Article - Processor Heat Dissipation

[CCW]

Processor Heat Dissipation:

Processors draw power. This inevitably ends up as heat. This article is to educate you and make you aware of the amount of heat that common processors dissipate and to compare how this has changed over time from generations of processors to different cores.

Processors execute instructions. These instructions are electronic and the faster a processor is the more it can execute and therefore the hotter it gets. This is why all recent processor are cooled using a HSF (Heatsink and Fan). Heatsinks are made of a metal that conducts heat well such as copper and aluminium. Copper is the better of the two at conducting heat, in fact it conducts heat well when it’s impure.

(Picture 1) This graph shows that Copper is a much better conductor of heat than Aluminium and is therefore the better metal to use as heatsinks. Silver however is better than Copper but only by 24 W/m*C. This increase is negligible considering the extra cost of making silver heatsinks.

A heatsink consists on as base and fins connected to the base. The base is often made of copper even if the rest of the heatsink is aluminium; it is the base that touches the Processors core. The fins take the hat up away from the base. All modern Heatsinks have a fan attached to the top. This is to push the heat away from the heatsink. The fan can even be pushing or pulling. It is better for the air to be pushing onto the heatsink, as in blowing down over the fins as this cools better. If the fan is pulling more pressure is needed to pull the heat away and thus it does not cool as effectively. Copper contains loads of “batches” of ions each “batch” with a free electron. Basically, a free electron will collide with an ion at the base on the heatsink. This electron after colliding speeds up (its gains kinetic energy during the collision.) It then moves to the colder part of the copper which in this case would be the fins of the heatsink. This keeps happening until the cold end becomes hot. This is why a fan is used to cool the heatsink.

Due to the way that processors are designed, the assumption that a 100MHz processor makes 10 watts of heat then a 1GHz processor has to dissipate 100 watts of heat is not true. Over the years as clock frequencies have increased the way in which processors are made has changed. Along with the way in which the Processor is now kept cool. The way modern Processors are designed means that a 1GHz Processor can be more than 10 times the speed of a 100 MHz Processor. To be able to run at this clock speed, the processor needs voltage put through it. Naturally, the more voltage there is being put through the processor the more heat it is going to create. This is vital in choosing a cooler for the processor especially if you are Overclocking.

The following formula is used to work out how much heat a Processor is dissipating:
Overclocked Watts = Default Watts * (Overclocked MHz \ Default MHz) * (Overclocked Vcore \ Default Vcore) ²

Let’s say I have an Athlon XP 1700+. Its default wattage of heat dissipation is 49.35 watts. I am running it at 2145MHz and the stock speed is 1466 and I am running it at its stock Vcore (the amount of voltage being supplied to the chip) of 1.5.

Overclocked Watts = 49.35 * (2145 \ 1466) * (1.5 \ 1.5) ² = 72.2071962054

Cooling systems such as the Vapochill by Asetek can handle up to 160 watts of heat which is why these systems are ideal for cooling Overclocked chips where the Vcore is much higher than it would be at stock. The calculation above shows you that even at the stock Vcore with a 679MHz Overclock the chip is producing over 47% more heat.
This graph shows the difference between the heat dissipation of the Athlon XP Palamino Core and the Athlon XP Thoroughbred Core. The Thoroughbred is a better Overclocker than the Palamino because it requires less voltage to reach higher clock speeds and because it requires less voltage it also creates less heat.

[CCW]

Palamino Cores:

[CCW]

Thoroughbred Core:

[CCW]

Comparison of the two cores (not every rating):



So what do you guys think of my first proper article? Also, can you please rate this thread, Thanks,

Craig

[Holst]

Very interesting.

I think you got your % increase wrong. 72/49 = 1.47 not 1.6..

Interesting to see that according to AMD the 2400, 2600 and 2700 all have approximately the same wattage.... can you give the exact numbers they quote..
Strange that the B types are all about the same where as the palomino's are pretty linear in there wattage increase.

Its also important to note that those wattages are maximum, in the past people have released programs to calculate CPU wattage.. they all seem to introduce a factor of 0.8 to drop the wattage but ive never worked out where that number came from.

I dont suppose you know Thunderbird wattages as well they are pretty hot running beasts.

Although id take AMD's numbers with a slight pinch of salt anyway. I dont suppose you have an ammeter to test the real power consumption of different CPUs ?

