Confused by "Power Factor"

[google]

I recently bought one of those plug-in electricity consumption meters (Brennenstuhl PM230) with a view to measuring how much it costs to run various devices.

Yesterday I measured the household email/file server, a mini-itx EPIA5000 with a single hard drive, which uses a laptop-style external SMPS "brick". The reported consumption is about 1W (!). This is obviously completely wrong. Switching to the amps setting, it reports that the device is drawing 0.12A with a power factor of 0.04 (ie. 240V x 0.12A x 0.04PF = 1.15W). Ignoring the power factor would give about

29W, which seems much more in the right ball park.

My objective is to get a feel for how much electricity devices are actually consuming. Should I ignore the power factor and just multiply volts by amps? Should I also then multiply by about 0.77 to arrive at RMS power?

[Ian Stirling]

These decices IME, are completely unsuitable for measuring computers, or anything else with advanced PSUs. I've got one from the same maker, and comparing with integrating the power used on a scope, it gets some PSUs 100% wrong. (reads 180W vs 90W).

The problem is that it only measures the mains several times a cycle, and extrapolates the current waveform between. This it gets right, for loads like motors, fans, and many simple power supplies. But, for more advanced stuff, it really requires sampling ten times faster than these devices do.

Power factor is unrelated. Power factor is now synchronised to the voltage the current draw is. With AC lines, you can have 10A of 240V current flowing, but if it's out of phase, you can get negligable power actually usable.

[Dave Fawthrop]

|I recently bought one of those plug-in electricity consumption meters |(Brennenstuhl PM230) with a view to measuring how much it costs to run |various devices. | |Yesterday I measured the household email/file server, a mini-itx |EPIA5000 with a single hard drive, which uses a laptop-style external |SMPS "brick". The reported consumption is about 1W (!). This is |obviously completely wrong. Switching to the amps setting, it reports |that the device is drawing 0.12A with a power factor of 0.04 (ie. 240V |x 0.12A x 0.04PF = 1.15W). Ignoring the power factor would give about |29W, which seems much more in the right ball park.

Without doing the maths this looks OK, and quite possible, to me.

|My objective is to get a feel for how much electricity devices are |actually consuming. Should I ignore the power factor and just multiply |volts by amps? Should I also then multiply by about 0.77 to arrive at |RMS power?

Concentrate on the *watts* consumed by AC electrical equipment that is what you *pay* for. Volts*amps and power factor are best left to the professionals.

This link explains everything.

[google]

So are you saying that I can't even trust the current draw (Amps) that it's reporting? That there's no way with this device to figure out how many watts (the ones I'm paying for) are being consumed?

[Ian Stirling]

But, these meters will not accurately (or even within throwing distance of it) measure many loads unfortunately.

[Ian Stirling]

Exactly. All the figures are a rough guess, on many loads, making bad assumptions about the shape of the current draw. There is no way, without knowing something about the design of the SMPS that you can even guess if it's accurate. I don't know of any of these that sample at a decent rate, ones I bought from LiDl exhibit exactly the same behaviour.

[Brian Sharrock]

Once upon a time; DC (Direct current) was generated using transformers, Rectifiers inductors and big capacitors to sequentially transform the incoming mains AC to a different more appropriate voltage then only accept the positive going parts of the AC waveform then 'smooth' out the ripples by opposing some ripples and 'shorting' some to earth. For these old style rectifiers power supplies a power factor measurement was appropriate. Nowadays; manufacturers utilise 'Switched Mode Power Supplies'; essentially the cost of semiconductor has fallen so much that a more satisfactory and more efficient way of acquiring DC from an AC source is available. The 'power factor' meter simply hasn't been designed to cater for these devices ... so " ..., these meters will not accurately (or even within throwing distance of it) measure many loads unfortunately"

BTW, while you were pondering the readings and wondering about the discrepancy between 1W (observed) and 30W (intuitive) ... did you have a cup of tea .... from a 3Kw kettle ?

[google]

OK, thanks. I'll chalk this up as another example of live and learn.

