(retro!) GeForce 7800 GS, does it really have 8 ROPs?

[Hi-Fi Man]

Blast from the past people, I just obtained a BFG 7800 GS OC AGP and most of the internet says this card has 8 ROPs which is half of what a full G70 has. However it has a 256 bit bus just like a full G70, How is this possible? From what I understand the size of the memory bus is tied to how many memory controllers a chip has which is (usually) tied to how many ROPs a chip has.

Funny thing is it never loses to a GeForce 6800 Ultra which has 16 ROPs, no matter the amount of AA or pixels. Nvidia didn't change the ROPs from NV40 to G70 so I would've expected slightly less performance at higher resolutions and AA from the 7800 GS compared to the 6800 Ultra but that's not the case.

Does this chip really have 8 ROPs or 16? Or some kind of memory interleaving or something else I'm missing (memory controllers independent of ROP count)?

 

[Soulkeeper]

http://www.beyond3d.com/content/reviews/38/2
This link has some good info on the subject.
The picture at the bottom seems to indicate that the number of ROPs are not directly tied to the "crossbars" for the G70.
I've forgotten pretty much everything from the pre-unified shader era, but that conversion G80 and on might be the reason for what's going on here.

http://pclinks.xtreemhost.com/video_g.htm
Also note here, how all G70 based vid cards share the same die and same memory width.
Only the PS/VS count differs between them.

I owned a 6800 GS, I believe, which did the same thing.
It had the same mem bandwidth as the full 6800, but disabled ROPs PS/VS

 

[stahlhart]

Let's try having this discussion again, without the previous rudeness.
-- stahlhart

 

[SPBHM]

this card used the 7900, not the 7800 die, right?

 

[Hi-Fi Man]

Nice info Soulkeeper, that's what I kind of figured but wasn't totally sure.

SPBHM, it uses G70 which is the 110nm die. The 79xx cards use the newer G71 die which is on a 90nm node.