QTS/OB explanation needed

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I'd like to use the Eminence Alpha15A in pairs to go with my AER MD2Bs that will soon be arriving. I must confess I don't understand what 'Q' is really and why a high 'Q' driver will give me more bass in OB. I've found that generally it's the woofers with small magnets that have a high Q. Does this mean that the small motor exerts less control over the moving mass, allowing it to flop around more or something? Is a high Q driver merely a high-distortion driver? I would think a driver with a massive motor applied to the same mass would give far better transients, no? What am I really giving up by using a high Q driver?

Also, the AERMD2B has an insanely strong motor and low moving mass yet has a high .7 Q. I also don't understand this.
 
You can either choose between a high Q driver or eq the bass up, or both....

A high Q (meaning less mechanical and electrical dampening) driver will (for what I know about it) distort more in an enclosure. But as any driver will have much less distortion in an open baffle it will mostly result in a system well within distortion limits....

Qts is electrical Q (Qes) and mechanical Q (Qms) together and therefore looking only at the motor of the driver won't tell everything about the Qts...

I hope this helps...:)
 
Target Qtc:
Best transient response Q = .5
Critically Damped Q= .7
Over-damped Q < .5 Base will roll off before F3
Under-damped Q > .7

Since Low frequency driver design balances efficiency, low F3 and Qts, most drivers choose to allow the enclosure to raise Qtc, usually by base reflex loading.

Basically, smaller magnets required to raise Qts result in either lower efficiency, higher F3, or both. it does not directly affect distortion, as a first order effect.

Open baffle speakers Qts = Qtc, so the choices are either choose a driver in the .5 to .7 range and accept the penalties, or use equalization to restore the frequency response lost when a driver of less than Qts of .5 is used. The equalization has its own penalties.

HTH.

Doug
 
Definition and Implications of Q

FWIW, I'm fairly novice, but will try to distill what I think I've learned into something useful. :D

You are correct that q goes down with larger motor, lighter cone, and greater control over the speaker mass. Q is the ratio of stored energy to damping, and the magnet contributes to damping, so bigger magnet means higher damping and lower q. But no, higher q does not imply greater distortion, just a different tuning of the resonant system.

Oh, and a small sealed box adds resonance and increases q. As a sealed box get larger, it's resonance becomes less, its damping greater. and the system's q greater. An infinitely large box would add nothing to Qts, and act just like an infinite baffle! Isn't life grand?

There are two crucial consequences of Q (no extra charge for the alliteration), and Qts in indeed the useful measure in an open baffel system.

The consequences are frequency response, and stored energy--the amount of resonance.

To expand on Doug's useful distinctions:

Over-damped: Q < .7, Bass will slowly roll off before F3, and continue to slowly roll off below F3

Critically Damped: Q = .7, flattest response down to F3, then moderate rolloff

Under-damped: Q > .7, Bass begins to *rise* before F3, peaks at F3, then rolls off quickly.

The frequency-response upshot is underdamped systems have a peak, a boom, at F3, and little bass below. Critically damped systems have flat bass down to F3, with moderate rolloff below. Overdamped systems start to roll off before F3, but do so very slowly, so they will actually have more bass at frequencies significantly below F3.

The other factor is resonance, which, as it increases, tends to store energy and bleed if off gradually, smearing transient response. This effect leads to the "one note bass" complaint about tuned systems--whenever there is a note anywhere near resonance, the system hums for a while at its resonant frequency. And this "hum" smears transients. Think of the resonant systems you know--a tuning fork or a wine glass--they would make lousy speakers. Efficient, but lousy.

One of the things people like about open baffle bass is this lack of enclosure-induced resonance and overhang, and the corresponding ability to follow the bass clearly as it changes frequency. A resonant driver would degrade that.

Some people think that resonance affects the perception of "fast" or "slow" bass. Fast bass being damped and dying out quickly, and slow bass bring resonant with a lot of overhang.

So underdamped systems produce more bass with less power, but roll off quickly and have a resonant peak at F3. Hum hum hum--and equalization can flatten the total amount of energy, but can't stop the overhang/humming.

( servo-control or dynamic computer-controlled equalization might)

Critically damped systems have flat frequency response, store a moderate amount of energy, and roll off moderately.

Overdamped systems have bass that dips early but ultimately goes lower, and hold little energy. For this reason, some people like to pick overdamped bass drivers and then use equalization to give themselves flat bass that goes deeper, with less smearing (faster). And this requires more power from the amp.

All of these are fairly new thoughts to me, and some may be wrong. I welcome all comments and corrections, and hope this is useful.

