Thermal Design on P4....?!?!?

[[NH]Naughtyboy]

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

[Mikey Boo]

well for a dumbazzed reponse hotter than 57.8W... sorry about that i just had to do it

[[NH]Naughtyboy]

Originally posted by Mikey Boo
well for a dumbazzed reponse hotter than 57.8W... sorry about that i just had to do it


HAHAHA......Well your at least not wrong...it sure is hotter.

Anyone knows how do calculate this...any mathematic formula..or something???

[hara]

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.

[hara]

That'll be 92.8W... quite hot.

[bigben2k]

Originally posted by hara
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.
Hum... P=VI, which would make power proportional to voltage...no?

[hara]

but I=V/R

therefore P=V*V/R

P=V^2/R.

By changine V, r remainst constant.

There, you have it

[bigben2k]

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?

[[NH]Naughtyboy]

OK...thnx alot fellas.
Really apreciate the help ;) :)

Another question though...does this respond to all cpu:s..??..AMD as well..??

[hara]

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?.

It's a pretty good approximation :)

[hara]

OK...thnx alot fellas.
Really apreciate the help

Another question though...does this respond to all cpu:s..??..AMD as well..??

In general, at least on desktop CPUs, yes.

[[NH]Naughtyboy]

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 ?

[Balinju]

V : Voltage
R : Resistance
I : Current
P : Power (you guessed)

[Balinju]

Originally posted by hara
It's a pretty good approximation :)

It's not!! :cry: :cry:

The resistance vary with the Temperature.

Think of a bulb and the exeriments we did in the physics lab about 2 years ago :cool:

Since silicon chips are non ohmic, then the resistance changes with temp. You can't predict by how much unless you do some experimentations

[Balinju]

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

[hara]

Originally posted by Balinju
It's not!! :cry: :cry:

The resistance vary with the Temperature.

Think of a bulb and the exeriments we did in the physics lab about 2 years ago :cool:

Since silicon chips are non ohmic, then the resistance changes with temp. You can't predict by how much unless you do some experimentations

The power used in a cpu is to switch the transistors on and off. The resistance variation due to temperature is neglidgable compared to the variations due to other things. It isn't as simple as comparing a CPU with a bulb

[Balinju]

Originally posted by hara
The power used in a cpu is to switch the transistors on and off. The resistance variation due to temperature is neglidgable compared to the variations due to other things. It isn't as simple as comparing a CPU with a bulb

i know that it is not as simpls as a bulb, i said so to just give an example.

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.

[[NH]Naughtyboy]

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..????

[hara]

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.

[hara]

Originally posted by Balinju
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

The Problem is that what you're saying applies to non ohmic conductors. The CPU isn't a simple "conductor" so I doubt if the rule will hold there.

[satanicoo]

check this link (http://computernerd.com/cgi-bin/thermalcalculator.cgi) for a calculator

[hara]

I find this conflicting :shrug:

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

Maybe Intel states also a maximum dissipation value?

[Balinju]

Originally posted by hara
The Problem is that what you're saying applies to non ohmic conductors. The CPU isn't a simple "conductor" so I doubt if the rule will hold there.

holy shit how difficult it is to make you understand something,
The CPU isn't a simple "conductor"

you said it yourself, you can't assume that R is constant just because the cpu isn't a simple conductor.
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

[Balinju]

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 :evilaugh:

[hara]

Originally posted by Balinju
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 :evilaugh:

Don't make fun of yourself. You're the only one debating this subject. Are you saying Everyone is wrong by disapproving these approximations:

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 :rolleyes: )

Think Twice before babbling out something

if you want to know exactly what is the power dissipation, then you need to find the VI graph for the CPU core.

If only it was so simple...

[satanicoo]

Originally posted by hara
I find this conflicting :shrug:

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

Maybe Intel states also a maximum dissipation value?

its 57.8 with trotle enabled. with it disabled, it goes to 69.9.

dont ask me why anyway

:rolleyes:

[Balinju]

Don't make fun of yourself. You're the only one debating this subject. Are you saying Everyone is wrong by disapproving these approximations:

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 )


1. i am by no means making fun of myself
2. i am not saying that power is not proportional to clock frequency and to voltage squared
3. My debate is that power is not linearly proportion to voltage squared. It is in somekind proportion, but not LINEAR. Understood??? If not, forget everything.
4. If power was linearly proportion to voltage squared, this would imply that the cpu is an ohmic resistor. With all the transistors inside and all the other circuits, i think that it is by no means ohmic.

I could be wrong :mad: , i could be right :evilaugh:

Anyone can help??

[hara]

I've researched this a bit further. If you don't believe the approximations look here:

http://www.overclockers.com/tips40/



Heat Production = CPU Spec Watts * (New FSB/Spec FSB) * ((New Volts/Spec Volts)^2)

[Balinju]

Originally posted by hara
I've researched this a bit further. If you don't believe the approximations look here:

http://www.overclockers.com/tips40/

ok you wanted it, there you got it.

look at the part: BAD NEWS.

You want to disproove a LAW??????

[hara]

Originally posted by Balinju
ok you wanted it, there you got it.

look at the part: BAD NEWS.

You want to disproove a LAW??????

Hehe, I nearly shit myself reading this.

Do you even have a vague idea what is moore's law?

http://arstechnica.com/paedia/m/moore/moore-1.html#part1

http://arstechnica.com/paedia/m/moore/moore-4.html#part2

http://arstechnica.com/paedia/m/moore/moore-6.html#part3

Read a couple of these articles FIRST, then ask questions later. I suggest you cease spreading misinformation.

[Balinju]

of course i do, we can talk about this tomorrow at school, i don't have time now.

[utabintarbo]

You guys crack me up (hara & balinju). How long have you two been married?:p

[Balinju]

Originally posted by utabintarbo
You guys crack me up (hara & balinju). How long have you two been married?:p

lol for about 7 years :cry:

joking :D

[hara]

We never agree on anything :)

[Balinju]

Originally posted by hara
We never agree on anything :)

you are wrong, we agree on FMZ :evilaugh:

[hara]

FOXX MARK ZAMMIT!!!! :evilaugh: :evilaugh: :evilaugh: :evilaugh:

[Balinju]

Originally posted by hara
FOXX MARK ZAMMIT!!!! :evilaugh: :evilaugh: :evilaugh: :evilaugh:


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