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The capacity of 2K × 16 PROM is to be expanded to 16 K × 16. Find the number of PROM chips required and the number of address lines in the expanded memory.
Ans. Capacity required =16k x 16Available chip (PROM) =2k x 16The number of chip =(16k x 16)/(2k x 16) = 8Total word capacity in the chip = 2 x 210Hence the address line required for single chip = 11The word capacity in the expanded memory 16k = 214There is the address lines required are 14. Between then 11 will be general and 3 will be connected to 3x8 decoder.
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