Charlie Demerjian uses IDC's die-size estimate for Fermi

cusideabelincoln

Diamond Member
Aug 3, 2008
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Knowing IDC, he must have had a method for coming up with that number. Perhaps Charlie is using the same method? He could have just as easily used Anand's number of 460mm.
 

deputc26

Senior member
Nov 7, 2008
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Originally posted by: cusideabelincoln
Knowing IDC, he must have had a method for coming up with that number. Perhaps Charlie is using the same method? He could have just as easily used Anand's number of 460mm.

Yes but as pointed out in the above thread, Anand's estimation method was flawed. And I believe IDC stated his method plainly.
 

Blazer7

Golden Member
Jun 26, 2007
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I'd say that there 're times that even Charlie feels the need to post some decent info and it's likely that IDC is a better source than most of his. :D

That said I won't be surprised if Charlie indeed used IDC's numbers.
 

mmnno

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Jan 24, 2008
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Originally posted by: deputc26
Oct 3, IDC guesses 530mm^2 for Fermi die size

http://forums.anandtech.com/me...AR_FORUMVIEWTMP=Linear

Oct 6, our favorite ex-Inqer uses the same number as the lower bound for a die-size guess.

http://www.semiaccurate.com/20...d-and-high-end-market/

Coincidence?:laugh:

http://www.semiaccurate.com/20...appen-gt300-tapes-out/

Charlie said 530 square mm in July. He linked that article in the article you just linked.

Probably coincidence, I don't think IDC reads Chuck's articles.
 

deputc26

Senior member
Nov 7, 2008
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Originally posted by: mmnno
Originally posted by: deputc26
Oct 3, IDC guesses 530mm^2 for Fermi die size

http://forums.anandtech.com/me...AR_FORUMVIEWTMP=Linear

Oct 6, our favorite ex-Inqer uses the same number as the lower bound for a die-size guess.

http://www.semiaccurate.com/20...d-and-high-end-market/

Coincidence?:laugh:

http://www.semiaccurate.com/20...appen-gt300-tapes-out/

Charlie said 530 square mm in July. He linked that article in the article you just linked.

Probably coincidence, I don't think IDC reads Chuck's articles.

Oh well :) there goes my theory, it would have been hilarious if true,
rose.gif
maybe it was vice-versa?
 

Idontcare

Elite Member
Oct 10, 1999
21,110
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Originally posted by: deputc26
Oct 3, IDC guesses 530mm^2 for Fermi die size

http://forums.anandtech.com/me...AR_FORUMVIEWTMP=Linear

Oct 6, our favorite ex-Inqer uses the same number as the lower bound for a die-size guess.

http://www.semiaccurate.com/20...d-and-high-end-market/

Coincidence?:laugh:

Actually I hope to god that the 530mm^2 number was already in the public domain somewhere prior to my having stated it in these forums...not going to say jack about cause-and-effect here but I will say I very much intentionally delayed saying anything about Fermi specs (and Cypress) in all my posts until I was pretty sure I had read the info elsewhere in the public domain. :eek:

Since we are sort of on the topic, these items might generate some cerebral discourse.
 

Lonyo

Lifer
Aug 10, 2002
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Originally posted by: Idontcare
Originally posted by: deputc26
Oct 3, IDC guesses 530mm^2 for Fermi die size

http://forums.anandtech.com/me...AR_FORUMVIEWTMP=Linear

Oct 6, our favorite ex-Inqer uses the same number as the lower bound for a die-size guess.

http://www.semiaccurate.com/20...d-and-high-end-market/

Coincidence?:laugh:

Actually I hope to god that the 530mm^2 number was already in the public domain somewhere prior to my having stated it in these forums...not going to say jack about cause-and-effect here but I will say I very much intentionally delayed saying anything about Fermi specs (and Cypress) in all my posts until I was pretty sure I had read the info elsewhere in the public domain. :eek:

Since we are sort of on the topic, these items might generate some cerebral discourse.

What's the optimal sort of yield that someone is likely to see after x years on a process? (obviously not 100%).
I'm sure there have been AT articles discussing yields for Intel and I might be inventing an 80% number because I don't remember them all that well, but given a an 80% RV740 yield, that would roughly translate to a theoretical sub-50% yield for the GT300, which isn't that great.

