Power Factor Measurement

[CarnivalBen]

Is there a way to accurately measure the power factor of an 240v LED lamp? The only test equipment I have is a cheap-ish multimeter, and a Rigol DS1054Z, with standard probes and a current probe...

[NiHaoMike]

Just get a Kill-a-Watt. Cheap and accurate enough for most hobby uses.

[coppice]

Just get a Kill-a-Watt. Cheap and accurate enough for most hobby uses.

A lot of people use the term Kill-a-Watt like they would use hoover to describe a generic vacuum cleaner. The actual Kill-a-Watt products give pretty accurate results. Some of their competitors seem to just make up the readings as they go along.

[IconicPCB]

IF you have a ( isolated ) current probe for the scope and a high voltage probe then measure phase difference between current and voltage.

power factor  = cos ( phase difference)

[coppice]

IF you have a ( isolated ) current probe for the scope and a high voltage probe then measure phase difference between current and voltage.

power factor  = cos ( phase difference)

That works well if the current waveform is a nice sine wave, but with load distortion it quickly becomes very hard to judge power factor on an oscilloscope screen.

If you are trying to be really accurate about power factor you should note that some CTs and other current sensors you might to sense the current signal can introduce a few degrees of delay.

[Seekonk]

Just happened to have a screen save on my desktop, a 9W E27  LED Just imagine hoe the utility company will feel when their load is filled with these.   

[coppice]

Just happened to have a screen save on my desktop, a 9W E27  LED Just imagine hoe the utility company will feel when their load is filled with these.

Hard to say really. The power factor is fairly good, which is what the utilities tend to emphasise. They complain less about horrible distortion. Many active power factor correctors result is far worse distortion than when you disable them.

[CarnivalBen]

Thanks for your replies...

So a kill-a-watt is an easy way to get the PF... but its kinda cheating and I'm not gonna learn much from doing it that way! As well as actually needing to know the PF, I also want to learn how to use my new scope, and this seems like an interesting thing to do... so maybe I'll try to measure it with a scope, and then use a kill-a-watt to see how accurate I was 

IF you have a ( isolated ) current probe for the scope and a high voltage probe then measure phase difference between current and voltage.

power factor  = cos ( phase difference)

When you say isolated current probe... do you mean a current clamp? I have an active current clamp probe, although I'm a bit concerned that the resolution will be too low as it can measure up to 600A, and a little 5W LED bulb isn't going to draw much current...

Can I make sure I've got the maths right: If I measure the time delay between the voltage and current in seconds (t), frequency = 50Hz (f), then find the phase angle (pa): 2 * pi * f * t. Then PF = cos (pa). Is that right?

If you are trying to be really accurate about power factor you should note that some CTs and other current sensors you might to sense the current signal can introduce a few degrees of delay.

If I compare the voltage and current of a linear load, like an incandesant bulb, I'm guessing that if my probe does introduce a few degrees of delay then I'd be able to see a phase difference? Then I can use correct my calculations by this offset?

[coppice]

If you are trying to be really accurate about power factor you should note that some CTs and other current sensors you might to sense the current signal can introduce a few degrees of delay.

If I compare the voltage and current of a linear load, like an incandesant bulb, I'm guessing that if my probe does introduce a few degrees of delay then I'd be able to see a phase difference? Then I can use correct my calculations by this offset?

You conflated two things there. An incandescent light bulb is far from linear. The filament varies so much in temperature during each half cycle of the mains that you will see 15% to 20% THD in the current waveform. It will, however, not shift the current significantly in phase, so you should be able to check for phase alignment between the voltage and current waveforms.

[garre]

There seems to be some confusion about Total Harmonic Distortion (THD) and Power Factor (PF).  An incandescent light bulb is primarily a resistance load and will have almost no effect on the PF.  The THD comes from some resistance changes as the filament heats and cools in the bulb.  What causes the PF are the inductive and capacitance loads on the circuit, not resistance loads.  Since you clip-on ammeter is an inductively couple meter it will induce a PF in the measurement on the scope, so you will need to compensate by determining the PF introduced with measuring a purely resistance load vs the power supply within the LED bulb that is not a purely resitanc load.  The difference will be the PF of the LED bulb. 

The PF is most important to a power utility to optimism there power transfer and billing. On a power line most of the loads are inductive and thus the current will lag the voltage.  Since the limitation on capacity of a power line is the current capacity of the wire,  you don't want get to much current that is out of phase with the the voltage (VARS) or you will have current that will limit the capacity of the wire and not give you billable power capacity.  Power companies add a capacitance load to keep their systems just slightly inductive with a PF of a about (.9).  Since a power factor of (one) would be at the resonance of the system the system would become somewhat unstable.  Since the system is inherently an inductive load, the power company will keep the power factor just on the inductive side with a slightly inductive PF, but not to far as to reduce the power billing capacity of the system.   

