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#1 |
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diyAudio Member
Join Date: Jan 2011
Location: aptos, california
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to begin, i don't need help, i'm just sharing an awesome project
so, i've had the behrninger k450fx for a very long time and it is excellent and i would recommend it to anyone looking for a PA/keyboard speaker, as it does indeed perform considerably for both, and has fantastic sound quality though the stock bugera speaker is lacking a little on the low end. anyways, i've decided to go mad scientist on that little thing. basically the idea is, instead of simply replacing the stock speaker, i'm going to fit 11 speakers into the stock enclosure, all hooked up to the stock amp. yes, correct impedence, i love wiring |
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#2 | |
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diyAudio Member
Join Date: Dec 2002
Location: u.k
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Quote:
so youve managed to rewrite Mr ohms laws then? only i thought in a series circuit the current was the same through them all,and the voltage shared,so the overal power was the same wether you have 1 8 ohm driver or 96 0.83 ohm drivers in series |
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#3 |
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diyAudio Member
Join Date: Jan 2011
Location: aptos, california
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XD that's what i mean, i was talking in terms of impedence. if you have two four ohm speakers wired in serial, they count as one 8 ohm speaker, and the current etc. is passed the same through all of them, as though they are one speaker. sorry for the inclarity!
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#4 |
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diyAudio Member
Join Date: Jan 2011
Location: aptos, california
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Also! i have decided to move the 10" subs down to 8" lanzar max8's, simply because the 10's wouldn't fit without almost completely rebuilding the cabinet! and that completely defeats the purpose! XD i want to have a supremely modified k450fx, not a k450fx amp in a new box!
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#5 |
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diyAudio Member
Join Date: Jan 2011
Location: aptos, california
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me and my silly mistakes. two 4ohm speakers in serial would end with 2ohm. damn, i have to stop that.
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#6 |
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diyAudio Member
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They'd end up (in series) at 8ohm. In parallel, they'd be 2ohm.
__________________
"Throwing parts at a failure is like throwing sponges at a rainstorm." - Enzo My setup: http://www.diyaudio.com/forums/multi...tang-band.html
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#7 |
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diyAudio Member
Join Date: Jan 2011
Location: aptos, california
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are you sure? i thought in series the impedence of the speakers was multiplied by the number of speakers in the series (i always use speakers of the same impedence) and in parallel if the impedence of the two speakers is the same, you divide the impedence of one of them by 2 to get the final impedence. i also never use more than two speakers in a parallel config. it just makes things easier.
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#8 | |
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diyAudio Member
Join Date: Jan 2011
Location: aptos, california
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Quote:
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