1 votes 1 votes How many address lines and data lines are required to provide a memory capacity of $16 K \times 16$? $10, \:4$ $16, \: 16$ $14, \:16$ $4, \:16$ CO and Architecture ugcnetcse-june2019-paper2 co-and-architecture memory + – Arjun asked Jul 2, 2019 • retagged Jul 5, 2022 by Shubham Sharma 2 Arjun 8.4k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments Verma Ashish commented Jul 3, 2019 reply Follow Share I am not doing n/2 or 16/8 anywhere. It is just $2^{14}\times \color{blue} {16}\implies \color{blue}{16}$ data lines 0 votes 0 votes srestha commented Jul 3, 2019 reply Follow Share @Verma Ashish Why $16$ also cannot be converted in $2^{x}$ format? while for address bit we are just taking a value, which is power of $2?$ 0 votes 0 votes Verma Ashish commented Jul 3, 2019 i edited by Verma Ashish Jul 3, 2019 reply Follow Share Address is used for locating a particular word. And we want entire word so all bits of word are fetched parallely. We can also think as memory is an array of horizontal and vertical lines.. then out of 2^n hz lines we have to select one line. (Using n to 2^n decoder we can do it) So n bits are enough for selecting a particular line. And there are 16 vt. lines which are used for data transfer. Word addressable==> entire word has to be transferred (not some bits)==> data bus size= word size 1 votes 1 votes Please log in or register to add a comment.
5 votes 5 votes ROM memory size = $2^{m}$ × n m = no. of address lines n = no. of data lines Given, 16K × 16 = $2^{14}$ × 16 Address lines = 14 Data lines = 16 So, option C is the right answer. Lakshmikanta answered Jul 7, 2019 Lakshmikanta comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 2^m will decide address line and n will decide data line So here m address line and n data line 16K x 16------>2^14 x 16-----> 14 address line and 16 data line option 3 is correct Arnabh Gangwar answered Jul 11, 2019 Arnabh Gangwar comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Option:(C) 14,16 since the memory size =16K i.e., =(2^4)*(2^10) Therefore, the number of address bits=4+10=14 And,given already given that size of one register=16 bits Therefore,number of data bits=16 Shagun Singh answered Sep 5, 2019 Shagun Singh comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Memory capacity = 16K x 16 lets decode the above (16k x16 ) notation "16K" is the total number of location mean 16 x 1024 = 16384 locations "16" means at each location 16 bit is stored so therefore 16 data line(data bus width) are required So sum up everthing total memory capacity = 16x1024(k)x16 =262144 bit or 32768 byte or 32kbyte total address line required = ${\color{Red} \log_2 16384 = 14}$ total data line required = 16 Above result indicate option (C) is the right answer tron46 answered Nov 22, 2019 tron46 comment Share Follow See all 0 reply Please log in or register to add a comment.