DRC

edited February 2018 in Layout
Sorry to ask some basic questions.

Are the DRC functions designed for whole chip or for a specified small area ?
Does it support hierarchy ?

Thank you.

Comments

  • edited February 2018

    Hello,

    basically the DRC functions can work on a whole chip. It can also work on a subcell or a certain region.

    Hierarchy is supported in a sense that DRC can digest a hierarchical layout, but the computation will render a flat layer. So you can run a DRC on a hierarchy, but the result will be a flat set of error or output shapes.

    The flat approach usually requires a lot of resources. KLayout offers a solution to reduce the resource requirements by providing "tiling": this means, the layout is cut into tiles with a configurable size and overlap. Each tile is worked on in a flat fashion. Tile computation can be run on multiple cores. Depending on the tile size, the resource requirements are small and the parallel execution on multiple cores provides an easy and scalable way to improve performance.

    This is in contrast to tools utilizing a hierarchical geometry processing approach where to tool tries to do as much computation inside a cell hence saving the effort of replicating the computation and results for each cell instance. In some applications hierarchical computation algorithms are very effective (like memory banks). Other applications benefit very little from hierarchy (like huge routing layers). In applications, where long-range interactions are important (for example, density computation), hierarchical processing does not render true benefits. In addition to this, the area of hierarchical algorithms is covered with IP claims. This is the reason why I still favour the flat + tiling approach.

    Personally, I like the scalability and predictability of the performance. I'd also like to add that in times of cloud computing compute power is cheap while licenses are expensive. So a free and scalable solution is cost effective even if the algorithmic performance is not leading edge.

    Best regards,

    Matthias

  • edited November -1
    Matthias

    Thank you very much for your proper answers.

    Best regards
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