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Power Factor Correction for a small 60W AC Induction motor

  1. Apr 26, 2016 #1

    I have a light AC fan, that I wish to correct its power factor.
    According to my measurements, it has around 0.7 LAG power factor.

    Here are my digital measurements:

    Vsupply = 204VAC
    I(load) = 0.237Amps
    phi = 43 , pf = 0.7 LAG

    My target power factor (pf) is to be 0.85 , and to neutralize (reasonably) the lagging power factor, I inject it with leading VAR supplied by the capacitor bank.

    The supply here, have a maximum value of 230VAC , so I got capacitor with Vrated = 450VAC > Sqrt(2)*230

    For the capacitance this is the formulas/algorithm I've used to get C:

    Q1 = Vrms * Irms * pf1 * tan(43)
    Q2 = Vrms * Irms * pf1 * tan(32) //since cos^-1(32) = ~0.85
    Qcap = Q1 - Q2 = Vrms ^ 2 / Xc
    Xc = Vrms ^2 / Qcap = 1 / 2 * pi * f * C //hence C is calculated

    *Note*: I connected two 100K 2W in series across capacitor terminals, as bleeding resistor for electrical safety

    1) I branched a 6uF 450VAC in parallel with the load, I noticed that the load drew more current of 0.4Amps where originally it draw 0.2Amps // This is referred to over-compensation
    2) I decreased the 6uF to 4uF , still were an over-compensation, but lighter than before, and the motor drew 0.3Amps
    3) I kept on decreasing the capacitance until 1.6uF there I noticed that the motor got back drawing its original amount of current
    4) Still decreasing, until 0.5uF what I noticed is that the current is as was at original ~0.237Amps and no improvement in reducing this current, not reducing the phase angle and pf improvement

    My questions are the followings:
    A) Is power factor correction applicable for such a small 60W load?
    B) What is the value of capacitance C to improve its power factor?
    C) After revising formulas, why I am not having an answer that intersect with real hardware working?

  2. jcsd
  3. Apr 26, 2016 #2
    Last edited: Apr 26, 2016
  4. Apr 27, 2016 #3

    So you're saying the principle of power factor correction, by using shunt capacitance is not applicable for a 60W inductive load (fan)?

    According to the Eaton link you gave, the smallest motor was rated at 1hp that is more than ten times the rated power of my load.

    Should capacitor bank be ranked in pF instead of uF -if applicable- for such a small load?

  5. Apr 27, 2016 #4
    Basically, correcting the PF of a single fractional-hp motor, within the larger picture of total home PF -- is like a raindrop affecting ocean salinity -- there are much larger PF-sources (refrigerator, A/C, fans, power company transformers, etc., etc..) in the big picture.

    I believe you'll find that selecting the "...just right..." capacitor value will only be possible using a capacitor box, where you can add/shunt in SMALL capacitor values until you "null" the PF for the motor...but, hopefully, while you're doing that testing no other motors come on-line elsewhere in your house (nor on the distribution line INTO your house).
  6. Apr 27, 2016 #5

    Amazing comparison!

    So you're saying that the true C[uF] can only be get through trial and error? Couldn't it have a range at least?

    Thank you
  7. Apr 28, 2016 #6

    jim hardy

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    Shut capacitance should work.

    You should be able to calculate and get a result that's as good as your measurement.
    What is the nature of your measuring instrument ?

    I tried to follow your formulas but decided, being math challenged i have to stick closer to Pythagoras and Ohm

    Taking your VA2 as hypotenuse2 of power triangle
    and subtracting from it (0.7 of your VA)2 the adjacent or real side2
    i should be left with reactive side2
    taking √of that result
    i get 34.5 VA reactive for your motor.
    So if you shunt that motor with a capacitor sized to draw 34.5VA you'd compensate to pf of 1

    Repeating that sequence for PF = .85 instead of .7 i get 25.5 VA
    so you want a capacitor that draws the difference between 34.5 and 25.5 VA, 10 VA

    at 204 volts that's 49 milliamps
    Xc =204/0.049 = 4163 ohms ,
    C = 1/2πfXc
    are you at 60hz?
    C= 1/(377X4163) = 0.63uf

    Closest you tried was 0.5uf , then you threw in the towel. But you may have been closer than you think especially if you're working in the last digit on your digital meter.

    One of us has a mistake in our arithmetic else we'd be closer together on our estimate for shunt capacitor.. Can you point out mine? I am notorious for mis-keying into Windows calculator....and for swapping arithmetic steps midstream, wouldnt surprise me a bit if i'm wrong.

