ISRO2014-17

Question

If each address space represents one byte of storage space, how many address lines are needed to access RAM chips arranged in a $4 \times 6$ array, where each chip is $8\;\text{K} \times 4$ bits?

• A. $13$
• B. $15$
• C. $16$
• D. $17$

Answer 1

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

Comments

Your approach to the question is wrong.We have to solve this question keeping in mind the memory structure as 4*6 array.

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Correct sol is:

First it is required to select one row.For this 2 bits are required as there are 4 rows.Then in each row we have to to select 1byte block , thus require 2 bit again(As in row there are 3 blocks of 1 byte each).Then in each chip we require 13 bits.Thus total address bits are 17bits.

Answer 2

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

Comments

first  divide ram chip (8k * 4)  by 8 to convert into bytes as it has been  said in question that each address space can save up to one byte now there are 4 rows to choose amongst those 4 rows 2 bits required and there are 6 columns it would require 3 bit to locate in which row and which column the ram chip is there therefore 12+ 2 + 3 = 17 bits required tell if this approach is correct or not

Answer 3

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.

Comments

how is the memory capacity 8k/2...can u please elaborate it?

Answer 4

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

Comments

You are missing one thing here: How do you decide which chip to select with in the row?

With in each row there are 6 8k*4 chips

if we form 8k*8 chips there will be total 3 8k*8 chips

So, two adress these 3 chips you will 2 more bits.

So the total sum will be 15+2 = 17

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thanx