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Old 8th March 2008, 10:37 PM   #1
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Default Capacitance

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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Old 9th March 2008, 12:14 AM   #2
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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?
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Old 9th March 2008, 04:28 PM   #3
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I never thought about that. I would think the diaphragms movement physically needs at least a few watts.
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Old 12th March 2008, 02:36 PM   #4
oshifis is offline oshifis  Hungary
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Quote:
Originally posted by Conrad Hoffman
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.
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Old 12th March 2008, 03:53 PM   #5
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Absolutely right. I'm just trying to make people aware that the amplifier is now "seeing" a lossy capacitance, not the pure capacitance that one might initially suspect.
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Old 12th March 2008, 08:57 PM   #6
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Is there any way to calculate the power being dissipated by the diaphragm in this matter?
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Old 13th March 2008, 08:02 AM   #7
DamageG is offline DamageG  United States
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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%.
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Old 18th March 2008, 12:41 PM   #8
oshifis is offline oshifis  Hungary
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Quote:
Originally posted by DamageG
Typical speakers have an efficiency of a percent or 2 ...
... or rather 0.1%
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Old 19th March 2008, 10:22 PM   #9
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Quote:
Originally posted by ak_47_boy
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...
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Old 20th March 2008, 02:31 PM   #10
oshifis is offline oshifis  Hungary
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Quote:
Originally posted by Brian Beck
By the way, it is interesting to note that there are conditions and frequencies in which the impedance at the drive terminals goes inductive...
Could it be because the mass of the diaphragm converts to inductance when mechanical-to-electrical conversion is made?
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