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Discussion of the "Pano" method for estimating amplifier power requirements

pjug

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There was some interesting (but misplaced) discussion of amplifier power requirements in a recently closed thread. One point of disagreement had to do with the validity of the Pano method for estimating how much amplifier power you need. You can find the description of this here:
http://archimago.blogspot.com/2017/09/musings-how-much-amplifier-power-do-you.html
https://www.diyaudio.com/forums/multi-way/204857-test-voltage-power-speakers.html

If you use very dynamic music with a peak near 0dB, then it seems to me that this is pretty much as good as using a scope to get an idea. Of course, it is only going to give a ballpark idea either way when you estimate using dynamic music that you would listen to. It is not worst case and I think those of us who think this method is valid would still want a good amount of extra power for peace of mind.

I hope those who have criticized this methodology will take the time to carefully read through the @Archimago description before discussing. If there are problems with this it would be good to understand what these are. On the other hand, if this proves to be OK, then it will be useful to be able to point people to the links above. The question of how much amplifier we need seems to come up over and over again on this site.
 

NTK

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Archimago's analysis says, in his case, with speakers having sensitivity of 92 dB at 2.83 Vrms, he needs 70 W. Therefore, if a person has speakers of the more typical sensitivity of 86 dB at 2.83 Vrms, he/she will need 4 times the power, which is 280 W. I'll buy that.

[Edit]
In Archimago's analysis, he used a single frequency sine wave of -12 dBFS. Therefore, his assumption is that the music or movies you will be playing will have a crest factor of no more than 12 dB. If you desire more headroom for music or movies with higher crest factor (as it was pointed out in the other thread, gun shots have CF of more than 30 dB), then you'll need higher (short term) power capability.
 
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pozz

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Blumlein 88

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Archimago's analysis says, in his case, with speakers having sensitivity of 92 dB at 2.83 Vrms, he needs 70 W. Therefore, if a person has speakers of the more typical sensitivity of 86 dB at 2.83 Vrms, he/she will need 4 times the power, which is 280 W. I'll buy that.

[Edit]
In Archimago's analysis, he used a single frequency sine wave of -12 dBFS. Therefore, his assumption is that the music or movies you will be playing will have a crest factor of no more than 12 dB. If you desire more headroom for music or movies with higher crest factor (as it was pointed out in the other thread, gun shots have CF of more than 30 dB), then you'll need higher (short term) power capability.
No, this is a misunderstanding of what the test shows you.

You can do the test with a -20 db signal if you wish. It'll predict the same number as if you use a -12 db signal. The multiplying factor will be different, but the resulting estimate is the same.

Don't have time at the moment. Later I'll see if I can present an example that is clear on this showing why the level of the test signal has no relation to crest factor. Note there is a relation of crest factor of the music track you choose to set your max listening volume, but not the test signal.
 
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Blumlein 88

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I made a straightforward model of power requirements before: https://www.audiosciencereview.com/...n-do-amplifiers-clip.12757/page-2#post-380060

No considerations of spectrum or room effects beyond distance.
What about critical distance? In most rooms at some point being twice as far from the speaker no longer results in -6 db sound level. Reflections in the room come into play that create the diffuse sound field.

http://www.moultonlabs.com/more/acoustical_measurements_for_the_rest_of_us/P1/

http://www.bnoack.com/index.html?http&&&www.bnoack.com/acoustic/criticaldistance.html

1597156638800.png
 
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NTK

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No, this is a misunderstanding of what the test shows you.

You can do the test with a -20 db signal if you wish. It'll predict the same number as if you use a -12 db signal. The multiplying factor will be different, but the resulting estimate is the same.

Don't have time at the moment. Later I'll see if I can present an example that is clear on this showing why the level of the test signal has no relation to crest factor. Note there is a relation of crest factor of the music track you choose to set your max listening volume, but not the test signal.
Crest factor is absolutely critical in this discussion. CF is the ratio between peak voltage and RMS voltage. Average power is calculated from RMS voltage. Clipping occurs when the amplifier cannot output the required peak voltage, not RMS voltage. (The relationship between voltage and current is determined by Ohm's law. If the amplifier cannot supply the required current, voltage will sag, therefore it will also be manifested as voltage clipping.)