What does your predicted yield take into account though, since surely clock speed targets could have an impact. Are you saying yield rate based on defects, or yields for dies working at a target clock speed?
50% of full fledged GT300 dies and then 30% more which fail to hit the required high end clock rate wouldn't be terrible, but 50% of functional dies (no defects) of any given (reasonable) clock speed would be terrible. i.e. 50% that can either be high end GT300, or the slower clocked variant (ex. GTX280 and GTX260 cards if they were only different clocks).
 

deputc26

Senior member
Nov 7, 2008
548
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Originally posted by: Idontcare
Originally posted by: deputc26
Oct 3, IDC guesses 530mm^2 for Fermi die size

http://forums.anandtech.com/me...AR_FORUMVIEWTMP=Linear

Oct 6, our favorite ex-Inqer uses the same number as the lower bound for a die-size guess.

http://www.semiaccurate.com/20...d-and-high-end-market/

Coincidence?:laugh:

Actually I hope to god that the 530mm^2 number was already in the public domain somewhere prior to my having stated it in these forums...not going to say jack about cause-and-effect here but I will say I very much intentionally delayed saying anything about Fermi specs (and Cypress) in all my posts until I was pretty sure I had read the info elsewhere in the public domain. :eek:

Since we are sort of on the topic, these items might generate some cerebral discourse.

lol, given the content of the this thread i think we can conclude with a high degree of certainty that Fermi will be 530mm^2, or 8.7% smaller than GT200 was on 65nm which isn't too bad.

Looking at IDC's photos i can't tell if they are using transistor count or die size as the relevant metric for the relative yield data, anyone care to elaborate?

Edit: Should have looked at all the pics. it looks like they are using die size as the relevant metric for the relative yield data, it seems to me like transistor count would be more relevant, anyone care to elaborate?

 

Forumpanda

Member
Apr 8, 2009
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Originally posted by: deputc26
Looking at IDC's photos i can't tell if they are using transistor count or die size as the relevant metric for the relative yield data, anyone care to elaborate?
'AreaGT300' and 'AreaRV740' ;)
 

bfdd

Lifer
Feb 3, 2007
13,312
1
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Originally posted by: Forumpanda
Originally posted by: deputc26
Looking at IDC's photos i can't tell if they are using transistor count or die size as the relevant metric for the relative yield data, anyone care to elaborate?
'AreaGT300' and 'AreaRV740' ;)

what he really meant was Arena. he dropped the n on accident, you know cause they are totally going to fight to the death for our amusement.
 

nyker96

Diamond Member
Apr 19, 2005
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Originally posted by: cusideabelincoln
Knowing IDC, he must have had a method for coming up with that number. Perhaps Charlie is using the same method? He could have just as easily used Anand's number of 460mm.

460mm is correct in terms of transister count. 500+ is too big for the transister count, i doubt it's correct. besides at that size, it will be very hard to make at good yield.
 

Idontcare

Elite Member
Oct 10, 1999
21,110
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Originally posted by: nyker96
Originally posted by: cusideabelincoln
Knowing IDC, he must have had a method for coming up with that number. Perhaps Charlie is using the same method? He could have just as easily used Anand's number of 460mm.

460mm is correct in terms of transister count. 500+ is too big for the transister count, i doubt it's correct. besides at that size, it will be very hard to make at good yield.

Using xtor counts and xtor density as an means of estimation is not without merit but when you do that you make certain assumptions regarding the things you are attempting to estimate and those assumptions may actually invalidate the very reasoning you are making the estimates to begin with.

My post in another Fermi thread:

Originally posted by: MODEL3
With 2,249X scaling for transistor number, the 5870 (40nm) was only 1,284X bigger (die size) in relation with 4870 (55nm) (334mm2/260mm2).

So the hypothetical GTX380 (40nm) with 3,15 billion transistors will have also 1,284X scalling regarding die size.

So GTX380 (3,15 billion transistors) will be 603mm2. (470mm2*1,284)

So the real GTX380 with 3 billion transistors (with the logic that you are using) is going to be less than 603mm2.

In the best case scenario will be 603mm2/3,15*3=575mm2.

I don't want to come across as trying to harass you over your estimations and logic here as it is all plausible, but I did want to quote you as a segue to my interjecting the following info regarding xtor density and why comparing xtor density between two architectures, and even comparing it to two IC's designed by the same company and using the same architecture, is a rather tricky thing to do with any degree of confidence because there are so many unknowns that we must make assumptions in regards to in order to perform such an analysis.

It is the absence of this info, the trade-off decisions made by project managers, that can result in such estimations being critically flawed because at best they will be right for all the wrong reasons.

Not that we shouldn't try, but we should be clear(er) on what it is we are assuming must be true when making xtor density comparisons. (such as "assuming the design optimization budget was the same for RV770 and Cypress, xtor density for Cypress was as equally optimized for 40nm as the xtor density for RV770 was optimized for 55nm", etc)

The following is dialogue I had with another forum member in a pm, it seems generically applicable enough to the current topic that I thought it would add value in my reposting it here in-toto. It is not meant to be all-encompassing of the subject matter of xtor density, a 600pg book could be written in pursuit of that, but it meant to communicate the salient points regarding such metrics

Die Density: Will more tightly packed transistors create more heat. Require less voltage to run? And how was ATI able to cram 900+ million transistors into such a small space and Nvidia, with 50% more transistors took up more than 50% larger die space?

xtor density doesn't really equate to power-consumption (heat) directly. Transistors are two dimensional creatures, they have a length and a width.

