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.
Adarsh said:
4 months ago
what is meant by 2218?
@All.
it is just the model number of the given RAM, you need to check the chip size they have given it is 16k x 1.
Hence, we know that chip size is equal to a multiple of the depth size of the chip (16k) and the width of the chip (1bit).
Hence it is equal to 16k only, we know that 2^n, where n is the number of address lines used, 16k (16*1024) can be written as (2^4) + (2^10) after solving we get 2^ (4+10) equals to 2^14 after refer this to 2^n we get the n=14 as the answer.
@All.
it is just the model number of the given RAM, you need to check the chip size they have given it is 16k x 1.
Hence, we know that chip size is equal to a multiple of the depth size of the chip (16k) and the width of the chip (1bit).
Hence it is equal to 16k only, we know that 2^n, where n is the number of address lines used, 16k (16*1024) can be written as (2^4) + (2^10) after solving we get 2^ (4+10) equals to 2^14 after refer this to 2^n we get the n=14 as the answer.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers