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Error in Capacitance obtained between theoretical and simulated values between two parallel wires
Posted 9 dic 2011, 10:03 GMT-5 MEMS & Nanotechnology, MEMS & Piezoelectric Devices, Geometry, Results & Visualization Version 3.5a 1 Reply
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Hi.
I am working on COMSOL 3.5.a and as a basic case, I am interested in calculating the capacitance between two wires.
As I am presently working on the simulation in 2D case ( front view of the wires), the results obtained would not be in Farads but Farads/metre.
Theoretically, there is a formula for two cylindrical wires that are separated by a distance of dand the length of the wires is l. The circumference of the wires is 2r. The formula used to calculate is C= epsilon0*pi*l/ (ln(l/r)*(sqrt(sqrt((l*l)+(2d*2d))-l)/((sqrt((l*l)+(2d*2d))+l)). The theoretical results that I obtained using this formula is 1.82fF
Now, in the practical case, I am using a square instead of a circle (in front view), as my actual project is simulation of capacitance between plates. I do not think using a square instead of a circle is going to make a huge difference to the results.
Now, in COMSOL,I have covered the two squares with a huge box of permittivity 1 and boundary conditions of zero charge. The squares are given material conditions of Copper. The boundary settings are arranged in such a way that we have 1V in one square and 0V in the second square. These conditions give me a capacitance of 24pF/m.
(During the theoretical calculation I considered length l to be 1mm and so when I substitute this length l of 1 mm in the result obtained from COMSOL, I get 24fF) This is a huge difference to the value obtained theoretically and therfore, I would be happy if someone could point out the errors in my simulation model.
Thanks.
I am working on COMSOL 3.5.a and as a basic case, I am interested in calculating the capacitance between two wires.
As I am presently working on the simulation in 2D case ( front view of the wires), the results obtained would not be in Farads but Farads/metre.
Theoretically, there is a formula for two cylindrical wires that are separated by a distance of dand the length of the wires is l. The circumference of the wires is 2r. The formula used to calculate is C= epsilon0*pi*l/ (ln(l/r)*(sqrt(sqrt((l*l)+(2d*2d))-l)/((sqrt((l*l)+(2d*2d))+l)). The theoretical results that I obtained using this formula is 1.82fF
Now, in the practical case, I am using a square instead of a circle (in front view), as my actual project is simulation of capacitance between plates. I do not think using a square instead of a circle is going to make a huge difference to the results.
Now, in COMSOL,I have covered the two squares with a huge box of permittivity 1 and boundary conditions of zero charge. The squares are given material conditions of Copper. The boundary settings are arranged in such a way that we have 1V in one square and 0V in the second square. These conditions give me a capacitance of 24pF/m.
(During the theoretical calculation I considered length l to be 1mm and so when I substitute this length l of 1 mm in the result obtained from COMSOL, I get 24fF) This is a huge difference to the value obtained theoretically and therfore, I would be happy if someone could point out the errors in my simulation model.
Thanks.
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1 Reply Last Post 9 dic 2011, 10:48 GMT-5
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Posted:
1 decade ago
Hi Anjana,
with your geometry (d/r = 2) I get 21.1 fF/mm with the formula in en.wikipedia.org/wiki/Capacitance for the circular wire.
A square wire should give a little higher capacitance, so the 24 pf/m doesn't look bad. You should check your formula. From a practical view the 21 pf/m look quite realistic too.
Cheers
Edgar
with your geometry (d/r = 2) I get 21.1 fF/mm with the formula in en.wikipedia.org/wiki/Capacitance for the circular wire.
A square wire should give a little higher capacitance, so the 24 pf/m doesn't look bad. You should check your formula. From a practical view the 21 pf/m look quite realistic too.
Cheers
Edgar