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+14 votes
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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 $4K \times 16$?

  1. $10$ address, $16$ data lines
  2. $11$ address, $8$ data lines
  3. $12$ address, $16$ data lines
  4. $12$ address, $12$ data lines
asked in Digital Logic by Veteran (59.6k points)
edited by | 2.2k views

4 Answers

+19 votes
Best answer
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$
answered by Active (4.3k points)
edited by
+7 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.
answered by Active (4.8k points)
+6 votes
ans c)
answered by Loyal (5.2k points)
+3 votes
Ans: C

add lines, 2^n= 4K= (2^2)*(2^10)= 2^12... ; n=12

data lines, m=16
answered by Loyal (7.4k points)


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