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

[NTK]

But you don't want to test with an average case. You want to test with an extreme case.

In this case, I agree with the logic as given by the standards committee working group. The math is straight forward, so I am in no compulsion to run any test. Anyway, I currently don't have access any of my audio gear so I can't do any even if I wanted to.

Another angle to look at this problem is from the source signal. If you are using a digital signal source, the maximum level is deterministic. You add a margin of 3 dB (or 6 dB if you feel you want to be ultra conservative) for inter-sample overs. Figure out the gain setings, and you'll know what the maximum required output level is.

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.

In Archimago's step 4, he measured the voltage at the speaker terminals with a single frequency sine wave (at 120 or 220 Hz) using his voltmeter. The result is 8.35 Vrms.

In step 5, he squared 8.35 to get 70 W at 8 ohm. He gave his reasons but I'll explain it in relation to crest factor.

8.35 Vrms into 8 ohm gives 8.7 W, not 70 W. At 70 W and at 8 ohms, Vrms is 23.7 V. If it is a single frequency sine wave, V_peak will be 33.5 V. Then the requirement is set such that the amp only needs to provide a peak voltage of 33.5 V.

With a CF of 12 dB, the ratio of V_peak to Vrms is 4. Therefore, a signal with CF of 12 dB and V_peak of 33.5 V with give a Vrms of 33.5/4 = 8.375 V (which should be 8.35 V but is off by 0.3% due to round off errors). Now you see where the 12 dB CF came in?

Because both signals have the same Vrms, therefore (average) powers and sound pressure levels are the same. However, the 12 dB CF signal will require an amp that can output a peak voltage of 33.5 V, which is "equivalent" to an amp outputing 70 W into 8 ohm with a single frequency sine wave.

 

[pjug]

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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.

Doesn't your concern about impedance pretty much work its way out? Really this is a test for how much voltage you need with your system (your speakers). The conversion to power is just because we are talking about amplifier ratings.

I think the reason for low values in the survey are because CF in the test tracks is not too extreme and also because Archimago did not emphasize turning the volume to an uncomfortable level. Anyway, I would be to take the result and pad it by a factor of two or three just in case.

 

[NTK]

Lately I haven't seen any speakers other than say bookshelves that have such low sensitivities. Most of them clock near 90db.

86 dB sensitivity is on the low side, but not too unusual. For example: Revel M126Be
[Edit] Yes. Bookshelves usually have low sensitivity.

 

[Blumlein 88]

In this case, I agree with the logic as given by the standards committee working group. The math is straight forward, so I am in no compulsion to run any test. Anyway, I currently don't have access any of my audio gear so I can't do any even if I wanted to.

Another angle to look at this problem is from the source signal. If you are using a digital signal source, the maximum level is deterministic. You add a margin of 3 dB (or 6 dB if you feel you want to be ultra conservative) for inter-sample overs. Figure out the gain setings, and you'll know what the maximum required output level is.


In Archimago's step 4, he measured the voltage at the speaker terminals with a single frequency sine wave (at 120 or 220 Hz) using his voltmeter. The result is 8.35 Vrms.

In step 5, he squared 8.35 to get 70 W at 8 ohm. He gave his reasons but I'll explain it in relation to crest factor.

8.35 Vrms into 8 ohm gives 8.7 W, not 70 W. At 70 W and at 8 ohms, Vrms is 23.7 V. If it is a single frequency sine wave, V_peak will be 33.5 V. Then the requirement is set such that the amp only needs to provide a peak voltage of 33.5 V.

With a CF of 12 dB, the ratio of V_peak to Vrms is 4. Therefore, a signal with CF of 12 dB and V_peak of 33.5 V with give a Vrms of 33.5/4 = 8.375 V (which should be 8.35 V but is off by 0.3% due to round off errors). Now you see where the 12 dB CF came in?

Because both signals have the same Vrms, therefore (average) powers and sound pressure levels are the same. However, the 12 dB CF signal will require an amp that can output a peak voltage of 33.5 V, which is "equivalent" to an amp outputing 70 W into 8 ohm with a single frequency sine wave.

