By how much can the maximum absolute voltage of 2.3V for DDR2 DRAM chips be exceeded? I don't intend to do it, but I have a motherboard that lets the memoy voltage be set as high as 2.839V, and it's maker says that's OK, although they won't guarantee against damage.
Nothing. That would be why they call it "absolute maximum" eh? Maybe you get lucky and nothing bad *apparently* happens, maybe it goes Pffft!, maybe it just slowly corrupts your files.
The motherboard manufacturer or the memory manufacturer? An engineering rep or some guy in an off-shore call center?
-- Rich Webb Norfolk, VA
One way to get that info, is if the manufacturer writes up an ISSCC (International Solid State Circuits Conference) paper. Sometimes, if they've done a new semiconductor process, the details will be given in the paper, including the "real" absolute max (i.e. the value that killed the device in the lab, something akin to breakdown voltage). The only time I read and compared such a value using a paper like that, it was about 200mV above the datasheet value, which isn't much margin at all (and that was for a non-Intel processor).
Or, you can look in an enthusiast forum, and see what they're up to. Some Micron D9 tested at 2.51V. Try to find a thread where they burned some up :-)
Paul
All the RAM chip makers and engineers who used to work for local chip companies all agreed with you -- emphatically. It was the motherboard maker, BioStar that said it was OK to exceed 2.3V, even though they wouldn't guarantee against damage to the memory, and they said the memory would very likely burn up at 2.4V. I don't think any of those people at BioStar knew anything -- too pretentious and authoritarian.
The fine geometry MOS these days is nearly like eeprom of the dark ages. There is a threshold shift of the fets as they are clocked. My recollection is threshold voltages get larger over time, but don't quote me. The shift is incorporated into the spice models (or should be). [If the threshold increases, then the weak model.] If you increase the voltage, you probably won't get the 10 year life for which the process is probably specd.
Stick with Asus or Gigabyte. I'm leery of 2nd tier mobos.
If they won't guarantee it, then they know it is too high. The numbers you cite are almost 20% higher than spec.
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Read your own question again, with emphasis as shown.
In other words, pretty much zero, if you want it to operate as advertised.
Good Luck! Rich
The world may never know:
The motherboard manufacturer, BioStar, wouldn't guarantee against damage even if I never exceeded the chip maker's maximum absolute of
2.3V, nor would they guarantee that the voltage applied to the memory would never go above that by itself, such as because of a glitch."larry moe 'n curly" wrote in news:a6b3aff5- snipped-for-privacy@f20g2000pro.googlegroups.com:
Actually I think they just mean that you're welcome to try. They tell you what they think is right, but they're quite happy to let people thrash test at their own expense and blog about it. After all, what better publicity can there be for them? Beats any possible claim they could make for themselves. When they tell you they think it will burn up at n.nn volts, they're just giving you fair warning.
These days, a piece of string is as long as a million tech-happy bloggers say it is, so long as enough of them form a consensus about it.
By as much as you want!
For the most part though, voltage is directly related to temperature. (simply ohms law) Increasing the voltage generally increases the core temperature. The life expectancy decreases exponentially in many devices as the voltage is increased past it's absolute maximums. The maximums also tend to be "ideal"(hence the usage of absolute). Because of variations of manufacturing, usage, etc, the maximums are a statistical result. Your usage may vary.
"Jeff Johnson" wrote in news:i75mh8$2gc$ snipped-for-privacy@news.eternal-september.org:
Ohm's law has nothing to say about temperature!
While a resistance or voltage drop might vary with temperature, Ohm's law does not deal with that.
"Jeff Johnson" wrote in news:i75mh8$2gc$ snipped-for-privacy@news.eternal-september.org:
Which, because it deals with the increase in power, is actually Watt's law (P=V/I), but that doesn't say anything about heat and temperature either. :)
P=V/I ?
That should turn up the heat....
mike
m II wrote in news: snipped-for-privacy@news.x-privat.org:
Oh yeah, it would, wouldn't it? :)
P=V*I....
Huh?
-- "For a successful technology, reality must take precedence over public relations, for nature cannot be fooled." (Richard Feynman)
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