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A case study on acoustic mismatch between cone and air


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#1 of 29 Vaughan Odendaal

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Posted August 15 2007 - 07:45 AM

In my pursuit to improve my knowledge on audio, I ask more questions on the physical operation of cone speakers and the "coupling" that takes place between it and the air.

As I understand it, cone speakers are extremely inefficient and a small percentage of electrical power is converted to mechanical energy. The reasons for this, again, according to my knowledge, are many. One, because the magnetic field and electrical fields are not 100 % efficient, there are eddy currents that reduce overall efficiency.

The suspension has resistance properties, the voice coil has power handling limits and too much input power can result in thermal compression. We also have the enclosure which dictates to a large degree how much overall resistance (or lack thereof) there is going to be within the system.

But the transfer of energy from cone to air is highly inefficent because the cone has mass and it needs to be large to excite a given amount of air. The air also has resistance to motion. Large cones are needed to excite large amounts of air.

Larger drivers can move more air easier because the effective piston area is larger and can "punch" a greater amount of air out in the time required compared to smaller drivers.

But is there something that I'm missing ? Horn loaded speakers have a good impedence match to the air because a large proportion of electrical energy is converted to acoustical energy.

But if someone more knowledgeable than myself (and I'm sure there are here ) can explain to me how cone speakers have a poor impedence match to the air (and if the reasons I cited have a large influence on it) then I would appreciate it.

Thank you.

--Regards,

#2 of 29 ChrisWiggles

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Posted August 16 2007 - 02:17 PM

Yes speaker efficiency varies, but why is that a concern? The point is sound quality, not transducer efficiency. Amplifiers of various designs vary significantly in their efficiency too.

Horn loaded speakers can increase the sensitivity(but not necessarily the electrical efficiency, sensitivity and efficiency are distinct, and I'd say that sensitivity is not overly relevant unless loudness is a goal, and efficiency really is completely irrelevant) but they also decrease dispersion which you may not want. And they may not sound as good.

Other speaker designs like planar designs can be horribly electrically inefficient, but they can sound incredible.

Efficiency can certainly be an issue in driver design if a designer is concerned about heat dispersal and avoiding compression and that kind of thing, but when buying speakers not only are you not going to find any information on efficiency, it's really a pretty pointless spec since it tells you absolutely nothing about the speaker except its electrical efficiency. And unless you are SERIOUSLY trying to conserve energy I don't see why anyone would care in the least about this.

#3 of 29 Vaughan Odendaal

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Posted August 17 2007 - 08:22 AM

Chris, I thought I would bring this topic up because it is fascinating, at least to me, and I wish to discuss it further.

In the other thread you also question the need for me to bring up these questions. I don't know why, but again, I find it fascinating and I want to learn further. I'm not saying that efficiency means anything other than being an electrical metric. I just want to understand the physics of what is happening.

I know that speaker efficiency is on the order of a couple of percent. Horn loaded speakers are far more efficient. But I would like to understand how the air is coupled better, or rather, why the air is not coupled to the driver well in a normal monopole speaker.

The reasons I cited I assume were correct for the inefficency ?

--Regards,

#4 of 29 Vaughan Odendaal

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Posted August 17 2007 - 08:25 AM

When you mention that efficiency is irrelevant, what do you mean ? Although technically speaking, efficiency and sensitivity are defined differently and hence can't be the same based on definition, is there not an inverse relationship between the two ?

A speaker with a high sensitivity also happens to be more efficient because it converts more electrical power to mechanical. Is that not correct ? Please correct me if I'm wrong.

--Regards,

#5 of 29 ChrisWiggles

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Posted August 17 2007 - 09:09 AM

Quote:
Originally Posted by Vaughan Odendaal
Chris, I thought I would bring this topic up because it is fascinating, at least to me, and I wish to discuss it further.

In the other thread you also question the need for me to bring up these questions. I don't know why, but again, I find it fascinating and I want to learn further. I'm not saying that efficiency means anything other than being an electrical metric. I just want to understand the physics of what is happening.

I know that speaker efficiency is on the order of a couple of percent. Horn loaded speakers are far more efficient. But I would like to understand how the air is coupled better, or rather, why the air is not coupled to the driver well in a normal monopole speaker.

