Answer:
26 address bits are needed to address 2∧26 locations.
There are 24 = 16 subarrays.
Step-by-step explanation:
Considering a 512Mb DRAM, it means it can store 512*2∧20 bits = 2∧9X2∧20 = 2∧29. Since it is a 8 bit word, at a time, 8 bits are accessed. Therefore, the capacity can be re-written as 2∧26X 2∧3 = 2∧26X8
![It simply means 26 address bits are needed to address 2∧26 locations, since there are 8 banks, 3 address bits are needed to select a bank ( 8 = 2∧3 )]()
![Each subarray is 256K which means 19 bits are needed to select one address from this 256K subarray ( since 256K = 2∧19 )]()
![22 addresses are used up out of a total of 26 addresses (19 for locating a location in 256K subarray and 3 for banks). The remaining addresses are 4 ( 26 - 22 ). These 4 addresses are used to locate a subarray. thus there are 2∧4 = 16 subarrays]()
There are a total of 8 banks and within a bank, there are 32 subarrays.
Within each subarray there are 256K locations each of 8 bit size.
Note: " ∧" means raise to power