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Debunking the Myth that a Lighter Diaphragm Enables a "Faster" Speaker Driver

NTK

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"A faster driver needs a light cone. Or does it?" is the title of a blog post on the Purifi web site.

It is not a very long post, but if you are really busy, here is a TL;DR summary:
  • Displacement in speaker diaphragm does not generate the pressure wave in air that is the sound we hear. Acceleration does.
  • What causes the diaphragm to accelerate (and therefore the air in front of it to pressurize) is force. A lower acceleration only results in lesser sound pressure level (SPL), i.e. less loud, not how quickly it appears, i.e. the lack of speed.
  • The speed at which we can modulate "force" is unrelated to the mass of the diaphragm.
I'll expand a little further on Purifi's blog post, since someone will inevitability ignore the last point above and will insist that acceleration is force divided by mass, and therefore lower mass gives higher acceleration. So how do we find how much acceleration we need?

The late Siegfried Linkwitz (RIP) gave us a very handy formula to predict the free field SPL generated by a speaker driver, given its size, diaphragm travel, and frequency. [Link, see the box "Theory Behind the Nomographs"] It is:
SPL = 94.3 + 20 log10(x) + 40 log10(f) + 40 log10(d) - 20 log10(r) where: x is the peak-to-peak diaphragm travel in meters, f is frequency in Hz, d is the effective diameter of the diaphragm in meters (d = sqrt(4 * Sd / pi), with Sd = effective area in m^2) r is the listening distance in meters

Now, say we want to generate the same SPL at two different frequencies, f1 and f2, what will the diaphragm travels (x1 and x2) be?
SPL1 = 94.3 + 20 log10(x1) + 40 log10(f1) + 40 log10(d) - 20 log10(r)
and
SPL2 = 94.3 + 20 log10(x2) + 40 log10(f2) + 40 log10(d) - 20 log10(r)

Since we want SPL1 = SPL2 ,and "d" and "r" remain the same, we have:
20 log10(x1) + 40 log10(f1) = 20 log10(x2) + 40 log10(f2)
or
log10(x1) + 2 log10(f1) = log10(x2) + 2 log10(f2)
or
x1 * f1^2 = x2 * f2^2
or
x2 = x1 * f1^2/f2^2

So, the amount of travel the diaphragm needs to produce the same SPL in inversely proportional to the ratio of the frequencies squared (i.e. with the same diaphragm travel, SPL goes up/down by 12 dB/octave).

How about acceleration? Well, given the displacement amplitude x, acceleration = x * (2*pi*frequency)^2. Which means acceleration goes up by frequency squared. Since, for the same SPL:
x2 = x1 * f1^2 / f2^2
and
a2 = x2 * (2*pi * f2)^2 = x1 * (f1^2 / f2^2) * (2*pi * f2)^2 = x1 * (2*pi * f1)^2 = a1

The acceleration amplitudes are the same! And therefore forces. Amazingly we need the same force amplitude to produce the same SPL regardless of frequency. Of course, the rate of fluctuation of force is higher with higher frequencies, but the force magnitude is independent of frequency. We need to wiggle the diaphragm more frequently, but that is completely countered by the fact that we need to wiggle it less far.

There are plenty of other reasons why a woofer is not suitable to produce treble. Mass of the diaphragm ain't one.
 

Jim Shaw

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There are plenty of other reasons why a woofer is not suitable to produce treble. Mass of the diaphragm ain't one.
If in reviewing classical physics and kinetics, we glance at the formulas for force and momentum, both show a direct connection to mass. F=m*a and m*v^2. Speaker cones/voice coils/suspensions are not without mass.

F=ma should tell us that moving a CVS (cone/voice coil/suspension takes force. As the mass of the system increases, more force is needed to make the move. Momentum specifies that a CVS in motion will continue in motion unless a force is applied to stop it. Again, mass raises its proportional head. If F=mv^2, again mass appears. Audiophiles greatly simplify describing that chore with the amplifier "damping factor" (a simplified measure of the effective driving point impedance of the amplifier output circuits compared to the load impedance at that dynamic).

It is unclear to me how these physical issues directly involving mass can be ignored. Kindly illuminate.
 
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NTK

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If the travel of the diaphragm is kept the same, raising the frequency will result in higher acceleration and therefore need higher force. In fact, the acceleration and therefore force required will go up proportional to the square of frequency.

