Can clocks as opposed to voltage damage components?

Heat resulting from the increased clock speed could but the clock speed alone, no.
 
Jokester said:
Assuming no change in voltage, if you increase the clock speed by 50% you'll increase heat by 50%.

Jokester
Click to expand...

So assuming thats correct a Conroe running @ 3.0ghtz with a temp of 40C with a set fan speed, then unclocking it by 50% will naturally give half then?..... 20C?......

Not sure wether i agree or not..... :confused:
 
He doesn't mean the direct temperature reading, he means the heat given out by the CPU. The temp reading you get from your motherboard isn't directly comparable to the heat output of your cpu. Down clocking quite clearly won't work in the same way, not to mention the fact that ambient air temperature is 20degrees.

Then again I did get 17 degrees with my 3000+ once :cool: Cool winters day + open window + ski gear = so much fun.
 
t31os said:
So assuming thats correct a Conroe running @ 3.0ghtz with a temp of 40C with a set fan speed, then unclocking it by 50% will naturally give half then?..... 20C?......

Not sure wether i agree or not..... :confused:
Click to expand...
If you think about it in terms of power output (watts) then I think it is true. When you start translating the Watts into temperature you get things like ambient temperatures getting involved, and it might not be as simple as a linear relationship.
 
No, say at 3GHz its generating 100W of heat at full load, reducing it's speed to 1.5GHz will mean it only produces 50W of heat.

Jokester
 
it depends on the cooling and the cpu as to the heat increase from overclocking. if done properly overclocking shouldnt damage any components.

MW
 
Jokester said:
No, say at 3GHz its generating 100W of heat at full load, reducing it's speed to 1.5GHz will mean it only produces 50W of heat.

Jokester
Click to expand...
So, If you measured the current of a CPU, it would use half the amps at 1.5ghz right? If the voltage (VCore) is constant? If so, cool. :cool:
 
Generally it follows

Overclocked Watts = Default Watts * (Overclocked Mhz \ Default Mhz) * (Overclocked Vcore \ Default Vcore)²

So if voltage is constant then

Overclocked Watts = Default Watts * (Overclocked Mhz \ Default Mhz)

Using Jokester example

Half speed gives

W = 100 x 1.5/3 = 50w

---------

For temperature, that's dependent on the thermal resistance of the cooling measured in C/W. Tc is the variable to watch. Follow the example below, as the thermal resistance is a constant then the Tc will increase in direct proportion to the CPU wattage. Assuming a constant chassis temp, then its a standard y=mx+C linear scale. Up until the cooling can dissipate no more.

TR = RCS + RSA = (Tc - Ta)/Pd

or Tc = Pd*TR + Ta

where:
TR = Total Thermal Resistance
Tc = Tcase, cpu case temperature
Ta = chassis ambient temperature
Pd = total cpu power dissipation
RCS = thermal resistance, case-to-sink
RSA = thermal resistance, sink-to-ambient

e.g. A C2D E6600 at stock speeds

Say a Tuniq Tower at 2000RPm with AS5
TR = 0.16 + 0.0045 = 0.1645 C/W

Ta = 25C (thermometer inside case)

Pd = 65W stock settings as per Intel TDP @ 2.4GHz 1.325v

So put in the numbers

0.1645 = (Tc - 25)/65 Therefore Tc = 35.7C
 
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