[PIC] Power factor measurement using PIC18f4520

[KhaledOsmani]

Right, your real-life circuit will not need a resistor in the supply leads.

Hello Brad,

What about power rating of the ~5R resistor?

What about voltage rating of the inductor?

What about wiring and mm² concernings of wiring? For ~3Amps current drawing, thickness of wire loading the 220/50VAC should not be less than 3mm² , and interconnection between cap and RL circuit should also have a significant thickness.

I mean, soldering the total circuit using 0.8mm soldering wire wouldn't be a good idea. Right?

 

[BradtheRad]

Hello Brad,

What about power rating of the ~5R resistor?

It represents ohmic resistance in the inductor. It's not a separate component. I looked at the 20 ohms DC resistance of my 120VAC window fan, and then I took an arbitrary fraction of that. It's the resistance of many turns of thin wire inside the motor (inductor).

What about voltage rating of the inductor?

What about wiring and mm² concernings of wiring? For ~3Amps current drawing, thickness of wire loading the 220/50VAC should not be less than 3mm² , and interconnection between cap and RL circuit should also have a significant thickness.

I mean, soldering the total circuit using 0.8mm soldering wire wouldn't be a good idea. Right?

As for power rating, it is probably similar to the secondary of a low voltage transformer. You may be able to locate construction guidelines.

See these links to charts of wire gauges showing safe amp-carrying capacity:

http://amasci.com/tesla/wire1.html



http://www.powerstream.com/Wire_Size.htm

 

[c_mitra]

Most domestic fans are AC induction motors and they are designed to start on a single phase using a capacitor. The capacitor shifts the phase sufficiently so that the current in the start windings *appear* to be coming from another phase. Although the shift in phase is not enough (90 vs 120 deg) it is sufficient to start the motor and keep the sense of rotation. This is simpler and cheaper than using the concept of *shaded pole* motors.

The capacitor is often chosen very conservatively and the net result is that the AC fan appears to the power supply as a capacitor rather than an inductor. This is fortunate because this excess capacitive reactance can compensate for other small inductive reactances, coming from, say small items that use universal motors (mixers, drills, some power tools etc).

I believe that your fan is already fully compensated for the inductive reactance. The phase angle that you are observing is actually coming from the capacitor and cannot be reduced further. If the fan has a speed regulator (most of the small speed regulators are basically series resistors), the phase angle will *increase* at lower speeds.

 

[KlausST]

Hi,

I mean, soldering the total circuit using 0.8mm soldering wire wouldn't be a good idea. Right?

Solderng wire sounds like the wire that melts when you touch it with the soldering iron tip.

--> it is not meant to build a "wire" to connect two points in your circuit.

Klaus

 

[KhaledOsmani]

The circuit makes no sense.
I've bought an 220/50VAC to have a supply of 50VAC, but how to shunt connect at it an inductor? When inductor is fully charged, it represents a wire, so this circuit is an act of short-circuit.

Plus, how to get a 50mH inductor that supports 50volts, without another coil to transform magnetic field to another output voltage?

Initially Brad thought to replace big inductive loads with lower ac voltage loads (in this case 50VAC), so he shunt put a coil across 50VAC supply, this was in simulator.

In real circuit you cannot connect across any supply directly a coil; plus how can a coil draw 3Amps without manifesting at any operation? The coil is not driving a shaft to turn, nor transforming any voltage, how is it supposed to draw 3 Amps doing nothing.

I thought I couls monitor power factor behavior using this circuit, it seems simply this must be thrown away. It is not working in any form!!

 

[BradtheRad]

In real circuit you cannot connect across any supply directly a coil; plus how can a coil draw 3Amps without manifesting at any operation? The coil is not driving a shaft to turn, nor transforming any voltage, how is it supposed to draw 3 Amps doing nothing.

Inductors are commonly referred to as chokes. They restrict AC.

The 50mH inductor gives the right amount of choke action, so that it has a reactive impedance of 15.7 ohms at 50 Hz (from the formula XL = 2 * Pi * f * L).

 

[FvM]

The only relevant restriction is that the choke must be operated within it's voltage and current ratings. If "supports 50volts" means that it's designed to work up to 50 VAC, you shouldn't exceed the voltage level much to avoid core saturation.

 

[c_mitra]

The only relevant restriction is that the choke must be operated within it's voltage and current ratings. If "supports 50volts" means that it's designed to work up to 50 VAC, you shouldn't exceed the voltage level much to avoid core saturation.

An ideal inductor has only a current rating because a choke cannot stand a voltage because it has a low resistance. The reactance of an inductor depends on the frequency and the voltage specification, if any, just means that it may spark over due to insulation breakdown.

 

[KhaledOsmani]

I`m very unfamiliar with inductors unfortunately.

