I always wanted to ask...why Bias voltage on ESL?

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OK, since I am not an Electrical Engineer (Mechanical) I have always been puzzled why we need bias voltage on ESL diaphragm.

The AC amp voltage out (assume 10V) enters single step up tranny (assume 70:1), creating voltage on the stators of +350VAC and -350VAC (700VAC across stators). The diaphragm in between the stators is at +5000VDC. Doesn't the stator at -350VAC have a bigger delta (5350V) than the +350VAC stator (4650V) causing more "pull" than "push", thus holding back the diaphragm? Why not a much lesser voltage to reduce the delta, or Zero V (magnetized diaphragm?) for true push / pull? This example is when amp's speaker out is referenced to ground. Should ESLs work "better" when your amp outs are bridged (both speaker outs are opposite to each other and not zero)

Why not make the stators fixed voltage at +5000VDC and -5000VDC and the Diaphragm voltage stepped up from amp for true push pull?

Am I thinking of this all wrong? Am I just an electrical retard? Sorry for the lame question...

There I feel better now....I can continue to fart with my ML CLS and Acoustat panels with a little more understanding after someone here clues me in.
 
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Oh hell, I am just going to ask. To eliminate the above "push / pull imbalance" situation, I was thinking of doing the following:

+5000VDC and -5000VDC Bias on stators (or much less voltage). Mylar in between was to have magnepan style wire running up and down the mylar (equiv 4 ohm) and the amp drives the mylar/wire like a magnepan speaker between teh two stators. I have CLS Panels and was going to try a few experiments. I know I need to be careful of the panels and make sure the I don't arc...

Crazy? Maybe I should just stick with building Diesel Hydrotreaters (I work in the oil industry)
 
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Hello john65b,

Lots of good and inter-related questions.
I won't have time till this afternoon to properly cover them and put together some pics to illustrate.
I will try and cover the following topics which should provide the answers you want.

1) Push-Pull forces in a constant-voltage ESL (ie low resistance diaphragm)
2) Forces in a constant-charge ESL (ie high resistance diaphragm)
3) What happens to the forces in 1) & 2) when diaphragm is positioned away from center.
4) Comparison between 1) and the configuration you suggest with the equal and opposite bias voltages on the stators.

BTW, the equal and opposite biased stator configuration has been used commercially by FINAL, and in a modified form by Beveridge. It is sometimes called an inverted ESL. Attached FINAL white paper and "patent"(cough, cough). Not sure how you get a patent on something that is public knowledge and discussed in patents as far back as the 1929 Kellog patent. In any case...some reading material on the subject.
 

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At a very basic level you can reconcile the apparent imbalance in push vs pull by considering only the push or pull “increment” relative to the zero signal forces on the diaphragm. Take a look at Attachment #1 and notice that in Figure (A) with 5kV applied to the diaphragm and 0kV on both stators, the diaphragm feels an attractive force for both front and rear stators. If the diaphragm is centered between the stators the forces will be equal and thus will cancel out since they act in opposing directions. Now if we applied +1kV to the front stator and -1kV to the rear stator as in Figure (B), you would get the apparent imbalance in push vs pull shown by the blue arrows. But if you break the blue arrows down into the zero signal force(red) and the increments(green) you can see that there are equal push and pull “increments”

1) A more in depth look at the Push-Pull constant voltage ESL:
The constant voltage ESL uses a highly conductive diaphragm and a lower value charging resistor between the HV supply and the diaphragm to keep the voltage on the diaphragm constant. The electrostatic attractive force between two objects like the diaphragm and a stator is proportional to the square of the difference in voltage potential and inversely proportional to the square of the distance between them. If the diaphragm stayed centered between the stators as it would when playing mid and high frequencies, we can plot relative forces on the diaphragm from the stators. Notice in Attachment #2 the static force generated by each of the stators does not increase linearly with increasing stator voltage. It varies with the square of the applied stator voltage. But, notice what happens when we sum the forces on the diaphragm from both stators…the resultant shown in green varies linearly with stator voltage. So, this one of the key reasons the push-pull ESL is the standard configuration. The push pull arrangements cancels out the distortion caused by the square-law static forces.

