watts=heat?

lets say a GPU needs 23 watts to power it, and the next gen GPU needs 23 watts as well.

does this mean they will both run at about the same temps? or is it more than that, since some of the new 40nm nvidia chips are 23 watt and so is the 9600M GT

 

No, just because it requires 23watts to power them both doesn't mean they will chunk out the same amount of heat. It depends on how efficient these parts are. Meaning it depends on how much of that 23W will go to do something useful instead of coming out the other end as heat.

Watts is a measurement of power, it is the amouont of energy per unit of time. :wink:

 

johnny is right.

nevertheless, wattage can be an more-than-vague indicator of heat. in general, more watts does mean more heat, especially if you are comparing cards from the same line of technology.

if you have two gpu's running at 23 watts, it is really anyone's game as to which is hotter.

but if you have one gpu at 23 watts, and another at 35+ watts... you can bet with confidence the 23 watt gpu is MUCH cooler.

 

Heat = wasted energy.

So, more efficient designs should produce less heat if at the same wattage.

 

No, they will probably produce the same heat at the same wattage if thermal resistance is the same. The idea of a more efficient design is to produce less wattage for the same work.

23 Watts is 23 Watts, your temperatures are going to be dependent on thermal conductivity, not efficiency.

 

We are not talking about TDP. We are talking about the power required to run the chip by the way. Even then, different companies defines TDP differently. :wink:

Simply speaking, we are talking about what goes in, not what comes out.

 

This is true, but you seem to be missing the fact that what goes in must be exactly what comes out or else the temperature of the part will increase without bound. The laws of thermodynamics do not depend on what you do with the energy in between; it doesn't matter how efficiently the 23W are being used (although as a general rule, next gen computer parts tend to give you greater performance per watt than the previous gen).

However, if your part is drawing 23W of power, it must somehow dissipate or store these 23W. In theory, it is possible to dissipate the energy in a variety of ways (heat, electric current, light, etc.), but in practice, it is dissipated almost entirely as heat. So indeed, two parts that draw the same amount of power will produce the same amount of heat.

 

Surely some of the energy must go to doing the calculations itself. Otherwise, couldn't you theoretically have an infinitely powerful computer running on virtually no energy?

 

Of course some of the energy must go into doing the calculations -- ideally, all of it would go that way (though this is impossible to accomplish in reality). However, what happens to that energy once the calculations are done? Energy can be neither created nor destroyed so it doesn't vanish into nothingness and still has to be dissipated as heat.

 

So you are saying all the power you put into a light bulb comes back out as heat? Sure a massive percentage of it does come out as heat energy. Some of it does go towards making photons shoot out the other end. (simply speaking )

Of course you might know how these chips works better than myself. And of course I know the total amount of energy remains constant in a closed loop system. But surely these chips aren't that inefficient such that most of the energy thrown in comes out of the other side as heat? Surely some of it must go towards pushing the electrons around these micro circuits?

 

IINM Johnny is saying a more efficient card will for instance only require 10% of its power to do the same task as a less efficient card which may for instance require 20% of its power. So in this case with both cards being rated at 23W one will generate less heat because its using 2.3W as opposed to the other card at 4.6W.

Rather than saying both cards will be running at their full power or the same wattage, which was my initial understanding.

 

No. A typical light bulb converts only about 90% of the energy it uses into heat (see here and here). However, a light bulb actually produces light whereas graphics cards generally do not.

Yes, of course. But where does the energy go after that? Electrons moving faster through micro circuits corresponds to a higher temperature of the card and this must somehow be remedied.

 

I am not trying to pick a argument or anything. Plus I am heavily jet lagged. And it really doesn't seem we are talking on the same wave length either. lol I shall come back to this when my head is alittle clearer. ^^

 

I'm not trying to pick an argument either; I'm just curious where the energy would go. I guess it might depend on how we define what the 23W that a graphics card draws actually means.

 

i think more wattage implies more heat but if the die is shrinked, then it may not result to that much heat.

 

In general, I consider a "more efficient" chip to do one of two things:

1) Provide higher performance at the same wattage, in which case the heat output would remain unchanged because the amount of energy required is unchanged, or

2) Provide the same performance at less wattage, in which case it would produce less heat due to the lower wattage required.

 

The temperature of the processor is dependent upon the cooling system and the contact area between the cpu and the heatsink.

For example, my Opteron 185 uses 125 watts of power and peaks at around 125F under full load
My T7400 uses 34 watts and peaks around 140F.
Both systems have their fans on full, so no watts doesn't equal heat unless the same heatsink and surface area for contact is the same.

