[davidsh]

From another thread:
 

 Quote:

  Hi davidsh,
 
When you want to change the efficiency of the electrostats, you either change the bias voltage or the spacer thickness.  The capacitance on the film has nothing to do with the efficiency. 
 
What you want to achieve when coating the diaphragm is that you want it to be conductive and hold the charge for as long as possible.  The coating materials that I recommended in my earlier posts will give the resistance of around 10-100 M ohm per square.  Think of the diaphragm as a capacitor and the coating as a bleeding resistor connecting on it.  The bleeding resistor slowly bleeds away the charge.
 
Wachara C.

What puts me off is that coloumbs law states that the force exerted on a static object in an electric field depends on the charge of the object and the magnitude of the electric field (the electric field depends on spacer thickness and the voltage potential between the stators).
The charge that the diaphragm will hold is dependent on the capacitance of the diaphragm as well as bias voltage as I see it. In other words the number of electrons that you can stuff onto the diaphragm.
 
More charge = more electrons = more capacitance and voltage = greater force

Can anyone elaborate? As I see it the capacitance of the diaphragm must one of the determining factors of sensitivity.

 

[wakibaki]

More charge = more electrons = same capacitance and more voltage = more capacitance and same voltage.

 

[davidsh]

Bonus question: What determines the self capacitance of the film (aside from size)?

 

[analogsurviver]

Bonus question: What determines the self capacitance of the film (aside from size)?

The film itself has next to zero, but certainly negligible capacitance compared to the driver without the film. For say a Stax Lambda Pro driver the gap size is 0.5 mm on each side - meaning distance between the two stators is 1 mm of air. The film is approx 1 micro meter thick - that is to say 1000 times thinner. Although the dialectric constant http://hyperphysics.phy-astr.gsu.edu/hbase/tables/diel.html for solids is higher than for air, it is next to zero or negligible in the case of (push-pull) electrostatic driver .

 

[davidsh]

Bonus question: What determines the self capacitance of the film (aside from size)?

The film itself has next to zero, but certainly negligible capacitance compared to the driver without the film. For say a Stax Lambda Pro driver the gap size is 0.5 mm on each side - meaning distance between the two stators is 1 mm of air. The film is approx 1 micro meter thick - that is to say 1000 times thinner. Although the dialectric constant http://hyperphysics.phy-astr.gsu.edu/hbase/tables/diel.html for solids is higher than for air, it is next to zero or negligible in the case of (push-pull) electrostatic driver .

Yes I'm aware, though that was not what I meant
I was thinking about the self capacitance of the film ie. the ability of the film to store a charge when a bias voltage is applied. No charge on film = no sound

 

[wakibaki]

Self capacitance is pretty meaningless in this case. We're interested in mutual capacitance. You can only drive a charge onto something with respect to something else. Ground if all else fails. I mean you have to have something to connect the other side of the PSU to. Even if the connection is invisible (surface resistivity).

Unfortunately you don't have an accurate enough grasp of the meaning of charge, voltage and capacitance to make meaningful discussion possible. We're more concerned with area and separation than the characteristics of the film initially. Go read up.

 

[jcx]

the ideal for ES headphone/loudspeaker is for the charge just to be stuck to the film - to be a good capacitor the charge has to move on and off the electrode in response to the V changing across the terminals
 
for ES the practical implementation is to have a very high resistance coating on the moving diaphragm and charge it up with a bias V
 
the coating does form the plate of a capacitor with each Stator, Q=C*V works for the calculation of Q "trapped" on the diaphragm, C is pretty close to ~A*e_0/t with some edge effects
 
the high sheet resistance of the coating on the diaphragm is intended to make it hard for the charges to move with changing audio frequency signal voltage
 
if you try to make a AC measurement of the high resistance coated diaphragm to Stator capacitance you won't succeed when the meter's test frequency is much higher than the R*C time constant(s)
 
 
technically the dielectric constant of the diaphragm material does enter into the diaphragm to Stator C calc on one side - but the correction is negligible since the film thickness is fractional % of the air spacing

 

[davidsh]

Self capacitance is pretty meaningless in this case. We're interested in mutual capacitance. You can only drive a charge onto something with respect to something else. Ground if all else fails. I mean you have to have something to connect the other side of the PSU to. Even if the connection is invisible (surface resistivity).

