
Home  Forums  Rules  Articles  diyAudio Store  Blogs  Gallery  Wiki  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Planars & Exotics ESL's, planars, and alternative technologies 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
20th August 2008, 09:16 AM  #31 
R.I.P.
Join Date: May 2006
Location: Netherlands

Bryan,
I'd still be interested to see your model (by email). When you have time, can you reply to post #27? AJ 
20th August 2008, 10:50 AM  #32 
diyAudio Member

bshaw147
I mostly agree, but the science of acoustics isn't sorted out completely yet. arendjan Your frequency response curves got me quite confused.. I don't see effects of interference of signals coming from different loactions on strip. If you follow the Baxandall's article, at 1 cm, with given panel sizes, you'll get 10db/decade rolloff at ~30Hz, assuming current drive, and assuming that beginnig of rolloff if 3db, and a 20db/decade rolloff at 8.5kHz. At 2m, 10db rolloff would begin at about 5.5kHz, 20db rolloff would be too far into high frequencies. What I see is essentially a flat response, with the only nonflatness noticeable from 50cm, at LF, till 3kHz, it's somewhat too short to determine slope of rolloff. Of course, I don't know how the panel is actually driven above xformer  constantcurrent, constantvoltage, or constantpower. However, it's still quite logical that character of drive can't depend on listening distance. Or I just don't see something? (By the way, if it wouldn't bother you much, can you get current response after transformer, so it would be possible to derive response for panel itself, without transformer?) 
22nd August 2008, 01:38 AM  #33 
diyAudio Member
Join Date: Aug 2008

