[isw]

In article ,
Greg Berchin wrote:

On 6 Mar 2017 12:56:47 GMT, abbeynormal wrote:

can you tell me your impression of what electrostats would sound like if
somehow they were driven differently?

Beats me. I have found that trying to predict how something will sound by
analyzing the math alone generally doesn't work very well.

I can say this, though. High school physics tells us that, when driven by
voltage, the force on the diaphragm is:

(Vbias+Vaudio)^2 (Vbias-Vaudio)^2
F = e0A--------------------- - e0A---------------------,
2[(|dFR|/2)-deltaD]^2 2[(|dFR|/2)+deltaD]^2

where e0 = the permittivity of free space
A = the area of the diaphragm
Vbias = the bias voltage applied to the stators
Vaudio = the audio signal voltage
|dFR| = the distance (gap) between the front and rear stators
deltaD = how far the diaphragm has moved from its center rest position

Note that the force changes as deltaD changes, i.e., when the diaphragm moves
closer to one of the stators and farther from the other, even when Vbias and
Vaudio are held constant.

But when driven by charge, the force on the diaphragm is:

|qbias*qaudio|
|F| = 2--------------
e0A

where qbias = the bias charge applied to the stators
qaudio = the charge applied by the audio signal
and the direction of the force can be determined by context.

Note that the force is constant as long as the charges are constant,
regardless
of the position of the diaphragm.

Now, how do we design a charge amplifier? In theory, it's not all that
difficult. Hint: electrostatic loudspeakers are capacitors, capacitors store
charge, and current integrated over time equals charge.

Some electrostatic speakers use a material with extremely high surface
restivity for the diaphragm (made e.g. by lightly rubbing the plastic
sheet with a carbon-dust-coated pad). The claim is that doing it that
way puts the diaphragm in a "constant charge" mode (IIRC).

Isaac