Question

a computer employs ram chips of 256*8 and rom chips of 1024*8.the computer needs 2k bytes of ram,4k bytes of rom and 4 integer unit with 4 register each. the two highest order bits of address bus are assigned 00 for ram 01 for rom and 10 for integer register 1. how many ram chips will be needed .2 .draw the memory address map for the system.3.give the address range in hexadecimal for ram,rom and integer

Answer 1

Answer:

a) 8 RAM chips and 4 ROM chips are required

RAM chip size = 256*8

Required memory size = 2k bytes = 2*1024 bytes = 2*1024*8 bits

Total number of RAM chip required -

(2*1024*8)/(256*8) = 8 chips

ROM chip size = 1024*8

Required memory size = 4k bytes = 4*1024 bytes = 4*1024*8 bits

Total number of ROM chips required -

(4*1024*8)/(1024*8) = 4 chips

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