Single-phase power systems
Depicted above is a very simple AC
circuit. If the load resistor's power dissipation were substantial, we
might call this a "power circuit" or "power system" instead of regarding
it as just a regular circuit. The distinction between a "power circuit"
and a "regular circuit" may seem arbitrary, but the practical concerns
are definitely not.
One such concern is the size and cost of
wiring necessary to deliver power from the AC source to the load.
Normally, we do not give much thought to this type of concern if we're
merely analyzing a circuit for the sake of learning about the laws of
electricity. However, in the real world it can be a major concern. If we
give the source in the above circuit a voltage value and also give power
dissipation values to the two load resistors, we can determine the
wiring needs for this particular circuit:
83.33 amps for each load resistor adds up
to 166.66 amps total circuit current. This is no small amount of
current, and would necessitate copper wire conductors of at least 1/0
gage. Such wire is well over 1/4 inch in diameter, weighing over 300
pounds per thousand feet. Bear in mind that copper is not cheap either!
It would be in our best interest to find ways to minimize such costs if
we were designing a power system with long conductor lengths.
One way to do this would be to increase
the voltage of the power source and use loads built to dissipate 10 kW
each at this higher voltage. The loads, of course, would have to have
greater resistance values to dissipate the same power as before (10 kW
each) at a greater voltage than before. The advantage would be less
current required, permitting the use of smaller, lighter, and cheaper
wire:
Now our total circuit current is
83.33 amps, half of what it was before. We can now use number 4 gage
wire, which weighs less than half of what 1/0 gage wire does per unit
length. This is a considerable reduction in system cost with no
degradation in performance. This is why power distribution system
designers elect to transmit electric power using very high voltages
(many thousands of volts): to capitalize on the savings realized by the
use of smaller, lighter, cheaper wire.
However, this solution is not without
disadvantages. Another practical concern with power circuits is the
danger of electric shock from high voltages. Again, this is not usually
the sort of thing we concentrate on while learning about the laws of
electricity, but it is a very valid concern in the real world,
especially when large amounts of power are being dealt with. The gain in
efficiency realized by stepping up the circuit voltage presents us with
increased danger of electric shock. Power distribution companies tackle
this problem by stringing their power lines along high poles or towers,
and insulating the lines from the supporting structures with large,
porcelain insulators.
At the point of use (the electric power
customer), there is still the issue of what voltage to use for powering
loads. High voltage gives greater system efficiency by means of reduced
conductor current, but it might not always be practical to keep power
wiring out of reach at the point of use the way it can be elevated out
of reach in distribution systems. This tradeoff between efficiency and
danger is one that European power system designers have decided to risk,
all their households and appliances operating at a nominal voltage of
240 volts instead of 120 volts as it is in North America. That is why
tourists from America visiting Europe must carry small step-down
transformers for their portable appliances, to step the 240 VAC (volts
AC) power down to a more suitable 120 VAC.
Is there any way to realize the
advantages of both increased efficiency and reduced safety hazard at the
same time? One solution would be to install step-down transformers at
the end-point of power use, just as the American tourist must do while
in Europe. However, this would be expensive and inconvenient for
anything but very small loads (where the transformers can be built
cheaply) or very large loads (where the expense of thick copper wires
would exceed the expense of a transformer).
An alternative solution would be to use a
higher voltage supply to provide power to two lower voltage loads in
series. This approach combines the efficiency of a high-voltage system
with the safety of a low-voltage system:
Notice the polarity markings (+ and -)
for each voltage shown, as well as the unidirectional arrows for
current. For the most part, I've avoided labeling "polarities" in the AC
circuits we've been analyzing, even though the notation is valid to
provide a frame of reference for phase. In later sections of this
chapter, phase relationships will become very important, so I'm
introducing this notation early on in the chapter for your familiarity.
The current through each load is the same
as it was in the simple 120 volt circuit, but the currents are not
additive because the loads are in series rather than parallel. The
voltage across each load is only 120 volts, not 240, so the safety
factor is better. Mind you, we still have a full 240 volts across the
power system wires, but each load is operating at a reduced
voltage. If anyone is going to get shocked, the odds are that it will be
from coming into contact with the conductors of a particular load rather
than from contact across the main wires of a power system.
There's only one disadvantage to this
design: the consequences of one load failing open, or being turned off
(assuming each load has a series on/off switch to interrupt current) are
not good. Being a series circuit, if either load were to open, current
would stop in the other load as well. For this reason, we need to modify
the design a bit:
Instead of a single 240 volt power
supply, we use two 120 volt supplies (in phase with each other!) in
series to produce 240 volts, then run a third wire to the connection
point between the loads to handle the eventuality of one load opening.
This is called a split-phase power system. Three smaller wires
are still cheaper than the two wires needed with the simple parallel
design, so we're still ahead on efficiency. The astute observer will
note that the neutral wire only has to carry the difference of
current between the two loads back to the source. In the above case,
with perfectly "balanced" loads consuming equal amounts of power, the
neutral wire carries zero current.
Notice how the neutral wire is connected
to earth ground at the power supply end. This is a common feature in
power systems containing "neutral" wires, since grounding the neutral
wire ensures the least possible voltage at any given time between any
"hot" wire and earth ground.
An essential component to a split-phase
power system is the dual AC voltage source. Fortunately, designing and
building one is not difficult. Since most AC systems receive their power
from a step-down transformer anyway (stepping voltage down from high
distribution levels to a user-level voltage like 120 or 240), that
transformer can be built with a center-tapped secondary winding:
If the AC power comes directly from a
generator (alternator), the coils can be similarly center-tapped for the
same effect. The extra expense to include a center-tap connection in a
transformer or alternator winding is minimal.
Here is where the (+) and (-) polarity
markings really become important. This notation is often used to
reference the phasings of multiple AC voltage sources, so it is
clear whether they are aiding ("boosting") each other or opposing
("bucking") each other. If not for these polarity markings, phase
relations between multiple AC sources might be very confusing. Note that
the split-phase sources in the schematic (each one 120 volts ∠ 0o),
with polarity marks (+) to (-) just like series-aiding batteries can
alternatively be represented as such:
To mathematically calculate voltage
between "hot" wires, we must subtract voltages, because their
polarity marks show them to be opposed to each other:
If we mark the two sources' common
connection point (the neutral wire) with the same polarity mark (-), we
must express their relative phase shifts as being 180o apart.
Otherwise, we'd be denoting two voltage sources in direct opposition
with each other, which would give 0 volts between the two "hot"
conductors. Why am I taking the time to elaborate on polarity marks and
phase angles? It will make more sense in the next section!
Power systems in American households and
light industry are most often of the split-phase variety, providing
so-called 120/240 VAC power. The term "split-phase" merely refers to the
split-voltage supply in such a system. In a more general sense, this
kind of AC power supply is called single phase because both
voltage waveforms are in phase, or in step, with each other.
The term "single phase" is a counterpoint
to another kind of power system called "polyphase" which we are about to
investigate in detail. Apologies for the long introduction leading up to
the title-topic of this chapter. The advantages of polyphase power
systems are more obvious if one first has a good understanding of single
phase systems.
- REVIEW:
- Single phase
power systems are defined by having an AC source with only one voltage
waveform.
- A split-phase power system is
one with multiple (in-phase) AC voltage sources connected in series,
delivering power to loads at more than one voltage, with more than two
wires. They are used primarily to achieve balance between system
efficiency (low conductor currents) and safety (low load voltages).
- Split-phase AC sources can be easily
created by center-tapping the coil windings of transformers or
alternators.
Three-phase power systems
Split-phase power systems achieve their
high conductor efficiency and low safety risk by splitting up the
total voltage into lesser parts and powering multiple loads at those
lesser voltages, while drawing currents at levels typical of a
full-voltage system. This technique, by the way, works just as well for
DC power systems as it does for single-phase AC systems. Such systems
are usually referred to as three-wire systems rather than
split-phase because "phase" is a concept restricted to AC.
But we know from our experience with
vectors and complex numbers that AC voltages don't always add up as we
think they would if they are out of phase with each other. This
principle, applied to power systems, can be put to use to make power
systems with even greater conductor efficiencies and lower shock hazard
than with split-phase.
Suppose that we had two sources of AC
voltage connected in series just like the split-phase system we saw
before, except that each voltage source was 120o out of phase
with the other:
Since each voltage source is 120 volts,
and each load resistor is connected directly in parallel with its
respective source, the voltage across each load must be 120 volts
as well. Given load currents of 83.33 amps, each load must still be
dissipating 10 kilowatts of power. However, voltage between the two
"hot" wires is not 240 volts (120 ∠ 0o - 120 ∠ 180o)
because the phase difference between the two sources is not 180o.
Instead, the voltage is:
Nominally, we say that the voltage
between "hot" conductors is 208 volts (rounding up), and thus the power
system voltage is designated as 120/208.
If we calculate the current through the
"neutral" conductor, we find that it is not zero, even with
balanced load resistances. Kirchhoff's Current Law tells us that the
currents entering and exiting the node between the two loads must be
zero:
So, we find that the "neutral" wire is
carrying a full 83.33 amps, just like each "hot" wire.
Note that we are still conveying 20 kW of
total power to the two loads, with each load's "hot" wire carrying 83.33
amps as before. With the same amount of current through each "hot" wire,
we must use the same gage copper conductors, so we haven't reduced
system cost over the split-phase 120/240 system. However, we have
realized a gain in safety, because the overall voltage between the two
"hot" conductors is 32 volts lower than it was in the split-phase system
(208 volts instead of 240 volts).
The fact that the neutral wire is
carrying 83.33 amps of current raises an interesting possibility: since
it's carrying current anyway, why not use that third wire as another
"hot" conductor, powering another load resistor with a third 120 volt
source having a phase angle of 240o? That way, we could
transmit more power (another 10 kW) without having to add any
more conductors. Let's see how this might look:
A full mathematical analysis of all the
voltages and currents in this circuit would necessitate the use of a
network theorem, the easiest being the Superposition Theorem. I'll spare
you the long, drawn-out calculations because you should be able to
intuitively understand that the three voltage sources at three different
phase angles will deliver 120 volts each to a balanced triad of load
resistors. For proof of this, we can use SPICE to do the math for us:
120/208 polyphase power system
v1 1 0 ac 120 0 sin
v2 2 0 ac 120 120 sin
v3 3 0 ac 120 240 sin
r1 1 4 1.44
r2 2 4 1.44
r3 3 4 1.44
.ac lin 1 60 60
.print ac v(1,4) v(2,4) v(3,4)
.print ac v(1,2) v(2,3) v(3,1)
.print ac i(v1) i(v2) i(v3)
.end
VOLTAGE ACROSS EACH LOAD
freq v(1,4) v(2,4) v(3,4)
6.000E+01 1.200E+02 1.200E+02 1.200E+02
VOLTAGE BETWEEN "HOT" CONDUCTORS
freq v(1,2) v(2,3) v(3,1)
6.000E+01 2.078E+02 2.078E+02 2.078E+02
CURRENT THROUGH EACH VOLTAGE SOURCE
freq i(v1) i(v2) i(v3)
6.000E+01 8.333E+01 8.333E+01 8.333E+01
Sure enough, we get 120 volts across each
load resistor, with (approximately) 208 volts between any two "hot"
conductors and conductor currents equal to 83.33 amps. At that current
and voltage, each load will be dissipating 10 kW of power. Notice that
this circuit has no "neutral" conductor to ensure stable voltage to all
loads if one should open. What we have here is a situation similar to
our split-phase power circuit with no "neutral" conductor: if one load
should happen to fail open, the voltage drops across the remaining
load(s) will change. To ensure load voltage stability in the even of
another load opening, we need a neutral wire to connect the source node
and load node together:
So long as the loads remain balanced
(equal resistance, equal currents), the neutral wire will not have to
carry any current at all. It is there just in case one or more load
resistors should fail open (or be shut off through a disconnecting
switch).
This circuit we've been analyzing with
three voltage sources is called a polyphase circuit. The prefix
"poly" simply means "more than one," as in "polytheism" (belief
in more than one deity), polygon" (a geometrical shape made of
multiple line segments: for example, pentagon and hexagon),
and "polyatomic" (a substance composed of multiple types of
atoms). Since the voltage sources are all at different phase angles (in
this case, three different phase angles), this is a "polyphase"
circuit. More specifically, it is a three-phase circuit, the kind
used predominantly in large power distribution systems.
