Let us take a concrete example.
A given rhythm processor has 2400 MHz, having a TDP of 60 Watts and a tension of origin of 1.4v.
If one compares it with another processor, whose characteristics will be the same ones except the frequency, 2200 MHz for example, one calculates a coefficient of different adjustment for these two processors.
- Processor 1:
k1 = p (V ² *f)
k1 = 60 (1.4 ² *2400)
k1 = 0,012755102040816326530612244897959
- Processor 2:
k2 = p (V ² *f)
k2 = 60 (1.4 ² *2200)
k2 = 0,013914656771799628942486085343228
It is observed that coeffcient it adjustment of processor 2 is higher than that of processor 1.
Let us see now what occurs in several common cases from overclocking.
Overclocking case N°1. Increase in the frequency of 500 MHz.
By calculation,
P = (f+500) *V ² *k
processor 1 develops ~ 72.49 Watts
processor 2 develops ~ 73.63 Watts
Obervations: One can notice that the tension was not modified, and that however the processors dissipate more Watts. The temperature thus will increase.
Overclocking case N°2. Increase in the frequency of 500 MHz and the tension of 0.2v.
By calculation,
P = (f+500) * (V+0.2) ² *k
processor 1 develops ~ 94.69 Watts
processor 2 develops ~ 96.17 Watts
Obervations: One can notice that the processor with the lowest coefficient of adjustment dissipates less. And this, although its frequency of operation is higher. Note retain the importance of K here.
Overclocking case N°3. Increase in the two processors at a frequency of 3000 MHz with a tension of 0.3v.
By calculation,
P1 = (f+600) * (V+0.3) ² *k
P2 = (f+800) * (V+0.3) ² *k
processor 1 develops ~ 110.58 Watts
processor 2 develops ~ 120.17 Watts
Obervations: One notes a variation of 10 Watts between these two processors has equal frequencies/tensions. One can deduce from it, that more the difference between the coefficients of adjustment (and thus instantaneous frequency deviation, but not tension and TDP of origin) will be large, more the equal variation with overclocking will be tall.
A strongly overclocké processor will consume always more than one little overclock processor, has frequency/tension equal, it is what will not interest us the continuation.
- Why the temperature, and why necessity increases to provide a system with powerful cooling?
Each system of cooling in our use will have a thermal resistance.
Thermal resistance notes °C/W, the design and the composition of our system of cooling will determine its thermal resistance. More it is low and best cooling will be. It is its capacity to dissipate Watts produces by the processor, and more precisely, to bring its temperature closer to the ambient temperature. That is to say x°C/W increase in x°C per dissipated Watt.
It is noted that the elementary materials have a more or less raised thermal conductivity.
That is to say 401 W (m*K) (Watt by meter-Kelvin) for Copper, 429 W (m*K) for the Money, and 237 W (m*K) for Aluminium, these values are given has 20°c. The more this value is raised, the more the material is conductive thermically.
Also let us note that the values however above vary according to the temperature and from moisture, and that thermal conductivity is directly in connection with electric conductivity, for example, the conductivity of a diamond is close to the null one.
With similar design a Copper ventirad will be more éfficace that a ventirad in Aluminiu.
Two points:
it does not use money, or diamond (between 895-2300 W (m*K), because they are too much expensive, and that their propritétés, although strongly intéréssantes, would not allow much better results than those obtained with copper. Let us recall that the limiting factor is the temperature of the ambient air.
it ventilator which blows on the radiator, of watercooling, or the ventirad, is also an important factor. Its capacity has renouveller the air on the surface of the exchanger is a factor determining in the capacity of this one to evacuate heat. It as should be known as when the air velocity increases, it cooled.
In a given system, thermal resistance is defined by the ambient temperature and the temperature of the processor, and its load to be dissipated.
For a system whose thermal resistance is of 0.15°C/W, with an ambient temperature of 20°c and one 60 Watts has to dissipate, one will have a temperature of 9°c to the top of the ambient temperature,
either 29°c, or (0.15*60) +20 = 29.
With a load of 100 W, one obtains 35°c, but while trying to dissipate this same load with a less powerful system half, let us say whose thermal resistance is of 0.30°C/W, one obtains 50°C. That packs quickly in summer, with an ambient temperature of 32°c for example, that would give us 62°C.
This data is non-néglieable in a passive system for example, knowing that thermal resistance will be relatively high.
By increasing the load to be dissipated, one thus increases the temperature, by decreasing thermal resistance, one decreases the temperature (in certain proportions).
Only the systems with phase shift propose negative thermal resistances, however to speak about negative resistance is scientifically false.