GATE CSE 2019 | Question: 2

Question

The chip select logic for a certain DRAM chip in a memory system design is shown below. Assume that the memory system has $16$ address lines denoted by $A_{15}$ to $A_0$. What is the range of address (in hexadecimal) of the memory system that can get enabled by the chip select (CS) signal?

• A. C800 to CFFF
• B. CA00 to CAFF
• C. C800 to C8FF
• D. DA00 to DFFF

Comments

Isn't RAM and ROM removed since 2016?

---

this question is from GATE 2019

---

I think we need to assume that the chip is consuming 11 address lines here

Answer 1

$(A_{15} \:  A_{14} \:  A_{13} \:  A_{12} \:  A_{11} \:  A_{10} \:  A_9 \:  A_ 8 \: A_7 \: A_6 \: A_5 \:  A_4 \: A_3 \: A_2 \: A_1 \: A_0)$

According to question:

$A_{15} = 1, \: A_{14} = 1, \: A_{13} = 0, \:  A_{12} = 0, \: A_{11} = 1$

So the possible range in binary:

$(\bf{1 1 0 0 1}$$ 0 0 0 0 0 0 0 0 0 0 0) \text{ to } (\bf{1 1 0 0 1}$$ 1 1 1 1 1 1 1 1 1 1 1)$

Converting to Hexadecimal:

$(C800) \text{ to } (CFFF)$

Option A.

Comments

@Ruturaj Mohanty

Why can't you assume the opposite?

I mean $0$ for $1$ and $1$ for $0$?

Is it a norm that its followed like this only in the DRAM?

---

Chip will be selected if output will be 1 of CS logic which is shown. So for that we have to given values accordingly which will decide which value could be addressed and which will not

---

How you came to know that no. Of chip select bits are 5 from A15-A11

---

See the figure, AND gate is given which outputs CS means chips select

Answer 2

The memory system that can be enabled by the chip select signal

$\rightarrow$ From this we can understood that, $CS$ should be $1$.

$\rightarrow$ Note that it is $AND$ gate, $\implies$ all input lines should be $1$.

$\rightarrow$ $A_{15}$.$A_{14}$.$\overline{A_{13}}$.$\overline{A_{12}}$.$A_{11}$ = $1\implies$ $(A_{15} = 1$ and $A_{14} = 1$ and $\overline{A_{13}} = 1$ and $\overline{A_{12}} = 1$ and $A_{11} = 1)$

$\rightarrow$ $\overline{A_{13}} = 1$ $\implies$ ${A_{13}} = 0$

$\rightarrow$ $\overline{A_{12}} = 1$ $\implies {A_{12}} = 0$

$\therefore$ The address denoted by A$_{15}$ to A$_0$ is , (Note A$_{15}$ as MSB and A$_{0}$ as LSB)

 (1100) (1__ __ __) (__ __ __ __ ) ( __ __ __ __ )

$\rightarrow$ For starting address,keep all $\bf{0}$'s in the blanks, and for ending address keep all $\bf{1}$'s in the blanks.

$\rightarrow$ Starting address :- 1100 1$\bf{000}$ $\bf{0000}$ $\bf{0000}$$\implies$(C800)$_H$

$\rightarrow$ Ending address :- 1100 1$\bf{111}$ $\bf{1111}$ $\bf{1111}$$\implies$(CFFF)$_H$

Comments

If you have such a doubt, then check with options... Then you will easily understand what they want

---

Yes, that is what I actually do.

I just wanted to confirm.

Thanks.

---

how we are assuming that a15 is msb, a0 is lsb it can be vice versa it is not mentioned ?

Answer 3

Answer = Option [A]

Answer 4

C800 to CFFF  option A is right

Answer 5

11001 (00000..000 to 11111....111)

which is C800 to CFFF

Answer 6

• A. Range of address will be from C800 to CFFF