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.
Yoghalakshmi said:
1 decade ago
The size of the memory is N*M
where N is the address lines and M is word length
no of registers/memory location required is 2^N
Given memory capacity is 16k
thus 2^N=16K
1K=1024 memory locations
thus16k=16*1024=16384
now 2^N=16384
After factorising 16384 by 2 we ll get N AS 14
SO ADDRESS LINE REGUIRED IS 14.
where N is the address lines and M is word length
no of registers/memory location required is 2^N
Given memory capacity is 16k
thus 2^N=16K
1K=1024 memory locations
thus16k=16*1024=16384
now 2^N=16384
After factorising 16384 by 2 we ll get N AS 14
SO ADDRESS LINE REGUIRED IS 14.
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