Qb and Vb in acoustic suspension systems

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The volume Vb of an acoustic suspension system in terms of its damping factor Qb is typically given by:

Vb = Vas / [(Qb^2/Qts^2) - 1]

Division by zero results in an undefined quantity, so [(Qb^2/Qts^2) - 1] cannot be zero. Thus (Qb^2/Qts^2) cannot be 1. In turn, Qb^2 cannot equal Qts^2, so Qb cannot equal Qts. Also, if Qb < Qts, Vb is negative, which cannot be so in reality.

For practical use we must restrict the above equation to values of Qb in which Qb > Qts. So we never see a driver with Qts equal to or greater than 0.707. In a sealed box, such a driver will never be able to attain a "flat" response!

For drivers with a very high Qts, say 0.68, we can never achieve a Qb of less than that value in a "real" enclosure. So how do we get slightly lean bass response (the result of a Qb less than 0.707)? Never!

Comments?
 
Pjotr said:
Your'e right Qb cannot be lower than Qt of the speaker. When you want Qb == Qt you'll need an infinite box volume. So use a very, very big box :clown: or better an open baffle.

The only other option is to use horn loading or a LW transform.

;)


In practice, that's all we can do. The volume provided by an open baffle is theoretically "infinite," but take into consideration the dimensions of the surrounding space (i.e. room) and it is no longer. Of course, one can always take his open-baffle speaker outside, where the volume of air is truly infinite. ;)
 
coolkhoa said:
The volume Vb of an acoustic suspension system in terms of its damping factor Qb is typically given by:

Vb = Vas / [(Qb^2/Qts^2) - 1]

Division by zero results in an undefined quantity, so [(Qb^2/Qts^2) - 1] cannot be zero. Thus (Qb^2/Qts^2) cannot be 1. In turn, Qb^2 cannot equal Qts^2, so Qb cannot equal Qts. Also, if Qb < Qts, Vb is negative, which cannot be so in reality.

For practical use we must restrict the above equation to values of Qb in which Qb > Qts. So we never see a driver with Qts equal to or greater than 0.707. In a sealed box, such a driver will never be able to attain a "flat" response!

For drivers with a very high Qts, say 0.68, we can never achieve a Qb of less than that value in a "real" enclosure. So how do we get slightly lean bass response (the result of a Qb less than 0.707)? Never!

Comments?

Qb is always greater than Qts.

Your other suppositions are entirely incorrect,
because for a start your equation is wrong :

Qb/Qts = root (Vas/Vb+1)

(Qb/Qts)^2 = Vas/Vb +1
Vas/Vb = (Qb/Qts)^2 - 1
Vb= Vas/ [(Qb/Qts)^2- 1]

the first equation tells you :
for Vb = Vas Qb = 1.414Qts
for Vb = 1/3 Vas Qb = 2Qts.

the other rearrangements tell you the same thing in reverse.

There are no restrictions on the value of Qts,
which in practise can vary between 0.1 and 2,
Though for bass units the range is usually 0.20
to impractically high values above 0.7 up to 1.
(unless open baffle or car free air mounting)

:) sreten.
 
diyAudio Moderator Emeritus
Joined 2001
Not sure about this discussion. What do you mean by Qb? Do you mean a speaker with Qts put into a closed box to yield Qtc?

If that is the case, I just thought I would post the formula for how we arrive at Qtc. Hope this helps.
 

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diyAudio Moderator Emeritus
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Re: Re: Qb and Vb in acoustic suspension systems

sreten said:



There are no restrictions on the value of Qts,
which in practise can vary between 0.1 and 2,

I wondered about that too.

I am guessing that when KBK said that, he was assuming that it was understood that we were not interested in speaker/box combos that did not yield absolutely flat response. That "acoustic suspension" system implied flat response.

My understanding is the "classic acoustic suspension " system is as follows:

Qts = 0.38
Vb = 1/3 Vas

Since that time, I believe "acoustic suspension" system has broadened a little to mean a sealed box where the box is appreciably smaller than Vas.
 
Re: Re: Qb and Vb in acoustic suspension systems

sreten said:


Vb= Vas/ [(Qb/Qts)^2- 1]

the first equation tells you :
for Vb = Vas Qb = 1.414Qts
for Vb = 1/3 Vas Qb = 2Qts.

Errh... How did you get from

Vb= Vas/ [(Qb/Qts)^2- 1]
to
Vb = Vas Qb = 1.414Qts ??

sreten said:

There are no restrictions on the value of Qts,
which in practise can vary between 0.1 and 2,
Though for bass units the range is usually 0.20
to impractically high values above 0.7 up to 1.
(unless open baffle or car free air mounting)

:) sreten.

I think coolkhoa assumed butterworth response and closed box. In that case, since Qb>Qts Qts cannot be larger than 0.7, unless Qts is altered with negative resistance or stuffing in the box.
 
Re: Re: Re: Qb and Vb in acoustic suspension systems

Svante said:


Errh... How did you get from

Vb= Vas/ [(Qb/Qts)^2- 1]
to
Vb = Vas Qb = 1.414Qts ??



I think coolkhoa assumed butterworth response and closed box. In that case, since Qb>Qts Qts cannot be larger than 0.7, unless Qts is altered with negative resistance or stuffing in the box.

For a Butterworth alignment Qb=0.707, end of story.
It is a maximally flat alignment with no peaking.

Other Q's give other alignments.

The first equation :

Qb/Qts=root(Vas/Vb-1)

If Vb=Vas, then Qb = root2 x Qts.

