Need some advice please

[over288]

Im currently using a X2 5600+, and Im not quite happy with it =s

Called all the pc stores near me, and they're all out of E8x00s

Im planning to change mobo for another 3870 for crossfire

Im seeking some advice on whether to get an E6850 or Q6600/6700 or maybe the best Phenom i can get? Basically anything around the price of a E8500.

Also, do Xeons work the same as Core 2 duos on 775? (planning to get X38 mobo)

 

[CTho9305]

Im currently using a X2 5600+, and Im not quite happy with it =s

What's wrong with it? Based on my experience with an X2 4400+ and an ATI Radeon HD 2900 XT, your CPU wouldn't be holding you back even in Crysis.

 

[Idontcare]

Originally posted by: over288
Im currently using a X2 5600+, and Im not quite happy with it =s

Called all the pc stores near me, and they're all out of E8x00s

Im planning to change mobo for another 3870 for crossfire

Im seeking some advice on whether to get an E6850 or Q6600/6700 or maybe the best Phenom i can get? Basically anything around the price of a E8500.

Also, do Xeons work the same as Core 2 duos on 775? (planning to get X38 mobo)

Along the lines of what CTho9305 is asking, if we know more about (1) what you intend to do with your computer, and (2) what specifically makes you unhappy about your current system, then we can better guide you away from options that are likely to continue to make you unhappy.

From what everyone on the web says the socket 775 XEONs perform exactly the same way as the desktop socket 775 chips (at same clock, etc).

However this is contrary to what Intel says...according to Intel the XEONs are produced from a uniquely different stepping (mask set) from their desktop counter-parts as the XEONs are supposedly wired up just a little bit differently inside to make them faster for their intended tasks (server stuff).

I have no idea whether there is a difference (don't own a XEON) but I am inclined to believe Intel given the lawsuit they open themselves up to should they be lying.

 

[Markfw]

Well, aside from the fact that my X3210 doesn't OC nearly as well as my Q6600's (same stepping, 2 B3's and a B3 X3210) , I can;t see a difference. Except that at a lower vcore, it produces more heat, thats odd.

 

[Idontcare]

Originally posted by: Markfw900
Well, aside from the fact that my X3210 doesn't OC nearly as well as my Q6600's (same stepping, 2 B3's and a B3 X3210) , I can;t see a difference. Except that at a lower vcore, it produces more heat, thats odd.

(loosely here, please no one get overly pedantic on me)

V = I x R

P = I x R^2

If you have same V but different P for two otherwise identical CPU's then the system-level resistance is of one CPU must be higher than the other.

This can come from simple process variations that happened to each chip (wafer) during manufacturing.

Linewidths (copper interconnect) can be slightly narrower (or wider) for one chip, but still in spec. Or CMP endpoint on one wafer may have been notably lagged versus another (thicker versus thinner metal lines for just one metal layer, but still in spec)

That's just one example, for each electrical element presented in the chip and there are hundreds of discreet electrical elements.

In the equations above, I (current, amps) will vary too of course but that is intentionally controlled by power regulation whereas R is hard-wired to each chip and doesn't change unless you blow-out a circuit element (over volting, etc).

 

[CTho9305]

Originally posted by: Idontcare

Originally posted by: Markfw900
Well, aside from the fact that my X3210 doesn't OC nearly as well as my Q6600's (same stepping, 2 B3's and a B3 X3210) , I can;t see a difference. Except that at a lower vcore, it produces more heat, thats odd.

(loosely here, please no one get overly pedantic on me)

V = I x R

P = I x R^2

If you have same V but different P for two otherwise identical CPU's then the system-level resistance is of one CPU must be higher than the other.

No, if you have the same V and higher P, the resistance is lower or the capacitance is higher. Higher R only causes higher P if V also goes up.

In the equations above, I (current, amps) will vary too of course but that is intentionally controlled by power regulation whereas R is hard-wired to each chip and doesn't change unless you blow-out a circuit element (over volting, etc).

Power regulators for processors supply a constant voltage, not a constant current.

 

[betasub]

Originally posted by: Idontcare
Along the lines of what CTho9305 is asking, if we know more about (1) what you intend to do with your computer, and (2) what specifically makes you unhappy about your current system, then we can better guide you away from options that are likely to continue to make you unhappy.

No response from OP.

Sometimes it's hard to help out...

 

[Idontcare]

Originally posted by: CTho9305

Originally posted by: Idontcare
(loosely here, please no one get overly pedantic on me)

V = I x R

P = I x R^2

If you have same V but different P for two otherwise identical CPU's then the system-level resistance is of one CPU must be higher than the other.

No, if you have the same V and higher P, the resistance is lower or the capacitance is higher.

How does your statement differ from mine? If one has lower resistance (as you state) does that not require the other to have higher resistance (as I state)?

Originally posted by: CTho9305
Higher R only causes higher P if V also goes up.

Riddle me this, how is it that a 100W lightbulb consumes more power than a 60W lightbulb...considering that both use the same voltage... (hint, the answer is resistance)

Originally posted by: CTho9305

Originally posted by: Idontcare
In the equations above, I (current, amps) will vary too of course but that is intentionally controlled by power regulation whereas R is hard-wired to each chip and doesn't change unless you blow-out a circuit element (over volting, etc).

Power regulators for processors supply a constant voltage, not a constant current.

Again, how does your statement differ from mine?

I'm just a little confused here

 

[CTho9305]

Originally posted by: Idontcare

Originally posted by: CTho9305

Originally posted by: Idontcare
(loosely here, please no one get overly pedantic on me)

V = I x R

P = I x R^2

If you have same V but different P for two otherwise identical CPU's then the system-level resistance is of one CPU must be higher than the other.

No, if you have the same V and higher P, the resistance is lower or the capacitance is higher.

How does your statement differ from mine? If one has lower resistance (as you state) does that not require the other to have higher resistance (as I state)?

I read yours as implying that the higher resistance one would be higher power. The form you wrote the power equation in also implies that higher R = higher P.

Originally posted by: CTho9305
Higher R only causes higher P if V also goes up.

Riddle me this, how is it that a 100W lightbulb consumes more power than a 60W lightbulb...considering that both use the same voltage... (hint, the answer is resistance)

Right...

Originally posted by: CTho9305

Originally posted by: Idontcare
In the equations above, I (current, amps) will vary too of course but that is intentionally controlled by power regulation whereas R is hard-wired to each chip and doesn't change unless you blow-out a circuit element (over volting, etc).

Power regulators for processors supply a constant voltage, not a constant current.

Again, how does your statement differ from mine?

I'm just a little confused here

I read your statement as saying that the power regulator acts like a current source source rather than a voltage source. Even though you can do Thevenin/Norton equivalents, I think it's very misleading to talk about current. Your statement sounds like "R varies a lot, and I varies slightly but the regulator attempts to keep I constant" to me. While you're technically not wrong, for the purposes of talking to the audience of OCer's, I think it leads people to incorrect conclusions.

 

[Martimus]

P = I x R^2

Don't you mean P = I^2 x R?

 

[DSF]

Originally posted by: Martimus

P = I x R^2

Don't you mean P = I^2 x R?

Sounds right to me if we're using the equation P = IV.

 

[Idontcare]

Originally posted by: Martimus

P = I x R^2

Don't you mean P = I^2 x R?

Yep, I had it backwards in my post. This is also the reason I was not understanding what CTho9305 was disagreeing about in my post. I had higher R = higher P which is wrong, as the equation correction bears out.

It is really lower R at constant V means higher I, which in turn means much higher P.

To be fair though, anytime someone asks others to not get pedantic that is usually a flag to say "half of the following is probably wrong".