s2ibis2 question?

Hello,

  I am a new learner for IBIS models version 2.1 and is using s2ibis2.
Now I am facing a problem: while I characterize for a simple 3-state I/O
pad (refer to attachment), the generated .spi file will not function
correctly. After examining the .spi file, I found the TN pin will NOT
be set while characterizing for the A pin (and vice versa). So I need
to correct this error by hand. Is there any way to make this process
more automatically? Or I make a wrong .s2i file?

  Besides, if I just want the bd4t model for IO pin, could I just not
create for the other 2 models for A and TN?

  Any help or comments will be appreciated.

best regards,
Arthur Huang

ps. my s2ibis2 is s2ibis2.sparc, version no. v0.91BETA

[attachment]
(first is bd4t.s2i, and second is bd4t)

| TEST IBIS characterization
[IBIS Ver] 2.1
[File rev] 1.0
[Date] Mon Oct 14 15:56:30 1996
[Disclaimer] for characterization purpose only
[Spice type] hspice
| [Iterate]
| [Cleanup]
[Component] bd4t
[Manufacturer] Arthur
| [Package model] pkg_type
[Spice file] bd4t
[Temperature range] 50 100 0
[Voltage range] 3.3 3.0 3.6
[R_pkg] 1.0m 0.5m 1.5m
[L_pkg] 2.5nH 2.0nH 3.0nH
[C_pkg] 4.0pF 3.5pF 4.5pF
| [Diff pin]
| describe the differential pin list

| [Pin mapping]
| currently just neglect this command

[Pin]
A A s_A A
TN TN s_TN TN
IO IO s_IO bd4t
-> A TN
VDD vdd s_p POWER
pGND g GND GND

[Model] A
[Model type] Input
[Polarity] Non-Inverting
[Vinl] 0
[Vinh] 5.0

[Model] TN
[Model type] Input
[Polarity] Non-Inverting
[Vinl] 0
[Vinh] 5.0

[Model] bd4t
[Model type] I/O
[Polarity] Non-Inverting

* testing spice deck for bd4t IBIS 2.1 model characterization

* change the following line to your library file
.inc "bd4t.lib"
.global vdd gnd vpp

xCKT A IO ZI TN bd4t

.subckt bd4t a io zi tn
xnd1 a tn n1 nand
xnd2 n1 tn n2 nand
xiv1 n2 n3 inv
mpio io n1 vdd vdd p w=60u l=1u m=1
mnio io n3 gnd gnd n w=30u l=1u m=1
xiv2 n4 zi inv
xiv3 io n4 inv
.ends bd4t

.subckt nand a b z
mpa z a vdd vdd p w=6u l=1u m=1
mpb z b vdd vdd p w=6u l=1u m=1
mna z a mid gnd n w=3u l=1u m=1
mnb mid b gnd gnd n w=3u l=1u m=1
.ends nand

.subckt inv a z
mp z a vdd vdd p w=6u l=1u m=1
mn z a gnd gnd n w=3u l=1u m=1
.ends inv
Received on Mon Oct 14 02:36:21 1996