On Tue, Oct 25, 2011 at 5:20 PM, Mike Harrison wrote= : > The clue is in the word "Block". > > The blocks are not a bucket of single-bit locations, but chunks that can > be used in specific > configurations, with powers of 2 as the address inputs. > Therefore 2049x8 uses 2 blocks. > Larger widths have parity bits available, so 2048x9 can be done in one > block. > Makes sense, thank you! > If all else fails, read the datasheet! Believe me, if the datasheet contained all the answers, a mailing list wouldn't be necessary. Sometimes the answers you're looking for evade you and you really need another human being to provide them. --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .