At the risk of exposing myself as an absolute novice, here goes. If I build a 2 way speaker using 2 8ohm speakers and a parallel circuit, would I effectively have about a 4ohm speaker?
Really simple crossover question.
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assuming you're talking 8ohm tweeter + 8ohm mid-woofer:
no
you'd have an 8ohm speakerdiVine Sound - my DIY speaker designs at diVine Audio- Bottom
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No, the load does not go by the woofer, but rather by the combined impedance of the woofer, tweeter and the crossover. But even then, the impedance may be 6 ohms over a good protion of the response and still be considered an 8 ohm impedance.- Bottom
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Thanks for the response. So if I wired two 8ohm drivers in parallel, no crossover, the overall resistance is halved right? So in a single first order crossover in parallel, there's just one cap and one coil. There's enough resistance to bring what is a 4ohm speaker up to 8ohm? Is this relationship linear with increased cap value and coil value?- Bottom
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Once you get past the big picture idea of how crossovers work, you need to model the impedance and SPL in software. Big picture: the crossover cuts highs or lows by increasing the impedance at those frequencies so the two drivers are never truly in parallel at any given frequency. Pesky details: an "8 ohm" driver may have an impedance that varies between 5 ohms and 30 ohms at different frequencies. How a particular crossover config will react with that is complicated enough that you really need crossover design software along with accurate SPL and impedance curves of the drivers to see what's going on. Textbook formulas for SPL and impedance are pretty worthless in the real world.- Bottom
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Originally posted by crazybastardThanks for the response. So if I wired two 8ohm drivers in parallel, no crossover, the overall resistance is halved right? So in a single first order crossover in parallel, there's just one cap and one coil. There's enough resistance to bring what is a 4ohm speaker up to 8ohm? Is this relationship linear with increased cap value and coil value?
CdiVine Sound - my DIY speaker designs at diVine Audio- Bottom
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Originally posted by buggers"So if I wired two 8ohm drivers in parallel, no crossover, the overall resistance is halved right"
No, its doubled. In series halves the Ohms.
CdiVine Sound - my DIY speaker designs at diVine Audio- Bottom
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Originally posted by Dennis HOnce you get past the big picture idea of how crossovers work, you need to model the impedance and SPL in software. Big picture: the crossover cuts highs or lows by increasing the impedance at those frequencies so the two drivers are never truly in parallel at any given frequency. Pesky details: an "8 ohm" driver may have an impedance that varies between 5 ohms and 30 ohms at different frequencies. How a particular crossover config will react with that is complicated enough that you really need crossover design software along with accurate SPL and impedance curves of the drivers to see what's going on. Textbook formulas for SPL and impedance are pretty worthless in the real world.
Most drivers come with an impedance graph. Can you go by this graph or do you measure your drivers individually. I have a pair of Hippi@#@ something tweeters, and I heard that they were tested and matched individually before leaving the factory. However, most other manufacturers publish a standard graph for a driver line. Can you trust them?- Bottom
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Originally posted by cjdnoooo, in parallel = halved (if you've got two equal impedance items wired in parallel). Series doubles.
C
8 and 8 in parallel is 1/( 1/8 + 1/8 ) = 4
8 and 4 in parallel is 1/( 1/8 + 1/4 ) = 2.67Santino
The only true wisdom is in knowing you know nothing.- Bottom
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Originally posted by crazybastardMost drivers come with an impedance graph. Can you go by this graph or do you measure your drivers individually. I have a pair of Hippi@#@ something tweeters, and I heard that they were tested and matched individually before leaving the factory. However, most other manufacturers publish a standard graph for a driver line. Can you trust them?- Bottom
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Originally posted by littlesaintIt's technically not halved. It's the inverse of the sum of the inverse of the individual impedances. Though from the math, if they are equal it is indeed halved.
8 and 8 in parallel is 1/( 1/8 + 1/8 ) = 4
8 and 4 in parallel is 1/( 1/8 + 1/4 ) = 2.67diVine Sound - my DIY speaker designs at diVine Audio- Bottom
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I'm thinking about getting the new parts express woofer tester. I'm pretty sure it measures impedance as well. Will it work with mid/tweets as well?- Bottom
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