Mains Harmonics improves with longer dead time in input current waveform

Hello,
The attached shows two slightly different mains input currents to 140W offline lighting products. (Class C products for Mains harmonics)
Strangely, the one with a longer “dead time” gives a better pass of mains harmonic current regulations. (by dead time I mean the time that the current is zero).
Both currents are equally in phase with the mains voltage as you can see.
Why does the one with the longer dead time more closely conform to mains harmonics? Should we make the dead time greater to improve mains harmonics further still?

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The one with longer dead time fails on the 5th harmonic only
The one with the shorter dead time fails on the 5th and 27th harmonics.

You don't tell anything about the involved switcher topology, how should we know what "dead time" means in this regard?

Hello,
Thanks, i see your point, and i will now ask a more poignant question....

This is a closely related post:

Mains harmonic current levels with a five step "stairstep" current


Rearding Class C regulatory limits for mains harmonic currents (EN61000-3-2)…..is it possible for a mains input current to a ~150W offline product to pass Class C regulations when it has no more than five steps in a “stairstep” mains input current waveform?
Attached is the LTspice simulation of a “product” which produces a five stage “stairstep” mains input current waveform. If you run the FFT on the input current waveform, you will see that it fails Class C regulations for Mains harmonic current levels.
(The excel doc shows that it fails)
What I am asking is…… is it ever possible for a five step “stairstep” mains current to ever pass Class C harmonic regulations?

Page 7 of the following shows the Class C mains harmonic current limits….
**broken link removed**

(EN61000-3-2)

Well, ill be forthright and declare that it is impossible for a product that draws a five stage "stairstep" mains input current to pass EN61000-3-2 regulations on mains harmonic input currents.
Here is the proof:

As an addition, the red waveform in the attached scope shot shows the mains input current to a offline 148W Class C product. As you can see, it is about as close as it can get to imitating a sinusoid, -you can see this by looking at the way it “shadows” the equivalent sinusoidal waveform shown here in green.
However, the product with the red waveform fails EN61000-3-2 Class C regulations on Mains Harmonics Current levels…..it fails on the 19th and 21st harmonics(950Hz & 1150Hz).
This is pretty conclusive proof that an offline product that draws input current from the mains with a five step “stairstep” waveform cannot possibly ever pass EN61000-3-2 Mains harmonic current regulations…do you agree?

Admittedly this analysis was done in LTspice but i think LTspice is very accurate with such low frequency analysis as mains harmonics.

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The 19th harmonics is 100% over the limit
The 21st harmonic is 16% over the limit
All other mains harmonics are passes, but that is irrelevant, overall, the product is a fail.

Your observations are correct but the interpretations are not convincing.

A square wave is a combination of sine functions of odd frequency and weights; sum {1 over n {sin(n x)}} where n is odd.

The process is valid except at the discontinuities; we see Gibbs phenomena at these points (you also see in the graphs)- https://en.wikipedia.org/wiki/Gibbs_phenomenon

What we are trying is to make a sine curve out of square waves. How many square waves we will need to get a good approximation?

The fundamental problem is that your square waves are bandwidth limited; they themselves are approximate square waves made up of 5 or 7 or 11 sine waves (depends on the sampling).

If you change the dead time from zero (any finite value) you are essentially having a square wave with sloping sides (trapezoidal) - they are actually better when you have limited bandwidth.

I cannot do the mathematics but you can certainly figure it out by trial and error. Look out for harmonics that are prime (or relatively prime??) multiples...

You have shown that the modeled waveform doesn't meet power quality requirements. But how about the waveform of the actual product?

What's their specification for harmonics? May be they use some kind of filter that reduces at least higher order harmonics?

May be they use some kind of filter that reduces at least higher order harmonics?

Click to expand...

Thanks, The product is made up of "sequential linear current regulators", and it passes EMC without needing a filter, other than a 220nF X2 capacitor.

You have shown that the modeled waveform doesn't meet power quality requirements. But how about the waveform of the actual product?

Click to expand...

Thanks, if this idealised simulation waveform can not be made to pass the regulations, then we are thinking that the actual product will never be able to pass the harmonic regulations. The actual product has a similar waveform, though to be honest, its not quite as “neat” as that.

What's their specification for harmonics?

Click to expand...

Thanks, we don’t have a particular spec on that, other than we want to try and pass the regulations on harmonics.
I believe a five step “stairstep” waveform simply cannot pass Class C regulations on harmonics, do you agree?

Hi,

A 5 step voltage waveform may create a lot of harmonic currents. It is likely that it will not pass regulations.

But you say current regulator. This may introduce less harmonics, because the current is limited... and therfore the influence on voltage waveform is limited.

Klaus

Thanks, the thing is the mains input current waveform in #4 above fails mains harmonics (class c regs).
As you can see, it is as close as it could be to a sinusoid for a 5 step waveform.
...and yet it still fails...this surely is conclusive proof that a 5 step current waveform cannot ever pass mains harmonic regs to class c?

Hi,

I just simulated a 5 step calculated sine wave.
The FFT shows those results.


Does this meet your results?

Klaus

Thanks Klaus, all your results are less than 3%, so yes, you have a pass, but then i dont understand how the waveform of #4 above is a fail?

The one in #4 is as sinusoidal as its possible and yet it fails.

Hi,

I see a difference: My stepsize in Y is equal.

Klaus

As far as I understood, the discussion is about the harmonics generated by a real LED driver circuit similar to DT3001, not a hypothetical stair waveform.

In the real circuit, the transition between current steps is rounded by LED I/V characteristic. Thus I prefer to believe that none of the presented simulations with time controlled switched current is modeling the actual application waveform correctly.

As far as I understood, the discussion is about the harmonics generated by a real LED driver circuit similar to DT3001, not a hypothetical stair waveform.
In the real circuit, the transition between current steps is rounded by LED I/V characteristic. Thus I prefer to believe that none of the presented simulations with time controlled switched current is modeling the actual application waveform correctly.

Click to expand...

Thanks, yes you are right, the transition in the DT3001 based circuit is slightly rounded, however, that rounding mainly affects frequencies which are well above the mains harmonic frequencies, which are just up to 2kHz. –And in the simulation of the idealised stairstep waveform in #4 above, the steps are not abrupt , but happen as a slope over 100us. So I think the simulation from the top post, which kind of led to the waveform of #4, is a good way to look at the situation.
The simulation, the actual one, is attached here.