Question

A computer employs ram chips of 256 x 8 and rom chips of 1024 x 8. the computer system needs 2k bytes of ram, 4k bytes of rom, and four interface units, each with four registers. a memory-mapped 1/0 configuration is used. the two highest-order bits of the address bus are assigned 00 for ram, 01 for rom, and 10 for interface registers.

a. how many ram and rom chips are needed?

b. draw a memory-address map for the system.

c. give the address range in hexadecimal for ram, rom, and interface.

Answer 1

Answer:

a) 8 RAM chips and 4 ROM chips are required

Explanation:

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

Answer 2

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