Unfortunately you don't have an accurate enough grasp of the meaning of charge, voltage and capacitance to make meaningful discussion possible. We're more concerned with area and separation than the characteristics of the film initially. Go read up.

Of course we are talking relative to ground. Is that enough answer?
What exactly am I to read up on? I am aware of how area and separation affects the sensitivity etc. I am well aware of the relations between voltage, charge and capacitance afaik. I am new to this subject, though, and I will make mistakes wrt terminology.
I don't expect that the characteristics of the film will be of importance per se, but I would like to know more about what affects the capacitance of the film and some typical values etc.

the ideal for ES headphone/loudspeaker is for the charge just to be stuck to the film - to be a good capacitor the charge has to move on and off the electrode in response to the V changing across the terminals

for ES the practical implementation is to have a very high resistance coating and charge with a bias V

the coating does form the plate of a capacitor with each Stator, Q=C*V works for the calculation of Q "trapped" on the diaphragm, C is pretty close to ~A*e_0/t with some edge effects

the high sheet resistance of the coating on the diaphragm is intended to make it hard for the charges to move with changing audio frequency signal voltage

if you try to make a AC measurement of the high resistance coated diaphragm to Stator capacitance you won't succeed when the meter's test frequency is much higher than the R*C time constant(s)


technically the dielectric constant of the diaphragm material does enter into the diaphragm to Stator C calc on one side - but the correction is negligible since the film thickness is fractional % of the air spacing

Informative answer.
You say A•e_0/t
What do you mean by t?
Thanks.

 

[Don Hills]

Time.

 

[jcx]

t = thickness or distance in this case - its just the parallel plate capacitance formula: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html uses d
 
that is a capacitance from the Stator-diaphragm geometry but with the high resistivity coating on the ES diaphragm forming one plate it isn't behaving like a typical capacitor at audio frequencies
 
mostly what you see electrically is the capacitance between the 2 Stators since they are good conductors - although the detailed equations for ES transducer operation do have to account for the displacement current from the charged diaphragm when it moves
 
a noticeable fraction of the C you can measure at the amp end of the cable in ES headphones is the wire-to-wire parasitic C of the cable itself - often special foamed dielectrics are used to keep the cable C down to "only" ~30% of the Stator-to-Stator C
 
 
in any system there are multiple capacitances - the larger ones of concern are those between good conductors in close proximity - but in addition any isolated conductor can also be viewed as having a capacitance to "earth" - hopefully small enough to ignore
 
but the full description is a matrix of capacitances between each possible pairing of however many conductors you are analyzing, including the default "earth" or even "infinity" if you find yourself adrift in space far, far away

 

[davidsh]

I appreciate your reply, but let's assume we've got a diaphragm with perfectly static charges (as that is what we want in an ideal world, right?) for now and no voltage potential between the stators. What would determine the charge on the diaphragm, eg. the capacitance of the diaphragm? This is what I meant with self capacitance by the way. Not sure if wrong..

After all we can't make a proper description of the ESL without knowing the charge on the diaphragm as I see it, whether it being static or varying slightly with frequency/amplitude between stator-stator as seen in the real world.

 

[wakibaki]

There is normally no voltage between the stators, since both are connected to the same side of the PSU. Until you apply a signal.
 
