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I'm trying to calibrate the IC ATM90E26. This is an energy metering IC, it measures current through a shunt resistor and voltage through a voltage divider circuit. The IC will be part of a 220V AC circuit, with a maximum current of 0.7A.

The thing is that in order to calibrate this IC you need, according to datasheet, to:

  • Calibrate gain at unity power factor
  • Calibrate phase angle compensation at 0.5 inductive power factor.

And here is where I get a bit lost. To achieve a power factor of 1 the load in the calibration circuit has to be purely resistive, this I can get with a light bulb (non-LED). But to get a power factor equal to 0.5 I'm no really sure how.

So here comes the question, how can I make a test / calibration circuit that generates a power factor equal to 0.5? I mean with what type of every day use device can I achieve this? will a common fan(220V AC) do the trick?

My current set-up for measurement of electronic "things" consist of: 2-channel oscilloscope, multimeter (a regular one, and a amp-clamp one), and a variable DC power supply (30V - 5A max). How can I measure the apparent, real and reactive power to calculate the power factor?

I've been reading a lot on this subject on the internet but I get a bit confused on how to make the circuit and how to do a proper measurement of it.

Thank you for your time :D

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  • \$\begingroup\$ Do you know how to make an inductive load that has a PF of 0.5? \$\endgroup\$
    – Andy aka
    Mar 9 at 18:30
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But to get a power factor equal to 0.5 I'm no really sure how.

Start by examining the power triangle: -

enter image description here

Then define how much current you want to flow. As an example, I've chosen 0.5 amps. That means that the impedance required (the hypotenuse on the power triangle) is 220 volts divided by 0.5 amps = 440 Ω.

That impedance is made up from a resistor in series with an inductor and together they form the right angle in the power triangle above.

Given that \$\Phi\$ is 60° for a power factor of 0.5, reactance (\$X\$) can be found by re-arranging this: -

$$ \sin(60°) = \dfrac{X}{Z}\text{ or } X = Z\cdot\sin(30°) = 381.05\text{ Ω}$$

And, using the cosine relationship for sides and angles, R = 220 Ω.

So, if your mains AC frequency is 50 Hz, the inductance required to be in series with the 381.05 Ω resistor is: -

$$L = \dfrac{381.05}{2\cdot\pi\cdot 50} = 1.2129 \text{ henries}$$

So here comes the question, how can I make a test / calibration circuit that generates a power factor equal to 0.5? I mean with what type of every day use device can I achieve this? will a common fan(220V AC) do the trick?

Well, to most EEs making an inductor can be done from commonplace stuff but if you mean making it from stuff that anyone has common access to my advice is forget it. Do the job properly if you want an accurate calibration.

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  • \$\begingroup\$ Hello Andy, thank you for answering. Can I ask you something to confirm that I understood you? You say that, according to calculations, I need a 381.05 Ω resistor in series with a 0.7 Henries inductor to get a Power Factor of 0.5 in the load. And a current of 0.5Amps at 220V AC. This means that I need a lot of new equipment to achieve this, but if I get a common inductor with known inductance (wich I'm not really sure how) . Can I select a resistor that, knowing voltage and current RMS, will make the circuit have a 0.5 power factor? \$\endgroup\$ Mar 10 at 13:20
  • \$\begingroup\$ @IvánGerber I think I made an error in one of my calcs in my answer so bear with me as I check things.... \$\endgroup\$
    – Andy aka
    Mar 10 at 13:30
  • \$\begingroup\$ OK I fixed my calcs - you would need a 1.2129 henry inductor and a 220 ohm resistor to draw a current of 0.5 amps at 50 Hz from a 220 volt supply. Sorry for the mix-up. \$\endgroup\$
    – Andy aka
    Mar 10 at 13:35
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    \$\begingroup\$ @IvánGerber if you chose an inductor that is lower in inductance, the resistor will also be proportionately lower to achieve the same power factor but, then, the problem will be that the overall current draw may be very much higher than 0.5 amps and this could create smoke/fire big time. In other words, if you choose an inductor that is only 121.29 mH (as opposed to 1.2129 H), the resistor would need to be 22 ohms and you would be taking 5 amps from a 220 volt supply. \$\endgroup\$
    – Andy aka
    Mar 10 at 13:36
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    \$\begingroup\$ @IvánGerber if you want to properly calibrate your chip you need to provide a precision load. I can't see any way around this. If you can buy a 1 henry inductor, then put it in series with a 180 mH inductor then add another series inductor of 33 mH. Total inductance is 1.213 henries. Do the same for the resistor if you can't get this value in the wattage you want. You will also need a resistor that can handle a power of 55 watts. \$\endgroup\$
    – Andy aka
    Mar 10 at 14:42

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