I have a Micron SDRAM (MT16KTF1G64HZ-8GB). The size of the memory is 8GB. I did some calucaltions and 8GB of data means 2^36 bits capacity. Now when I look in the Micron data sheet, the row address is 16 bits wide and the column address is 10 bits wide. That will give us a total of (2^16)*(2^10) bits which is 2^26 bits. But the memory width should be 2^36. How are the rest of the bits in the memory addressed?
The data sheet also mentions that this is a dual rank memory. Does this mean that there are two sets of 2^26 bits? Even then, that would result in 2*(2^26)=2^27 bits. We're still not at 2^36 bits.
The data sheet also mentions that there are 8 device banks. Does this mean that each bank has 2^27 bits? But even then it would give us 2^30 bit. Still not at 2^36.
I'm really confused. Could someone please help me?