1000 whats!

[ratdude747]

Re: 1000 whats!

idk.... if i ever get a really good graphics card it might crank it up...

[ratdude747]

Re: 1000 whats!

depends on if i ever upgrade the GPU in it...

[Rulycat]

Re: 1000 whats!

Maybe when you get those GeForce 780GTXUltraMegaToasterGTi in SLi ratdude?

[370forlife]

Re: 1000 whats!

Even with a updated graphics card, it takes a lot to get close to 400W.

http://www.jonnyguru.com/forums/showthread.php?t=7526

He has a Q9550 @ 3.4GHz and two 4870's in crossfire, and even in 3dmark06 he dosen't get over about 350W from the wall.

[PCBONEZ]

Re: 1000 whats!

Is 950 watt power supply too much for a 400 watt computer. Perhaps there are other adverse effects. Heat may or may not be one of them.

Big PSU's are mostly advertising hype for people that like shinny things.
Assuming a modern Intel CPU... [I dunno about AMD..]
Few people need more than a 350-400 watt PSU.
Even most 'gaming' machines don't need more than 500-600w.

The Core architecture uses a lot less power than the old Netburst -plus- most of the Core based CPU's have 'Speed Step' like laptops [might not call it that but it's effectively the same thing].
- Most C2D are 65 watt TDP max.

DDR2 & DDR3 use less power [per stick] than DDR -and- the sticks tend to be bigger [more RAM] so fewer sticks are needed.
[2x2GB sticks uses much less power than 4x1Gb sticks.]

Most of the high watts PSU's simply add tons of +12v leaving +3.3 & +5 in the range it always was.
The -only- purpose in that is to support Video Cards.

If you look up the TDP of your video card[s] then add 350 watts [add 400w if your CPU[s] use more than 100w] to it that's roughly how big a PSU you need.
[But you do need to check prespective PSU's specs to make sure the amps are on the correct rails.]

~~
Too big a PSU just means you spent too much money on the PSU and it will take more electricity to run the system because you will be below the PSUs optimum efficiency range. In some cases there might not be enough load on it for it to regulate voltage properly.

[This is ball-park generalization]
Peak efficiency on older single silicon type PSU's falls at about 70-80% of rated watts. On the newer PSU's that use silicon in parallel the peak is around 50% [and the curve is flatter so the good range 'sweet spot' is wider, probably something like 40-70%]. Mid-range and lower grade PSU's can have some single silicon and some in parallel so it's too complex to have a simple rule of thumb that's always right.
.

[PCBONEZ]

Re: 1000 whats!

Good luck pulling 400W from that computer. It will probably have a hard time hitting 250W.

Agree.

I played with a Kill-A-Watts on my wife's machine.
-
Abit IP35-E
E4500
nVidia 7300GS[or GT?]
DVICO Tuner.
2x2GB.
3x WD 500Gb in Mirrored RAID w/spare.
2X Optical.
-
It usually sits at 90-120w [from the wall] and I couldn't get it over 140w - as I recall....
It wasn't really an experiment about the PC load.
Just making sure I wasn't overloading the wall socket by checking each device.
.

[Trinite]

Re: 1000 whats!

False: and do not follow Ohm's Law

Diodes follow Ohm's law _perfectly_. At a given temperature and current, you will get exactly the same forward voltage drop, from which you can calculate resistance.
So do lightbulbs.

Okay, so you make the claim that diodes follow Ohm's law perfectly, which would imply that they act just as a resistor (further backed by your lightbulb reference).

Two diodes in paralel - halving operating current -> says nothing about halving resistance, you are jumping to (wrong) conclusions.
Diodes do not exhibit a !! Linear !! resistance, like ordinary resistors do.
Therefore halving current does not halve voltage drop !

Diodes are NOT pure resistive loads. They have their forward voltage at rated current and there is no way to avoid that.

Yet, here and here you completely contradict yourself. In order to follow Ohm's Law, a load must have a linear resistance otherwise it cannot follow I = V/R. You don't know what you are talking about. Game over for you, man. You blew it and you are done. You've proven yourself wrong by your own words. Unless you can show why the derivative of a resistance curve involving two diodes is not half that of of a single diode, then what you say has no merit. I suggest you take some calc courses before you come pretending to know everything. Derivatives work just the same for both nonlinear relationships and linear relationships. The derivative is always a function of the slope of a line or curve.

[Pyr0Beast]

Re: 1000 whats!

Okay, so you make the claim that diodes follow Ohm's law perfectly, which would imply that they act just as a resistor (further backed by your lightbulb reference).

You made that claim.

But yes. At same operating conditions - same resistance.

Just not linear relation at any other conditions.

Yet, here and here you completely contradict yourself. In order to follow Ohm's Law, a load must have a linear resistance otherwise it cannot follow I = V/R.

Uh ?

You can calculate diode's resistance any time you want from drop voltage and current.

But, are you saying that I =/= V/R ?

I = V/R any time you want. Just that resistance changes as well.

[Trinite]

Re: 1000 whats!

You made that claim.

But yes. At same operating conditions - same resistance.

Just not linear relation at any other conditions.

I said diodes did not follow Ohm's Law.

Uh ?

You can calculate diode's resistance any time you want from drop voltage and current.

But, are you saying that I =/= V/R ?

I = V/R any time you want. Just that resistance changes as well.

That's the instantaneous value, which does work with the formula. However, I=V/R is not a formula that is compatible with a diode's resistance curve.


"Unlike fundamental Coulombs and Gauss Laws which are always true. Ohm's Law is not universally true. For example, for semi-conducting elements like diodes Ohm's Law is not obeyed at all. Elements which obey Ohm's Law are called resistors."

Suggest you read the whole thing: