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+20 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 $4K \times 16$?

  1. $10$ address, $16$ data lines
  2. $11$ address, $8$ data lines
  3. $12$ address, $16$ data lines
  4. $12$ address, $12$ data lines
in Digital Logic by Veteran
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4 Answers

+32 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$

Correct Answer: $C$
by Active
edited by
In a ROM Matrix

address lines means - horizontal line

data line means - vertical line

correct me if I'm wrong
+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.
by Active
+6 votes
ans c)
by Active
+3 votes
Ans: C

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

data lines, m=16
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