Harmonics in polyphase power systems
In the chapter on mixed-frequency signals,
we explored the concept of harmonics in AC systems:
frequencies that are integer multiples of the fundamental
source frequency. With AC power systems where the source
voltage waveform coming from an AC generator (alternator) is
supposed to be a single-frequency sine wave, undistorted,
there should be no harmonic content . . . ideally.
This would be true were it not for
nonlinear components. Nonlinear components draw current
disproportionately with respect to the source voltage,
causing non-sinusoidal current waveforms. Examples of
nonlinear components include gas-discharge lamps,
semiconductor power-control devices (diodes, transistors,
SCRs, TRIACs), transformers (primary winding magnetization
current is usually non-sinusoidal due to the B/H saturation
curve of the core), and electric motors (again, when
magnetic fields within the motor's core operate near
saturation levels). Even incandescent lamps generate
slightly nonsinusoidal currents, as the filament resistance
changes throughout the cycle due to rapid fluctuations in
temperature. As we learned in the mixed-frequency chapter,
any distortion of an otherwise sine-wave shaped
waveform constitutes the presence of harmonic frequencies.
When the nonsinusoidal waveform in question
is symmetrical above and below its average centerline, the
harmonic frequencies will be odd integer multiples of the
fundamental source frequency only, with no even integer
multiples. Most nonlinear loads produce current waveforms
like this, and so even-numbered harmonics (2nd, 4th, 6th,
8th, 10th, 12th, etc.) are absent or only minimally present
in most AC power systems.
Examples of symmetrical waveforms -- odd
harmonics only:
Examples of nonsymmetrical waveforms --
even harmonics present:
Even though half of the possible harmonic
frequencies are eliminated by the typically symmetrical
distortion of nonlinear loads, the odd harmonics can still
cause problems. Some of these problems are general to all
power systems, single-phase or otherwise. Transformer
overheating due to eddy current losses, for example, can
occur in any AC power system where there is
significant harmonic content. However, there are some
problems caused by harmonic currents that are specific to
polyphase power systems, and it is these problems to which
this section is specifically devoted.
It is helpful to be able to simulate
nonlinear loads in SPICE so as to avoid a lot of complex
mathematics and obtain a more intuitive understanding of
harmonic effects. First, we'll begin our simulation with a
very simple AC circuit: a single sine-wave voltage source
with a purely linear load and all associated resistances:
The Rsource and Rline
resistances in this circuit do more than just mimic the real
world: they also provide convenient shunt resistances for
measuring currents in the SPICE simulation: by reading
voltage across a 1 Ω resistance, you obtain a direct
indication of current through it, since E = IR.
A SPICE simulation of this circuit with
Fourier analysis on the voltage measured across Rline
should show us the harmonic content of this circuit's line
current. Being completely linear in nature, we should expect
no harmonics other than the 1st (fundamental) of 60 Hz,
assuming a 60 Hz source:
linear load simulation
vsource 1 0 sin(0 120 60 0 0)
rsource 1 2 1
rline 2 3 1
rload 3 0 1k
.options itl5=0
.tran 0.5m 30m 0 1u
.plot tran v(2,3)
.four 60 v(2,3)
.end
fourier components of transient response v(2,3)
dc component = 4.028E-12
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.198E-01 1.000000 -72.000 0.000
2 1.200E+02 5.793E-12 0.000000 51.122 123.122
3 1.800E+02 7.407E-12 0.000000 -34.624 37.376
4 2.400E+02 9.056E-12 0.000000 4.267 76.267
5 3.000E+02 1.651E-11 0.000000 -83.461 -11.461
6 3.600E+02 3.931E-11 0.000000 36.399 108.399
7 4.200E+02 2.338E-11 0.000000 -41.343 30.657
8 4.800E+02 4.716E-11 0.000000 53.324 125.324
9 5.400E+02 3.453E-11 0.000000 21.691 93.691
total harmonic distortion = 0.000000 percent
A .plot command appears in the
SPICE netlist, and normally this would result in a sine-wave
graph output. In this case, however, I've purposely omitted
the waveform display for brevity's sake -- the .plot
command is in the netlist simply to satisfy a quirk of
SPICE's Fourier transform function.
No discrete Fourier transform is perfect,
and so we see very small harmonic currents indicated (in the
pico-amp range!) for all frequencies up to the 9th harmonic,
which is as far as SPICE goes in performing Fourier
analysis. We show 0.1198 amps (1.198E-01) for the "fourier
component" of the 1st harmonic, or the fundamental
frequency, which is our expected load current: about 120 mA,
given a source voltage of 120 volts and a load resistance of
1 kΩ.
Next, I'd like to simulate a nonlinear load
so as to generate harmonic currents. This can be done in two
fundamentally different ways. One way is to design a load
using nonlinear components such as diodes or other
semiconductor devices which as easy to simulate with SPICE.
