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5 votes
5 votes

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? 
    1. 128, 16, 11, 5, 5 to 32 
    2. 256, 16, 11, 5, 5 to 32 
    3. 128, 16, 10, 6, 6 to 64 
    4. 512, 15, 10, 5, 5 to 32 

2 Answers

Best answer
11 votes
11 votes

(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. 

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