However, as it is shown in the formula Linkwitz gave, to produce the same SPL, the travel required by the diaphragm goes down proportional to the square of frequency. As acceleration is the second time derivative of displacement, the magnitude of acceleration increases/decreases linearly with the magnitude of displacement. Therefore, the effects cancel each other, and the magnitude of acceleration required to produce the same SPL is independent of frequency. Since acceleration magnitude is the same and mass is the same, thus force required from the motor is the same.

Of course, when the mass of the system is different, to get the same system responses (i.e. FR, step/impulse responses, etc.), you'll need to adjust the stiffness, damping, and driving force to compensate. What post #1 is about is that you don't need generate a higher magnitude force to produce the same SPL at a higher frequency than at a lower frequency. Therefore, the bandwidth of the system (which is really what gives it its speed) is largely independent of the mass of the diaphragm (or, more precisely, the moving mass).
 
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dfuller

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You don't, but for a higher Mms you need more motor power for the same control over its movement. So when you see companies using these gargantuan ceramic magnet motors, it's because they're compensating for the increased mass of the voice coil and diaphragm. That, I think, is the thing that is overlooked here.

Actually this is a case where I can apply some guitar-side knowledge. The Celestion G12M-25 (the "Greenback") and the G12H-30 use the same cone and very similar voice coils (quite possibly the same, actually... I can't find anything different between them that would be different). The only difference is the motor magnet, which is significantly larger on the H30 (50oz) than the M25 (35oz). The H30 is quite a bit brighter, extends lower even with the same Fs (tl;dr there are both 55hz and 75hz versions), and much (+3-4dB) more sensitive. The motor is better able to control the moving mass.
 

thorvat

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Of course, when the mass of the system is different, to get the same system responses (i.e. FR, step/impulse responses, etc.), you'll need to adjust the stiffness, damping, and driving force to compensate. What post #1 is about is that you don't need generate a higher magnitude force to produce the same SPL at a higher frequency than at a lower frequency. Therefore, the bandwidth of the system (which is really what gives it its speed) is largely independent of the mass of the diaphragm (or, more precisely, the moving mass).

While post #1 is correct your last statement (in bold) is false and is contradictory to the previous statement (in bold), which is correct.
Bandwidth of the system is indeed a function of the driving force, membrane mass, surface area, stiffness and damping, and as such is dependent on all of these factors.
 
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NTK

NTK

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Post #1 really was a trick post. It doesn't make sense. How can the force required to move the driver diaphragm be the same for the same SPL at all frequencies? Why then do sub drivers have massive magnets, if they don't need to generate massive amounts of force?

This question is in itself illuminating. Why do sub-woofer drivers need massive magnets and tweeters don't? Most subs are not to produce sound above ~80 Hz anyway, otherwise they may betray their presence. They hardly need to be "fast", unlike tweeters. Why don't tweeters have huge motors to give them the huge amount of force so they can accelerate their diaphragms really really fast?

So what was amiss? Well, the calculations in post 1 only considered the amount of force to move the diaphragm, as if the speaker driver was operating inside a vacuum chamber. Air loading was ignored. An exercise for those who are interested: If you have a sealed sub, how much pressure loading the diaphragm will have to support when you slam it from +xmax to -xmax? The motor will have to supply the force to do that. Also, add acoustic loading on top of that. These loads are displacement, and therefore frequency, dependent (for the same SPL), and they increase when frequency goes down!

These pressure and acoustic loads are independent of the mass of the diaphragm. It doesn't matter if the diaphragm is massless, they are not affected. And that's why sub drivers have huge motors, and tweeters don't, while being able to output as high or higher SPL.
 

sq225917

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Bunkum. To accelerate a higher mass you need greater force, end of. Assuming the maximum force and rate of acceleration is limited a lighter driver on the same motor will achieve a greater acceleration.it's power vs torque.

Excursion demands diminishing with increasing frequency for the same spl doesn't come into it, higher mass is higher mass and less mass equals increased acceleration. A lighter driver can be moved faster with the same force. It you don't believe it try walking up the stairs with 200lbs in your back.
 
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Destination: Moon

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Much of the OP is beyond my full understanding. That said, much of the discussion only considers acceleration as if a voice coil only moved in one direction. That there is both acceleration and deceleration in 2 opposite directions seems like it makes mass doubly important in determining the forces needed to produce sound energy
 

dshreter

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There’s a nuance to this which makes it counterintuitive and slightly misleading. A heavy cone indeed would generate lower SPL. But assuming the right stiffness and linearity, it can still deliver accurate frequency response.