    Can you try a spreadsheet and see what currents you would get for shunt capacitance of 0, 0.1 ,0.2 ,0.3, etc microfarads ?

    old jim
  8. Apr 28, 2016 #7
    Dear old Jim,

    Thanks for your sharing.

    My measurement instrument is a self-made Phase, P, V, I, pf micro-controller based (PIC18) with a 20x4 LCD as the output:

    For a 200W incandescent resistive light bulb, the instrument shows a phase of 0.000 and hence an ideal pf of 1.
    For the 60W fan, it shows a phase angle around 43 deg and hence a pf around 0.737 LAG.

    The first trial was with a 6uF cap that provoked an over-compensation, where the load current increased to 0.4Amps where originally it was 0.237Amps. Then I used a 2.5uF using this calculation:
    204v x 0.237 A = 48.35VA (this is only true if there is un-distorted sine in current and voltage, and both values are RMS values)
    with a power factor of 0.737 this means it is about 48.35VA x 0.737 = 35.6W of true power
    and 32.7 VAr (according pythagoras)
    Currents: 174mA true current and 160.3mA reactive (imaginary) current.

    The reactive current from the fan is 160.3mA.
    The compensation capacitor needs to generate exactely the same current (but in opposite direction).
    Therefore Xc should be 204V/160.3mA = 1273 Ohms.
    C = 1/ (2 * pi * 50Hz * 1273 Ohms) = 2.50uF

    //Also there wasn't any phase improvement (being closer more to zero) and the load current returned into its original value of 0.2Amps where it should have been decreased if capacitance is correct

    What I have noticed is that heading to goal is done with decreasing capacitance, especially that spreadsheets you were mentioning have higher capacitance for higher KW usage of load.

    I was down for the 0.6uF you had calculated, and once again, I did not see any current decreasing (I`m not waiting the current to step down to a critical value, no, but at least go from 0.2Amps to 0.18Amps for example)

    Final note to be taken into consideration is that Vsupply in my first post had a value of 204VAC, but this is not constant, it could be changing to have a maximum of 230VAC and a minimum of around 200VAC depending on source, and on load demand etc...

    If a 60W is too negligible to have clear results, I can buy tomorrow a 200W industrial fan, and do all my tests on it.

    Much thanks and appreciation
  9. Apr 28, 2016 #8

    jim hardy

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    That's the capacitance to get to unity power factor not 0.85 ....

    A cheap fan motor likely draws a really nonlinear magnetizing current because they skimp on iron to keep it cheap.

    To rule out waveshape a test at reduced voltage would be interesting.... Do you have a Variac ? Or 115 volts ? At half voltage the torque falls to one quarter , so it'll draw starting current so feel the windings for overheat.....

    As one approaches the knee of the magnetization curve,
    magnetizing current becomes rich in 3rd harmonics -
    maybe that's why it appears to take less capacitance than expected ? I'm just guessing, haven't thought through that thought experiment yet.... run it past your "built in doubter" ....
    Reducing applied voltage will pull flux down into the linear region of the BH curve reducing distortion of magnetizing current waveshape..
    Last edited: Apr 28, 2016
  10. Apr 29, 2016 #9
    Dear old Jim,

    Can you confirm that If 6uF over-compensated the motor, and 0.2uF did not affect toward VAR compensation process, then the "true" capacitance value lies in between the extremum of this range?
  11. Apr 29, 2016 #10

    jim hardy

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    Sorry for the length of this post. If you want, just go to second part.
    This is how i plod along, exploring "what if" 's until one sorta jumps out.

    What about our arithmetic ?
    Well, 0.2 uf being 15915 ohms should pass 12.8 ma of current
    taking your post #7 values
    reactive current of 160.3 - 12.8 leaves 147.5 ma
    which with 174ma real gives 228 ma total at angle tan-1 ## \frac{147.5}{174} ## = 40.3° , pf = 174/228 = .76

    So the 0.2uf should not have changed things much, leaving you 40.3° in the lag.

    But the 6 uf surely overcompensated...

    I think that's what you said you observed.....
    So If that's indeed what you observed
    and if your instruments are not fooling you

    the answer to your question is "yes , the capacitance to drive PF to 0.85 lies between 6 and 0.2 uf .

    Please understand this statement is really confusing :
    ................................................ PART TWO...................................

    What if the instruments are fooling us ?

    Two things are worrisome
    1. Your "source" you said is affected by load. Does it interact with the motor, or with the capacitors? Is it stout like a wall receptacle or is it some sort of inverter or ferroresonant regulating supply ? They can apply distorted sine wave power.

