M
mauricio
Guest
Is there any way to get the resistance of a selected polygon or does
anybody has a rutine for that?
Thanks
anybody has a rutine for that?
Thanks
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T
tattvamasi@gmail.com
Guest
It is pretty simple to write your own, you can get the length and the
width of the selected rectangle, and grab the resistivity of the layer
from the technology files, to do the necessary calculation to get the
resistance - rho*l/a
I beleieve there is an example of this in the ROD documentation as part
of the installs
Partha
width of the selected rectangle, and grab the resistivity of the layer
from the technology files, to do the necessary calculation to get the
resistance - rho*l/a
I beleieve there is an example of this in the ROD documentation as part
of the installs
Partha
M
mauricio
Guest
If it is a rectangle, then it's as simple as you said, but if I have a
polygon then I think the simples way to do it is to reduce it to
rectangles and that is where I'm stuck. I found some papers talking about
it, but I hope someone has a rutine already written or an example of how
to do it.
Anyway, thanks Partha, I'm going to check in the ROD doc to see if I can
find something.
polygon then I think the simples way to do it is to reduce it to
rectangles and that is where I'm stuck. I found some papers talking about
it, but I hope someone has a rutine already written or an example of how
to do it.
Anyway, thanks Partha, I'm going to check in the ROD doc to see if I can
find something.
T
Tim
Guest
The problem I see with a skill routine is that the function won't know
which edges of the shape are the 2 ends of the resistor unless if you
do a partial select of the 2 edges that define the ends. Then any shape
can be broken up into rectangles and/or use the fomula for
non-rectangular shapes for non 0/90 shapes.
Tim
which edges of the shape are the 2 ends of the resistor unless if you
do a partial select of the 2 edges that define the ends. Then any shape
can be broken up into rectangles and/or use the fomula for
non-rectangular shapes for non 0/90 shapes.
Tim
G
Gerry Vandevalk
Guest
Of course not.
Resistance ( a 3d phenomena abstracted into a 2d problem. )
Your assumption that a "polygon" has resistance makes the assumption that
you identify the 2 ( hopefully not more ;^)
end points.
What you probably want is some sort of idealized routine to calculate an
approximate DC resistance of a rectilinear shape when
you assume that the current flow is between your two "ideal ends"
Even this is a tricky problem.
I have created path based resistor structures that were mostly rectilinear
and that computed the "number of squares" of each segment and
added (a user over-ride-able ) corner factor of ( 0.56 or 0.566 squares ...
I think this was a fab choice !? )
and then multiplied the total # of squares with the rho of the material to
get you a first (or second)
order resistance ( when added to a Fab supplied end correction factor that
was proportional to the width ... )
something like R = RHo * ( #squares ) + endfactor / WidthOfEnd1 + endfactor
/ widthofEnd2
assuming that the resistor can be broken up into rectangular sections and
corners ...
and each segment contributes Lseg/Wseg to #squares and each corner adds a
cornerfactor of 0.56square.
This was good enough in most cases ... ( unless you really needed a more
exact 3d sim run ... )
But if you require better accuracy ( or more complex structures ) either you
understand this problem much better or
you will not get the accuracy you are expecting ...
( Rant ...
Try calculating the real value of a corner in an ideal case ( say a 90
degree bent resistor with each leg being 10 times longer
than wide and then calculate the ideal corner factor. ( assume everything is
perfect ... )
Then Try rounding the corners by 1/10 of the grid and do an exact mesh
extraction that models this corner.
The accuracy of this corner factor is very suspect !!!
But I have spent many hours at different times being asked to include a very
precise cornerfactor.
Note that even though the new buzzwords at DAC of DFM and DFY are implicitly
only hooked into DSM issues,
I still faced these problems in 5u technologies .. and they had a hard time
telling me the corner rounding variations then!
.... Rant off
)
If you want accurate resistors (or matched ones ) making them rectangular
and larger than minimum helps/
-- Gerry
"mauricio" <mcontaldo@gmail.com> wrote in message
news:eb9b4633a06b41995255963e1d9bc05a@localhost.talkaboutcad.com...
Resistance ( a 3d phenomena abstracted into a 2d problem. )
Your assumption that a "polygon" has resistance makes the assumption that
you identify the 2 ( hopefully not more ;^)
end points.
What you probably want is some sort of idealized routine to calculate an
approximate DC resistance of a rectilinear shape when
you assume that the current flow is between your two "ideal ends"
Even this is a tricky problem.
I have created path based resistor structures that were mostly rectilinear
and that computed the "number of squares" of each segment and
added (a user over-ride-able ) corner factor of ( 0.56 or 0.566 squares ...
I think this was a fab choice !? )
and then multiplied the total # of squares with the rho of the material to
get you a first (or second)
order resistance ( when added to a Fab supplied end correction factor that
was proportional to the width ... )
something like R = RHo * ( #squares ) + endfactor / WidthOfEnd1 + endfactor
/ widthofEnd2
assuming that the resistor can be broken up into rectangular sections and
corners ...
and each segment contributes Lseg/Wseg to #squares and each corner adds a
cornerfactor of 0.56square.
This was good enough in most cases ... ( unless you really needed a more
exact 3d sim run ... )
But if you require better accuracy ( or more complex structures ) either you
understand this problem much better or
you will not get the accuracy you are expecting ...
( Rant ...
Try calculating the real value of a corner in an ideal case ( say a 90
degree bent resistor with each leg being 10 times longer
than wide and then calculate the ideal corner factor. ( assume everything is
perfect ... )
Then Try rounding the corners by 1/10 of the grid and do an exact mesh
extraction that models this corner.
The accuracy of this corner factor is very suspect !!!
But I have spent many hours at different times being asked to include a very
precise cornerfactor.
Note that even though the new buzzwords at DAC of DFM and DFY are implicitly
only hooked into DSM issues,
I still faced these problems in 5u technologies .. and they had a hard time
telling me the corner rounding variations then!
.... Rant off
)
If you want accurate resistors (or matched ones ) making them rectangular
and larger than minimum helps/
-- Gerry
"mauricio" <mcontaldo@gmail.com> wrote in message
news:eb9b4633a06b41995255963e1d9bc05a@localhost.talkaboutcad.com...
Is there any way to get the resistance of a selected polygon or does
anybody has a rutine for that?
Thanks
Guest
On Thursday, September 15, 2005 6:32:26 PM UTC+5:30, mauricio wrote:
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Hi can anybody solve this problem???Is there any way to get the resistance of a selected polygon or does
anybody has a rutine for that?
Thanks
--
Disclaimer : Privileged and Confidential information is contained in this e
mail and/or attachments and is subject to legal privilege. This e mail is
intended solely for the recipient/s mentioned above. If you are not the
intended recipient or responsible for delivery of the message to the
intended recipient you may not use, copy, distribute or deliver to anyone
this e mail and/or any attachments or any part of its contents or take any
action in reliance on it. In such case, you should destroy this message,
and notify us immediately. All reasonable precautions have been taken to
ensure no viruses are present in this e-mail. Our company is not liable for
any loss arising from unauthorized access to our company's email account by
any third party and/or any loss caused due to the transmission of any
viruses. The views, opinions, conclusions and other information expressed
in this electronic mail that do not relate to the company's business shall
be deemed to be the personal opinion of the sender and not endorsed by the
company.