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Best way to determine pad pitch

GMaggy

What is the best way to find the pad pitch when the design has 27000+ pads?  Using the (combine relation1 relation2) operator crashes my session as Ravel attempts to store 729 million+ combinations.  Even if I break the design into quadrants to reduce the number of pads being looked at, Ravel still has to deal with 48 million+ combinations.  Also, looking only at the pads nearest the design edge doesn't give me the result I'm looking for.  I need to find the minimum pad pitch in the design.

Do you have any suggestions?

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  • After some more investigation, I have found a more reliable method of finding a good cutoff value as per above:

    ;; Form pairs of x-y coordinates corresponding to each bump.
    (define x_y
      (transform (bump) bumppad
        ((xcoordinate (center bump)) (ycoordinate (center bump)))))

    ;; Group y coordinates corresponding to the same x coordinate into
    ;; collections. They correspond to y coordinates of bumps on the same
    ;; "row", where each row has the same x coordinate.
    (define ys
      (PAIR_2 (contract y (x y) x_y)))

    ;; Pick the "row" of y coordinates with the largest size. (Because the
    ;; probability is high that there is a pair with small pitch, giving a
    ;; small cutoff value.) Expand the collection to form a relation of
    ;; the y coordinates on this "row" (for a given x coordinate).
    (define yrow
      (expand ys (ys)
        (PAIR_1 (CHOOSE_MAX_VALUE (transform (ys) ys
                                    (ys (count ys)))))))

    ;; Find the cutoff value by finding the smallest different between y
    ;; coordinates in the "row" we are working with.
    (define cutoff
      (CHOOSE_MIN (transform (y1 y2)
                      (select (y1 y2) (combine yrow yrow)
                        (ordered? y1 y2))
                    ((abs (difference y1 y2))))
                  IDENTITY))

    It assumes that pads for the most part are placed in a grid-like pattern, and is fast and efficient even for a large number of pads.

    Regards,

    Björn

     

Reply

  • After some more investigation, I have found a more reliable method of finding a good cutoff value as per above:

    ;; Form pairs of x-y coordinates corresponding to each bump.
    (define x_y
      (transform (bump) bumppad
        ((xcoordinate (center bump)) (ycoordinate (center bump)))))

    ;; Group y coordinates corresponding to the same x coordinate into
    ;; collections. They correspond to y coordinates of bumps on the same
    ;; "row", where each row has the same x coordinate.
    (define ys
      (PAIR_2 (contract y (x y) x_y)))

    ;; Pick the "row" of y coordinates with the largest size. (Because the
    ;; probability is high that there is a pair with small pitch, giving a
    ;; small cutoff value.) Expand the collection to form a relation of
    ;; the y coordinates on this "row" (for a given x coordinate).
    (define yrow
      (expand ys (ys)
        (PAIR_1 (CHOOSE_MAX_VALUE (transform (ys) ys
                                    (ys (count ys)))))))

    ;; Find the cutoff value by finding the smallest different between y
    ;; coordinates in the "row" we are working with.
    (define cutoff
      (CHOOSE_MIN (transform (y1 y2)
                      (select (y1 y2) (combine yrow yrow)
                        (ordered? y1 y2))
                    ((abs (difference y1 y2))))
                  IDENTITY))

    It assumes that pads for the most part are placed in a grid-like pattern, and is fast and efficient even for a large number of pads.

    Regards,

    Björn

     

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