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6 Answers

Best answer
22 votes
22 votes

see in the array there are 6*4 =24 chip so to address them we need 5 bit .

and this is bye addressable "each address space represents one byte of storage space" .

so , for each chip 8k*4 bit = (213+2)/8 Byte =212 

so for addressing we need (12 + 5 ) =17 bits line

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6 votes
6 votes
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4
8kx4 8kx4

Since each address space is 1byte, if you visualize 4x6 array as above. instead of 6 chip in each row.

Here 12 rows are there each row 1byte.[ 4bits +4bits]

to select any one row out of 12 rows - 4 bit

then once we select the row we need to select any 1 line among 8K address lines

so 8K = $2^{3}$ x $2^{10}$

4bit + 13bit = 17 bit

1 votes
1 votes

Whenever we are given chip description as $xK\times y$, address lines required are $logxK$ and data lines required are $y$.

In this question, the size of chip is $8K\times 4$, which is a nibble, meaning one memory cell is storing only half of data. We'll need two simultaneous/consecutive cell access to output one byte of data, effectively halving storage capacity of the chip.

This means the memory capacity of chip is actually $\frac{8K}{2}=4K$. For $4K$ memory we need $log4K = 12$ address lines.

There are $4\times6 = 24$ such chips. We need $\left \lceil log 24 \right \rceil=5$ bits to select one out of $24$ chips.

Total address lines are $12 + 5 = 17$. OPTION D.

0 votes
0 votes
option B seems  correct answer,

each chip require 13 address lines as there are 8K rows,

also there 4 chips lined one below another, creating 4 rows o chips, that would take 2 bit to select one chip out of 4

hence in total 13 + 2 = 15 address lines
Answer:

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