# Speaker power

## Main Question or Discussion Point

Hello,

I have a question that I've been struggling to find an answer to / understand. Probably because my understanding of physics is quite limited :)

Before I ask the question, these are the things I take as facts, so please correct me if something is wrong:

1. A speaker in half space (close to a wall), measures at 6 dB SPL higher than the same speaker in free field - because pressure doubles (True for the omnidirectional part of the spectrum). Neumann LINK

2. Two speakers, relatively close to each other, playing the same signal, measure at 6 dB SPL higher than each speaker individually - again, because pressure doubles. This is not true for uncorrelated signals. Sengpiel audio LINK

3. If I have one speaker, and I want to achieve +6 dB SPL, I need to quadruple the power (twice the power = +3 dB), this means using 4 times more electricity.

CONFUSION: But if we have two speakers side-by-side, as in one of the examples above, we get the same 6 dB boost by using only twice the electrical power.

To further clarify. The inverse-square-law explains that for a doubling of sound pressure we need 4 times the intensity. With intensity being "power per unit of area", and power being the rate of energy "usage" - this makes me conclude that having two speakers on simultaneously - uses twice the electrical power, but "creates" 4 times sound intensity, which, to me, seems paradoxical.

I'm sure I'm either misunderstanding something, or conflating some terms, but I hope that I've at least made clear what I'm misunderstanding. Any clarification would be much appreciated.

Related General Engineering News on Phys.org
Overthinking it further :) :

Turning on two speakers is 6dB SPL more (assuming perfect summation) than each playing individually.

"Facts":

1. Driving two speakers requires two times electrical power.

2. 6 dBSPL increase is 4 times the intensity at the measurement device, which is 4 times the acoustic power ---> Maybe this is the wrong assumption, that power and intensity are linearly correlated. But i don't see why they wouldn't be. Even though intensity is "power per unit area", to have 4 times the power in a unit of area, we need 4 times the power at source. Even though not all the power reaches the "unit area", it is still a linearly proportional increase.

So, how can 4 times acoustic power be a result of 2 times electrical power?

Last edited:
3. If I have one speaker, and I want to achieve +6 dB SPL, I need to quadruple the power (twice the power = +3 dB), this means using 4 times more electricity.

CONFUSION: But if we have two speakers side-by-side, as in one of the examples above, we get the same 6 dB boost by using only twice the electrical power. ...
I believe you are confusing power and pressure (SPL).

See: http://www.indiana.edu/~emusic/etext/acoustics/chapter1_amplitude4.shtml
A doubling of power equals an increase of +3dB.
a doubling of amplitude from one source to another equals an increase of +6 dB

Idyit
Speakers work by moving air, if you have a rectified speaker setup then one speaker pushes and one speaker pulls (i.e. The negative and positive on the drivers are reversed), this moves proportionately more air without room compression for the same sine wave being reproduced.

This only works on a single channel and is solely dependent on the wiring within a cabinet for that single channel and will only work for an even number of speakers.
It is commonly used in guitar speaker cabinets.

Most quad boxes are wired this way and some guitar amps have a rectified multi speaker output as well.
I can't tell you the technical or algorithmic expressions, but I can tell you it is much louder and has a lot more punch.

I can't say I've ever seen it in stereo setups.

Hi NTL2009 and Idyit, thanks for the input, but that wasn't really what I was having issues with. I spent some time researching, and what I found was this:

Intro facts:
1. Double the energy = + 3dB
2. Double the amplitude = + 6 dB

Real-life experiment results:

Situation 1: One active speaker, measured @ 1m, @ 1 W of power, creates 90 dB SPL. To achieve + 6 dB, quadruple the power is necessary, meaning 4W. Measured to be true. Checks out.

Situation 2: Two times the same active speaker (meaning 2 speakers). Each speaker 1 W of power, SPL measured in the "sweet spot" (equal distance between both) is + 6 dB SPL. Measured to be true. Checks out.

My main "problem" was -> How come in situation 1 I am using 4 W total to achieve 96 dB, and in situation 2 I am using 2 W total to achieve the same 96 dBSPL.

In the meantime I found that "Situation 2" behaves like this.

+ 3 dB for double the power (remember, two active speakers)
+ 3 dB for double the cone size.

So the total is + 6dB.

This still doesn't account for the smaller energy bill in Situation 2, but I'm sure it has something to do with air impedance and what not, I just can't wrap my head around it yet :)

What is the efficiency of a typical speaker? Are two more efficient than one? I'm not sure.

Cheers

Idyit
Just out of interest, how are you wiring the speakers and how do you achieve the same resistance for the total circuit ?

I suspect you are using more power in the two speaker system because the resistance is halved ?

