How many 6.5in driver equals one 8in driver


If I am doing a simple math equation of a surface of a circle, pi * r^2, then:

8in. :  3.14 * 4 ^2 = 50.24 in^2

6.5in:  3.14 * 3.25^2 = 33.16 ^2

 

So it seems like it will take 2 6.5in drivers to equal an 8 in. driver.

andy2

andy, how do you define driver impedance of 4 Ohms?

It's the nominal impedance within the usage freq. range.  For woofer, it is about up to 3KHz, after that the voice coil impedance kicks in to a gradual increase as freq. go up.  The plot you have above probably comes from the impedance plot of a woofer.  For tweeter, it is 4ohm almost flat up to 20KHz because obviously tweeter has to operate up to 20Khz.  But you will have the xover which has its own impedance.

to get 4 Ohms total four driver spkr system I used 16 Ohms drivers, in parallel

It is very difficult to find a 16ohm drivers.  Most are either 8ohm or 4ohm.  The one you show there is probably 8ohm woofer.

parallel connected drivers have also direct connection to low output amplifier impedance, to dampen resonances. 

Maybe for tube amplifiers but in which case the xover can use a zobel network to linearize the impedance curve.  Remember it is not just the impedance of the drivers, it is also the impedance of the xover too.

For solid state amplifiers which has very high damping factor so the impedance of the drivers probably won't be an issue.  So you won't have "dampen resonances" issues.

Here is the impedance plot of SB Satori Textreme 5in. woofer.  It stays pretty flat.

one driver plot, probably the best selected one, out many manufactured!

SPL (f) plot will be messy for many drivers in series..

btw., pl. try to measure 10x of those and compare.. that’s what I did many years ago..

andy2

The Math tells us 2.5 is the answer. I would up the ante to (4) 6.5 inch drivers equaling a single 8" driver. When designed and properly enclosed, an 8 inch driver/woofer can sound outstanding!  Same goes for running (2) 8-inch driver/woofers in series.

 

Happy Listening!

@jonwolfpell 

"The smaller driver has to move faster & further than the big driver to produce low notes at higher volumes"

please clarify, I get 'long throw' to move more air, or move it more forcefully, but wouldn't faster be higher frequencies?