The reasons I cited I assume were correct for the inefficency ?

--Regards,

I don't question why you asked to be dismissive, but merely because if you have a particular reason for understanding this, maybe there are particular answers that can help or resources we can direct you to. It sounded like perhaps you were embarked on a project or something, and maybe we could help with that.

The efficiency is an electrical question, not any kind of coupling to the air. Efficiency is basically related to how much of the energy is lost to heat and the like in the circuit. Sensitivity is related to how much volume you get for a certain input, which is why you see sensitivity specs, since that is relevant. Obviously you can't have a high sensitivity if the whole circuit is very inefficient, but you can have drivers with the same electrical efficiency with very different sensitivities, depending on the way that the driver is used in an enclosure, horn or whatnot. A sensitive speaker can be driven very loud with very little power, a speaker with low sensitivity requires more power to reach the same volume output. But this doesn't tell you how good the speaker sounds, but it does tell you how much amplification you may need to drive the speaker properly.

So, to differentiate efficiency and sensitivity, let's just take some driver. The efficiency is basically just considering how much of the electrical current is being lost to heat and friction in the driver and things like that, and isn't actually doing any work. The fewer losses you have in the circuit, the more efficient it is, basically. But it's not relative to how much sound you're getting. That same driver could be used in a sealed box or in say a ported box, and the latter may allow you to get more volume from that same driver and thus be more sensitive, but the efficiency of the driver isn't any different.

#6 of 29 ChrisWiggles

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Posted August 17 2007 - 09:15 AM

Quote:
Originally Posted by Vaughan Odendaal
A speaker with a high sensitivity also happens to be more efficient because it converts more electrical power to mechanical. Is that not correct ? Please correct me if I'm wrong.

--Regards,

Inevitably so, but you could have a very efficient driver used in a way that leads to low sensitivity. The point is that efficiency isn't relevant to your buying choices. IT doesn't tell you anything interesting about the speaker. Why would one want to know the efficiency of their speaker drivers or their speaker? I don't know any reason that would be of any interest. Sensitivity tells you something interesting, which is how much power you may need to drive the speaker to the desired volume. Efficiency doesn't tell you that.

So for example, a motor, say a car motor, will have some mechanical efficiency based solely on how the motor itself is designed and how it works. You're going to have so much energy being created by burning gasoline, but not all of that is going to be captured by the motor, some of that is going to be lost in the process to heat, and friction and stuff like that. The efficiency of the motor doesn't change. It isn't particularly interesting except to the engineers. What interests you may be the gas mileage of your car, which is more like sensitivity, it's actually related to something that interests you. You put that motor in a small, lightweight sports car, the efficiency is the same. You put that motor into a dumptruck or something, the motor's efficiency is still the same, but since the motor now has to move around a gargantuan vehicle, the gas mileage is going to be a lot lower.

#7 of 29 Vaughan Odendaal

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Posted August 17 2007 - 09:49 AM

Chris, thank you for having the time to answer my questions ! Most kind. I guess then that the efficiency part of the equation answers part of my question.

So the suspension system is not completely linear, there will be resistive losses, the coil itself and magnet will not maintain a strong magnetic field at all excursion levels which reduces motor strength, and there are eddy currents created which will further reduce motor strength.

But when I hear people say that cone speakers have a poor impedence match to air, what are they talking about ? I imagine that it's difficult for cone drivers to enegerize the air because of the physical surface area required and the size needed to reproduce low frequencies has to be large.

Is this more or less what is an impedence mismatch ? I'm not entirely sure but I'm willing to find out.

Thanks again.

(btw, your post answered the gist of what I wanted, so thanks)

#8 of 29 ChrisWiggles

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Posted August 17 2007 - 01:05 PM

Quote:
But when I hear people say that cone speakers have a poor impedence match to air, what are they talking about ?

I have no idea. That statement sounds nonsensical. Can you link to a source?

Quote:
I imagine that it's difficult for cone drivers to enegerize the air because of the physical surface area required and the size needed to reproduce low frequencies has to be large.


