Power factor, UPS, computer

I see. Apart from being nice to the electricity supply, what advantage is there that makes Corsair do it? It doesn't improve the output voltage for the computer presumably? And it'll cost more to make.

Reply to
Uncle Peter
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When it comes down to it, it means they can sell the thing into regulatory environments that require SMPSUs be nice to the electrical supply!

On a practical level, there are so many SMPSUs in use now, that if they don't have PFC it becomes a real problem for the supply.

No

Indeed

Reply to
John Rumm

But they're sold to home users for gaming machines. Business computers don't have big powerful supplies like that.

Reply to
Uncle Peter

You can do fancy stuff on simply rectified mains, without smoothing it to DC first, but then all that does is shift the smoothing problem to the other side of the inverter so to speak. Its very hard to match 'constant draw' from a AC to match 'constant supply' for DC..

The whole grid was designed to run resistors and electric motors, never DC devices.

What you really need is a fridge sized inductor/capacitor arrangement to filter out current peaks from being propagated upstream of a typical domestic or industrial situation.

Those already exist for really large industrial plant as power factor correctors, but they could be extended.

Reply to
The Natural Philosopher

If they can make a variable ratio voltage convertor like John Rumm was suggesting, then that should sort the problem completely.

Reply to
Uncle Peter

Zero power factor mean zero power. Exactly so.

A current is flowing in to an energy store Eg capacitor in the first quarter cycles and then flows back out in the next quarter.

So the energy is stored and then released.

No power is consumed because there is no resistance.

Think of the analogy of pumping water uphill. If there were no gravity, no energy would be required though water was moving (Neglecting friction)

The USAian way of looking at this business is quite confusing. It takes something that is quite simple & makes it look complicated.

There is a better explanation here that shows both ways of looking at the business

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The second (vector) method is looking at what is happening in an instant of time. The problem can then all be solved by simple trigonometry. Sines, cosines and pythagorus. Are even by with a scale diagram and measuring with a ruler.

Reply to
harryagain

Pure Capacitor.

Think about banging a pure capacitor straight onto say a normal12V battery. In the first instant of time it is a dead short, infinite current flows (there being no resistance in our theoretical capacitor) but voltage is zero. As time passes the capacitor charges, current becomes (almost) zero and voltage returns to 12V.

So the current is leading the voltage.

This reverse and happens many times a second on our AC power supply.

Pure Inductor Think about banging a pure inductor on a 12V battery. In the first instant of time the voltage is 12 volts but current is zero. As the magnetic field builds the current builds limited by the hysterisis of ther magnetic field As time passes the current becomes infinity (there being no resistance to limit it in our theoretical inductor)

So the current is lagging the voltage.

Reply to
harryagain

I mean't it wouldn't be acceptable if a signicant load was to be that bad. It would mean that supply cables, switchgear and transformers would all have to be that much bigger (and hence more costly)

Reply to
harryagain

Yes I know all that. I was just saying I don't agree with CIVIL. It stops people understanding what's actually going on.

Reply to
Uncle Peter

Electric motors and transformers work most efficiently on a sine wave.

Reply to
harryagain

The link showed something not far off a sine wave.

Reply to
Uncle Peter

Class A (Computer over 600W) applies to my PC power supplies (750 and 850W). Which gives this table of harmonics: Harmonic order Maximum permissible harmonic current Odd:

3 2.3 5 1.4 7 0.77 9 0.40 11 0.33 13 0.21 15?n?39 0.158/n Even: 2 1.08 4 0.43 6 0.30 8?n?40 0.238/n

So er..... what's that in a PF reading?

These are not related to power factor. These are describing the shape of the wave by Fourier anylaysis,

It describes the shape of any wave by superimposed even and odd harmonics. (Sine waves)

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So in theory any wave form can be described by superimposing harmonics sine waves on top of the basic frequency. This enables subsidiary calculations in designing electrical equipment.

The above is describing the maximum degree of distortion permitted. Odd harmonics =50 x odd numbers Hz (makes symetrical wave forms Even harmonics = 50 x even numbers Hz (makes assymetrical wave forms)

In practice it would take an infinite number of harmonics to describe say a square wave.

Reply to
harryagain

When I said power factor, I meant the reading that my energy meter is gi= ving me, which we were assuming was not actually power factor, but disto= rtion.

-- =

Lysdexia: a peech imspediment we live to learn with...

Reply to
Uncle Peter

What sort of PC needs a near-kilowatt PSU? The thirstiest one I've encountered only took 150-200 W (mains input) and that was a brute of a thing. I can't help thinking that there's some specmanship going on here, but that's another story.

The standard gives maximum values of input current harmonics for EMC purposes. The PF for a particular device will depend on its (rms-summed) actual harmonic currents and its actual mean power consumption. Measuring the individual fundamental and harmonic current components would be a rather pointless exercise. Much easier to measure the PF directly using a wattmeter that can measure both apparent and true mean power.

The plug-in power & energy monitor that Maplin (and others) sell is quite good down to low PFs (0.3 or less, IME). In contrast those whole-house 'energy monitor' devices that just use a small current transformer that clips over a meter tail are invariably useless at low PF. They sample current alone, so have insufficient data to indicate true power.

Reply to
Andy Wade

All of this conventional AC theory power factor wibbling is all well and good, but not actually relevant to the issues of poor power factor exhibited by SMPSUs anyway.

Reply to
John Rumm

Harmonic currents convey no average power to a load but do contribute to apparent power and hence to I^2*R losses in the supply - so they are very much related to power factor.

Actually they aren't doing that since only the amplitude of the harmonics is given. You'd need the phase information too to synthesise (not analyse, BTW) a particular waveshape.

Reply to
Andy Wade

A Radeon R9 290 (or a 7970) uses 375W. Add an i7 extreme using 130W, and you're up to 505W already. Add the rest of the computer in and you're over 600W, even before you add a second graphics card.

It's a plug-in Maplin type. What PF would it show at the EU limit?

Reply to
Uncle Peter

Well, we'll have to make some assumptions. Let's assume that the input power is only a tad over the 600 W figure that brings it into Class A. At 230 V that gives us an input current of just over 2.6 A, ignoring harmonics. Next I'll consider only odd harmonics up to the 13th: the contribution of higher odd harmonics is likely to be negligible and we'll be using full-wave rectification so the level of even harmonics should also be negligible.

Now take the max. permitted odd harmonic currents in EN 61000-3-2: 2.3,

1.4, 0.77 A etc. Add up their squares to get the total mean-squared harmonic current - I make it ~8.2 A^2. Next add the square of the fundamental current and then take the square root to get the total RMS input current: 3.9 A (900 VA input).

So the power factor is 2.6/3.9, i.e. just under 0.7. That's fairly poor, but this is a rather artificial worst-case example. For higher input powers the PF will get better, since the harmonic levels allowed are constant. In practice I'd expect the distortion to be much lower and the PF to be at least in the 0.8 to 0.9 range.

Reply to
Andy Wade

Here's an interesting article that goes into a bit more detail:

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Reply to
Andy Wade

I get PF 0.71, but this was an old stereo at full blast (with a transformer PSU in it), two LCD monitors fairly old, and my CIT 850W PSU (which quotes only 672 watts on the 12V line, which is where the majority of computer current is drawn) heavily loaded (as in as heavy as I can load it before the output voltage makes the graphics card crash). I'm drawing about 500W from it I think.

Reply to
Uncle Peter

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