Harmonic aggregation
The key idea
A harmonic source injects a current with a magnitude and an angle. Two drives inject 30 A of 5th harmonic into the same bus. They make 60 A only if their angles agree. If the angles are opposite, the two currents cancel and the bus receives nothing.
The idea
Most lessons introduce harmonics one source at a time. A drive draws a pulsed current, so it injects a 5th, a 7th and an 11th. Put several of those sources on one bus and a new question appears. A single-source picture never raises it. The question is how the injections combine.
They do not add as amps. At a given order, each source's current is a phasor. The phasor is 30 A at some angle, and the angle gives the point in the 250 Hz cycle where the pulse occurs. Currents that meet at a bus obey Kirchhoff's law, and Kirchhoff's law adds phasors.
So two 30 A fifth-harmonic sources give 60 A when they are in phase. They give zero when they are in opposition, and about 42 A when they are 90° apart. This is the same right-angle arithmetic that governs THD. Here you apply it across sources instead of across orders.
The diversity factor compares the phasor sum with the arithmetic sum of the magnitudes. Two identical sources in phase give 1.0. At 90° they give 0.71, and in opposition they give 0. Real plants give a value between these limits, and no one designs that value. The angles come from the load on each drive, the length of its cable and the firing time of its converter. Nobody sets the angles, and the angles change during the day.
The aggregate current then becomes a voltage problem. The total current on the bus flows into the harmonic impedance of the system upstream. The voltage drop across that impedance is the distortion that every other customer on the bus receives. For an inductive system the impedance increases with frequency. Therefore the same current causes more distortion at a higher order.
Try it
The widget shows two drives, one bus and one harmonic order. Hold both currents constant and move only the angle. The amps do not change, but the bus current and the distortion change from the full sum to zero.
V₅ at the bus: 3.00% · typical per-order limit 3%
Vector sum
60.0 A
at 0°
Arithmetic sum
60 A
the sum of the two nameplate values
Diversity factor
1.000
vector sum ÷ arithmetic sum
V₅ at the bus
3.00%
12.0% of rated current × 5 × 0.05 pu
- 1 drive 1
- 2 drive 2, dashed as a wave
- sum what the bus carries
Both drives inject at the 5th on a bus rated 500 A. The system behind the bus is a pure inductance of 0.05 pu, so the distortion is V₅% = I₅ (% of rated) × 5 × 0.05. Move the angle and the amps do not change, but the bus current falls from 60 A to zero. A planner therefore cannot add nameplate figures to find the harmonic current. The answer depends on angles that nobody controls, and those angles change with the load, the cable lengths and the converter firing.
Why it matters
- Straight addition overstates the problem and the cure. If you size a filter for the arithmetic sum, you buy equipment for a current that the bus may never carry. For this reason, standards use diversity exponents instead of plain sums.
- Cancellation is a real design tool. Phase-shifting transformers cancel harmonics on purpose. Feed one drive through a delta winding and a second drive through a wye winding. The 30° shift puts their 5th and 7th harmonics in opposition. A 12-pulse arrangement does exactly this.
- Diversity is temporary. The angles that cancel today depend on the load. A plant that measures clean at part load can exceed a limit when one drive trips or a process changes. A study that assumed diversity then has no margin.
- The voltage answer needs the impedance too. The same aggregate current is harmless on a stiff bus and severe on a weak bus. The result is worse if a resonance occurs at the order that the sources inject.
The math, if you want itOptional — the page reads completely without it
At one order, the aggregate is the complex sum of the source currents:
phasor sum at order h
Ih = Σk Ik ∠ θk
Two sources reduce to the cosine rule. At 90° the cross term becomes zero and gives the right-triangle result. 30 A and 30 A make 42.4 A, not 60 A:
two sources at right angles
Ih = √( I₁² + I₂² ) = 42.4 A
The widget divides that result by the arithmetic sum to get the diversity factor:
diversity factor
kdiv = IhΣk |Ik|
Nobody knows the real angles. Therefore standards replace the phasor sum with a summation law. The law gives a result between the sum of the magnitudes and the sum of the squares. Each order has its own exponent. The exponent is 1 at the low orders, where the angles usually agree. It is approximately 1.4 for the 5th and the orders near it, and 2 at the high orders, where the angles are random:
summation with a diversity exponent
Ih = ( Σk Ikα )1/α
The exponent is a compromise, not a law of physics. α = 1 gives plain addition and α = 2 gives the root-sum-square. The values between them show how much cancellation experience permits.
The aggregate current then meets the system impedance. If the impedance is purely inductive, the reactance increases with the order:
voltage distortion at the bus
Vh = Ih · h · X₁
The factor h makes a high order more severe than its current alone suggests. A 13th-harmonic current makes more than twice the voltage of a 5th-harmonic current of the same size. A small injection high in the spectrum can therefore matter as much as the obvious injection at the 5th.
See it in Phasor
Phasor's harmonic penetration study takes the spectrum and the angles of each source. It sums them at every bus and reports the voltage distortion order by order. The angles are inputs, not assumptions. You can therefore run the case where the diversity is not present. Switch one drive off, or align every source, and see the result at the bus.