Power Factor Correction for a small 60W AC Induction motor

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Main Question or Discussion Point

Hello,

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

Results:
*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?

Thanks
 

Answers and Replies

  • #3
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Might this article help? http://www.eaton.com/ecm/groups/public/@pub/@electrical/documents/content/sa02607001e.pdf

But, also, be aware of this aspect of "home" PF-correction: http://www.nlcpr.com/Deceptions1.php
Hello,

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?

Thanks
 
  • #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).
 
  • #5
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Hello,

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
 
  • #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
 
  • #7
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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
 
  • #8
jim hardy
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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
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.
http://demonstrations.wolfram.com/MagnetizingCurrentWaveformInAnIdealSaturableInductor/

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..
 
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  • #9
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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?
 
  • #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 :
//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
................................................ 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

ShadedPole1.jpg

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
 
  • #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)
http://www.jocet.org/papers/69-A30009.pdf

upload_2016-4-29_20-51-0.png


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

old jim
 
  • #12
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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
 
  • #13
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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
 
  • #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
http://www.tcf.com/docs/fan-engineering-letters/single-phase-ac-induction-squirrel-cage-motors---fe-1100.pdf?Status=Master

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

-----> 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.
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
shadedpole2.jpg

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
http://www.electronics-tutorials.ws/transformer/current-transformer.html

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

what do you think ?
 
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  • #15
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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?

Thanks
 
  • #16
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Here is the current sensor link:

http://ekt2.com/products/productdetails?ProductId=bed34f47-f444-4fc0-96a9-a52a09eb677a

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
 
  • #17
jim hardy
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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?
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.....
upload_2016-4-30_10-29-12.png

you'll need a CMOS opamp that can handle differential input voltage

here's a TI appnote see fig 5
http://www.ti.com/lit/an/sboa068/sboa068.pdf


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
 

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  • #18
jim hardy
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Perfect it's for a 200 ohm burden
and note phase shift is specified


upload_2016-4-30_11-14-57.png


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
 
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  • #19
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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

Thanks
 
  • #20
NascentOxygen
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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)
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?
 
  • #21
jim hardy
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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
I must gave an addictive personality , or maybe adhd...... this is a interesting project.
I want to learn PIC's but only have an Arduino ; got it to count time of day in Roman Numerals but that project is on hold for a decoder circuit board to drive my VFD display segments... Takes a lot of characters to do hour, minute and second in 24 hr format.

I spent some time looking into absolute value circuits. They're a lot easier with dual supply
but i sure like simplicity of single supply.



Here's an introduction to them
http://sound.westhost.com/appnotes/an001.htm
and a more formal paper
http://www.ti.com/lit/an/sboa068/sboa068.pdf

The challenge is to amplify your signal up to a couple volts
but to place that gain in the absolute value converter like i mentioned in post 17 amplifies the converter's error too...
so i am thinking to boost the CT signal as AC before handing it to an absolute value circuit might be a good idea.
but it's hard to amplify AC with single supply opamps..

So i went outside and mixed some cement . Subconscious does its best work while conscious isn't around.

Of course - the transformer puts out a small AC signal
and we'd like it to become a large AC before we hand it to absolute value circuit.

Notice that the absolute value circuits described have unequal input impedance during alternate half cycles
so they recommend driving with a "low impedance source" , ie one that's electrically stout.
So to keep things simple we need an AC amplifier that's single supply, low output impedance and predictable gain.
Sounds to me like a job for LM386. Readily available, very cheap, and easy to use.

Here's its datasheet
http://www.ti.com/lit/ds/symlink/lm386.pdf
gain is 20 to 200
made to drive a speaker so has plenty of current drive= low output impedance

that current transformer of your looks like a really handy piece of test gear , if this works out i'll build one too.

i'll be back

old jim
 
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  • #22
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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?
Hello NascentOxygen,

The manufacturer tolerance is of 5%.
The problem is that here in the market, the smallest value for C is of 1uF, there isn't any in <1uF range. I have to add in series to decrease capacitance to requested value. Adding small 0.1uF in shunt by turn, is very difficult since 0.1uF requires a huge set of capacitors to have that value.

I've bought 1uF (x2), 1.5uF (x3), 4uF, 5uF and 6uF and combined them in a way that I got a minimum of 0.2uF as far as I remember.

Attached in a photo of a 1.5uF capacitor
 

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  • #23
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I must gave an addictive personality , or maybe adhd...... this is a interesting project.
I want to learn PIC's but only have an Arduino ; got it to count time of day in Roman Numerals but that project is on hold for a decoder circuit board to drive my VFD display segments... Takes a lot of characters to do hour, minute and second in 24 hr format.

I spent some time looking into absolute value circuits. They're a lot easier with dual supply
but i sure like simplicity of single supply.



Here's an introduction to them
http://sound.westhost.com/appnotes/an001.htm
and a more formal paper
http://www.ti.com/lit/an/sboa068/sboa068.pdf

The challenge is to amplify your signal up to a couple volts
but to place that gain in the absolute value converter like i mentioned in post 17 amplifies the converter's error too...
so i am thinking to boost the CT signal as AC before handing it to an absolute value circuit might be a good idea.
but it's hard to amplify AC with single supply opamps..

So i went outside and mixed some cement . Subconscious does its best work while conscious isn't around.

Of course - the transformer puts out a small AC signal
and we'd like it to become a large AC before we hand it to absolute value circuit.

