We will start with the relation between maximum input voltage (V) and maximum displacement (X). Since the closed box differential equation is the same as driver differential equation, we can use (D6) of Driver Analysis and just replace t with s and Qt with Qs.

        Bl         /   2  s         2 \
      ----- V = X |-  + -- j  + s  |                    (O1)
      Re Mm        \      Qs           /

Our next relation is (A3) of Driver Analysis, it relates the maximum acceleration (Z) of the cone to the acoustic output ().

             2    2
       = --- A  |Z|                                        (O2)
          2c

The final relation, Z=-(^2)X, relates the maximum displacement (X) and maximum acceleration (Z). Substitute this into (O1) and then (O1) into (O2).

             2 4   2
       = --- A  |X|                                       (O3)
          2c

                2   2  2                        -2
             A  Bl  V  /      s       s^2 \
       = --- ----------  |-1 + ---- j  + ---- |            (O4)
          2c     2   2   \     Qs       ^2  /
                Re  Mm

This is still in complex form, it has to be converted to amplitude form:

                2   2  2                             -1
             A  Bl  V / / s^2     \2  /  s  \2 \
       = --- ---------- | | ---- - 1 | + | ---- |  |       (O5)
          2c     2   2  \ \ ^2      /   \ Qs  /  /
                Re  Mm

This result is only good for frequencies that are low enough that voice coil inductance is negligible. We can see a graph of acoustic output below.

Defined as the ratio of acoustic intensity output () to the electrical power input (Pe). First we assume that the sound wavelengths are very large with respect to the diameter of the driver. Secondly we take the measurement at some frequency above, and away from, the effects of resonance. Third, the driver in is an infinite baffle. And fourth, we assume that the coil inductance is negligible. This is a graph of acoustic output () with respect to frequency at constant voltage input (V). Three Qs values are shown: 1, .707, and .5. Notice that, in the resonant region, the Qs effects the acoustic output. Outside that range Qs makes little difference. If the point for measuring acoustic output was too high, then the voice coil inductance (Le) would result in a lower measurement. You can see the reduction of acoustic output due to Le on the leftmost part of the graph. Our point of reference is a point that is neither affected by Le or resonant effects. Luckily, (sub)woofers have such a point; Ze approximately equal to Re (see Z14 of Driver Analysis).

Use (O4) as a starting point; it already assumes the frequency is low enough to not be affected by inductive effects of the voice coil.

                2   2  2                        -2
             A  Bl  V  /      s       s^2 \
       = --- ----------  |-1 + ---- j  + ---- |            (N1)
          2c     2   2   \     Qs       ^2  /
                Re  Mm

Above the resonant region the resonant term can be simplified to -1.

                2   2  2
             A  Bl  V
       = --- ----------                                    (N2)
          2c     2   2   
                Re  Mm

The total power going into the driver is given by Pe = V^2/Re.

                2  2
            Bl  A
=>     = --- --- -- Pe                                     (N3)
          2c   2 Re  
              Mm

As mentioned at the beginning, the ratio of the power in (Pe) to power out () is the efficiency (n).

                      2  2
                 Bl  A
=>    n = -- = --- --- --                                  (N7)
           Pe   2c   2 Re  
                    Mm

We can simplify this to Theile-Small parameters. Rv=(Bl^2)/Re, Vas=(B*A^2)/Km, Qes=sqrt(Mm*Km)/Rv, t=sqrt(Km/Mm).

Reference Efficiency:

                                     3                  3
              Km             1   t Rv Vas      1   t
=>    n = ---- --- Rv Vas = ----- ---------- = ----- --- Vas (N8)
           2cB   2          2c^3 sqr(Mm Km)   2c^3 Qe   
                Mm

All that work and we have to conclude that the reference efficiency is EXACTLY the same as for the driver in free air. The enclosure that the box is placed makes no difference in the upper frequencies, away from resonance.

The acoustic output () of the driver is, n*Pe. But this is only valid in the small range of frequencies above resonance, where impedance (Ze) is equal to coil resistance (Re). We want to find the frequency (3) where the acoustic intensity is down by 1/2 or 3dB. We use (N7), setting =n*Pe/2=n*V^2/Re/2.

                       2   2  2 
          nPe      A  Bl  V 
       = ---- = --- ----------                             (F1)
           2     4c     2   2  
                       Re  Mm

And substitute for  using (O5).

            2  2   2         2  2  2 / /      2 \2     2  \-1 
        Bl  A  V      Bl  A  V | |    s  |    s   | 
=>    --- ---------- = --- --------- | |1 - --- | + ----- |     (F2)
      4c     2   2    2c    2   2  | |     2  |    2  2 |
            Mm  Re          Mm  Re   \ \       /    Qs  /

Remember that we are solving for 3=.

