Thank you for all answers, so Q is is the change....
Inductor
To be more specific:
Actual main system
2 tops with eighteensound 15cx1000 with fd250 inside
2 subx with 15mb1000
dbx crossing subs and tops at 250 hz and 2 amps
cd, virtualdj pc pioneer mixer etc ...
Backup system
2 Eighteensound woofer and 2 compression unit with horn and with the same crossover (ordinary 500w rms speaker)
same amp rack but no dbx just a equalizer with a similar eq curve as the dbx.
I´ve also tested both systems with crossover flat eq setting
.... and my brother says the second one has a more "crystal sound", of course it has also less power...
I would build and test as you say, but Eighteeensound doesn´t build crossovers, I´ve asked them, and You know how it´s diffilcult to get inductors or inductor formers here in our poor land, but I will try those 16 ohm ones and then compare the sound with the dbx crossed at 1 khz at 2khz and at 4khz, and my ear will tell me something...
By the way the speakers are performing fine and the coils didn´t get burned yet.
Thank you once again.
Inductor
To be more specific:
Actual main system
2 tops with eighteensound 15cx1000 with fd250 inside
2 subx with 15mb1000
dbx crossing subs and tops at 250 hz and 2 amps
cd, virtualdj pc pioneer mixer etc ...
Backup system
2 Eighteensound woofer and 2 compression unit with horn and with the same crossover (ordinary 500w rms speaker)
same amp rack but no dbx just a equalizer with a similar eq curve as the dbx.
I´ve also tested both systems with crossover flat eq setting
.... and my brother says the second one has a more "crystal sound", of course it has also less power...
I would build and test as you say, but Eighteeensound doesn´t build crossovers, I´ve asked them, and You know how it´s diffilcult to get inductors or inductor formers here in our poor land, but I will try those 16 ohm ones and then compare the sound with the dbx crossed at 1 khz at 2khz and at 4khz, and my ear will tell me something...
By the way the speakers are performing fine and the coils didn´t get burned yet.
Thank you once again.
Wolf,
Okay, I read your post # 20, so let me ask you this. Assume an ideal, two-way, second order cross-over network, correctly implemented. That is, we are only considering what we want to occur in terms of an electrical network. How much cut-off, that is attenuation, do each of the high-pass and low-pass filters of the network produce at the cross-over frequency Fc? (answer in decibels).
Answer correctly and you win a vacation in the Caribbean (and also prove my point).
Regards,
Pete
Okay, I read your post # 20, so let me ask you this. Assume an ideal, two-way, second order cross-over network, correctly implemented. That is, we are only considering what we want to occur in terms of an electrical network. How much cut-off, that is attenuation, do each of the high-pass and low-pass filters of the network produce at the cross-over frequency Fc? (answer in decibels).
Answer correctly and you win a vacation in the Caribbean (and also prove my point).
Regards,
Pete
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Hey Wolf,
Now I understand, I'm pretty sure. For a second order cross-over, the cross-over frequency is the frequency at which inductive reactance equals capacitive reactance. So no matter the load impedance that is connected to the filter, and the capacitance and inductance of the filter is unchanged, then the cross-over frequency stays the same.
However, I am sure that you are aware that the load impedance enters into the equations for solving for capacitance and inductance of second order high/ low pass filters. There is a reason for that- to insure that cut by the filters is -6 dB at the crossover frequency.
So, going by the definition of the cross-over frequency that you are using, I can connect a 8 Ohm resistor in place of the 16 Ohm resistor that both filters of the cross-over network have been configured to work with, and the crossover frequency stays the same. However then, frequency response through the cross-over region is no longer flat, as the cut of the filter is dependent on resistance of the load.
-Pete
Now I understand, I'm pretty sure. For a second order cross-over, the cross-over frequency is the frequency at which inductive reactance equals capacitive reactance. So no matter the load impedance that is connected to the filter, and the capacitance and inductance of the filter is unchanged, then the cross-over frequency stays the same.
However, I am sure that you are aware that the load impedance enters into the equations for solving for capacitance and inductance of second order high/ low pass filters. There is a reason for that- to insure that cut by the filters is -6 dB at the crossover frequency.
