0 0 votes A computer with 32-bit wide data bus uses 4 K x 8 static RAM memory chips. The smallest memory this computer can have is: (a) 32 kb (b) 16 kb (c) 8 kb (d) 24 kb CO & Architecture ram co-and-architecture out-of-gatecse-syllabus + – spawndon 4.6k views answer comment Share Follow Print See all 2 Comments 2 2 Comments reply Riya Roy(Arayana) commented Jan 13, 2016 reply Follow flag is the ans 16KB ? 0 0 replyShare spawndon commented Jan 13, 2016 reply Follow flag yes, but how, plz explain? 0 0 replyShare Please log in or register to add a comment.
Best answer 2 2 votes Its a 32bit data bus so you need minimum 4 chips because each chip has 8 bits of data Input/Output (8*4) = 32 bits . Since with 4 chips the smallest memory is : 4 * ( 4kB * 8 ) / 8 = 2 ^14 = 16 KB Riya Roy(Arayana) answered Jan 13, 2016 • selected Jan 14, 2016 by spawndon Riya Roy(Arayana) comment Share Follow See all 2 Comments 2 2 Comments reply spawndon commented Jan 14, 2016 reply Follow flag From my question, where did you get the information: "each chip has 8 bits of Data I/O" ? What does "4K*8" actually mean? 0 0 replyShare Riya Roy(Arayana) commented Jan 14, 2016 reply Follow flag See they said that 4K*8 so the 8 here is no of data lines coming out and 4k is the addressable lines. Since you have 32bit bus you need to have 4 chips minimum. 0 0 replyShare Please log in or register to add a comment.
0 0 votes 32 bit data bus means 4B data is transferred per sec (means memory word size) Given ram - 4K* 8 = 4K*2*4= 8K*4 So memory size = 8Kb Address bus bits= 13 govind answered Jan 13, 2016 govind comment Share Follow See 1 comment 1 1 comment reply spawndon commented Jan 13, 2016 reply Follow flag Hi, please explain a bit, 1] Why did you break 4K*8 into 4K*2*4 ? 2] How did you get 13 bit address bus? Where is 2^13 coming from? 0 0 replyShare Please log in or register to add a comment.