Let's survey the advantages of a
three-phase power system over a single-phase system of equivalent load
voltage and power capacity. A single-phase system with three loads
connected directly in parallel would have a very high total current
(83.33 times 3, or 250 amps:
This would necessitate 3/0 gage copper
wire (very large!), at about 510 pounds per thousand feet, and
with a considerable price tag attached. If the distance from source to
load was 1000 feet, we would need over a half-ton of copper wire to do
the job. On the other hand, we could build a split-phase system with two
15 kW, 120 volt loads:
Our current is half of what it was with
the simple parallel circuit, which is a great improvement. We could get
away with using number 2 gage copper wire at a total mass of about 600
pounds, figuring about 200 pounds per thousand feet with three runs of
1000 feet each between source and loads. However, we also have to
consider the increased safety hazard of having 240 volts present in the
system, even though each load only receives 120 volts. Overall, there is
greater potential for dangerous electric shock to occur.
When we contrast these two examples
against our three-phase system, the advantages are quite clear. First,
the conductor currents are quite a bit less (83.33 amps versus 125 or
250 amps), permitting the use of much thinner and lighter wire. We can
use number 4 gage wire at about 125 pounds per thousand feet, which will
total 500 pounds (four runs of 1000 feet each) for our example circuit.
This represents a significant cost savings over the split-phase system,
with the additional benefit that the maximum voltage in the system is
lower (208 versus 240).
One question remains to be answered: how
in the world do we get three AC voltage sources whose phase angles are
exactly 120o apart? Obviously we can't center-tap a
transformer or alternator winding like we did in the split-phase system,
since that can only give us voltage waveforms that are either in phase
or 180o out of phase. Perhaps we could figure out some way to
use capacitors and inductors to create phase shifts of 120o,
but then those phase shifts would depend on the phase angles of our load
impedances as well (substituting a capacitive or inductive load for a
resistive load would change everything!).
The best way to get the phase shifts
we're looking for is to generate it at the source: construct the AC
generator (alternator) providing the power in such a way that the
rotating magnetic field passes by three sets of wire windings, each set
spaced 120o apart around the circumference of the machine:
Together, the six "pole" windings of a
three-phase alternator are connected to comprise three winding pairs,
each pair producing AC voltage with a phase angle 120o
shifted from either of the other two winding pairs. The interconnections
between pairs of windings (as shown for the single-phase alternator: the
jumper wire between windings 1a and 1b) have been omitted from the
three-phase alternator drawing for simplicity.
In our example circuit, we showed the
three voltage sources connected together in a "Y" configuration
(sometimes called the "star" configuration), with one lead of each
source tied to a common point (the node where we attached the "neutral"
conductor). The common way to depict this connection scheme is to draw
the windings in the shape of a "Y" like this:
The "Y" configuration is not the only
option open to us, but it is probably the easiest to understand at
first. More to come on this subject later in the chapter.
- REVIEW:
- A single-phase power system is
one where there is only one AC voltage source (one source voltage
waveform).
- A split-phase power system is
one where there are two voltage sources, 180o phase-shifted
from each other, powering a two series-connected loads. The advantage
of this is the ability to have lower conductor currents while
maintaining low load voltages for safety reasons.
- A polyphase power system uses
multiple voltage sources at different phase angles from each other
(many "phases" of voltage waveforms at work). A polyphase power system
can deliver more power at less voltage with smaller-gage conductors
than single- or split-phase systems.
- The phase-shifted voltage sources
necessary for a polyphase power system are created in alternators with
multiple sets of wire windings. These winding sets are spaced around
the circumference of the rotor's rotation at the desired angle(s).
Phase rotation
Let's take the three-phase alternator
design laid out earlier and watch what happens as the magnet rotates:
The phase angle shift of 120o
is a function of the actual rotational angle shift of the three pairs of
windings. If the magnet is rotating clockwise, winding 3 will generate
its peak instantaneous voltage exactly 120o (of alternator
shaft rotation) after winding 2, which will hits its peak 120o
after winding 1. The magnet passes by each pole pair at different
positions in the rotational movement of the shaft. Where we decide to
place the windings will dictate the amount of phase shift between the
windings' AC voltage waveforms. If we make winding 1 our "reference"
voltage source for phase angle (0o), then winding 2 will have
a phase angle of -120o (120o lagging, or 240o
leading) and winding 3 an angle of -240o (or 120o
leading).