It's the voltage differential between the diaphragm and the stators that maintains the charge. If the stators are connected to -ve, then electrons are pushed onto the stators and pulled off the diaphragm (connected to +ve). Take the applied voltage away and the charge will slowly leak away through surface resistivity and ionic conductivity. As long as the charge doesn't leak away, the voltage between the stators and diaphragm remains. That's the meaning of voltage. Lots of charge, big voltage. Small charge (coulombs), small voltage. In the same geometry, that is.
 
Because the stators have the same voltage, they both attract the diaphragm equally and with no signal it's in equilibrium. When the signal makes the stators alternately more positive or negative in opposition, the diaphragm is pushed and pulled to one side or the other.
 
Capacitors are normally made of conductors. Leaving aside the question of dielectric, the principal feature affecting the capacity of a structure to store charge is geometry. In order to get as far as possible from each other, like charges distribute themselves on the surface of a conductor. That's why we use foil to make caps. The more surface area, the bigger the capacitance, all other things being equal.
 
You'll make progress faster if you just ask questions rather than putting up your assumptions and assertions and waiting to get them shot down.

 

[davidsh]

So as the diaphragm charges the stators act as ground when no signal is passing through?

 

[analogsurviver]

So as the diaphragm charges the stators act as ground when no signal is passing through?

No. Stators (or anything else ) of an electrostatic driver are not acting as ground - the closest to that would be stators connected to ground potential via some pretty high resistance - coupled to high voltage amp trough capacitor, in order not to have the usual half of the high voltage power supply voltage directly on the stators.  Stax SRM1MK2 amp, which is a DC coupled design troughout, places approx 450 V DC directly on the stators - half of its supply of 900 V DC or so.
 
But, in order to understand ESL better, please DO read this first : http://www.hembrow.eu/personal/widerangeelectrostatic.html

 

[wakibaki]

'Ground' or 'Earth' is one of the most troubling concepts for budding engineers to come to terms with. Frequently understanding only dawns when it is required to design an antenna system or RF output amplifier. Earth is not just a place to send large unwanted currents in an emergency.
 
The planet is a conductor. It acts like a huge metal ball. Sometimes we connect our circuits to the Earth, sometimes not, even though they may have a symbol on the schematic that is called 'Ground'.
 
We can design an amplifier to drive a dipole antenna. This is one with 2 prongs. The feeder splits at the junction to the antenna and one side connects to one prong, the other to the other prong. The prongs don't touch. You should be able to visualise a horizontal antenna, 2 equal chunks of conductor in a straight line with a 2 conductor feed wire connected at the middle point. This would be oriented (say) horizontally. Now you can operate this antenna close to the Earth (but not too close, more than a few wavelengths), or faraway in outer space. Charge is pulled off one prong and forced onto the other prong, then this reverses very rapidly. The moving charge causes the casting off of loops of magnetic flux.
 
We can alternatively design an amplifier to drive a (vertical) monopole antenna. In this case one wire of the feeder is connected to the base of the antenna, the other wire is connected to Earth. The connection to earth is usually established by burying numerous conductors in the ground in a circular pattern radiating out from the base of the antenna like spokes of a wheel. Charge rushes onto and off the vertical monopole (and out of and into Earth). Local to the antenna the Earth approximates to a flat plane. The effect is that there is a virtual reflection of the monopole vs. the ground which is like a big shiny ballbearing to RF. You CAN'T operate this antenna in outer space.
 
The ESL assembly is like the dipole antenna, but one of the conductors encloses the other. It operates independent of Ground and it's not really useful to focus on the stray capacitance to Ground but rather we focus on the mutual capacitance between the stators and diaphragm.
 
As jcx has pointed out, for the system to function, it is necessary that the charge on the diaphragm is less mobile than the charge on the stator. For this reason the assembly (described) differs from a capacitor, in that the conductive film has a high resistivity. This means that the time constants of the DC EHT circuit and the AC stator-stator circuit are different, it's quicker to shunt charge from stator to stator than it is from stator(s) to diaphragm. Other approaches have been tried, such as aluminized foil fed through a Megohm+ resistor.