Another is to add some AC current sources in parallel with
the load resistor. The latter method is often preferred by
engineers for simulating harmonics, since current sources of
known value lend themselves better to mathematical network
analysis than components with highly complex response
characteristics. Since we're letting SPICE do all the math
work, the complexity of a semiconductor component would
cause no trouble for us, but since current sources can be
fine-tuned to produce any arbitrary amount of current (a
convenient feature), I'll choose the latter approach:
Nonlinear load simulation
vsource 1 0 sin(0 120 60 0 0)
rsource 1 2 1
rline 2 3 1
rload 3 0 1k
i3har 3 0 sin(0 50m 180 0 0)
.options itl5=0
.tran 0.5m 30m 0 1u
.plot tran v(2,3)
.four 60 v(2,3)
.end
In this circuit, we have a current source of
50 mA magnitude and a frequency of 180 Hz, which is three
times the source frequency of 60 Hz. Connected in parallel
with the 1 kΩ load resistor, its current will add with the
resistor's to make a nonsinusoidal total line current. I'll
show the waveform plot here just so you can see the effects
of this 3rd-harmonic current on the total current, which
would ordinarily be a plain sine wave:
time v(2,3)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 0.000E+00 . . * . .
5.000E-04 4.918E-02 . . . * . .
1.000E-03 8.924E-02 . . . * . .
1.500E-03 1.137E-01 . . . . * .
2.000E-03 1.204E-01 . . . . * .
2.500E-03 1.123E-01 . . . . * .
3.000E-03 9.595E-02 . . . *. .
3.500E-03 7.962E-02 . . . * . .
4.000E-03 7.051E-02 . . . * . .
4.500E-03 7.242E-02 . . . * . .
5.000E-03 8.457E-02 . . . * . .
5.500E-03 1.018E-01 . . . * .
6.000E-03 1.163E-01 . . . . * .
6.500E-03 1.201E-01 . . . . * .
7.000E-03 1.075E-01 . . . .* .
7.500E-03 7.738E-02 . . . * . .
8.000E-03 3.338E-02 . . . * . .
8.500E-03 -1.687E-02 . . * . . .
9.000E-03 -6.394E-02 . . * . . .
9.500E-03 -9.932E-02 . * . . .
1.000E-02 -1.179E-01 . * . . . .
1.050E-02 -1.191E-01 . * . . . .
1.100E-02 -1.074E-01 . *. . . .
1.150E-02 -9.010E-02 . .* . . .
1.200E-02 -7.551E-02 . . * . . .
1.250E-02 -6.986E-02 . . * . . .
1.300E-02 -7.551E-02 . . * . . .
1.350E-02 -9.010E-02 . .* . . .
1.400E-02 -1.074E-01 . *. . . .
1.450E-02 -1.191E-01 . * . . . .
1.500E-02 -1.179E-01 . * . . . .
1.550E-02 -9.932E-02 . * . . .
1.600E-02 -6.394E-02 . . * . . .
1.650E-02 -1.687E-02 . . * . . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
fourier components of transient response v(2,3)
dc component = 1.349E-11
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.198E-01 1.000000 -72.000 0.000
2 1.200E+02 1.609E-11 0.000000 67.570 139.570
3 1.800E+02 4.990E-02 0.416667 144.000 216.000
4 2.400E+02 1.074E-10 0.000000 -169.546 -97.546
5 3.000E+02 3.871E-11 0.000000 169.582 241.582
6 3.600E+02 5.736E-11 0.000000 140.845 212.845
7 4.200E+02 8.407E-11 0.000000 177.071 249.071
8 4.800E+02 1.329E-10 0.000000 156.772 228.772
9 5.400E+02 2.619E-10 0.000000 160.498 232.498
total harmonic distortion = 41.666663 percent
In the Fourier analysis, the mixed
frequencies are unmixed and presented separately. Here we
see the same 0.1198 amps of 60 Hz (fundamental) current as
we did in the first simulation, but appearing in the 3rd
harmonic row we see 49.9 mA: our 50 mA, 180 Hz current
source at work. Why don't we see the entire 50 mA through
the line? Because that current source is connected across
the 1 kΩ load resistor, so some of its current is shunted
through the load and never goes through the line back to the
source. It's an inevitable consequence of this type of
simulation, where one part of the load is "normal" (a
resistor) and the other part is imitated by a current
source.