This sequence of phase shifts has a
definite order. For clockwise rotation of the shaft, the order is 1-2-3
(winding 1 peaks first, them winding 2, then winding 3). This order
keeps repeating itself as long as we continue to rotate the alternator's
shaft:
However, if we reverse the
rotation of the alternator's shaft (turn it counter-clockwise), the
magnet will pass by the pole pairs in the opposite sequence. Instead of
1-2-3, we'll have 3-2-1. Now, winding 2's waveform will be leading
120o ahead of 1 instead of lagging, and 3 will be another 120o
ahead of 2:
The order of voltage waveform sequences
in a polyphase system is called phase rotation or phase
sequence. If we're using a polyphase voltage source to power
resistive loads, phase rotation will make no difference at all. Whether
1-2-3 or 3-2-1, the voltage and current magnitudes will all be the same.
There are some applications of three-phase power, as we will see
shortly, that depend on having phase rotation being one way or the
other. Since voltmeters and ammeters would be useless in telling us what
the phase rotation of an operating power system is, we need to have some
other kind of instrument capable of doing the job.
One ingenious circuit design uses a
capacitor to introduce a phase shift between voltage and current, which
is then used to detect the sequence by way of comparison between the
brightness of two indicator lamps:
The two lamps are of equal filament
resistance and wattage. The capacitor is sized to have approximately the
same amount of reactance at system frequency as each lamp's resistance.
If the capacitor were to be replaced by a resistor of equal value to the
lamps' resistance, the two lamps would glow at equal brightness, the
circuit being balanced. However, the capacitor introduces a phase shift
between voltage and current in the third leg of the circuit equal to 90o.
This phase shift, greater than 0o but less than 120o,
skews the voltage and current values across the two lamps according to
their phase shifts relative to phase 3. The following SPICE analysis
demonstrates what will happen:
phase rotation detector -- sequence = v1-v2-v3
v1 1 0 ac 120 0 sin
v2 2 0 ac 120 120 sin
v3 3 0 ac 120 240 sin
r1 1 4 2650
r2 2 4 2650
c1 3 4 1u
.ac lin 1 60 60
.print ac v(1,4) v(2,4) v(3,4)
.end
freq v(1,4) v(2,4) v(3,4)
6.000E+01 4.810E+01 1.795E+02 1.610E+02
The resulting phase shift from the
capacitor causes the voltage across phase 1 lamp (between nodes 1 and 4)
to fall to 48.1 volts and the voltage across phase 2 lamp (between nodes
2 and 4) to rise to 179.5 volts, making the first lamp dim and the
second lamp bright. Just the opposite will happen if the phase sequence
is reversed:
phase rotation detector -- sequence = v3-v2-v1
v1 1 0 ac 120 240 sin
v2 2 0 ac 120 120 sin
v3 3 0 ac 120 0 sin
r1 1 4 2650
r2 2 4 2650
c1 3 4 1u
.ac lin 1 60 60
.print ac v(1,4) v(2,4) v(3,4)
.end
freq v(1,4) v(2,4) v(3,4)
6.000E+01 1.795E+02 4.810E+01 1.610E+02
Here, the first lamp receives 179.5 volts
while the second receives only 48.1 volts.
We've investigated how phase rotation is
produced (the order in which pole pairs get passed by the alternator's
rotating magnet) and how it can be changed by reversing the alternator's
shaft rotation. However, reversal of the alternator's shaft rotation is
not usually an option open to an end-user of electrical power supplied
by a nationwide grid ("the" alternator actually being the combined total
of all alternators in all power plants feeding the grid). There is a
much easier way to reverse phase sequence than reversing alternator
rotation: just exchange any two of the three "hot" wires going to a
three-phase load.
This trick makes more sense if we take
another look at a running phase sequence of a three-phase voltage
source:
1-2-3 rotation: 1-2-3-1-2-3-1-2-3-1-2-3-1-2-3 . . .
3-2-1 rotation: 3-2-1-3-2-1-3-2-1-3-2-1-3-2-1 . . .
