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?
8
10
14
16
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
21 comments Page 2 of 3.

Enver said:   9 years ago
Are address bits and address lines are the same thing ?

Then what is the address location?

Raj said:   10 years ago
2^10 is always referred as a kilo, right so i.e. K.

16 = 2^4 so we add up 10+4 = 14 address bits.
(1)

A Purohit said:   10 years ago
Here K means Kilobytes.

So, 1 K = 1024 bytes.

Ravi said:   1 decade ago
1024 bits is equal to the 1 k

Ravikiran said:   1 decade ago
Why did you use 1024 as 1 K?

Nandhini said:   1 decade ago
Why did you use 1024?

Jeffrey maghinay said:   1 decade ago
2^4= 2 x 2 x 2 x 2 = 16.
2^10= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024.
2 to the power of 4 = 16.
2 to the power of 10 = 1024.
So: add the two power 4 + 10 = 14.

Sri said:   1 decade ago
How to factorising 16384 by 2?

Priyanka said:   1 decade ago
16k = 16*1024 = 2^4*2^10.

So total 14 address lines require.

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.


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