Answer to Question #273196 in Electrical Engineering for Nandini

Question #273196

Describe how one can adjust the address lines,data line if 4096 bytes of RAM is required by a system.The lowest chip size is 256* 8 give complete diagram

1
Expert's answer
2021-11-30T15:04:02-0500

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.

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