I still cannot understand how I can branch in parallel a choke to a voltage source without making a fault. Where is the reason in 3Amps current drawn from simulation?

This is the only 50mH choke I've found, look how small, it barely supports 1VAC. What must I do, why things are getting more complex from subject to another?!



See how small it is with respect to a small resistor. If this was big enough, I can directly connect it to 50VAC leads?

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Inductors are commonly referred to as chokes. They restrict AC.

The 50mH inductor gives the right amount of choke action, so that it has a reactive impedance of 15.7 ohms at 50 Hz (from the formula XL = 2 * Pi * f * L).

Brad,

Could you make a pencil made schematic, or post a picture of that "choke" that I`m supposed to use please? Can you tell why it can safely be connected to 50VAC, without any series resistor on?

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An ideal inductor has only a current rating because a choke cannot stand a voltage because it has a low resistance.

Then how come I`m supposed to feed it directly from the 50VAC? Circuit is either incomplete, either true only for simulation.

 

[c_mitra]

It will have enough reactance at 50Hz and that is the factor that will restrict the current. It will have only around 15ohm reactance that may not be sufficient at 220V (it will go up in flame) but at 50VAC it will take 3-4A current (that is still too much for the small size).

 

[KhaledOsmani]

It will have enough reactance at 50Hz and that is the factor that will restrict the current. It will have only around 15ohm reactance that may not be sufficient at 220V (it will go up in flame) but at 50VAC it will take 3-4A current (that is still too much for the small size).

How can I get a bigger one? What about connecting the secondary of a 220/50 transformer to a source of 50VAC? With 5Amps rating, can I see 3 amps circulation and correction of phase using pre-calculated caps?

 

[BradtheRad]

How can I get a bigger one? What about connecting the secondary of a 220/50 transformer to a source of 50VAC? With 5Amps rating, can I see 3 amps circulation and correction of phase using pre-calculated caps?

Yes, the secondary of a transformer is a suitable load. I was experimenting with a few transformers in my parts box. Normally these are for step-down, and carry a few A. However what I did was to apply AC to the secondary thru a variac. They admitted current, which I measured as voltage across a 1 ohm resistor rated for 10W.

You need a choke/transformer which can handle 100 or 200W. It needs to be much larger than your photo in post #369. Expect it to be a few inches on each side. Perhaps it can be a twin of your step-down transformer.

Ideally you would have an assortment of chokes/transformers to try in your lab.

 

[KhaledOsmani]

Yes, the secondary of a transformer is a suitable load. I was experimenting with a few transformers in my parts box. Normally these are for step-down, and carry a few A. However what I did was to apply AC to the secondary thru a variac. They admitted current, which I measured as voltage across a 1 ohm resistor rated for 10W.

You need a choke/transformer which can handle 100 or 200W. It needs to be much larger than your photo in post #369. Expect it to be a few inches on each side. Perhaps it can be a twin of your step-down transformer.

Ideally you would have an assortment of chokes/transformers to try in your lab.

Thanks Brad.

I have no lab unfortunately, and I need to buy every component on my own, and test it inside my room.

So you were saying that using a variac (autotransformer) feeding the secondary of 5A rated 220/50VAC with/without capacitor would let me see power factor behavior?

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Hello,

This is the secondary of 220/50VAC transformer, that I intend to use as a choke to be fed from a 50VAC source.



There still one problem: I do not know its inductance in Henry, and my multi-meter doesn't have an inductance function. How can one know the inductance of a coil? What if its inductance is larger/smaller than 50mH little much or too much? What are the consequences (beside that its impedance would have a different value).

Shall I try to feed it from the 50VAC and try by trial and error the replies for adding/removing caps?

thanks

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I've applied 220VAC on the primary of this transformer, to know its L:

I've used Ohm's law to know its impedance Z.

XL = √(Z^2 - R^2)

Then: L = XL / 2πf >>>>>>>>> I got around 8H. Does this have to mean that also the secondary have a henry value way bigger than mH? I shall do this test tomorrow when I have access to a variac.

But if the secondary Henry are tens, this experiment won't work because Impedance will be so much big, that only few milli-amps will travel from source to choke and thereby I won't be able to see current reduction using caps.

Right?!

 

[BradtheRad]

I've applied 220VAC on the primary of this transformer, to know its L:

I've used Ohm's law to know its impedance Z.

XL = √(Z^2 - R^2)

Then: L = XL / 2πf >>>>>>>>> I got around 8H.

Yes, 8H is reasonable for the high voltage side. (Especially if you have 230V house voltage.)

The step-down secondary has a lesser Henry value.

There still one problem: I do not know its inductance in Henry, and my multi-meter doesn't have an inductance function.

How can one know the inductance of a coil?

There is a method, by hooking up a known capacitor across the winding, so you have an LC tank circuit. Cause it to ring for a few cycles. Measure the frequency. Frequency is based on C and L values.

