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Ivy Bridge Temperatures Could Be Linked To TIM Inside Integrated Heatspreader: Report

a lot of people are making comments in here based on past removal of IHS, These processors are different. Even I removed the IHS on my Opteron 939. I wouldnt do it on an Ivybridge though, Unless you had a cooler that wouldnt crush the chip. And the standard coolers wouldnt work as the lever wouldnt press onto anything...As the IHS which it would usually touch would no longer be there.

I'll wait on a very good guide hitting the net :p

As for it not being a factor with regards to heat, The thermal paste isnt as good as the solder technique used on Sandy, how can people say its not factor =\ ofcourse it is
 
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I'm going to disagree with the the following. Jokester is correct in that the IHS is there for mechanical protection, but I believe the cooling effect is more relevant to these boards. The IHS makes the chip run colder, taking it off will make it run hotter.

mmj_uk said:
... IHS for the most part hinder cooling...
Click to expand...

anyon said:
Having a large copper heatsink directly touching is even better than a tiny IHS
Click to expand...

Jokester said:
The IHS is there for mechanical protection.
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The copper block soldered to the core improves heat transfer. The reasoning is pretty straight-forward. Doubling the surface area through which heat is transferred halves the associated temperature gradient. The IHS is much bigger than the core.

There is a temperature drop associated with any interface between materials. With TIM at these wattages and the usual mounting area, it's 2 - 5 degrees. There is a temperature drop across the 3mm of copper as well, but it's of the order of a thousandth of a degree. It's tiny, all the thermal resistance is across the boundary. Then there's another drop of <1 degree across the cpu-ihs soldered interface.

The temperature drop between the copper ihs and the processor is significant if using tim, and negligible if using solder. The temperature drop across the ihs and the waterblock is big, easily over 10 degrees but probably under 20.

If you replace the solder under the ihs with tim, you should expect at least 10 degrees temperature increase. Tim is **** compared to solder. Intel will know this, the decision will be related to cost.

And the crucial part- if you take a soldered IHS off a processor, you've removed one temperature drop. The thousandth-of-a-degree one. In exchange, you've reduced the area by a factor of 20 or so. It's a mistake unless you're willing to solder a waterblock to the core, and even then I'd suggest soldering the waterblock to the ihs instead.

The above isn't magic- please feel free to Google thermal resistance and have a bit of a think about the problem.
 
I don't see it like that

if you had a single lane road, that then went to 5 lanes and into a car park, would it fill faster that just one lane straight into the car park?

The area of the core is small, and i can't see how widening the area after the restriction would help

the car park analagy has flaws but I'm sure u see what i mean
 
You are presupposing soldered core. We aren't talking about that.

In context we were replying to a completely different set of comments.

As for what you are actually saying, I agree it won't be clear cut.
 
The IHS doesn't improve cooling, purely on the basis that if you remove it the base of your cooler does exactly the same function as it. The main benfit of removing the IHS is that you remove one of the boundaries, two TIM based boundaries in the case of the A64 and IB chips.
 
Jokester said:
The IHS doesn't improve cooling, purely on the basis that if you remove it the base of your cooler does exactly the same function as it. The main benfit of removing the IHS is that you remove one of the boundaries, two TIM based boundaries in the case of the A64 and IB chips.
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If you take the extreme, where the die has something larger fused to it increasing the surface area for a further heat transfer to take place, then it will be better?

The argument is that a soldered IHS does something similar.
 
Thermal conductivity of copper is 400 in si units. Assume a pessimistic 200w from the cpu.

Conductivity = power/(thickness*temp delta)

temp delta = conductivity*thickness/power

temp delta = (400*3/1000)/200 = 1.2/200 = 0.006 degrees.

The internal conductance just doesn't signify, all the thermal resistance is at the boundary. Keep the ihs on, provided the ihs is soldered and you're not willing to solder the waterblock to the cpu.

Liquid metal and similar complicate this, as the resistance at the boundary shouldn't be much worse than solder.
 
The past few heatsinks I've bought don't allow you to apply near unlimited pressure, they're all quite carefully done to reach a maximum pressure against a CPU IHS and the mounting mechanism.

The Archon I'm using for example I've fully tightened with a socket wrench, and it's not as tight as the old style coolers which just dropped onto 4 screw pillars coming from behind the motherboard. If it had that type of mechanism, and I was confident the socket itself wouldn't hinder the heatsink (I've not looked at this) then I'd consider cutting off the IHS.

As it stands now though, when I see reports that IHS removal doesn't help, then I'm going to be very suspicious it's down to less pressure on the heatsink against the IHS.
 
JonJ678 said:
Thermal conductivity of copper is 400 in si units. Assume a pessimistic 200w from the cpu.

Conductivity = power/(thickness*temp delta)

temp delta = conductivity*thickness/power

temp delta = (400*3/1000)/200 = 1.2/200 = 0.006 degrees.

.
Click to expand...


I see a rather significant error in your first formula: you have ignored the cross sectional area. Without that you could have the 200W travelling down a 1 atom thick wire and still get the same temp delta.

You have then rearranged the formula incorrectly.

If you look at your final formula, you would get a SMALLER temperature drop with either a LOWER conductivity or GREATER power.
 
namnoc said:
I see a rather significant error in your first formula: you have ignored the cross sectional area. Without that you could have the 200W travelling down a 1 atom thick wire and still get the same temp delta.

You have then rearranged the formula incorrectly.

If you look at your final formula, you would get a SMALLER temperature drop with either a LOWER conductivity or GREATER power.
Click to expand...

You make good points here. Rather than check the equation i wrote it down from the units- it should have been p*x/(a*temp). Too used to per-unit-area calculation i suppose.

No excuses regarding rearranging the expression though, i'm out by a reciprocal. Perhaps you'd like to post a numerically correct version?
 
I think I have been working too hard, I read that as "out by a receptacle" and wondered if you were standing be a toilet. I guess that could be a new way of taking the **** :)

Anyway, the assumed starting values may be out, but:

Delta T = Power * Length / (Conductivity * Area)

Assuming:
Power = 200 W
Length = 0.003m (3mm)
Area = 0.0009m2 (3cm x 3cm)
Conductivity = 400 W/(m.k)

Delta T = 200 * 0.003 /(400 * 0.0009) = 1.6 degrees

Which is a lot higher than I guessed.
 
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