Acoustics

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3.7. Examples of transducer calculations

Example 3.6. An ideal moving-coil loudspeaker produces 2 W of acoustic power into an acoustic load of 4 × 10⁴ N s/m⁵ when driven from an amplifier with a constant voltage output of 1.0 V rms. The area of the diaphragm is 100 cm². What open-circuit voltage will it produce when operated as a microphone with an rms diaphragm velocity of 10 cm/s?

Solution. From Fig. 3.35 we see that, always,

\tilde{e} = B l \tilde{u} .

$\tilde{e} = B l \tilde{u} .$

The power dissipated W gives us the rms volume velocity of the diaphragm U _rms:

U_{rms} = \sqrt{\frac{W}{R_{A}}} = \sqrt{\frac{2}{4 \times 10^{4}}} = 7.07 \times 10^{- 3} m^{3} /s

$U_{rms} = \sqrt{\frac{W}{R_{A}}} = \sqrt{\frac{2}{4 \times 10^{4}}} = 7.07 \times 10^{- 3} m^{3} /s$

u_{rms} = 0.707 m / s,

$u_{rms} = 0.707 m / s,$

B l = \frac{e_{rms}}{u_{rms}} = \frac{1}{0.707} = 1.414 T \cdot m

$B l = \frac{e_{rms}}{u_{rms}} = \frac{1}{0.707} = 1.414 T \cdot m$

Hence, the open-circuit voltage for an rms velocity of 0.1 m/s is

e_{rms} = 1.414 \times 0.1 = 0.1414 V .

$e_{rms} = 1.414 \times 0.1 = 0.1414 V .$

Example 3.7. A lead magnesium niobate–lead titanate (PMN-PT) crystal as shown in Fig. 3.36 with w = 0.5 mm, d = 2 mm, and h = 2 mm has the following mechanical and electrical properties:

d ₃₁ = 750 × 10⁻¹² C/N or m/V
Y = 20 × 10⁹ N/m²
ρ = 8000 kg/m³
ε ₀ = 8.85 × 10⁻¹² F/m
ε _r = 6500

k_{31} = d_{31} \sqrt{\frac{Y}{ε_{0} ε_{r}}} = 0.442

$k_{31} = d_{31} \sqrt{\frac{Y}{ε_{0} ε_{r}}} = 0.442$

This crystal is to be used in a microphone with a circular (weightless) diaphragm. Determine the diameter of the diaphragm if the microphone is to yield an open-circuit voltage of −70 dB re 1 V rms for a sound pressure level of 74 dB re 20 μPa at 100 kHz.

Solution. The circuit for this transducer with the transformer removed is shown in Fig. 3.39, where the circuit elements are defined by

{C^{'}}_{M} = (1 - k_{31}^{2}) \frac{h}{Y w d} = \frac{1 - {0.442}^{2}}{20 \times 0.5 \times 10^{6}} = 8 \times 10^{- 8} m / N,

${C^{'}}_{M} = (1 - k_{31}^{2}) \frac{h}{Y w d} = \frac{1 - {0.442}^{2}}{20 \times 0.5 \times 10^{6}} = 8 \times 10^{- 8} m / N,$

M_{M} = \frac{4}{π^{2}} ρ w d h = \frac{4 \times 8 \times 0.5 \times 2 \times 2 \times 10^{- 6}}{π^{2}} = 6.5 \times 10^{- 6} kg / m^{3},

$M_{M} = \frac{4}{π^{2}} ρ w d h = \frac{4 \times 8 \times 0.5 \times 2 \times 2 \times 10^{- 6}}{π^{2}} = 6.5 \times 10^{- 6} kg / m^{3},$

C_{E} = \frac{ε_{0} ε_{r} w h}{d} = 8.85 \times 10^{- 12} \times 6.5 \times 0.5 = 28.8 \times 10^{- 12} F,

$C_{E} = \frac{ε_{0} ε_{r} w h}{d} = 8.85 \times 10^{- 12} \times 6.5 \times 0.5 = 28.8 \times 10^{- 12} F,$

R_{M} = negligibly small .

$R_{M} = negligibly small .$

Because only the open-circuit voltage is desired, C_E may be neglected in the calculations. f _rms is the total force applied to the crystal by the diaphragm. Solving for e _rms yields

e_{rms} = \frac{f_{rms} (d_{31} / C_{E})}{1 - ω^{2} M_{M} C_{M}^{'}} .

$e_{rms} = \frac{f_{rms} (d_{31} / C_{E})}{1 - ω^{2} M_{M} C_{M}^{'}} .$

Figure 3.39 Analogous circuit of the impedance type for a crystal microphone.

The force f equals the area of the diaphragm S times the sound pressure p. Solving for p,

\begin{matrix} p_{rms} = 20 \times 10^{- 6} \times 10^{74 / 20} \\ = 0.1 N / m^{2} \end{matrix}

$\begin{matrix} p_{rms} = 20 \times 10^{- 6} \times 10^{74 / 20} \\ = 0.1 N / m^{2} \end{matrix}$

Solving for e,

\frac{1}{e_{rms}} = 10^{70 / 20} = 3.16 \times 10^{3},

$\frac{1}{e_{rms}} = 10^{70 / 20} = 3.16 \times 10^{3},$

e_{rms} = 3.16 \times 10^{- 4} V .

$e_{rms} = 3.16 \times 10^{- 4} V .$

Hence,

\begin{matrix} S = \frac{f_{rms}}{p_{rms}} = \frac{3.16 \times 10^{- 4} (1 - {6.28}^{2} \times 10^{10} \times 6.5 \times 8 \times 10^{- 14})}{0.1 \times (750 / 28.8)} \\ =9 .65 \times 10^{- 5} m^{2} \\ S = 0.965 {cm}^{2} . \end{matrix}

$\begin{matrix} S = \frac{f_{rms}}{p_{rms}} = \frac{3.16 \times 10^{- 4} (1 - {6.28}^{2} \times 10^{10} \times 6.5 \times 8 \times 10^{- 14})}{0.1 \times (750 / 28.8)} \\ =9 .65 \times 10^{- 5} m^{2} \\ S = 0.965 {cm}^{2} . \end{matrix}$

This corresponds to a diaphragm with a diameter of about 1.1 cm.

Example 3.8. A loudspeaker diaphragm couples to the throat of an exponential horn that has an acoustic impedance of (300 + j300) N·s/m⁵. If the area of the loudspeaker diaphragm S _D is 0.08 m², determine the mechanical impedance load on the diaphragm because of the horn.

Solution. The analogous circuit is shown in Fig. 3.40. The mechanical impedance at terminals 1 and 2 represent the load on the diaphragm:

Figure 3.40 Example of a mechanoacoustic transducer. The acoustic impedance of a horn (at terminals 3 and 4) loads the diaphragm with a mechanical impedance S _D ²(300 + j300) N s/m.

\begin{matrix} Z_{M} = \frac{\tilde{f}}{\tilde{u}} = S_{D}^{2} (300 + j 300) \\ = 6 .4 \times 1 0^{- 3} (300 + j 300) \\ = 1 .92+ j 1 .92 N \cdot s / m . \end{matrix}

$\begin{matrix} Z_{M} = \frac{\tilde{f}}{\tilde{u}} = S_{D}^{2} (300 + j 300) \\ = 6 .4 \times 1 0^{- 3} (300 + j 300) \\ = 1 .92+ j 1 .92 N \cdot s / m . \end{matrix}$

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Table of Contents for Acoustics

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3.7. Examples of transducer calculations

Table of Contents for
Acoustics