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CPU Temp. VS voltage

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  • CPUs
  • Processors
Last response: in CPUs
October 4, 2003 8:55:05 PM

Is there any benefit of going to subzero operation of a processor that can not be accomplished through raising the core voltage ?
Then to overclocking means, the important thing is to keep heat away from the CPU, not getting it working to 0 ºC, is that right ?

More about : cpu temp voltage

a b à CPUs
October 5, 2003 2:07:57 AM

Yes. When you reach a certain temperature, your CPU won't clock any higher, regardless of voltage. The higher the temperature, the lower the clock limit. Lowering the temperature does 2 things:

1. Allows higher clocking
2. Allows higher voltage tolerance before heat limits clocking.

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October 5, 2003 10:02:43 AM

P = IV. Increasing the voltage will increase the power dissipation of the processor.

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October 5, 2003 12:00:46 PM

My barton acts wild to even a slight increase in voltage. Take a look at this:

2300mhz 1.65 volts = 39c idle 43c load
2400mhz 1.75 volts = 45c idle 50c load
2424mhz 1.8 volts = 50c idle 55c load

Even just a 24mhz speed increase and .05 volts adds 5c to the temperature! Heres my 2100xp:

2200mhz 1.75v = 44 idle 48 load
2245mhz 1.8v = 46 idle 50 load
2300mhz 1.9v = 50 idle 54 load
2335mhz 2v = 54 idle 59 load

I could run a whole .1v more through my 2100 and run it at the same temperature (2100 1.9v same as 2500 1.8v). And cpu speed doesnt realy make a difference. I tried stock speed 1.75v on my barton and 2400mhz 1.75 and there was only 2c difference.....

<A HREF="http://service.futuremark.com/compare?2k1=7000747" target="_new"> 3D-2001 </A>
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October 5, 2003 2:34:53 PM

OK, thank you
But (always but ...) my real question is if that "limit temperature" is above ambient temperature or below (-4ºC or something like that ...). To go to -4 you need to "refrigerate", to keep less than 50ºC you only need to cool (using my processor to heat the water for my coffee machine. i.e. ;-) ).
Is the percentage earned going to -4ºC really important above an "ideal" cooling solution able to take away as heat as you need ?
Are then Peltier to consider ? Could be useful to cool the air before it is blown to the heatsink, in place of putting the Peltier in the heatsink ? the Peltier would do with fewer power...
why do fan buses feed 18W if a fan usually takes 2 ?
October 5, 2003 7:10:16 PM

After all my reading on transistors and processors the basic answer is: processors work better at lower temperatures.

The problem with heat is that it causes leaks so the cpu will operate incorrectly.

So there are two ways to ensure that your cpu works better: better heat disippation or cooling and lowering the voltage (but lowering the voltage too much will not allow the cpu to work correctly).

Extreme cooling techniques are a way to to be pretty sure that heat is not a factor.
a b à CPUs
October 5, 2003 10:21:55 PM

Tell me something I didn't know. That simple formula doesn't tell you what affect cooling will have on achievable frequency. Processors tend to be able to clock higher at lower temperatures.

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a b à CPUs
October 5, 2003 10:31:27 PM

I think the biggest problem is that a processor has hot spots, that you're not always going to see reported by the CPU temperature probes. CPUs don't conduct heat ideally across the die. And as another member mentioned, at a certain temperature the transistor no longer works (he said "leaks"). Lowering the temperature across the entire contact surface assures that those hot spots are adjacent to even colder spots, so they will cool more efficiently. That's my take on it. And better cooling means transistors that work, but more voltage means more heat. So it's a viscious cycle of trying to remove the heat so you can increase the voltage more, to get better clocks, which also increases heat to be removed.

It's usually not worth it for most hobbiest to cool below ambient temperatures. Which is why water coolers are popular, inexpensive cooling to near ambient temperatures.

Peltier coolers are not exactly ideal as they have a tendancy to turn into hotplates when they fail, burning everything they contact!

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October 6, 2003 12:55:14 AM

Raising the voltage gives a squared relationship with regards to thermal output.

max OC output (W) = max stock output (w) * (OC Mhz / stock Mhz) * (OC Voltage squared/ Stock voltage squared)

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