Hi AJ
Looking back at post 27... 1. Do you model the diaphragm as constant voltage or constant charge? I modeled the diaphragm as having constant surface charge density in operation. In reality, as Streng's indepth paper shows, charge will tend to move on the surface. This is bad if we are interested in creating a linear acoustic transducer. This is exactly why some designers choose high resistant coatings like graphite powder. Some designers (me included) add a very large value resistor inline with the diaphragm contact as to oppose the flow of charges on the surface. It may take a little longer for the diaphragm to be in equilibrium with the bias power supply (t=RC). I don’t think it matters. 2. "The electrical field would be pretty much Vsig/2d on a macro scale (the diaphragm would have to take on the potential of the electric field at that point)." If we ignore the fact that our stators have holes then yes, the electric field anywhere in between the plates is the above simple expression (parallel plate capacitor, where 2d is distance between plates and Vsig is the potential difference maintained between them). If you are familiar with Laplace's equation for determining the scalar electric potential you will see that I get the same thing in the notes I provided in the email. Once we have the potential determining the electric field is trivial, E=grad (potential). Since the potential only varies in the zdirection (I am ref. my notes) this amounts to taking the derivative of the electric scalar potential with respect to the z spatial variable. You will see that it is simply a constant. Calculating the forces requires one more step. The electrostatic force acting on a charged body is just F=qE, instead I find force density so f=sigma*E, where f=Newtons/m^2 and sigma=charge (measured in Coulombs) per meter squared. As you realize we have a distribution of charge on both sides of the diaphragm, which implies that we have a force on the left and right side, adding will give the net force density acting on the diaphragm; this is what we are interested in. We need to actually take the average of the force densities acting on the left and right side. The reason is we have to remove the contribution of the force exerted on itself by itself. To understand this, ‘you can not stand in a basket and expect to pull yourself outside of it by pulling up on the handles above your head.’ Now, the big difference with coating one side as opposed to coating both sides is that the coated side is a free charge distribution while the uncoated side will develop an induced surface charge distribution; due entirely by the electric field polarizing the molecules in the film (stretching them and so establishing a dipole moment distribution). I have to look back at my notes when I was using the derived expressions and finding order of magnitude estimates for the free and bound charge distributions. I just redid the calculation and they were on the same order of magnitude (10^6 Coulombs/m^2). However, I still have to prove that they will always be of the same magnitude for different values of bias potential and drive potential (you call it Vsig). I am not sure of this right now, I will be tonight. The reason I am unsure is that I did this calculations with different values for the potentials and I did see an order of magnitude change. If you are ready I can share with you the force density when we coat just one side, its not that much more difficult then the first derivation. It just relies on theory of how electrostatic fields can polarize materials. All of this is in any standard college level Electromagnetism book (I recommend Griffiths Introduction to Electricity and Magnetism and the graduate version by Jackson). If you want to get deep we can consider the electric field produced not by solid plates but by perforated ones. Jackson solves for the fields around a circular aperture. It involves specialty functions that are too esoteric for me. I think there may be a more “elegant” way to attack this problem. One of the nice things about electrostatics is that it is a linear theory, and thus obeys the principal of superposition (one of the deepest things I discovered in College). We know the field of a solid plate and we know the field that would be produced by a circular disk (everywhere) that is at the same potential as the solid plate, do you think we can just subtract the solid plate field from the disk field? I started to do this derivation and it’s pretty clean. Now we just have to sum over all the perforations, specific down to what pattern is used, i.e. 1/8” diameters on 3/16” staggered centers. I plan on finishing this sometime in the near future, but I must leave time to build my own ESLs . I waited about five years to get building, putting the anal in analytics. 3. Does your model incorporate the negative stiffness introduced by induced charge on the stators? If I understand your question correctly, this is hard to account for and I think its small potatoes for correcting the above force expression. I started to solve this problem; the only way I know how to tackle this one is to determine the Green's Function. To be extremely brief and do no justice to the theory of Greens functions.... A Green's function is particular solution we look for. It is the systems response due to a unit input to the system. Once you have the functional form of it you can solve almost any boundary value problem you can think with any kind of distribution of sources. Here our unit source is a point charge, somewhere between the plates. You can get the function by using the method of images; you will need an infinite number of image charges to satisfy the homogeneous boundary conditions of grounded plates. Now that we have the Green's Function we can go back to the original problem of a sheet of charge in between the plates and solve for the electric scalar potential. But, wait the problem is even more complicated in that the surface distribution depends on the shape of the diaphragm, which is known by solving the wave equation. But, to solve the wave equation you need the forcing, to know the forcing you need the potential. You see the problem is coupled you have to solve them together. Coupled partial differential equations are one of the hardest things to solve for in mathematical physics. The potential via Green's thm. is related to the just founded Green Function. All of this is in chapter 1 of Jackson's Electrodynamics book. The expressions are anything but tidy. So I got my plates rolled into a nice arc, I will attach pictures shortly. I don’t have a digital camera but my girlfriend does. Today I worked on the frames for the panels. They are coming out very nicely. I learned a great an easy way to curve wood! Really gives it a professional look. All you do is take a wet paper towel and wrap the wood with it, then put in your microwave. Mine only needed 3050 seconds; you have plenty of time to put the now compliant wood into whatever shape you like. I just made an arc, which parallels the plate’s arc. This will serve as my top and bottom pieces. If you never tried it, do it, you’ll love it. For bigger pieces you’ll need a steamer (or a bigA$$ microwave). Hope the above helps you. If you never studied electrodynamics or only just a little it really is worth all those late nights and early mornings. I would recommend mastering Intro to Electricity and Magnetism by Griffiths first then move on to the more rigorous and ambitious Jackson text. Have you had any success with using your stretcher jig for curved tensioning? I was thinking that we could just tension the film when it is flat then with another jig compress the steel curved stators together. With a curved design we should only tension the film in one direction, this way our film will act as a continuum of violin strings. Do you have some thoughts you care to share on techniques for getting a uniform coating, film type, coating materials, and optimal methods? Take Care, Bryan Shaw Bjs2@njit.edu 
22nd August 2008, 09:27 AM  #34 
diyAudio Member
Join Date: Aug 2008

AJ,
Believe it or not I did manage to forget something; you expressed interests in seeing a computer simulation. My coding is nothing special and ill probably lose my hair trying...I was compsci major when i first when to school, then i left the dept. to study math. I love LabView, makes writing programs that have to interface to other devices extremely straightforward (except for the occasional proprietary stuff). Anyway, getting off subject. Mathematic a, would be what i would use to write up the sim. Imagine seeing what your music looks like in your room. Of course our model makes a few assumptions. Did you catch mine in the notes; I only evaluate the force density at the average position of the membrane. This avoids that coupled problem I mentioned earlier. Believe it or not, I don’t think that this will hurt us any when we assume that the forcing has only a timedependence, uniform in space. We usually find the solution when the drive voltage signal only contains one frequency and we do a little Fourier analysis to find the solution for noise, basically. The time average of the amplitude response to just one frequency component is deadcenter, which is where we evaluate the force. You can look at the coupled scenario, made me too queasy. Your earlier comment of using some kind of FEM routine wont be necessary since the solution for the damped forced wave equation with clamped edges is well known, and can be visualized nicely. I suppose each element would be a differential area element of the film whose motion is governed by the simple harmonic oscillator (mx..=kx x..= second order time derivative of amplitude, should include a damping term on RHS proportional to Beta*x.) Having these models connect with the acoustic routine is where all the fun will happen... Take Care, Bryan 
22nd August 2008, 07:31 PM  #35  
R.I.P.
Join Date: May 2006
Location: Netherlands