    2. The nature of your PIC phase measurement. It works fine for resistive load where everything is (hopefully) sinusoidal.
    How does it measure phase?
    I googled PIC Power Factor Measurement
    and found only tutorials that use zero crossing to determine phase difference.
    That works fine for sine waves
    but recall my remark that nonlinearity of BH curve distorts magnetizing current ?
    And your qualifying remark in post 7 about "undistorted sine in current" ?
    Now I think we both are on to your trouble..
    A Sine wave can be distorted around its zero crossing as well as its peak.
    Look at this magnetizing current wave, the bottom one, and observe the voltage and current zero crossings are not 90 degrees out of phase... Current and flux zero crossings are pushed apart by hysteresis

    Nonlinearity and hysteresis of the iron distort your current wave.

    Reality Bites. Again.... I'm no stranger to that.

    of course added to that magnetizing current is a more sinusoidal load current wave.
    Point being - i do not know what points on the voltage and current waves to pick for phase difference measurement.
    Maybe the inflection points of current and voltage waves ? Intuitively that seems right but i haven't attempted a math cross-check on it.
    Do you have an oscilloscope with which you can observe your current and voltage waves ?

    I suspect the popularized zero crossing technique is , let's say anemic .The true phase difference has to be some calculable function like centroids or something, maybe inflection points i dont know. There are plenty of people here more academic than i perhaps we'll get some help. (Where's Bozorth when you need him ?)
    That's why i suggest a test at half voltage where your iron stays in the linear part of its BH curve.
    It'll greatly lessen the iron deficiency.

    Worthy experiment ?

    old jim
  12. Apr 29, 2016 #11

    jim hardy

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    Couldnt let go.

    Here's a paper on magnetizing current in unloaded transformers. Of course your motor is more akin to partially loaded transformer......
    (he's studying harmonics so ignore his references to "Power Factor" it's not the same one we know)


    I'd round up a scope and some opamps , see what 1st and 2nd derivatives of your current wave form reveal.

    old jim
  13. Apr 30, 2016 #12
    Dear old Jim,

    It was a nice and vast explanation, thank you.

    I did not use the Zero Crossing Detection Method to calculate phase between V and I wave forms and hence the power factor.
    I had troubles while doing it, so I switched into an analogue method that consists of using an XOR gate, with V and I(load) as inputs: The output [referring to XOR logic] would produce a 'logic' proportional to phase angle. Fudge factor must be eliminated (in coding).

    Here is a hand-drawn circuit used to V, I, pf measurements headed to analog PORTA port in PIC18F4520:

    I used v,i,p,pf,final.png
  14. Apr 30, 2016 #13
    I used ADC to convert voltage (analog) values into a digital one (10-bit internal ADC of PIC18F) to get V, and I and pf.

    I have used the RC circuit to stabilize and filter the outputs.

    I have an efficient V, and I measurements output, but yesterday I noticed that there is a problem with pf measurement, because I branched a 12uF caps with 200K shunt bleeding resistor across it, and logged it into the system to see V, I, pf:
    V, and I were efficient but the phase angle showed 0.000 deg instead of 90 deg and I still couldn't know why.

    -----> Regardless of this system, I have a handy clamp meter, that shows around 0.24Amps when fan is connected without shunt capacitance.
    Using same clamp meter, and assuming power factor is improved, I must see a decreasing from 0.24Amps to a lower value.
    I`m not seeing this decrements at all, therefore no power factor was improved, because if so, its effect should be clear from current measurement.

    Thank you
  15. Apr 30, 2016 #14

    jim hardy

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    here's as good a quick introduction to motors as i've run across
    i'm saving a copy on my "for pf" folder

    hmmm so it has difficulty with leading power factor ? I'm chewing on your XOR , back later....

    Now the problem statement is getting more precise...
    "A question well stated is half answered"
    And a picture is worth a thousand words......

    aha i think i see a trouble
    Is that pass through current transformer a simple transformer or an electronic gizmo with external power that you didnt show?
    I suspect the former. If so it's not used correctly.
    you are asking your current transformer to drive a load of several thousand ohms plus two diode drops.
    There is a term "Burden" in metering jargon that refers to the load which a current transformer is asked to drive.
    A current transformer should be operated with a load as close to zero ohms as you can achieve. You measure its output with an ammeter not a voltmeter.
    The higher the "burden" the less accurate the current transformer. And the greater its phase shift. Here's why.
    It has certain inductance even with only a one turn primary
    and voltage across its one turn primary is L di/dt of course
    and voltage across its secondary is larger by the turns ratio.
    Now if current is Isin(wt) , voltage is Ldi/dt = L X I X wcos(wt) and that's a 90 degree error in your phase measurement.