If an AC signal has an RMS voltage of 1 V, and a peak voltage of 4 V, the CF is 12 dB. If the load is 8 ohm, average power is 0.125 W, and instantaneous peak power is 2 W.

For a sine wave with the same RMS voltage of 1 V (peak voltage = 1.414 V, CF = 3 dB), average power is the same (0.125 W), but instantaneous peak power is only 0.25 W.

If an amplifier is rated at 0.25 W output as tested by a single frequency sine wave, what is its clipping voltage? A clipping voltage threshold of 1.414 V will be sufficient to accurately reproduce the sine wave, but the signal with 12 dB CF will be severely clipped. The 12 dB CF signal needs 4 V, which is 2.83 times higher!

As amp power rating is measured (or calculated) with single frequency sine waves, we will therefore need to add a very healthy margin for music to avoid clipping. 1 W average power of music is not the same as 1 W average power of a sine wave. You can easily clip an amp rated at 5 W continuous power with 1 W (average power) of music if the amp cannot deliver significantly higher short term power than its rated continuous power, while you thought you had a 5X margin.
 

levimax

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Reading through the Arichimago article and the linked thread and this thread it appears that the "peak" power of an amp is just as important if not more important than the "continuous" power rating. This is the opposite of what I previously thought which was that "peak" power was a bogus measurement. As far as I know there are no "standards" to measure "peak" power although Amir does measure it according to his standard. How long does an amp need to deliver "peak power" before sagging (assuming unregulated power supple) to handle peaks in music?
 
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pjug

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Crest factor is absolutely critical in this discussion. CF is the ratio between peak voltage and RMS voltage.
Yes, you should do the test with a track having the greatest CF you expect to play on your system (and also make sure the peak is near 0dB). Then this should give you a good estimate of the power you need to do that.
 

NTK

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Yes, you should do the test with a track having the greatest CF you expect to play on your system (and also make sure the peak is near 0dB). Then this should give you a good estimate of the power you need to do that.
Not really necessary. CF of 12 dB is what ANSI/CTA-2034A has already standardized on.

CF.JPG
 

Blumlein 88

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Crest factor is absolutely critical in this discussion. CF is the ratio between peak voltage and RMS voltage. Average power is calculated from RMS voltage. Clipping occurs when the amplifier cannot output the required peak voltage, not RMS voltage. (The relationship between voltage and current is determined by Ohm's law. If the amplifier cannot supply the required current, voltage will sag, therefore it will also be manifested as voltage clipping.)

If an AC signal has an RMS voltage of 1 V, and a peak voltage of 4 V, the CF is 12 dB. If the load is 8 ohm, average power is 0.125 W, and instantaneous peak power is 2 W.

For a sine wave with the same RMS voltage of 1 V (peak voltage = 1.414 V, CF = 3 dB), average power is the same (0.125 W), but instantaneous peak power is only 0.25 W.

If an amplifier is rated at 0.25 W output as tested by a single frequency sine wave, what is its clipping voltage? A clipping voltage threshold of 1.414 V will be sufficient to accurately reproduce the sine wave, but the signal with 12 dB CF will be severely clipped. The 12 dB CF signal needs 4 V, which is 2.83 times higher!

As amp power rating is measured (or calculated) with single frequency sine waves, we will therefore need to add a very healthy margin for music to avoid clipping. 1 W average power of music is not the same as 1 W average power of a sine wave. You can easily clip an amp rated at 5 W continuous power with 1 W (average power) of music if the amp cannot deliver significantly higher short term power than its rated continuous power, while you thought you had a 5X margin.
Yet the fact Pano picked -12 db is in no way at all related to a crest factor of 12 db. Zero connection. He could have picked -40 db, or -3 db or 0 db. The latter two might be dangerous with the test tone for blowing something up. But none of it has anything to do with crest factor. You will only confuse yourself to think so.
 

Blumlein 88

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Reading through the Arichimago article and the linked thread and this thread it appears that the "peak" power of an amp is just as important if not more important than the "continuous" power rating. This is the opposite of what I previously thought which was that "peak" power was a bogus measurement. As far as I know there are no "standards" to measure "peak" power although Amir does measure it according to his standard. How long does an amp need to deliver "peak power" before sagging (assuming unregulated power supple) to handle peaks in music?

