pmr@drutx.UUCP (Rastocny) (04/03/85)
Subject: How/why do Bose speakers work? "It's all done with mirrors! And microprocessors use micromirrors." Before my terminal died, I began ramblings on why Bose gets bass from a 4.5" driver. Someone said that they use long-throw drivers, and they do. But to achieve any kind of acoustic efficiency, you need a large piston area (the area of the piston in square inches that is performing the electrical-to-mechanical conversion). Rule of thumb: the lower a driver's resonant frequency (Fs), the lower the note it can reproduce (not always true, but let us assume that it is for this explanation without getting into why). Bose uses nine drivers to achieve an effective piston area equivalent to that of about a standard 10" acoustic suspension driver. In order to make a 4.5" driver possess a low Fs, you can do two things: 1) increase the compliance of the driver's suspension (make the suspension less stiff), and 2) increase the mass of the cone. This is fine, but you know you never get anything for nothing! Let's see what happens in each case. When you increase the driver's compliance you must also change the physical characteristics of the linear motor. You can increase the size of the magnetic gap where the voice coil operates. This is the best but most expen$ive solution and is therefore rarely used. \<---------- cone ---------->/ \ / \ / \ / \ / \________/ | | <------ Voice Coil and Former (bobbin) | | ___ | ____ | ___ |\\\|.||\\\\||. |\\\| |\\\|.||\\\\||. |\\\|<----- Magnetic focusing gap |\\\| |\\\\| |\\\| |\| |\\| |\| |\| |\\| |\| |\|_____|\\|______|\| |\\\\\\\\\\\\\\\\\\\| |\\\\\\\\\\\\\\\\\\\| |\\\\\\\\\\\\\\\\\\\| |_______magnet______| The other way around this dilemma is to make the voice coil longer so at least some of it always is within the magnetic gap. This is the most common solution but it also reduces the efficiency of the driver by creating a magnetic field outside of the magnetic gap. \<---------- cone ---------->/ \ / \ / \ / \ / \________/ .| |.<------ Voice Coil and Former (bobbin) ___ .| ____ |. ___ |\\\|.||\\\\||. |\\\|<----- Magnetic focusing gap |\| .| |\\| |. |\| |\| |\\| |\| |\| |\\| |\| |\|_____|\\|______|\| |\\\\\\\\\\\\\\\\\\\| |_______magnet______| Cone pistons also possess a physical limitation of reproducing wavelengths shorter than their diameters. (That's why good sounding tweeters have small diameter pistons). As mentioned in another article, as the wavelength of the signal increases and approaches the diameter of the piston, the dispersion pattern of the driver changes from a hemispherical pattern (well dispersed) to a cardioid pattern (poorly dispersed). ^ .... | . . | / . . | ||/ . . | dispersion pattern of wavelengths much larger ||\ . . | than the diameter of the piston \ . . | . . | .... | v / . ........ . ^ ||/ . . | Dispersion pattern of wavelengths much smaller ||\ . . | than the diameter of the piston \ . ........ . v Since there is a practical limit to the throw of a driver (especially a small diameter driver), you must also increase its mass to again reduce the Fs. Increasing its mass reduces the drivers transient response (f=ma). (This is why the best sounding tweeters are also low mass.) Increasing the mass also increases the power handling requirement of the voice coil. Usually, increasing the diameter of the voice coil (VC) is this solution (larger heat sink) but this also increases the VC inductance and reduces its high-frequency reproducing ability. (This is why the best sounding tweeters have small VCs.) Without getting into a detailed discussion of woofer alignments, basically Bose compromised and chose not to reproduce deep base and settle for a roll-off somewhere around 40Hz, and keep the driver's mass low for reasonable transient response in the bass/midbass regions. The trick Bose used to get around these problems is by building a speaker- specific equalizer that electrically boosts the signal for the low and high frequency roll-offs. Again, there is a practical limit to the amount of equalization you can apply to a driver in the power-handling limit of the VC. scales: SPL (vertical); Freq. (horiz) ................. SPL curve of a Bose speaker (no EQ) . . . . . . . . . . . . ................. EQ curve of Bose compensation unit ......................... Resulting SPL curve of Bose EQ on Bose speaker . . Dispersion is still a major problem with a 4.5" driver reproducing very high frequencies. Bose simply aimed eight of the drivers rearward and used the natural surface reflections of the room to improve dispersion. These are the major topics Bose speakers address. There are many many more interrelated and intricate problems in properly designing loudspeakers. As the song goes, "We've only just begun.." Yours for higher fidelity, Phil Rastocny AT&T-ISL ihnp4!drutx!pmr
