Harmonic aggregation
The key idea
A harmonic source injects a current with a magnitude and an angle. Two drives each pushing 30 A of 5th harmonic into the same bus make 60 A only if their angles agree. If the angles are opposite, the currents cancel and the bus receives nothing at all.
The idea
Most lessons introduce harmonics one source at a time: a drive draws a pulsed current, so it injects a 5th, a 7th, an 11th. Put several of those sources on one bus and a question appears that the single-source picture never raises: how do the injections combine?
Not as amps. At a given order, each source's current is a phasor: 30 A at some angle, the angle marking where in the 250 Hz cycle its pulse lands. Currents meeting 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, zero when they are in opposition, and about 42 A when they are 90° apart. It is the same right-angle arithmetic that governs THD, applied across sources instead of across orders.
The diversity factor compares the phasor sum with the plain arithmetic sum of the magnitudes: 1.0 for two identical sources in phase, 0.71 at 90°, 0 in opposition. Real plants land somewhere between — and nobody designs that value. The angles come from the load on each drive, the length of its cable, the firing time of its converter. Nobody sets them, and they drift through the day.
The aggregate current then becomes a voltage problem. Whatever total survives the summation flows into the harmonic impedance of the system upstream, and the voltage drop across that impedance is the distortion every other customer on the bus receives. For an inductive system the impedance grows with frequency, so the same current does more damage at a higher order.
Try it
Two drives, one bus, one harmonic order. Hold both currents constant and move only the angle: the amps never change, but the bus current and the distortion swing 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 both the problem and the cure. Size a filter for the arithmetic sum and you are buying equipment for a current the bus may never carry, which is why 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 another through a wye winding, and 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, and a study that leaned on diversity is left with no margin.
- The voltage answer needs the impedance too. The same aggregate current is harmless on a stiff bus and severe on a weak one, and worse still if a resonance sits at the order 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 vanishes and leaves 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|
In practice nobody knows the real angles, so standards replace the phasor sum with a summation law that lands between the sum of magnitudes and the sum of squares, with its own exponent per order. The exponent is 1 at the low orders, where the angles usually agree; about 1.4 for the 5th and its neighbors; and 2 at the high orders, where the angles are effectively random:
summation with a diversity exponent
Ih = ( Σk Ikα )1/α
The exponent is a compromise, not a law of physics: α = 1 is plain addition, α = 2 is root-sum-square, and the values between record how much cancellation experience permits.
The aggregate current then meets the system impedance. For a purely inductive system, the reactance grows with the order:
voltage distortion at the bus
Vh = Ih · h · X₁
That factor of 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 the same size. A small injection high in the spectrum can matter as much as the obvious one at the 5th.
See it in Phasor
Phasor's harmonic penetration study takes the spectrum and the angles of each source, sums them at every bus, and reports the voltage distortion order by order. The angles are inputs, not assumptions, so you can run the case where the diversity is not there. Switch one drive off, or align every source, and see what the bus receives.