[Dave Fawthrop]

|Dave Fawthrop wrote: |> On 23 Mar 2006 01:59:27 -0800, [e-mail] wrote: |> |> |I recently bought one of those plug-in electricity consumption meters |> |(Brennenstuhl PM230) with a view to measuring how much it costs to run |> |various devices. | |> Concentrate on the *watts* consumed by AC electrical equipment that is what |> you *pay* for. Volts*amps and power factor are best left to the |> professionals. | |But, these meters will not accurately (or even within throwing distance |of it) measure many loads unfortunately.

Who cares! OP is talking about roughly *1* watt so the cost to run it is

2/3 of knack all.

[Ian Stirling]

The fact that the power factor for the power supply will probably be about unity, and it's reporting a moderately significant current draw means that the 1W figure is likely to be completely, utterly inaccurate.

10, 20, or 30W is more liklely.

These things measure the current 4 or 5 times a cycle, then extrapolate out, making assumptions. They do not directly measure power use. They calculate it from measured currents at several points, then extrapolate out. Because of the design of many recent SMPS, this can be very, very inaccurate.

[Ian Stirling]

Oh if it were that simple...

Many SMPSs, like old linear power supplies draw current only near the peak of the mains voltage, in a wave that looks like a slightly left-skewed sine, with the peak a hair to the left of the peak voltage.

For these, the meters that I've looked at work just fine. More modern and larger PSUs may do it slightly differently. The PC power supply that I mentioned earlier in the thread produced a current waveform that looked more like (over the top 1/10th or so of voltage peak)

|\ | \__ __| \___

Which understandably would utterly screw up a meter expecting something like: __ / \ | \ __/ \___

Especially if it's trying to extrapolate the 'real' shape from one or two points.

This is for several reasons - changes in legislation with regards to some sorts of noise that you can inject back into the mains, as well as advances in technology.

[Ian Stirling]

Oops - I forgot to give you my solution, that I sometimes use. Take an airtight largish box, or cupboard, a 100W bulb.

Put the bulb in the box, and measure the temperature rise till it plateus. Do the same with the thing to test. The result is approximately simply the ratio of the rises, multiplied by

100W.

A small computer fan, or something can make this more accurate.

[John Rumm]

What the others said re the ability of these things to cope with a SMPS.

Where are you getting your current reading from though?

IME you can get reasonably accurate current measurements using a true RMS clamp meter round a few turns of the live conductor in the cable feeding the SMPS (I have a short extension lead made with a section of out sheath removed just for the purpose). You can then usually assume they will tend to be a somewhat capacitive load. So multiply your observed VA by about 0.6 or 0.7 and you should be in the right ballpark.

[john]

John Rumm wrote: ms much more in the right ball park.

I've been hit by expensive bills recently so I installed a DIY recording consumer current meter made up of a commercial 100/400A clamp meter enhanced with 33 turns of insulated wire wrapped round one leg. This coil is connected to a Maplin digital current meter set on mA. This has an RS232 interface so I sample the household curent every 30s on an old laptop. It's not a very linear arrangement but does the job after calibration with an independent digital clamp meter. Now I can work out where all the energy bill are coming from over a 24 hour period. The PF of all the individual items in the house can be measured with a plug in power meter as described in the OP. I haven't measured the PC's PF yet.

john2

[<me9]

Best (and probably cheaper) way is to use one of these, my server uses about

55WH.

[Mike Barnes]

In case I ever need to do anything like this, how accurate would it be to measure the current using a moving-coil meter? With the obvious safety precautions, of course.

[manatbandq]

There may also be power factor correction built into the power supply.

MBQ

[Ian Stirling]

Very, very accurate - assuming it's a true-RMS meter. If it's not, then essentially not. The earlier suggestion of a reconditioned electricity meter is a good one.

[Derek ^]

OTOH if it's a moving coil (milli) -ammeter, it won't measure AC at all.

Reasonable for low budget experiments.

Or one of these:

imagine cheap ones are a bit thin on the ground.

DG

[Derek ^]

Forgive me and all that if I've got it wrong, but wasn't that the indication he got on a cheap wattmeter presumed to be totally erronius because of the way the meter (sampling at a low frequency) and the power supply (drawing pulses of current at a different frequency) happen to work.

DG