You are way over thinking this. Look at my example using a -20 db signal which gives the same answer. Crest factor isn't part of this.

 

[NTK]

You are way over thinking this. Look at my example using a -20 db signal which gives the same answer. Crest factor isn't part of this.

OK. I give up

 

[pjug]

In this case, I agree with the logic as given by the standards committee working group. The math is straight forward, so I am in no compulsion to run any test. Anyway, I currently don't have access any of my audio gear so I can't do any even if I wanted to.

Another angle to look at this problem is from the source signal. If you are using a digital signal source, the maximum level is deterministic. You add a margin of 3 dB (or 6 dB if you feel you want to be ultra conservative) for inter-sample overs. Figure out the gain setings, and you'll know what the maximum required output level is.


In Archimago's step 4, he measured the voltage at the speaker terminals with a single frequency sine wave (at 120 or 220 Hz) using his voltmeter. The result is 8.35 Vrms.

In step 5, he squared 8.35 to get 70 W at 8 ohm. He gave his reasons but I'll explain it in relation to crest factor.

8.35 Vrms into 8 ohm gives 8.7 W, not 70 W. At 70 W and at 8 ohms, Vrms is 23.7 V. If it is a single frequency sine wave, V_peak will be 33.5 V. Then the requirement is set such that the amp only needs to provide a peak voltage of 33.5 V.

With a CF of 12 dB, the ratio of V_peak to Vrms is 4. Therefore, a signal with CF of 12 dB and V_peak of 33.5 V with give a Vrms of 33.5/4 = 8.375 V (which should be 8.35 V but is off by 0.3% due to round off errors). Now you see where the 12 dB CF came in?

Because both signals have the same Vrms, therefore (average) powers and sound pressure levels are the same. However, the 12 dB CF signal will require an amp that can output a peak voltage of 33.5 V, which is "equivalent" to an amp outputing 70 W into 8 ohm with a single frequency sine wave.

I agree with @Blumlein 88 that it is too bad that -12dB was used, and also Archimago's shortcut calculation of simply squaring the measured voltage also adds confusion. Here is how I would do it, from the point of Arch measuring 8.7V.

- The -12dB test tone has a peak that is -9dB down, or a factor of 2.82. Therefore, a sine wave peaking at 0dB has amplitude 2.82 greater than the test tone.

- Multiplying 8.7V X 2.82 gives 24.5V. Computing power into 8 ohms from this: (24.5^2)/8 = 75W.

No crest factor involved with the test tone.

 

[PeteL]

[

Doesn't your concern about impedance pretty much work its way out? Really this is a test for how much voltage you need with your system (your speakers). The conversion to power is just because we are talking about amplifier ratings.

I think the reason for low values in the survey are because CF in the test tracks is not too extreme and also because Archimago did not emphasize turning the volume to an uncomfortable level. Anyway, I would be to take the result and pad it by a factor of two or three just in case.

[

Doesn't your concern about impedance pretty much work its way out? Really this is a test for how much voltage you need with your system (your speakers). The conversion to power is just because we are talking about amplifier ratings.

I think the reason for low values in the survey are because CF in the test tracks is not too extreme and also because Archimago did not emphasize turning the volume to an uncomfortable level. Anyway, I would be to take the result and pad it by a factor of two or three just in case.

Not really, Pano itself in the comment says that it should be the measured impedance, not the nominal one AND if the tone chosen will show different voltages depending wher this frequency is on the impedance curve

 

[pjug]

Not really, Pano itself in the comment says that it should be the measured impedance, not the nominal one AND if the tone chosen will show different voltages depending wher this frequency is on the impedance curve

Ah, good point. It would be a good idea to know your speaker impedance at the tone frequency and use that value to calculate better results.

Edit: Wait... the level has been set before the test tone come into play. So a good amplifier will put out the same voltage with a test tone regardless of frequency. So why would the value measured by the voltmeter change if the meter reads true RMS over that range?

