Michael Chan
Compaq
-----Original Message-----
From: Mellitz, Richard [mailto:[email protected]]
Sent: Wednesday, August 04, 1999 4:03 PM
To: '[email protected]'
Subject: [SI-LIST] : RE: RESULTS to Proposal: Rs
correlation/collaboration for W-Eleme nts
I think I've got to augment my table. Apparently XFX reports Rs in terms of
radians. If we multiply the Rs from XFX (with lambda = 1) times the
sqrt(2*pi) we get: Rs=1.314E-3 Ohms/(sqrt(Hz).meter) which is roughly the
same as every one else. The farthest one off of the group seems to be the
HSPICE field solver.
So this means if you used XFX to determine Rs for HSPICE,, you need to
multiply the Rs value (in the tlp file) by the sqrt(2*pi).
I still don't have a good answer for ground plane losses yet. I have someone
using yet another modeler to do both lossy and lossless ground plane. I'll
post a new table one I get some new data.
Richard Mellitz
Intel
-----Original Message-----
From: Mellitz, Richard
Sent: Wednesday, August 04, 1999 1:13 PM
To: '[email protected]'
Subject: RESULTS to Proposal: Rs
correlation/collaboration for W-Elements
Rs Tabulations so far (8/4/99):
______________________________
Equation: signal + ground
Rs= 1.806e-3 Ohms/(sqrt(Hz).meter)
Equation: signal alone
Rs= 1.455e-3 Ohms/(sqrt(Hz).meter)
R0= 4.454 Ohms/meter
SIMPEST (COMPAC)
Rs= 1.3555e-3 Ohms/(sqrt(Hz).meter)
RO= 4.519685 Ohms/meter
AP SIM
Rs= 1.245e-3 Ohms/(sqrt(Hz).meter)
R0= 5.346 Ohms/meter
Quad Design's XFX (universe =50 mils, Lambda =1, Integral
Mode)
Rs= 5.241E-4 Ohms/(sqrt(Hz).meter)
R0= 4.456 Ohms/meter
Quad Design's XFX (universe =50 mils, Lambda =6, Integral
Mode)
Rs= 1.284E-3 Ohms/(sqrt(Hz).meter)
R0= 4.456 Ohms/meter
HSPICE 98.4 field solver
RS= 7.26284e-04 Ohms/(sqrt(Hz).meter)
RO= 4.37069 Ohms/meter
It looks like SIMPEST, AP SIM, XFX lambda=6, and Rs equation
for the signal line are all close.
Questions to resolve for accurate W element usage.
1) Should Rs include return plane skin effects? Michael Chan used a PEC
in SIMPEST and because it seems close to others, I might assume the other
did not include ground losses.
2) Why do I need to set lambda= 5 to 6 in Quad Designs XFX to get Rs
results that are similar to other modeling programs?
Richard Mellitz,
Intel
-----Original Message-----
From: Mellitz, Richard
[mailto:[email protected]]
Sent: Monday, August 02, 1999 10:51 AM
To: '[email protected]'
Subject: [SI-LIST] : Proposal: Rs
correlation/collaboration for W-Elements
<< File: Mathcad - ms_loss_eq.pdf >>
Apparently the W element model uses a
pseudo-propagation function with the following form.
P(f)= exp{-sqrt[
(G0+f*Gd+j*2*pi*f*C)*(R0+sqrt(f)(1+j)Rs+j*2*pi*f*L) ]*len }
(From HSPICE application note "Boosting
Accuracy of W Element for Transmission Lines with Nonzero Rs or Gd Values")
Let's assume that this is valid for some
conditions. It would be nice to know what the assumptions are.(geometry,
frequency, etc.) We can talk about the validity of the above in another
thread.
I would like to make a proposal. I would
like to know what various field solvers report in regards to the above
propagation function. Let's start with a microstrip first (and only look at
skin effect). The geometry follows.
Height over ground: 0.004"
Width of conductor: 0.006"
Thickness of conductor: 0.001"
Conductivity: 0.58E8 mho/meter
Let's all use the same units for Rs. Say:
Ohms/(sqrt(Hz)*meter)
Now, A colleague of mine has supplied a
formula that is used in microwave
design. I have attached a PDF file with
details. (Too tough for text, TTFT
:-)), I remember foobar)
The answer, using the closed form formula
for Rs is:
1.806E-03 ohms/(sqrt(Hz)*meter)
If this is the magnitude of complex Rs, then
Re(Rs) would be
1.277E-03 ohms/(sqrt(Hz)*meter)
I have received sidebar results from some of
you folks, but I don't want to post other people answers. However I will
compile a table of posted results. There are issues of complex number
involved. Remember I'm looking for the Rs for the above propagation
formula.
Step 2 will be to do same for a strip line
geometry where:
Height over ground: 0.005"
Width of conductor: 0.0025"
Thickness of conductor: 0.0005"
Distance between ground planes: 0.0105
It would be appreciated if we could find out
what "tricks" people are using to get Rs from their field solvers.
Regards,
Richard Mellitz
Intel
<<Mathcad - ms_loss_eq.pdf>>
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