DDR3 标准/数据表中的 CK (tCK, nCK) 单位歧义？

Question

I am designing a simplistic memory controller and PHY on an Artix-7 FPGA but am having problems reading the datasheet. The timings in the memory part's datasheet (and in the JEDEC JSD79-3F doc) are expressed in CK/tCK/nCK units, which are in my opinion ambiguous if not running the memory at the nominal frequency (eg lower than 666 MHz clock for a 1333 MT/s module).

If I run a 1333 MT/s module at a frequency of 300 MHz -- still allowed with DLL on, as per the datasheet speed bins, -- is the CK/tCK/nCK unit equal to 1.5 ns (from the module's native 666 MHz), or 3.33 ns (from the frequency it is actually run at)? On one hand it makes sense that certain delays are constant, but then again some delays are expressed relative to the clock edges on the CK/CK# pins (like CL or CWL).

That is to say, some timing parameters in the datasheet only change when changing speed bins. Eg tRP is 13.5 ns for a 1333 part, which is also backwards compatible with the tRP of 13.125 ns of a 1066 part -- no matter the chosen operating frequency of the physical clock pins of the device.

But then, running a DDR3 module at 300 MHz only allows usage of CL = CWL = 5, which is again expressed in "CK" units. To my understanding, this means 5 periods of the input clock, ie 5 * 3.33 ns.

I suppose all I am asking is whether the "CK" (or nCK or tCK) unit is tied to the chosen speed bin (tCK = 1.5 ns when choosing DDR3-1333) or the actual frequency of the clock signal provided to the memory module by the controlling hardware (eg 3.3 ns for the 600 MT/s mode)?

Answer 1

This is the response of u/Allan-H on reddit who has helped me reach a conclusion:

When you set the CL in the mode register, that's the number of clocks that the chip will wait before putting the data on the pins. That clock is the clock that your controller is providing to the chip (it's S DRAM, after all).
It's your responsibility to ensure that the number of clocks you program (eg CL=5) when multiplied by the clock period (eg 1.875ns) is at least as long as the access time of the RAM. Note that you program a number of clocks, but the important parameter is actually time. The RAM must have the data ready before it can send it to the output buffers.
Now let's run the RAM at a lower speed, say 312.5MHz (3.2ns period). We now have the option of programming CL to be as low as 3, since 3 x 3.2ns > 5 x 1.875ns.
BTW, since we are dealing with fractions of a ns, we also need to take the clock jitter into account.

Counterintuitively, the DRAM chip doesn't know how fast it is; it must be programmed with that information by the DRAM controller. That information might be hard coded into the controller (eg for an FPGA implementation) or by software which would typically read the SPD EEPROM on the DIMM to work out the speed grade then write the appropriate values into the DRAM controller.

This also explains timing values defined as eg "Greater of 3CK or 5ns". In this case, the memory chip cannot respond faster than 5 ns, but the internal logic also needs 3 positive clock edges on the input CK pins to complete the action defined by this example parameter.