|T O P I C R E V I E W
||Posted - Jun 29 2022 : 20:30:19
I have a doubt with respect to the imaginary part of the output capacities in the matrix C. Shouldn't they be affected by a frequency value? What is the analogy considering electro-quasi-static?
I understand that, traditionally, a capacitance estimation results in Cap = C -j(G/w).
For example, let's say I'm modelling a dielectric with e_abs = (3.5 -j0.1)e_0. The capacitance value output between two conductors results C_12 = 9.0e-12 -j3.2e-10. Do I obtain directly a conductance (of value 3.2e-10) in parallel to the capacitance between the conductors?
|2 L A T E S T R E P L I E S (Newest First)
||Posted - Jul 04 2022 : 21:20:58
Alright, it makes sense. I believe it is clear now.
||Posted - Jul 04 2022 : 18:13:32
we should have published a white paper on that point.
FasterCap does NOT implement any dielectric model for simulating lossy dielectric over a frequency range. It is up to you to provide the right value of the complex dielectric permittivity at the frequency of interest. Of course this would vary according to the frequency - so you would get different substrate conductivities for the different permittivity values at the different frequencies by running multiple simulations.
Said that, if you directly obtain a conductance value depends only on how you specify the complex part of the permittivity. In fact, if you already scale this part by omega (2*pigreek*frequency) you will directly get a conductivity value. If you instead use the 'standard' imaginary permittivity part, you will need to multiply by omega the result to get the conductivity (conductance to be used in parallel to the capacitance).
You may check the sample file "circular_wire_over_gnd_plane_2d_lossy.lst" and the referenced paper R. F. Harrington, C. Wei, "Losses on Multiconductor Transmission Lines in Multilayered Dielectric Media", IEEE Transactions on MTT, Vol. 33, No. 7, Jul 1984 for further insights.