Acoustics

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6.11. Examples of loudspeaker calculations

Example 6.1. Given the efficiency of Eq. (6.47) for a loudspeaker in an infinite baffle, determine the reference sound pressure equivalent to the efficiency assuming that the directivity factor Q (for radiation to one side) equals 2.

Solution. The sound pressure at distance r, assuming no directivity, is related to the acoustic power radiated to one side as follows (see Eqs. 6.40 and 6.46):

p_{rms} (r) = \sqrt{ρ_{0} c I} = \sqrt{\frac{ρ_{0} c W_{1}}{2 π r^{2}}}

$p_{rms} (r) = \sqrt{ρ_{0} c I} = \sqrt{\frac{ρ_{0} c W_{1}}{2 π r^{2}}}$

where I is intensity at distance r. W ₁ = W/2 is total acoustic power radiated from one side of the diaphragm.

The reference sound pressure is

\begin{matrix} p_{rms} (r) = \sqrt{\frac{ρ_{0} c}{2 π r^{2}}} \sqrt{\frac{E_{f f}}{200} W_{E}} \\ = \frac{e_{g (rms)} ρ_{0} B l S_{D}}{2 π r R_{E} (M_{M D} + 2 M_{M 1})} \end{matrix}

$\begin{matrix} p_{rms} (r) = \sqrt{\frac{ρ_{0} c}{2 π r^{2}}} \sqrt{\frac{E_{f f}}{200} W_{E}} \\ = \frac{e_{g (rms)} ρ_{0} B l S_{D}}{2 π r R_{E} (M_{M D} + 2 M_{M 1})} \end{matrix}$

which is that given by Eq. (6.32) in the passband where α _c ≈ 1 and it is assumed that R _g ≪ R _E.

Example 6.2. As an example of the efficiency to be expected from an electrodynamic loudspeaker of conventional design mounted in an infinite baffle and radiating directly from both sides of the baffle, let us calculate the reference efficiency E _ff from Eq. (6.48) for the case of a commercial loudspeaker with an advertised diameter of 10 cm. In addition, let us calculate the ratio of the efficiency at the suspension resonance frequency to the reference efficiency. The values of the six Thiele–Small parameters are as follows:

R _E = 7 Ω
Q _ES = 2.2
Q _MS = 5
f _S = 125 Hz
S _D = 56 cm²
V _AS = 0.002 m³ (2 L)

Also

a = \sqrt{S_{D} / π} = 42.22 mm

$a = \sqrt{S_{D} / π} = 42.22 mm$

Q_{T S} = \frac{Q_{E S} Q_{M S}}{Q_{E S} + Q_{M S}} = 1.528

$Q_{T S} = \frac{Q_{E S} Q_{M S}}{Q_{E S} + Q_{M S}} = 1.528$

ρ ₀, density of air = 1.18 kg/m³
c, speed of sound = 344.8 m/s.

Solution. From Eq. (6.48), we obtain

E_{f f} = 100 \frac{8 \times {(3.14)}^{2} \times 0.002 \times {(125)}^{3}}{2 \times {(344.8)}^{3}} = 0.342 %

$E_{f f} = 100 \frac{8 \times {(3.14)}^{2} \times 0.002 \times {(125)}^{3}}{2 \times {(344.8)}^{3}} = 0.342 %$

For radiation from one side of the loudspeaker only, divide this figure by 2. Hence, only 0.171% of the available electrical power is radiated to one side of the diaphragm at mid to low frequencies. This illustrates the statement made at the beginning of this chapter that the efficiency of this type of loudspeaker is usually low.

For our example, the transition frequency at which the loudspeaker starts to become more directional and the efficiency decreases rapidly occurs when ka lies approximately between 1 and 2. For our example, ka = 1 corresponds to a frequency of

f = \frac{c}{2 π a} = \frac{344.8}{2 π \times 0.04222} = 1.3 kHz

$f = \frac{c}{2 π a} = \frac{344.8}{2 π \times 0.04222} = 1.3 kHz$

Obviously, a smaller diaphragm of lighter weight would result in this transition extending to a higher frequency. However, a reduction in the mass M _MD occasioned by a smaller diaphragm will cause an increase in the first resonance frequency with a resulting loss in bass response. A further disadvantage of a smaller diaphragm is that, for a given sound pressure, a greater voice-coil velocity

{\tilde{u}}_{c}

${\tilde{u}}_{c}$

is needed. A longer air gap and a larger magnet structure must therefore be provided.

From Eq. (6.54), we obtain the efficiency at the suspension resonance:

E_{f f S} = 1.528 \times 5 \times 0.342 = 2.61 %

$E_{f f S} = 1.528 \times 5 \times 0.342 = 2.61 %$

Hence, the ratio E _ffS/E _ff equals 7.64, and efficiency at f _S equals 1.3% for radiation from one side only.

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6.11. Examples of loudspeaker calculations

Table of Contents for
Acoustics