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Measuring INPUT VA

H

HooFS

n00b
Joined
Apr 5, 2004
Messages
5
I just had a few questions I couldn't seem to get google to answer today:

1- Will a PSU use every bit of it's VA rating? So if a PC has a PSU that req. 460 VA, will often use this much, if the system doesn't house many HDDs or such would it use less? I know this to be true of watts and such.

2- Is the V*I really always accurate to determine VAs? I say this because I have 5 sony systems of recent design (all past 2.4 P4's) with an average of 421 VAs, then I open up my first Dell D266 (a P2-266) and it's rated at 7a! that's 805 VA! That doesn't seem right, anyone help?
 
a psu will not use more wattage than neccessary. A 250W psu and a 500W psu will use the same wattage on the same system assuming that efficiency is equal, and that the 250W supply is sufficient to power the system. Of course the figures will vary slightly in real-world scenarios due to varying efficiencies of psu's.

the volts * current formula is both accurate and misinformative. volts * real_current_draw is a very accurate way to determine the wattage being used by a psu. However, it is also misinformative because you cannot multiply the maximum current rating by the voltage to get what the psu can handle -- it doesn't work like that...
 
Thank you, but this would be regarding INPUT, that is VA required for the PSU before it transforms anything into watts. This is actually regarding a UPS purchase. What I'm doing right now is taking the Volts (115) and multiplying by the Amps to get the VA value. UPS systems are rated by this. However I'm wondering if they will always use this calculated value. And in some cases I think this is misleading because there is no way a slot 1 P2-266 NEEDS 805 VA, but I'm thinking even if so, the PSU might just need that much. I'm no electrician obviously, will it absolutely require this much? I don't need exact values, just safe estimates.
 
VA is a strange rating. You need to intimately know the topology of a power circuit to determine its VA rating.

"Volts" times "Amps" equals "Watts," right? To oversimplify, yes. For most users, that's fine. The power meter in homes measures by units of kilowatts per hour, not kVA/h. That's because VA takes into consideration this strange concept of "imaginary power." In purely resistive circuits (circuits with only resistors or resistor-like components), VA equals watts. But once you include capacitors, transistors, and inductors (i.e. 99% of circuits), you get into this thing called impedence, which takes into account resistance, inductance, and capacitance. It's very complicated for this discussion, but I'll leave you with one final concept.

Imagine a right triangle. One leg of the triangle, the bottom leg, is real power (watts), one leg of the triangle is imaginary power (VARs is the units). The hypotenuse is total power (VA). That's why the VA rating always seems bigger than you would expect--it's the hypotenuse, where wattage is a leg of the right triangle. The arctangent of the angle between real power and total power is the power factor. Thus, if you know the power factor and the total power (VA), then you can find out the real power draw.
 
Now for my more practical post...

Take the peak power output rating shown on your PC power supply, then multiply it by 1.5 for a very generous margin of power.
 
I really do appreciate the insight, but I need to know more about just the input, will this be constant? All the info I have on some of these PSUs is the input (100-120V ~/7A) or (200-240~/3.5A) 50-60Hz. I have no peak rating. If this is a much larger issue than I think it is or if this is insufficient data, just say so please =)
 
http://www.silentpcreview.com/article177-page4.html (see the yellow chart)

hope that clears it up a bit

to actually figure the load on the power supply
first you would determine the theoretical maximum of all the components
http://takaman.jp/D/index.html?english
then guess at the difference between theory and actual use
for instance a HDD will have a spinup amp draw 4 times as high as its runtime value
or a CPU under load vs one at idle

the AC draw will vary to meet the DC demand plus the inefficiency of the supply which will vary depending on the load
for a specific operating temperature, another variable

its alot easier to simply build a system and then measure the average AC draw with an watt meter, the" kill a watt" being popular and readily available
(that link is just the first return not a source recommendation)
 
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