Discussion in 'General Electronics Chat' started by almotions, Aug 6, 2009.

1. almotions Thread Starter Active Member

Feb 6, 2009
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0
Hi all,
Assuming a sinosoidal AC supply to a non-linear load which draws distorted current. My question is
1. how does people define non-linear load in general,is it purely resistive or it contains reactive components?

2.Assuming the distorted current waveform is in phase with the voltage,the displacement factor=1 and distortion factor equals to less than 1,which gives a Total power factor of less than 1 and therefore draws reactive power right? Can we conclude that harmonics draws reactive power? or does it draw active power as well?

3.how do we calculate the active and reactive power drawn by these harmonics?

Thanks

Regards,

2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
1.
Non-linear loads can take many forms. An incandescent lamp is a simple example. A semiconductor phase-controlled (e.g. triac) resistive load is another. The magnetizing (exciting) current for a power transformer may contain many harmonic components. The list goes on and on .....

2. & 3.
Harmonics will contribute to the circuit losses but may or may not be significant - at the very least harmonic current in circuit conductors will contribute some losses. Harmonics in variable speed motor drive systems might add to magnetic circuit losses - an so on.

Take the rather artificial, but relatively easily analyzed example of a sinusoidally driven circuit having a square wave current in phase with the driving voltage. The displacement factor is 1 and the distortion factor is 0.9 which leads to the power factor = 0.9

For more complex load current waveforms the analysis may not be as straightforward. Try analyzing a sinusoidally driven, phase-controlled thyristor circuit with a purely resistive load - both the displacement and distortion factors are less than unity for firing delays greater than 0°. The analysis usually involves some integration.

Presumably one would always apply the relationship S^2=P^2+Q^2 to find the unknown reactive power Q - since S and P are readily derived.

For arbitrary complex waveforms one would have to make direct measurements where an adequate mathematical analysis is not achievable.

3. almotions Thread Starter Active Member

Feb 6, 2009
46
0
hi tnk,thanks for you explanations
So what you are trying to say is that even when current and voltage is in phase,the PF may not be 1 due to harmonics and thus harmonics draws reactive power even in purely resistive loads?THanks

4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Keeping in mind I have consumed some alcohol and may be a bit foggy ....

A qualified "yes" to your statement. I would prefer to say that, even though the fundamental components of the voltage and current are in phase the pf can be less than unity owing to the presence of harmonics.

5. The Electrician AAC Fanatic!

Oct 9, 2007
2,724
496
If the circuit contains only passive elements, and no capacitors or inductors, then it draws no reactive power.

A search of the web turns up a bunch of sites that don't correctly describe the modern view of Power Factor. A non-linear load, without reactive components, can create a Power Factor less than unity. The modern way of thinking is to consider Power Factor to be composed of a Displacement Factor and a Distortion Factor.

The Displacement Factori is the cosine of the phase shift between the *fundamental* component of the applied voltage and the *fundamental* component of the current. A number of the web sites I found say that it is the angle between the voltage and current, without specifying that it is the fundamental frequency components of voltage and current that should be used. Some PF meters apparently just look at zero crossings of voltage and current and take that to be the Displacement Angle. That is wrong.

The other factor involved is the Distortion Factor, which is the ratio of the RMS fundamental component of current to the RMS total current.

The Power Factor (PF) is given by Displacement Factor * Distortion Factor.

To determine by measurement if you have any reactive power, you could use a scope with trace math to display the instantaneous product of voltage and current. This would be the instantaneous power, and if it ever changes sign, then you have reactive power, otherwise not.

6. almotions Thread Starter Active Member

Feb 6, 2009
46
0
Question,you are talking about linear loads only right? Because in nonlinear loads,PF≠1,and Q=VIsinθ which gives reactive power.

Sorry,i don't get it.Why is taking the zero crossings of voltage and current be wrong?Isn't that taking the angle between the fundamental voltage and current?

I haven't actually use a reactive power meter before.But i'm a bit curious when you said the instantaneous power changes sign,that indicates there is reactive power.Why is that? Isn't the instantaneous power measurement always changing in sign even when measuring active power?
Forgive me for asking too many questions

7. almotions Thread Starter Active Member

Feb 6, 2009
46
0
I'm sorry,ignore the last question...

8. The Electrician AAC Fanatic!

Oct 9, 2007
2,724
496
In what I say here, I'm assuming that the exciting voltage is a sinusoid.

I was talking about passive loads, which means no active devices are involved such as transistors, tubes, opamps, etc.

It is entirely possible for a passive load to be non-linear. A good example would be an MOV transient suppressor. Typical non-linear loads don't draw reactive power, they only create harmonics.

It is possible to have a non-linear capacitor, or more commonly, a non-linear inductor. These would create harmonics, as well as drawing reactive power.

But the typical capacitor won't be non-linear and will draw a sinusoidal current which will have no harmonics, but will constitute reactive power. (If the applied voltage is not a clean sinusoid, but is distorted, a capacitor will draw a current proportional to the derivative of the applied voltage, and the current will contain harmonics. But this won't be due to any non-linearity of the capacitor; it will be due to the distortion of the applied voltage)

And, you can have a combination of all these things.

Because if the load is non-linear and/or reactive, the current will not have the same waveshape as the voltage; it will be non-sinusoidal, and the zero crossings of the current (which will contain harmonics) may not be the same as the zero crossings of the fundamental component of the current.

The Displacement Factor is defined using the fundamental components of the voltage and current.

The instantaneous power isn't always changing sign when there is no reactive power because when the voltage goes negative, so does the current, and their product is therefore always of the same sign, taken by convention to be positive.

I'm attaching 3 scope captures.

The first shows the voltage and current from a transformer with a (linear) 300Ω load.

The voltage is the stepped-down line voltage, and it's somewhat flat-topped, but you can see that the current has the same waveshape as the voltage.

The yellow trace is the voltage, the green trace is the current and the purple trace is the instantaneous product of the two (instantaneous power, in other wods).

The second capture shows the voltage, current and instantaneous power when a 22 volt MOV is added in parallel with the 300Ω resistor. You can see that the current becomes peaky and no longer has the same waveshape as the applied voltage; the MOV is a non-linear resistor.

Notice that the instantaneous power (purple trace) does not go negative; there is no reactive power drawn with this load, but there are harmonics.

The third capture shows the voltage, current and instantaneous power when a 15μF capacitor is connected in parallel with the 300Ω resistor and the 22 volt MOV.

The capacitor draws a current proportional to the derivative of the applied voltage (which is added to the current drawn by the resistor and the MOV; the green trace shows the total current), and you can see some ringing here and there, probably due to the many switching power supplies in people's homes in the big city.

What is important to notice is that the instantaneous power (purple trace) goes negative during part of the cycle. That's how you know that the load is drawing reactive power.

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Last edited: Aug 13, 2009