If we were to add more current sources to
the "load," we would see further distortion of the line
current waveform from the ideal sine-wave shape, and each of
those harmonic currents would appear in the Fourier analysis
breakdown:
Nonlinear load simulation
vsource 1 0 sin(0 120 60 0 0)
rsource 1 2 1
rline 2 3 1
rload 3 0 1k
i3har 3 0 sin(0 50m 180 0 0)
i5har 3 0 sin(0 50m 300 0 0)
i7har 3 0 sin(0 50m 420 0 0)
i9har 3 0 sin(0 50m 540 0 0)
.options itl5=0
.tran 0.5m 30m 0 1u
.plot tran v(2,3)
.four 60 v(2,3)
.end
fourier components of transient response v(2,3)
dc component = 6.299E-11
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.198E-01 1.000000 -72.000 0.000
2 1.200E+02 1.900E-09 0.000000 -93.908 -21.908
3 1.800E+02 4.990E-02 0.416667 144.000 216.000
4 2.400E+02 5.469E-09 0.000000 -116.873 -44.873
5 3.000E+02 4.990E-02 0.416667 0.000 72.000
6 3.600E+02 6.271E-09 0.000000 85.062 157.062
7 4.200E+02 4.990E-02 0.416666 -144.000 -72.000
8 4.800E+02 2.742E-09 0.000000 -38.781 33.219
9 5.400E+02 4.990E-02 0.416666 72.000 144.000
total harmonic distortion = 83.333296 percent
As you can see from the Fourier analysis,
every harmonic current source is equally represented in the
line current, at 49.9 mA each. So far, this is just a
single-phase power system simulation. Things get more
interesting when we make it a three-phase simulation. Two
Fourier analyses will be performed: one for the voltage
across a line resistor, and one for the voltage across the
neutral resistor. As before, reading voltages across fixed
resistances of 1 Ω each gives direct indications of current
through those resistors:
Y-Y source/load 4-wire system with harmonics
*
* phase1 voltage source and r (120 v /_ 0 deg)
vsource1 1 0 sin(0 120 60 0 0)
rsource1 1 2 1
*
* phase2 voltage source and r (120 v /_ 120 deg)
vsource2 3 0 sin(0 120 60 5.55555m 0)
rsource2 3 4 1
*
* phase3 voltage source and r (120 v /_ 240 deg)
vsource3 5 0 sin(0 120 60 11.1111m 0)
rsource3 5 6 1
*
* line and neutral wire resistances
rline1 2 8 1
rline2 4 9 1
rline3 6 10 1
rneutral 0 7 1
*
* phase 1 of load
rload1 8 7 1k
i3har1 8 7 sin(0 50m 180 0 0)
i5har1 8 7 sin(0 50m 300 0 0)
i7har1 8 7 sin(0 50m 420 0 0)
i9har1 8 7 sin(0 50m 540 0 0)
*
* phase 2 of load
rload2 9 7 1k
i3har2 9 7 sin(0 50m 180 5.55555m 0)
i5har2 9 7 sin(0 50m 300 5.55555m 0)
i7har2 9 7 sin(0 50m 420 5.55555m 0)
i9har2 9 7 sin(0 50m 540 5.55555m 0)
*
* phase 3 of load
rload3 10 7 1k
i3har3 10 7 sin(0 50m 180 11.1111m 0)
i5har3 10 7 sin(0 50m 300 11.1111m 0)
i7har3 10 7 sin(0 50m 420 11.1111m 0)
i9har3 10 7 sin(0 50m 540 11.1111m 0)
*
* analysis stuff
.options itl5=0
.tran 0.5m 100m 12m 1u
.plot tran v(2,8)
.four 60 v(2,8)
.plot tran v(0,7)
.four 60 v(0,7)
.end
Fourier analysis of line current:
fourier components of transient response v(2,8)
dc component = -6.404E-12
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.198E-01 1.000000 0.000 0.000
2 1.200E+02 2.218E-10 0.000000 172.985 172.985
3 1.800E+02 4.975E-02 0.415423 0.000 0.000
4 2.400E+02 4.236E-10 0.000000 166.990 166.990
5 3.000E+02 4.990E-02 0.416667 0.000 0.000
6 3.600E+02 1.877E-10 0.000000 -147.146 -147.146
7 4.200E+02 4.990E-02 0.416666 0.000 0.000
8 4.800E+02 2.784E-10 0.000000 -148.811 -148.811
9 5.400E+02 4.975E-02 0.415422 0.000 0.000
total harmonic distortion = 83.209009 percent
Fourier analysis of neutral current:
fourier components of transient response v(0,7)
dc component = 1.819E-10
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 4.337E-07 1.000000 60.018 0.000
2 1.200E+02 1.869E-10 0.000431 91.206 31.188
3 1.800E+02 1.493E-01 344147.7638 -180.000 -240.018
4 2.400E+02 1.257E-09 0.002898 -21.103 -81.121
5 3.000E+02 9.023E-07 2.080596 119.981 59.963
6 3.600E+02 3.396E-10 0.000783 15.882 -44.136
7 4.200E+02 1.264E-06 2.913955 59.993 -0.025
8 4.800E+02 5.975E-10 0.001378 35.584 -24.434
9 5.400E+02 1.493E-01 344147.4889 -179.999 -240.017
This is a balanced Y-Y power system, each
phase identical to the single-phase AC system simulated
earlier. Consequently, it should come as no surprise that
the Fourier analysis for line current in one phase of the
3-phase system is nearly identical to the Fourier analysis
for line current in the single-phase system: a fundamental
(60 Hz) line current of 0.1198 amps, and odd harmonic
currents of approximately 50 mA each.
