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 µA 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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