How are you calculating the total power ?

Just out of interest, how are you wiring the speakers and how do you achieve the same resistance for the total circuit ?

I suspect you are using more power in the two speaker system because the resistance is halved ?

How are you calculating the total power ?
I think these points are key.

From the OP's lnk, a different page, compares Electrical Parameters to Sound Parameters. This makes it clearer for someone like me with background in electronics:

http://www.sengpielaudio.com/calculator-ak-ohm.htm

Sound pressure p N/m² = Pa ≡ V or E voltage
Particle velocity v m/s ≡ I current
Acoustic impedance Z N·s/m³ ≡ R resistance
Sound intensity J or I
W/m² ≡ P power

When you double the voltage level into the same R, you get 4x the power (watts), as P = E2/R

See the above equivalencies, and with an acoustic impedance that remains constant (free field), doubling the sound pressure will give 4x the acoustic power.

It can be very difficult to make even roughly accurate acoustic measurements, due to phase differences, reflections, room interactions, and the interactions of two sources with supporting or cancelling wave points. It is normally done in a large an-echoic chamber, and it is still difficult.

But your issues may be simpler than that - I also am curious how you know you are inputting 1 W and 4 W. You really need to measure both Voltage and current, and take phase into account. Or simpler, you will be close dealing with ratios, and assuming speaker impedance remains fairly constant - and remember that if you double the Voltage to the speaker, the watts will be 4x (Z remained constant). For double the power, increase voltage by 1.414x.

Isn't this just an interference pattern? At some places you get 4 times the intensity, at other places you get nothing. If you integrate over a surface around both the speakers you should still get twice the power on average.

Just out of interest, how are you wiring the speakers and how do you achieve the same resistance for the total circuit ?

I suspect you are using more power in the two speaker system because the resistance is halved ?

How are you calculating the total power ?
Disclaimer: I'm neither an engineer, nor a physicist, so just a note that the answer is probably really simple, or I am making some stupid mistake. Take that into account please :) And I really appreciate the effort.

To answer your above question. I am not actually measuring anything except SPL.

I "know" that I am using "double" electrical power in the two speaker setup because, well... I turn on the other speaker :) And at the same distance from both, each individually measure an SPL value 6 dB lower than both together. Is this the dumb assumption? :)

That's what meant when I wrote that I measured the results. But to be honest, people smarter then me have told me that if I have a single speaker and I boost the volume until I measure a 6 dB increase in SPL - electrical power consumption has increased four times.

And this:

"In the meantime I found that "Situation 2" behaves like this.

+ 3 dB for double the power (remember, two active speakers)
+ 3 dB for double the cone size.

So the total is + 6dB."

This is the info I found in literature... Not by experimentation. I can link the text here if it helps...

Idyit
Yup I'd agree with that ;-)

Yup I'd agree with that ;-)
With what exactly? :)

Idyit
+3db from the halved impedance (i.e twice the current)
+3db from twice the cone size (i.e twice the amount of air movement)

+3db from the halved impedance (i.e twice the current)
+3db from twice the cone size (i.e twice the amount of air movement)
Yes, but the part I'm struggling to understand is how is it possible that I have half the electricity bill with the 2 speaker setup, vs. the one speaker setup at same SPL.

Because with the 2 speaker setup I double the electrical power (turn on two speakers), but quadruple the acoustic power (a measured 6 dB increase = 4 times power). Wouldn't this mean more energy out than it came in? :)

Last edited:
Idyit
You electricity bill will double if you are running the same volume on the amp.

The mysteries of more air movement have been around along time, I guess the only answer is you are converting the power much more efficiently using two speakers.

Yeah, thinking off it... I think the answer to the following sub-question would put me out of my misery :)

With power being constant, simply doubling the speaker size (cone area), increases the output for 3 dB. But doesn't the increased size come with downsides? Heavier, harder to move, more air resistance.... Which would require more power to compensate?

Idyit
I doubt air resistance would make any difference, The magnets in the drivers might make a small difference.

All I can tell you is more speakers equals a much bugger sound, working with a Band for 20years, If I had two quad boxes plugged into the same amp I could run the amp at a much lower volume and get the same sound on stage, I can't tell you the fluid dynamics / physics equations for it, but I know it from standing in front of it :-)

Yeah, thinking off it... I think the answer to the following sub-question would put me out of my misery :)

With power being constant, simply doubling the speaker size (cone area), increases the output for 3 dB. But doesn't the increased size come with downsides? Heavier, harder to move, more air resistance.... Which would require more power to compensate?
But power wouldn't be constant. If you double the cone size, and keep the motion distance the same, you are moving more air, and that takes more power.

If you double the cone size, and then reduce the motion distance to get to the same SPL, you would have brought the power back to the original level.

