29 votes 29 votes The capacity of a memory unit is defined by the number of words multiplied by the number of bits/word. How many separate address and data lines are needed for a memory of $4K \times 16$? $10$ address, $16$ data lines $11$ address, $8$ data lines $12$ address, $16$ data lines $12$ address, $12$ data lines Digital Logic gate1995 digital-logic memory-interfacing normal + – Kathleen asked Oct 8, 2014 recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 11.6k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 43 votes 43 votes ROM memory size $=2^{m} \times n$ $m=$ no. of address lines $n=$ no. of data lines Given, $4K \times 16$ $= 2^{2} \times 2^{10} \times 16$ $= 2^{12}\times 16$ Address lines $=12$ Data lines$ = 16$ Correct Answer: $C$ Sanket_ answered May 27, 2016 edited May 21, 2019 by Naveen Kumar 3 Sanket_ comment Share Follow See all 2 Comments See all 2 2 Comments reply mrinmoyh commented Dec 17, 2018 reply Follow Share In a ROM Matrix address lines means - horizontal line data line means - vertical line correct me if I'm wrong 7 votes 7 votes prithatiti commented Oct 26, 2022 reply Follow Share @mrinmoyh You are right 0 votes 0 votes Please log in or register to add a comment.
8 votes 8 votes Ans C) The size of memory = 4K x 16 bits = 2^12 x 16 bits So therefore, 12 address lines and 16 data lines. Prasanna answered Dec 19, 2015 Prasanna comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes Ans: C add lines, 2^n= 4K= (2^2)*(2^10)= 2^12... ; n=12 data lines, m=16 rishu_darkshadow answered Oct 7, 2017 rishu_darkshadow comment Share Follow See all 0 reply Please log in or register to add a comment.