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A starter's question regarding capacitance calculation
Posted 17.09.2010, 11:58 GMT-4 Low-Frequency Electromagnetics Version 5.2 9 Replies
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Hi, there,
I just began to learn COMSOL and tried to calculate the capacitance of parallel-plate capacitor. I think it is a very simple simulation: I have read the tutorial and repeat the examples given by COMSOL website. However, I met problems which I dont know why. My simulation is ( I use COMSOL 3.5):
1. choose space dimension: 3D
2. choose Application Modes: AC/DC module/Statics, Electric/Electrostatics
3. built two parallel block with size (1x1x0.1)
4. In physics/subdomain setting, choose copper from the material library and then use the default constitutive relation, which set the relative permitivity is 1 for copper.
5. In physics/boundary setting, choose the BCs are electrical potential for one plate and ground for the other one.
6. meshing and solve this module
My questions are:
1. why the relative permmitivity for copper and other metals are 1 in COMSOL?
2. When postprocessing the results, I plot electric field with arrows. I expect the electric field lines should be parallel between the two metal plates and be curves around the boundaries. But, as I see , the electric field lines penetrate into the metal and no lines between the plates. Why?
I searched the capacitance calculations with Comsol and saw some people added a big block to contain the two plates. Then, I followed the same steps as above and added one big block (5x5x5) . In the physics/subdomain setting, I tried to set the material for this big block is aire and the ones for the two plates are copper. But, I am not allowed to set the big one as aire. Furthermore, in boundary settings, the boundaries for the two plates become gray and can not be changed. Why?
Does anyone can give me a hint what's wrong with my simulation?
Thanks a million.
best,
a confused beginner
I just began to learn COMSOL and tried to calculate the capacitance of parallel-plate capacitor. I think it is a very simple simulation: I have read the tutorial and repeat the examples given by COMSOL website. However, I met problems which I dont know why. My simulation is ( I use COMSOL 3.5):
1. choose space dimension: 3D
2. choose Application Modes: AC/DC module/Statics, Electric/Electrostatics
3. built two parallel block with size (1x1x0.1)
4. In physics/subdomain setting, choose copper from the material library and then use the default constitutive relation, which set the relative permitivity is 1 for copper.
5. In physics/boundary setting, choose the BCs are electrical potential for one plate and ground for the other one.
6. meshing and solve this module
My questions are:
1. why the relative permmitivity for copper and other metals are 1 in COMSOL?
2. When postprocessing the results, I plot electric field with arrows. I expect the electric field lines should be parallel between the two metal plates and be curves around the boundaries. But, as I see , the electric field lines penetrate into the metal and no lines between the plates. Why?
I searched the capacitance calculations with Comsol and saw some people added a big block to contain the two plates. Then, I followed the same steps as above and added one big block (5x5x5) . In the physics/subdomain setting, I tried to set the material for this big block is aire and the ones for the two plates are copper. But, I am not allowed to set the big one as aire. Furthermore, in boundary settings, the boundaries for the two plates become gray and can not be changed. Why?
Does anyone can give me a hint what's wrong with my simulation?
Thanks a million.
best,
a confused beginner
9 Replies Last Post 22.02.2017, 17:05 GMT-5
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Posted:
1 decade ago
1. First, you need to recognize that you can't just model the physical parts, you also have to model the volume around them. For a capacitor, this is because that volume is filled with electric fields, and you need to let Comsol compute those fields.
2. Comsol grayed out the boundary settings for the plates because you immersed them in a surrounding volume. It always does this with internal boundaries, since often such boundaries are very often (but not in this case) surfaces upon which you would not want to set any conditions. The solution to this is very simple, fortunately. There's a checkbox in the boundary setting dialog box that lets you turn on the settings for the internal boundaries. Just do that, and you'll be able to set those conditions as you wish.
3. In Comsol 3.5, you really don't need to set the subdomain properties of your air block. Just leave it at the default of having unity dielectric constant. However, you will need to set boundary conditions on the external boundaries of that block. Think about the physics for a while and you should be able to make intelligent choices there. (Hint, consider the symmetries in the problem).
Good luck.
2. Comsol grayed out the boundary settings for the plates because you immersed them in a surrounding volume. It always does this with internal boundaries, since often such boundaries are very often (but not in this case) surfaces upon which you would not want to set any conditions. The solution to this is very simple, fortunately. There's a checkbox in the boundary setting dialog box that lets you turn on the settings for the internal boundaries. Just do that, and you'll be able to set those conditions as you wish.