2) Now a brief look at the Push-Pull constant Charge ESL:
Rather than trying to keep the voltage on the diaphragm constant, this arrangement keeps the charge on the diaphragm constant by using a highly resistive coating. Another basic law of electrostatic forces concerns charges placed in electrostatic field. When the two stators receive voltages of opposite polarity an electrostatic field is generated between them with strength directly proportional to the voltage and inversely proportional to the distance between them. The field strength is the same across the entire space between the stators. A charge placed in the field will be acted on by a force proportional to the field strength and the charge value. If a highly resistive diaphragm was charged up with the same bias voltage used in the constant voltage example above, the charge deposited on the diaphragm would be such that the calculated force on the diaphragm centered in the gap would be the same as the green summation in the first example. See Attachment #3.

3) What happens to the forces when the diaphragm moves away from center?
From 1) & 2) above it may look like there is no need for the inherently linear force generated by the constant charge arrangement since the non-linearities of the constant voltage configuration cancel out. Well, that is only true if the diaphragm is centered. As soon as the diaphragm moves away from center the constant charge configuration comes into its own. Remember that the electrostatic field strength is constant across the whole gap. With a highly resistive diaphragm, the HV supply cannot act to change the diaphragm charge during an audio cycle motion of the diaphragm. So with charge constant, and field strength constant, the force on the diaphragm will be constant for any position of the diaphragm in the gap. For example suppose a 2kV midrange signal was applied while the diaphragm was also undergoing a large low frequency motion. Attachment #4 shows that the force due to the 2kV midrange signal is the same for all diaphragm positions as it was when centered. Now look at what happens for the constant voltage ESL when handling the same situation. Attachment #5 shows that as the diaphragm moves closer to a given stator the associated force increases dramatically since it is inversely proportional to the square of the distance between the diaphragm and the stator. The resulting summation of forces shown in green is non-linear as well.

Let me know if any of this doesn’t make sense, and I’ll try to clarify before moving on to the inverted ESL configuration.
 

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I think I am getting this, seems maybe the lower resistance offers up the ability to restore charge quicker on constant voltage, while the higher resistance constant slows down charge refresh (and loss of charge)?

But I don't see how the constant charge can be any different from constant voltage. Doesn't a stator at +/-5KV have the same affinity of attraction to +1000V of voltage and of +1000V of charge on a diaphragm resulting in the same force on diaphragm?? Regardless of location between stators?

I think what you are saying is the closer the constant voltage diaphragm gets to a stator A, the force continues to increase toward stator A and lessens on Stator B, while constant charge the forces are always equal on both stator A and B? The imbalance I am trying to wrap my head around only exist on constant voltage and constant charge is somewhat immune to the imbalance?
 
I think I am getting this, seems maybe the lower resistance offers up the ability to restore charge quicker on constant voltage, while the higher resistance constant slows down charge refresh (and loss of charge)?

But I don't see how the constant charge can be any different from constant voltage. Doesn't a stator at +/-5KV have the same affinity of attraction to +1000V of voltage and of +1000V of charge on a diaphragm resulting in the same force on diaphragm?? Regardless of location between stators?

I think what you are saying is the closer the constant voltage diaphragm gets to a stator A, the force continues to increase toward stator A and lessens on Stator B, while constant charge the forces are always equal on both stator A and B? The imbalance I am trying to wrap my head around only exist on constant voltage and constant charge is somewhat immune to the imbalance?

There is a BIG difference between constant voltage and constant charge. Since the operation of the ESL depends on the charge on the diaphragm, and the resulting force of attraction and replusion to the stators, it is desirable to keep the charge as constant as possible to maintain best linearity (i.e. low distortion). You will find that most (if not all) commercial ESL's use constant charge, not constant voltage. This is the purpose of the typical large value resistor between the bias power supply and the diaphragm (500 megohms for Acoustat). This resistor converts the constant voltage of the power supply to a constant charge on the high-resistance coating of the diaphragm. You do NOT want fast changes in the charge on the diaphragm, as this will result in higher distortion.