A good example of this is when you apply a tuning fork to a piece of furniture or whatever. It can still resonate due to the vibration input even though it is very heavy.
 
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thorvat

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These pressure and acoustic loads are independent of the mass of the diaphragm. It doesn't matter if the diaphragm is massless, they are not affected. And that's why sub drivers have huge motors, and tweeters don't, while being able to output as high or higher SPL.

Eh, that is only one factor.

The reason why sub driver has larger motor than tweeter is because sub motor has to do more work to produce same SPL, and the motor is the one doing that work. It has to move larger mass, it has to move it further away (to produce same SPL as tweeter) and it has to fight higher resistance (airflow resistance and dumping) while moving it.

You seem to be missing the fact that, although woofer and tweeter are producing same SPL, woofer motor is doing much more WORK and thus is asking much more power from the amp to do it.

Making large membrane of the woofer to produce high freequencies of the tweeter would be technologycally extremely challenging and impractical price-wise for many reasons (most of them we didn't even touch, like cone break-up for example) and that is the reason why we don't have well performing loudspeakers with a single driver.
 
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pozz

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Guys, a speaker is a transducer. The driver and air form a single system, the mechanical parts of the driver being intermediaries between the supplied electrical variation and output acoustic variation. This is why suspension compliance and all the talk of burn in is not an audible factor. The work involved is in what it takes to move the air, not what it takes to move the driver independently. And a huge motor, mainly, allows for current control.

@NTK showed that focusing on the mass of the diaphragm is lay intuition. This holds not just for subs, but for the ongoing thread about electrostatic speakers, too. No such thing as fast sound.

You don't have to take my word for it, either.
1629450631834.png

For context, the blue curve represents a roughly doubled diaphragm mass for 6.5" driver. Despite that, the transient response barely changes. The added inductance produces a greater comparative effect as seen in the green curve.

Here's the frequency response:
1629451191493.png

Around 1kHz the response is the same for all curves. The added inductance causes the high frequencies to dip earlier, while the added mass changes the damping and resonant modes of the driver, affecting not HF but LF efficiency. Cool, right? At higher frequencies the driver diaphragm does move faster. But intuition about weight would say it's harder to repetitively cycle a heavier object at a higher rate than a lower one.

Really, I'm just paraphrasing what @NTK wrote here:
Of course, when the mass of the system is different, to get the same system responses (i.e. FR, step/impulse responses, etc.), you'll need to adjust the stiffness, damping, and driving force to compensate. What post #1 is about is that you don't need generate a higher magnitude force to produce the same SPL at a higher frequency than at a lower frequency. Therefore, the bandwidth of the system (which is really what gives it its speed) is largely independent of the mass of the diaphragm (or, more precisely, the moving mass).
 

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AnalogSteph

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So what was amiss? Well, the calculations in post 1 only considered the amount of force to move the diaphragm, as if the speaker driver was operating inside a vacuum chamber. Air loading was ignored.
Typical cone drivers are actually quite inefficient. They would move much the same in a vacuum as in free air - note how Qts calculation involves zero parameters relating to air, for example. Almost all input power ends up simply heating the voice coil. This is also why acoustic resonances only tend to reflect back on impedance response as tiny blips. You have to go horn loading for acoustic effects to become relevant, at which point you'll see a correspondingly wiggly impedance response as well.
 

thorvat

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Around 1kHz the response is the same for all curves. The added inductance causes the high frequencies to dip earlier, while the added mass changes the damping and resonant modes of the driver, affecting not HF but LF efficiency. Cool, right? At higher frequencies the driver diaphragm does move faster. But intuition about weight would say it's harder to repetitively cycle a heavier object at a higher rate than a lower one.

It is hardly surprising that adding weight affects LF response and not response at 1kHz as moving mass is what matters with LF. If you wanted to affect response of that same driver at high frequencies you should leave the mass as it is and enlarge or reduce the diameter of the driver as air resistance is what matters more with higher frequencies as speed of the membrane increases and amplitude gets lower for the same SPL.

It's not about intuition - there are a number of reasons all related to phyisics why subs have large diameter drivers and tweeter small. Both of them should have the least possible mass and the maximum possible stiffnes.
 