It's the same difficulty for any kind of transducer. They all have to move the same amount of air to get the same result. How they do that depends on what kind of transducer you're talking about, and each kind of design has its own strengths and weaknesses. None of the drivers are "energizing" air, they are simply displacing air to create pressure changes, and these pressure changes travel through the air and we hear them as sound.

#9 of 29 Vaughan Odendaal

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Posted August 17 2007 - 06:17 PM

Chris, I'll try to link you to multiple sources concerning the "impedence match" issue. I'm going out now but I would like to continue this discussion if you will.

Thank you again and I will speak soon.

--Regards,

#10 of 29 Jacob C

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Posted August 18 2007 - 02:58 AM

Quote:
Efficiency is basically related to how much of the energy is lost to heat and the like in the circuit. Sensitivity is related to how much volume you get for a certain input, which is why you see sensitivity specs, since that is relevant.

So, if the efficiency is high but the sensitivity is low where is the left over energy going? According to the information above it cannot go into heat since that is related to efficiency. It has to go somewhere.
Jacob
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#11 of 29 Vaughan Odendaal

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Posted August 18 2007 - 03:38 AM

A few links to the impedence mismatch issue :

http://hyperphysics.....o/spk2.html#c4

http://www.rocketrob...echart/spkr.htm (Read the section on "speaker efficiency" which is near the end of the article)

http://www.du.edu/~j.../tech/speak.htm (Read the section on "Horns" which is near the end of the article)

And lastly :

http://epic.mcmaster...en/speaker.html

I think that is enough to give you the idea of what I mean. I just need to understand this more. Thanks.

We'll wait. . .Posted Image

--Regards,

#12 of 29 ChrisWiggles

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Posted August 18 2007 - 06:01 AM

Quote:
Originally Posted by Jacob C
So, if the efficiency is high but the sensitivity is low where is the left over energy going? According to the information above it cannot go into heat since that is related to efficiency. It has to go somewhere.

Well, the circuit's efficiency itself isn't changed, but it can be dissipated elsewhere, as I attempted to describe depending on how the driver is used. For instance, if you put the driver in a sealed box, the back-wave energy from the driver which is equal to the front-wave energy, is contained in the enclosure, so you only get half the output in terms of sound. If you put that driver in a ported enclosure, now you have re-captured a great deal of that back-wave energy, so you have more output. The driver's electrical efficiency hasn't really changed, but the speaker's sensitivity has significantly.

#13 of 29 Vaughan Odendaal

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Posted August 18 2007 - 06:07 AM

With ported speakers, the sensitivity only increases significantly at and around port tune. Posted Image

--Regards,

#14 of 29 ChrisWiggles

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Posted August 18 2007 - 06:10 AM

Quote:
Originally Posted by Vaughan Odendaal
A few links to the impedence mismatch issue :

http://hyperphysics.....o/spk2.html#c4

http://www.rocketrob...echart/spkr.htm (Read the section on "speaker efficiency" which is near the end of the article)

http://www.du.edu/~j.../tech/speak.htm (Read the section on "Horns" which is near the end of the article)

And lastly :

http://epic.mcmaster...en/speaker.html

I think that is enough to give you the idea of what I mean. I just need to understand this more. Thanks.

We'll wait. . .Posted Image

--Regards,

Ah yes, well the basics of that are to restrict the air movement basically, and restrict the propagation of the wave, so you have more of the energy being radiated in one direction, rather than all over the place. But again, why the questions? What's your project?

#15 of 29 ChrisWiggles

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Posted August 18 2007 - 06:12 AM

Quote:
Originally Posted by Vaughan Odendaal
With ported speakers, the sensitivity only increases significantly at and around port tune. Posted Image

--Regards,

No, it increases overall. You have output of that back-wave energy at all frequencies, not just at the port/enclosure tune.

#16 of 29 Vaughan Odendaal

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Posted August 18 2007 - 07:14 AM

As far as I know, that is completely untrue. Ported subwoofers are helmholtz resonators by definition, so they only work under a narrow band of frequencies.

Besides, the port itself isn't 'activated' until the driver is moving near or at the frequency tuned by the port. How can the air be moved out the port if the cone is moving too fast above tuning ?

