Does the system use the actual needed wattage or does it use the PSUs rated wattage?

[ZZDMAN]

Hey people,
i'm guessing the answer is the system only uses the wattage it needs (in my case around 280w peak load). My psu is rated at 450w. It only uses what it needs right?

thanks

[TDOGZZ]

yup ,it uses only what it needs afaik,lol(the amps are probably different,i,d assume it would use all it,s amps)

[P.I.M.P_SLi]

I'm sure it only uses what it needs, although I'm not completely sure if the same goes for older psu's as well.

[ZZDMAN]

Ok and how bout the amps as TDOGZZ mentioned?

thanks btw

[Susquehannock]

Amps = same thing. The unit will only used as much as is required.

Amps * voltage = Watts

[ZZDMAN]

Thanks Guys :P

[ozymandias]

Actually, it uses more than the actual current needed - this is where efficiency comes into play. The best psu's manage about 80%; of all the power pumped into it, 80% goes to your components, the remaining 20% is waisted.

[TDOGZZ]

Originally Posted by Susquehannock

Amps = same thing. The unit will only used as much as is required.

Amps * voltage = Watts

good to know,thx for the info!

i assumed amps is what gives the push for the voltage so the amperage would be at max just for efficiancy like if a heavy source load(tornado fan,raptor hdd...)needed it.<---i shoulda ,lol.

[Stallion]

If a PSU is 550w, with a 650w max. Is that 550w the "80% efficiency" range? I'm wondering about something like this but I don't know how to phrase the question.

[Josie Wales]

Ozymandias went off on a tangent, but I am happy to explain. The rating on the PSU should be for continous output and the max rating is for momentary peak output. I say "should be" because some power supplies do not give continous ratings like they should.

Lets use your example of a 80% efficient power supply that has a 550w continous rating with the ability to provide 650w at peak. Now a computer system only draws the wattage/amperage it needs from the power supply. If the computer is using 320 watts then the power supply must draw 400w of power from the wall outlet to produce the 320 watts (400w * 80% = 320w). If there was a sudden demand for 600 watts then the power supply must draw 750w of power from the wall outlet to produce the 600 watts (750w * 80% = 600w). However, the power supply would only be able to produce 600 watts for a few seconds and it would start to overload if the demand did not drop below 550w. If demand went above 650 watts the power supply would overload.

[afrakes]

Yeah, there will always be power lost in the conversion, so the PSU will actually need more power on the AC side to break even. Like others have mentioned, 75-80% is the kind of efficiency you'd expect to see in an ATX2.x power supply at typical loads. This means they would require somewhere between 120-125% of the actual DC power requirement of your computer.

But power supplies are rated for maximum DC load, so you don't need to worry about efficiency unless you want to look at something on the AC side (such as how much it costs to run your computer, sizing a UPS, etc.).

[larrymoencurly]

Originally Posted by ozymandias

Actually, it uses more than the actual current needed - this is where efficiency comes into play.

Any waste isn't used by "it" but by the PSU itself.

If a computer used all the power the PSU could send to it, then the computer would run a lot hotter if a PSU with double the capacity was installed.