I always wanted to ask...why Bias voltage on ESL? - diyAudio
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Old 8th August 2013, 12:24 AM   #1
john65b is offline john65b  United States
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Default I always wanted to ask...why Bias voltage on ESL?

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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Last edited by john65b; 8th August 2013 at 12:47 AM.
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Old 8th August 2013, 12:46 AM   #2
john65b is offline john65b  United States
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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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Old 8th August 2013, 01:26 PM   #3
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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.
Attached Files
File Type: pdf WP-Inverter0905.pdf (875.6 KB, 45 views)
File Type: pdf Final_7054456.pdf (136.5 KB, 39 views)
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Old 9th August 2013, 03:15 AM   #4
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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.
Attached Images
File Type: gif push-pull_simple.gif (48.3 KB, 296 views)
File Type: gif CV_centered.gif (26.4 KB, 293 views)
File Type: gif CQ_centered.gif (21.0 KB, 291 views)
File Type: gif CQ_offcenter.gif (18.3 KB, 287 views)
File Type: gif CV_offcenter.gif (26.6 KB, 289 views)
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Old 9th August 2013, 03:40 AM   #5
john65b is offline john65b  United States
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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...
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Old 9th August 2013, 12:31 PM   #6
john65b is offline john65b  United States
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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?
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Old 9th August 2013, 06:36 PM   #7
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Quote:
Originally Posted by john65b View Post
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.
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Old 9th August 2013, 08:53 PM   #8
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Quote:
Originally Posted by john65b View Post
…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)

Quote:
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.

Last edited by bolserst; 9th August 2013 at 09:16 PM.
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Old 9th August 2013, 09:01 PM   #9
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Quote:
Originally Posted by AcoustatAnswerMan View Post
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.

Quote:
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.
measuring sensitivity of ESLs

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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Old 10th August 2013, 02:19 PM   #10
john65b is offline john65b  United States
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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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Last edited by john65b; 10th August 2013 at 02:39 PM.
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