What is commonly designated as a "1-2-3"
phase rotation could just as well be called "2-3-1" or "3-1-2," going
from left to right in the number string above. Likewise, the opposite
rotation (3-2-1) could just as easily be called "2-1-3" or "1-3-2."
Starting out with a phase rotation of
3-2-1, we can try all the possibilities for swapping any two of the
wires at a time and see what happens to the resulting sequence:
No matter which pair of "hot" wires out
of the three we choose to swap, the phase rotation ends up being
reversed (1-2-3 gets changed to 2-1-3, 1-3-2 or 3-2-1, all equivalent).
- REVIEW:
- Phase rotation,
or phase sequence, is the order in which the voltage waveforms
of a polyphase AC source reach their respective peaks. For a
three-phase system, there are only two possible phase sequences: 1-2-3
and 3-2-1, corresponding to the two possible directions of alternator
rotation.
- Phase rotation has no impact on
resistive loads, but it will have impact on unbalanced reactive loads,
as shown in the operation of a phase rotation detector circuit.
- Phase rotation can be reversed by
swapping any two of the three "hot" leads supplying three-phase power
to a three-phase load.
Polyphase motor
design
Perhaps the most important benefit of
polyphase AC power over single-phase is the design and operation of AC
motors. As we studied in the first chapter of this book, some types of
AC motors are virtually identical in construction to their alternator
(generator) counterparts, consisting of stationary wire windings and a
rotating magnet assembly. (Other AC motor designs are not quite this
simple, but we will leave those details to another lesson).
If the rotating magnet is able to keep up
with the frequency of the alternating current energizing the
electromagnet windings (coils), it will continue to be pulled around
clockwise. However, clockwise is not the only valid direction for this
motor's shaft to spin. It could just as easily be powered in a
counter-clockwise direction by the same AC voltage waveform:
Notice that with the exact same sequence
of polarity cycles (voltage, current, and magnetic poles produced by the
coils), the magnetic rotor can spin in either direction. This is a
common trait of all single-phase AC "induction" and "synchronous"
motors: they have no normal or "correct" direction of rotation. The
natural question should arise at this point: how can the motor get
started in the intended direction if it can run either way just as well?
The answer is that these motors need a little help getting started. Once
helped to spin in a particular direction. they will continue to spin
that way as long as AC power is maintained to the windings.
Where that "help" comes from for a
single-phase AC motor to get going in one direction can vary. Usually,
it comes from an additional set of windings positioned differently from
the main set, and energized with an AC voltage that is out of phase with
the main power:
These supplementary coils are typically
connected in series with a capacitor to introduce a phase shift in
current between the two sets of windings:
That phase shift creates magnetic fields
from coils 2a and 2b that are equally out of step with the fields from
coils 1a and 1b. The result is a set of magnetic fields with a definite
phase rotation. It is this phase rotation that pulls the rotating magnet
around in a definite direction.
Polyphase AC motors require no such
trickery to spin in a definite direction. Because their supply voltage
waveforms already have a definite rotation sequence, so do the
respective magnetic fields generated by the motor's stationary windings.
In fact, the combination of all three phase winding sets working
together creates what is often called a rotating magnetic field.
It was this concept of a rotating magnetic field that inspired Nikola
Tesla to design the world's first polyphase electrical systems (simply
to make simpler, more efficient motors). The line current and safety
advantages of polyphase power over single phase power were discovered
later.
What can be a confusing concept is made
much clearer through analogy. Have you ever seen a row of blinking light
bulbs such as the kind used in Christmas decorations? Some strings
appear to "move" in a definite direction as the bulbs alternately glow
and darken in sequence. Other strings just blink on and off with no
apparent motion. What makes the difference between the two types of bulb
strings? Answer: phase shift!
Examine a string of lights where every
other bulb is lit at any given time:
When all of the "1" bulbs are lit, the
"2" bulbs are dark, and visa-versa. With this blinking sequence, there
is no definite "motion" to the bulbs' light. Your eyes could follow a
"motion" from left to right just as easily as from right to left.
Technically, the "1" and "2" bulb blinking sequences are 180o
out of phase (exactly opposite each other). This is analogous to the
single-phase AC motor, which can run just as easily in either direction,
but which cannot start on its own because its magnetic field alternation
lacks a definite "rotation."
