Acoustics

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9.3. Reference efficiency

In the middle-frequency range many approximations usually can be made to simplify the analogous circuit of Fig. 9.3. Because the drive unit is very small, the mass of the diaphragm and the voice coil M _MD is very small. This in turn usually means that the compliance of the suspension C _MS is large to keep the resonance frequency low. Also, the conductance of the suspension G _MS usually is large, and the reactance ωC _M1 is small. Hence, in this frequency range, the circuit reduces essentially to that of Fig. 9.4a, where the conductance behind the diaphragm is

G_{M B} \equiv \frac{1}{ρ_{0} c S_{D}} m \cdot N/s .

$G_{M B} \equiv \frac{1}{ρ_{0} c S_{D}} m \cdot N/s .$

(9.2)

With the area-changing and electromechanical transformers removed, we get Fig. 9.4b, where the radiation conductance at the throat is

G_{M T} \equiv \frac{S_{T}}{ρ_{0} c S_{D}^{2}} m \cdot N/s .

$G_{M T} \equiv \frac{S_{T}}{ρ_{0} c S_{D}^{2}} m \cdot N/s .$

(9.3)

Figure 9.4 Simplified analogous circuits of the admittance type for the drive unit in the region where the motion of the diaphragm is resistance-controlled by the horn (a) and with the transformers removed (b).

As before, S _T is the area of the throat and S _D is the area of the diaphragm in m². We have assumed here that the cavity behind the diaphragm in this frequency range is nearly perfectly absorbing, which may not always be true. Usually, however, this circuit is valid over a considerable frequency range because of the heavy damping provided by the conductance of the horn G _MT. Also, G _MT usually is smaller than G _MB so that most of the power supplied by the diaphragm goes into the horn.

Solution of Fig. 9.4b gives us

{\tilde{i}}_{2} = \frac{G_{M B}}{G_{M B} + G_{M T}} \tilde{i} .

${\tilde{i}}_{2} = \frac{G_{M B}}{G_{M B} + G_{M T}} \tilde{i} .$

(9.4)

Assuming that the output resistance R _g of the generator is small compared with the coil resistance R _E of the drive unit, the total electrical power supplied from the generator is

Total power supplied = {| \frac{\tilde{i}}{\sqrt{2}} |}^{2} (R_{E} + B^{2} l^{2} \frac{G_{M B} G_{M T}}{G_{M B} + G_{M T}}) .

$Total power supplied = {| \frac{\tilde{i}}{\sqrt{2}} |}^{2} (R_{E} + B^{2} l^{2} \frac{G_{M B} G_{M T}}{G_{M B} + G_{M T}}) .$

(9.5)

Using the solution of Eq. (9.4), the reference efficiency E _ff is equal to the power delivered to the horn,

{| {\tilde{i}}_{2} / \sqrt{2} |}^{2} B^{2} l^{2} G_{M T},

${| {\tilde{i}}_{2} / \sqrt{2} |}^{2} B^{2} l^{2} G_{M T},$

times 100 divided by the total power supplied:

E_{f f} = \frac{{(G_{M B} / (G_{M B} + G_{M T}))}^{2} B^{2} l^{2} G_{M T}}{R_{E} + B^{2} l^{2} G_{M B} G_{M T} / (G_{M B} + G_{M T})} \times 100 .

