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I am trying to determine what force is needed for the membrane of an electrostatic speaker (headphones, etc) to move the air at a particular frequency. I know that the formula for the force acting on the plates is

$F=\frac{\epsilon_0AV^2}{2d^2}$

but this is for a "static" approach, as in the plates are polarized, kept at a distance, and the force is determined. It does not account for both the polarizing voltage and the signal, whih is the case for an electrostatic speaker/headphone.

Sometime ago (unfortunately I no longer have the bookmark) I found that Baxandall derived this formula to include the two:

$F=\epsilon_0A\left(\frac{2V_{pol}^2x}{d^3}+\frac{V_{pol}V_{signal}}{d^2}\right)$

where x is the excursion of the membrane for a generic setup like this:

ESL

Also Baxandall determined that for a generic spacing of 2mm, the theoretical maximum force that can be achieved is about 70N, but more likely 50N, in practice.

Now, let's assume the headphones need to have an SPL of $80\text{dB}$, so $0.2\text{Pa}$ [#1]. How do I calculate the force needed to achieve this? Do I simply divide the pressure to the surface area? If so, for i.e. 10x10cm, the force would be $20\text{N}/\text{m}^2$? I know that there is a stretched membrane which adds stiffness, which needs more voltage for more force, all limited by the maximum field intensity, but for the bare calculations, is this enough? Does this not need the mass of the membrane (if it should matter), or the air? If so, where does the frequency term fit in, or is it no longer needed because it's considered a flat response? If so, what if it's voltage driven, or current? A lot of questions seem to diverge. If the way I asked needs clarifications, please tell me before reaching for tomatoes.

[#1] I don't know if this is different for headphones and for speakers, since the former are calculated for a 1mW input and the pressure is taken at the ear level, while the latter is calculated for a 1W input and SPL taken at 1m.

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