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I have a RAM-using application that

  • would occasionally find extra RAM helpful, and
  • can afford the extra hardware, but
  • has a tight power budget.

Thus, I would include the extra RAM only if this extra RAM did not draw too much power when unused.

How can I calculate the power the RAM draws when unused?

I am trying to calculate this power as follows. Please tell me if and where I err.

Here is a sample datasheet for a 4-gigabit DDR3 chip: Micron's MT41K1GM4 DDR3L-RS-1600. Relevant specs:

  • Burst refresh current: 215 mA.
  • +45-degree-C temperature self refresh: 6.0 mA.
  • VDD = VDDQ = 1.35 V.

As far as I know (though I could be wrong), the "Burst refresh current" is irrelevant to my question. If this is right, then I believe that I need a power-overhead budget of (6.0 mA)(1.35 V) = 8.1 mW to carry this chip when my device is not using it, which will be most of the time. By, unused, as @Rocketmagnet has observed, I mean that no reading or writing is going on; I do not mean that the RAM can forget its data.

If all this is right, I can afford the 8.1 mW. However, am I overlooking something important, please? I ask because I cannot afford much more power just to carry hardware that won't much be used.

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Don't forget about VTT termination power. Also they have some excel power calculators: micron.com/products/support/power-calc A quick look at their calc showed standby power was around 100mW. The calc seems to have really good info. – Some Hardware Guy Oct 3 '12 at 16:16
1  
When you say "unused"? Do you mean that you could power down the RAM and lose the contents? – Rocketmagnet Oct 3 '12 at 16:22
   
@Rocketmagnet: No. Good question. The RAM's contents must be maintained. Otherwise I'd just gate the power, and then the power overhead would be zero. – thb Oct 3 '12 at 17:30
@SomeHardwareGuy: You coult make your comment an answer. I would upvote it. – thb Oct 3 '12 at 19:11
@SomeHardwareGuy: If you answer and feel like extending your earlier, useful comment: what is VTT termnination power, please? I understand VDD/VCC, but VTT does not mean anything to me, nor do I understand what is terminated at VTT, nor does a search-engine search seem to turn up anything about VTT that relates to the present question. Aren't all the DDR's terminations transmission-gated when not in use? As far as I know, a CMOS transmission gate neither dissipates power, nor transmits power, nor even uses any control power when open and not switching. – thb Oct 3 '12 at 19:21

1 Answer

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Don't forget about VTT termination power. Also they have some excel power calculators: micron.com/products/support/power-calc A quick look at their calc showed standby power was around 100mW. The calc seems to have really good info.

Re: VTT Term I believe you're correct when you're not doing anything the power consumed is probably negligible. In DDR you need to provide VTT termination for the signals, basically a resistor tied to 1/2 your VDD voltage. Again it probably doesn't mean anything for power.

Don't forget that you have to refresh the DDR periodically, setting it into self refresh mode should do that for you. That should be your lowest power mode, and you don't have to provide a clock to it at that point.

Oh and if you can, consider LPDDR, the lower power stuff is around 1.35V now I think. I just read an article the other day that low power dram orders have now surpassed regular DDR so it might even be cheaper for you.

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