Switch-mode power supplies and Current Harmonics

[permanent1]

Ok, a room containing about thirty pcs is connected into the distribution panel of a building. Voltage harmonic distortion averages at approx 4% and the current distortion in the neutral is above 100% at full load. One of the current phases lies idle so distortion is neglected for this phase. The current wave-form displays the large 3rd and 5th harmonic frequencies created by Switch Mode Power Supplies.
Firstly: In the larger scheme can all eventualites be covered by specifying the neutral conductor to be the same as the phase conductors ie triple-n problems?
Secondly: What other issues should i be looking for in the context of this room as part of a larger building?
Thirdly: Is there a rule of thumb to take into account SMPS when designing distribution based on this loading?
Thanks All!!!

 

[davidbeach]

For loads that are largely SMPS, you can see neutral currents in the range of 150-200% of phase currents and would be well served by having the neutral sized at 200% of the phase conductors. If you have several rooms of this type, and your facility is served at a higher voltage, you might want to consider harmonic mitigation transformers to introduce a 30% phase shift on half your harmonic loads to cancel much of the 5th and 7th harmonics.

 

[itsmoked]

Hey David,
How can you exceed one phase in the neutral? Can you give me a napkin description?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[permanent1]

itsmoked: From reading around the topic combinations of the 3rd, 6th and the 9th Current Harmonic known as the Triple-n Harmonics can add in phase thus producing large currents in the neutral.

Ok so my next question is:
Has anyone come across a Current Harmonic problem that manifests itself as a 200% neutral current with regard to the phase current and needed to find a solution to correct the problem? A solution other than oversized neutral conductors?
If so what was the best solution for the problem and how did you go about solving the problem?
Thanks All!!!

 

[jghrist]

If one of the phases is idle, and the other two phases are equal, then the neutral current would equal the phase current even without any harmonics.

 

[davidbeach]

Looking at just the fundamental, the third harmonic and the ninth harmonic - and the same current in each phase except shifted by 120 degrees - you can get some interesting results. If you have cos([θ])+1/3*cos(3*[θ])+1/9*cos(9*[θ]) for the phases, the rms value of the neutral current is essentially the same as the rms value of the phases (it's lower by the missing 15th, 21st, 27th, etc.). If you raise the third and ninth to cos([θ])+2/3*cos(3*[θ])+4/9*cos(9*[θ]) the rms of the neutral is now 1.876 times the rms of the phase.

 

[ozmosis]

If you are looking for a rule of thumb guideline, I once read somewhere that a typical PC SMPS is capable of producing up to 1amp of current onto the neutral

 

[itsmoked]

Thanks folks.

permanent1;
How about just pulling two neutrals? That way all of the wire is the same size no odd man out.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[waross]

Hi itsmoked;
In theory, fine.
In practice, I am always concerned that loads are connected to the proper neutral whenever there is more than one neutral in a junction box.
There is less labour pulling one wire than two wires even if it is larger. Also, there will be more lost time explaining to the workers why we aren't doing it like we always do it.
That said, there have been a few special cases when I have pulled two neutrals. There is a reason, but it is economic and not technical. You can do it if you think the situation warrants it.
Quiz Keith, Single phase, two 120 volt loads, shared neutral.
Currents L1, 9 amps, L2, 7 amps, Neutral 6 amps. Why?
(No harmonics)
respectfully

 

[davidbeach]

One neutral per phase conductor works well for the branch circuits, but once you have collected all those branch circuits in a panel, the neutral to that panel will have to be over sized to accommodate the additional current. You will also have 4 current carrying conductors in the conduit and need to account for that with derating (80% per NEC).

 

[itsmoked]

warosswaross; Why? Because the two loads have different power factors. This causes them to have more than an algebraic difference which would've been the obvious 2A thru the neutral.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[sitl]

First of all, you should distribute your load evenly through all three phases, since the unbalance in your case is causing unnecessary load in neutral conductor.

Second, the reason why triplen harmonics (3rd, 9th, 15th etc, since even ones are negligible) are always endangering neutral conductor is that the triplen-harmonic waveforms of three 120o phase-shifted fundamental-frequency waveforms are actually in phase with each other. The phase shift figure of 120o generally assumed in three-phase AC systems applies only to the fundamental frequencies, not to their triplen harmonic multiples. Therefore all triplen harmonics add with each other in the neutral conductor of a 4-wire Y-connected load.

There are ways of rectifying this situation. The best thing to do is to feed non-linear loads independently of the linear loads and also try to contain the harmonics as low in the distribution system as possible (as low means as close to the non linear loads as possible). This is done by using transformers that feed those loads with primary connected in delta, or using passive or active harmonic filters, etc. Triplen harmonics propagation upstream of the network is in this way stopped. Here one must bear in mind that transformer could be overloaded due to the losses generated by triplen harmonics that are trapped in his delta winding. Also the neutral conductor from this transformer to the non-linear load(s) is still at risk from triplen harmonics, and therefore must be adequately sized.

 

[waross]

itsmoked
exactly
This is a situation that is not anticipated in the NEC or CEC in regards to the number of current carrying conductors in a raceway or cable. Fortunately it is not that common a situation, and the loads tend to be less than the maximum for the circuits, reducing the heating effect.
respectfully

 

[RalphChristie]

Little of track from the original question, but in addition to sitl's reply,

Download this video clip at:

http://www.brainfiller.com/documents/HarmonicSample.wmv

(It is a rather big file)

The guy (Jim Phillips) explains the harmonic issue quite good.





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[itsmoked]

Thanks waross. I can see why it isn't in the NEC as I took a look at solving the exact equations it got rather ugly quickly.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com