So, going by the definition of the cross-over frequency that you are using, I can connect a 8 Ohm resistor in place of the 16 Ohm resistor that both filters of the cross-over network have been configured to work with, and the crossover frequency stays the same. However then, frequency response through the cross-over region is no longer flat, as the cut of the filter is dependent on resistance of the load.
-Pete
CT-
Paragraph 1, absolutely!
Paragraph 2:
Not all xovers are -6dB at the xover frequency. Some actually do not sum flat, so remember that too. This is the reason for 'spread factor' as Vance Dickason calls it in the LDC. (Since we are speaking in terms of textbook functions.)
Paragraph 3:
The Q of the filter dictates the damping at the knee that the xover applies to the acoustic responses. Higher Q means it can peak at the knee before rolloff, and low Q means the knee will rolloff sooner and further from Fc of the xover. You mean; "the Q of the filter is dependant on the resistance of the load." And no, it will no longer be flat in pretty much all cases.
The whole reason I'm stating what I am is that FR, and other criterion were not present in the question. Just the xover point, and the 'resistances' in use. There is technically no way to say what the acoustic xover will then be as different drivers operate differently, and then you have the Q shift which changes things pretty dramatically. Therefore- you only have the Fc as the xover sees it. So- you get a most accurate answer in saying that the xover freq does not change, which does not mean that the acoustic rolloff frequency won't change. You change drivers, you change your acoustic xover point, but only a simulation, measurement, or complex math can show or dictate what that new acoustic xover point will be. Obviously- any swap of this sort would be suboptimal, and a total redesign should be completed for best results anyway.
I answered the best I could with the info supplied, and the answer I gave is likely the one the OP would have wanted in terms with how he asked it.
Later,
Wolf
Paragraph 1, absolutely!
Paragraph 2:
Not all xovers are -6dB at the xover frequency. Some actually do not sum flat, so remember that too. This is the reason for 'spread factor' as Vance Dickason calls it in the LDC. (Since we are speaking in terms of textbook functions.)
Paragraph 3:
The Q of the filter dictates the damping at the knee that the xover applies to the acoustic responses. Higher Q means it can peak at the knee before rolloff, and low Q means the knee will rolloff sooner and further from Fc of the xover. You mean; "the Q of the filter is dependant on the resistance of the load." And no, it will no longer be flat in pretty much all cases.
The whole reason I'm stating what I am is that FR, and other criterion were not present in the question. Just the xover point, and the 'resistances' in use. There is technically no way to say what the acoustic xover will then be as different drivers operate differently, and then you have the Q shift which changes things pretty dramatically. Therefore- you only have the Fc as the xover sees it. So- you get a most accurate answer in saying that the xover freq does not change, which does not mean that the acoustic rolloff frequency won't change. You change drivers, you change your acoustic xover point, but only a simulation, measurement, or complex math can show or dictate what that new acoustic xover point will be. Obviously- any swap of this sort would be suboptimal, and a total redesign should be completed for best results anyway.
I answered the best I could with the info supplied, and the answer I gave is likely the one the OP would have wanted in terms with how he asked it.
Later,
Wolf
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More of the same... no references, no models (again).
Maybe because the COAX is more critical and frequency output curve not so smooth, if you notice in my pot #10 and compare it to other woofer drivers. Probably the tweeter CD is the same but horn different and also maybe better for crossover.
I don't know where/why you find a problem since there's mail (and e-mail) for national or international orders, for components and drivers. So that's not an excuse.
Of course Eighteeensound is not designing crossovers for all the drivers they have and maybe some like Beyma have them for their COAX only. A crossover (Beyma FD250) is not something that you can use with all the manufactured drivers from all the manufacturers. It's like using a bicycle wheel in your car (stupid example).
Only if it was a similar layout to the one that was posted (to adapt) and it's not, because it's a 12dB, second order, 2KHz. Besides their drivers have different sensitivities, outputs, impedances, phases, Fs and roll offs to name a few, only SPL can be dialed in in the crossover (shunts).
The new crossovers have to be designed from scratch by me or any other person with that ability. The model is based in my link on the same post above. I mentioned that the third order crossover frequency needs alterations from 1.5KHz to 1KHz modeled in the right software simulator and new crossover components specified as for new attenuation of the tweeter CD. A new crossover from Beyma might suffer off the same problems and very likely might have different and unexpected repercussions in the sound system. Not a very nice prognostic, but hey, who said you can not try...
Have a nice time.
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