When had a T2500, the die size was smaller than my T5300. The T2500 uses less power, 31 watts versus 34 watts, but it ran noticeably hotter, because there was less contact between the heatsink an the cpu. The more contact area, the more efficient the heat can be dissipated from the processor. Upgrading to the T7400 brought a bigger die size and slightly lower temperatures than my T5300

K-TRON

 

But what if the die size of two chips are equal, but one user less power than the other?

 

That energy/unit of time is Joules over seconds.

E.g. 23W = 23J/s or 26J/2s..etc

 

Thanks for reminding me of grade 10 physics...-.-'''''

Anyways, so where do the energy go?

So, in a light bulb it goes towards pushing electrons around the high resistance wire (tungsten filament), to warm the wire up to a point where the atoms starts to spit out photons. The heat you feel will be from the infra-red emitted along with the visible light.

In a micro circuit the heat is also coming from resistance within the little wires as you are pushing electrons through the circuit. The atoms within a wire will vibrate at normal temperatures, and as the electrons tries to get through the wire it's going to pump into these atoms (resistance), thus losing some of it's kinetic energy. Of course if you cool these wires down to "absolute zero" technically there will be no resistance (the vibrating atoms stops).

The point is you are giving the electrons kinetic energy. Making it move through the circuit (processing the signals you throw at it). Same happens in a light bulb except you are pushing it through a high resistance wire as that is what you want. The faster you push these electron through, the harder they are going to bump into the atoms (losing more energy, more heat comes out the other side), therefore your chip gets hotter.

If anyones know better (and I know some of you that do ^^) feel free to correct/add in!

 

Since we are talking about chips, we can assume that CMOS technology is being used. The heat generated in a chip can be caused by the IR heating (any resistance with current running through it), and by the current going through the CMOS transistors. The latter is the much larger percentage. The CMOS transistors act like switches with finite resistances, and they dissipate power, especially when they are required to switch fast, since they do not go from perfectly open to perfectly closed (zero resistance) instantaneously.

Just about all the power consumed in a chip is dissipated as heat, since there is no light or sound emitted. However, depending on die sizes and thermal resistances between the silicon junction and ambient, the actual chip temperatures may be different for the same power consumed. We calculate the junction temperature Tj by the simple formula

Tj = Power dissipated x thermal resistance from junction to ambient + Ta

where Ta= ambient temperature.

Thermal resistance is a function of die size and how well the die is attached to the package, and how the package is attached to the PC board (or other assemblies). Also note that Ta is the ambient temperature the PC board sees, and that depends on how good the cooling system of the computer/laptop is.

 

Again, what if you have 2 different 45nm chips in 2 EXACT systems?

1 chips uses 35w, while the other uses say 25w.

All variables remain the same. Same cooling systems, etc. Same die size. In this case, in terms of heat production, which runs cooler under the same loads?

I think this was the OP's original point.

 

Yes less wattage means cooler system. Basically all energy pumped into a semiconductor ends up as heat.

 

How do you know the die sizes are the same?

The other variables not controlled are: (a) the actual power dissipated on the chip (note that the specs do not tell you the actual power used in a given situation), and (b) if there is any throttling used. A chip rated at higher power may be more efficient, i.e., it can do more for a given clock frequency.

In general, of course, the higher wattage being actually dissipated will lead to a higher chip temperature, all other things being equal.

 

fysics, beautiful (if you know it)

 

LOL.. it's the jumping of electrons between orbitals which release the photons, not the warming up of the material. hahah. it's all part of the wave mechanics.

 

Yes...the electrons gain energy thus jumping up in orbitals. I was trying to avoid some of the details. Nor do I want anymore memories of my chem course. :laugh:

 

From thermodynamics, we know that total energy is always conserved.

dE/dt=dQ/dt+dW/dt

where dE/dt is power, dQ/dt is heat rate and dW/dt is work (which usually includes mechanical work, emission of photons, or other energy releases away from a control volume, etc).

In the grand scheme, this means that unless your power usage you are quoting involves emission of photons (e.g. LEDs or lasers) or moving a motor (i.e. mechanical work), then that power figure DOES mean heat dissipation. For most electronic devices involving circuit boards etc, this IS pretty much all heat dissipitation. For a laptop, a part of this total power is not dissipated as heat which includes your monitor (energy accounting for the emission of photons) and your HDD (energy for mechanical work)

in terms of temperature, that depends on the nature of heat transfer. since most of the energy IS dissipated as heat, but depending how heat transfer is for a given device, will end up actually affecting what temperature this is.

 

And wave mechanics is considered to be Physics. LOL.. :laugh:

 

Yeah I know it all gets alittle messed up. I did that in physics in HS, then again in chem...>.> Stupid really.

 

Yes the 23W of power drawn by the graphics processor unit is equal to 23W of heat energy emitted into your room (given your room is a closed system).

 

No, the power drawn by the GPU is used to do work. Heat is wasted energy and hence will always be less than the amount drawn.

 

thats wrong.

2nd law of thermodynamics. entropy always increases in the universe. all energy capable of doing work will essentially become heat at some point. so its true that if the gpu is drawing this much power, then this much heat is emitted, effectively. only the motor from the fan in a gpu is spinning, but that mechanical work basically turns into heat too for the room, given that the room is the control volume

 

From the 1st law, the difference in energy (lost in the form of heat) is from inefficient processes (since there is never 100% transfer of energy to useful work). The energy transferred is already used to do useful work (not just mechanical, mostly electrical) so it will not be lost as heat.

 

yea first law is basically what i said before:

dE/dt=dQ/dt+dW/dt

thats only half the story. you look at the entire process. what does it ultimately become of this energy capable of doing work for the room.

if the room was your thermodynamic control volume, and your laptop is fed this much power, then all of that power will essentially be heat emitted into the room after all is said and done.

we are talking about the gpu here, so theres a fan doing mechanical work and then the rest is safe to say will be heat dissipation as joule heating. joule heating is direct heat generation. mechanical work from the fan too will move fluid, which when accounting energy for the entire room, means its heat generated for the room also. therefore all power is essentially turned to heat for a room as the control volume

 

Not all of the electricity used by the processor becomes heat. Some of the electricity is used to transmit signals. If the signals are lost, you would have a defective processor.

 

no. if we can somehow measure the power drawn specifically by the gpu. then that is the power drawn by just the gpu.

most electrical devices are ohmic resistors, essentially. from joule heating of ohmic resistors we know all the power will become heat. the only question is whether this total power input into the gpu has other forms of work done, say mechanical work or emission of photons. in this case it has mechanical work with the fan. but for the room, this mechanical work energy will be translated into heat also.

you can only say for example if you have a monitor taking 50W. and you are pointing the monitor out a window. Then you can say that not all 50W will be heat generation in the room, because some of this power will be used to generate photons that are shot out a window through a glass pane, but the rest will be heat emitted to the room

 

If it takes you 10W move a block on a frictioned surface and 1W of energy is lost from friction as heat, then 9W is used to move the block. That 9W used to move the block is not in the form of heat.

 

power is a little tricky sometimes because its time rate of energy.

think energy first. if you need to move a block up a frictionless inclined plane then total energy will always be conserved. you expend kinetic energy to move an object up and by the time you stop your expended kinetic enregy is equal to the potential energy gain of the object.

KE=PE, where KE is power input in this case

when you allow the object to drop back down, the PE becomes KE. work done around a closed path for frictionless system is zero.

take a friction inclined plane. you move the object and now you have to account for friction that works against the direction of your applied kinetic energy

so KE=PE-friction

If you let the object fall back down, then friction also exist. Essentially, work done by friction around a closed path is non-zero. if you want to think of friction as heat as a mechanical analogy, then over time, the energy system cannot continue to move continually based on energy conservation because friction is a disspitation mechanism leading to increased entropy

but thats all mechanical analogy and quasi unrelated. lets talk about the electronic device, and specifically a graphics card like the OP suggested.

here you have ohmic resistors and perhaps a mechanical fan at most. We set the room as the thermodynamic control volume. lets say this GPU draws 50W. the question is, what is the heat generation for the room due to the GPU.

in ohmic resistors, power=VI. This power is also heat because the potential drop across an ohmic resistor is due to irreversible inelastic collision processes involving charge carriers with other charge carriers, phonons etc (the mechanical analogy). anyway, the voltage you measure is essentially a measure of this propensity to charge carrier collision (or friction if you will). current is the amount of these charge carriers per unit time passing through the ohmic conductor. this is fundamentally what P=VI means.

Ok but since the 50W input doesnt all get wasted in ohmic losses. some of it is used to power a fan that mechanically moves fluid to cool the device. all fan does here is rearrange heat spatially in terms of the room away from the device. it doesnt destroy or remove heat from a room. but in the end, the mechanical work put into the room by the GPU fan is itself turned into heat through friction when the fluid runs out of momentum again due to friction.

therefore all 50W into the GPU will be 50W of heat for the room, room being the control volume

EDIT: minus sign