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Author Topic: 6870 x2 vs 6950 x2 for mining  (Read 2670 times)
Panther Modern (OP)
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May 01, 2011, 07:04:40 PM
 #1

I'm looking to order graphics cards for my new build soon which I intend to use for both mining (on free electricity) and gaming. I'm wondering in regards to mining alone, how the two compete in terms of mhash/s I will get in real world conditions, and if the $100 increase on the 6950s (unlocked most likely) is worth whatever increase they give in performance.
I've seen the hardware comparison wiki, but some additional info on the matter would be greatly appreciated.
JorgePasada
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May 01, 2011, 09:09:43 PM
 #2

From my (albeit limited) understanding.

Once you unlock the 6950's, they're the exact same card as a 6970.
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May 01, 2011, 10:45:21 PM
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I'm talking about 6870s not 6970s Wink
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May 01, 2011, 10:49:29 PM
 #4

From my (albeit limited) understanding.

Once you unlock the 6950's, they're the exact same card as a 6970.
This is not the case.  There are two kinds of unlocks, one that unlocks additional shaders and one that is a complete flash to the 6970 bios.  The second method of unlocking is riskier and could damage your card.

I'm not even sure if you can buy unlockable 6950s these days.   

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JorgePasada
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May 02, 2011, 03:53:10 AM
 #5

From my (albeit limited) understanding.

Once you unlock the 6950's, they're the exact same card as a 6970.
This is not the case.  There are two kinds of unlocks, one that unlocks additional shaders and one that is a complete flash to the 6970 bios.  The second method of unlocking is riskier and could damage your card.

I'm not even sure if you can buy unlockable 6950s these days.   

Thanks for the heads up.

http://www.newegg.com/Product/Product.aspx?Item=N82E16814102932

Anyone have any luck with those? Thinking about shoving 4 of them into a system, seems about the cheapest per shader I can get.
TurboK
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May 02, 2011, 05:05:59 AM
 #6

Go for 4x 5850s instead. Way more mhash per $, if you can still find that Sapphire Extreme version that recently sold for something like 140$.

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