Attempting to make an 'inverted' ESL using the more desirable constant-charge method will be very difficult. I don't know how would you make the stators a high resistance circuit, in order to keep the charge constant. Better to stick with the more traditional approach of keeping a constant charge on a high-resistance diaphragm, and applying the push-pull audio to the stators of your choice.

This same concept is also true for conventional magnet/voice coil speakers. Designers take great pains to keep the magnetic force as constant as possible for all positions of the voice coil, and hence lowering distortion. Constant force is a good thing, whether it be due to magnetic or electrostatic forces.
 
…I think what you are saying is the closer the constant voltage diaphragm gets to a stator A, the force continues to increase toward stator A and lessens on Stator B, while constant charge the forces are always equal on both stator A and B?
Yes, with constant voltage on the diaphragm the closer it gets to stator A the greater the rate of increase of the force towards stator A, and the greater the rate of decrease of the force toward stator B. A decidedly nonlinear situation.

With constant charge on the diaphragm, it is best to visualize the force on it as coming from the electrostatic field between the stators rather than from each stator separately. The field strength is constant at any position between the stators so the force on the charged diaphragm is the same no matter its position in the gap.
Basically the electrostatic equivalent of magnetic lines of flux in a woofer voice coil gap.
Field strength = (stator voltage difference)/(stator separation distance)

I am missing the difference between "Charge" and "Voltage" on the Diaphragm....maybe a quick explanation here before going on to Inverted would be good...

Google and Wikipedia will provide you with a plethora of scientific/technical definitions for charge and voltage.
I will try to give a simplified description of them that will relate easily to ESLs.

Most all materials have a natural balance of negative charged electrons and positively charged protons. If electrons are added to a material by some means it will have an excess of electrons and will be negatively charged. If some electrons are removed from the material by some means it will then have a shortage of electrons and will be positively charged. In short, charge is a measure of the imbalance in the number of electrons and protons in a material.

Voltage potential can be thought of as an electrical property measuring the ability to make charges(ie electrons) move. The higher the voltage the more charges that can be moved in a given time, or alternatively the shorter the time it takes to move a given amount of charges.

It so happens that since an ESL diaphragm and stator system form a capacitor there is a direct relationship between charge and voltage.
They can be thought of as different sides of the same coin.
Q = V x C or in words…
(charge on diaphragm) = (stator to diaphragm voltage) x (stator to diaphragm capacitance)

C = k x A / d
(capacitance) = (constant) x (stator area) / (stator to diaphragm distance)

This charge/voltage relationship has the following consequence:
1) For constant voltage diaphragm, as the diaphragm is moved towards a stator the charge on it increases.
2) For constant charge diaphragm, as the diaphragm is moved towards a stator the voltage on it decreases.

Taking this relationship one step further:
1) For constant voltage diaphragm, we can calculate the charge on the diaphragm for any position and use the charge/field strength relationship to calculate the resulting force which will give exactly the expected non-linear trend for a constant voltage diaphragm ESL.
2) For a constant charge diaphragm, we can calculate the voltage on the diaphragm for any position and use the stator/diaphragm voltage and distance relationship to calculate the force on the diaphragm for each stator. Summing these two forces will give the expected linear trend for a constant charge diaphragm ESL.
 
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You will find that most (if not all) commercial ESL's use constant charge, not constant voltage. This is the purpose of the typical large value resistor between the bias power supply and the diaphragm (500 megohms for Acoustat). This resistor converts the constant voltage of the power supply to a constant charge on the high-resistance coating of the diaphragm.
The large value resistor does keep the total charge on the diaphragm constant, but does not keep the charge on the diaphragm from moving around to areas closest to stators. This is most obvious when the diaphragm is operating in the bass range near a resonance mode and standing waves exist on the diaphragm. Besides distortion reduction, an equally important purpose of the large value resistor was first pointed out by Hunt I believe. It provides stability for operation at low frequencies. Without it, a push-pull ESL with low resistance diaphragm exhibits diaphragm to stator collapse when diaphragm motion exceeded roughly half the gap.