What should be surprising here is the
analysis for the neutral conductor's current, as determined
by the voltage drop across the Rneutral resistor
between SPICE nodes 0 and 7. In a balanced 3-phase Y load,
we would expect the neutral current to be zero. Each phase
current -- which by itself would go through the neutral wire
back to the supplying phase on the source Y -- should cancel
each other in regard to the neutral conductor because
they're all the same magnitude and all shifted 120o
apart. In a system with no harmonic currents, this is
what happens, leaving zero current through the neutral
conductor. However, we cannot say the same for harmonic
currents in the same system.
Note that the fundamental frequency (60 Hz,
or the 1st harmonic) current is virtually absent from the
neutral conductor. Our Fourier analysis shows only 0.4337 �A
of 1st harmonic when reading voltage across Rneutral.
The same may be said about the 5th and 7th harmonics, both
of those currents having negligible magnitude. In contrast,
the 3rd and 9th harmonics are strongly represented within
the neutral conductor, with 149.3 mA (1.493E-01 volts across
1 Ω) each! This is very nearly 150 mA, or three times the
current sources' values, individually. With three sources
per harmonic frequency in the load, it appears our 3rd and
9th harmonic currents in each phase are adding to
form the neutral current.
This is exactly what's happening, though it
might not be apparent why this is so. The key to
understanding this is made clear in a time-domain graph of
phase currents. Examine this plot of balanced phase currents
over time, with a phase sequence of 1-2-3:
With the three fundamental waveforms equally
shifted across the time axis of the graph, it is easy to see
how they would cancel each other to give a resultant current
of zero in the neutral conductor. Let's consider, though,
what a 3rd harmonic waveform for phase 1 would look like
superimposed on the graph:
Observe how this harmonic waveform has the
same phase relationship to the 2nd and 3rd fundamental
waveforms as it does with the 1st: in each positive
half-cycle of any of the fundamental waveforms, you
will find exactly two positive half-cycles and one negative
half-cycle of the harmonic waveform. What this means is that
the 3rd-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
harmonic multiples!
If we were to plot all three 3rd-harmonic
waveforms on the same graph, we would see them precisely
overlap and appear as a single, unified waveform (shown here
in bold):
For the more mathematically inclined, this
principle may be expressed symbolically. Suppose that A
represents one waveform and B another, both at the
same frequency, but shifted 120o from each other
in terms of phase. Let's call the 3rd harmonic of each
waveform A' and B', respectively. The phase
shift between A' and B' is not 120o
(that is the phase shift between A and B), but
3 times that, because the A' and B' waveforms
alternate three times as fast as A and B. The
shift between waveforms is only accurately expressed in
terms of phase angle when the same angular velocity
is assumed. When relating waveforms of different frequency,
the most accurate way to represent phase shift is in terms
of time; and the time-shift between A'
and B' is equivalent to 120o at a
frequency three times lower, or 360o at the
frequency of A' and B'. A phase shift of 360o
is the same as a phase shift of 0o, which is to
say no phase shift at all. Thus, A' and B'
must be in phase with each other:
This characteristic of the 3rd harmonic in a
three-phase system also holds true for any integer multiples
of the 3rd harmonic. So, not only are the 3rd harmonic
waveforms of each fundamental waveform in phase with each
other, but so are the 6th harmonics, the 9th harmonics, the
12th harmonics, the 15th harmonics, the 18th harmonics, the
21st harmonics, and so on. Since only odd harmonics appear
in systems where waveform distortion is symmetrical about
the centerline -- and most nonlinear loads create
symmetrical distortion -- even-numbered multiples of the 3rd
harmonic (6th, 12th, 18th, etc.) are generally not
significant, leaving only the odd-numbered multiples (3rd,
9th, 21st, etc.) to significantly contribute to neutral
currents.
In polyphase power systems with some number
of phases other than three, this effect occurs with
harmonics of the same multiple. For instance, the harmonic
currents that add in the neutral conductor of a
star-connected 4-phase system where the phase shift between
fundamental waveforms is 90o would be the 4th,
8th, 12th, 16th, 20th, and so on.
Due to their abundance and significance in
three-phase power systems, the 3rd harmonic and its
multiples have their own special name: triplen harmonics.
All triplen harmonics add with each other in the neutral
conductor of a 4-wire Y-connected load. In power systems
containing substantial nonlinear loading, the triplen
harmonic currents may be of great enough magnitude to cause
neutral conductors to overheat. This is very problematic, as
other safety concerns prohibit neutral conductors from
having overcurrent protection, and thus there is no
provision for automatic interruption of these high currents.
The following illustration shows how triplen
harmonic currents created at the load add within the neutral
conductor. The symbol "ω" is used to represent angular
velocity, and is mathematically equivalent to 2πf. So, "ω"
represents the fundamental frequency, "3ω " represents the
3rd harmonic, "5ω" represents the 5th harmonic, and so on:
In an effort to mitigate these additive
triplen currents, one might be tempted to remove the neutral
wire entirely. If there is no neutral wire in which triplen
currents can flow together, then they won't, right?
