Triplen harmonics in the neutral
The key idea
A balanced three-phase load returns no current through its neutral. That property is one of the main reasons for three-phase systems. It is not true at the 3rd harmonic. The 3rd, 9th and 15th arrive in step in all three phases, and they add instead of cancel. A perfectly balanced group of electronic loads can therefore run its neutral hotter than any of its phases.
The idea
Every phase carries the same current waveform. Each phase starts a third of a cycle after the phase before it. That delay makes the three currents sum to zero at the star point, so a balanced load needs no return conductor. Harmonics do not change the delay. Phase B is still phase A one third of a cycle later. Harmonics change the effect of that delay.
A third of a cycle is 120° of the fundamental. It is 360° of the 3rd harmonic, because the 3rd completes three cycles in the time the fundamental completes one. A rotation of 360° returns a wave to its start point. The 3rd harmonic of phase B therefore has no displacement from the 3rd harmonic of phase A. All three phases carry the same 3rd-harmonic wave.
Now add the three currents in the neutral. The neutral carries the current that the three phases do not cancel. The fundamentals are 120° apart and sum to zero, as always. The three 3rd harmonics are identical, so they sum to three times one of them. The result is a clean 150 Hz current on a 50 Hz system, or 180 Hz on a 60 Hz system. The designer sized that conductor for a current near zero.
The same arithmetic gives the general rule. The h-th harmonic of phase B has a displacement of h × 120°. Orders 1, 4, 7 and 10 leave a remainder of 1 after division by three. These orders put the three phases 120° apart in ABC order, which is a positive-sequence set. Orders 2, 5, 8 and 11 leave a remainder of 2, and they put the phases 120° apart in the reverse order. That set is negative sequence.
Both sequences are three equal phasors at even angles, so both sum to zero in the neutral. Only the multiples of three add together. These orders are the triplen harmonics. They form a zero-sequence set: three phasors that point the same way.
This rule would be only a curiosity if no load produced much 3rd harmonic. Single-phase rectifier loads produce a large 3rd harmonic. The switch-mode supply in every computer, LED driver and charger draws current in a short pulse near the voltage peak. The 3rd is the largest harmonic in that pulse, and it is commonly 30% to 70% of the fundamental. A building with many of these loads has many triplen sources. All of these sources are in phase with each other.
Try it
Increase the 3rd harmonic and watch the neutral waveform grow. Then increase the 5th harmonic to its maximum. The neutral waveform does not change at all.
neutral rms = 90% · fundamental + 5th cancel · 3rd adds ×3
| Order | In each phase | Sequence | In the neutral |
|---|---|---|---|
| fundamental | 100% | positive | cancels: the three sum to zero |
| 3th | 30% | zero | adds ×3: 90% of the fundamental |
| 5th | 20% | negative | cancels: the three sum to zero |
The three phases here are perfectly balanced. They have the same load and the same spectrum, and only the third-of-a-cycle delay separates them. The fundamental and the 5th cancel in the neutral at any size. The 3rd is identical in all three phases. The neutral therefore carries a clean 150 Hz wave three times the size of the 3rd harmonic of one phase. Real loads are never exactly balanced, and that unbalance adds its own fundamental-frequency current to the neutral.
Why it matters
- The neutral can be the most heavily loaded conductor in the cable. The old guideline that a neutral may be smaller than the phases comes from a time of linear loads. With a 33% 3rd harmonic, the neutral already carries about 90% of the fundamental phase current. The neutral also has no overcurrent device on it in most installations.
- A balanced load does not remove this current. An equal load on each phase corrects every other neutral problem. It does not correct this one, because triplen harmonics add in phase whether or not the phases are balanced. Real load unbalance adds its own fundamental-frequency current to the triplen current.
- Transformers see triplen current as zero sequence. A delta winding forms a closed loop. The triplen current circulates in that loop instead of a path upstream. A delta–wye transformer therefore protects the network above it. The same circulating current also heats the winding that contains it.
- Triplen current also produces voltage distortion. Triplen current in the neutral and the earthing impedance produces a 3rd-harmonic voltage. That voltage moves the star point. It also adds to the total distortion that every load on the bus receives.
The math, if you want itOptional — the page reads completely without it
Phase B carries the waveform of phase A with a delay of a third of a cycle. The h-th harmonic term therefore has a displacement of h × 120°. At the 3rd harmonic, that displacement goes to zero:
a third of a cycle, seen by the 3rd harmonic
sin( 3(θ − 120°) ) = sin( 3θ − 360° ) = sin( 3θ )
Only the remainder of h ÷ 3 decides the behavior of a set of three. Every order therefore belongs to one of three families:
sequence of the h-th harmonic
h mod 3 = 1 → positive · = 2 → negative · = 0 → zero (triplen)
A positive-sequence set and a negative-sequence set are three equal phasors 120° apart. Each set sums to zero at the star point at any size. A zero-sequence set is three identical phasors, so it sums to three times one phasor. With balanced phases, the neutral therefore carries the triplen harmonics and nothing else. The orders are orthogonal, so their RMS values add in squares:
neutral rms, balanced phases
IN = √( Σh = 3, 9, 15… (3·Îh)²2 )
Îh is the peak of the h-th harmonic in one phase. The result depends on two assumptions. The three phases are balanced, and they have the same spectrum. Both assumptions are close to true for a group of identical rectifier loads, which also have the largest 3rd harmonic. The widget models orders 1, 3 and 5 only. The 9th and the 15th would raise the neutral current further, and they would never lower it.
See it in Phasor
A harmonic penetration study in Phasor carries each order through the network with its own sequence. Triplen current therefore follows the zero-sequence path, down earthed star points and around delta windings. The 5th and the 7th take a different route. Neutral conductors and earthing conductors are part of the model. The study therefore reports the current that the neutral carries, so you do not have to assume that it is empty.