peak voltage wrong. you forgot to divide multiple it by 0.707

[DonKu]

162 watts peak * (0.707)^2 = 80.98 watts RMS.

[BrianHG]

162 watts peak * (0.707)^2 = 80.98 watts RMS.

Really?

Google: how to convert peak watts to true RMS watts
Response:

To convert peak watts to true RMS watts, multiply the peak watts by 0.707.

Remember, the square root of 2 = 1.41421356237 and 1/1.41421356237 = 0.7071067812.
You are not supposed to * by .707 to the power of 2, you are supposed to take the peak power and divide by the square root of 2, hence, 162/(sqr-root2)  or 162/1.41421356237 = 115 watts, not 81 watts.

True RMS wattage is not 50% peak wattage like with you 81 watt calculation.  That is just plain wrong!

162 watts peak would mean a square wave at 72vp-p.
115 watts RMS would means a sine wave with 72vp-p.
81 watts would mean a square wave with 50% on duty cycle.  Sine waves deliver more effective power than that.

[TimFox]

The peak power for a sine wave is twice the mean power (rms power is a vulgar error).
72 V pk-pk = 36 V peak = 25.46 V rms.
(25.46)² / 8 = 81 W mean power.

[magic]

You end up multiplying RMS voltage by peak current, should be RMS voltage by RMS current for in-phase sinewave.

Google: how to convert peak watts to true RMS watts
Response:

To convert peak watts to true RMS watts, multiply the peak watts by 0.707.

Also ask it how many genders there are if you are at that

[fourfathom]

162 watts peak * (0.707)^2 = 80.98 watts RMS.

Really?

Yes, really. 
Assuming a sinewave:
V peak-to-peak / (2 SQRT(2)) = VRMS
VRMS^2 / R = power

Of course you can juggle the factors with Ohm's Law and algebra, but it comes out the same.

[BrianHG]

The peak power for a sine wave is twice the mean power (rms power is a vulgar error).
72 V pk-pk = 36 V peak = 25.46 V rms.
(25.46)² / 8 = 81 W mean power.

Hmmm, make me wonder.  I have an amp I measured with a clean 95vp-p output.  The manufacturer claims a 200w RMS output.  According to TimFox, they are off by a huge amount since:

95/2=47.5v, = 33.59v
(33.59)² / 8 = 141 W mean power.

On the other hand, if I fed the amp a square wave, then:
(47.50)² / 8 = 202 W mean power.
But, this is a square wave, so, always all 47.5v will be seen by the 8 ohm resistor.  (Assuming infinite slew-rate)
That's 5.9375 amps seen by the 8 ohm load.
5.9375 amps X 47.5v = 282 watts?  Double of TimFox's 'mean' power output formula.

Ok, am I to assume by TimFox's results that a sine wave's 'mean' power output into a resistive load is ~50% of a square wave at the same peak voltage?

[rsjsouza]

Vrms = 0.707 × Vpk for sinewave Voltage not Power.
Because of the quadratic relationship between P, V and R, this follows:
Prms = 0.5 × Ppk

[IanB]

This is what I calculate:
72v p-p into 8 ohm = 36v / 8ohm = 4.5 amp.
4.5 amp X 36v = 162 watts peak.
Or 162 watts peak * 0.707 = 115 watts RMS.
A far cry from what the presenter listed as 81 watts.

It is me that wrong with my 162/115 watts, or the presenter with his half peak power watts of 81?

The rest of his figures all all off by the same amount...

As others have said, it is 81 W.

Assuming the voltage is a sine wave, then the amplitude is 36 V. The RMS voltage is 36 / √2 = 25.5 V. The power is V_RMS² / R = 25.5² / 8 = 81 W.

A simpler way to do it is to take half the peak power. The peak power is 36² / 8 = 162 W. The average power is half that, which is 162 / 2 = 81 W.

[fourfathom]

Ok, am I to assume by TimFox's results that a sine wave's 'mean' power output into a resistive load is ~50% of a square wave at the same peak voltage?

