Rich,
In HSPICE 99.2 frequency dependent resistance modeling
was changed.
Previous versions of HSPICE model frequency dependent
resistance as:
R(f) = Ro + Rs*sqrt(f)
The changes occured after Dmitri left.
regards,
scott
-- Scott McMorrow Principal Engineer SiQual, Signal Quality Engineering 18735 SW Boones Ferry Road Tualatin, OR 97062-3090 (503) 885-1231 http://www.siqual.com <http://www.siqual.com>
"Mellitz, Richard" wrote:
I need a little clarification here. I must be thinking too slow.
Do HSPICE versions prior to 99.2 not use the Sqrt(f)*(1+j)*Rs?
Do Quad Design's W models use the Sqrt(f)*(1+j)*Rs?
If the Sqrt(f)*(1+j)*Rs are only valid for cylindrical conductors, then should we not use W elements for microstrip and strip line or are they just less accurate. If so, how much less and at for what frequencies?
Dmitri you said you used Rs*Sqrt(f) in the W element. Is this Rs the curve fit coefficient for Re(rs) or the |rs|?
...Richard Mellitz Intel
-----Original Message----- From: Dmitri Kuznetsov [ mailto:[email protected] <mailto:[email protected]> ] Sent: Monday, August 02, 1999 3:02 PM To: [email protected] Subject: Re: [SI-LIST] : Proposal: Rs correlation/collaboration for W-Elements
I agree with Richard, the really important factor is accuracy of Rs and Gd values. As they are multiplied by frequency, a small error makes a huge difference. But for a given Rs and Gd, my algorithm in
W element will give you mathematically accurate answer.
I would like to comment on the Sqrt(f)*(1+j)*Rs skin-effect equation introduced in Hspice 99.2. It has mathematically correct imaginary part, and does not require frequency-response correction. But this equation is only valid for cylindrical conductors and only at higher frequencies.
This skin-effect equation produces the corresponding inductive component Ls(f)=Rs/(2*Pi*Sqrt(f)). Thus, the inductance becomes infinite at dc. This substantially alters the waveforms especially when Rs is large.
This was the reason I used Rs*Sqrt(f) in W element. This equation gives asymptotically correct loss, and I was restoring the correct
imaginary part for any, not just cylindrical, configuration by applying frequency-response correction.
Regards, Dmitri Kuznetsov
======================================================= Dmitri Kuznetsov, Ph.D. Principal Engineer
ViewLogic Systems, Inc. e-mail: [email protected] 1369 Del Norte Rd. Tel: (805)278-6824 Camarillo, CA 93010 Fax: (805)988-8259 =======================================================
"Mellitz, Richard" wrote: > > 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>> > > ------------------------------------------------------------------------ > Name: Mathcad - ms_loss_eq.pdf > Mathcad - ms_loss_eq.pdf Type: Acrobat (application/pdf) > Encoding: base64
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