| Author | Message | |||||
     Michael Daly (mdaly) Member Username: mdaly Post Number: 6 Registered: 12-2004 |
Hi, I am trying to run a simulation where two electrodes are attached to opposites sides of a dielectric and there is a gap in the middle. I am interested in seeing the electric field if 500V AC is applied. I looked at the sample AC simulations, including Backstrom's, but they are only if the current density for some area is specified. I would like instead to specify the voltage on the plate electrodes. Any help is appreciated. | |||||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 435 Registered: 06-2003 |
I guess I don't understand the problem. The procedure is: 1) select your variables. 2) write a set of mathematically consistent PDE's determining the variables. 3) impose appropriate value (Dirichlet) boundary conditions with the VALUE statement or derivative conditions with the NATURAL statement. (See the documentation of NATURAL to determine the exact meaning of derivative boundary conditions.) For example, if you choose voltage V as a variable and write div(eps*grad(V))= rho, then VALUE boundary conditions will impose values on V and NATURAL boundary conditions will impose values of NORMAL(eps*grad(V)), ie NORMAL(D), at the boundary. All this is derived from the text of your equations. There are no built-in interpretations of the boundary condition forms. | |||||
| Michael Daly (mdaly) Member Username: mdaly Post Number: 7 Registered: 12-2004 |
I think my question is how to run an AC simulation with voltage V as my variable. I just care about steady state, but I'm not sure how to define V with the BCs if it's time-varying but I don't want a time-dependent simulation. | |||||
| Takeyuki Ojima (tac) New member Username: tac Post Number: 1 Registered: 07-2005 |
My suggestion is to break V into the real and imaginary parts because "AC" means V has some "Phase". And for steady state, use dt(V)=j.omega.V and I=dt(rho) which is alse a complex current. Then you will have four variables with two equations and be required two more equations to solve. What do you want as answer of your simulation? | |||||
| Takeyuki Ojima (tac) New member Username: tac Post Number: 2 Registered: 07-2005 |
In addition for an "AC" case, electromagnetic fields are in general not expressed only by V. A vector potential A must be included because magnetic field is produced by the accompanied current. | |||||
| Michael Daly (mdaly) Member Username: mdaly Post Number: 8 Registered: 12-2004 |
Sorry, I'm still having trouble with this simulation, as it's not my area of expertise. What I specifically would like to know is, what boundary condition should I assign to the electrode regions to give an AC voltage? And Takeyuki, thank you for the advice but should I really need a time-dependent equation (dt(V)=j*omega*v) for a steady state problem? I'm not sure how exactly to formulate my equations either. | |||||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 438 Registered: 06-2003 |
There seems to be some semantical confusion here. "AC" means alternating current, which strictly speaking is a time-dependent problem. You can solve for cases of "steady oscillation" by assuming V=V0*exp(j*omega*t). If you perform the differentiation and substitution in the time-dependent equation and divide out the exponential, you get a "steady-state" (ie, not requiring a time integration) equation in the amplitude V0. Provided you have done this, you then apply a VALUE boundary condition to V0, implying a steady oscillating voltage with amplitude V0. Full implementation of Maxwell's equations in this mode can be messy, and appropriate simplifications depend on what your system looks like and what you can assume about its behavior. This is the area of Mr Ojima's expertise, and the context in which his remarks were made. Professor Backstrom discusses these issues at length in "Waves by Finite Element Analysis", available through the link on our bookstore page www.pdesolutions.com/bookstore.html. There are examples in the Backstrom_Books/Waves folder in the FlexPDE distribution. | |||||
| Michael Daly (mdaly) Member Username: mdaly Post Number: 9 Registered: 12-2004 |
Thank you both for your help. I am still not seeing what I should in my simulation. I know from experiment that the open circular cavity should exhibit an "hourglass" shape for the electric field, and currently in the simulation the field is turned 90 deg so that the strongest parts are on the side when they should be on the top and bottom. I was not able to implement all of your suggestions so I am posting the code I created and would very much appreciate it if one of you looked it over sometime. Thank you again -- I would be entirely lost without your help.
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 442 Registered: 06-2003 |
The equations of electromagnetics admit of an astonishing variety of formulations and interpretation. The most important factor in the decision about formulation is the frequency range in which you operate. But other factors are 1) what you know, what you can assume, and what you want to find out; 2) what kinds of materials are in the system. As Takeyuki has pointed out, electrical currents generate magnetic fields. Do you need to calculate these? I don't know - this is not my field of expertise. At this point, you have not described the structure of your system, what you have and what you need. As far as I can see, you have two options: 1) search the literature and textbooks for a mathematical analysis of a problem similar to yours. 2) hire an EE professor from your local college as a consultant to help you formulate your equations. | |||||
| Takeyuki Ojima (tac) Member Username: tac Post Number: 5 Registered: 07-2005 |
I think what you want will be achieved by treating as an electrostatic problem, because your computational domain is far smaller than the wavelength at 20 khz. Attached is my modification for your reference.
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| Takeyuki Ojima (tac) Member Username: tac Post Number: 6 Registered: 07-2005 |
In addition, electrodes must be zero-thickness in this case for your correct simulation, because otherwise the boundary condition at the interface between the electrode and dielectric would be Natural(V)=0, that means normal(E)=0, which is not acceptable on the surface. Attached is probably a correct description.
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| Takeyuki Ojima (tac) Member Username: tac Post Number: 7 Registered: 07-2005 |
"Cavity_statics2.pde" again
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| Michael Daly (mdaly) Member Username: mdaly Post Number: 10 Registered: 12-2004 |
Thank you for the corrections to my code. The results with this new code are the same as what I was getting when I did a simple DC simulation. I assume you know what you're talking about so maybe it is also the result for AC, even though we observed in an experiment the field being stronger in the vertical direction. Thanks for all your help. |
Circular_cavity2