Question

Consider a 512 Mb DRAM with 8 banks, 256k subarrays, and 8 bit word. How are the address bits allocated? How many subarrays are there and how are they arranged?

Answer 1

Answer:

26 address bits are needed to address 2∧26 locations.

There are 24 = 16 subarrays.

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