8800 GTX for 540$

[Track]

http://www.newegg.com/Product/ProductList.asp?Submit=ENE&N=2010380048&Subcategory=48&description=&Ntk=&srchInDesc=8800

 

[lewbaseball07]

great

 

[prozac26]

Nice.

If they were in Canada, I would of buy one.

 

[djplanet]

Sweet, but I want a GTS for under $350.

No, I can't be satisfied.

 

[s-d]

Quote:

"I heard was going to produce a Quad GPU version of these (2x8850GX2)"


my question is this:

1) a standard residencial circuit runs on a 15A breaker
2) a top of the line power supply unit is 1000W and has a peak input requirement of 16A
3) a system setup of 2 of these "quad" gpu's must certainly require a 1000W power supply


wouldn't such a system continuously trip the circuit breaker in a session of F.E.A.R (even if all that is pulled into the outlet is the power supply for this unit only)?

 

[dean7]

They will have to make it more power efficient I guess.

 

[sarsoft]

http://www.newegg.com/Product/ProductList.asp?Submit=ENE&N=2010380048&Subcategory=48&description=&Ntk=&srchInDesc=8800

Man i got them for 630 when they got out. 90 bucks down the drain. Oh well thats the price you pay. 8O

 

[Valtiel]

hmm whats that syndrome again? the one that causes people to buy stuff early? just kidding. Its a good thing those prices are going down now I can start planning on buying one.

 

[HDDFreak]

Quote:

"I heard was going to produce a Quad GPU version of these (2x8850GX2)"


my question is this:

1) a standard residencial circuit runs on a 15A breaker
2) a top of the line power supply unit is 1000W and has a peak input requirement of 16A
3) a system setup of 2 of these "quad" gpu's must certainly require a 1000W power supply


wouldn't such a system continuously trip the circuit breaker in a session of F.E.A.R (even if all that is pulled into the outlet is the power supply for this unit only)?

I'm not going to pretend that i'm an expert on electronics but wouldn't you get more amps by dropping the voltage? You get less current by upping the voltage, so why can't you get more by lowering it. Isn't that what v=ir (voltage = current x resistance) means?