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