Design tradeoffs

2076 Views 10 Replies 5 Participants Last post by looneybomber, Nov 3, 2009

fredk

Discussion Starter · Oct 18, 2009

Posts in another thread got me to thinking about design tradeoffs to achieve an objective.

Out of curiosity, I modeled two 15" CSS SDX15 vs 1. The gain is reduced excursion (and thus distortion) and no need for something like the Rekhorn1 for a highpass filter. It also reduces the needed amp power for a given spl level. The cost is slightly more money and a lot more space.

I am curious about the reduction in power. The dual driver model requires half the power to deliver the max spl of a single driver setup. I know that you gain system efficiency as you increase box volume. I also remember reading that you gain ~6db from a second sub if co-located [driver coupling] vs 3db if not co-located.

The dual driver model gains 6db for a signle box of twice the volume. I am wondering the following:

Will co-locating the drivers in a single box give you even more gain, or will you lose some of that 6db if drivers are spaced far apart? You could build a 7' box with a driver top and bottom vs two drivers at the bottom.

Here is the excursion difference at equal volume 1,000w single vs 500w dual:

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looneybomber

· #2 · Oct 18, 2009

Here's how it works,
+3db efficiency due to doubling the drivers (2x cone, motors, and enclosure volume)
+3db sensitivity due to cutting resistance in half, meaning you get 2x more power (so 500w to 1000w)
Together, because you have 2x drivers and 2x power, you get +6db.

Co-located gives you the full SPL benefit of the 2nd driver...but it doesn't do anything for smoothing out the in-room frequency response.

Having them scattered around usually does not give you the full 6db gain because Sub B is helping to boost nulls created from Sub A and sub B isn't increasing peaks created by Sub A.

Think of it as having two engineers in the same room; You can get twice the designing done (assuming they stop arguing over who's right). Then put an engineer and an artist in the same room; You only get the designing power of the one engineer, but the artist helps smooth things and make things pretty. Wow, that's actually not a very good analogy...

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fredk

Discussion Starter · #3 · Oct 18, 2009

+3db sensitivity due to cutting resistance in half, meaning you get 2x more power (so 500w to 1000w)

OK, I live in the land of the electronically challenged. I'm guessing this is the same thing as, or closely related to, impedance? Hmm... So, if it was a dual voice coil driver and I wired it such that the impedance remained the same as for a single driver I would lose the extra power handling and this 3db??

The interesting part of modeling two drivers vs one was that I could get the max spl of the one driver configuration on half the power. This left me wondering if this had something to do with box size increasing, or if it was just due to the increase in cone size.

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DrWho

· #4 · Oct 18, 2009

fredk said:
The interesting part of modeling two drivers vs one was that I could get the max spl of the one driver configuration on half the power. This left me wondering if this had something to do with box size increasing, or if it was just due to the increase in cone size.

I would suggest ignoring the +3dB that comes from halving the impedance that the amplifier sees, but rather think of the amount of power going into each driver separately...and really it's best to think of the voltage being delivered and not the power (since a speaker impedance isn't flat). The amplifier is always putting out the same voltage, but the amount of current that flows is dependant on how the speakers are wired. In parallel, there are two equal paths so you get double the current (and thus double the power). When in series, you get half the current and each driver sees half the voltage, so you end up with the same SPL for a dual driver versus a single when the amplifier voltage is the same. However, if you can increase your voltage so that each driver sees its max power, then you still end up with +6dB maxSPL versus the single driver. You just gotta make sure your amp has the voltage to drive it.

So +3dB comes from doubling the amount of electrical power being delivered and then the other +3dB comes from halving the acoustic space for each driver. Two drivers in 1/4 space will play 3dB louder than a single driver in 1/8 space. Two drivers in 1/2 space will play the same as a single driver in 1/8 space.
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fredk

Discussion Starter · #5 · Oct 19, 2009

Thanks Mike. I am usually good with technical stuff, but I seem to find electrical theory a challenge. It took me a bit of suplemental reading to understand the impedance/current/voltage stuff. I'll have to work on the 14 space half space stuff another day.

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SteveCallas

· #6 · Oct 19, 2009

Better to go with two discrete subs - more placement options to smooth out thr FR and a LLT large enough to suit two high excursion drivers will become extremely large and extremely heavy.

You don't necessarily have to have the subwoofers right next to each other to yield 6db gain.

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Blaser

· #7 · Oct 19, 2009

Placing them 4-5 ft close to each other garantees 6 db all the way up to 80 Hz. The problem of placing subs far apart is you may also have cancelation at higher frequencies, which is not necessarily a good thing. There are free modeling programs for speakers placement in a shoe-boxed room. They are very good IMO.

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looneybomber

· #8 · Nov 1, 2009

DrWho said:
So +3dB comes from doubling the amount of electrical power being delivered and then the other +3dB comes from halving the acoustic space for each driver.

It's not from reducing the acoustical space of the driver, because they both have the same acoustic space. As taken from http://fulcrum-acoustic.com/wordpress/wp-content/uploads/2008/07/comments-on-half-space.pdf

"The presence of...an identical loudspeaker results in a change in the mass loading (the effective mass of the air in front of a diaphragm), as well as the acoustical resistance (the resistance to diaphragm motion presented by the air). An increase in efficiency implies an increase in the acoustical resistance."

So it should be a doubling of cone area while maintaning an equally proportional BL.

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fredk

Discussion Starter · #9 · Nov 2, 2009

While you fellas are discussing the finer points of acoustic space and mass loading, I am still pondering the application of simple electronics.

I still don't understand why the power, as measured in watts is halved when I use two drivers in the model above. I went back and re-read some basic electronics stuff. IF I understand correctly, there is a defined relationship between watts, resistance and current. As the number of drivers is doubled, given the wiring scheme WinISD assumes, the resistance is halved and current is doubled leaving the number of watts consumed for a given spl level should remain the same.

What am I missing?? :huh:

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DrWho

· #10 · Nov 2, 2009

looneybomber said:
It's not from reducing the acoustical space of the driver, because they both have the same acoustic space.

...

So it should be a doubling of cone area while maintaning an equally proportional BL.

Put two drivers in a field and then place a wall equidistant between them. The acoustical behavior does not change if you remove the wall. I don't think it's a contradiction to claim that there is more than one way to model the behavior.

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looneybomber

· #11 · Nov 3, 2009

Well a wall would have quite a drastic change on it's sound; one woofer would be in one room and the other would be in another room. If they were in the same room, that "wall" would not be a wall, but instead could be considered part of the baffle and a baffle change will not have much effect on a subwoofer.

What you're proposing is the same as having opposed firing drivers, one on opposite sides of an enclosure. That's really no different than the drivers together on the same baffle because of how long wave lengths are.

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