Adding series resistor to sub

Diy subwooofer.org article on dipole setups.

Let's say we want to build a dipole bass system using four 12" drivers with the following specifications: Vas: 164 litres., Fs=30Hz, Qts=1.10, Qes=1.30, Qms=7.0, R=8 ohms, Xmax=8mm, Sd=0.0547m^2 .

To maximize efficiency, the drivers will be wired in parallel, giving an effective Re of 2 ohms. We also want to know what SPL levels we can expect if we drive the system with 100W of power.

To increase the Qts to the target value, we can use a series resistor Rs, and calculate its value as follows:

Qes'=Qts'*Qms/(Qms-Qts')
Qes' = 1.75*7/(7-1.75)
Qes' = 12.25/5.25
Qes' = 2.33

Rs = Re*(Qes'-Qes)/Qes
Rs = 2*(2.33-1.30)/1.30
Rs = 2*1.03/1.30
Rs = 1.6 ohms

As we plan to drive the system with 100W of power, assuming 10:1 differences between average and peak levels, we can use a 10W or greater resistor for Rs.

As the total resistance, Rs+Re, will be 3.6 ohms, the amplifier will have to be capable of driving at least a 3.6 ohm load.

Edit: could be figured out from here, but this link has the direct equation.
 

kelticwizard

diyAudio Moderator Emeritus
2001-09-18 2:33 am
Connecticut, The Nutmeg State
With all respect, rcw, I tried that and the Qts for the drivers stayed the same.

The bass shape stayed the same in reference to the midband, which is what Qts controls.

Two woofers in series doubles the nominal impedance, but keeps Qts identical.

Two woofers in parallel halves the impedance, but keeps the Qts identical. It will also increase your SPL per 2.83 Volts by 6 dB.