3. In Comsol 3.5, you really don't need to set the subdomain properties of your air block. Just leave it at the default of having unity dielectric constant. However, you will need to set boundary conditions on the external boundaries of that block. Think about the physics for a while and you should be able to make intelligent choices there. (Hint, consider the symmetries in the problem).
Good luck.
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Posted:
1 decade ago
Dear Mr. Koslover,
Thank you so much for the suggestions.
Now, I can set up the interner boundary conditions.
Another question is: if I set the boundary condition for one plate is port and set the forced voltage as Vin=1 V , set the BC for the other plate is also port with the voltage equals zero, and in Options/Integration Coupling Variable, I choose domain 1 which is the big box containing the two plates and set a variable Cpp with expression 2*We_emes/V^2 to calculate the capacitance, then in Expression/Global Expressions, set Cpp as the global variable. After meshing and solving this problem, in postprocessing procudure, I try to look at this global value, but the value is significantly different with my expectation. Could you tell me what's wrong with my simulation?
Furthermore, if I choose the input for port with energy method, I needn't to set up the capacitance global variable, instead, I can look at the C11_emes value in the postprocedure. Why?
Thanks a lot!
Thank you so much for the suggestions.
Now, I can set up the interner boundary conditions.
Another question is: if I set the boundary condition for one plate is port and set the forced voltage as Vin=1 V , set the BC for the other plate is also port with the voltage equals zero, and in Options/Integration Coupling Variable, I choose domain 1 which is the big box containing the two plates and set a variable Cpp with expression 2*We_emes/V^2 to calculate the capacitance, then in Expression/Global Expressions, set Cpp as the global variable. After meshing and solving this problem, in postprocessing procudure, I try to look at this global value, but the value is significantly different with my expectation. Could you tell me what's wrong with my simulation?
Furthermore, if I choose the input for port with energy method, I needn't to set up the capacitance global variable, instead, I can look at the C11_emes value in the postprocedure. Why?
Thanks a lot!
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Posted:
1 decade ago
Your stated approach is very different from what I would do. I would certainly not set the plates to "port" boundary conditions, but rather to fixed voltages. Capacitance is a purely electrostatic problem, so I would use the electrostatics application mode. Set the plate voltages symmetrically (+V and -V) and surround them by a large box, with the box boundary condition set as potential = 0 (aka, ground). With symmetry in geometry and potentials, the field lines will go from plate to plate, rather than from plate to box, which is important if you are to compute the capacitance correctly.
Your energy method (i.e., setting the electrostatic energy = 0.5*CV^2, and then solving for C) should then work fine. Just make sure the box is much larger than the capacitor, that the plates have equal and opposite voltages, that the boundary on the box is at zero potential, and that you have a sufficiently fine mesh.
Your energy method (i.e., setting the electrostatic energy = 0.5*CV^2, and then solving for C) should then work fine. Just make sure the box is much larger than the capacitor, that the plates have equal and opposite voltages, that the boundary on the box is at zero potential, and that you have a sufficiently fine mesh.
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Posted:
1 decade ago
excuse me, i saw your discussion and i remember my problem
i think i can write my question here after you discussion because maybe mine related to your
i have to simulation a varactor( variable capacitor by voltage) , but i have some problem
1. i don't know how i can calculate beam's spring constant
2. i want to calculate capacitance between 2 plate one of them is grounded and other have voltage which is used for displacing the upper plate, my design has another plate which is used as electrod and is grounded. voltage between electrod and upper plate is used for providin electro static fors to displace upper plate, and capacitance between upper plate and bottom plate must be calculated as varactor, how i can calculate that capacitance?
i hope you understood my question si i can sedn you my design
thanks a lot very very much
i think i can write my question here after you discussion because maybe mine related to your
i have to simulation a varactor( variable capacitor by voltage) , but i have some problem
1. i don't know how i can calculate beam's spring constant
2. i want to calculate capacitance between 2 plate one of them is grounded and other have voltage which is used for displacing the upper plate, my design has another plate which is used as electrod and is grounded. voltage between electrod and upper plate is used for providin electro static fors to displace upper plate, and capacitance between upper plate and bottom plate must be calculated as varactor, how i can calculate that capacitance?
i hope you understood my question si i can sedn you my design
thanks a lot very very much
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Posted:
1 decade ago
Hi
a fixed-free beam spring constant is (first order approcimatin) 3*E*I/L^3 with E the Young modulus, I the inertia around the rotation axis and L the length.
But you get it fully out from a COMSOL structural model defining material constants (E,nu,rho) and its geometrical shape (which ends up to define I and L). Then you need to couple the electric force as a surface force along the beam edges/boundaries.
I believe there is a model in the model library.