Voltage increases are the most interesting bit, a 40% increase in voltage will double your max wattage... even at the same clock... scary stuff

Nice work CCW, next step is explaining about die size and P4 heatspreaders :D

[CCW]

Originally posted by Holst
I dont suppose you know Thunderbird wattages as well they are pretty hot running beasts.

Although id take AMD's numbers with a slight pinch of salt anyway. I dont suppose you have an ammeter to test the real power consumption of different CPUs ?

Voltage increases are the most interesting bit, a 40% increase in voltage will double your max wattage... even at the same clock... scary stuff

Nice work CCW, next step is explaining about die size and P4 heatspreaders :D

I have the wattages for the thunderbirds, didnt think any one would be interested in them now but I wil lthrow a few graphs together later. I stil lhavent finshed comparing the pallies to the thoroughbreds

i need to check a few of the calculations as well i know im gonan explain about die size when i get onto bartons, i know little about P4s whcih is why I havent mentioned them.

Thanks Holst.

Craig

EDIT - :doh: changed it to 47% :D CHeers Holst

[CCW]

Holst, a graph showing Thunderbirds heat dissipation :toast:


EDIT - Some more Barton stuff coming soon and a comparison between the Pallie and thoroughbred results.

[jmke]

great info CCW !

[Penguin4x4]

In addition to CCW, to find out the Electrical and Thermal properties of your proccesor, check out this page:

http://users.erols.com/chare/elec.htm

:)

[CCW]

Yeah, I was going to add that link and a few other later. Glad you guys like the info! :D

Craig

[Holst]

Nice link penguin.

Thats where the 0.8 factor comes in when calculating wattage.

0.8 of the max wattage is the approximate Typical Power Dissipation.

Just out of interest my barton at 3gig is 167watts max and 131 Typical.
Pretty damed hot :D

[CCW]

Have fun cooking your eggs and sausages on that thing in the morning Holst! :D

Craig

[saaya]

167W :eek:

[BrainStorm]

CCW, nice article. I think you ought to highlight a bit more the exponential effect on heat of increased voltages vs the linear increase in heat from pure clock speed increases. That is to say that voltage increases cause considerably more heat than clock speed increases (44% increase in speed = 44% increase in heat. 20% increase in voltage = 44% increase in heat).

[AMD Engineer]

GW & GJ CCW!

Nice reading BTW, I have tried to study the effect of die size, and the ability to cool (pr. cm^2 if you what I mean)
But I found very little change when comparing palomino vs Thoroughbreds.

And I think is caused by the effeciency of Solid to solid heattransfer.
But it is required that you have a good colling block, because of the heat-distribution (Good distribution makes core size less important)

Just my 2 cents :)

[gobbo]

1800 + @ 25.5mhz @ 1.95v
141.115w at load!
128w at idle

C/w rating of cooler = 0.116

BTW CCW, and everyone else, i found this little calculator on the web http://newstuff.orcon.net.nz/wCalc.html that takes all the maths work out!

Little java application that uses the same equation u mentioned in your atricle.

Very informative btw, but a bit too much info on the old physicls for meh brain to take in, only a network engineer you see :P

[CCW]

Thanks guys. Due to some very incitive and constructive critiscism Ive had th articele is to be re-written. Every thing will be expanded upon.

Thanks all for you time and help!

Craig

[CCW]

*bump*

[macci]

The one and only Overclocking Wattage Calculator can be found over at BenchTest.com (http://www.benchtest.com/calc.html)
According to this calc. I've at one point been running a chip that produced 240W of heat :D (1.8A@3563MHz 2.2V LN2 cooled)

[CCW]

The one and only? It uses the samne formula outlined in my article. That lin kwill go in the article when it gets re-written.

Craig

[Penti]

Originally posted by gobbo
1800 + @ 25.5mhz @ 1.95v
141.115w at load!
128w at idle

C/w rating of cooler = 0.116

BTW CCW, and everyone else, i found this little calculator on the web http://newstuff.orcon.net.nz/wCalc.html that takes all the maths work out!

Little java application that uses the same equation u mentioned in your atricle.

Very informative btw, but a bit too much info on the old physicls for meh brain to take in, only a network engineer you see :P

Haha, so you got an in-socket sensor i see.

[gobbo]

Nope, i jsut used the on die temp reader and then used his equation to work it out with!

[Holst]

Originally posted by macci
The one and only Overclocking Wattage Calculator can be found over at BenchTest.com (http://www.benchtest.com/calc.html)
According to this calc. I've at one point been running a chip that produced 240W of heat :D (1.8A@3563MHz 2.2V LN2 cooled)

ROFL macci, thats pretty damed funny.