$E_{f f} = \frac{{(G_{M B} / (G_{M B} + G_{M T}))}^{2} B^{2} l^{2} G_{M T}}{R_{E} + B^{2} l^{2} G_{M B} G_{M T} / (G_{M B} + G_{M T})} \times 100 .$

(9.6)

From Eqs. (9.2, 9.3 and 9.6) we get

E_{f f} = \frac{100 B^{2} l^{2} (S_{T} / S_{D})}{(1 + S_{T} / S_{D}) ((B^{2} l^{2} + S_{D} ρ_{0} c R_{E}) (S_{T} / S_{D}) + S_{D} ρ_{0} c R_{E})}

$E_{f f} = \frac{100 B^{2} l^{2} (S_{T} / S_{D})}{(1 + S_{T} / S_{D}) ((B^{2} l^{2} + S_{D} ρ_{0} c R_{E}) (S_{T} / S_{D}) + S_{D} ρ_{0} c R_{E})}$

(9.7)

or, in terms of Thiele–Small parameters,

E_{f f} = \frac{100 (S_{T} / S_{D}) S_{D} c}{(1 + S_{T} / S_{D}) ((S_{D} c + ω_{S} Q_{E S} V_{A S}) (S_{T} / S_{D}) + ω_{S} Q_{E S} V_{A S})},

$E_{f f} = \frac{100 (S_{T} / S_{D}) S_{D} c}{(1 + S_{T} / S_{D}) ((S_{D} c + ω_{S} Q_{E S} V_{A S}) (S_{T} / S_{D}) + ω_{S} Q_{E S} V_{A S})},$

(9.8)

where we have used Eqs. (6.27) and (6.30) for the Bl factor. We note that the value of G _MT, and hence the ratio S _T/S _D would seem to need to be large for high efficiency. However, if S _T/S _D becomes too large, reference to Fig. 9.4b shows that too much power will be dissipated in G _MB and the efficiency will be low. To optimize the efficiency, let us now differentiate the above with respect to (S _T/S _D) and equate the result to zero. Hence maximum efficiency occurs when

\frac{S_{T}}{S_{D}} = \sqrt{\frac{S_{D} ρ_{0} c R_{E}}{B^{2} l^{2} + S_{D} ρ_{0} c R_{E}}} = \sqrt{\frac{ω_{S} Q_{E S} V_{A S}}{S_{D} c + ω_{S} Q_{E S} V_{A S}}},

$\frac{S_{T}}{S_{D}} = \sqrt{\frac{S_{D} ρ_{0} c R_{E}}{B^{2} l^{2} + S_{D} ρ_{0} c R_{E}}} = \sqrt{\frac{ω_{S} Q_{E S} V_{A S}}{S_{D} c + ω_{S} Q_{E S} V_{A S}}},$

(9.9)

so that the maximum efficiency is

E_{f f} (max) = \frac{100 B^{2} l^{2}}{{(\sqrt{B^{2} l^{2} + S_{D} ρ_{0} c R_{E}} + \sqrt{S_{D} ρ_{0} c R_{E}})}^{2}},

$E_{f f} (max) = \frac{100 B^{2} l^{2}}{{(\sqrt{B^{2} l^{2} + S_{D} ρ_{0} c R_{E}} + \sqrt{S_{D} ρ_{0} c R_{E}})}^{2}},$

(9.10)

which can also be given in terms of Thiele–Small parameters:

E_{f f} (max) = \frac{100 S_{D} c}{{(\sqrt{S_{D} c + ω_{S} Q_{E S} V_{A S}} + \sqrt{ω_{S} Q_{E S} V_{A S}})}^{2}} .

$E_{f f} (max) = \frac{100 S_{D} c}{{(\sqrt{S_{D} c + ω_{S} Q_{E S} V_{A S}} + \sqrt{ω_{S} Q_{E S} V_{A S}})}^{2}} .$

(9.11)

To increase the efficiency further, it is seen from Eq. (9.10) that the length l of the wire on the voice coil should be increased as much as possible without altering electrical resistance R _E. Within given space limitations, this can be done by winding the voice coil from wire with a rectangular cross section rather than with a circular cross section. This means that the voice-coil mass will be increased. Increasing l further will demand a wire of larger cross section, which will require a larger air gap, with a corresponding reduction in B or increase in magnet size. Also, the voice coil must not become too large as its mass will limit the high-frequency response.

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9.3. Reference efficiency

Table of Contents for
Acoustics