Well, the RMS voltage of a 2VPP squarewave is 1VRMS. Power: 1V^2 / 1 Ohm = 1W.
The RMS voltage of a 2VPP sinewave is 1/SQRT(2) = 0.707...  Power: 0.707^2 / 1 Ohm = 0.5W

[IanB]

Ok, am I to assume by TimFox's results that a sine wave's 'mean' power output into a resistive load is ~50% of a square wave at the same peak voltage?

It's actually 50% of the DC power at the peak voltage. And it's equal to the DC power at the RMS voltage.

Prms = 0.5 × Ppk

I think if we were being precise would say average power rather than RMS power? If you total up (integrate) the power delivered to the load over a time interval, and then divide the total energy by the time interval, you would get average power over that interval.

[BrianHG]

Thanks everyone, I was confusing RMS voltage with mean power.
'Mean Power' from a sine wave with a resistive load means 50% peak power if it were a source square wave.

However, it appears that a number of high end amps I've seen also have falsely advertised their true RMS capabilities, though they are fairly accurate with their peak power ratings.

[Zero999]

This is what I calculate:
72v p-p into 8 ohm = 36v / 8ohm = 4.5 amp.
4.5 amp X 36v = 162 watts peak.
Or 162 watts peak * 0.707 = 115 watts RMS.
A far cry from what the presenter listed as 81 watts.

It is me that wrong with my 162/115 watts, or the presenter with his half peak power watts of 81?

The rest of his figures all all off by the same amount...

As others have said, it is 81 W.

Assuming the voltage is a sine wave, then the amplitude is 36 V. The RMS voltage is 36 / √2 = 25.5 V. The power is V_RMS² / R = 25.5² / 8 = 81 W.

A simpler way to do it is to take half the peak power. The peak power is 36² / 8 = 162 W. The average power is half that, which is 162 / 2 = 81 W.

I just use the following formula:

P = V_P²/(2R)

[pqass]

See this on how to calculate amplifier power.
Basically, supply 1KHz sine input (at just below clipping), then measure across the speaker terminals with your DMM on VAC, square the result, and divide by speaker ohms.      P=V_rms²/R

[coppice]

Google: how to convert peak watts to true RMS watts
Response:

To convert peak watts to true RMS watts, multiply the peak watts by 0.707.

You do realise that when you Google something you need to ignore that first response that comes from the AI, right? The current Google AI is the only AI in the world you can rely on. So far it has been 100% wrong in every search I have seen. Nothing else comes close to that consistency.

[TimFox]

One can calculate the "rms power" of a sinusoidal voltage across a constant resistor, but that parameter has no practical use.
The result is that P_rms = P_mean x (3^1/2) = P_mean x (1.225) .

[TimFox]

Google: how to convert peak watts to true RMS watts
Response:

To convert peak watts to true RMS watts, multiply the peak watts by 0.707.

You do realise that when you Google something you need to ignore that first response that comes from the AI, right? The current Google AI is the only AI in the world you can rely on. So far it has been 100% wrong in every search I have seen. Nothing else comes close to that consistency.

That first response that Google gives to a technical question seems to be an AI-generated summary from actual responses cited below, but the summary appears to be merely grammatical with no logical constraints.

[IanB]

That first response that Google gives to a technical question seems to be an AI-generated summary from actual responses cited below, but the summary appears to be merely grammatical with no logical constraints.

That is the general character of large language models. They generate a word salad that statistically resembles what might be generated by a real human*. However, no logical, mathematical or technical content should be assumed.

* This was predicted by the remarkably prescient Douglas Adams, when he described the product of an AI vending machine as something that was "Almost, but not quite, entirely unlike tea." I find it quite telling that AI generated text is almost, but not quite, entirely unlike anything a real person would write.

[coppice]

Google: how to convert peak watts to true RMS watts
Response:

To convert peak watts to true RMS watts, multiply the peak watts by 0.707.

You do realise that when you Google something you need to ignore that first response that comes from the AI, right? The current Google AI is the only AI in the world you can rely on. So far it has been 100% wrong in every search I have seen. Nothing else comes close to that consistency.

