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

[ferrellms]

"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.

Wow, interesting stuff and some mechanical engineering that is beyond me. But getting back to speaker drivers and how they affect sound...

How can "speed" be relevant if a driver can reproduce all the frequencies it needs to at the volume needed? Intuitively, you'd think that a lighter tweeter could follow the sound of a cymbal better than one heavier. But if the measurements don't show a different response (in the audible frequency range, of course) there will be no difference.

"Speed" is audiofoolery as far as sound goes if the drivers are appropriate for their required frequency response.

 

[sq225917]

So, heavier equals slower, equals quieter, independent of frequency because acceleration equals spl, not frequency. Check, idiots guide to fast sounding drivers myth filed in memory banks.

 

[Jim Shaw]

"Speed" is audiofoolery as far as sound goes if the drivers are appropriate for their required frequency response.

We can also say (with joking impunity) that sound power levels have nothing to do with power levels... IF they are appropriate for the power levels we're listening to...

Huh? Whadidja just say? F=m*a doesn't matter if you don't care about F?
It only takes a little 9th-grade middle school calculus to 'get' the relationship between displacement (x), velocity (dx/dt), acceleration (dv/dt), and the force required (F=m*a), and the integrals over time to get right back to a, v, and x as a result of force.

Mass doesn't go away. If it did, our world would be a much different place.

 

[tuga]

"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.

I would be interested in seeing comparative measurements at low levels, though not sure which measurements would reveal significative/meaningful differences.

 

[sq225917]

All that said, you put a lead dome on a tweeter it ain't reaching 20k

 

[NTK]

Acceleration = Force/Mass

Same force, high mass → low acceleration = low SPL
Same force, low mass → high acceleration = high SPL

The relationship (for first order effects) is independent of frequency.

 

[dshreter]

All that said, you put a lead dome on a tweeter it ain't reaching 20k

It will, just VERY VERY quietly

 

[alex-z]

All that said, you put a lead dome on a tweeter it ain't reaching 20k

That depends on how much motor strength you are willing to throw at the problem.

 

[Soria Moria]

It's just an alternative construction at hand which tends to have good high frequency properties.

Now that this thread was revived can I ask you what you mean by high frequency properties?

 

[sq225917]

That depends on how much motor strength you are willing to throw at the problem.

Indeed, it's a entirely counter intuitive thing, where it either moves or cannot and if it can then it can move at frequency but perhaps almost infinitely little....

 

[briskly]

Now that this thread was revived can I ask you what you mean by high frequency properties?

Extended high-frequency response on-axis and minimization of resonances well beyond the audio range. These have to do with the shape of the radiator and how the motor transfers force to the radiator.
The first is inherent to flat transducers. If you break the membrane apart into microscopic elements, the radiated pressures sum constructively on the central axis. The symmetry of the system breaks apart off-axis. As high-frequency wavelengths approach emitter size, path length differences over the membrane induce destructive interference.
The second comes about by motor coupling to the modes of the membrane. A standing wave has positive and negative components, and we can obtain a more even frequency response by driving the membrane evenly so that the sum cancels out. Similar techniques are applied in multi-subwoofer setups to reduce the coupling to room modes.
The reduced inertial forces from the low mass of the membrane also lend themselves to some degree of air damping of the remaining resonances.

 

[Newman]

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

Huh? Whadidja just say? F=m*a doesn't matter if you don't care about F? It only takes a little 9th-grade middle school calculus to 'get' the relationship between displacement (x), velocity (dx/dt), acceleration (dv/dt), and the force required (F=m*a), and the integrals over time to get right back to a, v, and x as a result of force. Mass doesn't go away. If it did, our world would be a much different place.

I gather from your objections ever since post #2 in this thread, that you think that reducing the mass of a diaphragm confers some benefit in the sound of a driver, other than a tiny efficiency gain? Kindly illuminate.

Look, let me demonstrate what Purifi and NTK are saying, in another way.

Here is the Purifi illustration ( © Purifi, fair use claimed):-

The key point is that the sound pressure -- the bit we hear -- tracks the acceleration exactly, even with a perfectly instantaneous pulse. It doesn't look like the velocity, it doesn't look like the displacement.

Notice especially how quickly acceleration and sound pressure responded to the instantaneous signal: they responded instantly, and to their full value.

The velocity takes time to respond to the impulse, the displacement takes time to respond, but the sound pressure doesn't.

So let's look only at acceleration and sound pressure, when we add mass to the diaphragm:-

And notice what DID NOT HAPPEN:-

Perhaps this makes it a little clearer to the reader why adding mass doesn't change the speed and responsiveness of the sound pressure.

cheers

(PS apologies for terrible red markup by hand, my writing hand's wrist is broken and in a cast since Monday)

 

[restorer-john]

Here is the Purifi illustration ( © Purifi, fair use claimed):-

How ironic Purifi choose to use square waves when their amplifiers produce such ugly ones...

Anyway, we'll see how rapidly they change their tune when they start attempting to produce treble units with a modicum of efficiency, unlike their bass driver efforts.