Wrong wattage calculation on a review of an audio amp.

[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.