A Test. How much Voltage (power) do your speakers need?

I measured the test tone at:

  • 2 volts or less

    Votes: 334 40.6%
  • Between 2-5 volts

    Votes: 252 30.7%
  • Between 5-10 volts

    Votes: 106 12.9%
  • Between 10-20 volts

    Votes: 55 6.7%
  • Over 20 volts.

    Votes: 75 9.1%

  • Total voters
    822
It definitely works. For my test (details in the full range section), I recorded 0.75V with my 98dB @ 1W speakers. Squared gives 0.5625W. If you work it out long hand, 0.75V @ 8 ohms gives 0.0703W (V^2 / R) at -12dB. Each 3dB is double power, so 0.1406W at -9dB, 0.28125W at -6dB and 0.5625W at -3dB (biggest sine wave possible).

So what happened there, well to work it out Pano's way, I squared the voltage. To do it long hand, I squared the voltage, divided by 8 (speaker impedance) to work out power, then multiplied by eight (3 doubles) to get from -12dB to -3dB. Cool, wish I had thought of that.

Brian.
 
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And we have a winner, ladies and gentleman! Please give a nice round of applause to Brian - Brig001. :wave2:

He explained it very well and succinctly, but I'll go over it again. Mooly was getting there, too.
  1. You set your playback level with a digital source. In this case I asked you to set the loudest level you ever use.
  2. You measure the -12dB (below full scale) test tone and get a voltage
  3. Maximum sine level on digital is -3dB, or 9dB higher than the test tone
  4. 9dB is a voltage gain of 2.83 Does that number look familiar?
  5. Multiply your measured voltage by 2.83 to reach the maximum possible sine level
  6. You now have the highest RMS voltage a sine wave will ever play on your system at that volume setting.
  7. Find the power of that sine wave.
  8. Measured voltage X 2.83 squared/8 Voila! That's how much power an amp needs to supply at 8 ohms for your music not to clip.
Let's look at that with example numbers:
  • Your measured voltage is 5 volts.
  • 5x2.83=14.15 (the 9dB offset)
  • 14.15x14.15=200.22
  • 200.22/8=25 watts
  • Or... 5x5=25 (watts) Voila!
Why divide by 8? Because most amplifiers state a power rating into 8 ohms. They may also give other ratings, but you'll always see an 8 ohm rating. It's just a matter of convenience for you to know the standard power rating of the amp you need. 5W, 25W, 250W.

Your speakers may not be exactly 8 ohms, but the voltage you measured can be tracked back to an amplifier's 8 ohm rating.
If you have 4 ohm speakers you might want to go by the amplifier's 4 ohm rating. Simply take your measured voltage, square it, then double that to find the 4 ohm power rating you need.

If an amplifier is honestly rated*, for example to AES or IEC standards, then it can output a sine wave at the rated power into the specified load. 8 ohms, 4 ohms, whatever. Peak voltage will be 3dB higher, or twice the power. The amp may be able to supply a little more voltage, but it's going to soon start distorting and clipping.

This little test is simple, it lets you measure a voltage, then multiply it to get a reference level. E.G. "I measured 3V on the test tone -OK, I need at least a 9 watt amp to play that loud".

*Yes, we all know that the marketing department writes the specs, but you are smart enough to see thru that. Look for IEC ratings, or better yet, measure for yourself!
 
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And 6.5" Tannoy dual concentrics in a Voight pipe gave me 4.0V. That would mean a peak of 16V is required. By your equation, that means a 250W amp for clean peaks. Hmmm... kinda high.
You went a step too far. All you needed to do is 4x4=16. That's the power rating of the amp you need. (see my post above) You made the mistake of squaring that again to get to your result. Good effort, but one step too far. :D
 
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Like Ra7, you are making it too complicated.
since i already know my speakers lowest impedance(have measured it)
it would be.
7x7=49/2.7(ohm)=18.14AX49Volt=888.86watts.
If you measured 7V, then your maximum sine level would be 19.8V (9dB higher). You need that much voltage at the maximum sine wave level. Peak will be 3dB higher.

If you want to know how many watts that is into your 2.7 ohm load, do this:
19.8x19.8/2.7=145 You need an amp that can supply about 150 watts RMS into a 3 ohm load. See?
 
2.5v rms into 4 ohm load.

