"error loading operating system" and sudden power-down

Joined
Apr 24, 2008
Messages
38
Okay, so I just recently upgraded my Video card and Motherboard.

Previous:
EVGA GeForce 8800
Asus p5b

New:
EVGA GTX 260
EVGA nforce 760i

I am having two problems, one of which I have (mostly) figured out.

The "error loading OS" is due to the mobo swap, as far as i can tell, and can be resolved with a repair installation. (Please correct me if i am wrong).

However, the sudden power down issue is worrying me. I figured it was a power supply problem, so I swapped out the 260 for the 8800. However, the power down still occurred. I'm going crazy. Any help is appreciated.
 
Can you run through exactly what the issue is? I'm unsure of when this power-down is occurring.
 
Sure.
The first time:
1) Power on
2) Press DEL to enter BIOS
3) Check boot priority settings
4) Rearrange settings to my preference (CD>Removable>HDD)
5) Press F10, "Save settings to CMOS? Y/N"
6) Before I can press Y > Enter, Power Down.

After Swapping out the 260 for the 8800

1) Power switch
2) Insert XP CD (To begin repair installation)
3) POST
4) "Ready to boot from CD. Press any key to continue..." >Press key
5) XP CD begins to load, showing various prompts for RAID and the like
6) Power down w/o warning.
 
It could be a power supply issue, or it could be the motherboard. What PSU are you using?
 
Rosewill RX950-D-B
I had thought it was a PSU issue but the PSU worked before the switch. I do know, however, that it is extremely close to the tolerances of the 260 in amperage.
What I could try is putting the old mobo in and testing the new card...

(Side note: The two 6pin power connectors are on the same +12v 20A circuit. Dumb design idea.)
 
(Side note: The two 6pin power connectors are on the same +12v 20A circuit. Dumb design idea.)

Actually this is a good idea, each card (both 6 and 8 pin connectors) should be connected to the same rail for stability purposes. Being on the same rail reduces the voltage differential for both inputs. It's in the docs.
 
Actually this is a good idea, each card (both 6 and 8 pin connectors) should be connected to the same rail for stability purposes. Being on the same rail reduces the voltage differential for both inputs. It's in the docs.

Okay, but then if they're on the same rail they don't have enough Amperage to satisfy the card's needs. Having them on different rails would mean that the amperage would increase from 20A to 40A because the resulting circuit would be in parallel.


Aaaaanyway, back on topic. Any other suggestions or ideas?
 
Okay, but then if they're on the same rail they don't have enough Amperage to satisfy the card's needs.

I understand - but sometimes you can't change the laws of physics. A few good midrange power supplies use a single high amperage rail. Several +20A rails would be better. My thermaltake 1200W outputs 106A over four rails (20 36 30 20).
 
I understand - but sometimes you can't change the laws of physics.).

What I just stated above was Kirchhoff's Junction Rule.

What's annoying about my PSU is that both of the 6pin plugs are on the same 20A circuit and I need at min 35A to run the GPU.
 
Rosewill RX950-D-B
I had thought it was a PSU issue but the PSU worked before the switch. I do know, however, that it is extremely close to the tolerances of the 260 in amperage.
What I could try is putting the old mobo in and testing the new card...
If the PSU is actually a good unit, then it's more than powerful enough to handle even a triple-SLI setup. However, Rosewill's quality fluctuates quite a bit, so I don't know if the one you have is actually a quality unit. The 750W and 850W versions are pretty good though, so I would give this one the benefit of the doubt. If you have another PSU you can test with though, I suggest that you do so.
(Side note: The two 6pin power connectors are on the same +12v 20A circuit. Dumb design idea.)

Not really. Each 6-pin connector is rated for a maximum output of 75W, so the combined output of 150W for both connectors is still well below the rail's maximum of 240W.
 
Not really. Each 6-pin connector is rated for a maximum output of 75W, so the combined output of 150W for both connectors is still well below the rail's maximum of 240W.

But at the same time the maximum rating for that circuit in amperage is 20A. The GPU requires a 550(about)W psu and 35A minimum.


I don't have another PSU at the kind of rating i need, but I'm going to try running things with the old mobo again. Probably tomorrow though. As for this current discussion, however, I'm happy to continue.
 
Glad you know so much about Kirchhoff's circuit laws. I guess if that was there before I quoted ya, we would have known. Use an adapter from the other rail. You shouldn't use both rails like that but it's better than nothin' - right?
 
Glad you know so much about Kirchhoff's circuit laws. I guess if that was there before I quoted ya, we would have known. Use an adapter from the other rail. You shouldn't use both rails like that but it's better than nothin' - right?

Yeah, I'm taking college physics.

Sadly, I tried that already too. (Though it took longer for it to happen -_- ) It's really looking like a motherboard issue, isn't it?
 
yep. sounds like the mb. I think ya left it in a bad boot state too long.
 
The GPU requires a 550(about)W psu and 35A minimum.
No. First of all, nVidia exaggerates all their power requirements because most cheap PSUs aren't actually capable of putting out their rated wattage. Secondly, that rating is also for an entire system, not just for the card itself.

The power specification for a 6-pin PCI-E connector mandates that each one can supply a maximum of 75W. Therefore, two 6-pin cables can provide at most 150W total, or 12.5A combined. The power provided by the PCI-E slot, which is a maximum of 75W due to the fact that the GTX260 is backwards-compatible with PCI-E 1.x (PCI-E 1.x provides up to 75W through the slot, while 2.0 provides up to 150W), comes from the rail that supplies the 24-pin ATX connector, which is a different rail than the one that supplies power to the PCI-E power cables. Therefore, there is absolutely no risk of your video card coming anywhere near the limits of the PSU to supply power. In fact, although the card could theoretically draw up to 225W of power, its actual requirements are more like 125-150W.
Glad you know so much about Kirchhoff's circuit laws.

This actually has about fuck all to do with Kirchoff's laws. This is all Ohm's law ;).
 
UPDATE:
It was the motherboard. I'm currently running off of my old p5b and the 260, no problems. Time to contact EVGA for support/RMA
 
This actually has about fuck all to do with Kirchoff's laws. This is all Ohm's law ;).

The sum of all currents to and from a node equal zero. That is exactly what we are talking about. Ohms law is another method but is not being applied here. I = E/R No one mentioned voltage or resistance - except you.
 
The sum of all currents to and from a no That is exactly what we are talking about.

Actually, Kirchoff's junction rule is that the sum of all currents entering a junction equals the sum of all current exiting the junction, not that both equal zero (although I'm sure that you meant the difference between the sum of all entering current and the sum of all exiting current is equal to zero ;)). However, that does not really apply here because the current supplied by the power supply is variable and depends on the load, so the actual current has very little to do with the maximum that the PSU is rated for. Plus, the actual requirements for each rail are given by P = V*I (or P = (I^2)*R, but that form of the equation is not pertinent here because the resistance is not known), which is a permutation of Ohm's law. Since there is an actual maximum power that can be drawn from the PCI-E connectors based on the specification (which is 75W per 6-pin connector), as long as that maximum power is within the rail's capabilities, the maximum current it can provide is not an issue.
 
Actually, Kirchoff's junction rule is that the sum of all currents entering a junction equals the sum of all current exiting the junction, not that both equal zero (although I'm sure that you meant the difference between the sum of all entering current and the sum of all exiting current is equal to zero ;)).

It's the the same thing mathematically. Of course depends on your definition of + and - for polarity's sake but the result is the same.

I1 = I2 + I3

0 = I2 + I3 - I1
 
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