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.