UGC NET CSE | June 2019 | Part 2 | Question: 19

Question

How many address lines and data lines are required to provide a memory capacity of $16 K \times 16$?

• A. $10, \:4$
• B. $16, \: 16$
• C. $14, \:16$
• D. $4, \:16$

Comments

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

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@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 $2?$

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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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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