Problem:

I have always taken it on faith that the back EMF of a driver coil is equal to Bl*v, where v is the velocity of the voice coil, and Bl is that mysterious driver spec. Curiosity got to me, so I had to see why this was true. First I will calculate the force-current ratio (Bl) of a overhung voice coil and then relate the velocity of the coil to its back EMF. This gives rise to the concept of virtually equivalent mechanical resistance

Summary of Given Terms:
      Rc           Voice coil radius
      Re           Total resistance of the coil
      Hg           Height of the gap
      B            The magnetic field vector field within the gap (in Telsas)
      n            Density of windings in coil (windings/meter)
      I            Current flowing through the voice coil
      x            The length of coil that is below the magnetic gap
      v            Velocity of the coil (+v is upwards)
      V            Voltage across to coil

Summary of Calculated Terms:
      F            Force on the voice coil
      Bl           Force-current ratio of the driver
      N            Number of windings below the magnetic gap
                  The total magnetic flux through the Gap
                  The Motional EMF induced in moving coil
      Rv           Virtually equivalent mechanical resistance

Force-Current Ratio (Bl):

We will assume that the magnetic field (B) in the gap is uniform, and that the windings on the voice coil are evenly spaced and tightly packed (n).

Assume the perspective of looking at the voice coil from the top. The voice coil is wound from top to bottom, counter-clockwise. The flow of current (I) is also counter-clockwise. The orientation of the magnetic field also has to be taken into account. Let the north poles be outside the voice coil, and the south inside. With this, the magnetic field (B) is directed inward. Now we can calculate the amount of force (F) that is applied by the interaction of the magnetic field and current.

      b                 
                         a is the beginning of the coil entering B
F = I  ds x B    where   b is the end of voice coil exiting B
                        ds is an element in the direction of the winding
      a

Our integration is symmetric around the center and B is obviously perpendicular to ds. We can calculate |ds| = Rc * da, where Rc is the radius of the voice coil, and da is the angle element. Since there are n*Hg windings inside the magnetic gap we integrate a total angle of 2*pi*n*Hg radians.

      b               b                2**n*Hg
                                        
F = I (ds x B) = z I |B| |ds| = z I |B|  Rc da = 2  I |B| n Hg Rc z
                                        
      a               a                   0

where

z is a unit vector pointing upward

So we see the force on the coil is upward and is directly proportional to the current (I). We also see what a designer can do to increase the strength of the motor in a driver, but we don't see the relationships between these values; |B| is proportional to 1/(Hg Rc). So increasing the density of windings in the magnetic gap is about the only option. We will define a constant, Bl = 2 n Hg Rc, that takes into the effect of these design parameters. Therefore |F| = Bl*I.

Motional EMF

We now have to make the further assumption that the total flux () of B, passing through the magnetic gap, is also contained totally in the pole. This is an idealized diagram, in reality the flux is not so well contained. The total cylindrical area (A) of the gap is given by A = 2  Hg Rc, so

 = |B| A = |B| 2  Hg Rc

Let x be the total length of the coil below the magnetic gap. The flux that travels down the pole has to cut through the planes made by each of the windings that it passes; the windings below the gap. The number of windings that surround this flux is N = x*n. So, now we have an expression for the Motional EMF () of the coil:

      d(N )   d(x n |B| 2  Hg Rc)      dx
|| = ------ = -------------------- = Bl -- = Bl * |-v|
        dt              dt               dt

We would like to orient this EMF and get rid of the absolute value signs. Let a positive  imply that the induced EMF voltage is positive at the top of the coil. Then a downwards velocity (-v) of a the driver induces a back EMF that is in a clockwise direction, making the top of the coil a higher potential (+).

= -Bl*v

Virtually Equivalent Mechanical Resistance

If a driver's terminals are not connected then there is a resistive force that opposes the velocity of the cone. But when the terminals are shorted, this resistive force is much greater. Obviously the electrical part of the driver is opposing this velocity. Without the knowledge that this extra resistance is electrical, you might assume it is just a mechanical resistance. Lets calculate the force generated by the back EMF of the driver, and see if it really acts like a mechanical resistive force.

We will take the general case, and assume the driver is hooked up to a voltage source. The voltage across the coil is V, and is dependent on time. What resistive force does the cone experience if is moving at speed dx/dt? We know that the back EMF () generated by the coil moving through the magnetic gap is:  = -Bl(dx/dt). Since the voltage source has no resistance, we can use the resistance of the coil (Re) to find the current (I) that is induced by this EMF; I = (V+)/Re. This current will apply a force on the cone according to F = Bl*I. Therefore,

                               2   
              (V+)   Bl     Bl   dx
F = Bl I = Bl ----- = -- V - -- * --
               Re     Re     Re   dt

There are two terms on the right hand side. The first is the direct result of applying a voltage to the coil. The second is the resistive force, it is indistinguishable from a regular mechanical resistive force, so it is logical to treat it the same. Let Rv = (Bl^2)/Re, and call it the "virtually equivalent mechanical resistance".

    Bl        dx                     dx
F = -- V - Rv --     OR     F = - Rv --  (when V=0)
    Re        dt                     dt

When the voltage (V) is zero, the terminals of the driver are effectively shorted. The extra resistive force felt when pushing on the cone is shown by the second formula above in this case.