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PFC circuit.

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alphacat

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Harmonic distortion

Hello friends.

I'd like to build this summer a PFC circuit and currently started reading about factors that reduce power factor.
I'm trying to understand something about harmonic distortion.
The current provided by the power line is a pure sine wave that has only one frequency, the fundamental frequency (50Hz or 60Hz), assuming that there's no voltage distortion in the power line, right?

Only non-linear components cause harmonic distortion to the wave.
What does it mean to cause harmonic distortion?
Does it mean to produce harmonics that have frequency which is an integer multiple of the current's fundamental frequecny?
If these non-linear devices only distort the already existing harmonics, then I dont understand where these harmonic have come from in the first place?

Moreover, I wanted to ask please how do harmonics' existence reduce power factor?

I hope someone could throw light on this matter.

Thank you.
 
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Hero999

Banned
Some appliances don't draw a sinusoidal waveform, for example rectifiers draw their current in pulses. This causes the waveform to become distorted because the resistance in the cables causes the voltage to drop.
 

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crutschow

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Most Helpful Member
The power line voltage remains essentially a sine wave but the current can have harmonics due to such things as rectifier-capacitor power supplies and SCR/Triac dimmers, which draw power in pulses.

These pulses have the same frequency as the power line which introduces new harmonic frequencies in the current that are a multiple of the line frequency, thus it is called harmonic distortion. Since the current is now not a sinusoid in phase with the voltage, the power factor is no longer 1. This causes higher peak currents in the power line for a given load power then a pure sinusoidal current would, increasing IR losses in the lines (making the power company unhappy).
 

alphacat

New Member
Thanks guys :)

crutschow, you said that the power line creates new current harmoics due to the drawn high current pulses.
Could you please detail about that?
As Hero999 showed in his example, there're no current harmonics drawn from the power line, only the current pulses which, as you said, have the fundamental frequecny.
 

Hero999

Banned
The current pulses contain harmonics because they are non-sinusoidal.

Here's a Fourier transform of the above wavefrom.

You clearly see the harmonics of 50Hz.
 

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alphacat

New Member
Oh now I get it,
It relays on the principle that you can describe any periodic waveform with a Fourier series, and if the periodic waveform is non-sinusoidal, then the Fourier series would contain more harmonics than just the first one which has the fundamental frequecny.
Thank you :)

Is it correct that the reason that the harmonic reduce power factor, is becaues many 'Pt's values are close to zero, where Pt is the instantaneous current? (Pt = Vt*It).

How exactly harmonics travelling in the neutral line disrupts other devices connected to the neutral line?

Thanks alot :)
 

Hero999

Banned
Odd harmonics on a polyphase supply add together in the neutral which can cause it to overheat.

Harmonics also increase heating in transformers due to eddy, hysteresis and skin effect losses. Odd harmonics are also reversed clocked and try to drive motors in reverse causing increased heating and vibration.
 

alphacat

New Member
Thank you very much Hero.

I read about PFC device claims it saves up to 50% in the watt consumption (active power consumption).
I wanted to ask please, the appliance itself would consume less watt with the PFC used, or this PFC only saves I^2*R losses that are due to wires resistance?

Thanks.
 

crutschow

Well-Known Member
Most Helpful Member
Some of these devices reduce the voltage to the device to just the minimum necessary to power the motor. This improves the power factor by minimizing the magnetizing current (and consequent I²R loss), thus increasing the efficiency of the motor.

I would expect a possible 50% only for a large, lightly loaded or low duty-cycle motor, not for a typical household appliance.
 

alphacat

New Member
Some of these devices reduce the voltage to the device to just the minimum necessary to power the motor. This improves the power factor by minimizing the magnetizing current (and consequent I²R loss), thus increasing the efficiency of the motor.

I would expect a possible 50% only for a large, lightly loaded or low duty-cycle motor, not for a typical household appliance.

Hey Crutschow.
Thanks for the answer.

I tried asking you in another thread about how do these devices find out the minimun voltage necessary for the motor to drive its load, and how do they actually reduce the voltgae once that minimum value was found?

Thanks again.
 

crutschow

Well-Known Member
Most Helpful Member
Hey Crutschow.
Thanks for the answer.

I tried asking you in another thread about how do these devices find out the minimun voltage necessary for the motor to drive its load, and how do they actually reduce the voltgae once that minimum value was found?
Well, I may not have replied since I don't really know the answer.

Perhaps it monitors the current as it lowers the voltage. The current will initially drop as the losses are being reduced, but below a certain voltage the current will start to increase so the motor can continue to provide the required load output power. The point at which the current starts to increase would be the optimum operating voltage.

You would need a logic circuit (probably a µP) to do this determination since it would continually have to raise and lower the voltage slightly to maintain the optimum point, which may change due to line voltage or motor load changes.

The voltage can be controlled by a triac, similar to a light dimmer.
 
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alphacat

New Member
Thank you very much crutschow.

There's something I didnt quite understand in what you said.
You said that when the voltage begins to drop, there's a point in which the current will start rising in order to keep the motor driving its load.
According to |V| = |I| * |Z|, since |Z| stays constants (It is the motor's impedance), then when |V| drops down, then I must also drop down, isnt it?
 

crutschow

Well-Known Member
Most Helpful Member
The motor impedance (and thus input current) is not constant, it is determined by the load it is driving and the motor losses. The motor input power equals the motor power losses plus the output power. Since P = V * I, as the voltage drops below the optimum point the current will have to increase to keep the motor output power constant.
 
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