All that is an over-simplification, as speakers/drivers are optimized to work in a certain range, and you just can't change things like tha , they need to be designed for it. But yes, connecting two identical speakers in parallel, with an amp capably of driving both loads equally well, you should have double the watts in (same voltage 2x current so 2x watts), and double the sound power out. Again, I think you are confusing pressure levels (~ voltage) with power levels (~ watts). Pressure only needs to increase 1.414x for a doubling of power. Assuming constant acoustic impedance.

Again, I think you are confusing pressure levels (~ voltage) with power levels (~ watts). Pressure only needs to increase 1.414x for a doubling of power. Assuming constant acoustic impedance.
Well, I'm not sure I'm confused about what you say. I understand that + 6dB is twice the voltage (pressure), but 4 times the power.

But power wouldn't be constant. If you double the cone size, and keep the motion distance the same, you are moving more air, and that takes more power.

If you double the cone size, and then reduce the motion distance to get to the same SPL, you would have brought the power back to the original level.
Well, exactly! But, to account for the + 6 dB when we add "another active speaker", the sources say that + 3 dB is for double the electrical power used, and the additional + 3 dB is for double the cone size (6 dB total). And I don't understand that logic for the reason you stated (if I understand correctly) -> Increasing the cone size uses more energy, requires more power to "move", if dB gain is required.

Meaning, all other parameters remaining constant, I don't see how simply increasing cone size gives me "free 3 dB". I'm sure it isn't free, I just don't understand where it comes from :)

A qoute from the web as an example: "Recently upgraded (more low end) and downsized my subs, going from a horn loaded design to ported cabs. Reduced size by 25%, decreased weight by 5%, increased LF output by 6dB, using 3dB more power, which happened to be available from the amps I was using.

Re-built 4 cabinets with 2 Lab 12s in each. As predicted, doubling the cones adds 3dB sensitivity, and was verified by tons of real world testing slogging cabinets across the snow, mud and coal dust. "

Last edited:
Idyit
One thing I'll add is an outdoor setting may change your results, In all the outdoor gigs I've done the levels dropped radically for the same power amps.
I think NTL2009 is on the right track, twice the cone size in an enclosed space is twice the air movement/pressure being generated, outdoors the equation will be very different.

One thing I'll add is an outdoor setting may change your results, In all the outdoor gigs I've done the levels dropped radically for the same power amps.
I think NTL2009 is on the right track, twice the cone size in an enclosed space is twice the air movement/pressure being generated, outdoors the equation will be very different.

Exactly, because having a wall behind the speakers adds 6 dB - because it radiates into half space. Placing a sub on the floor in the corner (quarter space) can add 12 dB or more... Having other reflections also adds "volume". And considering 10 dB is subjectively evaluated as doubling the loudness, it's clear why you need more power outdoors.

But, adding a second active speaker adds 6 dB even in free field. That's what literature says.

Last edited:
I think I found the answer, but the problem is that I don't understand it :) :)

It's confusing to me too, but Linkwitz is an expert in the field (I'm familiar with his work) so I'll accept that explanation that it is the doubling of surface area that is increasing the efficiency of electrical to acoustic conversion.

So if we accept that, it's clear that in the case with a single speaker, you need to double the electrical power in to double the acoustic power out, because you didn't change the surface area. Then the comparison works, right?

Though I'm still not grasping how two individual, identical speakers which put out Y acoustic power each with X electrical power in can somehow create 4x the acoustic power. His formulas say so, but I don't get it. How can they act differently together versus separate (unless they are being coupled somehow, but I see no mention of that?).

Though I'm still not grasping how two individual, identical speakers which put out Y acoustic power each with X electrical power in can somehow create 4x the acoustic power.
Exactly, that's the thing that's bothering me.

If we look at it from another angle...

1. One, perfectly omnidirectional speaker in free field radiates 100 dB in all directions (point source - spherically) at a low enough frequency.
2. If we now add one dimension, mount the speaker into a wall, it now radiates all the energy into half the space (Let's assume no deconstructive interference - which is true at a low enough frequency).
3. This now means that we have doubled the radiating power in front of the wall, which would mean + 3dB.
4. But we get + 6 dB!! That's measured fact, and all sources say it behaves like so.

And most simply say, "yes because we are adding pressures and not power".... But if you check my above example, you can understand why this explanation is a bit handwavy. We are always talking about energy in some form, and in this case, it is DOUBLE the energy, not FOUR TIMES, which would be necessary for doubling the pressure.

So I'm sure there is a reason for this increased efficiency, I just don't get it yet :) And this reason is not "because pressure is doubled", yes, pressure is doubled because power is quadrupled.

Last edited: