Dual Voice Coil Woofers and Resistive Damping

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    This is a brief discussion of how the second voice coil of a dual voice coild wooofer may be used to tune a sealed box design where the amplifier is connected to only one voice coil. It is well known that with a dual voice coil woofer Qts varies depending on whether a single voice coil is connected to the amplifer or the two voice coils are connected in series or parallel. When both voice coils are connected, either in series or in parallel ,Qts is approximately 1/2 or less than that when only a single coice coil is connected. However, when a single voice coil is connected, the second voice coil can be loaded by a resistance to provide additional damping. How this comes about is shown here.

First, we start with Newtons Law, the sum of the forces acting on the moving mass are equalt to the moving mass times the acceleration of the mass:


(1)         
F = ma = mx"

where
F is the sum of the forces, m is the mass, and a is acceleration which is equal to the second derivative of position, x".
Looking at the driver the forces include the motor force, that due to the suspension resistance and that due to the suspension compliance;


(2)          
  Fm -Rms X' - X / Cms= MmsX"         

Here Fm is the motor force, Rms is the mechanical resistance of the driver suspension, and Cms is the suspension compliance. The motor force is also composed of several parts. For the voice coil connected to the driving amplifier


(3)        
    Fm1 = (Bl) x I


where I is the current flowing through the voice coil. Assuming the amplifier has zero output impedance for simplicity, and neclecting the effects of the voice coil inductance which is negligable near the driver's resonance


(4)             
I = (Es - Eb) / Re


In the equation above, Es is the source voltage provided by the amplifier and Eb is the back EMF generated in the driver motor structure. Eb is given as


(5)        
   Eb = (Bl) X'


where X' is the velocity of voice coil in the gap.Using this realtionship we can rewrite the expression for the current as

    
(6)         
  I = (Es - (Bl) X') / Re


From this we can write the forces generated in the motor by the first voice coil as



(7)             
Fm1 = (Bl) ((Es - (Bl) X') / Re



For the second voice coil we again have a back EMF generated within the coil due to its movement in the gap. Assuming the properties of the second voice coil asre the same as those of the first the back EMF will be the same and the current flowing through the second voice coil is given as

    
(8)             
I = - (Bl) X' /(Re + Rl)


where Rl is the resistance of the load resistor connected across the second voice coil. Thus,


(9)    
   Fm2 = -(Bl)(Bl) X' / (Re + Rl)


Here any currents flowing in either voice coil due to mutual coupling are ignored. Summing Fm1 and Fm2, and rewriting we obtain the equation for the motion of the driver,



(10)        
(Bl)( Es/Re) = Mms X" + [Rms + (Bl)(Bl) / Re + (Bl)(Bl) / (Re + Rl)] + X / Cms



This equation is recognized as the basic equation for a 2nd order system,


(11)      
  F = m x" + b x' + k x

for which the natural frequency is given as

(12)           
wn = Sqrt (k/m)

and Q is given as


(13)       
Q = Sqrt(km) / b = wn m / b


With this observation we see that the resonant frequency of the driver is given as


(14)        
wn = Sqrt[1/(Cms Mms)]


For the driver Q we can break it down into three seperate Qs: Qms, Qes, and Qds. Qms is given as


(15)           
Qms = wn Mms / Rms

and is representative of the mechanical damping. Qes is given as

          
(16)     
  Qes = wn Re Mms / [(Bl)(Bl)]

and is representative of the electrical damping due to the back EMF generated in the first voice coil.Finally, Qds is given as


(17)          
Qds = wn (Re +  Rl) Mms / [(Bl)(Bl)]

which represents the damping due to the back EMF generated in the second voice coil. Filally we can write the equation of driver motion as


(18)         
(Bl) Es / (Re Mms) = X" + wn ( 1/Qms + 1/ Qes + 1/Qds) X' + wn wn X
      

Finally, we note that the seperate Qs can be combined as


(19)         
1/Qts  = (1/Qms + 1/Qes + 1 /Qds)

If the load across the second voice coil, Rl is high, or the voice coil is left open (Rl = infinity), 1/Qds goes to zero and re recover the standard relationshp for Qts,

(20)          
Qts = Qms Qes /(Qms + Qes)

On the other hand, if Rl is zero, or the second voice coil is shorted Qds = Qes and Qts is given as


(21)    
   Qts = Qms Qes / (2Qms + Qes)


We should note that


(22)          
Qds = Qes (Re + Rl) /Re


and with a little effort we can write


(23)         
Qts = Qms Qes / [Qes + Qms (1 + Re/(Re + Rl))]

or

(24)         Qts = Qms Qes / (Qes + c Qms)


where


(25)      
c = 1 + Re /(Re + Rl)


again we see that if Rl is zero we obtain the result of Eq (21 ) and if Rl is infinity that of Eq (20 ).


Qts given by Eq(21) is identical to the value of Qts we would obtain if the woofer were driven with both voice coils connected, either in parallel of in series. I reality there may be some differences between Qts obtained for the case where both voice coils are drivern and the case where Rl is large or infinite. These differences arise due to factors not considered in this simple analysis.
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