Notice that the absolute value circuits described have unequal input impedance during alternate half cycles
so they recommend driving with a "low impedance source" , ie one that's electrically stout.
So to keep things simple we need an AC amplifier that's single supply, low output impedance and predictable gain.
Sounds to me like a job for LM386. Readily available, very cheap, and easy to use.

Here's its datasheet
http://www.ti.com/lit/ds/symlink/lm386.pdf
gain is 20 to 200
made to drive a speaker so has plenty of current drive= low output impedance

that current transformer of your looks like a really handy piece of test gear , if this works out i'll build one too.

i'll be back

old jim

Dear old Jim,
Hi,

What about replacing the 10K burden in my old attached circuit with a 200Ohm, and make changes via software to know the current magnitude in the same time having no more than 10' phase shift?
 
  • #24
jim hardy
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What about replacing the 10K burden in my old attached circuit with a 200Ohm, and make changes via software to know the current magnitude in the same time having no more than 10' phase shift?

It's moving in the right direction
but i worry about thediode bridge
Mr CT expects not to be asked for more than 1 volt but you're asking him to jump two diode drops
I'm ready to scan that sketch




Is this a hobby or a serious industrial application? In other words, are you open to use commercial grade parts and one dollar general purpose amppifiers instead of ten dollar precision ones?

How is your breadboarding ? Do you mind some experimentation ?

I have a sketch worked out that just might do . If it still looks plausible after a night's sleep i'll scan and post it.
If it doesn't look too hard to build , try it. Uses ubiquitous LM324 .Those better CMOS opamps will cost around $10 each. If plain old 324's come close you know it can be done better. If you use sockets, upgrading opamps is easy.


back in a few (with any luck)
 
  • #25
jim hardy
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I think this would work
Start with R1 = 100 ohms, might make it 10 if amp behaves well at g=200
(or leave it alone and have a really sensitive current probe)

CT gives 0.5 ma/amp, X 100 Ω = 50mv/amp, X gain of 20 = 1volt/amp out of LM386
470uf removes the DC offset

A1 rectifies it as follows
During negative half cycles it's just an inverting amplifier with an extra diode. It makes the negative half cycles positive and hands them to A2.
During positive half cycles A1 output would like to go negative but can't because there's no negative supply, best it can do is zero.
So positive signal comes through both 10k's to "A" and the diode blocks A1 from reducing it to zero at node A.

Node A feeds A2 which is just a buffer with high input resistance, so there's no voltage drop across the 10K's and A2 receives the positive half cycles as if A1 weren't even there.
Result being A2 receives current signal of 1 volt per amp fully rectified. It only hands that result to A3 .
Only reason A2 is included is so that any input current drawn by the averager won't cause any voltage drop across the 10k's during positive half cycles.

A3 is an averager, called "Sallen Key" two pole low pass filter set for 11hz . The reason it's 11hz not 10 is that made the capacitor values come out even. Get good caps if you can, 5% or better and hand pick matched 100K resistors 1% or better.
It smooths the rectified peaks yet is still fast enough that your eye won't notice any delay - i doubt your PIC samples faster than that.
(i cheated and used an online calculator to figure capacitors because the formulas are intimidating)
currentransfrmr 001.jpg


A4 is a comparator and it inverts. Its output will be high during positive half cycles.
If you run this from the +5 volts supply you can wire his output direct to the XOR
if you run it from higher voltage put a 5.1 volt zener diode with cathode on A4's output and anode on his - input. That'll limit his output to about 5 volts.
The voltage divider on A4's + input assures that he recognizes A1's attempt to make zero volts as zero - A1 only has to achieve <0.05v .

Hmmm - an AC coupled coupled LED on A4's output would be a nice feature, it'd tell you when the gizmo is seeing a current signal large enough to measure..

Potential troubles to be found out by experimentation

1. LM324 is not a precision opamp. Any current that flows into A1's input will cause error in current measurement. I studied the datasheet real hard and i think that current will be only nanoamps, so that's why i used 10K resistors instead of 100k , to reduce that effect. The LM324 inputs are "soggy" only for negative , as i read the datasheet , and they don't go negative in this circuit.

2. The LM386 i believe to be DC coupled so it won't introduce phase shift except at the output decoupling capacitor. That's why it is so huge(470uf).
The unequal current on alternate half cycles drawn by A1 and his 10k's will be only 1/100th of what flows through the 100 ohm, that's why the 100 ohm and big cap - remember "output impedance" ? 10KΩ can't load down 100Ω very much.


What do you think ?

I am NOT any kind of expert circuit designer, this is a hobby level circuit like i would build for myself.
There are expert analog folks on PF and i welcome critique

If you build this we'll have to check for phase shift and accuracy.

The LM386 gain is increased to 200 by closing the gain switch. There may be improvement in CT's phase accuracy by lowering R1 to 10 ohms and running Mr LM386 at Gain of 200 :: we'd have to test, i do not know whether switching in that gain capacitor makes LM386 shift phase..

Anyhow i expect you can handle phase error in software - our goal is to make it constant , and lowering burden helps the CT in that regard.

What do you think? Only two IC packages, LM386 is 8 pin dip and LM324 is 14 pin dip , i'd use sockets...
buy the LM386 version that''s spec'd for 5 volts not 6 and you can run it all off same 5V power supply.
A higher supply will let you measure higher current - an amp will be near max because you only get about 3 volt swing with 5 volt supply and an amp RMS is 2.8 amps peak to peak.
lastly dont forget power to LM324 - i didnt show it; and if your power supply is noisy bypass pin 7 of the LM386.

Regards , and if you make this please post results . ( Buy extra parts - your friends will want one too.)
 
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