The change in shape will change the electrode distance, hence change the capacity you want out (if yu are in 2D do not forget there are quite some fringe effects neglected)
--
Good luck
Ivar
a fixed-free beam spring constant is (first order approcimatin) 3*E*I/L^3 with E the Young modulus, I the inertia around the rotation axis and L the length.
But you get it fully out from a COMSOL structural model defining material constants (E,nu,rho) and its geometrical shape (which ends up to define I and L). Then you need to couple the electric force as a surface force along the beam edges/boundaries.
I believe there is a model in the model library.
The change in shape will change the electrode distance, hence change the capacity you want out (if yu are in 2D do not forget there are quite some fringe effects neglected)
--
Good luck
Ivar
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Posted:
1 decade ago
thanks Ivar
one of the big problem that i face to it, is Residual stress, in Comsol documentation i red an example about that but it wasn't very useful for me
when i simulate a simple fixed-fixed beam in comsol spring constant of beam is much lower than theoretic solve, ant that different results is due to residual stress. i don't know how to apply residuall stress to a simple beam or plates of a variable capacitor?
another question is about calculation of capacitance.
i have attached a image of my design but in 2D view. but my design in comsol is 3D. i want calculate capacitance between plate 1 and 2 while i applied different voltage between plate 1 and 3. in fact difference voltage between plates 1 and 3 causes plate 1 move to down and change capacitance between plates 1 and 2. it is great problem for me to simulate capacitance between 1 and 2.
thanks a million
one of the big problem that i face to it, is Residual stress, in Comsol documentation i red an example about that but it wasn't very useful for me
when i simulate a simple fixed-fixed beam in comsol spring constant of beam is much lower than theoretic solve, ant that different results is due to residual stress. i don't know how to apply residuall stress to a simple beam or plates of a variable capacitor?
another question is about calculation of capacitance.
i have attached a image of my design but in 2D view. but my design in comsol is 3D. i want calculate capacitance between plate 1 and 2 while i applied different voltage between plate 1 and 3. in fact difference voltage between plates 1 and 3 causes plate 1 move to down and change capacitance between plates 1 and 2. it is great problem for me to simulate capacitance between 1 and 2.
thanks a million
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Posted:
1 decade ago
Hi
If I read you correctly you are in MEMS analysis and your MEMS manufacturing process is inducing permanent stress in the layers so that the overall stiffness of you system is dfferent from an unstressed model. This is rather common, I must say, and the only way around I know about is to build your model layer by layer as you do the techno-processes (and you need partial saamples to validate the reslts for each steps) and then at the end you can run your simualtion on the correctly stressed model with the desied voltage and hopefully the desired stiffness / deformations.
For the capacitance, this is rather a geometrical driven value (except for fringe effects which are not necesarily small) so if you have the correct deformations then COMSOL should calculate the "C" value correctly (including fringe effects if you are in 3D with a sufficiently large volume of "air" around. The 3D model is though very costly so sometimes estimating fringe effects from a 2D model and extrapolating to 3D is about as good, if possible.
--
Good luck
Ivar
If I read you correctly you are in MEMS analysis and your MEMS manufacturing process is inducing permanent stress in the layers so that the overall stiffness of you system is dfferent from an unstressed model. This is rather common, I must say, and the only way around I know about is to build your model layer by layer as you do the techno-processes (and you need partial saamples to validate the reslts for each steps) and then at the end you can run your simualtion on the correctly stressed model with the desied voltage and hopefully the desired stiffness / deformations.
For the capacitance, this is rather a geometrical driven value (except for fringe effects which are not necesarily small) so if you have the correct deformations then COMSOL should calculate the "C" value correctly (including fringe effects if you are in 3D with a sufficiently large volume of "air" around. The 3D model is though very costly so sometimes estimating fringe effects from a 2D model and extrapolating to 3D is about as good, if possible.
--
Good luck
Ivar
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Posted:
8 years ago
I have one question regarding capacitor.
i have done modeling and simulation of a simple plate capacitor and i got some ans but to cross check i did hand calculation and both ans are different. but when i am making capacitor plate very thin like 1mm then the ans of simulation matches with my hand calculation ans.but by increasing thickness of capacitor to 7mm capacitance value got changed.
why so?
because in equation it is not dependent on the thickness.
i have done modeling and simulation of a simple plate capacitor and i got some ans but to cross check i did hand calculation and both ans are different. but when i am making capacitor plate very thin like 1mm then the ans of simulation matches with my hand calculation ans.but by increasing thickness of capacitor to 7mm capacitance value got changed.
why so?
because in equation it is not dependent on the thickness.
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Posted:
8 years ago