            / /      2 \2     2   \-1 
       1    | |    s  |    s    | 
=>    --- = | |1 - --- | + ------ |                         (F3)
       2    | |      2 |     2  2 |
            \ \    3  /   3 Qt  /

            /         \2    2
         4  |  2    2 |   s  
=>    23 = |3 - s  | + ---                               (F4)
            |         |     2
            \         /   Qs 

                                  2
         4    4     2  2    4   s  
=>    23 = 3 - 23 s + s  + ---                         (F5)
                                  2
                                Qs 

        4  /     1   \   2  2     4
=>    3 + |2 - ---- | s 3  + s                          (F6)
           \    Qs^2 /

Use the quadratic formula to jump to possible solutions:

            /  1      \         / /   1     \2     \
            | ---- - 2| +/- sqrt| | ---- - 2|  + 4 |
        2   \ Qs^2    /         \ \ Qs^2    /      /   2
=>    3  = ---------------------------------------- s     (F7)
                              2

The "sqrt" term is always larger than the first term, therefore the "-" symbol does not make sense. So we have

                  / /  1      \       / /   1     \2     \ \
                  | | ---- - 2| + sqrt| | ---- - 2|  + 4 | |
                  | \ Qs^2    /       \ \ Qs^2    /      / |
=>    3 = s sqrt| -------------------------------------- | (F8)
                  \                   2                    /

The half-power point is very messy. We can do a simplification for when Qt=sqrt(1/2). This Qt is a common value among speaker builders. Note all the zeros occurring when substituting in to (F8). We can get three conclusions from this,

      3 > s    (when Qs < sqrt(1/2))      (F9)

      3 = s    (when Qs = sqrt(1/2))      (F10)

      3 < s    (when Qs > sqrt(1/2))      (F11)

Since most speaker builders what to get the most amount of bass from their driver, we should derive a formula for the box size that achieves the lowest (best) F3. We will use (F8) for finding F3 and use (R3) to substitute s=t*Qs/Qt into (F8).

                     / /  1      \       / /   1     \2     \ \
                     | | ---- - 2| + sqrt| | ---- - 2|  + 4 | |
           Qs        | \ Qs^2    /       \ \ Qs^2    /      / |
=>    3 = -- t sqrt| -------------------------------------- | (M1)
           Qt        \                   2                    /

                  / /        2\        / /        2\2     4\ \
           t     | \1 - 2 Qs / + sqrt \ \1 - 2 Qs / + 4Qs / |
=>    3 = -- sqrt| ---------------------------------------- | (M2)
           Qt     \                       2                  /

To minimize the above we notice that we only have to minimize the numerator inside the first sqrt.

              /        2\        / /        2\2     4\
=>    numer = \1 - 2 Qs / + sqrt \ \1 - 2 Qs / + 4Qs /      (M4)

                      2        /         2     4\
=>    numer = 1 - 2 Qs  + sqrt \ 1 - 4 Qs + 8Qs /           (M5)

Differentiate, and solve for Qs.

              2       -8 Qs + 32 Qs^3
=>    0 = 4 Qs  + -----------------------                   (M6)
                       /         2     4\
                  sqrt \ 1 - 4 Qs + 8Qs /

=>    Qs = sqrt(1/2) = .707                                 (M7)

Therefore Qs=sqrt(1/2) will minimize 3. Putting this back into (M1) we see that

                       t
=>    minimum 3 = ---------- (M8)
                   sqrt(2) Qt

From (F10) we recall that 3 = s when Qs = sqrt(1/2). The graph shows the effect of Qs on 3. Remember that Qt/t is just a constant. This does not tell the whole story though. It is natural to ask, how large should the enclosure be to achieve this low 3? A little bit of computer assisted math performed on (M1) will show us that,

                2 Qt^2
=>    Vb = Vas ----------  (box size for min 3)            (M9)
               1 - 2 Qt^2

Also notice that Qt is restricted to values below sqrt(1/2). (M9) is based on the fact that the system quality factor (Qs) is equal to sqrt(1/2). Since in (A8) we concluded that Qs>Qt, we see the restriction on Qt is valid.

It is interesting that in (M8) a driver with high Qt is required in order to get the lowest 3. Yet (M9) demands that Qt be small in order to have a small box. Even in the "simple" case of trying to get the most amount of bass out of your driver, a person still must make compromises between low 3 and small box size.

Start with the (O5), to relate acoustic output () to maximum voltage applied to the voice coil (V).

                2   2  2                             -1
             A  Bl  V / / s^2     \2  /  s  \2 \
       = --- ---------- | | ---- - 1 | + | ---- |  |       (TL1)
          2c     2   2  \ \ ^2      /   \ Qs  /  /
                Re  Mm

Use the reference efficiency term (n) (N7) to simplify the coefficient in (TL1).

              2                             -1
             V / / s^2     \2  /  s  \2 \
       = n -- | | ---- - 1 | + | ---- |  |                (TL2)
             Re \ \ ^2      /   \ Qs  /  /

By making the assumption that the voice coil impedance (Ze) is always equal to Re, we can make a simplifying assumption; the power dissipated by the coil (Pe) is given by Pe=V^2/Re. The increased impedance near resonance will only serve to reduce the amount of heat dissipated by the coil. Therefore  can only be under estimated.

                / / s^2     \2  /  s  \2 \-1
       = n Pe | | ---- - 1 | + | ---- |  |                (TL3)
                \ \ ^2      /   \ Qs  /  /

At what frequency is the acoustic output maximum? This is already answered in the Mass Spring Equations.