    The correct way to operate a current transformer is with its secondary so nearly short circuited that its voltage is insignificant.
    (Ideally it'd be absolutely shorted but zero ohm wire is too expensive)
    That lets it produce current in secondary that's in phase with primary current
    we usually measure that with an ammeter

    i think this approach will work
    diode clamps protect Mr.LM324
    0.1 ohm makes millivolts out of the current signal
    LM324 acts as comparator, 1K pullup resistor helps him drive output high (unaided he only gets within 1.5V of V+).
    That should fix your phase angle measurement.

    Why i think it should work :
    In any transformer, the Current ratio is almost but not quite equal to Turns ratio,
    (the mental step necessary to understand them is to imagine one with a 1::1 turns ratio)
    the primary and secondary currents differ by whatever is the magnetizing current required to make flux in the core
    and for sinewaves flux is in proportion to voltage
    so by minimizing voltage we minimize flux which minimizes magnetizing current

    and minimizes the fraction of your primary current getting wasted as magnetizing current.

    if you open circuit the secondary then all your primary current is wasted as magnetizing current and you get that 90 degree phase shift between voltage and current. Voltage out is proportional to primary current/reluctance of core and reluctance is highly nonlinear. Shorting secondary removes that nonlinearity, and removes the phase shift too.

    I think your phase measurement is off by 90 degrees because your CT secondary is effectively an open circuit.. And if that's where you take your amplitude measurement it's highly inaccurate for same reason.

    Here's a current transformer tutorial

    I think : "CT , cast down your burden ! "

    what do you think ?
    Last edited: Apr 30, 2016
  16. Apr 30, 2016 #15
    Dear old jim,

    I must spend time, analyzing your explanations, and demonstrations.
    I got your point about CT Burden, and problems related.

    At the moment, I`m concerned with your improved circuit of current measurement:

    I`m using an ADC to first of all, know the current value.

    Using your improved circuit, its output is directed to XOR gate, but what about its value measuring? Should I take from the same output, another wire heading to ADC?

  17. Apr 30, 2016 #16
    Here is the current sensor link:


    From another part, and in my previously attached image about measurements circuitry, you can notice that there is a bridge to rectify resultant output voltage from CT in order to test its value after code conversion in Amps.

    Using your circuit, how can I ADC the current value, knowing that the max Vin to PIC analog pin is 5V

    Thank you
  18. Apr 30, 2016 #17

    jim hardy

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    That output is logic level and will be useless for amplitude.

    There are plenty of absolute value circuits using opamps
    with dual supply it's easy
    but i dont know if you have negative available

    trouble with this one is you need gain. Might experiment and see how it does with 2nd stage a follower with gain.....
    you'll need a CMOS opamp that can handle differential input voltage

    here's a TI appnote see fig 5

    But i'd be mighty tempted to try out the above circuit with a couple more sections of an LM324
    change the 200 K resistors to maybe 10K
    and place 1 meg in series with pin 2 to raise Zin (strictly speaking should get one in pin 3 as well, )
    and add a 10:: 1 voltage divider between pins 7 & 8 for some gain

    What is the ratio current transformer you have ? How much current do we expect out of CT at say an amp of motor current ?
    CT datasheet should say what burden it wants, or how much voltage it can make with reasonable error.

    old jim

    Attached Files:

  19. Apr 30, 2016 #18

    jim hardy

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    Perfect it's for a 200 ohm burden
    and note phase shift is specified


    What is most sensitive range of your ADC ?

    give me a while.... we only have to handle 5 milliamps, that's great .

    Meantime, my 0.1 ohm can become 200
    one amp in primary will produce ~ 1/2000 = 500ua, which X 200 ohms = 100 millivolts
    that's more like it

    edit - fixed typo 200 not 100 ohm burden
    Last edited: Apr 30, 2016
  20. Apr 30, 2016 #19
    Dear old Jim,

    Thanks for all of your efforts.

    I`m interested in knowing the amplitude of the load current, as well as using current as input to one of XOR gate.

    What about mixing my old circuit with yours, in a way that I can sense current value, and from another part, lead it to XOR circuit phase shift freed to know phase angle, would that be possible?

    The maximum Vanalog input to PIC is 5V. The range is from 0 to 5V and my loads are usually <1.5Amps

  21. Apr 30, 2016 #20


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    Hi DjMadness. For the capacitors you are using, is there any indication of their accuracy? Does their label provide the manufacturer's tolerance? Can you attach a photo of the smaller cap?

    Knowing that the needed capacitance is < 1μF, perhaps power factor improvement would show up clearer if you monitored the current while adding small shunt capacitors in increments of 0.1μF?
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