Unfortunately, doing so just causes a different problem: the
load's "Y" center-point will no longer be at the same
potential as the source's, meaning that each phase of the
load will receive a different voltage than what is produced
by the source. We'll re-run the last SPICE simulation
without the 1 Ω Rneutral resistor and see what
happens:
Y-Y source/load (no neutral) with harmonics
*
* phase1 voltage source and r (120 v /_ 0 deg)
vsource1 1 0 sin(0 120 60 0 0)
rsource1 1 2 1
*
* phase2 voltage source and r (120 v /_ 120 deg)
vsource2 3 0 sin(0 120 60 5.55555m 0)
rsource2 3 4 1
*
* phase3 voltage source and r (120 v /_ 240 deg)
vsource3 5 0 sin(0 120 60 11.1111m 0)
rsource3 5 6 1
*
* line resistances
rline1 2 8 1
rline2 4 9 1
rline3 6 10 1
*
* phase 1 of load
rload1 8 7 1k
i3har1 8 7 sin(0 50m 180 0 0)
i5har1 8 7 sin(0 50m 300 0 0)
i7har1 8 7 sin(0 50m 420 0 0)
i9har1 8 7 sin(0 50m 540 0 0)
*
* phase 2 of load
rload2 9 7 1k
i3har2 9 7 sin(0 50m 180 5.55555m 0)
i5har2 9 7 sin(0 50m 300 5.55555m 0)
i7har2 9 7 sin(0 50m 420 5.55555m 0)
i9har2 9 7 sin(0 50m 540 5.55555m 0)
*
* phase 3 of load
rload3 10 7 1k
i3har3 10 7 sin(0 50m 180 11.1111m 0)
i5har3 10 7 sin(0 50m 300 11.1111m 0)
i7har3 10 7 sin(0 50m 420 11.1111m 0)
i9har3 10 7 sin(0 50m 540 11.1111m 0)
*
* analysis stuff
.options itl5=0
.tran 0.5m 100m 12m 1u
.plot tran v(2,8)
.four 60 v(2,8)
.plot tran v(0,7)
.four 60 v(0,7)
.plot tran v(8,7)
.four 60 v(8,7)
.end
Fourier analysis of line current:
fourier components of transient response v(2,8)
dc component = 5.423E-11
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.198E-01 1.000000 0.000 0.000
2 1.200E+02 2.388E-10 0.000000 158.016 158.016
3 1.800E+02 3.136E-07 0.000003 -90.009 -90.009
4 2.400E+02 5.963E-11 0.000000 -111.510 -111.510
5 3.000E+02 4.990E-02 0.416665 0.000 0.000
6 3.600E+02 8.606E-11 0.000000 -124.565 -124.565
7 4.200E+02 4.990E-02 0.416668 0.000 0.000
8 4.800E+02 8.126E-11 0.000000 -159.638 -159.638
9 5.400E+02 9.406E-07 0.000008 -90.005 -90.005
total harmonic distortion = 58.925539 percent
Fourier analysis of voltage between the
two "Y" center-points:
fourier components of transient response v(0,7)
dc component = 6.093E-08
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.453E-04 1.000000 60.018 0.000
2 1.200E+02 6.263E-08 0.000431 91.206 31.188
3 1.800E+02 5.000E+01 344147.7879 -180.000 -240.018
4 2.400E+02 4.210E-07 0.002898 -21.103 -81.121
5 3.000E+02 3.023E-04 2.080596 119.981 59.963
6 3.600E+02 1.138E-07 0.000783 15.882 -44.136
7 4.200E+02 4.234E-04 2.913955 59.993 -0.025
8 4.800E+02 2.001E-07 0.001378 35.584 -24.434
9 5.400E+02 5.000E+01 344147.4728 -179.999 -240.017
total harmonic distortion = ************ percent
Fourier analysis of load phase voltage:
fourier components of transient response v(8,7)
dc component = 6.070E-08
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.198E+02 1.000000 0.000 0.000
2 1.200E+02 6.231E-08 0.000000 90.473 90.473
3 1.800E+02 5.000E+01 0.417500 -180.000 -180.000
4 2.400E+02 4.278E-07 0.000000 -19.747 -19.747
5 3.000E+02 9.995E-02 0.000835 179.850 179.850
6 3.600E+02 1.023E-07 0.000000 13.485 13.485
7 4.200E+02 9.959E-02 0.000832 179.790 179.789
8 4.800E+02 1.991E-07 0.000000 35.462 35.462
9 5.400E+02 5.000E+01 0.417499 -179.999 -179.999
total harmonic distortion = 59.043467 percent
Strange things are happening, indeed. First,
we see that the triplen harmonic currents (3rd and 9th) all
but disappear in the lines connecting load to source. The
5th and 7th harmonic currents are present at their normal
levels (approximately 50 mA), but the 3rd and 9th harmonic
currents are of negligible magnitude. Second, we see that
there is substantial harmonic voltage between the two "Y"
center-points, between which the neutral conductor used to
connect. According to SPICE, there is 50 volts of both 3rd
and 9th harmonic frequency between these two points, which
is definitely not normal in a linear (no harmonics),
balanced Y system. Finally, the voltage as measured across
one of the load's phases (between nodes 8 and 7 in the SPICE
analysis) likewise shows strong triplen harmonic voltages of
50 volts each.