You can calculate inductance from the standard formula. (Ohmic resistance does not enter into the equation, although if it is too high it can alter the frequency.)

The C value should not be so small that the ringing is at too high a voltage. It should not be so large that too great current flows. Try several C values in an effort to get reasonable results.

What if its inductance is larger/smaller than 50mH little much or too much? What are the consequences (beside that its impedance would have a different value).

There is a formula which relates turns ratio, primary inductance and secondary inductance. You know the step-down ratio and primary inductance, so you can calculate (tentatively) the secondary inductance.

So you were saying that using a variac (autotransformer) feeding the secondary of 5A rated 220/50VAC with/without capacitor would let me see power factor behavior?

Shall I try to feed it from the 50VAC and try by trial and error the replies for adding/removing caps?

A variac allows you to increase power gradually. You can keep watch for heating, etc. The variac ought to be rated for the maximum expected Amperes.

My variac is rated 1.75A (purchased used on Ebay). I want to be sure I don't overload anything, so I step down 120V to 60V through a 2:1 autotransformer, then to the variac, then to my test transformer. It's all part of making do with the equipment we have available.

But if the secondary Henry are tens, this experiment won't work because Impedance will be so much big, that only few milli-amps will travel from source to choke and thereby I won't be able to see current reduction using caps.

Right?!

I'm pretty sure there can be a wide range of usable values. The idea is for you to be able to observe the effects of inductive reactance, at levels you can measure. You want inductive reactance to play a dominant role. You do not want ohmic resistance to be dominant.

 

[c_mitra]

You can measure the secondary inductance using the same method. First calculate the DC resistance using a multimeter. This is going to be your R. Next use 50VAC and measure the current into the secondary. You can use the DMM AC voltage and current functions for this measurements. Next you calculate Z (you can only get the magnitude of Z but that is all we need) by using V/I.

Use the same techniques to calculate the L. It will be much less than the primary inductance (because of the turns ratio).

 

[KhaledOsmani]

Hello,

So for a 220/50 > n: 4.4 Therefore secondary inductance would be around 8/4.4 = ~1.8181 <<< This is too high and results accordingly in a too high XL = 2 * pi * 50 * 1.8181 , and 50 / XL = few milli-amps.

How can I use the same secondary as the choke, but decrease the overall resistance, maybe using parallel connected resistors? Or this would distort seeing inductive behavior?

If adding parallel capacitors would not work, shall I order to make a 10,000/50VAC transformer, and use its secondary as the choke, by assuming that its first inductance is of 8H, then the secondary inductance is of few mH? Or at 10KV the inductance is far from 8H?

Are there any possible solution to use same secondary as the choke? From seeings, it is clear that its inductance is not in the milli Henries range. What to do in such case?

Finally, If this was used, and resultant ampere was in milli-amps. When branching a parallel capacitor, I would see less milli-amps, would that be an effective demonstration of controlling power factor, or I should keep thinking about several amps drawing such that its reduction would be more profound ?

Thanks

 

[KlausST]

Hi,

So for a 220/50 > n: 4.4 Therefore secondary inductance would be around 8/4.4 = ~1.8181 <<< This is too high and results accordingly in a too high XL = 2 * pi * 50 * 1.8181 , and 50 / XL = few milli-amps.

R, Z, L, C... is transferred from primary to secondary (or back) with the square of the winding ratio.

In your case: 4.4 ^2= 19.36

Therefore the secondary (theoretically) is 8H / 19.36 = 0,41H
(A real circuit deviates from the idealvalues)

I think you need a detailed study on how AC circuits with R, L, C accordig currents, voltages, phase reacts.
It seems you are not familiar with AC calculations and vector diagrams where you easily can see/construct/review magnitude and phase angles.
I really find those vector diagrams help a lot to understand the behaviour of an AC circuit...and how compensation works.

Klaus

 

[c_mitra]

R, Z, L, C... is transferred from primary to secondary (or back) with the square of the winding ratio.

In your case: 4.4 ^2= 19.36

Therefore the secondary (theoretically) is 8H / 19.36 = 0,41H
(A real circuit deviates from the idealvalues)

@KhaledOsmani, you can start with this value as a guide and this is a reasonable value. If you follow my suggestions, you will find the measured value close to this estimate.

 

[KhaledOsmani]

Hello,

Is there any switching circuitry, ruled by PIC that connect in/out cap banks when needed other than the open-relay switch that also provide load [caps] to be connected at the full cycle?

Are there any solid-state techniques that functions in the same way as an open-relay switch functions but with no moving mechanical parts?

if yes, any schematic, mode of operation please?

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[KlausST]

Hi,

Are there any solid-state techniques that functions in the same way as an open-relay switch functions but with no moving mechanical parts?

A solid state relay?

Klaus