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binding a layout to schematic and putting pins in the layout

Inas Mohammed

Hello,

I have a script which generates a layout transistor pattern.

Then i needed to bind this layout to a schematic in order to have it XL happy.

I searched in the older links here and i found these functions: lxSetConnRef & bndSetInstsBindingByName.

And i searched for them in the manuals to know how they are used, I found that i have to use "lxSetConnRef" first then use the second function. both functions have the condition to have the instances as leaf one, what does it mean??????

I tried that and it worked nicely with the transistor has only one multiplier, but it couldn't work with the other transistors which have more than one multiplier.

So can you please tell me:

1) what is meant by leaf instance?

2) How can i make it for transistors consisting of more than one multiplier?

About putting the pins in the layout, i thought about it like the following: i just made some brain storming with my self about that.

- determine the pins needed to be placed, and their layers (text and box)

- Identify every transistor will be connected to which pins.

- for a certain transistor, identify their Drains, sources, and gates will be connected to which pins and place them.

- check if this transistor if its a dummy so all their terminals will be set to vdd.

So please can you tell me what the followed procedures for placing the pins for a pattern using skill script, if you can give a simple sample code it will be nice

Thanks

Regards,

Inas

Parents

  • Inas,

    A leaf instance simply means that this is not a hierarchical instance (on the schematic side, primarily). In other words, if you have a hierarchical design with an instance of a block called I1, you can't bind everything in that block in one go - you've got to go down to the individual components.

    For binding one to many (e.g. m-factor) devices, you can either use:

    bndAddInstsBindingByName(schCv "M1" layCv "I1")
    bndAddInstsBindingByName(schCv "M1" layCv "I2")
    bndAddInstsBindingByName(schCv "M1" layCv "I3")

    etc. You may get warnings until you've bound enough instances. You can also use bndSetInstsBindingByName for the first call, but only the first for a particular schematic instance.

    Or you could use bndSetObjectsBinding:

    bndSetObjectsBinding(schInstId list(layInstId1 layInstId2 layInstId3))

    i.e. you use the database id of the instances instead, and pass a list for the  layout side.

    I can't really write your  pin placement code for you, sorry.

    Regards,

    Andrew.

Reply

  • Inas,

    A leaf instance simply means that this is not a hierarchical instance (on the schematic side, primarily). In other words, if you have a hierarchical design with an instance of a block called I1, you can't bind everything in that block in one go - you've got to go down to the individual components.

    For binding one to many (e.g. m-factor) devices, you can either use:

    bndAddInstsBindingByName(schCv "M1" layCv "I1")
    bndAddInstsBindingByName(schCv "M1" layCv "I2")
    bndAddInstsBindingByName(schCv "M1" layCv "I3")

    etc. You may get warnings until you've bound enough instances. You can also use bndSetInstsBindingByName for the first call, but only the first for a particular schematic instance.

    Or you could use bndSetObjectsBinding:

    bndSetObjectsBinding(schInstId list(layInstId1 layInstId2 layInstId3))

    i.e. you use the database id of the instances instead, and pass a list for the  layout side.

    I can't really write your  pin placement code for you, sorry.

    Regards,

    Andrew.

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