The length, or more specifically the minimum length possible for a given node, tends to be the metric that catches a lot of headlines. But the width is also important.

Drive currents are normalized per transistor width. nano-amps per micron.

http://www.realworldtech.com/p...ID=RWT072109003617&p=5

What is the relevance of drive current? It determines the amount of current that leaves the transistor which is then used to drive (turn on) subsequent transistors.

The higher your drive current is means the smaller (narrower in width) you can make the xtor (resulting in higher density) while achieving the same amount of amps coming out of the xtor to drive more xtors (the act of computing).

Now the architecture is what determines how many more xtors you need to drive to do your computation. This is where NV and ATI diverge and is why their xtor density can be so different.

Also drive current is voltage dependent, so you can use smaller (narrower) xtors but increase the operating voltage and get more drive current out of them that way.

Now increasing voltage will increase heat and power-consumption. So if you implement an architecture that needs lots of drive current but you want high xtor density (for lower manufacturing costs, higher yield) then you increase the voltage.

Or you could optimize your architecture to not need so much drive current and then you could use higher xtor density with lower volts.

ATI is able to cram so many xtors into such a small area because they use more narrower xtors, less net idrive per xtor, which means they either up the voltage to boost the net idrive per xtor or they implemented an architecture that is less demanding of drive currents.

(incidentally if you checkout Anand's article on Intel's choice of 8T sram on Nehalem versus 6T sram on Penryn you'll see it comes down to similar architecture vs. power-consumption vs. xtor density tradeoffs).

The architecture dependence is why you'll see me repeatedly stating in the forums that making xtor density comparisons between AMD and NV is pointless unless we know far more technical/intimate details about the architecture and design tradeoffs between Vcc and GHz that were made by the project managers.

I don't understand why if both ATI and NV use TSMC processes, why are the transistor densities so different? The transistor size of the ATI design and the Nvidia design are the same size on 40nm, correct?

The smallest xtor size (as in Lg or gate length as its called) is the same, as well as the Idrive (nA per um) for any given voltage. This is true because they both do use the same 40nm TSMC process tech.

But the architecture dictates the needs for Idrive, which then requires choices to be made in terms of voltage (power-consumption tradeoff) versus diesize (cost, yield tradeoff).

Also both voltage and the minimum gate length determine clockspeed.So if you find out you need a certain operating voltage to hit your targeted clockspeeds but you used needlessly wide xtors (too much Idrive generated from the voltage and your selected xtor width) then you are just needlessly generating power-consumption and heat. The design tools they have nowadays are good enough to eliminate much of this uncertainty though.
 

OCGuy

Lifer
Jul 12, 2000
27,224
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Im not suggesting anything here, but have you ever seen IDC and Charlie in the same room?


Me niether.






:Q
 

Idontcare

Elite Member
Oct 10, 1999
21,110
64
91
Originally posted by: Lonyo
What's the optimal sort of yield that someone is likely to see after x years on a process? (obviously not 100%).
I'm sure there have been AT articles discussing yields for Intel and I might be inventing an 80% number because I don't remember them all that well, but given a an 80% RV740 yield, that would roughly translate to a theoretical sub-50% yield for the GT300, which isn't that great.

What does your predicted yield take into account though, since surely clock speed targets could have an impact. Are you saying yield rate based on defects, or yields for dies working at a target clock speed?
50% of full fledged GT300 dies and then 30% more which fail to hit the required high end clock rate wouldn't be terrible, but 50% of functional dies (no defects) of any given (reasonable) clock speed would be terrible. i.e. 50% that can either be high end GT300, or the slower clocked variant (ex. GTX280 and GTX260 cards if they were only different clocks).

It's functional yield, not bin yield (which would be the product of functional and parametric yield...clockspeed and power-consumption being factors themselves in the parametric yield). And obviously it is functional yield sans DFM (design for manufacturing) such as harvesting and fusing. It simply speaks to "entitlement", that which the process node's intrinsic characteristics would seem to deliver at a minimum.

Alpha is the defect clustering factor, from a classical negative binomial distribution used to calculate yields, it's what distinguishes yield estimations based on areal properties from those created by the assumption of an entirely random Poisson distribution of killer defects (the simpler rule of thumb equation that you see in my post way above in which the limit of alpha goes to infinity).

You can read more about it here, although that paper is not in and of itself a defining paper on the topic, it just delves into the utility of the equation a little more deeply with the extensive examples and tables so it lends itself nicely to employment as an educational aid IMO.

Also don't believe the 530mm^2 number as gospel truth...it simply reflects the diesize that is referred to in the highest of circles I have the pleasure of interacting with, but those circles don't necessarily overlap with the circles of people who actually know the truth or are involved in creating it in the first place.

The fact that the same number has repeatedly shown up in a number of public domain locations means nothing more concrete than that some of the people involved in the circles are prolific at talking about it.

It will be interesting to see if the diesize number turns out to be correct.
 

MODEL3

Senior member
Jul 22, 2009
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Originally posted by: Idontcare
My post in another Fermi thread:

Originally posted by: MODEL3
With 2,249X scaling for transistor number, the 5870 (40nm) was only 1,284X bigger (die size) in relation with 4870 (55nm) (334mm2/260mm2).

So the hypothetical GTX380 (40nm) with 3,15 billion transistors will have also 1,284X scalling regarding die size.

So GTX380 (3,15 billion transistors) will be 603mm2. (470mm2*1,284)

So the real GTX380 with 3 billion transistors (with the logic that you are using) is going to be less than 603mm2.

In the best case scenario will be 603mm2/3,15*3=575mm2.

I don't want to come across as trying to harass you over your estimations and logic here as it is all plausible, but I did want to quote you as a segue to my interjecting the following info regarding xtor density and why comparing xtor density between two architectures, and even comparing it to two IC's designed by the same company and using the same architecture, is a rather tricky thing to do with any degree of confidence because there are so many unknowns that we must make assumptions in regards to in order to perform such an analysis.

............................................................

I don't exactly like someone to cut and past my post the way he likes:

Some users used a logic to calculate the transistor space and said that Fermi will be around 650mm2 or more.

I disagreed with their estimation and i used their logic in order to prove that,
even when we use that logic that they used, the design cannot be +650mm2.

I started my post with:

Originally posted by: MODEL3

With this logic:

GTX280 = 1400 million transistors 576mm2 at 65nm

GTX285 = 1400 million transistors 470mm2 at 55nm

4870 = 956 million transistors 260mm2 at 55nm
http://www.anandtech.com/video/showdoc.aspx?i=3405

5870 = 2,15 billion transistors 334mm2 at 40nm
http://www.anandtech.com/video/showdoc.aspx?i=3643&p=1

..................................................

and I didn't say that the design will be 575mm2 and stoped there....

I ended my post with:

Originally posted by: MODEL3

I guess (based on this logic) a good estimation is 576mm2 (same as GTX280)

The thing is that with a new architecture, new process technology and with the way that NV calculates the transistors (does the old numbers include cache transistors? does in this case the 3billion figure include cache transistors?) it is too difficult to say what the die size will be.

So i already was saying that this logic cannot bring accurate results because this will a new architecture and on a new process technology and the variables involved will be more.

Also one day (at least) before you or i posted a specific die size number for Fermi in that topic,
Nvidia had already confirmed that Fermi will be less than GTX280 (576mm2).

Example:

http://www.fudzilla.com/content/view/15782/65/

Since we know that GT200 had a massive size of 576 square millimeters we can tell you that Fermi is slightly smaller. This was confirmed by Nvidia but we could not get the exact size of the chip. We just learned that the chip has 3.1 billion transistors and that is the only number they wanted to give away.

 

Idontcare

Elite Member
Oct 10, 1999
21,110
64
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GF100 - 529.17mm2 for every GeForce GTX 465, 470, 480, 485[?]
Currently, GF100 is the largest monolithic chip in the world - no other competes with GF100 in terms of number of transistors packed on one square of silicon, interlined with copper interconnects. Finding the size of the die was not exactly an easy task - in order to find it out, you have to slice the sides on a heavy and large IHS [Integrated Heat Spreader] without knowing how large the die is - needless to say, a complex affair. Physically, GF100 has 16 clusters with 32 cores in each [512 total, 352/448/480 active in the shipping parts as of 2010/8/9], 48 ROPs and 60 Texture Address Units.

Long story short, GF100 die, the massive three billion transistors, 40nm part is approximately 23x23mm2 in size, i.e. 529.17mm2. Direct competitor: AMD Cypress, 337mm2 and Hemlock, 674mm2 [2x 337mm2].

http://www.brightsideofnews.com/news/2010/8/9/nvidia-fermi-geforce-die-sizes-exposed.aspx

lol...there was so much drama over this at the time that I just had to post a link to Theo's results. Sorry for the thread necro. Now I got to go buy some folks their beer :p