Question

A computer employs RAM chips of 512 x 8 and ROM chips of 2048 x 4. The computer
system needs 2M x 16 of RAM, and 4K x 16 of ROM and two interface units with 256
registers each. A memory-mapped I/O configuration is used.
a. How many RAM and ROM chips are needed?
b. Tell, how many lines of the address bus must be used to access total memory?
How many of these lines will be common to all chips?
c. Propose, the number of lines must be decoded for chip select? Specify the size of
the decoder
d. Devise a memory-address map for the system and Give the address range in
hexadecimal for RAM, ROM and interface.
e. Design memory chip layout for that address map.

Answer 1

Answer:

computer employs RAM chips of 512 x 8 and ROM chips of 2048 x 4. The computer

system needs 2M x 16 of RAM, and 4K x 16 of ROM and two interface units with 256

registers each. A memory-mapped I/O configuration is used.

a. How many RAM and ROM chips are needed?

b. Tell, how many lines of the address bus must be used to access total memory?

How many of these lines will be common to all chips?

c. Propose, the number of lines must be decoded for chip select? Specify the size of

the decoder

d. Devise a memory-address map for the system and Give the address range in

hexadecimal for RAM, ROM and interface.

e. Design memory chip layout for that address map. mean

Answer 2

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