In an SDRAM how do address rows/columns and rank width and bank width relate to the total memory size?

Question

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?

Answer 1

We're talking about an SODIMM module here. It has multiple chips on it, and has an overall format of 1G (2³⁰) locations of 64 (2⁶) bits each. (Total of 2³⁶ bits.)

The module contains 16 (2⁴) chips that contain 4G (2³²) bits each. (Total of 2³⁶ bits.)

The memory in each chip is organized as 8 (2³) banks, each with 64K (2¹⁶) rows and 1024 (2¹⁰) columns of locations that contain 8 (2³) bits each. (Total of 2³² bits per chip.)

Answer 2

This module has 16 chips in parallel, each chip holding 512 Mbytes. The chips are arranged in two ranks as two groups of 8 chips. Each chip is 4 Gbit in size. Each chip provides 8 data pins, so each group of 8 provides 64 data pins - these are connected in parallel to the module pins. So 2^32 bits per chip times 2^4 chips = 2^36 (8 GB). Now, addressing: 2^3 = 8 banks, 2^16 = 65536 rows, 2^10 = 1024 colums, and 2^3 = 8 data pins (bits per address) results in 2^(3+16+10+3) = 2^32 bits.