Question

Given: 2 MB of physical R/W memory, composed of multiple 256KB chips, a CPU with a 21 bit address bus and an 8 bit data bus. Answer the following questions. h. (1 pt) Suppose I replaced the 256K RAM chip at the highest address, with a 256K EPROM chip, what would the address be of the lowest byte in the EPROM?

Answer 1

--> no of 256KB chips in 2MB = 2MB / 256 KB

                                                = 2²¹ / 2¹⁸

                                                = 8 chips

--> can be addressed using 3 bits.

--> address of last chip is 111

--> address of lowest byte is all zeroes i.e. 18 zeroes

--> so total address is 1 1100 0000 0000 0000 0000

--> converted to hex gives, 0x1C000.