Amperage question

[blueonblack]

Ok, I was looking at the new Cooler Master Silent Pro M2 1000-watt PSU I got. It has a single 12-volt rail that is rated at 80 amps. (Wow!) My electrical knowledge is not what it shoudl be, so my question is, how can it supply 80 amps when it's drawing from a 30-amp circuit at the wall?

[OvRiDe]

It can provide up to 80 AMPs at 12V where as the wall socket is providing up to 30 AMPs at 110 Volts.

Also there is another aspect to take into consideration is that the PSU is producing DC and the wall socket is producing AC.. where as the formulas are very similar there is an added factor involved with AC called PowerFactor.

I believe the average power factor is around .8

So for DC the formula would be..

P(W) = V(V) × I(A)
Watts are equal to volts times amps:
watt = volt × amp

so.. 12V X 80A = 960W

When it comes to AC the formula is..

P(W) = PF × I(A) × V(V)
Watts are equal to power factor times amps times volts:
watt = PF × amp × volt

so.. .8 X 30A X 110V = 2640W

However most household circuits use 15A breakers which would be...

.8 X 15A x 110V = 1320W .. which is why most household space heaters/hairdryers etc etc are usually in the 1500W or less range.

Hope this helps make sense of it.. There is more to it of course .. I just tried to hit the high points.

[blueonblack]

Hope this helps make sense of it..

Nooooooot really, though I appreciate the effort. I was hoping to avoid wattage in the whole equation, hoping that the current discrepancy could be explained without it. If not, so be it. From what you said though it sounds like the difference in voltage can be used to generate the extra amperage?

Since it has nearly ten times the voltage to work with (110 AC vs 12 DC), the PSU can use this extra voltage to provide more DC output current (amperage) than it draws in AC current?

Am I anywhere close there?

[OvRiDe]

Kinda sorta.. you're in the right ball park. I don't really have a better way to explain it, but that is the gist of it.

[blueonblack]

Ball park will have to do. Thanks!

[Konrad]

It can provide up to 80 AMPs at 12V where as the wall socket is providing up to 30 AMPs at 110 Volts.

Sorry, friend, a correction.

Standard wall receptacles are fitted with 14-gauge wiring, safely rated for 110-120VAC @ 15A. A maximum safe power of 1800W (technically, actually <1800VA, as effective power is often reduced by line-phase impedance). Exceeding the safety factors will trip breakers and blow fuses, few consumer appliances will draw more than ~12A or ~1440W.

Heavy duty wall receptacles, the sort meant to service microwave ovens and such, are fitted with 12-gauge wiring, safely rated for 110-120VAC @ 20A. Capable of safely sustaining ~2400W. I've never heard of a consumer PC PSU which requires 20A.

But yes, 12V @ 80A = 960W

[Konrad]

Incidentally ...

It's wise to calculate the loads of each voltage rail (12V, 5V, 3.3V, etc) being used by each component in your system added together. Then compare these numbers vs the specs for your PSU, which are rarely ideal. You can gain a bit of slack when considering that not every hardware device in your system will simultaneously be operating at 100% power draw, at least not outside bursts which aren't likely representative of typical usage/activity.

You can always refer to the UL part number database (it's gotta be on a little UL sticker on every decent PSU) to see which manufacturer really built the thing before it was licensed/rebranded, and to see the actual detailed specs they engineered into it.

[OvRiDe]

It's all good, but no need for a correction.. I was just using his original question to try and explain the math behind it.


... from a 30-amp circuit at the wall?

Where as I didn't go into the methodology behind why house hold circuits are 15A instead of 30A. I did mention it in my original post.



However most household circuits use 15A breakers which would be...

.8 X 15A x 110V = 1320W .. which is why most household space heaters/hairdryers etc etc are usually in the 1500W or less range.

Hope this helps make sense of it.. There is more to it of course .. I just tried to hit the high points.

I also just used 110V in my example because its at the low end of the range, which kind of sorta means that should be the least amount of wattage you should expect to produce, mathematically speaking that is. Typically I have seen devices rated at 110VAC, 115VAC, and 120VAC, which I have always wondered in way, why they don't all use the same voltage rating. I do understand that house voltage can vary from house to house (I have measured as high as 126VAC at the wall socket at my house), but I think that devices should be rated at the same voltage. Just my 2 cents.

[blueonblack]

And I was obviously the source of the confusion, with my original incorrect number of 30 amps. For some reason I had it in my head that the room was on a 30-amp circuit when obviously it's not.

Apologies to all concerned.

[TheMainMan]

The tl;dr summary:

12V x 80A = 960W
.8 x 110V x 15A = 1320W
1320W > 960W = :up:

[blueonblack]

12V x 80A = 960W
.8 x 110V x 15A = 1320W
1320W > 960W = :up:

Not sure what it says about me but this did more to explain it than anything.

Too many words, OvRiDe! :D

[OvRiDe]

LOL! Whatever works is a good thing! :D

[Konrad]

Haha, sorry, was just feeling antagonistically pedantic that day. Math errors cannot be tolerated!

Household electrics rated as 110V, 115V, 120V, 125V, etc, are basically all identical aside from engineering/legal/regulatory preferences. The AC power could be calculated with different weights on RMS or duty-load or impedance factors, and is sometimes derated a little by product manufacturers (or electric utilities) for whatever reason.

The fact is that all but the cheapest PSUs can accept a fairly wide range of deviations from the household norm, and unless they're servicing loads near the maximum 1800W limits, they won't use the maximum theoretical power being offered because their entire purpose is to just provide regulated power to the system downline. An higher-rated PSU won't draw more power (or cost more to operate) when operating at lower-rated PSU loads (barring minor inefficiency losses caused by wattage-optimized engineering tradeoffs).