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  1. Speaker power consumption quadruples when producing sound with double amplitude.

  2. Two speakers consume two times the energy of single speaker.

  3. Two speakers playing signal in phase double the sound amplitude.

If all these statements are correct,it means two speakers will produce same sound amplitude as single speaker with half the power consumption.That seems impossible becose if single speaker had more than 50% efficiency,it will create more energy than it consumes thus violating laws about conservation of energy.

Maybe it have to do with fact that double velocity is quadruple energy and sound a amplitude is related to volume of displaced air so two speakers can move same amount of air,but with half the particle velocity so the total energy is half ( 1/4 for single speaker,half total for two )

What are your thought? Is it really possible that two speakers need half the power to produce same sound amplitude? Is there something that I am missing that would prevent violating the laws of energy?

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5 Answers 5

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Assume in all that follows each speaker arrangement treated here is properly impedance-matched to the power source driving it. This is essential to properly compare the performance of the different speaker arrangements you mentioned.

The error in your analysis is that "two speakers consume twice the energy as a single speaker". This is not true. What is true is that two speakers are CAPABLE OF DISSIPATING TWICE THE POWER of a single speaker. If the impedance of the two-speaker system is 8 ohms and the one-speaker system is 8 ohms and you drive both of them with the same amplifier, the same electrical power will be dissipated in each case and both systems will radiate the same amount of acoustic power. However, the two-speaker system will accomplish this with less cone displacement amplitude, as others here have pointed out.

Why run a system with two speakers instead of one? Because if one speaker is rated at 50 watts maximum input power, then two can handle 100 watts- but they will not dissipate 100 watts of power if they are being driven by a 50 watt amplifier.

Similarly, if we have an amplifier rated at 50 watts output power and we connect it to a speaker system with two speakers, then each individual speaker sees a 25 watt source (NOT 50 watts!), and a single speaker would of course see the full 50 watt output.

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The "in phase"-ness depends on the position of the receiver. While there are positions where you'd get twice the amplitude, there are also positions with zero amplitude.

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  • $\begingroup$ Not if the they are close enough or play low frequency,ie wavelenght 4x longer than driver diameter.If two speakers with shape of cube with 10cm lenght where one entire side is driver surface are playing side to side 20 Hz sinewave,there is no cancellation,no zero amplitude dead spots,they act like single speaker. $\endgroup$ Feb 8, 2018 at 19:27
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The question is not stated at all well. Nevertheless I think that the question is alluding to something that I am convinced is incorrect and yet is a common assumption with certain loudspeaker crossovers. In particular, with Linkwitz-Riley crossovers, each driver is at -6 dB at the crossover point (where SPL and acoustic power is identical for both). The claim that I often see made is that they sum to 0 dB, which supposedly happens because they are in phase. This is nonsense. Each is producing power at 25% of the nominal amount; the total will 50% or -3db. With any L-R crossover, the two drivers will be coherent but there will be a -3dB dip at the crossover point.

In a popular book on loudspeaker design that has gone through more than a dozen editions since first published several decades ago, the author claims that when you put two speakers in place of one the acoustic power increases by +6 dB. He is so sloppy in his claims that he doesn't even both to make clear whether this happens when electrical power doubles (they are wired in parallel) or what exactly. Regardless it is incorrect. He is presumably counting on +3 dB from the doubling of electrical power, and thinks that another +3 dB comes from something else. He is confused. The only way that acoustic power will quadruple when two drivers are wired in parallel is if power from each driver doubles, which requires increasing the signal level supplied to each driver by a factor of 1.41 (the square root of 2). If the voltage supplied to an individual driver increases by this factor, power will increase by the square of this factor (because current will similarly increase by that factor), i.e., power will double, and if you do the same for two drivers, combined power will increase by a factor of 4 (+6dB). The author of that popular book has created a whole lot of misunderstanding.

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There is more functional load on one person to do a task as compared to the same task needed to be done together by two persons.

In case of your problem, we can use the same analogy as above.

In order to produce double the amplitude, there is more load on a single speaker alone. More of it's components is utilised extensively, which results in more friction, heat loss, which results in more power consumption(quadrupled in your case). But when you divide the same task between two speakers, less components of each single speaker is utilised, and hence less friction, heat loss exists and hence less power is consumed(double in your case).

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It's funny you say it's a mystery, of course it is not. In fact, it is in perfect agreement with physics, and your three statements are perfectly correct.

Your confusion arises because you're confusing Amplitude and Intensity. Intensity is proportional to $A^2$, which can be easily seen in any wave chapter of mechanics.

Intensity is power divided by surface. The wave is meant to be a spherical wave, so the surfaces grows as $4\pi r^2$. In sum

$I= \frac{P}{4\pi r^2} \propto A^2$

So it is perfectly right that, if you double the amplitude, you'll have 4 times the power consumed. If you had three times the amplitude, you'd get 9 times the consumption of normal amplitude.

If you add two speakers, you are just summing an identical system. The energy is an extensive quantity, so they just add up. Two systems = two times the energy of one system (if they're equal).

Finally, your third point, yes, the sound is a wave, so amplitudes add up when they coincide in the same point. If two maxima coincide, then you'll have twice the amplitude (and thus 4 times the intensity)

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  • $\begingroup$ I still dont understand,why arent two speakers 3db louder when their combined energy is double? Why their combined output is 6db higher amplitude? $\endgroup$ Feb 8, 2018 at 20:00
  • $\begingroup$ dB scale is not linear. $k[dB]=10\cdot \log \left(\frac{k}{k_0}\right)$ for any quantity $k$, being $k_0$ the reference value. Undo that operation and you will see. $\endgroup$
    – FGSUZ
    Feb 8, 2018 at 20:14
  • $\begingroup$ 3db of power is double ... 6db of amplitude is double,to obtain double amplitude of sound from speaker,it requires quadruple energy.Doubling number of speakers,we double amplitude,but energy consumed is only doubled,not quadrupled as expected. $\endgroup$ Feb 8, 2018 at 20:49
  • $\begingroup$ Sure: if you double the amplitude, you'll get 4 times the power/energy consumed. If you send a doubled signal, it will require 4 times the energy. Now you say, if you put 2 speakers, amplitudes will add, and so you will have the same situation... yes: twice amplitude, which requires twice the energy. But you only sent twice the initial energy. How can be 4 tiems the energy if you only sent two? No, you didn't create energy. The key is that you REDISTRIBUTED IT. Amplitudes only add constructively in certain positions, where you have 2A, and 4E. But there are other points with destructive one. $\endgroup$
    – FGSUZ
    Feb 8, 2018 at 22:57
  • $\begingroup$ The comment was too long. What I mean is that, with two speakers, you'll have points where two maxima add up. That's a constructive I.F. and you have 2A, 4E. However, you have other points in space where a maxima meets a minimum and the amplitude is 0, energy 0 too. So you're not creating energy, you're jsut redistributing it. Before you had E+E = 2E everywhere. Interference gives you series of 4-0-4-0... $\endgroup$
    – FGSUZ
    Feb 8, 2018 at 22:59

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