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Old 6th August 2001, 05:08 PM   #1
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If you take two or even three drivers, with a nom. impedance = 8 ohm, and connect them in paralell, you will get a nom. impedance resp. 4ohm and 2,66 ohm.
But how much will the nominell sensitivity increase.
Is there an exact formula to calculate this.
I`m aware that impedance/sensitivity vary according to frequenses
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Old 13th August 2001, 06:58 PM   #2
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Its not going to be any more sensitive, it might be louder because it's a lower impedence load for you amp, but unless their identical drivers, you shouldn't get any higher sensitivity, thats what i ofund in my projects at least.
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Old 5th September 2001, 05:53 PM   #3
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The actual answer is that every time you connect identical drivers in parallel, the impedance halves and the sensitivity doubles, ie increases by 6dB. The combination however now draws twice the current, so the EFFICIENCY has increased only 3dB.
The incease in sensitivity is 20*log(n) where n is the number of drivers, so 2drivers gives 6db, 3 drivers gives 9.54db etc..
The increase in efficiency is 10*log(n)
Hope this helps
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