[coppice]

There seems to be some confusion about Total Harmonic Distortion (THD) and Power Factor (PF).

Who was confused? you?

The PF is most important to a power utility to optimism there power transfer and billing.

The biggest effect of a poor power utility is 0.5PF means double the current for the same power. Double the current means double the conductor size. Double the conductor size means double the copper costs for long runs of cable. This does tie in with THD. THD in the current waveform alone has no effect on consumed or measured power. Power is only consumed where the voltage and current waveforms correlate. A power factor worse than the cables were designed for results in greater ohmic drops, which convert current into modulation of the supply voltage. Now if the current waveform is highly distorted the voltage waveform starts to gain some harmonic distortion too. This correlates, so now the THD does have an impact on consumed power. This is the utility's fault for supplying a dirty voltage waveform, so should they be paying for the harmonic energy, or should the consumer? The debate over this is why you need fundamental power measurement features being added to some utility meters.

[Ian.M]

You *CAN* *NOT* measure the power factor safely or even view the current and voltage waveforms with just a basic scope and a couple of standard probes. 
See: http://www.eevblog.com/2012/05/18/eevblog-279-how-not-to-blow-up-your-oscilloscope/

Your Rigol has the capability to subtract pairs of channels so you could do a differential measurement and avoid blowing up your scope, but one mistake and your scope and possibly you as well are toast.

However if you add a 240V isolating transformer, it becomes much safer and you only have to worry about making sure the probe monitoring the voltage is on x10 as it doesn't have enough voltage rating on x1.   That also frees up some functionality as you no longer need to use pairs of probes differentially. However grounding the low side of the bulb via a  probe ground lead removes the operator safety provided by the isolation transformer, so *BE* *CAREFUL*.      If you add a 1R power resistor between the  low side of the transformer output and the low side of the bulb, where the voltage probe ground lead is connected, you can measure the voltage across the resistor on the second channel to get the current.  The DS1054Z can do waveform maths including multiplication and integration, so you can get it to compute I*V and integrate that over a cycle to get the real power. Divide by the VA, (RMS V * RMS A)and you've got your power factor.

However if you get it wrong its *NOT* going to be a happy experience so I strongly recommend using a low voltage transformer (not a PSU or switching supply, as you want line frequency AC) and low voltage bulbs (add a bridge rectifier if they are DC only LEDs) until you are comfortable with the measurement technique and fully understand it.

[joeqsmith]

I was attempting to make some AC measurements with a scope as well using a transformer and a current sensor.   It worked alright but there is room for improvements.   We use 120 here but I don't see a problem with doing this at 220.

Because the AC line frequency is so low and the scope is so fast, my plan is to try and improve the measurements by oversampling the two signals.  I plan to model the non-linearity of the scope's ADC and use this to compensate for it.  I posted a link where I show some of the results.    There are a few more tests I need to run and then the plan is to add these changes into the software.   

This is all just for fun so I doubt I will do much more with it.  The biggest problem I see is that our incoming AC quality is pretty poor.   Looking at harmonic distortion on a line that is already distorted and trying to back out the contribution of the device seems like it could be a problem.    One way to solve this may be be to use a very clean power source that would power the load.  For small light bulbs, this would not be much of a problem.   Guess it depends how far down the rabbit hole you want to travel...

https://www.eevblog.com/forum/testgear/making-accurate-low-frequency-measurements-with-your-scope/

I'll make a new video once I have something working.   This was my first crack at it:

[coppice]

IF you have a ( isolated ) current probe for the scope and a high voltage probe then measure phase difference between current and voltage.

power factor  = cos ( phase difference)

That works well if the current waveform is a nice sine wave, but with load distortion it quickly becomes very hard to judge power factor on an oscilloscope screen.

If you are trying to be really accurate about power factor you should note that some CTs and other current sensors you might to sense the current signal can introduce a few degrees of delay.

Use math function. Grab 3 measures: RMS(I), RMS(V), AVG(I*V), all aligned in N integer cycles with either voltage or current waveform.

What do you expect to get from those 3 values for distorted waveforms?

[coppice]

What do you expect to get from those 3 values for distorted waveforms?

According to definition, PF=W/VA, so I've got real power (integral(i*vdt)/t) and apparent power (rms(i)*rms(v)), then do division.

That definition of PF is for sine waves. It gets very iffy for distorted waves, and people argue about what PF even means. Also, scopes are pretty noisy rms(i) and rms(v) integrate the noise, but avg(v*i) doesn't, because the noise doesn't correlate. That can give you a pretty poor result, even for sine waves, with a typical scope.