don@umd5.UUCP (04/05/85)
> Subject: How/why do Bose speakers work? > "It's all done with mirrors! And microprocessors use micromirrors." > -- [ after reading what follows, I tend to believe it !] > > Bose uses nine drivers to achieve an effective piston area equivalent to that > of about a standard 10" acoustic suspension driver. > -- [ nice speaker graphics in his posting ] > > Cone pistons also possess a physical limitation of reproducing wavelengths > shorter than their diameters. (That's why good sounding tweeters have small > diameter pistons). As mentioned in another article, as the wavelength of the > signal increases and approaches the diameter of the piston, the dispersion > pattern of the driver changes from a hemispherical pattern (well dispersed) to > a cardioid pattern (poorly dispersed). > > -Phil Rastocny [] OK, now just hold on here a cotton-picking minute! The only formula in Phil's posting is F=ma, and I buy his statement that for a small mass the transient response of the speaker is better than for a speaker with a large mass. This wavelength stuff he talks about has got me confused ... A wavelength shorter than the speaker's effective 10 inch diameter is the speaker's physical limitation you say? OK, Let us calculate the wavelength of a 20kHz wave for starters -- (300,000 km per second) divided by (frequency in kHz) = 15,000 meters This is approximately 9.32 MILES. Somehow I don't think this figures into Phil's explanation very well, but let's go with it still -- The limiting frequency at a 10 inch wavelength is: (300,000 km/s) divided by (0.254 m) = 1,180,000 kHz (about 1.18 GHz !!) Something is still wrong here! If Phil's explanation was correct, Bose speakers would either have to have a 9.32 mile diameter, or they would still be good tranducers at 1.18 GHz (even your dog couldn't hear that!!). Well, maybe I haven't punched holes in the explanation big enough to sail the Nimitz through just yet -- How about waveguides? If I model the 10 inch speaker as a circular waveguide, the cutoff wavelength is 3.41 times the radius of the waveguide. Now to calculate: (300,000 km/s) divided by (3.41 times 0.127 m) = 693 MHz Closer to 20 kHz, but still 30,000 times 20 kHz ... Now I don't profess to know exactly what this dispersion mechanism that Phil is talking about really is, but I seriously doubt the explanation that he posted. Any takers ? -- ----------------------------------------------------------------------------- "Space, the final frontier .." Final, hell! It's the frontier of frontiers !! ----------------------------------------------------------------------------- -==- IDIC -==- (Thanks Bob!) SPOKEN: Chris Sylvain ARPA: don@umd5.ARPA BITNET: don%umd5@umd2 CSNET: don@umd5 UUCP: {seismo, rlgvax, allegra, brl-bmd, nrl-css}!umcp-cs!cvl!umd5!don
herbie@watdcsu.UUCP (Herb Chong [DCS]) (04/07/85)
In article <461@umd5.UUCP> don@umd5.UUCP writes: > OK, Let us calculate the wavelength of a 20kHz wave for starters -- > (300,000 km per second) divided by (frequency in kHz) = 15,000 meters >This is approximately 9.32 MILES. Somehow I don't think this figures into >Phil's explanation very well, but let's go with it still -- try about 300 m/s. >SPOKEN: Chris Sylvain
crandell@ut-sally.UUCP (Jim Crandell) (04/10/85)
> A wavelength shorter than the speaker's effective 10 inch diameter is the > speaker's physical limitation you say? > OK, Let us calculate the wavelength of a 20kHz wave for starters -- > (300,000 km per second) divided by (frequency in kHz) = 15,000 meters > This is approximately 9.32 MILES. I'm sorry that I've never visited the planet where you live, and I'm not absolutely sure I really want to, but I really would like to see some of the musical instruments you use. Or do you measure audio frequencies in GHz there? -- Jim Crandell, C. S. Dept., The University of Texas at Austin {ihnp4,seismo,ctvax}!ut-sally!crandell