Digital Electronics - Memory and Storage - Discussion
Discussion Forum : Memory and Storage - General Questions (Q.No. 1)
1.
How many address bits are needed to select all memory locations in the 2118 16K × 1 RAM?
Discussion:
21 comments Page 3 of 3.
Pankaj sethia said:
1 decade ago
2^14=(16*(1024))=16K
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