Question

The memory for a certain computer consists of 4 Mbytes of RAM. This is to be made up from 256 K by 4 bits RAM chips.

(a) What is the number of RAM chips are needed in total?


(b) What is the number of address bits that must be fed to an address decoder for this system?

Answer 1

a. we know

2^8  = 256
2^10 = 1024 = 1 KB
2^20 = 1 MB
2^30 = 1 GB
2^32 = 4,294,967,296 = 4 GB = 2^22

256 x 4 bits is the capacity of a chip which means there are 256 rows in the ram with 4 bits in each row (1 nibble).

256 x 1024 x 8 bits = 256KB

1M = 1024k hence size of new ram

So size of 4M = (4*1024k *8)

So number of chips required   = (4*1024*8)/(256*4) = 32

b. Memory needs to address each word.A word is 4 bytes . Also 1 chip 4 bits and for byte (8 bits) now we just have to know the size of decoder and since there's 4 RAMs we should connect each one with a wire , and we can differntiate between those RAMS 2^5 = 32 , then all we need is 5 bits to differintaie , so the decoder should be 5*32.

Now you need log2(n) bits to address n bytes.hence you need 20 bits to address every byte in a megabyte

1MB = 2^22 hence we need 22 bits address for a 4MB memory.