Speaker power

  • Thread starter PSuran
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25
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Could it be this simple...

When I turn one speaker on, it creates sound, but increases air impedance while doing so. When I turn on another speaker with the same power, it "works into" this increased impedance. This increase in impedance when two speakers are on is what accounts for the increased efficiency?
 
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I was thinking about it today from a different angle - the logarithmic nature of power efficiency.


Everything we've been saying implies that:

- Using two speakers with same total amount of power (2 x 50W) is more efficient than one speaker with same power applied (1 x 100W).
- Speakers are "driven by voltage". Not sure I can visualize this, but hey. I believe it :)

The formula which relates power and current is: U squared = P/R

Let's test what happens in the example with our two speakers.

If:

Two speakers, 500W each:

R = 10 (let's say) ohm
P = 500 W
U = 70.7 Volts

So we have two speakers driven by 70.7 V each.

One speaker driven with the same total amount of power (1000 W):

R = 10 (let's say) ohm
P = 100 W
U = 100 Volts

So the single speaker is driven by 100 V.

If we compare 141.4V and 100V using the dB formula (dB = 20 log V1/V2), we get:

dB = 20 log 141.4 / 100, which is...

EXACTLY 3 dB


I'm just not sure how to visualize all this :)

P.S.:

The analogy which got me thinking like this is:

Let's say you have to transfer an X number of passengers (more than fits in one car) from A to B in the smallest amount of total time, with most efficient fuel usage.

Is it better to use one car driving faster, or two cars driving slower?

Considering (let's say) 50 horsepower is used to drive 100kmph, and 100 horsepower is used to drive 120kmph, it is much more efficient to just drive slower, and use two cars! (If you have two "free" cars at your disposal, and fuel is the only concern)
The exact numbers used in the analogy are wrong, but the point is clear I guess. We used less total fuel AND transferred all the passengers sooner.
 
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Where the two pressure waves intersect you are adding the two pressures together (bright spots), but you are also subtracting when they are opposite giving no pressure (dead spots). This will balance out so there is conservation of energy. You will have double pressure (4 times power since it is proportional to the square of the pressure) in some spots and zero pressure in others. The average power of the system stays the same as your input power. The analogy should be closer to waves on water intersecting. Sound engineers need to be careful of speaker placement because it will cause bright spots and dead spots because of this.
 
25
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Where the two pressure waves intersect you are adding the two pressures together (bright spots), but you are also subtracting when they are opposite giving no pressure (dead spots). This will balance out so there is conservation of energy. You will have double pressure (4 times power since it is proportional to the square of the pressure) in some spots and zero pressure in others. The average power of the system stays the same as your input power. The analogy should be closer to waves on water intersecting. Sound engineers need to be careful of speaker placement because it will cause bright spots and dead spots because of this.
Thanks for the input. I fail to see how this can be the case. If we take into account the fact that we can never get more than +6 dB if two speakers are combinted, consider these scenarios:

Situation 1: Two subs one next to the other. Play a 30 Hz sine wave. You pretty much have +6 dB in all directions, everywhere (low enough wavelength, sources close enough).
Situation 2: Now place them 2.8m apart. You have +6 dB at equal distance from both, but huge dips at other locations.

If, as you say, the balance of dead spots to brighter spots was the only important thing, we would need spots "brighter" than 6 dB in situation 2. That just doesn't happen.
 
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Good point, let me see if I can figure it out in the math.
 

tech99

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Thanks for the input. I fail to see how this can be the case. If we take into account the fact that we can never get more than +6 dB if two speakers are combinted, consider these scenarios:

Situation 1: Two subs one next to the other. Play a 30 Hz sine wave. You pretty much have +6 dB in all directions, everywhere (low enough wavelength, sources close enough).
Situation 2: Now place them 2.8m apart. You have +6 dB at equal distance from both, but huge dips at other locations.

