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WC loop plans + lots of questions - Page 2

post #11 of 12
Quote:
Originally Posted by tom thumb 2 View Post
Really? I thought it was linear with clock speed.
Not quite. It depends a lot on the voltage that you put through the chip too. In fact more on the voltage than the clock speed.

Quote:
Originally Posted by tom thumb 2 View Post
How does it's temperature affect the power it uses? I think you are mixing cause and effect. If not please explain!
It is a second order effect to be sure, but the hotter a electric component is the more resistance it has. The higher the resistance the more power it draws. It makes quite a small difference overall, I was being a tiny bit facetious when I said that.

Edit:

Charliehorse describes it better than I can (be bothered to):

Quote:
Originally Posted by charliehorse55 View Post
This. I had a problem where I had a 125w CPU in a 95w motherboard, when under load it would handle it until the CPU got to 50C, then the power consumption would surpass 95w, thus the voltage going to the CPU would drop and the CPU would subsequently become unstable and crash.

Now I've changed my quad into a tri-core so that I don't have that problem. But it does show how a CPU uses more energy the hotter it is.

Here's why:

CPUs require a certain amount of current to be flowing through the transistors in order for them to operate with stability.

Now, as you may know, the amount of current flowing through an electrical system is equal to the voltage divided by the resistance.

I = V / R

Another thing to note about electricity is that the resistance of any given material can be described with this equation: R = kT

Where k is a constant relative to the material. T is for Temperature, in Kelvins. What are Kelvins? Kelvins are actually the same as Celcius, in that 1 Kelvin = 1 Celcius. However, the Kelvin scale is offset so that 0 Kelvin (absolute cold) = -273 C. Likewise, 0 C = 273K

Now that we understand the math, lets take a look at how increased temperature causes the CPU to use more power.

I = V/R
R = kT

Sub 1 --> 2

I = V/ (kT)

Therefore, to keep the current the same while increasing the temperature, the voltage must be increased. Lets show the difference in power between 30c and 70C

30c in K = 303k
70C in K = 343k

Lets assume that the CPU is using 125W at stock volts.

P = VI

Therefore the CPU has 100A of current flowing through it at full load.

100 = 1.25/kT

Now, lets increase the temperature. Going from 303k to 343k is a 13% increase. Subsequently, the Voltage has to be 13% higher as well.

1.25 * 1.13 = 1.41 V

Now, plug that back into the power equation:

P = VA
At 30 C
125w = 1.25 * 100
at 70c
141W = 1.41 * 100

For anyone really interested, the formula for the effect of ambient temperature is:

(New ambient Temp in C + 273) / (Old ambient temp in C + 273) * Old Load Temp


--Charliehorse55

Edited by GingerJohn - 4/29/11 at 9:38pm
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Main
(21 items)
 
HTPC
(10 items)
 
 
CPUMotherboardGraphicsRAM
i5 2550k P8P67 Pro Sapphire HD 7950 G.Skill RipJaws X 1600 Cas 9 
Hard DriveHard DriveHard DriveCooling
Corsair Force 120 WD Blue 500GB WD Caviar Green 1TB XSPC RayStorm 
CoolingCoolingCoolingCooling
RX240 MCR 220 EK 7950 Copper Acetal  DDC-1T 
OSMonitorMonitorKeyboard
Windows 7 64-bit Dell U2311H Oculus Rift DK2 Ducky Shine 3 MX Brown 
PowerCaseMouseAudio
Corsair TX 750W CoolerMaster CM690 II G500 Klipsch ProMedia 2.1 
Audio
Asus Xonar DX 
CPUMotherboardRAMHard Drive
A10-6800K Gigabyte GA-F2A85XN-WIFI G Skill 1600 CAS9 Kingston SSD Now 60GB 
Hard DriveOptical DriveCoolingOS
WD Caviar Blue 1TB LG Slim Blu-Ray player Silverstone NT06-PRO  Widows 7 Home Premium 
PowerCase
Silverstone Sfx Series ST45SF 450W Silverstone SG05 
  hide details  
Reply
post #12 of 12
Lol, mount that 240mm on the TOP of the antec 300, waste of radiator if you don't mount any fans on the other slot. What's the pump mounting method?
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