With what I understand with mechanical systems, the air in the port won't respond to the pressure fluctuations caused by the driver above port tune precisely because the driver is not moving at the correct speed to move air in and out the port. Because of this, the driver is doing most of the work until frequency is reduced at and around the port tuning frequency. Then, and only then will the port begin to suck in air through the internal air pressure and load the active driver.

This is what I understand about ported speakers/subwoofers. A port is a narrow band resonator. It will not be 'activated' otherwise and therefore efficiency above port tune will be exactly the same as it would be for a sealed subwoofer (all things being equal).

It is only at port tune that the driver excursion is minimized because the air pressure being sucked through the vent is counter-balancing the cones excursion so distortion will be lower and output will be increased, hence greater efficiency down low.

Just my 2 c.

--Regards,

#17 of 29 ChrisWiggles

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Posted August 18 2007 - 07:25 AM

Quote:
Originally Posted by Vaughan Odendaal
As far as I know, that is completely untrue. Ported subwoofers are helmholtz resonators by definition, so they only work under a narrow band of frequencies.

Besides, the port itself isn't 'activated' until the driver is moving near or at the frequency tuned by the port. How can the air be moved out the port if the cone is moving too fast above tuning ?

No, your understanding is erroneous.

Quote:
With what I understand with mechanical systems, the air in the port won't respond to the pressure fluctuations caused by the driver above port tune precisely because the driver is not moving at the correct speed to move air in and out the port. Because of this, the driver is doing most of the work until frequency is reduced at and around the port tuning frequency. Then, and only then will the port begin to suck in air through the internal air pressure and load the active driver.

This is what I understand about ported speakers/subwoofers. A port is a narrow band resonator. It will not be 'activated' otherwise and therefore efficiency above port tune will be exactly the same as it would be for a sealed subwoofer (all things being equal).

Well yes, but it still is an open port at other frequencies, and the back-wave pressure created by the driver still exits, regardless if it is the tune frequency.

Quote:
It is only at port tune that the driver excursion is minimized because the air pressure being sucked through the vent is counter-balancing the cones excursion so distortion will be lower and output will be increased, hence greater efficiency down low.

Just my 2 c.

--Regards,


#18 of 29 Vaughan Odendaal

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Posted August 18 2007 - 07:32 AM

Chris, I have appreciated your advice before. But please don't tell me that my understanding is erronous without explaining why or demonstrating it.

A ported subwoofer is a narrow band resonator. That is a fact. Above the tuned frequency, the port contributes extremely little to the total output, moving more or less in phase with the active driver. As the frequency reproduced moves down to the tuned frequency, the reflex system starts to dominate the output, as it's sucking energy so efficiently from the active driver.

I'm surprised you wouldn't know this.

Quote:
Well yes, but it still is an open port at other frequencies, and the back-wave pressure created by the driver still exits, regardless if it is the tune frequency.

What you are saying goes against how ported speakers work. If a port is tuned low enough, then frequencies above that are not going to be energized by the "back-wave pressure".

How could it ? How can this happen if the air in the port is not going to be moving in or out because the driver has not matched the tuning frequency of the port ? A vent, by definition, only works under narrow band frequencies. So I honestly don't understand where you are coming from with the above assertions.

But I would be willing to find out.

--Regards,

#19 of 29 ChrisWiggles

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Posted August 18 2007 - 07:40 AM

Quote:
A vent, by definition, only works under narrow band frequencies. So I honestly don't understand where you are coming from with the above assertions.

Huh? The opening is there, always. It doesn't just magically go away at all other frequencies. The way it interacts with and whether it loads the driver is different depending with whether you're at, above, or below the port tune, but it's always there.

#20 of 29 Vaughan Odendaal

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Posted August 18 2007 - 07:48 AM

I suggest you read this article on ported operation.

http://www.hometheat....ayporting.html

What you describe is a misconception on how ported speakers work. Above tuning, the port is doing little. Very little. That is a matter of physics and it's undisputable.

The air mass in the port resonates with the springiness of the air in the enclosure (which is what a helmholtz resonator is). At this resonance the driver has to do very little work to excite the cabinet-port resonance, so its excursion drops to very little and hence so does its reactance.

Above tuning little happens because the driver is moving too slow to do anything to the air in the port.

Please read that article.

--Regards,


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