That first response that Google gives to a technical question seems to be an AI-generated summary from actual responses cited below, but the summary appears to be merely grammatical with no logical constraints.

If it were merely that, wouldn't it give a mix of right and wrong answers? I've looked at quite a few of these responses now and they are all wrong. Perhaps they don't bias the result towards the most common answer, but mash up a bunch of answers in a way that always blends right and wrong answers, to produce a very dialectical form of answer - i.e. always bad.

[coppice]

This was predicted by the remarkably prescient Douglas Adams, when he described the product of an AI vending machine as something that was "Almost, but not quite, entirely unlike tea." I find it quite telling that AI generated text is almost, but not quite, entirely unlike anything a real person would write.[/size]

I think that misses the mark. They produce something exactly like a real person would write. A real person who doesn't know the answer, and is bullshitting because they have a pathological inability to just say "I don't know".

[TimFox]

Google: how to convert peak watts to true RMS watts
Response:

To convert peak watts to true RMS watts, multiply the peak watts by 0.707.

You do realise that when you Google something you need to ignore that first response that comes from the AI, right? The current Google AI is the only AI in the world you can rely on. So far it has been 100% wrong in every search I have seen. Nothing else comes close to that consistency.

That first response that Google gives to a technical question seems to be an AI-generated summary from actual responses cited below, but the summary appears to be merely grammatical with no logical constraints.

If it were merely that, wouldn't it give a mix of right and wrong answers? I've looked at quite a few of these responses now and they are all wrong. Perhaps they don't bias the result towards the most common answer, but mash up a bunch of answers in a way that always blends right and wrong answers, to produce a very dialectical form of answer - i.e. always bad.

For many legitimate technical questions, there is only one right answer, but many wrong answers:  this biases the outcome probabilities from random assembly of the words that would be used to answer the question.

[BrianHG]

One can calculate the "rms power" of a sinusoidal voltage across a constant resistor, but that parameter has no practical use.
The result is that P_rms = P_mean x (3^1/2) = P_mean x (1.225) .

Arrrrggggg....
What in the world are you doing?
Now you have generated a whole new figure which still describes some total of power, but, we cannot use it?

[TimFox]

One can calculate the "rms power" of a sinusoidal voltage across a constant resistor, but that parameter has no practical use.
The result is that P_rms = P_mean x (3^1/2) = P_mean x (1.225) .

Arrrrggggg....
What in the world are you doing?
Now you have generated a whole new figure which still describes some total of power, but, we cannot use it?

No, I did not generate a whole new figure.  I am objecting to the too-common mis-understanding of rms voltage and mean power by pointing out what "rms power" actually means.
The "rms" value of a variable means the square root of the (mean value of the variable²) .
That is,  for a variable V(t), the rms value V_rms = (<V²(t)>)^1/2,  where <X(t)> means the mean value of X(t) averaged over time.
If you apply that V(t) across a constant resistance R, then the instantaneous power is given by P(t) = V²(t) / R .
The mean power <P> is therefore (1/R) x <V²(t)> = (1/R) x (V_rms)² .

Exercise for the interested reader:  write the true equation for rms power and calculate it for a sinusoidal voltage across a resistor.

[bdunham7]

Arrrrggggg....
What in the world are you doing?
Now you have generated a whole new figure which still describes some total of power, but, we cannot use it?

It is entirely possible to concatenate terms to come up with terms that have real physical meaning but are either terribly unwieldy, unconventional or not particularly useful, such as a picofathom.  RMS can be added to any series of data.  You can have RMS temperature or RMS Dow Jones Index or even RMS frequency if you like.  You can also have RMS roughness for surface finishes, that might seem odd but is common and useful to those that use it.  RMS watts is pretty useless AFAIK, but the term does have a real definition.

[TimFox]

Some important features of these calculations:
1.  The relationship between rms voltage and mean power is true for any waveform or random noise or DC value.
2.  The relationship between peak-to-peak voltage and rms voltage depends on the waveform:  the common equation V_rms = V_pk-pk/(8^1/2) is true for sinusoidal voltage.
3.  The term "true rms" for voltage or current is used due to distinguish that value from an rms value calculated from the mean absolute value of a waveform, assuming a sinusoidal voltage.  Such "average-responding" voltmeters were especially common on analog voltmeters and cheaper digital voltmeters:  the value needs to be re-calculated if the waveform is, for example, a square wave or Gaussian noise.

[bdunham7]

1.  The relationship between rms voltage and mean power is true for any waveform or random noise or DC value.

And just to add, as you know, some will claim that this relationship is the actual definition of RMS voltage.  Of course it isn't, even if RMS voltage can be measured accurately this way (thermal conversion).

[IanB]

1.  The relationship between rms voltage and mean power is true for any waveform or random noise or DC value.

And just to add, as you know, some will claim that this relationship is the actual definition of RMS voltage.  Of course it isn't, even if RMS voltage can be measured accurately this way (thermal conversion).

Just for grins, why isn't it the actual definition?

[TimFox]

1.  The relationship between rms voltage and mean power is true for any waveform or random noise or DC value.

And just to add, as you know, some will claim that this relationship is the actual definition of RMS voltage.  Of course it isn't, even if RMS voltage can be measured accurately this way (thermal conversion).

Just for grins, why isn't it the actual definition?

RMS is a mathematical term and is defined for any variable V(t)  as V_rms = (<V²(t)>)^1/2, averaged over the independent variable t.
The power relationship (the basis of thermal conversion measurement) is a consequence of this definition, not the definition itself.

[vk6zgo]

Really?

Google: how to convert peak watts to true RMS watts
Response:

To convert peak watts to true RMS watts, multiply the peak watts by 0.707.

Remember, the square root of 2 = 1.41421356237 and 1/1.41421356237 = 0.7071067812.
You are not supposed to * by .707 to the power of 2, you are supposed to take the peak power and divide by the square root of 2, hence, 162/(sqr-root2)  or 162/1.41421356237 = 115 watts, not 81 watts.

True RMS wattage is not 50% peak wattage like with you 81 watt calculation.  That is just plain wrong!

162 watts peak would mean a square wave at 72vp-p.
115 watts RMS would means a sine wave with 72vp-p.
81 watts would mean a square wave with 50% on duty cycle.  Sine waves deliver more effective power than that.

There are no such things as "RMS Watts"!
It is a completely meaningless measurement.

The best you can measure is average power,, which is RMS volts x RMS amps.
RMS volts are 0.707 x peak volts, & RMS current is 0.707 x peak current.

The whole idea of RMS values of voltage & current was so electrical engineers, who in those days dealt with things that actually did stuff, like boiling water & lifting objects could use the existing power formulas with ac.

If you had a DC lift (elevator) motor for instance, it was important to be able to compare its performance with an ac motor to choose the right size motor of the latter type.

The nominal "230v"RMS mains has a peak voltage of around 325 volts, & will illuminate an incandescent light bulb to the same brightness level as 230v DC.

The voltage across that lightbulb is expressed as RMS voltage, & the current through it as RMS current.
Using P =VI, 0.707 is multiplied by 0.707, giving near as dammit 0.5.

Remember power is "the rate of doing work".
The value of VI measured instantaneously at various points across each halfway varies from zero at the zero crossing to Vp x Ip at each peak, so real usable power varies across the waveform, & average power is the only thing we have to "do work" in the real world.

[IanB]

1.  The relationship between rms voltage and mean power is true for any waveform or random noise or DC value.

And just to add, as you know, some will claim that this relationship is the actual definition of RMS voltage.  Of course it isn't, even if RMS voltage can be measured accurately this way (thermal conversion).

RMS is a mathematical term and is defined for any variable V(t)  as V_rms = (<V²(t)>)^1/2, averaged over the independent variable t.
The power relationship (the basis of thermal conversion measurement) is a consequence of this definition, not the definition itself.

As highlighted above, the claim was made for RMS voltage in particular, not for any arbitrary variable.

And for RMS voltage in particular, it is defined as being the equivalent DC voltage that would produce the same power dissipation in a resistive load. The fact that this coincides with the definition resulting from mathematics, in this particular case, does not invalidate the alternative definition.

[vk6zgo]

Google: how to convert peak watts to true RMS watts
Response:

To convert peak watts to true RMS watts, multiply the peak watts by 0.707.

You do realise that when you Google something you need to ignore that first response that comes from the AI, right? The current Google AI is the only AI in the world you can rely on. So far it has been 100% wrong in every search I have seen. Nothing else comes close to that consistency.

That first response that Google gives to a technical question seems to be an AI-generated summary from actual responses cited below, but the summary appears to be merely grammatical with no logical constraints.

If it were merely that, wouldn't it give a mix of right and wrong answers? I've looked at quite a few of these responses now and they are all wrong. Perhaps they don't bias the result towards the most common answer, but mash up a bunch of answers in a way that always blends right and wrong answers, to produce a very dialectical form of answer - i.e. always bad.

I asked ChatGPT about a Carpenter relay---something that was widely used for many decades.
Its results were like somebody at an interview trying to BS their way round a question, ending up describing a thermal relay.

I told it that was wrong & described a Carpenter relay
It tried again, then just regurgitated what I said, a bit shuffled around.

Thinking it would have learnt, I tried again a week later---this time it said it had no knowledge of a Carpenter relay!
About as useful as a Ouija board!

[bdunham7]

And for RMS voltage in particular, it is defined as being the equivalent DC voltage that would produce the same power dissipation in a resistive load. The fact that this coincides with the definition resulting from mathematics, in this particular case, does not invalidate the alternative definition.

It appears we are now having an argument over the definition of "definition"! 

"RMS" has an independent definition, a mathematical operation.  "voltage" also has a definition.  The two together have a definition, one where the former term mathematically modifies the latter.  If I said "half a volt" (0.5V), would you claim that the definition of half a volt is that voltage which creates 1/4 watt of power to be dissipated in a 1-ohm resistor?  I wouldn't.

[vk6zgo]

And for RMS voltage in particular, it is defined as being the equivalent DC voltage that would produce the same power dissipation in a resistive load. The fact that this coincides with the definition resulting from mathematics, in this particular case, does not invalidate the alternative definition.

It appears we are now having an argument over the definition of "definition"! 

"RMS" has an independent definition, a mathematical operation.  "voltage" also has a definition.  The two together have a definition, one where the former term mathematically modifies the latter.  If I said "half a volt" (0.5V), would you claim that the definition of half a volt is that voltage which creates 1/4 watt of power to be dissipated in a 1-ohm resistor?  I wouldn't.

If you didn't specify ac or DC that would be a fair definition.
RMS voltages were obviously predictable, but the relationship was historically proven by just such a process.

[TimFox]

And for RMS voltage in particular, it is defined as being the equivalent DC voltage that would produce the same power dissipation in a resistive load. The fact that this coincides with the definition resulting from mathematics, in this particular case, does not invalidate the alternative definition.

It appears we are now having an argument over the definition of "definition"! 

"RMS" has an independent definition, a mathematical operation.  "voltage" also has a definition.  The two together have a definition, one where the former term mathematically modifies the latter.  If I said "half a volt" (0.5V), would you claim that the definition of half a volt is that voltage which creates 1/4 watt of power to be dissipated in a 1-ohm resistor?  I wouldn't.

Exactly.
RMS has many other uses in engineering besides voltage measurement.
There is the usual statistical “standard deviation”, which is the rms deviation of a variable from its mean value:  the square is called “variance” and is somewhat analogous to power.  With incoherent voltage terms (such as harmonics or noise), the powers add just as variances add for statistically independent variables.
Besides noise voltage, there are other random variables such as temporal jitter and surface roughness (mentioned above).

[BrianHG]

The nominal "230v"RMS mains has a peak voltage of around 325 volts, & will illuminate an incandescent light bulb to the same brightness level as 230v DC.

The voltage across that lightbulb is expressed as RMS voltage, & the current through it as RMS current.
Using P =VI, 0.707 is multiplied by 0.707, giving near as dammit 0.5.

It's these 2 paragraphs which are driving me crazy.  It doesn't sound like it should work, but it does...

Say my light bulb is a perfect 1 ohm.
230v DC would deliver 230 amps, or 52900 watts.
230*1.41421             = 325.27v peaks on a 230v ac source.
325.27*325.27 * 0.5 = 52900 watts, so, basically ~ the same.

[IanB]

It appears we are now having an argument over the definition of "definition"! 

It's simple. Sometimes you can have more than one definition for a thing:

1. An equilateral triangle is a triangle where the three sides are of equal length.

2. An equilateral triangle is a triangle where the three interior angles have the same measure.

Both are valid and correct. Neither one is better than the other. The first implies the second, and the second implies the first.

[TimFox]

It appears we are now having an argument over the definition of "definition"! 

It's simple. Sometimes you can have more than one definition for a thing:

1. An equilateral triangle is a triangle where the three sides are of equal length.

2. An equilateral triangle is a triangle where the three interior angles have the same measure.

Both are valid and correct. Neither one is better than the other. The first implies the second, and the second implies the first.

The word “equilateral” means equal lengths.
That such a triangle has equal angles is a theorem of Euclid.
Which part of “root-mean-square” do you not understand?

[BrianHG]

The peak power for a sine wave is twice the mean power (rms power is a vulgar error).
72 V pk-pk = 36 V peak = 25.46 V rms.
(25.46)² / 8 = 81 W mean power.

Hmmm, make me wonder.  I have an amp I measured with a clean 95vp-p output.  The manufacturer claims a 200w RMS output.  According to TimFox, they are off by a huge amount since:

Oooopppsie, my pro amp output was 180vp-p, not 95vp-p...

Redoing the calculation....

90v*90v /8 * 0.5 = 506.25 watts.

The model name is a 'David Belles OCM 500 Soloist Amplifier'.

Though, the data sheet claims 800watt bridged mono into 8 ohm, or 200 watts stereo into 8 ohm.
Something is still fishy about the 8 ohm 200 watts.  It's speaker output current of 100 amps is enough to do more with that 180vp-p.  Though perhaps there is added distortion exceeding a sweet spot the amp is tuned for.  Though the amp is only rated 400w into 4 ohm, unbridged.

[fourfathom]

Though, the data sheet claims 800watt bridged mono into 8 ohm, or 200 watts stereo into 8 ohm.
Something is still fishy about the 8 ohm 200 watts.  It's speaker output current of 100 amps is enough to do more with that 180vp-p.  Though perhaps there is added distortion exceeding a sweet spot the amp is tuned for.

This reminds me of the "peak music power" amplifier ratings.  Don't assuming anything about amplifier power ratings unless the conditions are unambiguously stated.

[coppice]

This reminds me of the "peak music power" amplifier ratings.  Don't assuming anything about amplifier power ratings unless the conditions are unambiguously stated.

Yeah, 1500W from four C cells.

[TimFox]

A proper specification for an audio power amplifier includes the distortion at the rated power output.
One can obtain higher power at higher distortion, possibly going all the way to a square wave output (depending on the amplifier), where the THD would be 48.3% of the fundamental output.

[rsjsouza]

Prms = 0.5 × Ppk

I think if we were being precise would say average power rather than RMS power? If you total up (integrate) the power delivered to the load over a time interval, and then divide the total energy by the time interval, you would get average power over that interval.

Indeed your language is better.  It should be P_AVG
 I was writing from my cellphone and doing multi-tasking. 

[TimFox]

See my reply #23 above in this thread.

[BrianHG]

This reminds me of the "peak music power" amplifier ratings.  Don't assuming anything about amplifier power ratings unless the conditions are unambiguously stated.

Yeah, 1500W from four C cells.

Yeah, of course...
Spend 10 minutes having those 4 C cells charge up a 0.1f capacitor bank to 160v and power a single bass drum strike with a balanced output amp to a single 1500w spike.

