The Subwoofer DIY Page v1.1
Dipole Bass Systems: Dipole Design
last updated: 13 April 2011

dipole.gif (1670 bytes)



Design Example
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.

We decide to use a baffle with an effective diameter of 1.35m.

From this, Fpeak and Fequal can be calculated:

Fpeak  = c/(D)
Fpeak  = 344/1.35
Fpeak  = 255  Hz

Fequal = Fpeak/3
Fequal = 255/3
Fequal = 85 Hz

We can model the baffle loss by using a spreadsheet that I put together for the purpose, called dipole.xls.  The spreadsheet's simulation is accurate enough for our use below Fpeak. Above Fpeak, the response of the system is greatly influenced by the shape of the baffle, so no attempt is made here to include it in the simulation.

Below is shown the estimated response curve for the given driver mounted in the stated baffle:


dipole_1.gif (6583 bytes)

To compensate for the 6dB/oct rollof, we select to do two things: increase the Qts of the system to 1.75 to flatten the low end response, and use a line-level or active 18dB/oct LP filter at around 80Hz to reduce the high frequency response.  

The effect of the increase in Qts can be seen in the graph below:

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The graph below illustrates the effects of an 78Hz 18dB/oct filter on the response:

dipole_3.gif (6870 bytes)

As can be seen from the graph above, the increase in Qts and the addition of the filter will produce an on-axis frequency response that extends from 28Hz to 90Hz, 0,-3dB.

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.  There will also be an efficiency gain as we're using 4 drivers, and an efficiency loss because of the baffle loss and the filtering.  These need to be taken into consideration when working out the final efficiency of the system.

Efficiency
We are using FOUR of these drivers in parallel.  By using four drivers this way, the resulting efficiency will be 12dB higher than the reference efficiency for one driver. However, there will be some insertion loss because of the use of a resistor to increase Qts. In this case, dipole.xls predicts that the efficiency will be reduced by about 2.6dB.