Im quite dubious that that value is correct. If I had the time to mod a board id fit some low guage wire between the CPU and the voltage regulation so I could measure the amperage and find out the true(ish) wattage, im pretty confident it would be well below what those calculations state even when running the most agressive benchmark.

On the site you link to they use a factor of 88% for average...

[Holst]

What cooling you got gobbo, if you dont have pelts or phase change your temps are innacurate :P IMHO :P

[Penti]

Originally posted by gobbo
Nope, i jsut used the on die temp reader and then used his equation to work it out with!

Is it so, what board do you use? That measuring must be off, no cooler in the world performs 0.116°C/W not even White Water.

[Ragnarok]

rofl.... 140W heat from. from my 1700+ @ 2.58 2.08 vcore ;P

[KnightElite]

That formula that you gave will give results... but quite likely they're quite far off from the actual ones. You should contact pHaestus for more info on this (he's a member on this forum), he's done a great deal of research on it, and is/has attempted to find a method to deduce the actual wattage output of a CPU.

A few things need fixing in your article:

Your description of the way a heatsink works left something to be desired (then again, I've taken a heat transfer course... ;)) I'll come back to this.
typos...
Due to the way that processors are designed, the assumption that a 100MHz processor makes 10 watts of heat then a 1GHz processor has to dissipate 100 watts of heat is not true. Over the years as clock frequencies have increased the way in which processors are made has changed. Along with the way in which the Processor is now kept cool. The way modern Processors are designed means that a 1GHz Processor can be more than 10 times the speed of a 100 MHz Processor. To be able to run at this clock speed, the processor needs voltage put through it. Naturally, the more voltage there is being put through the processor the more heat it is going to create. This is vital in choosing a cooler for the processor especially if you are Overclocking. This paragraph needs some clarification. In fact, I think that 75% of it is irrelevant to what you're trying to say ;). Voltage is put accross something, not through it.


Sorry to rain on your parade somewhat, but there's lots of room for improvement.

Now, heat transfer:

First, we shall start with temperature. From dictionary.com :
A measure of the average kinetic energy of the particles in a sample of matter, expressed in terms of units or degrees designated on a standard scale.

So, temperature is a measure of the average kinetic energy of particles. Fairly straight forward. This means that at hotter temperatures, the particles that make up matter move faster.

This allows things to happen, such as: Your CPU is turned on and begins computing. Being that it is not ideal, there are a lot of ways in which electriity is wasted, and turned into heat. This means that the particles of the CPU are now vibrating faster than they were before. Your heatsink sits on top of the CPU, and it is still cool at this point. The rapidly moving particles in the microprocessor die strike the particles in the heatsink, imparting energy to them, and losing some of their own. These particles strike more particles further along in the heatsink. This sequence continues, and that's how heat "flows" from a warmer area to a cooler one.

In the case of copper (or aluminium, or any pure metal), understanding the way heat transfer works requires understanding the molecular structure of the metal. In essence, metals are composed of positive ions, surrounded by a "cloud" of negative electrons. These electrons are not attached to individual ions, so they are free to move about within the metal. This is the reason that metals are good at conducting electricity as well as heat. When part of a metal becomes hot, the ions vibrate back and forth, because of their increased kinetic energy. A vibrating ion will then strike a free moving electron, which will accelerate it away at high speed (high energy). This electron will then colide with another ion at some point (it will probably not go very far) and if the ion is lower energy than the electron, the ion will absorb some of the electron's energy, and vibrate faster, raising its temperature. If the electron strikes an ion that has higher energy than it does, then the ion will give the electron more energy, and the cycle will continue. So gradually, heat will "flow" in this fashion from the hot to the cold parts of a metal.

Now, the way a heatsink works is that these rapidly moving particles within the heatsink will strike molecules of gas in the air. This is similar to what happens within the metal. This will cause the gas molecule to vibrate faster, and will cause it to gain kinetic energy. This makes each molecule take up more space, and therefore hotter gases are less dense than cooler gases, and they rise. This is called free convection, and it is how passive cooling works.

The most important thing therefore, in a heatsink, is surface area, since the more places there are for fast moving particles in the heatsink to colide with air molecules, the better the heat transfer will be. There are two ways to increase the surface area: Increase the amount of air hitting a surface of the heatsink. This is basically adding a fan. More airflow = more air particles hitting the heatsink. This means that the incoming air will be cooler, and there will be a larger energy difference between the air particles and the heatsink particles, so the air will be able to absorb more energy than warmer air could.
increase the surface area of the heatsink. This accomplishes the same goal, more heatsink particles coming into contact with more air particles. This is why heatsinks have fins and pins, rather than being featureless blocks.

