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>>
**** To unsubscribe from si-list: send e-mail to [email protected] In the BODY of message put: UNSUBSCRIBE si-list, for more help, put HELP. si-list archives are accessible at http://www.qsl.net/wb6tpu/si-list ****