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15.12. Static membrane compliance

At very low frequencies, all inertial and resistive elements have very little effect and can therefore be removed from the analogous circuit of Fig. 15.16 to obtain that shown in Fig. 15.17.

To obtain the static membrane and negative compliances C _MD0 and − C _ME0 respectively, we solve the static (in vacuo) membrane wave equation for the displacement η(w) versus the radial ordinate w

(\frac{\partial^{2}}{\partial w^{2}} + \frac{1}{w} \frac{\partial}{\partial w} + \frac{1}{C_{M E} S_{D} T}) η (w) = - \frac{f_{D 0}}{S_{D} T}, η (a) = 0

$(\frac{\partial^{2}}{\partial w^{2}} + \frac{1}{w} \frac{\partial}{\partial w} + \frac{1}{C_{M E} S_{D} T}) η (w) = - \frac{f_{D 0}}{S_{D} T}, η (a) = 0$

(15.68)

where f _D0 is the static membrane driving force, which is given by

f_{D 0} = β e_{in},

$f_{D 0} = β e_{in},$

(15.69)

and β is the electromechanical conversion factor, which is given by

β = 2 C_{E} \frac{E_{P}}{d},

$β = 2 C_{E} \frac{E_{P}}{d},$

(15.70)

where

C_{E} = ε_{0} S_{D} / (2 d) .

$C_{E} = ε_{0} S_{D} / (2 d) .$

Figure 15.17 Simplified analogous circuit for very low frequencies referred to the mechanical side.

We have referred the negative capacitance −C _E to the mechanical side in Eq. (15.68) so that it becomes a negative mechanical compliance

- C_{M E} = - \frac{C_{E}}{β^{2}} = - \frac{d^{2}}{4 C_{E} E_{P}^{2}} .

$- C_{M E} = - \frac{C_{E}}{β^{2}} = - \frac{d^{2}}{4 C_{E} E_{P}^{2}} .$

(15.71)

We solve Eq. (15.68) to obtain

η (w) = \frac{2 f_{D 0} J_{0} (α_{0} w / a)}{π T α_{0} (α_{0}^{2} - {(C_{M E} S_{D} T)}^{- 1} a^{2}) J_{1} (α_{0})}

$η (w) = \frac{2 f_{D 0} J_{0} (α_{0} w / a)}{π T α_{0} (α_{0}^{2} - {(C_{M E} S_{D} T)}^{- 1} a^{2}) J_{1} (α_{0})}$

(15.72)

where α ₀ is the first zero of the Bessel function J ₀. From this we see that the displacement becomes indeterminate when

α_{0}^{2} = {(C_{M E} S_{D} T)}^{- 1} a^{2} .

$α_{0}^{2} = {(C_{M E} S_{D} T)}^{- 1} a^{2} .$

(15.73)

Hence for stability,

T > \frac{2 ε_{0} a^{2}}{α_{0}^{2} d} {(\frac{E_{P}}{d})}^{2} .

$T > \frac{2 ε_{0} a^{2}}{α_{0}^{2} d} {(\frac{E_{P}}{d})}^{2} .$

(15.74)

The average displacement is given by

\begin{matrix} η_{a v} = \frac{1}{π a^{2}} \int_{0}^{2 π} \int_{0}^{a} η (w) w d w d φ = \frac{4 f_{D 0}}{π T α_{0}^{2} (α_{0}^{2} - {(C_{M E} S_{D} T)}^{- 1} a^{2})} \\ = \frac{2 J_{1} (α_{0}) η (0)}{α_{0}} = \frac{η (0)}{2.316} \end{matrix}

$\begin{matrix} η_{a v} = \frac{1}{π a^{2}} \int_{0}^{2 π} \int_{0}^{a} η (w) w d w d φ = \frac{4 f_{D 0}}{π T α_{0}^{2} (α_{0}^{2} - {(C_{M E} S_{D} T)}^{- 1} a^{2})} \\ = \frac{2 J_{1} (α_{0}) η (0)}{α_{0}} = \frac{η (0)}{2.316} \end{matrix}$

(15.75)

Hence, the displacement at the center is about 2.3 times the average, and the total combined static compliance is given by

C_{M T 0} = \frac{η_{a v}}{f_{D 0}} = \frac{4}{π T α_{0}^{2} (α_{0}^{2} - {(C_{M E} S_{D} T)}^{- 1} a^{2})} = {(\frac{1}{C_{M D 0}} - \frac{1}{C_{M E 0}})}^{- 1},

$C_{M T 0} = \frac{η_{a v}}{f_{D 0}} = \frac{4}{π T α_{0}^{2} (α_{0}^{2} - {(C_{M E} S_{D} T)}^{- 1} a^{2})} = {(\frac{1}{C_{M D 0}} - \frac{1}{C_{M E 0}})}^{- 1},$