Obviously, in most heatsinks, both of these methods are used. The same basic principles apply to waterblocks.


Anyway, that's my spiel on heat transfer for the time being :D. I might come back and add more later.

EDIT: Man... that was a long post. Almost takes a whole screen at 1600x1200, lol :D.

[CCW]

Excellent, thanks for the critiscism! A lot has to be done on this article.

Craig

[KnightElite]

Alright, here's my later addition to this thread ;).

This part has to do with how heat is generated inside the processor, and why modern microprocessors have more heat output than older ones.

First off, we must understand the way that a MOSFET (Metal Oxide Semiconductor Field Effect Transistor) works. These are the kind of transistors that make up modern microprocessors, as well as most other integrated circuits. When in a given state (on or off) these transistors use almost no power.... which is good, and which means they produce negligible heat. When switching (going from off to on, or on to off) however, is where they almost all of the power they consume, and therefore output the largest amount of heat.

Obviously, this means that if the transistor is switching more rapidly, it will generate more heat. This is why at a higher clock, the CPU will have more heat output.

Voltage comes into the equation of heat as well, as CCW mentionned. The way that a MOSFET works is that a voltage is applied to a part of the transistor called the gate. When voltage is applied to the gate, depending on the type of MOSFET (P channel or N Channel), gate will either open or close a channel connecting the source to the drain (the other two leads on the transistor). That's where heat is output.

Now, we know from Ohm's law that Voltage = Current * Resistance (V=I*R). We also know that Power = Voltage * Current (P = V*I). From these two equations, it is fairly straightforward to find that Power = Voltage * (Voltage/Resistance), or P = V²/R. Therefore, as CCW has said, power output varies with voltage².

What this tells us in terms of heat output is that the increasing the clock frequency of our CPU will cause the temperature to increase in a more or less linear fashion, increasing the voltage will cause the temperature to increase more drastically.

For example, if we consider a CPU which has a stock voltage of 1.6V. We assume that it's basic heat output is 50W when run at stock speed and stock voltage. The following chart shows heat output at the same speed with different voltage.
http://homepage.usask.ca/~kdh349/cpupower.jpg

Now, after seeing this, it seems rather foolish to drastically increase the voltage on the CPU. But... it allows us to overclock higher ( I won't go into why right now) so we do it anyway. But the fact that power output varies with the square of the voltage is the reason that sometimes it does not provide any advantage to keep increasing the voltage, since the increase in temperature negates any benefits that greater voltage would have.


The reason that CCWs formula for heat output of a processor is not very good, is because it is very hard to agree on a "full load" or "no load" state for the CPU, so it's heat ouput, and consequently temperature, vary wildly. This is, of course, because different levels of "load" cause different transistors to switch on and off, or different numbers of transistors.

For example, if you read this article (http://www.procooling.com/articles/html/amd_thermal_diode_testing_cali.php) you can see graphs of how different programs affect the load and temperature of the CPU, and why a "full load" state is hard to quantify.

I would put a link up to my professor's heat transfer information that we used for my class, but unfortunately he password protects the PDF files, and he seems to have changed the password since I took the class in first semester, unfortunately.

[MrLavender]

Originally posted by Holst
Its also important to note that those wattages are maximum, in the past people have released programs to calculate CPU wattage.. they all seem to introduce a factor of 0.8 to drop the wattage but ive never worked out where that number came from.
I thought the factor was 0.88 and it's because AMD say that it's not possible for end-user software to load the cpu more than that.

[Penti]

KnightElite

Great contribute for the thread! ;)

BTW doesn't u use U for voltage in formulas in the US?, here in Sweden we do write it as P = U * I for example, same §§§§ but i wonder if the standards in US is V for voltage an not U, it could be good to know. I though U was an international standard, since i read and electronic/computer program at "gymnasiet" A form of 3 year school we go in between like 16-19 Years of age, don't realy give us any education either lol, so if we have to go to "högskola" that should be equal to university or something, u can study 120 points in 3 years or 180 points in 5 years and offcourse the 5 year education is to prefer :p, and then there nobody that can afford to hire u becuse you will have to get pay more becuse of the education and whitout the education u cant get a job anyway :p

[KnightElite]

Well, I believe that in North America we use V for Voltage. At least in Canada we do, and since my textbooks are frequently american, and they do the same thing, I think it's safe to say that V is used in the USA as well.

