How many 2K x 8 RAM chips are needed?

Question

We want to build a memory with 4-byte words and a capacity of 2²¹ bits. 

1. How many 2K x 8 RAM chips are needed? 
2. How many address lines are needed for the memory? 
3. How many of these address lines are connected to the address inputs of the RAM chips? 
4. How many of these address lines will be used to select the appropriate RAM chip(s)? 
5. What size decoder is needed to build the memory? 
   • A. 128, 16, 11, 5, 5 to 32 
   • B. 256, 16, 11, 5, 5 to 32 
   • C. 128, 16, 10, 6, 6 to 64 
   • D. 512, 15, 10, 5, 5 to 32 

Comments

this is not in syllabus now right ?

Answer 1

(1) Memory Capacity = 2²¹ bits

RAM chip capacity = 2K * 8 bits = 2¹⁴ bits

So, number of RAM chips required = 2²¹ / 2¹⁴ = 2⁷ = 128

(2) Memory needs to address each word.
A word is 4 bytes. So, number of possible words = 2²¹/(4*8) = 2²¹/2⁵ = 2¹⁶
So, number of address lines needed for memory = 16

(3) A RAM chip has 2K rows of cells to select (each row has 8 cells of 1 bit each which will always be selected together). 
So, we need 11 bits to select any of these 2K rows.

(4) 16-11 = 5 address lines will be used to select the appropriate RAM chip(s)- 5 address lines to select 128 chips doesn't seem logical but this is how multi-byte words can be fetched from memory in parallel. At any time 4 RAM chips will be selected together and a byte can be fetched in parallel from all these, resulting in 4 bytes being fetched in 1 memory cycle. (See (s) in the question)

(5) We use 5 address lines to select the RAM chips. So, decoder needs to be 5 to 32. 

(A) choice. 

Comments

Can you please clarify my doubt ?

   We have total 128 chips as per the calculations above. Now in order to address one particular chip from these 128 chips we must have 128 different addresses right? but here decoder gives only 32 addresses. I am little confused there.

Answer 2

2^21/(2 X 2^10 X 2^3)=2^7=128