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01-19-2003, 07:18 PM | #1 |
Cooling Neophyte
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Thermal Design on P4....?!?!?
Well I got an 2.4B and according to INTEL spec my cpu has a thermal design of 57.8W at default speed and default vcore(1.5v).
Now my question is if anyone of you guys how to figure out what wattage the thermal design cpu will have for ex. @3000mhz and vcore at 1.7v
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01-19-2003, 10:59 PM | #2 |
Cooling Neophyte
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well for a dumbazzed reponse hotter than 57.8W... sorry about that i just had to do it
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01-20-2003, 07:00 AM | #3 | |
Cooling Neophyte
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Quote:
HAHAHA......Well your at least not wrong...it sure is hotter. Anyone knows how do calculate this...any mathematic formula..or something???
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01-20-2003, 09:30 AM | #4 |
Cooling Savant
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Ok, here it comes.
power is proportional to the voltage squared power is proportional to the speed (mhz) If you know enough math, you can accurately calculate the heat emitted by the CPU. If you don't, I can work it out for you. |
01-20-2003, 09:32 AM | #5 |
Cooling Savant
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That'll be 92.8W... quite hot.
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01-20-2003, 09:45 AM | #6 | |
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01-20-2003, 09:47 AM | #7 |
Cooling Savant
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but I=V/R
therefore P=V*V/R P=V^2/R. By changine V, r remainst constant. There, you have it Last edited by hara; 04-09-2003 at 08:55 AM. |
01-20-2003, 09:56 AM | #8 |
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Correct.
My formulae assumes that the current remains the same (which is incorrect) and yours assumes that resistance remains the same. But is that correct? |
01-20-2003, 10:06 AM | #9 |
Cooling Neophyte
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OK...thnx alot fellas.
Really apreciate the help Another question though...does this respond to all cpu:s..??..AMD as well..??
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||Naughty|| Windows XP Ilyama 19" P4 2.4b @ 3014ghz Abit IT7 MAX 512mb TwinMos PC3200@222mhz Radeon 9700 Pro SoundBlaster Audigy 3Com TX 10/100 220gb hdd Last edited by [NH]Naughtyboy; 01-20-2003 at 10:26 AM. |
01-20-2003, 11:25 AM | #10 | |
Cooling Savant
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Quote:
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01-20-2003, 11:26 AM | #11 | |
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Quote:
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04-09-2003, 06:15 AM | #12 |
Cooling Neophyte
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OK..guys.
Been a while since I was here last time. Atm moment I needed the formula posted here...have some problems though. I do understand how to calculate it...but I cant figure out what the different prefixes means in the formula. This hiw it was put to me: I=V/R therefore P=V*V/R P=V^2/R. Can someone explain to me...what does the thing stand for?? V: ??? R: ??? I: ??? P: Power I assume ?
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04-09-2003, 06:43 AM | #13 |
Cooling Savant
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V : Voltage
R : Resistance I : Current P : Power (you guessed)
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04-09-2003, 06:45 AM | #14 | |
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Quote:
The resistance vary with the Temperature. Think of a bulb and the exeriments we did in the physics lab about 2 years ago Since silicon chips are non ohmic, then the resistance changes with temp. You can't predict by how much unless you do some experimentations
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04-09-2003, 06:48 AM | #15 |
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But i can lead you to some way, for silicon chips, i.e. semi-conductor, the resistance decreases with temperature (please correct me if i am wrong).
Therefore taking the formula P = V^2/R; when R Decreases, P increases, so when you increase the voltage of the cpu and making the cpu become hotter (V increases implies P increases), you are also varying the resistance wich makes the cpu become hotter even more. hope that you understood my poor physics
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So the bullet proof vest aint a $hit when d laser is pointed to your head Kid Last edited by Balinju; 04-09-2003 at 06:56 AM. |
04-09-2003, 07:04 AM | #16 | |
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Quote:
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04-09-2003, 07:22 AM | #17 | |
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Quote:
the resistance change due to temperature difference was negligible only if the temperature difference is very small. if your cpu as standard temp and voltage has a temp of for example 50degrees C, and when overclocked the temp remained 50degrees C, or changed to somewhere near 50 degrees C, then you can ignore the difference in resistance, otherwise you cannot. to find the exact amout of heat dissipated, you need to find the VI graph of the cpu core material.
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04-09-2003, 08:09 AM | #18 |
Cooling Neophyte
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OK..so now I know how to calculte the formula....and I know what the prfixes stands for.
BUT...I have to know the value of "R=resistance". How do I calculate to get that value then....since as you say...it varys due to the temp rising/faling....???? Hara: Sorry if I´m noobish....but could show me how you calculted my first example....and how you got all values to the formula..????
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04-09-2003, 08:39 AM | #19 |
Cooling Savant
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Simple Calculation:
What we know so far: P ~ F P ~ V^2 .'. P=kFV^2 ; k - Constant of proportionality => k=P/FV^2 = 57.8/(2400x(1.5)^2) = 0.010703703 Substituting k P=0.010703703 x F x V^2 In your case: P=0.010703703 x 3000 x 1.7^2 = 92.8W You can substitute F and V (Frequency and Voltage respectively) to your liking @Balinju: Don't make the mistake of oversimplyfying CPUs. They have circuitry to change power dissipation. Neither the 57.8W is an accurate number given out by Intel. Heat is negligable (Even tens of degrees) in comparision to CPU load and other factors. One CPU may vary with another.
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04-09-2003, 08:54 AM | #20 | |
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Quote:
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04-09-2003, 11:30 AM | #22 |
Cooling Savant
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I find this conflicting :shrug:
Code:
Intel Pentium 4 2.40B GHz sSpec SL67Z in FC-PGA2 478-pin PackageOEM or Boxed? Both CPUID = 0F24, Stepping = B0, Order Code = RK80532PE056512 Thermal Design Power = 57.8 W, Maximum Junction Temperature = 70 Degrees C Absolute Maximum Core Voltage = 1.75 V Multiplier = 18.00 Nominal (default) ratings:---Internal Clock Rate = 2400 MHz, FS Bus Frequency = 133 MHz---Core Voltage = 1.500 V Actual Settings:---Internal Clock Rate = 2394 MHz, FS Bus Frequency = 133 MHz---Core Voltage = 1.500 V Thermal Throttling Not Enabled Estimated Actual Maximum Dissipation: 69.9 Watts
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04-09-2003, 12:29 PM | #23 | ||
Cooling Savant
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Quote:
Quote:
The cpu is of couse not an ohmic conductor, so forget that R remains constant, maybe you want to assume so for calculations, but to be perfectly accurate, you cannot
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04-09-2003, 12:36 PM | #24 |
Cooling Savant
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if you want to know exactly what is the power dissipation, then you need to find the VI graph for the CPU core.
Ask intel for it. Mabye you know, they could give it to you
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04-09-2003, 04:00 PM | #25 | ||
Cooling Savant
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Quote:
Power is proportional to voltage squared Power is proportional to clock frequency The values given out by Intel of Power Dissipation of every processor (because they didn't take into account the temperature ) Think Twice before babbling out something Quote:
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