In fact, the standard is to use V for DC voltage, and v for AC voltage :D.

[phaestus]

0.88 came from the discussion of Jim @ Benchtest.com and the author of CPUBurn software a few years ago. BTW the majority of the info in this first post WAS posted by Jim long long ago on his site. I don't doubt that you put it together independently, but never hurts to cite it. Anyway on to more relevant comments. This equation:

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

Can't be correct can it? Let's take a 1600+ Palomino (62.8W MAX) and overclock it to 2100+ speed (72W MAX) by simply changing the multiplier. The resulting CPU is identical to a 2100+, and so should also generate 72W. Let's plug that into your equation though. You come out with 78W! So discard that; doesn't work.

Now let's look at the effects of the CPU loading program that we choose on the final CPU temperature.

http://www.procooling.com/~phaestus/deltats.jpg

Since a heatsink's C/W is constant for a given mounting/paste application, the change in Delta T from difft programs must be due to fluctuations in wattage. Ok this info is interesting, but maybe not so relevant. Let's look at this:

http://www.procooling.com/~phaestus/ramtimings.jpg

As you change memory timings in the BIOS, then the Power of the CPU is also changing under load (makes sense since the system ram becomes the bottleneck when the cache isn't big enough for the operations). This effect SHOULD be much larger for changing FSB and keeping the same MHz. So the watts the processor produces is not only dependent upon the program that one uses, but all the ram timings, the amount of ram, and the FSB!

The easiest way to accept or reject these silly "My CPU runs at 250W" results is to put a temperature probe (calibrated!) at the inlet and outlet of your waterblock. I will wager you don't see a water temp rise of more than 0.5C. No way that would be the case for a 200+W heat load.

The easiest way to actually make any sense of CPU power is to modify a motherboard to read the current draw and vcore. This isn't as easy as one might think though; the THG article where they just used a clamp on ammeter isn't nearly accurate enough.

I apologize if my tone is too short; drinking first cup of coffee now.

[Penti]

phaestus
Hey, becuse your here now i need to ask something, Would an uncalibrated MAX6658 measure the internal diode from a Thoroughbred processor right? I would like to know that before i assembly an on-die diode reader with the max6658 i got.

Is there any differens between measureing the internal diode on the Palo compared to Tbred? The document from maxim says that it wont need any calibration for palominos(but i can guess calibration is necessary if i want a good reading), but how would it be with an tbred processor.



Originally posted by KnightElite
Well, I believe that in North America we use V for Voltage. At least in Canada we do, and since my textbooks are frequently american, and they do the same thing, I think it's safe to say that V is used in the USA as well.

In fact, the standard is to use V for DC voltage, and v for AC voltage :D.

Well of course we have measure voltages in volts but when we write it in formulas we use U instead of V as u uses, hehe so now i know that u dont use U in canada/us =). If we use V for DC and v for AC i don't know, that isn't mention in the little i've read about AC...

[phaestus]

Is the 6658 a relatively new IC? Not familiar with it. Anyway from looking at the datasheet it should work similarly to the 6657 but over a wider range and with a programmable overtemp setting.

When you talk about calibration of the diode readings, there are a few factors. 1 is the variability in the AMD CPU diodes (prolly +/-1 C or so) from one CPU to another. The other is the variability in the Max6658s from one IC to another (also +/-1C according to them). The third issue is the effect of solder joints and imperfect diode reader design (narrow traces, jumping pcb layers, using wire with resistance, etc) on the reported temperature. This third problem is the one to look out for, because it will lead to temperature compression in your results. The first and second are going to be not such a big deal assuming you keep using the same CPU and diode reader and comparing how changing things around in your system affects the diode temp.

Tbreds use the same 100uA/10uA pulses and so should be fine.

[Penti]

phaestus

Thanks it was the pulses i was mainly after. I did know about the traces/wires and soldring problem and i also read your article about it, there iam mainly concern about what wires i should use to connect it to socketpins on the backside of the board, shielded is i must right? and of course they need to be twisted. I think that would go pretty good if i follow your article and the datasheet. Is there same problem on the side where i hook it in the smbus?
Well i probely wont put it tougher just yet, maybe in the end of the summer or something, if i got any questions then i PM you on procooling. Thanks again.


Its the same as 6657 but it can measure below 0°C was what i though anyway, it did seem like it when i read the datasheet. It is in the 6657 datasheet as u did noticed... so i did take the 6658

[CCW]

*bump*