I had it cranked up listening to some rock. I may not listen to it at this level for long because it is hard on the ears. The speakers are quite efficient in the 93 db range.

This DIY chipamp will do about 10v rms into 4 ohm resistors continuous, both channels driven before clipping.

I will have to try my other listening system. Fun test!
 
Like Ra7, you are making it too complicated.

If you measured 7V, then your maximum sine level would be 19.8V (9dB higher). You need that much voltage at the maximum sine wave level. Peak will be 3dB higher.

If you want to know how many watts that is into your 2.7 ohm load, do this:
19.8x19.8/2.7=145 You need an amp that can supply about 150 watts RMS into a 3 ohm load. See?


your way is wrong.

unless if an 150watt rms amp can give 890 watt peak power.
 
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Don't you need to use the same digital source to play the test tone as you used to play the high volume music?
Yes, you do.

your way is wrong.
No, it's not. Check your math and you will see. Your 890 watts is the result of a faulty calculation. You threw amperage in there were you didn't need it. See my previous reply.
 
Yes, you do.


No, it's not. Check your math and you will see. Your 890 watts is the result of a faulty calculation. You threw amperage in there were you didn't need it. See my previous reply.

your calculations say that the peak power is twice the rms power +3db
but this is not the case.

i guess it depends at the music but i can make a 600 watt rms amplifier

distort.

on the contrary my other amplifier 1200watt rms can drive the speakers

to the point the protection kicks in with no audible distortion.
 
I am not clear how this tells you “how much power” other than under the assumption made.

If one takes the AES style pink noise used for testing, one finds the peak to average ratio is 6dB, that is the peak Voltage is twice what the average Voltage is and so if one wanted to test a loudspeaker at a 100W AES pink noise rating, one needs at least a 400Watt amplifier to deliver that signal unclipped.

This is signal is only surpassed in heating value by the instrument amplification or subwoofer reproduction of bass keyboards etc which put out a more or less continuous tones.

If you have seen an FM radio station where the VU meter was stationary (full compression), one usually has a peak to average ratio between 6 to 10dB, that is the peak values were 4 to 10 times the average level.
As it is the average level which corresponds most closely with our perception of loudness (all other things equal), one can see the mechanism of compression, it brings up the average level by bringing up the lowest levels most.

In what many would consider audiophile recordings, the peak to average ratio is more like 10 to 20dB and occasionally into the 30’s (dB of peak to average). Now, producing music with a 20-30dB peak to average, at a fixed perceived average loudness, requires 100 to 1000 times more power than the average level.

I have a fireworks recording that has about a 40dB peak to average ratio, where the very short peaks are about 10,000 times more powerful than the average level. This recording eats up headroom like nobody’s business and so hard to reproduce with speakers I urge people to try good headphones first.

Lastly, everyone knows what clipping sounds like right?

Did you know that if the clipping is short enough you can’t hear it as a flaw and aren’t aware of it?

It is true, you usually can’t hear instantaneous clipping as a flaw, it is undetectable unless it lasts long enough ….OR you compare it directly to an unclipped signal. Then, the clipped signal sounds less dynamic but that’s all.

It is one thing to assume that because you can’t hear an overt flaw, that nothing in the signal chain is clipped and affecting the system dynamics, but it is another to take an oscilloscope and set it to look at the highest peaks coming out of the amplifier and see if they are clipped off at the top or bottom.
Best,
Tom Danley
Danley Sound Labs
 
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I think where many are getting really confused in all this is in the basic concept of thinking of an amplifier as "putting out" say 10 watts RMS when in fact it puts out a voltage that (hopefully) is unaffected by loading. We equate that voltage to so many watts into 8 ohms.

Tom... if the pink noise comes off CD (and same for the fireworks recording) the absolute maximum peak level is known and can be translated into an equivalent "xxx watts into 8 ohms" amplifier is needed.

And then we make assumptions... and that is what this thread is about, a simple ball park figure for the amplifier power rating.

So can that amplifier maintain that "voltage" into the speaker at any frequency ?

Approx 28vrms into 8 ohms is 100 watts.

But the speaker isn't 8 ohm it varies. If the amp can maintain 28 volts though how do we now describe the power at other impedancies that the amp sees as the frequency varies. 28 volts at 3 ohms (some nominally 8 ohm speaker go down to that) is around 262 watts if the amp maintains it's voltage. But is that real power dissipated in the speaker ? or part reactive or apparent power. How does that affect the amplifier that is trying to maintain that voltage.

