Power Consumption And Reality?

[puterg33k]

Last weekend I told my cousin Nicky (an electronics engineer), that my computer needed a 500+ watt power supply. He looked at my, sighed loudly and said; what the heck are you talking about?

 

My response to my cousin Nicky was: I have a 5000+ Black edition, AM2+ Mobo from ASUS, 8800GTX, 4 Gigs DDR800, 2x 36gig Raptors, 1x 500 gig storage drives, DVD -/+ RW, and the various fans and such you need to utilize such a system.

 

He laughed at me and told me to cut open a power supply cable. He told me to be careful that I didn

Edited March 6, 2008 by puterg33k

[jammin]

Firstly, that's probably a pretty typical result.

 

Components generally use less power at full load than a lot of people (or those that market power supplies would have you) believe.

 

As for graphics cards manufacturers suggesting certain amp ratings, that is only for the 12v line.

The recommendations also take into account other components in a typical system and a wide range of power supplies in general.

Not all power supplies are built equal, so some overhead is going to be built in there as well.

Recommendations on PSUs from graphics card manufacturers are meant for consumption by those not technically minded, as well as those that are.

It's a sensible way to do things as it ensures there are less problems of people buying PSUs which aren't suitable.

 

Your method of measuring I'm assuming is an overall measurement.

i.e. you are measuring what the total power draw is, which would be 120v x the number of amps (assuming US mains supply).

That may not be taking efficiency into account either (generally around 80% on good modern PSUs)

 

Make sense?

Edited March 6, 2008 by jammin

[hardnrg]

It's hard to argue with an electronics engineer, although I'm surprised he doesn't have a Kill-A-Watt (or other brand) power monitor... you should go buy one, they're not much money

[94Camaro]

The video card can be supplied with higher current because of the magic of tranformers. The construction of a power supply has a transformer to drop the voltage to the requisite 12V. Doing so actually causes a current increase. A reduction of voltage by 10 times will ideally increase current by 10 times. Now, I say ideally, because the quality of the materials used can greatly affect just how closely this ratio occurs. Even with the best materials, that perfect ratio doesn't happen.

 

 

This isn't going to make you happy, but did he take phase angles into account in his measurements? Multiplying the two voltage and current when dealing with AC only works when current and voltage are phase matched. I'd be willing to bet that the current and voltage you are drawing from the line aren't, and you're therefore actually using even less real power! Probably not much less, but it's something. What he measured was likely apparent power, which actually has a real and imaginary component.

Edited March 7, 2008 by 94Camaro

[Waco]

Most people are surprised when they see the actual draw rate on their computer. Even the high-end rigs with dual/quad video cards and quad-core CPUs rarely go over 500 real watts.

 

However, why do a lot of graphics cards suggest you need to be able to supply more Amps that a regular household 15 Amp breaker can supply? How exactialy does this work? If my computer is only consuming a maximum of 2.65 Amps under the most strenuous conditions, why would I need to be able to supply the

[puterg33k]

Now, Im even more confused than when I first asked the question

[exeter_acres]

do a search for Ohms Law and learn it

[Comp Dude2]

or P=VI would be more relevant imo

[exeter_acres]

psst.. that is ohms law

[94Camaro]

or P=VI would be more relevant imo

but only for DC

 

In AC, S=VI* (apparent power = voltage * current conjugate)

Like I alluded to earlier, AC power V and I are both complex numbers. I.E. they have a magnitude and a phase angle. Apparent power (S) is made up of real (P) and reactive (Q) power. (S = P+jQ) The OP measured and calculated the magnitude of the apparent power (S). The "real power" (what you would actually be billed for) would be lower than the apparent power, unless the voltage and current were phase matched, at which point they would be equal.

[iKillSteal]

Now, Im even more confused than when I first asked the question

The folks here at OCC are too smart for their own good.

 

Let me put it to you in layman's terms. Your computer doesn't, and most likely never will, use as much power as you think it does. Does this mean you should skimp out on the PSU? No. Never. Get a good(do NOT trust the reviews on e-tailer websites trying to sell you stuff) psu and roll with it. Don't worry too much about what it's pulling, just be happy that it's working and not exploding into a ball of fire. Although that could happen, too...

[romeo55]

but only for DC

 

In AC, S=VI* (apparent power = voltage * current conjugate)

Like I alluded to earlier, AC power V and I are both complex numbers. I.E. they have a magnitude and a phase angle. Apparent power (S) is made up of real (P) and reactive (Q) power. (S = P+jQ) The OP measured and calculated the magnitude of the apparent power (S). The "real power" (what you would actually be billed for) would be lower than the apparent power, unless the voltage and current were phase matched, at which point they would be equal.

 

Didn't know that about AC power, good stuff to know :thumbs-up:

 

As for the reading at the line, some of that power isn't even being consumed by your computer, a good 20-25% (depending on the efficiency of the PSU) of that is being "lost" in the voltage conversions (mainly from the transformers).