Originally posted by: Furen
Originally posted by: coldpower27
Dual Core K8 is now 183mm2 on the AM2 platform, so it's not as small anymore. As well you have got to factor in that K8L Quad Cores are supposed to come with 4x the cache available on the Dual Core Windsor. So the close to 300mm2 die figure hold some merit. As well an optical shrink with a processor typically doesn't yield 50% reduction on the whole, it's closer to 60% really given past die shrinks as a guide.
So K8L Quad Core will have something like 4x512kB plus the additional 2MB of LV3 cache.
If cache takes about 40mm2 for about 1MB of it on the 90nm node, since the cache area was about 80mm2 for the 2x1MB on Windsor, then on the 65nm node 40mm2 should account for the 2MB of LV3, so your looking at about 260mm2 then you have to add in K8L's additional complexity and close to 300mm2 will be around right.
Remember this is a Quad Core processor. 150mm2 is completely unrealistic even an optical shrink of Windsor would only yield 110mm2 with a 2x512KB die. Double that and you get the Quad Core at 220mm2. Now with that said 90mm2 is also too small for 65nm Dual Core, Brisbane if it's 2x512KB only will be around the size that San Diego was so 110mm2 or so which is very reasonable.
Calm down, my friend, I was just doing a bit of brainstorming, not claiming that I know everything absolutely or anything of the sort (all those "shoulds" and the like should show that I'm purely speculating). I said 50% simply to make the calculations easier. I'll try again.
Let's say AMD decided to make a quad-core K8 (not K8L). First off, you only need a single northbridge and memory controller. If you think this is small just look at the increase in die-size that AM2 brought by adding DDR2 support and (according to the inq) quad crossbar ports, I'd guess (a wild guess) that the stuff that does not need to be duplicated is around 50mm^2 of the die, which would mean that around 130mm^2 would need to be duplicated. So 2x130=260 + 50 = 310mm^2. That would be the die size for quad-core K8 on the 90nm node. Assuming a 57% die-size due to the 65nm shrink (a winchester is about 57% the die size of a newcastle) that would give us a 177mm^2 die size. I'm probably low-balling it a bit since a lot of the info needed to calculate it is purely speculation but I'm probably not off by a whole lot. Consider that a Clawhammer (at 130nm) has a die size of 193mm^2 and a Toledo (dual-core, 90nm) has a die size of 199mm^2. Same cache sizes, and Toledo added SSE3.
So that leaves a whole lot of space for the L3 cache and the K8L improvements. I'm not gonna claim that our estimates are wrong (I, too, calculated around 300mm^2 for the die size of those monsters), I'm just saying that those dice may not be final quad-core silicon.
EDIT: By the way, when I said "90mm^2 X2s" I was talking about 65nm Manchesters, lol... stupid AM2 die-sizes always screw me over (I used 150mm^2 * .6 to calculate that), so you're right about that. AMD does indeed have more production capacity but it also serves a much bigger market. AMD pretty much said that it didn't want to produce 2x1MB L2 X2s 'cause they're too big and strain its capacity, even with FAB36 and Chartered fabbing for AMD. Clawhammer was indeed huge when it came out but the process was very mature and 90nm was coming relatively soon after its introduction, and a cache-cut part came out as well after a bit. Cache was a huge chunk of Clawhammer, I dont think it's going to be that big of a part of K8L.
Just wanted to make sure.
Heh, I doubt there is that much unneeded circuitry on the Dual Core K8, if the memory controller was boosted to serve 4 cores instead it would indeed become larger. 50mm2 of the die doesn't need to be duplicated out of 183mm2, I am not going to call you wrong, but I seriously doubt that much of the die is saved. And yes I did factor in the fact that it was a Dual Core K8 which you were talking about.
I really don't think the SSE3 instruction set is really a significant part of the die space. The San Diego core at 114mm2 already had that and Toledo was 199mm2, which is 58% the size of Toledo, so they saved 8% die space from Single to Dual Core, using that as a baseline.
I get about 190mm2 for a Quad Core K8 on the 65nm node or 316mm2 on the 90nm node. Which is mildly higher then what you have.
Well that is pretty obvious that AMD wants to optimize it's production capacity, they want to make as much money as they can, even with Fab 36 online, and the impending transistion of Fab 30 to Fab 38 and Fab 7 of Chartered Semi COnductor, as there is now rumors of the 0.5x multiplier 4000+, 4400+, 4800+ instead of having 2x1MB of cache, with the die size I predicted these cores with only 2x512KB cache now will be around the same level as Allendale. They will still prove to be more expensive in production cost then Allendale, due to the fact that AMD uses DSL SOI and that they are at least 1 layer thicker, but they are more optimal.
Soon is a matter of perspective I guess, Clawhammer came out in September 2003, and the 90nm shrink of that core didn't come till April 2005 for the desktop with San Diego, but like I said a core that intorduced a new architecture tends to be at least 200mm2 or close to it on the AMD side and this is using 200mm wafers, with 300mm wafers a ~300mm2 die size to start wouldn't be surprising. Though AMD does do a cache cut if they find they need more room to compete.
On the Intel side of the fence it varies.
Willamette on 0.18 micron is 217mm2 (Big)
Pentium M on 0.13 micron is 82.8mm2 (Small)
Core 2 Duo on 65nm is 143mm2 (Medium)
Yeah, but I would say we should stick with these die size estimates until more concrete proof comes along that can negate the picture we have seen here, 300mm2 is pretty damn realistic IMO. 65nm Quad Cores will be more expensive to produce relative to even 90nm Dual Cores, it will take hte 45nm process to bring this down to a more reasonable level.
Regarding the optical shrink I would say 60% is fair. Since the 0.13 micron to 0.09 micron shrink should yield a die size of 48% if mathematically perfect, so 58% is 10% larger, 90nm to 65nm is 52% for mathematically perfect, so 60% is fine for that.
No if Intel has 2x4MB and that is around 40% of the die for Core Architecture, then (4x512KB & 2MB) is not going to be that significant on K8L, maybe somewhere in the 30% range. But it will be enough to bring the die size to ~300mm2 or so, with the Quad Core K8L.
EDIT: Spelling, and minor additions.