handmadeDDR2 | "Double Data Rate 2"
NOTE: With this article I am not providing academic study material but simply a summary that can be understood even by a novice. A detailed explanation would take several hours and days to understand.
You are supposed to know what the f..k a RAM is if you are reading this.
| Structure | Explanation |
|---|---|
| Memory Channel | The memory channel control the "set" of DIMM's |
| DIMM | The DIMM (Dual In-Line Memory Module) is the "RAM MODULE" |
| Rank | The Ranks are the faces of the DIMM where a set of DRAM chips are placed |
| Bank | Each DRAM chip is devided into several "sectors" that can be imagined as chips |
| Row | The row can be compared to the traditional "address" of a RAM chip or the row of a notebook |
| Column | The Column can be compared to an additional address that define the memory location in a "2D" place, as a matrix |
For this example we are going to use a winbond 512 Mbit DDR2 chip. This is a 64MB (512 Mbit) DDR2 chip with transfer rates up to 1066 Mbps (DDR2-1066). It store DATA in a 16 bit length so for a 64 bit CPU architecture we would just need 4 chip paired and we would archive a total of 256 MB DDR2 module. If we would had a 8 bit DATA length we would need to pair 8 chips.
| BANK | ROW | COLUMN |
|---|---|---|
| 4 BANKS (2 BIT ADDRESS) | 8192 ROWS (13 BIT ADDRESS) | 1024 COLUMNS (10 BIT ADDRESS) |
Simple math: total addressing is: 33'554'432 addresses = BANKSxROWSxCOLUMNS
WHY? BANKS->2 BIT | ROWS->13 BIT| COLUMNS->10 BIT| TOTAL->25 BIT ADDRESS-> 2^25=33'554'432 LOCATIONS
If we want to find the total memory size we just have to multiply locations per data width -> 33'554'432 x 16 = 536'870'912 bits -> /8 -> 67'108'864 Bytes -> 64 MB
Simple? Yes.