Flatened top sine wave, does it mean anything?

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120118-0927 EST

T.M.:

How can you provide a solution when at this point in time the cause of the problem, failed components, is not known.

Flat topped voltage waveforms almost certainly are not the cause of electronic ballast failures.

The failure of phase shifting capacitors on induction motors also seems unlikely from a flat top voltage. These capacitors are in series with an inductor and thus a series resonant circuit. The tuning of this circuit relative to harmonics in the excitation would be of concern. We don't know what this frequency is.

I have already tried to explain to you that in the United States the part of the grid supplying the final distribution transformers has a flat top in many areas. Thus, changing the pole transformer is not going to eliminate the flat top.

.
 
gar said:
120118-0927 EST

T.M.:

How can you provide a solution when at this point in time the cause of the problem, failed components, is not known.

Flat topped voltage waveforms almost certainly are not the cause of electronic ballast failures.

The failure of phase shifting capacitors on induction motors also seems unlikely from a flat top voltage. These capacitors are in series with an inductor and thus a series resonant circuit. The tuning of this circuit relative to harmonics in the excitation would be of concern. We don't know what this frequency is.

I have already tried to explain to you that in the United States the part of the grid supplying the final distribution transformers has a flat top in many areas. Thus, changing the pole transformer is not going to eliminate the flat top.

.
Click to expand...

Areas where there are heavy power electronic consumers almost always experience bad quality voltages. This problem cannot be solved by installing capacitors. In my other life, the utility (I was with) resolved a similar problem by inspecting and measuring voltage and current waveforms at the different installations in the area that are using non-linear loads. The utility required each power customer to install filters at each load they have and the voltage quality had since improved in the area. Unless you can prove that the power feed itself is dirty, my story can help this specific condition like it helped us in my other life. The only difference is that the poster is a customer himself, not the utility.
 
120118-0948 EST

iwire:

To continue on transient analysis.

Thru observation, making assumptions, and testing theories it has become possible to fairly accurately describe electrical circuits with mathematical models.

From the past history of this analysis we can say that the instantaneous sum of the voltages around any closed loop is zero. We can also say that at any node the instantaneous sum of the currents is zero.

Now to the series RC circuit I described to you in post #44. 10 V, switch, 1 megohm, 1 ufd.

The instantaneous voltage equation is

Vsource = vr + vc

What is the voltage across a capacitor?

vc = q / C where q is the charge on the capacitor, and C is the capacitance of the capacitor.

What is the voltage across a resistor?

vr = i * R , but i is how fast charge is moving, rate of change, speed. Using calculus i = dq/dt .

Substituting into the above loop equation the result is

Vsource = R * dq/dt + q / C

Now we have what is called a differential equation.

After doing a lot of additional math, and assuming the initial charge on the capacitor is 0 at t=0, and that the switch closes at t=0, then the result is

i = (Vsource/R) * e(-t/R*C) where e = 2.71828

when t = 0 the current is (10/106)*1 any number to the 0th power is 1.
when t = R*C, in this case 1 second, then the current is 2.71828 to the -1 times the initial current, or 36.8% of the original current. At 2 time constants, RC is called the time constant, the current is 36.8% times 36.8% or 13.5% of the initial current.

Intuitively and by what you know about a capacitor it should be obvious that the initial current after switch closure is V/R.

More comment sometime later.

.
 
120118-1425 EST

iwire:

If at any point in the transient discussion what I am saying is not clear, then please ask questions.

If anyone wants to draw the circuits for display that would be useful.

.
 
Another waveform presentation

Another waveform presentation

I've really enjoyed and learned from Besoeker and Gar comments and analyses. I do data acquisition and analysis, in particular to look at flows and pressures in hydraulic systems. I decided to use my equipment and software to take a look at the power in my house.

To do this, I used a "Wall-Wart" with AC output, pretending it was, if you will, a Potential Transformer or PT, to get a level I could monitor. It was still a little high, so a pair of resistors divided it down to under 20V P-P. I had the software do Fourier analysis to find frequencies and maybe relative levels ... I'm VERY inexperienced with the Fourier stuff.

At any rate, it is very interesting that there are no even harmonics seen, and odd harmonics extend up to at least the 19th. All this from a "sine" (not very<g>) wave. I'm posting about 2 cycles as a jpg, and a pdf with the analysis and some comments.

I frankly was surprised at how much harmonic content was present ... my data application allows calculation of THD (Total Harmonic Distortion) but I doubt the -25dB I got in one try, not knowing what I'm doing. I calculate that -25dB would be in the 0.3% range ... and I don't think the signal is that good.

While this is very much a DIY project, it is not, as I understand it, one that gives or asks advice, and should be interesting as another point of reference.

Oh, Duke Energy, residential, 2 homes on a pad mounted transformer marked with a big 25 (assume kVA), 200A service.
 

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GeorgeB said:
I've really enjoyed and learned from Besoeker and Gar comments and analyses. I do data acquisition and analysis, in particular to look at flows and pressures in hydraulic systems. I decided to use my equipment and software to take a look at the power in my house.

To do this, I used a "Wall-Wart" with AC output, pretending it was, if you will, a Potential Transformer or PT, to get a level I could monitor. It was still a little high, so a pair of resistors divided it down to under 20V P-P. I had the software do Fourier analysis to find frequencies and maybe relative levels ... I'm VERY inexperienced with the Fourier stuff.

At any rate, it is very interesting that there are no even harmonics seen, and odd harmonics extend up to at least the 19th. All this from a "sine" (not very<g>) wave. I'm posting about 2 cycles as a jpg, and a pdf with the analysis and some comments.

I frankly was surprised at how much harmonic content was present ... my data application allows calculation of THD (Total Harmonic Distortion) but I doubt the -25dB I got in one try, not knowing what I'm doing. I calculate that -25dB would be in the 0.3% range ... and I don't think the signal is that good.

While this is very much a DIY project, it is not, as I understand it, one that gives or asks advice, and should be interesting as another point of reference.

Oh, Duke Energy, residential, 2 homes on a pad mounted transformer marked with a big 25 (assume kVA), 200A service.
Click to expand...

Nice post, sir.
I take it as a great compliment that you have learned from myself and Gar.

The waveform you posted is remarkably similar to that in post#1.
The issue of harmonic voltage distortion has come more into prominence in recent years. But it isn't quite a recent phenomenon.
Mercury arc rectifiers were an early form of conversion from AC to DC. There is a paper by a Dr Reade from about 1948 that gives guidance on the matter of harmonic voltage distortion and figures that are still relevant today.
The sources change. The physics doesn't.
 
steve66 said:
Cool. What combination of harmonics did it take to get that?
Click to expand...
Mostly third and ninth.
I popped in a little bit of higher orders to replicate the noise but, TBH, I'm not sure if it was noise or the limited resolution in the OP.
:)
 
I, like GeorgeB, am enjoying this subject. Other than a few posts that have been a bit defensive in nature this topic has been one of the best if not the best one I?ve come across in a long time. Without knowing much about the distribution system of the OP and doing an actual power system evaluation study all that can be done is to make assumptions based upon the wave form as provided.
 
templdl said:
I, like GeorgeB, am enjoying this subject. Other than a few posts that have been a bit defensive in nature this topic has been one of the best if not the best one I?ve come across in a long time. Without knowing much about the distribution system of the OP and doing an actual power system evaluation study all that can be done is to make assumptions based upon the wave form as provided.
Click to expand...
You make an important point if I may say so.
A huge wealth of knowledge is shared on this forum.
That some chose to dispute it is the way of the world.
 
gar said:
i = (Vsource/R) * e(-t/R*C) where e = 2.71828
Click to expand...

Off topic a little.....and stuff I'm sure you know.


e is an important number in mathematics and the base for natural or Napierian logarithms (and part of Euler's Identity).
Napierian after John Napier of Merchiston, a Scottish mathematician credited with the invention of logarithms and the decimal point.
Merchiston is an area of Edinburgh and was home to Napier University which was built round what remained of Merchiston castle.
Fond memories of the place.......
 
gar said:
120118-0927 EST
We don't know what this frequency is.
Click to expand...

It should first be ascertained if the total building load is lagging or leading.If it is leading,something is wrong.There can be resonance with one of the harmonic frequencies in the incoming supply voltage.The resonant harmonic frequency can be determined and its presence in the incoming supply voltage can be checked.Once confirmed,suitable action may be taken such as providing a harmonic filter etc.,


gar said:
120118-0927 EST
I have already tried to explain to you that in the United States the part of the grid supplying the final distribution transformers has a flat top in many areas. Thus, changing the pole transformer is not going to eliminate the flat top.
Click to expand...

As you are fond of doing calculations to prove your point,I wonder why you are not doing that here.For that purpose, assume that not only load on the secondary side but also on the primary side of the transformer is reduced.You are invited to calculate if that produces any change in the voltage wave shape or not.
 
iwire said:
Most of it but that is a problem with me and not your explanations.:)
Click to expand...
IMHO,you need not understand any derivation of results.If you somehow understand the results to be able to apply in every day activity,that is more than enough............
 
iwire said:
Please, don't ruin this thread. Considering I am the OP I will have no problem just closing it.
Click to expand...

I was just trying to help you to solve the problem you mentioned in post # 6.
 
120119-0932 EST

iwire:

As a means of communication I need to know where you have questions, and then I can try to provide some way to clarify to you what I am saying.

For the moment I will continue on the simple series RC circuit. This approach is iterative without calculus.

From the calculus derived equation we expect the current to decrease to about 37% of its initial value in 1 time constant (1 second here). If we assumed a constant current to the capacitor of 10/1,000,000 or 10-5, then in 1 second the charge transferred to the capacitor would be 10-5. So using the equation for capacitor voltage in terms of capacitance and charge on the capacitor, v = q/C, we get a voltage at the end of 1 second of v1 sec = 10-5/10-6 = 1/10-1 = 10 V.

Thus, with a constant current to the capacitor equal to the initial current the capacitor is fully charged in 1 second.

But with the simple series RC circuit connected to a constant voltage the current to the capacitor constantly changes in a lowering direction as the capacitor charges because the capacitor voltage is increasing and thus the voltage across the resistor is constantly decreasing.

Suppose you calculate what happens in the first 1/100 second, then use the new resistor voltage (10 V - new capacitor voltage) to calculate a new current for the next 1/100 second, and continue this process until you reach 1 second. For each new 1/100 second less charge is transferred to the capacitor. The capacitor voltage with this method of calculation will come much closer to that predicted by the exponential equation. Effectively calculus provides a means to accomplish reducing the the time increment to essentially zero, and not require a very large number of individual calculations. With our a priori knowledge of what calculus can do we believe the previous calculus derived equation as valid for predicting what happens in the simple RC circuit.

.
 
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