Impedance / sensitivity

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
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
 
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
 
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.