This is the equation repeated by KW
with the helful addition of :

Fb/Fs= root(Vas/Vb-1)

For sealed boxes alignments range from Qb = 0.5 to Qb =1.1,
though most are between 0.6 and 0.9, with the 0.7 Butterworth
being the most common for builders.

For a sealed Butterworth alignment any driver with Qts
below 0.7 can be used though Vb gets silly with Qts
near 0.7 and Fb gets silly with very low Qts values.

A Qts of 0.35 indicates for Qb = 0.7, Vb = Vas/3.
A Qts of 0.50 indicates for Qb = 0.7, Vb = Vas.


:) sreten.
 
coolkhoa,

I now see what you where describing, apologies.....

simply put :

if the design target is an inbox Qb of 0.7,
a sealed Butterworth alignment,
Qts must be less than 0.7.

A driver with Qts > 0.7 cannot be used by definition.

Drivers with Qts near 0.7 will give silly box volumes.

:) sreten.
 
Re: Re: Qb and Vb in acoustic suspension systems

Lower Qts by a negative series resistance ...

Yes, that is also an option. But this is an electronic option to lower Qe. Did some experiments with it and yes that works. But it has a big drawback. The voice coil will heat up, rising Qe again (because Re rises). Since you are working now with the difference of resistances, this difference can change severe and is quite audible. A proper tuned system becomes boomy after playing some time at high level because of the heating of the voice coil.

Using a LW transform works much better then or otherwise the way to go can be MFB. But that is not an easy way.

Cheers ;)
 
Re: Re: Qb and Vb in acoustic suspension systems

sreten said:


Qb is always greater than Qts.

Your other suppositions are entirely incorrect,
because for a start your equation is wrong :

Qb/Qts = root (Vas/Vb+1)

(Qb/Qts)^2 = Vas/Vb +1
Vas/Vb = (Qb/Qts)^2 - 1
Vb= Vas/ [(Qb/Qts)^2- 1]

the first equation tells you :
for Vb = Vas Qb = 1.414Qts
for Vb = 1/3 Vas Qb = 2Qts.

the other rearrangements tell you the same thing in reverse.

There are no restrictions on the value of Qts,
which in practise can vary between 0.1 and 2,
Though for bass units the range is usually 0.20
to impractically high values above 0.7 up to 1.
(unless open baffle or car free air mounting)

:) sreten.


Both of our equations are one and the same:

Vb = Vas / [Qb/Qts)^2 - 1]

expands to...

Vb = Vas / [(Qb^2/Qts^2) - 1]

For a flat response the exact value of Qb = (square root 2)/2, which is about 0.707, the commonly accepted value.

:)
 
Re: Re: Re: Qb and Vb in acoustic suspension systems

coolkhoa said:



Both of our equations are one and the same:

Vb = Vas / [Qb/Qts)^2 - 1]

expands to...

Vb = Vas / [(Qb^2/Qts^2) - 1]

For a flat response the exact value of Qb = (square root 2)/2, which is about 0.707, the commonly accepted value.

:)

apologies again...., they are equivalent. :) sreten.
 
diyAudio Moderator Emeritus
Joined 2001
Having heard of negative series resistance, but not knowing what it is, I did a quick look on Google.

After all, it has been mentioned as a way to lower Qts and, by extension, Qtc.

I found that while it is possible to do, if also makes an amp vulnerable to oscillation.

Since the whole idea of lowering Qts is to give a more controlled, snappier sound, it would seem to me that the solution is just transferring the problem from the speaker side to the amplifier side.

Am I right on this?
 
kelticwizard said:
Having heard of negative series resistance, but not knowing what it is, I did a quick look on Google.

After all, it has been mentioned as a way to lower Qts and, by extension, Qtc.

I found that while it is possible to do, if also makes an amp vulnerable to oscillation.

Since the whole idea of lowering Qts is to give a more controlled, snappier sound, it would seem to me that the solution is just transferring the problem from the speaker side to the amplifier side.

Am I right on this?

You are right in that a negative output impedance of an amplifier may lead to stability problems, but usually it does not go that far. Obviously, this series resistance may *never* be lower than Re, but that is never required. The main problem is that the resistance of the voice coil is temperature dependent, this leads to a Qes that varies with the temperature (= power) of the voice coil. This is a problem in any speaker, but is worstened by a negative series resistance.

Still, you have the box stuffing option, if Qts is 0.7+ ;)
 
coolkhoa said:
So we never see a driver with Qts equal to or greater than 0.707. In a sealed box, such a driver will never be able to attain a "flat" response!


Allow me to correct myself. Such drivers do exist, but they are not practical for hi-fi, namely low-fi Goldwoods. That's why Parts Express says that their T/S parameters "make them great for automotive environments." ;)

Consider that some of Goldwood's woofers have Qts of 0.81, 0.83, 0.88, and (get this) 1.69!

But the statement still applies: these drivers will never be able to attain an optimally flat response.
 
You can make an amp with negative output impedance and make a "maximally flat" system out of a speaker with Qts greater than 0.7. You can also make a linkwitz transform and change the overall Q and F3 to whatever you want....within reason.

"Optimally flat" has a different connotation to an engineer. Optimum is a relative term, and it depends on the metric used. 10 different people, when asked to come up with an optimum solution, will likely make 10 different choices. I have heard a system with Qtc=1.3 in a car, and it sounded perfectly fine, especially with the techno the guy liked to listen to. The metric in this case was to maximize trunk space....

Expand your horizons....
 
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