The following illustration is a graphical
summary of the aforementioned effects:
In summary, removal of the neutral conductor
leads to a "hot" center-point on the load "Y", and also to
harmonic load phase voltages of equal magnitude, all
comprised of triplen frequencies. In the previous simulation
where we had a 4-wire, Y-connected system, the undesirable
effect from harmonics was excessive neutral current,
but at least each phase of the load received voltage nearly
free of harmonics.
Since removing the neutral wire didn't seem
to work in eliminating the problems caused by harmonics,
perhaps switching to a Δ configuration will. Let's try a Δ
source instead of a Y, keeping the load in its present Y
configuration, and see what happens. The measured parameters
will be line current (voltage across Rline, nodes
0 and 8), load phase voltage (nodes 8 and 7), and source
phase current (voltage across Rsource, nodes 1
and 2):
Delta-Y source/load with harmonics
*
* phase1 voltage source and r (120 v /_ 0 deg)
vsource1 1 0 sin(0 207.846 60 0 0)
rsource1 1 2 1
*
* phase2 voltage source and r (120 v /_ 120 deg)
vsource2 3 2 sin(0 207.846 60 5.55555m 0)
rsource2 3 4 1
*
* phase3 voltage source and r (120 v /_ 240 deg)
vsource3 5 4 sin(0 207.846 60 11.1111m 0)
rsource3 5 0 1
*
* line resistances
rline1 0 8 1
rline2 2 9 1
rline3 4 10 1
*
* phase 1 of load
rload1 8 7 1k
i3har1 8 7 sin(0 50m 180 9.72222m 0)
i5har1 8 7 sin(0 50m 300 9.72222m 0)
i7har1 8 7 sin(0 50m 420 9.72222m 0)
i9har1 8 7 sin(0 50m 540 9.72222m 0)
*
* phase 2 of load
rload2 9 7 1k
i3har2 9 7 sin(0 50m 180 15.2777m 0)
i5har2 9 7 sin(0 50m 300 15.2777m 0)
i7har2 9 7 sin(0 50m 420 15.2777m 0)
i9har2 9 7 sin(0 50m 540 15.2777m 0)
*
* phase 3 of load
rload3 10 7 1k
i3har3 10 7 sin(0 50m 180 4.16666m 0)
i5har3 10 7 sin(0 50m 300 4.16666m 0)
i7har3 10 7 sin(0 50m 420 4.16666m 0)
i9har3 10 7 sin(0 50m 540 4.16666m 0)
*
* analysis stuff
.options itl5=0
.tran 0.5m 100m 16m 1u
.plot tran v(0,8) v(8,7) v(1,2)
.four 60 v(0,8) v(8,7) v(1,2)
.end
Note: the following paragraph is for
those curious readers who follow every detail of my SPICE
netlists. If you just want to find out what happens in the
circuit, skip this paragraph! When simulating circuits
having AC sources of differing frequency and differing
phase, the only way to do it in SPICE is to set up the
sources with a delay time or phase offset
specified in seconds. Thus, the 0o source has
these five specifying figures: "(0 207.846 60 0 0)", which
means 0 volts DC offset, 207.846 volts peak amplitude (120
times the square root of three, to ensure the load phase
voltages remain at 120 volts each), 60 Hz, 0 time delay, and
0 damping factor. The 120o phase-shifted source
has these figures: "(0 207.846 60 5.55555m 0)", all the same
as the first except for the time delay factor of 5.55555
milliseconds, or 1/3 of the full period of 16.6667
milliseconds for a 60 Hz waveform. The 240o
source must be time-delayed twice that amount, equivalent to
a fraction of 240/360 of 16.6667 milliseconds, or 11.1111
milliseconds. This is for the Δ-connected source. The
Y-connected load, on the other hand, requires a different
set of time-delay figures for its harmonic current sources,
because the phase voltages in a Y load are not in phase with
the phase voltages of a Δ source. If Δ source voltages VAC,
VBA, and VCB are referenced at 0o,
120o, and 240o, respectively, then "Y"
load voltages VA, VB, and VC
will have phase angles of -30o, 90o,
and 210o, respectively. This is an intrinsic
property of all Δ-Y circuits and not a quirk of SPICE.
Therefore, when I specified the delay times for the harmonic
sources, I had to set them at 15.2777 milliseconds (-30o,
or +330o), 4.16666 milliseconds (90o),
and 9.72222 milliseconds (210o). One final note:
when delaying AC sources in SPICE, they don't "turn on"
until their delay time has elapsed, which means any
mathematical analysis up to that point in time will be in
error. Consequently, I set the .tran transient
analysis line to hold off analysis until 16 milliseconds
after start, which gives all sources in the netlist time to
engage before any analysis takes place.