That's where something like this test moves into more case specific for known and specified equipment.

I think a test like this is genuinely useful and highlights how for the most part just how little "power" or voltage swing is needed.

This is going to run and run isn't it :)
 
In what many would consider audiophile recordings, the peak to average ratio is more like 10 to 20dB and occasionally into the 30’s (dB of peak to average). Now, producing music with a 20-30dB peak to average, at a fixed perceived average loudness, requires 100 to 1000 times more power than the average level.
Many posts back I made a comment on dynamic range in recordings, based upon an old Bob Cordell test. I found the page in wayback here

"This demonstration was a real "Wow" for the attendees. The Rickie Lee Jones (RLJ) cut was played at realistic, but certainly not unpleasant, levels in the relatively small hotel exhibit room on speakers with an estimated sensitivity of about 89 dB. The average power typically read 1-2 Watts, while the power on peaks often topped 250 Watts (the power display monitored only one channel, so these numbers should be interpreted as Watts per channel). On this cut, most peaks occurred with an aggressive thwack to a snare drum positioned dead center.

While it is true that the RLJ track has an unusually large dynamic range, this data still suggests that many listeners may be clipping their amplifiers more often than they think. This may especially be the case for those with tube amplifiers who are not using extraordinarily efficient speakers."


I agree with your comment that many well recorded albums can have dynamic range of 20dB or more, and movies can certainly add 20-30dB from dialogue (~ average) to loud SFX so I think many here are seriously underestimating the power they need if they never want to clip.

If I were to assume the 6V (5.7W at Re=6.3Ω) I mentioned in an earlier post was actually an average to get a never clip 20dB peak above that, I'm looking in the 600W class for amps.
 
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your calculations say that the peak power is twice the rms power +3db but this is not the case.
Yes it is the case. A sine wave has a peak voltage that is 3dB higher than its average (RMS) values. There is no doubt about that.
Do you think you made a mistake in the test when you measured 7V on the test tone?

I am not clear how this tells you “how much power” other than under the assumption made.
Correct. See the assumption in the first two posts and what Mooly says above. If you have done the test correctly, then you have measured a test tone that is 12dB (RMS) below the highest peak voltage that you ever use. You have just determined your peak voltage - at least for digital playback. The level will never go higher than that peak, unless you turn up the volume.

What would be the highest, unclipped sine wave value at your volume setting?
9dB above where you measured. Simple.

You now know the maximum voltage, and thus power, you need to play at your loudest setting. So you could say "OK, my 25W amp won't clip even when I play really loud, but it is right at the limit".

How do you know you have a 25W amp? You could measure it. Or you could take its advertised rating and reduce it a little, or a lot, depending on high much you think the marketing department has inflated the figure.
Rod Elliot has a good and funny page about amp power ratings here: Amplifier Power Ratings
 
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If I were to assume the 6V (5.7W at Re=6.3Ω) I mentioned in an earlier post was actually an average to get a never clip 20dB peak above that, I'm looking in the 600W class for amps.
I don't follow. Where is the 6V coming from? Is that what you need to get -20dB RMS at your listening level? Seems rather high, but possible.

If so, that indicates 60V peak, or 450W peak into 8 ohms. Or an RMS rating of about 225W into 8 ohms.
 
What would be the highest, unclipped sine wave value at your volume setting?
9dB above where you measured. Simple.
Sorry, I don't get this. You could easily have set the RMS of your test tone at 30dB down, adjusted the gain to get it to the same level. Then used that measured voltage as an average and known you would still be capable of peaks nearly 30dB higher than that, as they would not have reached 0dBFS.

Setting the level to -12dB is not realistic for potential program material.

I have a jazz disc somewhere here (I don't like the music) that noodles along at a modest level, which if I set it for a comfortable level at the LP (average of 80dB or so) is nice. Then all the horns come in at once and the difference on that initial note is easily >20dB, based upon your 9-12dB difference, artificially set, then I would be easily into clipping.
 
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Brett, don't be misled by average music levels and the test tone. Only the test tone level is important for the calculations. I could have set it at -3dB or at -33dB. As long as you know what the level of the test tone is - you can calculate the maximum level.* It isn't complicated.


*at the volume setting you used to measure the tone.