[IBIS] RE: [IBIS-Users] EBD file syntax

From: <rrwolff_at_.....>
Date: Wed Aug 22 2007 - 13:07:31 PDT
Nam,
 
Your syntax looks fine to me.  Capacitor C3 would be to an ideal ground.
In fact, all capacitors are connected to an ideal ground, even in the
transmission line case, where the capacitance is part of a transmission
line that references ideal ground.
 
Regards,
Randy
 
Randy Wolff
SI Modeling Manager
Signal Integrity R&D Group
Micron Technology, Inc.



________________________________

From: owner-ibis-users@eda.org [mailto:owner-ibis-users@eda.org] On
Behalf Of Nguyen, Duynam
Sent: Wednesday, August 22, 2007 1:25 PM
To: ibis@eda.org
Cc: ibis-users@eda.org
Subject: [IBIS-Users] EBD file syntax



 

Dear expert,

I am trying to describe the following situation in an EBD file: a signal
net that goes from a die pad to a package pin A9. Physically, this
signal net consists of two traces. Each trace is on a different layer
and they are connected together by a via which is modeled as a small
capacitor of value c3 to ground. The die

pad and the package pin are modeled by a series L and a grounded C. 

My EBD file attempt is depicted as follows

 

[Path Description] SIGNAL_NAME

Pin A9

Len = 0            L = l1 C = c1               /

Len = length2   L=l2   C = c2    R=r2     /

Len = 0                     C = c3                /         <== Allowed
? 

Len = length4  L=l4    C = c4    R=r4     /

Len = 0           L = l5  C = c5                /

Node DIE.5085      

 

I have 2 questions:

1. About the representation of the capacitance c3 in the EBD file, do I
violate any EBD file specification by doing this ?

   Assuming the specification allows it, would the capacitance c3 be
interpreted as grounded or in-series  ?  

2. If the representation of the capacitance c3 as above is not allowed,
can I take out the offending line and the rest is still permitted ?

   In other words, there are still two traces with an electrical
discontinuity connecting them as follow:

[Path Description] SIGNAL_NAME

Pin A9

Len = 0            L = l1 C = c1               /

Len = length2   L=l2   C = c2    R=r2     /

Len = length4  L=l4    C = c4    R=r4     /

Len = 0           L = l5  C = c5                /

Node DIE.5085      

 

Thank you much,

Nam 


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Received on Wed Aug 22 13:07:53 2007

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