I think you are misreading Arch. He does talk about peak power you need, but is not really talking so much about peak power amp specs vs continuous. You might have 10,000 watts in the amp continuous, but your peak power needs for your listening levels and speakers could be very much less. And does not depend upon the peak power of the amp. Obviously if your listening conditions demand power well beyond the continuous rating of your amp, then you need a bigger amp. Safest thing for that is to get an amp that can continuously put out power equal to the peaks of your listening needs. It is the peaks of your listening levels that need considering. Not "peak power ratings" of an amp.
 

PeteL

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After reading it trough, there is no major problem with the test, it's valid, to find out a ballpark figure of what one would generally need for it's loudest level, but the main problem with it is, for me, that it will give you a "best scenario" result. I am more interested in the worst case. The first and main reason for this is that they use the nominal impedance for the calculations, where in my book the impedance use for calculating how much one would need, should be the lowest point in the impedance curve, and choose a tone at this frequency, and then convert what does it mean in usual Amplifier nominal impedance power spec. That's already more complicated and not always (rarely) available for your speaker. Now the thing is. even then, If one would go to the length of full measurments of say: What power do I need If I want to listen at at least x dB SPL, with the y model speaker, and allow for the full uncompressed dynamic range any recording that has a crest factor of at least z the result would be every time higher, mainly for the impedance thing, but also you ask the subject to choose dynamic music, but he doesn't know the most dynamic music he'll ever want to listen to. Also, an argument could be made that the sample music when listen at the loudest level for the user could already be clipped. Normally, if done right, you should have a cue of that by getting a result that is higher than the specified power of the amp, but not necessaily, and that's where there are loose ends in amp specifications so we can't know for sure if the tested amp went into some soft clipping circuit and beefed up it's specified output. So, yes, it's a quick and easy test, based on real rigorous science, but it's problematic in many ways, It's too Quick and easy, and could be argued that it's a ballpark figure, but invariably and consistently lower than the worst case. Even having said that. I find the poll results too low. The reason for this could be that: 1. many people listen to overly compressed music and think it's dynamic 2. Many people have digital attenuation in their chain and don't know it. 3. The 100dB + efficient speakers are very popular. 3. Other measurments error. So yeah, lots of room for inconsistencies.
 

pozz

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How is it not considering room effects if volume is set at the listening position in your listening room?
But you don't want to test with an average case. You want to test with an extreme case.
The charts I posted are extreme cases. They do not take signal level or spectrum into account. They assume no specific room volumes or reveberation characteristics, and do no account for modal effects or interference. This means that power requirements are way higher than they need to be.

I also did not account for psychoacoustics. There are people who like really small amps, like 25W continuous output, because clipping introduces a lot of nonlinearity which makes the power seem sufficient. A cleanly reproduced signal sounds a lot quieter than a heavily clipped one in terms of subjective loudness.
What about critical distance? In most rooms at some point being twice as far from the speaker no longer results in -6 db sound level. Reflections in the room come into play that create the diffuse sound field.

http://www.moultonlabs.com/more/acoustical_measurements_for_the_rest_of_us/P1/

http://www.bnoack.com/index.html?http&&&www.bnoack.com/acoustic/criticaldistance.html

View attachment 77570
There's no true diffuse sound in small rooms as you know, so yup, the inverse square law (-6dB per doubling of distance) doesn't apply and a better model would follow that red curve.
 

levimax

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I think you are misreading Arch. He does talk about peak power you need, but is not really talking so much about peak power amp specs vs continuous. You might have 10,000 watts in the amp continuous, but your peak power needs for your listening levels and speakers could be very much less. And does not depend upon the peak power of the amp. Obviously if your listening conditions demand power well beyond the continuous rating of your amp, then you need a bigger amp. Safest thing for that is to get an amp that can continuously put out power equal to the peaks of your listening needs. It is the peaks of your listening levels that need considering. Not "peak power ratings" of an amp.

