Digital Electronics - Memory and Storage - Discussion
Discussion Forum : Memory and Storage - General Questions (Q.No. 15)
15.
How many 2K × 8 ROM chips would be required to build a 16K × 8 memory system?
Discussion:
4 comments Page 1 of 1.
Madhvi sukhanandi said:
6 years ago
2^1 = 2k
2^4 = 16k
2^1-4 = 2^3 = 8k.
2^4 = 16k
2^1-4 = 2^3 = 8k.
(1)
Kousalya said:
7 years ago
16k/2k=8.
(1)
Ankit RAj said:
9 years ago
Simple method is (16K * 8 )/(2K * 8 ) = 8. This is handy for solving these type of problems.
(1)
Pankaj sethia said:
1 decade ago
Since the required data bits are same so, we need only 8, 2k x 8 ROM chips. These chips will be connected in parallel and only one chip will be active at a time through address lines (14 lines). Extra three lines will be used to active one chip among 8 chips.
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