Standard die size (surface area)

[Cathar]

Maybe others could give their experiences, but in dealing with thousands of disk drives in customer installed bases under fairly heavy use, I can honestly say that disk drive reliability has practically fallen through the floor in recent years, and modern SCSI drives are not excepted either.

Now, more so than ever, the three most important words in computing are "backup", "Backup", and "BACKUP!".

[Les]

Originally Posted by Groth

..
Given 32 mm square copper heatspreader 1 mm thick, heat it with a 10 mm square 100 watt uniform heat source, cool the opposite side with a heatsink with an h(eff) of 20,000 W/m^2, what's the heat flux density at the heatspreader/heatsink interface?

My interpretation of Waterloo suggests that the spreading would result in a reduction of the c/w , even with a layer of TIM between IHS and HS.





TIM layer is simulated as 0.1x32x32mm of 10w/m*c material.

Might prefer a circular Die simulation.
For this Waterloo, additionally, calculates for Parabolic Flux distribution.

Edited and corrected.
Edit2 : Corrected "Kryotherm" to "Waterloo"

[Les]

Oh Woe.
Errors in last post.
Will edit and correct.

[Groth]

Very interesting.

The amount of drop from adding the IHS seems excessive to me, though. Since a thicker baseplate can be pictured as TIM-less IHS, those graphs imply that most every waterblock could use a thicker base.

The first run though of the more-detailed-die-for-Bill model failed. Seems someone left out a single character when scripting the IHS-WB TIM, leaving the block and IHS not actually connected. Simulation stopped when the die hit 1.4e+14 Kelvin (2^47, the highest number it handles).

Hey! You pulled off the graphs while I was replying! Now I'm all confused.

[Les]

Having silly problems with my guess of TIM simulant(terminology?).

Hopefully have sorted and chosen 0.1x32x32mm of 10w/m*c material.
10w/m*c material gives an acceptable. C/W(TIM)=0.1c/w for 0.1x10x10mm layer.
Was in panicking .
Happy with graphs now and posted

[Groth]

Hmmm, Arctic Silver claims 35 W/(cm^2*K) TIM, yours works out to 10 W/(cm^2*K).

For the IHS-WB, I'm using 20 W/(cm^2*K), based on pH's and Cathar's estimates of 0.07 C/W (which would include the C/W of the silicon).

For die-IHS (applied under perfectly controlled conditions) I went with AS's 35 W/(cm^2*K).

[Cathar]

Only problem I have with graphs is apparant disjuncture from user experience. I would say that every person I've ever heard of removing the IHS has noticed at least some sort of temperature drop, which includes air-cooled heatsinks, having a significantly lower effective value for h than ~20K.

IMO, theoretically the IHS only helps those with aluminium based heatsinks, or for those heatsinks that use a way-too-thin amount of copper material between the core and the fins. In practise, I have not seen anyone report higher temperatures with the IHS removed.

[Les]

Originally Posted by Groth

The amount of drop from adding the IHS seems excessive to me, though. Since a thicker baseplate can be pictured as TIM-less IHS, those graphs imply that most every waterblock could use a thicker base.
.

Just dunno.
However Waterloo possibly underestimates the 1-D Resistance of 10x10mm sections of TIM and Cu.(0.1c/w and ~ 0.026c/w)
I take this to be the differences in Resistances extrapolated to infinite h(eff)



I am always confused.

Edit: Corrected "Kryotherm" to "Waterloo"

[Les]

Originally Posted by Cathar

.... which includes air-cooled heatsinks, having a significantly lower effective value for h than ~20K.

IMO, theoretically the IHS only helps those with aluminium based heatsinks, or for those heatsinks that use a way-too-thin amount of copper material between the core and the fins. In practise, I have not seen anyone report higher temperatures with the IHS removed.

Don't entirely agree.
" heff is the effective heat transfer coefficient acting on the surface" be it die, HS baseplate,or IHS.
For a given HS these 3 will have different values.
Certainly h(bp eff) may be < 20k, and possibly h(IHS eff) may be. However h(die eff) will be above to give a TIMless C/W below 0.5( 1cm sq die)

[Les]

Originally Posted by Les

Just dunno.
However Waterloo possibly underestimates the 1-D Resistance of 10x10mm sections of TIM and Cu.(0.1c/w and ~ 0.026c/w)
I take this to be the differences in Resistances extrapolated to infinite h(eff)

I am always confused.

Edit: Corrected "Kryotherm" to "Waterloo"

"I take this to be the differences in Resistances extrapolated to infinite h(eff)"
Am now very unsure about this.
Needs further thought.