Attempting to make an 'inverted' ESL using the more desirable constant-charge method will be very difficult. I don't know how you would make the stators a high resistance circuit, in order to keep the charge constant. Better to stick with the more traditional approach of keeping a constant charge on a high-resistance diaphragm, and applying the push-pull audio to the stators of your choice.
I certainly prefer the many benefits of constant charge ESLs.
However, inverted ESLs do have a few benefits; the biggest is the ability to get 6dB more output for the same transformer step-up ratio.
http://www.diyaudio.com/forums/planars-exotics/199943-measuring-sensitivity-esls.html#post2779302

Diaphragms do have to be highly conductive, so no chance of constant charge operation. However, distortion can be minimized in the critical frequency range > 400Hz by using a separate bass panel, so that the section producing the frequencies > 400Hz can operate with the diaphragm centered between the stators and minimal motion about the gap producing distortion.
 
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OK, I know the basic difference - using water analogy - Charge (gallons), Current (gallons per min), Voltage (water pressure), but was hung on thinking charge reduction would equate to a voltage reduction...I now understand much better as this is not necessarily the case.

Now since the Constant Voltage is 6db greater SPL than Constant Charge, what if I just delete one stator voltage on Constant Voltage? So have my 70:1 step up from amp on the high conductivity diaphragm (I have tons of aluminized 12micron Mylar) and have back stator at -3kv bias, front stator grounded / floating. From what I am reading, should have same SPL as constant charge ESL, and operate much like a Magnepan Single ended (not push pull). With a high conductivity diaphragm, I get around the "this coating is better than that coating" issue....

But I guess I would be removing the biggest benefit from Constant Voltage methodology....
 
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Now since the Constant Voltage is 6db greater SPL than Constant Charge, what if I just delete one stator voltage on Constant Voltage? So have my 70:1 step up from amp on the high conductivity diaphragm (I have tons of aluminized 12micron Mylar) and have back stator at -3kv, front stator grounded / floating. From what I am reading, should have same SPL as constant charge ESL, and operate much like a Magnepan Single ended (not push pull).

CV(constant voltage) will give the same SPL as the CQ(constant charge) configuration if operated with the same step-up transformer and bias voltage. It is the inverted CV configuration(with its equal but opposite bias voltages on the stators) that is 6dB louder than either the CV or CQ standard configuration. I'll put together some notes and a few pics on the inverted ESL later today.

If you removed one of the stators of a push-pull CV esl, you would be removing one of the large balancing forces in Attachment #1 and the diaphragm would be pulled toward the remaining stator and most likely collapse to it unless the bias voltage was significantly reduced to a point where the diaphragm tension could over ride the force. SPL output would reduce linearly as bias voltage is reduced.

The other issue is that with only one stator, the force on the diaphragm would be non-linear no matter what it's position.

As a precautionary note, if you experiment with metalized mylar (and imperfectly coated stators) be prepared for arcs, melting plastic, and sometimes even a little flame. In my experience, a large value resistor between the HV supply and the diaphragm goes a long way in reducing the severity of the arcing, but does not necessarily eliminate it.
 

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Inverted Push-Pull ESL

Attachment #1 provides a comparison of the basic setup for an Inverted Push-Pull ESL and the Standard Push-Pull ESL. As AcoustatAnswerMan pointed out, the Inverted ESL must be operated in CV mode to work properly over the whole audio band, so we will stick with CV mode for the comparison.

Using the same method as in Post#4 for plotting up relative levels of Force on the diaphragm, we can get an understanding of why the Inverted ESL produces 6dB more output for the same input. Sticking with a bias voltage of 5kV, lets suppose the step-up ratio of the transformer is 1:200. So, if we had a 200W amplifier, it would swing maximum peak voltages of about +/-50V resulting in the +/-5000V on the stators. Attachment #2 provides tabulated values for the diaphragm and stator voltages along with the relative Force vs Input voltage plot for the Standard CV push-pull ESL.

