[SI-LIST] Re: Specctraquest Model ?

  • From: Tadashi Arai <arap@xxxxxxxxxxxxxx>
  • To: "[SI-LIST]" <si-list@xxxxxxxxxxxxx>
  • Date: Fri, 15 Mar 2002 13:24:52 +0900

Hi Ken,

There are some hints in Appendix A of "DF/SigNoise User Guide".

According to this (only I have an old version, 10.0), name of subckt
represents as follows;

VIA_A_VIA_25C17CSM_L1A0W600L27A-135W300_FST_4978

A: Drawing name
VIA_25C17CSM: Padstack name
L1: Start layer
A0: Line angle
W600: Line width (design unit)
L27: End layer
A-135: Line angle (-135degree)
W300: Line width

and via model has geometry informations.
By default, SPECCTRAQuest uses "closed form" equivalent circuit for via,
however you can use EM solved model for via by clicking "Solve" button
with selecting via model in model browser.

Though I have never userd this function, this is my $0.02.

Regards,

//// /// // / /  /    /    /      /       /        /         /          /
Tadashi Arai//Platform Developing Dept.,Desktop Prd Div. Fujitsu Limited
arap@xxxxxxxxxxxxxx  TEL:+81-42-370-7624  Inagi-shi, Tokyo, Japan
/          /         /        /      /      /     /    /   /  / / // /// 
On Thu, 14 Mar 2002 16:18:45 -0500
"Ken Beach" <kenbeach@xxxxxxxxxxxxxxxxxx> wrote:

> 
> 
> Hi,
> 
> I have a question regarding the meaning of the Signal Explorer net list
> model for a via.  The model definition is immediately below:
> 
> .subckt VIA_A_VIA_25C17CSM_L1A0W600L27A-135W300_FST_4978
> +1 2
> CL1A0W600 1 0 2.18755e-13
> CL27A-135W300 2 0 2.18755e-13
> RLL1A0W600_L27A-135W300 1 2 1e-7 L=2.33327e-09
> .ends VIA_A_VIA_25C17CSM_L1A0W600L27A-135W300_FST_4978
> 
> 
> I think this model definition is defining a pi network with a 0.218pf
> capacitor on the left, a 0.218pf capacitor on the right and a resistor of
> 10^-7 in series with a 2.33nH inductor in the arm bridging between nodes 1
> and 2.  However, we can't find in, the help manual, a clear description of
> the RLL1A0W etc. bridging arm.  So, here we are guessing.  Does anybody know
> what that bridging arm statement is saying?
> 
> Thanks in advance
> 
> Ken
> 
> 

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