Question

A computer has a memory space of 16 GB. a) How many address lines are required to span this address space, assuming it is byte- addressed? b) This computer has a block of 4 GB 32-bit-wide memory built using 512 MB static RAM chips that are each 8 bits wide. How many RAM chips are required to implement the memory?

Answer 1

16 GB = 2 ^ 4 * 2 ^30 B = 2 ^ 34 B = log base 2 ( 2 ^ 34 B ) = 34 bits ---- [1 G = 2 ^ 30 1 M = 2 ^ 20 ]

a) 34 bits represent address lines each having size of 1 B making a size of 16 GB so total of 2 ^34 address lines are required .

b) Memory Block of 4 GB = 2 ^ 32 B = 2 ^ 30 * 2 ^ 2 B which means 2 ^ 30 address lines each having size of 2^2 i.e   (32 bit or 4 B) making a total of 4 GB .

Static RAM Chips of 512 MB each of 8 bits wide constitute 2^9 * 2^20 * 1 B = 2 ^29 * 1 B - - [ 512 = 2 ^ 9]

2 ^ 30 * 4 B / 2 ^ 29 * 1 B = 2 ^1 * 4 B which means total of 2 RAM chips each of size 4 B would be required to implement the memory .

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