Now, if we switched out the transformer for one with a 1:100 step-up ratio and used it to power an Inverted ESL with +/-5kV bias voltages on the stators, you would get the voltages and relative Forces shown in Attachment #3. Notice that at each input voltage level the voltage difference between the diaphragm and each stator match that of the Standard arrangement from Attachment #1. If you kept the 1:200 ratio transformer you would double the voltage swing on the diaphragm, which would double the total Force resulting in the 6dB higher output as shown in Attachment #4

The "extra" output is coming from the fact that having the 2nd equal but opposite biased stator allows us to leverage the entire output voltage from the step-up transformer against that bias voltage value twice rather than having it split in half and fed to the stators before leveraging against the diaphragm voltage. Compare the +/-50V input lines in the table from Attachment #2 with the +/-25V input lines from the table in Attachment #4

One other thing to notice in Attachment #4 is that even though the diaphragm voltage swings higher than the bias voltage on the stators the total force on the diaphragm from the push-pull configuration is still linear as long as the diaphragm is centered in the gap. This is the case for the standard configuration as well. However, with a single ended ESL your signal voltage needs to be limited to something less than the bias voltage to avoid gross distortion.
 

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Bolserst, I thank you for taking the time to reply in the manner you have (great attachments!!). I liken these linearity differences between inverted and conventional ESL to common arguments with Planar speakers - the further / closer the planar diaphragm moves to / away from the magnets, the bigger they stray from linearity.

But yet they sound very good....at least to me they do...big Maggie Fan here.

I remember testing some new coating on Martin Logan CLS panel with just the back stator connected. It made music, clean sounding, and was noticeably less SPL than with front stator (for obvious reasons) didn't think much of it then, but after I put it all back together, I thought isn't a single stator CV inverted ESL operating like a planar Magnepan / Apogee (what do you call it - Single Ended Push More - Push Less?)

Now the Magnepan 20.X series has the diaphragm between to sets of magnet poles, and approximates a Inverted CV ESL in its operation, correct?

The other issue is that with only one stator, the force on the diaphragm would be non-linear no matter what it's position.
Is this true with single pole planars too?

If you ride the signal over the bias voltage (attachment)?


Maybe all this is obvious to everyone reading, and I am a little slow?
 

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I am the builder of the metalized stats you mentioned earlier, and legis is the end user. There is a large resistor in series with the HV-supply and the membrane is heavily segmented and curved thereby controlling the resonance and beaming behavior. Distortion may be a little higher, although it most likely also has to do with the interaction between the transformer and the loading capacitance.

The largest advantage is the complete absence of coating and charging issues, it just always works.
 
Is this true with single pole planars too?

No this is not true for single sided planars.
The force exerted on a current carrying wire in a magnetic field is proportional to the magnitude of the current and the strength of the magnetic field it is surrounded by. There is no limit on the current magnitude relative to the magentic field strength. This is more comparable to CQ ESLs where the force is proportional to charge and strength of electrostatic field. Unlike electrostatic forces, the force on a current carrying wire is orthogonal to both the direction of the current and the direction of the magnetic field. Attachement #1 depicts this relationship where magnetic field direction is from N to S, and current direction is from (+) to (-).

Attachement #2 shows a cross section of a single sided planar magnetic showing the forces generated based on the direction of current flow and magnetic field. Note that the magnetic field strength is not uniform, getting weaker with distance away from the magnets. This results in non-linear force on the wires attached to the diaphragm as it move alternately closer and further from the magnets. However, the force is more linear than a single sided ESL which has squared terms in its force equation.

Attachement #3 shows a cross section of a “push-pull” planar magnetic. I prefer to call it a “symmetric” planar since it is a better description of what the second set of magnets does. They provide a substantially uniform and symmetric magnetic field for the wires to move back and forth in. The main drawback to this configuration(besides added cost) is the potential for a cavity resonance in the mid-to-high frequency range.
 

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