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Will RAM @1800mhz and 1.9v fry an i5?

idata
Employee
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Dear Community,

 

 

Recently, I managed to pick up some Kingston HyperX DDR3 RAM sticks at 1800mhz ( http://www.valueram.com/datasheets/KHX1800C8D3K2_4G.pdf http://www.valueram.com/datasheets [...] 3K2_4G.pdf ), but I was unaware of the implications and impacts that this will have on my new intel i5 750 CPU. I am aware that intel states that anything "above 1.65v" will cause damage to i5/i7 processors, and in this case, the RAM seems completely incompatible with mine. Would there be any other approaches to get the RAM working safe and sound? Would underclocking it work (and at the same time, lowering the voltage)?

I'm just really worried that the CPU will get fried before I even get to the BIOS in order to try and crank things down a notch... Many thanks in advance.

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idata
Employee
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The voltage thing is easy: don't do it. Remember power = Voltage**2 / Resistance, so the difference won't be 1.9/1.65 or about 15% more power (heat), it will be 1.9**2 / 1.6**2 or about 33% more power (heat).

Intel has really good brief data sheets as far as the processors go. The chip set data sheets are almost as good.

The processor and chip set have to agree on Front Side Bus speed, but it is between the chip set and the memory how fast that bus will go. In either case, the negotiated speed is the lower of the two, so if your chip set will only do 1.333 GHz, that's how fast your 1.800 GHz memory will go.

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idata
Employee
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Did a little research on the i5-750, which is a very slick little package. Forget about FSB, this thing seems to have the chip set embedded -- memory bus and PCIe included, making FSB is irrelevant. According to the data sheet, it supports 1.033GHz (1.066?) and 1.333GHz memory, so your memory will be negotiated down to 1333. The voltage caveat remains.

A little processor history. In the days of the big mainframes, the fastest logic was ECL -- a non-saturating logic. What that means is you could set the clock rate to 0 and it would still consume about the same power. CMOS is saturating logic, which means that the electrical state becomes static at the clock. The power is consumed going between states, so clock speed is critical: set the clock to 0 and it's power consumption will be pretty close to nil, depending on how the chip talks to the rest of the world. Clock it too fast and it's effective resistance goes down, probably in inverse proportion to the clock rate. P=E**2/R, so if the resistance goes down, the power goes up.

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