The result of this analysis is almost as
disappointing as the last. Line currents remain unchanged
(the only substantial harmonic content being the 5th and 7th
harmonics), and load phase voltages remain unchanged as
well, with a full 50 volts of triplen harmonic (3rd and 9th)
frequencies across each load component. Source phase current
is a fraction of the line current, which should come as no
surprise. Both 5th and 7th harmonics are represented there,
with negligible triplen harmonics:
Fourier analysis of line current:
fourier components of transient response v(0,8)
dc component = -6.850E-11
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.198E-01 1.000000 150.000 0.000
2 1.200E+02 2.491E-11 0.000000 159.723 9.722
3 1.800E+02 1.506E-06 0.000013 0.005 -149.996
4 2.400E+02 2.033E-11 0.000000 52.772 -97.228
5 3.000E+02 4.994E-02 0.416682 30.002 -119.998
6 3.600E+02 1.234E-11 0.000000 57.802 -92.198
7 4.200E+02 4.993E-02 0.416644 -29.998 -179.998
8 4.800E+02 8.024E-11 0.000000 -174.200 -324.200
9 5.400E+02 4.518E-06 0.000038 -179.995 -329.995
total harmonic distortion = 58.925038 percent
Fourier analysis of load phase voltage:
fourier components of transient response v(8,7)
dc component = 1.259E-08
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.198E+02 1.000000 150.000 0.000
2 1.200E+02 1.941E-07 0.000000 49.693 -100.307
3 1.800E+02 5.000E+01 0.417222 -89.998 -239.998
4 2.400E+02 1.519E-07 0.000000 66.397 -83.603
5 3.000E+02 6.466E-02 0.000540 -151.112 -301.112
6 3.600E+02 2.433E-07 0.000000 68.162 -81.838
7 4.200E+02 6.931E-02 0.000578 148.548 -1.453
8 4.800E+02 2.398E-07 0.000000 -174.897 -324.897
9 5.400E+02 5.000E+01 0.417221 90.006 -59.995
total harmonic distortion = 59.004109 percent
Fourier analysis of source phase current:
fourier components of transient response v(1,2)
dc component = 3.564E-11
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 6.906E-02 1.000000 -0.181 0.000
2 1.200E+02 1.525E-11 0.000000 -156.674 -156.493
3 1.800E+02 1.422E-06 0.000021 -179.996 -179.815
4 2.400E+02 2.949E-11 0.000000 -110.570 -110.390
5 3.000E+02 2.883E-02 0.417440 -179.996 -179.815
6 3.600E+02 2.324E-11 0.000000 -91.926 -91.745
7 4.200E+02 2.883E-02 0.417398 -179.994 -179.813
8 4.800E+02 4.140E-11 0.000000 -39.875 -39.694
9 5.400E+02 4.267E-06 0.000062 0.006 0.186
total harmonic distortion = 59.031969 percent
Really, the only advantage of the Δ-Y
configuration from the standpoint of harmonics is that there
is no longer a center-point at the load posing a shock
hazard. Otherwise, the load components receive the same
harmonically-rich voltages and the lines see the same
currents as in a three-wire Y system.
If we were to reconfigure the system into a
Δ-Δ arrangement, that should guarantee that each load
component receives non-harmonic voltage, since each load
phase would be directly connected in parallel with each
source phase. The complete lack of any neutral wires or
"center points" in a Δ-Δ system prevents strange voltages or
additive currents from occurring. It would seem to be the
ideal solution. Let's simulate and observe, analyzing line
current, load phase voltage, and source phase current:
Delta-Delta source/load with harmonics
*
* phase1 voltage source and r (120 v /_ 0 deg)
vsource1 1 0 sin(0 120 60 0 0)
rsource1 1 2 1
*
* phase2 voltage source and r (120 v /_ 120 deg)
vsource2 3 2 sin(0 120 60 5.55555m 0)
rsource2 3 4 1
*
* phase3 voltage source and r (120 v /_ 240 deg)
vsource3 5 4 sin(0 120 60 11.1111m 0)
rsource3 5 0 1
*
* line resistances
rline1 0 6 1
rline2 2 7 1
rline3 4 8 1
*
* phase 1 of load
rload1 7 6 1k
i3har1 7 6 sin(0 50m 180 0 0)
i5har1 7 6 sin(0 50m 300 0 0)
i7har1 7 6 sin(0 50m 420 0 0)
i9har1 7 6 sin(0 50m 540 0 0)
*
* phase 2 of load
rload2 8 7 1k
i3har2 8 7 sin(0 50m 180 5.55555m 0)
i5har2 8 7 sin(0 50m 300 5.55555m 0)
i7har2 8 7 sin(0 50m 420 5.55555m 0)
i9har2 8 7 sin(0 50m 540 5.55555m 0)
*
* phase 3 of load
rload3 6 8 1k
i3har3 6 8 sin(0 50m 180 11.1111m 0)
i5har3 6 8 sin(0 50m 300 11.1111m 0)
i7har3 6 8 sin(0 50m 420 11.1111m 0)
i9har3 6 8 sin(0 50m 540 11.1111m 0)
*
* analysis stuff
.options itl5=0
.tran 0.5m 100m 16m 1u
.plot tran v(0,6) v(7,6) v(2,1) i(3har1)
.four 60 v(0,6) v(7,6) v(2,1)
.end
