Capacitance

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The capacitance between the diaphragm and stators does not draw amplifier power because the high resistance diaphragm/resistor prevents current flow. The capacitance between the two stators is what the amplifier sees.

Correct?
 

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I don't have much knowledge of electrostatic speakers, but if your drawing is correct, so are you.

An interesting thing is that a pure reactance can't dissipate any power. Since the speaker does work, i.e., produces sound, it must absorb amplifier power. Therefore, the speaker must have a resistive component no matter how good the dielectric (air?) might be. IOW, it must have a small but not insignificant dissipation factor in operation. Just curious- has anybody attempted to measure this?
 
Conrad Hoffman said:
Since the speaker does work, i.e., produces sound, it must absorb amplifier power. Therefore, the speaker must have a resistive component no matter how good the dielectric (air?) might be. IOW, it must have a small but not insignificant dissipation factor in operation.
That disssipation factor is the acoustical resistance of air and the radiated acoustical energy is dissipated on it.
 
The acoustical power would be dissipated in the air and not in the diaphragm. Although it would appear to the amplifier like it was.

Typical speakers have an efficiency of a percent or 2 so the effective dissipation factor of the ESL capacitor would be fairly small.

Something to ponder...

If you were to drive an ESL with a Class D amplifier, you would draw almost no power from the power supply. Unlike a linear amplifier, the Class D amp can charge and discharge the capacitance with an ideal efficiency of 100%. Where the ideal efficiency of a linear amplifier driving and ESL is near 0%.
 
ak_47_boy said:
Is there any way to calculate the power being dissipated by the diaphragm in this matter?

Prof. F.V. Hunt covers the topic of ESL impedance across several dozen pages in his definitive book "Electroacoustics". Essentially, the radiation resistance is the resistive part of the air load on both sides of the diaphragm which is dependent upon panel shape and dimensions. It is the part which captures the "work" that is being done when sound is made, electrical energy being converted to air molecule movement. This radiation resistance is transformed back to the drive terminals as a resistive component to the generally capacitive impedance. If there were no resistive component to the impedance, there could be no sound. The math is pretty intense for most DIYers, and probably wouldn't help you design better ESLs anyway. Prof. Hunt measured ESLs in air and in a vacuum, and showed how the impedance changed because the radiation resistance was removed in a vacuum. By the way, it is interesting to note that there are conditions and frequencies in which the impedance at the drive terminals goes inductive...
 
The diaphragm may take a small amount of energy to make sound but the majority of work is done by the amplifier charging and discharging the panel capacitance at the audio frequency. This causes two types of losses in the amplifier: switching loss and conduction loss which account for the heat generated by an amp driving an ESL. An ESL is very efficient if you don't include the amplifier loss. Another way to look at it: the amplifier sends current to the ESL but the ESL returns most of it which the amp must dissipate. This is often referred to as circulating current.
 
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