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AnalogSteph

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We know that SPL ~ a, and F = m*a, so SPL ~ F/m.

So who's the culprit limiting HF response then?

It pretty much has to be the combination of voice coil inductance L and its parasitic parakeet parallel capacitance C forming a 2nd-order lowpass. Voice coils are often wound in multiple layers to get the required number of turns, at which point parasitic capacitance goes up dramatically. (If you've ever dabbled in loop antennas, you'll have found that widest tunable bandwidth is achieved when using a large single turn affair. The more turns and the close the spacing the worse things get. Nobody even uses multiple layers in this realm, but it's not uncommon in audio transformers.)

So if you insist on a single layer, it follows that you have to minimize moving mass to still get a decent amount of efficiency out of the whole mess. Plan B is increasing voice coil diameter, up to the point where it is actually on the outer perimeter. I think this is how all these little 32 and 37 mm wideband drivers are built (as are dome tweeters, of course) - a magnet system this large still is realistic.

Purifi used another approach, attacking L with a shorting ring instead. After all, all we need is turns (the L in BL is wire length), not inductance per se.
 
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thorvat

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We know that SPL ~ a, and F = m*a, so SPL ~ F/m.

So who's the culprit limiting HF response then?

It pretty much has to be the combination of voice coil inductance L and its parasitic parakeet parallel capacitance C forming a 2nd-order lowpass.

Mass and surface area of the membrane are major limiting factor for the HF response of the driver. Even if you would have incredibly large magnet that would require only 1 coil on the driver it still wouldn't help much as those 2 factors would still be the major limitation.
 

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I am not sure what the "speed" term means here, does it mean the velocity of diaphragm or how quick the diaphragm react to force? Those two are different.

For velocity of diaphragm, increase moving mass velocity will get lower. F=ma still applies here. Higher mass, lower acceleration, lower average velocity of the diaphragm. Due to lower acceleration, the sensitivity will be lower too.

For how quick the diaphragm react to force, different moving mass will have the same "quickness". Whenever there is force acting on mass, there is acceleration, force and acceleration happen at the same time. If both heavier stuff and lighter stuff experience force at the same time, they will move at the same time, the difference is only velocities of them moving. To put it in another word, the phase of heavier moving mass is the same as lighter moving mass. The heavier stuff does not lag in phase.

So the conclusion is heavier moving mass will have lower velocity resulting in less sensitivity, but same "quickness" reacting to force and same phase because acceleration and force happen simultaneously.
 

bigjacko

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It is a very nice analysis of the relationship between excursion, frequency and acceleration, but is it related to the discussion of relationship between speed and moving mass?
Now, say we want to generate the same SPL at two different frequencies, f1 and f2, what will the diaphragm travels (x1 and x2) be?
In this sentence I assume you are saying using a single driver to play those two frequencies, so in the end you are just comparing the relationship between excursion, frequency and acceleration, rather than two drivers with different moving mass and what will happen.
 

thorvat

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I am not sure what the "speed" term means here, does it mean the velocity of diaphragm or how quick the diaphragm react to force?

"Speed" is in this discussion related to bandwidth and that is why max effective frequency that driver can produce is being discussed here.
 

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There is not such a thing as a "faster" driver, and there never will be.

Everyone with an engineering background knows control theory and mass-spring-damper constructions.
(or they should it's part of any curriculum starting mostly from the 2nd year or so)
More mass simply means that something takes more force to accelerate at the same amount.
Or in other words, with the same force, things will accelerate less.
Things resist when they are heavy (you know, Newton ;) )

For loudspeakers and 2nd (and higher) order mass-spring-damper systems, this results in that the frequency response collapses.

In practice the inductance is more dominant with most subwoofers, so this is what most people see.
Unless a proper amount of demodulation rings is being used.

On the lower frequencies below Fs, mass is a constant and the compliance is the dominant factor.

There is something else called a Q-factor, which doesn't have much to the with the mass perse.
Depending if it is under-damped, criticality damped or overly-damped, this automatically translates to an hump in the frequency response.
Or like max flat frequency response etc
 
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abdo123

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I think the best example I can think of when it comes to ‘fast bass’ is ported vs sealed speakers/subwoofers.

sure the energy is the same, but with ports the energy is delayed and it deviates from the ideal ‘minimum phase’ system.

discussing the amount of energy is pointless, but discussing the energy delay is relavent and audible.
 
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