Fourier analysis of line current:
fourier components of transient response v(0,6)
dc component = -6.007E-11
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 2.070E-01 1.000000 150.000 0.000
2 1.200E+02 5.480E-11 0.000000 156.666 6.666
3 1.800E+02 6.257E-07 0.000003 89.990 -60.010
4 2.400E+02 4.911E-11 0.000000 8.187 -141.813
5 3.000E+02 8.626E-02 0.416664 -149.999 -300.000
6 3.600E+02 1.089E-10 0.000000 -31.997 -181.997
7 4.200E+02 8.626E-02 0.416669 150.001 0.001
8 4.800E+02 1.578E-10 0.000000 -63.940 -213.940
9 5.400E+02 1.877E-06 0.000009 89.987 -60.013
total harmonic distortion = 58.925538 percent
Fourier analysis of load phase voltage:
fourier components of transient response v(7,6)
dc component = -5.680E-10
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.195E+02 1.000000 0.000 0.000
2 1.200E+02 1.039E-09 0.000000 144.749 144.749
3 1.800E+02 1.251E-06 0.000000 89.974 89.974
4 2.400E+02 4.215E-10 0.000000 36.127 36.127
5 3.000E+02 1.992E-01 0.001667 -180.000 -180.000
6 3.600E+02 2.499E-09 0.000000 -4.760 -4.760
7 4.200E+02 1.992E-01 0.001667 -180.000 -180.000
8 4.800E+02 2.951E-09 0.000000 -151.385 -151.385
9 5.400E+02 3.752E-06 0.000000 89.905 89.905
total harmonic distortion = 0.235702 percent
Fourier analysis of source phase current:
fourier components of transient response v(2,1)
dc component = -1.923E-12
harmonic frequency fourier normalized phase normalized
no (hz) component component (deg) phase (deg)
1 6.000E+01 1.194E-01 1.000000 179.940 0.000
2 1.200E+02 2.569E-11 0.000000 133.491 -46.449
3 1.800E+02 3.129E-07 0.000003 89.985 -89.955
4 2.400E+02 2.657E-11 0.000000 23.368 -156.571
5 3.000E+02 4.980E-02 0.416918 -180.000 -359.939
6 3.600E+02 4.595E-11 0.000000 -22.475 -202.415
7 4.200E+02 4.980E-02 0.416921 -180.000 -359.939
8 4.800E+02 7.385E-11 0.000000 -63.759 -243.699
9 5.400E+02 9.385E-07 0.000008 89.991 -89.949
total harmonic distortion = 58.961298 percent
As predicted earlier, the load phase voltage
is almost a pure sine-wave, with negligible harmonic
content, thanks to the direct connection with the source
phases in a Δ-Δ system. But what happened to the triplen
harmonics? The 3rd and 9th harmonic frequencies don't appear
in any substantial amount in the line current, nor in the
load phase voltage, nor in the source phase current! We know
that triplen currents exist, because the 3rd and 9th
harmonic current sources are intentionally placed in the
phases of the load, but where did those currents go?
Remember that the triplen harmonics of 120o
phase-shifted fundamental frequencies are in phase with each
other. Note the directions that the arrows of the current
sources within the load phases are pointing, and think about
what would happen if the 3rd and 9th harmonic sources were
DC sources instead. What we would have is current
circulating within the loop formed by the Δ-connected phases.
This is where the triplen harmonic currents have gone: they
stay within the Δ of the load, never reaching the line
conductors or the windings of the source. These results may
be graphically summarized as such:
This is a major benefit of the Δ-Δ system
configuration: triplen harmonic currents remain confined in
whatever set of components create them, and do not "spread"
to other parts of the system.
-
REVIEW:
-
Nonlinear components are those that
draw a non-sinusoidal (non-sine-wave) current waveform
when energized by a sinusoidal (sine-wave) voltage. Since
any distortion of an originally pure sine-wave constitutes
harmonic frequencies, we can say that nonlinear components
generate harmonic currents.
-
When the sine-wave distortion is
symmetrical above and below the average centerline of the
waveform, the only harmonics present will be
odd-numbered, not even-numbered.
-
The 3rd harmonic, and integer multiples of
it (6th, 9th, 12th, 15th) are known as triplen
harmonics. They are in phase with each other, despite the
fact that their respective fundamental waveforms are 120o
out of phase with each other.
-
In a 4-wire Y-Y system, triplen harmonic
currents add within the neutral conductor.
-
Triplen harmonic currents in a Δ-connected
set of components circulate within the loop formed by the
Δ.
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