             /          \-1/2
             |      1   |
       = s | 1 - ---- |              (when Qs>sqrt(1/2))  (TL4)
             |        2 |
             \     2Qs  /

We can put this into the into (TL3) to find the maximum acoustic output () with respect to input power (Pe).

                     Qs^4 
      t = n Pe ------------          (when Qs>sqrt(1/2))  (TL5)
                 /   2      \
                 \ Qs - 1/4 /

When the box and driver is tuned to a quality factor (Qs) below sqrt(1/2) then the maximum acoustic output is at the "point of reference" used in the calculation of reference efficiency (n). In this case acoustic output is a simple formula:

      t = n Pe                       (when Qs<sqrt(1/2))  (TL6)

Using the same method used to find both the maximum displacement (X) due to applied peak voltage (V), and maximum displacement (X) due to forced peak current (I), in the Driver Analysis (eq D1~D10 and Z1~Z5). We will solve for the maximum displacement (X) with respect to forced current in the closed box system. Two equations are used. The first relates the force (Fm) applied to the cone to the inertial, resistive and compliance effects. The second relates this force to the current flowing through the voice coil.

               2
              d x             dx
      Fm = Mm --- + (Rm + Rb) -- + (Km + Kb) x               (G1)
                2             dt
              dt

      Fm = Bl I                                              (G2)

Set (D1) equal to (D2).

                 2
                d x             dx
      Bl I = Mm --- + (Rm + Rb) -- + (Km + Kb) x             (G3)
                  2             dt
                dt

Just like we have done many times before, we can get rid of the differentials by assuming current (I) and displacement (x) vary in a sinusoidal fashion. Let I=Iexp(jt) and x=X*exp(jt).

            jt       2  jt            jt          jt
=>    Bl Ie   = -Mm  Xe  + (Rm+Rb)jXe  + (Km+Kb)Xe        (G4)

Cancel the "e" terms.

                /      2                     \
=>    Bl I = X |- Mm + (Rm+Rb)j + (Km+Kb) |               (G5)
                \                            /

Convert to conventional parameters. But first we need to show an intermediate result that will make that goal manageable;

      s   t   sqrt(Km+Kb)  (Rm + Rv + Rb)    sqrt(Km)     Rv
      -- - -- = ----------- ---------------- - -------- -----------
      Qs   Qe     sqrt(Mm)  sqrt(Mm (Km+Kb))   sqrt(Mm) sqrt(Mm Km)

                Rm + Rv + Rb   Rv   Rm + Rb
              = ------------ - -- = -------
                     Mm        Mm     Mm

With this we can convert (G5).

      Bl I     /   2  / s   t \        2 \
=>    ----- = X |-  + | -- - -- | j + s  |                (G6)
        Mm      \      \ Qs   Qe /          /

              2 2 /            2             2   \-1
         2  Bl I | /   2   2 \   / s   t \  2 |
=>    |X| = ----- | | s -   | + | -- - -- |   |           (G7)
            Mm^2  \ \         /   \ Qs   Qe /    /

Since the electrical power dissipated (Pe) in the voice coil is Pe=I^2*Re we can relate displacement (X) to power dissipated in voice coil (Pe).

               2    /            2             2   \-1
         2   Bl Pe  | /   2   2 \   / s   t \  2 |
=>    |X| = ------- | | s -   | + | -- - -- |   |         (G8)
            Re Mm^2 \ \         /   \ Qs   Qe /    /

Substitute this into our acoustic output equation, (DL1).

                      2    /            2             2   \-1
             2 4  Bl Pe  | /   2   2 \   / s   t \  2 |
=>     = --- A   ------- | | s -   | + | -- - -- |   |  (G9)
          2c      Re Mm^2 \ \         /   \ Qs   Qe /    /

Use reference efficiency to clean up the coefficient:

                   /            2             2   \-1
                 4 | /   2   2 \   / s   t \  2 |
=>     = n Pe   | | s -   | + | -- - -- |   |          (G10)
                   \ \         /   \ Qs   Qe /    /

Simplify:

                /             2                2     \-1
                | / s^2     \   / 1     t   \ s^2 |
=>     = n Pe | | ---- - 1 | + | -- - ----- | ---- |       (G11)
                \ \ ^2      /   \ Qs   s Qe / ^2  /

This is a true representation of the thermally limited acoustic output. The graph shows the acoustic output with respect to frequency; the box is tuned to Qs=sqrt(1/2). The simple formula above does not predict the peak shown. Using (24) of Mass Spring Equations, we can find the maximum acoustic output:

                       / 1     t   \-4
                       | -- - ----- |
                       \ Qs   s Qe /      
      t = n Pe -------------------------   (G12)
                 / / 1     t   \-2   1  \
                 | | -- - ----- |  - --- |
                 \ \ Qs   s Qe /     4  /

This is only valid when

      / 1     t   \
      | -- - ----- | < sqrt(2)
      \ Qs   Qe s /

This longer method takes advantage of the fact that the impedance is higher near resonance. The voltage applied to the voice coil can be increased to compensate. Effectively increasing the output of the driver, without going past the maximum thermal limit.