Question

A computer employs RAM chips of 256 x8 and ROM chips of 1024 x8.The computer system needs 2KB of RAM,4 KB of ROM and four interface units,each with four registers.A memory-mapped I/O configuration is used .The two highest order bits of the address bus assigned 00 for RAM,01 for ROM and 10 for interface register.i) How many RAM and ROM chips are needed? ii) Draw a memory address map for system. iii)Give the address range in hexadecimal for RAM,ROM and interface.

Answer 1

A byte is the minimum addressable quantity in a system. In this case , you wrote that the specific memory is arranged as 16K locations of 8 bits each . Thus , there are 16KB .