Best,

George
 
This was posted in another thread recently and I found it interesting.

http://www.geocities.com/kreskovs/Box-Q.html

If you don't have time to read it I guess the short story is Q= 0.7 probably has the flattest frequency response in an anechoic chamber, not your room. Unless your room is a giant theater. I'd also note that a Q of 1.0 doesn't mean the driver rings on and on after a note ends, it's more like a muted half cycle bump, so at 40hz it moves for something like 1/100th of a second longer than it should.

With OB your baffle size is causing a bass roll off so a well placed bump in the bass response could counteract some of that.
 
If you don't have time to read it I guess the short story is Q= 0.7 probably has the flattest frequency response in an anechoic chamber, not your room. Unless your room is a giant theater. I'd also note that a Q of 1.0 doesn't mean the driver rings on and on after a note ends, it's more like a muted half cycle bump, so at 40hz it moves for something like 1/100th of a second longer than it should.

With OB your baffle size is causing a bass roll off so a well placed bump in the bass response could counteract some of that.

Poptart,

EXACTLY! I use the Eminence Alpha A15 with my Lowthers in an OB and there is no bloated ringing bass at all. I used a lower Qts driver and some EQ boost before getting the Alphas's and found the Alpha's to be a better solution for me in my room. The bass is tight and reaches 40 Hz easily, great for acoustic jazz. No boom in my room.
 
There is no one answer to your question. The bass response is detemined by the size of the baffle and the Qts of the driver. A higher Qts driver will allow a little bit smaller baffle for the same low frequency roll-off. I would look for a driver with a Qts between 0.7 and 1.2 and then design the baffle with the driver or drivers to see what bass you can expect for particular baffle sizes. It is a system with a number of trade-offs. If you go to my MathCad models page, there are two pdf files showing two different approaches to OB design with full range drivers and the Eminence Alpha A15 woofers.

Hope that helps,
 
I have not heard bass below 70hz in an open baffle only because the one time I tried it was with the only bass drivers I had sitting around which were the JBL 2020h. These seemed to be far less than ideal for bass given that I didn't hear anything lower than about 70hz but the midbass was remarkable, if terribly meagre in output. That midbass was so clean and tight I had the impression that nothing had been added to the source and nothing had been taken away - I was thinking 'is this what bass really is?' So now I'm headlong chasing that sound. I'm more than curious what 40hz open baffle bass could sound like...
 
Hi Folks,

Thanks for the , and compliment, corrections, and clarifications. For some reason I can't seem to edit my previous post (again ;), perhaps the Statue of Limitations on stupidity has expired. :dead: )--if i can, I'll update it.

For now, let me just add these corrections:

1. My previous post's paragraph-three-sentence-one should say that a sealed-box system's q gets *smaller* as increasing box volume increases damping. I keep forgetting that *greater* damping means *smaller* q. D'oh!

2. Yes, the q-related frequency response curves I mentioned were for an ideal room, and virtually no-one's room is ideal. That was an oversight on my part, and an ironic one, for I believe that room effects/treatment/compensation is the aspect most overlooked by audiophiles. My bad.

3. And yes, because I ignored the room's effect on frequency response, I mistakenly correlated speaker q with the overall amount of overhang/ringing, when the room plays a big part. Is it correct to say that the *overall* resonances of the speaker-room system--the interaction of speaker resonance(s) with room resonance(s)--that determine overhang and blurring?

My understanding is that room resonances even vary by position within the room, and that most rooms really wreak havoc on speaker response from 200Hz down.

On a positive note, some say that open baffle bass is less likely to get the room ringing, because it radiates most to the front and back, and less to the sides, so it is less likely to excite sideways room resonances than a sealed box's omnidirectional pattern.

Again, I'm still learning. I hope this helps others--it educates me to write it. Apologies in advance for any mistakes, and thanks for any corrections.

Best,

George
 
1. My previous post's paragraph-three-sentence-one should say that a sealed-box system's q gets *smaller* as increasing box volume increases damping. I keep forgetting that *greater* damping means *smaller* q. D'oh!

Actually, the increased sealed box size reduces the spring supplied by the box air volume hence reduces the fc of the system. The impact on system damping is minimal.
 
I've been able to get solid OB bass down to ~50Hz with an 8" driver (Pioneer B20) and some EQ. Small baffle too (16"x20"). Can get closer to 40Hz if I overdo the EQ (trust me, you know when you've overdone it ;-p ). I think that flat-ish to 40Hz on OB shouldn't be a problem if you're willing to use helper bass and large-ish baffles/some EQ. Just look at JAMO's big OB unit. Its basically just a pair of 15"ers, a wide band and a tweeter in a baffle that isn't much more than what it takes to hold those woofers (but then they're also charging what, $10-$15K for a pair; sheesh! this is one of the simpliest DIY clone projects ever which we here on the fullrange forum would 1-up by using a fullranger uptop).

Anyway, yeah, you'll not want to go back to box speakers once you've heard this. I'm having that issue right now, albiet nearfield in my office. Far field in my living room isn't quite the same thing, but then I've not ever set up OB that would be adequate for that listening distance or room size in there before. That would be a bad move WAF-wise and also my sanity wise (pretty sure I wouldn't want to go back my current, poorly placed sealed units ;-p)

Kensai
 
MJK said:


Actually, the increased sealed box size reduces the spring supplied by the box air volume hence reduces the fc of the system. The impact on system damping is minimal.

But q goes down as a sealed box gets larger, right? The design calculators I'm using (like http://www.mhsoft.nl/spk_calc.asp ) seem to show lower q with a larger sealed box.

So is the smaller q value of a larger sealed box due to the decreased energy storage of the weaker spring, rather than increased damping?



Thanks!

George
 
OK, if we start with a driver that has a Vas, Qts, and fs and put it into a closed box of volume Vb

alpha = Vas / Vb

For the closed box system

Qtc = Qts (1 + alpha)^2

and

fc = fs (Qtc/Qts)

As the volume of the box increases, alpha decreases so that Qtc and fc also drop. As the volume of the box decreases, alpha increases so that Qtc and fc also rise. The volume of the box, the air spring acting on the back of the driver's moving cone, is what sets the Qtc and fc values. There is no change to the physical damping property in the driver's suspension.

Hope that helps,
 
MJK said:
OK, if we start with a driver that has a Vas, Qts, and fs and put it into a closed box of volume Vb

alpha = Vas / Vb

For the closed box system

Qtc = Qts (1 + alpha)^2

and

fc = fs (Qtc/Qts)

As the volume of the box increases, alpha decreases so that Qtc and fc also drop. As the volume of the box decreases, alpha increases so that Qtc and fc also rise. The volume of the box, the air spring acting on the back of the driver's moving cone, is what sets the Qtc and fc values. There is no change to the physical damping property in the driver's suspension.

Hope that helps,

Very much, thanks!

I realized that a sealed enclosure does not affect the [damping in the] driver's suspension, but does it provide any damping of it's own, beyond that of free air itself? From your equations, the answer seems to be "no".

If so, is it then correct to say that a box increases q by storing energy, in its air spring? (Another way of asking my original question.)

Your equation matches my experience with simulations: as Vb goes to infinity, Qtc --> Qts and an infinite volume box has the same effect as an infinite baffle, correct?

That would be why people generally look for higher Qts in speakers for open baffle use, since a baffle will not raise Qtc the way an enclosure will?

And how do damping materials affect Qts? By increasing damping, reducing the spring effect, or both?

(You can stop replying anytime you get sick of this. ;) )

Thanks again,

George
 
I realized that a sealed enclosure does not affect the [damping in the] driver's suspension, but does it provide any damping of it's own, beyond that of free air itself? From your equations, the answer seems to be "no".

If so, is it then correct to say that a box increases q by storing energy, in its air spring? (Another way of asking my original question.)

Yes, I think that is a good way of looking at the problem.

Your equation matches my experience with simulations: as Vb goes to infinity, Qtc --> Qts and an infinite volume box has the same effect as an infinite baffle, correct?

That would be why people generally look for higher Qts in speakers for open baffle use, since a baffle will not raise Qtc the way an enclosure will?

I have used drivers with similar fs but different Qts values, ~0.4 and ~1.2, in my open baffle speakers and found the higher Qts worked better. However with an active crossover and the capability to boost the inputs to a woofer a lot of different Qts values can be made to work.

And how do damping materials affect Qts? By increasing damping, reducing the spring effect, or both?

There are two sources of energy dissipation in a driver, the voice coil resistance Re and the suspension's internal mechanical damping Rms.

Qms = 2 pi fs Cmes Res where Res = (BL)^2 / Rms

Qms = 2 pi fs Cmes (BL)^2 / Rms

Qes = 2 pi fs Cmes Re

The suspension's damping is not very great hence high Qms values for most drivers. The voice coil resistance Re determines Qes which is really the dominant source of damping. I don't know of an easy way to increase damping in a driver hence lowering Qts without a major redesign. It would be hard to reduce Re and increase Rms without a significant alterations. You might want ot look at a different driver. Are you trying to lower the Qts of some specific driver? That may prove to be difficult.

(You can stop replying anytime you get sick of this. )

Not yet. :)

Hope that helps,
 
MJK said:


George asked: "And how do damping materials affect Qts? By increasing damping, reducing the spring effect, or both?"


Martin said: There are two sources of energy dissipation in a driver, [etc.]


Thanks again Martin--very helpful. Actually, I was unclear--I meant damping materials in the *enclosure*--fiber filling, etc. Do those affect Qts, and do they do it by increasing damping, reducing spring effect, both, or something else?

Thanks again for educating me.

Best,

George
 
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