Coil Inductance and Sound Quality (1 Viewer)

[Greg_R]

I recently read this white paper on Adire's website about woofer speed. Since a low Le is so great, why isn't everyone using Shivas (lowest Le driver)? I realize drivers with a flatter BL curve (XBL^2 technology) will have improved sound quality. However, why not multiple Shivas instead of a Tempest (if SQ is the primary concern and cost isn't an issue)?

 

[Greg_R]

What am I missing here?

 

[Anthony_Gomez]

I personally would use a pair of shivas over a tempest (greater displacement), but it would cost more and would require different amp loads. I would actually want to use 4 shivas and wired to 2ohms per channel on a power amp.

 

[ThomasW]

The answer is 'because it's all relative'.

Drivers like the original HE-15 and BluePrint had pretty high Le's. But they certainly sound good to most that have heard them.

In absolute terms, yes for a sub a low Le is 'best'. But comparatively speaking, drivers with a higher Le don't actually sound 'bad'

 

[Anthony_Gomez]

....it just usually limits how high they can be XO'd and the amount of equalization that is needed

 

[Greg_R]

Thanks!

 

[Dennis XYZ]

Apologies to Dan, who I consider a pretty smart guy, but the whole argument is bogus IMO. He says mass and motor strength are assumed to be constant. Well, that's true for any given driver but it's not true when comparing two different drivers. Increase BL and you have a driver capable of more acceleration with a given current. Same thing if you decrease mass. I think Dan boxed himself into a mental mind game and isn't seeing the forest for the trees in his white paper.

 

[Dennis XYZ]

Okay, let's work through the math. Quoting Dan:

"BLi = ma (eq 2)"

"a is the acceleration. This is what we're after. After all, acceleration IS the transient response - it's what dictates how fast the driver can change speed, which also means it dictates how fast the driver can move from position to position."

So we want to solve for a.

a = Bl*i/m

calculate the current i
i = V/Z
where V is volts and Z is the impedance

calculate Z, ignoring capacitive reactance
Z = Re + Le*2*pi*f
where Re is DC resistance, Le is inductance and f is frequency

Put it all together
a = (Bl*V)/(m*(Re + Le*2*pi*f))

The input signal is determined by V and f so we can assume those are constant for our test. But the driver design parameters Bl, m, Re, and Le will all affect how fast the cone will accelerate when we apply our signal.

 

[Mark Hayenga]

It's also not so much the nominal inductance value (usually measured at 1kHz) but the variation of inductance with excursion that can cause modulation of high frequency components. This is why all the high-end driver companies (Scan-Speak, Seas, Peerless, Lambda, etc) use extended pole pieces and faraday loops. Klippel has a great paper on this if anyone is interested.

 

[Mark Hayenga]

Here's the link:

http://www.klippel.de/pubs/aes2000/k...aper%20109.htm

Also remember that the more limited the bandwidth of your driver, the less inductance matters. Subwoofers are an excellent example. If you wanted a 12" or 15" for below 80Hz stuff, you could pick just about any driver on the market. If you wanted to run that driver up to 200-300Hz, options would become much more limited.

 

[Mark_E_Smith]

Consider this, if the voice coil is/has inductance then it will act like an inductor, rolling off/absorbing the high frequencies but it will start to do this before the actual speaker begins to roll off. Therefore inductance of the vc CAN play a roll in the detail IF the inductance Q is before the x-over for the next driver. ie the Dayton 305 with 1.4 mH of vc inductance but crossed above 1.2K