Quote:
At shorter distances you can clearly see the dips around 3.2 kHz and 4.5kHz, I assume as a result of destructive interference. Quote:


22nd August 2008, 08:50 PM  #36  
R.I.P.
Join Date: May 2006
Location: Netherlands

Hi Bryan,
thanks for your response. Before I forget to mention it, I did not receive your model or notes by email. You address so many point that I think it's easier and clearer if I quote you and comment on it. Quote:
A resistor in the HT line will keep the total charge on the diaphragm more or less constant, but it can not prevent charge movement on the diaphragm. So it's not an equal solution. Quote:
Quote:
Quote:
Fortunately this is expression is linear with the diaphragm displacement 'x' so no trouble there. The minus sign indicates that it is actually enforcing the deflection of the diaphragm. The force is the same order of magnitude as the drive force so we can't ignore it. If the charge is not constant then the force becomes highly non linear. Quote:
Quote:
I found a really nice differential expression on page two of this document by Edo Hulsebos. I think you will like it. Regards, ArendJan 

23rd August 2008, 03:23 PM  #37 
R.I.P.
Join Date: May 2006
Location: Netherlands

ESL simulator
As a result of this discussion I decided to build a macrosimulator for electrostatic loudspeakers. It can calculate the frequency response for flat, rectangular ESL panels. You can freely choose the position of the listener as well.
It is web based, and 100% free :) Just punch in your numbers and go! You can find it on the link below. ESL simulator Enjoy, ArendJan 
23rd August 2008, 08:38 PM  #38 
diyAudio Member
Join Date: Aug 2008

Hi Bryan,
I resent the email, with the pics. Your suggestion on not using graphite will be applied. I never was comfortable with rubbing it in anyway, at least your can measure its resistance in several places to verify that it is uniformish. S/R, stands for? (signal/resistance?) Your right, the large resistor keeps total charge pretty much steady but does not keep the charge density steady. The link you provided had some nice solutions that i will probably give a try. When you say that your numerical model would consider coupled springs, do you mean springs attached vertical up and down the film and springs that run normal to the films surface, or both. It is a really good idea to do this. We can model hard surfaces that will reflect the pressure waves as springs that have a very large spring constant (infinite even). Spring constant, from Hooke's Law says that the restoring force a spring would feel as it is displaced by a displacement of x is just F=kx, where k is the spring constant. I’m sure we can model the air the same way. For any kind of accuracy will need a lot of springs, and hence computations. It may take a long time to run it. If we take the limit of some kind of continuum of springs than our equations may look similar to equations you would see in the linear theory of acoustics. This may give a good way to check our routine. I don’t see the film being displaced vertically that much. This would help reduce the dimensions of the problem and keep computation time shorter. The equation on page two of Hulsebos includes an electrostatic part (to express the forcing) a mechanical part (this was the wave equation, last 3 terms of his equation), and an acoustic part. His second term includes the effect that the films charge will induce a stator charge. He does include a damping of the film by including his very last term. Take Care, Bryan 
17th June 2009, 06:07 AM  #39 
diyAudio Member
Join Date: Jun 2009
Location: Dunedin, New Zealand

Stator material
Calvin wrote: "ESLs are high impedance devices."
Would carbon fibre be suitable for stator or is the conductivity too low? I love cf and have done "some other than audio stuff" with it. If cf is possible material for stator it would open amazing possibilities! I'm starting to build small hybrid with stator in two sections, going to make frame from fibreglass and just started to think about conductivity of cf and can it be used in this project... 
17th June 2009, 07:47 AM  #40 
diyAudio Member
Join Date: Nov 2004
Location: close to Basel

Hi,
actually I don´t know if CF could be a good material for stators. From the viewpoint of conductivity it would, but conductivity of a stator is not the point to worry about with ´standard´ conductors having less than say 1kOhms of resistance. The stator´s behaviour under HVconditions has to be investigated. (There have been accidents with helicopter rotor flaps beeing destroyed by lightnings. ESLarcing is the same on a minor scale) There will always be arcing since You operate a ESL close to the field breakdown limits of air. How does carbon behave under this condition? I am afraid, that it ages quickly and might even be destroyed. Even if insulated impurities within the material might lead to accelerated aging. Besides beeing a strong but lightweight material I can´t see any advantages for carbon against Steel, Copper or Aluminium. Cost, Handling and probably Performance will be inferior. jauu Calvin 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Where did you get your Stator material?  jzh797s  Planars & Exotics  20  25th October 2006 11:15 PM 
Heatsink Preparation  dhole  Pass Labs  4  6th May 2005 12:41 PM 
Creating new PCB Contacts  bazooka  Parts  2  30th January 2005 11:45 PM 
New To Site?  Need Help? 