Unfortunately fabricadetadaco, your arguments are proving to be futile.
After all, a speaker with only one woofer would not be so grand and impressive in appearance as one with two!
However, your latest suggestion of paralleling four 16 ohm drivers may have appeal to presscot for the very reason I've just stated!
Let me explain my mindset on this project.
I’m aiming to build a pair of large floor-standing speakers which is intended to contain a pair of 12” woofers, a 6” midrange and a 1” tweeter per cabinet; where all drivers are what I already have, except the woofers.
The concept of the design is to use as much as benefit from those 6” midranges. For the woofers, actually, it is an idea of integrated subwoofer. Thus, x/o point on mid-to-low will be set as lowest as the midrange can do, hence, around 100 Hz.
AFAIK, a formula shows that an inductance value (L) will vary depending on driver’s resistance (Re). Therefore, to save the cost of inductors; also to reduce the mutual resistance, the lower Re is better, IMHO. However, I’m not going to build an amp-killer. So, 2 Ohms is not my goal, but, 4 Ohms will be.
In order to achieve 4 Ohms for the woofers, the best solution is to use two 8 Ohms in parallel. For the worse case, if 12” woofers with 8 Ohms can’t be found, alternative ways are introduced on previous posts — arrange drivers and a resistor in 2 configurations.
I’m now have an idea after seeing some of successfully commercial speakers using large and heatsink-like resistors in their crossover network. I, therefore, ask everybody for the opinions.
I’m aiming to build a pair of large floor-standing speakers which is intended to contain a pair of 12” woofers, a 6” midrange and a 1” tweeter per cabinet; where all drivers are what I already have, except the woofers.
The concept of the design is to use as much as benefit from those 6” midranges. For the woofers, actually, it is an idea of integrated subwoofer. Thus, x/o point on mid-to-low will be set as lowest as the midrange can do, hence, around 100 Hz.
AFAIK, a formula shows that an inductance value (L) will vary depending on driver’s resistance (Re). Therefore, to save the cost of inductors; also to reduce the mutual resistance, the lower Re is better, IMHO. However, I’m not going to build an amp-killer. So, 2 Ohms is not my goal, but, 4 Ohms will be.
In order to achieve 4 Ohms for the woofers, the best solution is to use two 8 Ohms in parallel. For the worse case, if 12” woofers with 8 Ohms can’t be found, alternative ways are introduced on previous posts — arrange drivers and a resistor in 2 configurations.
I’m now have an idea after seeing some of successfully commercial speakers using large and heatsink-like resistors in their crossover network. I, therefore, ask everybody for the opinions.
Do youself a service - drop the idea of that resistor. You burn half your amp power in that resistor, until you blow the resistor.
Get yourself an extra, cheap amp, a minidsp or similar and run the woofers active. The cost savings of not having to buy all the crossover parts for the woofer lowpass and midrange highpass will pay a good part of the extra electronics costs. In addition, you will not have to design baffle step compensation or other eq.
Johan-Kr
Get yourself an extra, cheap amp, a minidsp or similar and run the woofers active. The cost savings of not having to buy all the crossover parts for the woofer lowpass and midrange highpass will pay a good part of the extra electronics costs. In addition, you will not have to design baffle step compensation or other eq.
Johan-Kr
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