That's 1 beat per every 5 additional minutes after that for around 30 minutes completely draining those 4 C cells.  Or, in other words, around 5 full drum beats at 1500 watts over a 30 minute period.  I can get into music like that...

[coppice]

This reminds me of the "peak music power" amplifier ratings.  Don't assuming anything about amplifier power ratings unless the conditions are unambiguously stated.

Yeah, 1500W from four C cells.

Yeah, of course...
Spend 10 minutes having those 4 C cells charge up a 0.1f capacitor bank to 160v and power a single bass drum strike with a balanced output amp to a single 1500w spike.

That's 1 beat per every 5 additional minutes after that for around 30 minutes completely draining those 4 C cells.  Or, in other words, around 5 full drum beats at 1500 watts over a 30 minute period.  I can get into music like that...

Well, I guess those boom boxes claiming 1500W PMPO might have a supercap in them, but a 160V rating for it seems unlikely.

[BrianHG]

This reminds me of the "peak music power" amplifier ratings.  Don't assuming anything about amplifier power ratings unless the conditions are unambiguously stated.

Yeah, 1500W from four C cells.

Yeah, of course...
Spend 10 minutes having those 4 C cells charge up a 0.1f capacitor bank to 160v and power a single bass drum strike with a balanced output amp to a single 1500w spike.

That's 1 beat per every 5 additional minutes after that for around 30 minutes completely draining those 4 C cells.  Or, in other words, around 5 full drum beats at 1500 watts over a 30 minute period.  I can get into music like that...

Well, I guess those boom boxes claiming 1500W PMPO might have a supercap in them, but a 160V rating for it seems unlikely.

They were actually 1.5w RMS, or, 1500mw.

[coppice]

This reminds me of the "peak music power" amplifier ratings.  Don't assuming anything about amplifier power ratings unless the conditions are unambiguously stated.

Yeah, 1500W from four C cells.

Yeah, of course...
Spend 10 minutes having those 4 C cells charge up a 0.1f capacitor bank to 160v and power a single bass drum strike with a balanced output amp to a single 1500w spike.

That's 1 beat per every 5 additional minutes after that for around 30 minutes completely draining those 4 C cells.  Or, in other words, around 5 full drum beats at 1500 watts over a 30 minute period.  I can get into music like that...

Well, I guess those boom boxes claiming 1500W PMPO might have a supercap in them, but a 160V rating for it seems unlikely.

They were actually 1.5w RMS, or, 1500mw.

Actually some of them did put out quite a bit more than that when they started using class D amps. A colleague played around with some in the lab, helping with the class D chip implementation. Some of them actually had pretty good speakers, and could sound good. The only thing dumb about them was the big banner across in the packaging claiming 1500W (not mW) PMPO.

[BrianHG]

packaging claiming 1500W (not mW) PMPO.

Ahhhhh, peak momentary power output...
If you speakers are 8 ohms, you still need all that voltage range to drive them.
But 4 C batteries, that's only 6 volts.
Some serious voltage step up needs to happen.

[CatalinaWOW]

All was (is?) fair in the amplifier power rating wars.  You can't assume 8 ohm speakers either.  Some were rated into lower impedance loads.  2 ohms in one case I remember.  Not even a complete fabrication as the thumping crowd often parallels many speakers.

[BrianHG]

I remember years ago, like 35 years ago I made my own amp using OP27s powered to the max at +/-22v, driving matched pair darlington emitter follower stage.  The amp having a balanced output, heavily biased for the heat and warm sound, gave me a clean 40v p-p output.  (The transistors collectors had an unregulated extra ~5v headroom and the base drive bias had a charged bootstrapped cap2 allowing the bases to operate above and below the op-amp's output maintaining that class AB drive up to the limits of the opamp's output.)

This means that that amp delivered me just about 100 watts into 8 ohm.  I guess this wasn't too bad for a first design amp using just 2 opamps per channel with a few driver transistors and resistors + a few caps.

I believe the transistors were MJ11028 and MJ11029.  50 amps, 60v, or MJ11032, MJ11033.  50 amps, 120v in TO-3.