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Terminal charge in the electrostatic simulation

Posted 2011年11月22日 GMT-5 10:42 Low-Frequency Electromagnetics, MEMS & Nanotechnology, MEMS & Piezoelectric Devices, Results & Visualization Version 4.2 3 Replies

eastsea
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I am working on a 2-D electrostatic simulation now. I was confused on the derived valuse of terminal charge.

As we can get the line integration terminal charge (es.nD) with the unit C/m or the line average terminal charge (es.Q0_1) with the unit of C, my problem is, what is the difference bewteen them? I can't find the exact explaination on these values in the help documentation. Can someone tell me about this problem?

Thanks in advance!
3 Replies Last Post 2014年1月8日 GMT-5 03:15
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 2011年11月23日 GMT-5 01:34
Hi

You should not forget that COMSOL is ALWAYS calculating in 3D. So when we do a 2D representation of an object, COMSOL considers by default 1m depth (along z, and for most physics, as always there are exceptions check your doc) which means that all edges = boundaries in 2D are in fact a surface of length equivalent to the edge, but of depth of 1 m. Hence the units "per meter".

I.e. in solid you need to multiply by "solid.d" to get rid of the depth.

Note that you can always change the depth value in the main physics entry (for most physics) but you still need to multiply most variables. Use the units as check it helps to get the numbers correct.

Another thing one must adapt to (at least I needed to ;) is that msot entries are represented in the "element" dimensions of dx*dy*dz or corresponding boundaries dx*dy or dy*dz or dx*dz ...units hence the per /m^3 for domains and the per /m^2 for fluxes on boundaries

--
Good luck
Ivar
Hi You should not forget that COMSOL is ALWAYS calculating in 3D. So when we do a 2D representation of an object, COMSOL considers by default 1m depth (along z, and for most physics, as always there are exceptions check your doc) which means that all edges = boundaries in 2D are in fact a surface of length equivalent to the edge, but of depth of 1 m. Hence the units "per meter". I.e. in solid you need to multiply by "solid.d" to get rid of the depth. Note that you can always change the depth value in the main physics entry (for most physics) but you still need to multiply most variables. Use the units as check it helps to get the numbers correct. Another thing one must adapt to (at least I needed to ;) is that msot entries are represented in the "element" dimensions of dx*dy*dz or corresponding boundaries dx*dy or dy*dz or dx*dz ...units hence the per /m^3 for domains and the per /m^2 for fluxes on boundaries -- Good luck Ivar
eastsea
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Posted: 1 decade ago 2011年11月23日 GMT-5 02:52

Thanks a lot for the kindly explanation.
Have a nice day!
Thanks a lot for the kindly explanation. Have a nice day!
Zhun Wei
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Posted: 1 decade ago 2014年1月8日 GMT-5 03:15
Hi Ivar,

I have encountered a problem when solving capacitance in electrostatic model. I compute a simple parallel plate capacitance. I choose the upper plate as terminal, and set the voltage as 1V. But when I integral the surface charges on this surface, I found that it differs with the value of terminal charge or terminal capacitance.

Do you know the difference between them, thank you very much!!

Best Regards,
zhun
Hi Ivar, I have encountered a problem when solving capacitance in electrostatic model. I compute a simple parallel plate capacitance. I choose the upper plate as terminal, and set the voltage as 1V. But when I integral the surface charges on this surface, I found that it differs with the value of terminal charge or terminal capacitance. Do you know the difference between them, thank you very much!! Best Regards, zhun

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