Answer to Question #271491 in Electrical Engineering for Saurav kumar

Question

Describe how one can adjust the address lines, data

lines, if 4096 bytes of RAM is required by a system.

The lowest chip size is 256*8.

Give complet diagrams.[5 marks]

Answer 1

1.

a) 8x16

• Number of words = 8
• Number ofbits per word= 16
• So, in 8x16, the number of address lines is obtainedby number of words, that is,
• 8 = 2^3
• Therefore, it requires 3 address lines.
• The input-outputlines are calculated as, sum of address lines and the number of bits, that is,
• = 3 + 16
• = 19
• Therefore, it requires 191/0 lines.

b) 2Gx8

• Number of words=2G
• Number ofbits per word= 8
• So, in 2G x 8, the number of address lines is obtainedby number of words, that is,
• 2G = 2x 2^30
• =2^31
• Therefore, it requires 31 address lines.
• The input-outputlines are calculated as, sum of address lines and the number of bits(data lines), that is,
• = 31 + 8
• = 39
• Therefore, it requires 391/0 lines.

c) 16M x 32

• Number of words= 16M
• Number ofbits per word= 32
• So, in 16M x 32, the number of address lines is obtainedby number of words, that is,
• 16M =2^4 x 2^20
• =2^24
• Therefore, it requires 24 address lines.
• The input-outputlines are calculated as. sum of address lines and the number of bits(data lines), that is,
• = 24 + 32
• = 76
• Therefore, it requires 76 I/0 lines.

d) 256Kx 64

• Number of words= 256K
• Number ofbits per word= 64
• So, in 256K x 64, the number of address lines is obtainedby number of words, that is,
• 256K = 2^8 x 2^10
• = 2^18
• Therefore, it requires 18 address lines.
• The input-outputlines are calculated as, sum of address lines and the number of bits(data lines), that is,
• = 18 + 64
• = 82

Therefore, it requires 82 I/0 lines.

2.

a)

• To calculate number of bytes stored in the memory, 8 words require 16 bits.
• But 8 bits is one byte and 16 bits = 2 bytes.
• Therefore, it requires 8 x 2 = 16 bytes of memory to store in each memory.

b)

To calculate number of bytes stored in the memory, 2G words require 32 bits.

• 2G = 2x 2' = So, 2'= 2147483648 words.
• But a byte is 8 bits, so 8 bits =1 byte.
• Therefore, it requires 2147483648 x 1= 214748 3648 bytes of memory to store in each memory.

c)

To calculate number of bytes stored in the memory, 2G words require 32 bits.

• 16M =2^4 x 2^20
• =2^24
• So, 2^24 = 16777216 words.
• But a byte is 8 bits, so 32 bits = 4 bytes.
• Therefore, it requires 16777216 x 4 = 67108864 bytes of memory to store in each memory.

d)

To calculate number of bytes stored in the memory, 256K words require 64 bits.

• 256K = 2^8 x 2^10
• = 2^18
• So, = 262144 words But a byte is 8 bits,
• so 64 bits = 8 bytes.
• Therefore, it requires 262144 x 8 = 2097152 bytes of memory to store in each memory.