If, as you say, the balance of dead spots to brighter spots was the only important thing, we would need spots "brighter" than 6 dB in situation 2. That just doesn't happen.
1)The interference pattern is peaked, not sinusoidal, so the area above 0dB is less than half the total.
2)If you place two speakers close together, they are pushing the same piece of air. So each one has an easier time and absorbs less power from the generator. For this reason, no increase in sound field is obtained. Another way of looking at it is to say that each speaker is receiving power from the other, and when you measure the speaker impedance it is altered so that each speaker takes less power.
 
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1)The interference pattern is peaked, not sinusoidal, so the area above 0dB is less than half the total.
Sorry, I don't understand this :( Please clarify.

2)If you place two speakers close together, they are pushing the same piece of air. So each one has an easier time and absorbs less power from the generator. For this reason, no increase in sound field is obtained. Another way of looking at it is to say that each speaker is receiving power from the other, and when you measure the speaker impedance it is altered so that each speaker takes less power.
This makes perfect sense, and was my initial understanding. The only reason why I'm having issues with this explanation is the following:

I have measured (in live sound and studio situations) that the ONLY REQUIREMENT for the + 6 dB SPL increase is that the measurement device (SPL meter) is in the triangle apex (same distance, sweet spot) between the speakers. It doesn't matter how far away the speakers are one from the other.

So I would be inclined to conclude that what you wrote above is true, but that this impedance change is inherent to the point of summation (measurement), regardless of the source. What do you think?
 

tech99

Gold Member
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The in-phase addition will always occur along the centre line, and in other places depending on geometry.
The impedance change will occur only when the speakers are a small fraction of a wavelength apart.
 
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The in-phase addition will always occur along the centre line, and in other places depending on geometry.
The impedance change will occur only when the speakers are a small fraction of a wavelength apart.
OK then, try visualizing this:

Free field -> We have two speakers 2m apart. The SPL meter is at the same distance from both, at the triangle apex. If I understand your point, for a low enough frequency, we should be measuring + 6 dB (double power + impedance change). Also, if we just scale this setup up -> increase the distance between the speakers (to, let's say, 15 meters), and keep the SPL meter in the triangle apex - we will lose the "impedance" effect?
 

sophiecentaur

Science Advisor
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Having read most of these posts I have to conclude that sound must be one of the worst possible phenomena to discuss to help you get an understanding of Interference. The programme material covers a range of around ten octaves. The fields at the location of the listener are a combination of very near field, intermediate field and far field. The simple two-source interference pattern of Young's Slits just doesn't apply.
The only thing one can rely on is that two identical sources of Power will produce 3dB more total power than a single source and up to 6dB more in some places - as long as the mutual effects between the two sources are negligible. The power cannot be greater and it all 'has to go somewhere'. There is no Magic at work.

One of the best ways to learn about the behaviour of Waves is to start with the most ideal situation and that is almost certainly what you get with short Radio Frequency Dipoles, at a single frequency, with fields sampled with a short dipole probe. Dipoles are not ideal (omnidirectional and polarised) but neither are loudspeakers or microphones (or ears) ideal. If you get familiar with this stuff then you are less likely to make the sort of gaffs that are so likely when talking about sound phenomena.
 
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One post covered it well, 2 identical speakers in parallel driven from the same voltage will give you 6dB gain, 3 db from the double power, 3 db from the doubled cone/radiating area. 2 identical speakers regardless of connection, series or parallel will give you a gain of 3dB when driven the with same power. This gain comes from an efficiency increase due to the larger coupling area of the two cones.

Consider a speaker, when the cone moves it displaces a volume of air, but the mass of that air is a tiny portion of the mass the linear motor in the speaker has to move, which is the cone itself. So the motor power is being mostly used to move itself.

A good way to look at a speaker is an impedance matching problem. Most of the energy being put into the speaker in free air is used to move the cone. If we could come up with some sort of "air impedance matching transformer", so that we could increase the proportional amount of air mass the speaker moves relative to the cone mass, then you get a massive gain in efficiency. Doubling the cone area gives that increase in radiating efficiency, but also doubles the mass of the cones.

Enter the horn. This is an "air impedance matching transformer", now the speaker is able to use the electrical energy to more efficiently move air by matching the high pressure low displacement volume capability of the motor assembly to the large volume displacement low pressure mouth, voila way more noise.
 

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