 

[Blumlein 88]

OK. I give up

I don't want you to give up, I want you to understand this.

Where crest factor matters is choosing the music track.

Suppose I pick an orchestral piece with 20 db peaks above average level. I play as loud as I dare. Theoretically the peaks will be as loud as I ever ask my system to play. But with such a high crest factor I might have misjudged, I might be allowing momentary clipping every now and again.

If instead I choose a raging track from AC/DC with peaks only 6 db above average level and play as loud as I dare the chances I misjudge are reduced. Because it would be clipping nearly all the time if I set it too loud.

In either case however, once I've picked the loudest I'll advance my volume control the test procedure is the same for both pieces of music. And if I don't misjudge on the wider dynamic range recording I'll get the same answers on estimated power requirements. And that is whether I use the suggested -12 db signal or my own -20 db signal.

 

[pjug]

You are starting to over think this another way now. A good amp will maintain the voltage over a varying impedance and we are using the voltage as a benchmark against 8 ohm power ratings. You really only need to look at voltage and ignore the rest for this quick and dirty test.

Yes you caught me before I did my edit!

 

[PeteL]

Ah, good point. It would be a good idea to know your speaker impedance at the tone frequency and use that value to calculate better results.

Edit: Wait... the level has been set before the test tone come into play. So a good amplifier will put out the same voltage with a test tone regardless of frequency. So why would the value measured by the voltmeter change if the meter reads true RMS over that range?

The amp will not change with the frequency, that is true, but how many amps are linear regarding the LOAD, in the end that's what matters. If the impedance drops to 3 ohms, or 2 ohms, for a 4 ohm nominal speaker, that matters.
Edit, but you are right the voltage reading already considered the frequency content of the music and the impedance curve, but it's still an argument on using the actual measured impedance to assess the power, rather than the nominal one

 

[pjug]

The amp will not change with the frequency, that is true, but how many amps are linear regarding the LOAD, in the end that's what matters. If the impedance drops to 3 ohms, or 2 ohms, for a 4 ohm nominal speaker, that matters.

I would say difficult loads are outside of what this procedure is intended to tell you about amplifier requirements

 

[PeteL]

I would say difficult loads are outside of what this procedure is intended to tell you about amplifier requirements

You where right on this one. Again, I have not questioned the validity of the test, just it's limits.

 

[pjug]

OK I just finally tried this and it is a smart trick. Just really fast and easy. But it is hard to decide how loud is too loud and how extreme to go with test track CF. For example, if I use Rickie Lee Jones Ghetto Of My Mind (0dB peak, -24.7dB RMS), I don't have to play much louder than normal to calculate that 200W (8-ohms) is not enough power.

Also, it should be said that you should be careful about clipping with tracks like this! Better to start not too loud and see what you get otherwise you might go well into clipping without noticing.

 

[NTK]

I don't want you to give up ...

Alright. One more attempt to explain.

I think there is some misunderstanding of what crest factor is. It is not the same as dynamic range. It comes about when we have multitone signals. Below is an illustration.

Here we have a summation of 8 sine waves of different frequencies and phase (the solid black curve at the top is the sum). All of the individual sine waves have V_peak = 1 V, and they are all steady signals. In this case, the CF of the summation signal rose to 11.2 dB. (Phase of the individual signals were calculated to maximize the CF to make this a worst case.) We can see that a multitone signal made up of pure sine waves can have CF far greater than that of the original sine waves.

Assuming an 8 ohm load, the average power of each of the individual tones is 0.0625 W, and the combined average power is 0.5 W (as indicated by the 2 V rms). Now the V_peak of our signal is 7.24 V. If I have an amp rated for 0.5 W with a single frequency sine wave, a V_peak of 2.83 V will be good enough to meet the power rating, but way insufficient to reproduce our signal of 0.5 W average power.

All this is in addition to dynamic range of the program. That's why, if one wants to play music at an average power of 1 W, one may want an amp rated at 100 W (10 dB for the crest factor consideration, plus 10 dB for program peaks) to avoid clipping.