I understand that but if "peak power" is demanded only for very short periods of time is it really "cost effective / good engineering" to build an amp that can deliver continuous power at a peak level when it is only needed for transient peaks? Right now I am building a linear PS for a DIY amp.... I can chose to spec the transformer and caps several different ways. For instance I can chose a voltage and current rating that can supply enough continuous power to not over heat the output stage based on continuous power or I could chose a higher voltage transformer with a lower current rating that could deliver higher "peak" output and then "sag" to a safe level for the output stage. Yes of course I could build an output stage with more transistors and get a bigger heat sink and bigger transformer etc. but again why not design for higher peak voltage rather than continuous since that is what music demands?
 

pozz

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PeteL

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Crest factor is absolutely critical in this discussion. CF is the ratio between peak voltage and RMS voltage. Average power is calculated from RMS voltage. Clipping occurs when the amplifier cannot output the required peak voltage, not RMS voltage. (The relationship between voltage and current is determined by Ohm's law. If the amplifier cannot supply the required current, voltage will sag, therefore it will also be manifested as voltage clipping.)

If an AC signal has an RMS voltage of 1 V, and a peak voltage of 4 V, the CF is 12 dB. If the load is 8 ohm, average power is 0.125 W, and instantaneous peak power is 2 W.

For a sine wave with the same RMS voltage of 1 V (peak voltage = 1.414 V, CF = 3 dB), average power is the same (0.125 W), but instantaneous peak power is only 0.25 W.

If an amplifier is rated at 0.25 W output as tested by a single frequency sine wave, what is its clipping voltage? A clipping voltage threshold of 1.414 V will be sufficient to accurately reproduce the sine wave, but the signal with 12 dB CF will be severely clipped. The 12 dB CF signal needs 4 V, which is 2.83 times higher!

As amp power rating is measured (or calculated) with single frequency sine waves, we will therefore need to add a very healthy margin for music to avoid clipping. 1 W average power of music is not the same as 1 W average power of a sine wave. You can easily clip an amp rated at 5 W continuous power with 1 W (average power) of music if the amp cannot deliver significantly higher short term power than its rated continuous power, while you thought you had a 5X margin.
Crest factor is taken into account in the test, but it's the crest factor of the test tracks (music) that will prevail.
 

Blumlein 88

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Okay, 12 db was chosen mostly so you can take the voltage reading you get and square it for an easy way to the answer. I also think that is what is confusing to people.

I'd choose -20 db which happens to be an even 10 to 1 ratio.

So let me use Arch's example.

Using a -20 db test signal after setting my max volume I'd get 3.32 volts instead of Arch's 8.35 volts.

So if this were translated to 0 db or max ever voltage level you just multiply the voltage reading you get by 10x. In this case 33.2 volts. Use ohms law to find power.

33.2 squared divided by 8 ohms. You'll get 137.78 watts. I've not yet adjusted for an RMS reading. I'll do that here at the end where halving the calculated amount gives me RMS power rather than peak power. 68.89 watts. Vs the other methods 70 watts, and the difference just comes from rounding errors.

So set volume with the loudest music at the loudest you'll ever listen.

Send a -20 db 220hz tone thru and measure voltage.

Then 10x your voltage reading.

Then square divided by 8 and this divided by 2. You'll have the max RMS power needs for your system.

Oh, and if it were me, I wouldn't divide by 2 at the end. I'd leave that 3 db headroom in for intersample overs since this whole thing is predicated on digital inputs having a defined max 0 db level. And they do, but intersample overs can often be 3 to 3.5 db.
 

garbulky

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Archimago's analysis says, in his case, with speakers having sensitivity of 92 dB at 2.83 Vrms, he needs 70 W. Therefore, if a person has speakers of the more typical sensitivity of 86 dB at 2.83 Vrms, he/she will need 4 times the power, which is 280 W. I'll buy that.

[Edit]
In Archimago's analysis, he used a single frequency sine wave of -12 dBFS. Therefore, his assumption is that the music or movies you will be playing will have a crest factor of no more than 12 dB. If you desire more headroom for music or movies with higher crest factor (as it was pointed out in the other thread, gun shots have CF of more than 30 dB), then you'll need higher (short term) power capability.
Lately I haven't seen any speakers other than say bookshelves that have such low sensitivities. Most of them clock near 90db.
 
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