DRAM Power calculator

[Ahmad Hassan]

Hi there,

May I ask, what is the size of DRAM that snipper takes into account while modelling the DRAM energy in mcpat.py?

As per Micron datasheet, 1GB DRAM Chip has approximately 0.17J energy per sec which equals to 0.17 Watt power:

= IDD2N * Vdd * 14(ms) + IDD2P0 * Vdd * 986(ms) + 128205 * 27.375(nJ) * 8 // 8 DRAM chips per DIMM
= 0.178 Watt/sec

But in snipper, it is multiplied by (DRAM_CHIPS_PER_DIMM * DRAM_DIMMS_PER_SOCKET  + DRAM_POWER_STATIC_OFFCHIP_INTERFACE) which gives the power of around 4W. This is too high for 1GB DDR3 as per micron data sheet. As per my understanding the power of DDR3 should be around 0.1-0.4 W depending on percentage of time banks are open.

Am I overlooking something?

Thanks.

[Wim Heirman]

Hi Ahmad,

I don't remember the size of the DRAM chip I used. Probably it was for 4 Gbit chips. What does the 0.17 W that you quote cover exactly?

The value in mcpat.py is the idle power (includes leakage, other static power, and refresh) and is configured as 0.111W per device (mcpat.py line 10). The 4W number you observe includes idle power but also the DDR bus static power (0.022W/device, line 16), for all chips in the system. At 8 chips per DIMM, (.111W+.022W)*8 = 1W per DIMM, and 4 DIMMs per socket (line 22) yields around 4W.

If you configure larger systems, the number of sockets will go up: the default configuration is a quad-core socket, so if you specify -n8 you'll get a two-socket system with two DRAM controllers and 4 DIMMs each which should bring the total idle DRAM power to 8W.

Regards,

Wim

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