 

[Blumlein 88]

Alright. One more attempt to explain.

I think there is some misunderstanding of what crest factor is. It is not the same as dynamic range. It comes about when we have multitone signals. Below is an illustration.

View attachment 77659

Here we have a summation of 8 sine waves of different frequencies and phase (the solid black curve at the top is the sum). All of the individual sine waves have V_peak = 1 V, and they are all steady signals. In this case, the CF of the summation signal rose to 11.2 dB. (Phase of the individual signals were calculated to maximize the CF to make this a worst case.) We can see that a multitone signal made up of pure sine waves can have CF far greater than that of the original sine waves.

Assuming an 8 ohm load, the average power of each of the individual tones is 0.0625 W, and the combined average power is 0.5 W (as indicated by the 2 V rms). Now the V_peak of our signal is 7.24 V. If I have an amp rated for 0.5 W with a single frequency sine wave, a V_peak of 2.83 V will be good enough to meet the power rating, but way insufficient to reproduce our signal of 0.5 W average power.

All this is in addition to dynamic range of the program. That's why, if one wants to play music at an average power of 1 W, one may want an amp rated at 100 W (10 dB for the crest factor consideration, plus 10 dB for program peaks) to avoid clipping.

You gave a good example and definition of what crest factor is and where it comes from.

But it has nothing to do with why this test method uses a -12 or -20 db test signal.

It is also true dynamic range isn't directly the same as crest factor. The effects are similar however in that your max signal level is much greater than your average signal level when you have a high CF and when you have a recording with high dynamic range.

 

[NTK]

You gave a good example and definition of what crest factor is and where it comes from.

But it has nothing to do with why this test method uses a -12 or -20 db test signal.

It is also true dynamic range isn't directly the same as crest factor. The effects are similar however in that your max signal level is much greater than your average signal level when you have a high CF and when you have a recording with high dynamic range.

OK. A cool shower made all the difference. I did over-think (or not thinking at all ).

Now I think I get what Arhimago was doing. What he did was to set his volume knob on his amp to the maximum setting he'd ever use. Then find out with a sine wave input at reference level (0 dB), or something scalable to it, what the output V rms is, and figure out the never-to-exceed power.

The only minor issue I can see is that one will need to make sure the amp doesn't clip during the first step (playing music at the max listening level to determine the volume knob setting). For that a scope or ADC will be required.

Therefore, current score --> Blumlien88 +1, NTK -1

 

[Blumlein 88]

OK. A cool shower made all the difference. I did over-think (or not thinking at all ).

Now I think I get what Arhimago was doing. What he did was to set his volume knob on his amp to the maximum setting he'd ever use. Then find out with a sine wave input at reference level (0 dB), or something scalable to it, what the output V rms is, and figure out the never-to-exceed power.

The only minor issue I can see is that one will need to make sure the amp doesn't clip during the first step (playing music at the max listening level to determine the volume knob setting). For that a scope or ADC will be required.

Therefore, current score --> Blumlien88 +1, NTK -1

I wasn't keeping score, but yes you have it exactly now.

I think gain staging your system is useful. And I've helped others do that numerous times so it is something I've considered in depth. The Pano test is quick and dirty and simple for a ballpark figure.

 

[RayDunzl]

The amp will not change with the frequency, that is true, but how many amps are linear regarding the LOAD

A similar Speaker:

My antique amplifier at 5W into speaker:

 

[restorer-john]

If the amplifier cannot supply the required current, voltage will sag, therefore it will also be manifested as voltage clipping.

That is not the case for transient events. The voltage does not sag immediately. The filter capacitors along with the TXF capability will determine for how long, along with the current capability and design of the output stage.

Amplifiers can current limit, voltage limit or power limit (based on integration).

Short term, the bottom line is you need enough available voltage swing (the internal +/- PSU rails minus losses) to accurately reproduce any shape waveform within the confines of the gain you have set or designed. Long term, you need to be able to maintain the available voltage swing in the face of changing current demands from wildly fluctuating loads and frequencies, within the gain setting you have set/designed.