Opie, the efficiency of the driver will be unchanged with either connection, but the sensitivity will be different.
Suppose each coil is four ohms, and driven from a voltage source providing 2volts. With the coils in parallel, each will draw a current of 2/4=0.5 amps. If the product of flux density and coil length (Bl) is say 5 for each coil, then the force produced is F=Bli x2= 2.5 x 2 =5
Now wire them in series. Each will now only have 1volt across the coil, so the current in each coil is now only 1/4=0.25 amps. The total force is now 5 x .25 x 2 = 2.5.
The force on the cone is now one half that of the parallel connection, so the output is 6dB down, though the impedance is 8ohms rather than 2ohms.
Andrew
Suppose each coil is four ohms, and driven from a voltage source providing 2volts. With the coils in parallel, each will draw a current of 2/4=0.5 amps. If the product of flux density and coil length (Bl) is say 5 for each coil, then the force produced is F=Bli x2= 2.5 x 2 =5
Now wire them in series. Each will now only have 1volt across the coil, so the current in each coil is now only 1/4=0.25 amps. The total force is now 5 x .25 x 2 = 2.5.
The force on the cone is now one half that of the parallel connection, so the output is 6dB down, though the impedance is 8ohms rather than 2ohms.
Andrew
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.