Robert Koslover
Certified Consultant
Please login with a confirmed email address before reporting spam
Posted:
6 months ago
04.11.2025, 16:20 GMT-5
Updated:
6 months ago
04.11.2025, 16:22 GMT-5
I may not be understanding your problem completely, but it seems to me that you can represent the behavior of the resistive layer by assigning it a conductivity (S/m). You can relate DC resistance to the conductivity by the usual formula,
where
is the length (distance along the current path),
is the cross-sectional area to the current flow, and
is the conductivity. Then for the RF behavior, set an impedance boundary condition and tell it to use the properties from the associated material, the latter which then does not have to meshed volumetrically. This approach should be acceptable if your layer thickness is much greater than a skin depth. If not, then either mesh the volume of the layer itself, which may be computationally intensive (and don't use the impedance boundary condition) or replace the layer by an external resistor equivalent circuit (via the .ec physics in the AC/DC module, if you have that), if that is consistent with how your resistive layer is actually behaving.
-------------------
Scientific Applications & Research Associates (SARA) Inc.
www.comsol.com/partners-consultants/certified-consultants/sara
I may not be understanding your problem completely, but it seems to me that you can represent the behavior of the resistive layer by assigning it a conductivity (S/m). You can relate DC resistance to the conductivity by the usual formula, R = l/A\sigma where l is the length (distance along the current path), A is the cross-sectional area to the current flow, and \sigma is the conductivity. Then for the RF behavior, set an impedance boundary condition and tell it to use the properties from the associated material, the latter which then does not have to meshed volumetrically. This approach should be acceptable *if your layer thickness is much greater than a skin depth*. If not, then either mesh the volume of the layer itself, which may be computationally intensive (and don't use the impedance boundary condition) or replace the layer by an external resistor equivalent circuit (via the .ec physics in the AC/DC module, if you have that), if that is consistent with how your resistive layer is actually *behaving*.