Now let's examine a string of lights
where there are three sets of bulbs to be sequenced instead of just two,
and these three sets are equally out of phase with each other:
If the lighting sequence is 1-2-3 (the
sequence shown), the bulbs will appear to "move" from left to right. Now
imagine this blinking string of bulbs arranged into a circle:
Now the lights appear to be "moving" in a
clockwise direction because they are arranged around a circle instead of
a straight line. It should come as no surprise that the appearance of
motion will reverse if the phase sequence of the bulbs is reversed.
The blinking pattern will either appear
to move clockwise or counter-clockwise depending on the phase sequence.
This is analogous to a three-phase AC motor with three sets of windings
energized by voltage sources of three different phase shifts:
With phase shifts of less than 180o
we get true rotation of the magnetic field. With single-phase motors,
the rotating magnetic field necessary for self-starting must to be
created by way of capacitive phase shift. With polyphase motors, the
necessary phase shifts are there already. Plus, the direction of shaft
rotation for polyphase motors is very easily reversed: just swap any two
"hot" wires going to the motor, and it will run in the opposite
direction!
- REVIEW:
- AC "induction" and "synchronous"
motors work by having a rotating magnet follow the alternating
magnetic fields produced by stationary wire windings.
- Single-phase AC motors of this type
need help to get started spinning in a particular direction.
- By introducing a phase shift of less
than 180o to the magnetic fields in such a motor, a
definite direction of shaft rotation can be established.
- Single-phase induction motors often
use an auxiliary winding connected in series with a capacitor to
create the necessary phase shift.
- Polyphase motors don't need such
measures; their direction of rotation is fixed by the phase sequence
of the voltage they're powered by.
- Swapping any two "hot" wires on a
polyphase AC motor will reverse its phase sequence, thus reversing its
shaft rotation.
Three-phase Y and Δ configurations
Initially we explored the idea of
three-phase power systems by connecting three voltage sources together
in what is commonly known as the "Y" (or "star") configuration. This
configuration of voltage sources is characterized by a common connection
point joining one side of each source:
If we draw a circuit showing each voltage
source to be a coil of wire (alternator or transformer winding) and do
some slight rearranging, the "Y" configuration becomes more obvious:
The three conductors leading away from
the voltage sources (windings) toward a load are typically called
lines, while the windings themselves are typically called phases.
In a Y-connected system, there may or may not be a neutral wire attached
at the junction point in the middle, although it certainly helps
alleviate potential problems should one element of a three-phase load
fail open, as discussed earlier:
When we measure voltage and current in
three-phase systems, we need to be specific as to where we're
measuring. Line voltage refers to the amount of voltage measured
between any two line conductors in a balanced three-phase system. With
the above circuit, the line voltage is roughly 208 volts. Phase
voltage refers to the voltage measured across any one component
(source winding or load impedance) in a balanced three-phase source or
load. For the circuit shown above, the phase voltage is 120 volts. The
terms line current and phase current follow the same
logic: the former referring to current through any one line conductor,
and the latter to current through any one component.
Y-connected sources and loads always have
line voltages greater than phase voltages, and line currents equal to
phase currents. If the Y-connected source or load is balanced, the line
voltage will be equal to the phase voltage times the square root of 3:
However, the "Y" configuration is not the
only valid one for connecting three-phase voltage source or load
elements together. Another configuration is known as the "Delta," for
its geometric resemblance to the Greek letter of the same name (Δ). Take
close notice of the polarity for each winding in the drawing below:
At first glance it seems as though three
voltage sources like this would create a short-circuit, electrons
flowing around the triangle with nothing but the internal impedance of
the windings to hold them back. Due to the phase angles of these three
voltage sources, however, this is not the case.
One quick check of this is to use
Kirchhoff's Voltage Law to see if the three voltages around the loop add
up to zero. If they do, then there will be no voltage available to push
current around and around that loop, and consequently there will be no
circulating current. Starting with the top winding and progressing
counter-clockwise, our KVL expression looks something like this:
Indeed, if we add these three vector
quantities together, they do add up to zero. Another way to verify the
fact that these three voltage sources can be connected together in a
loop without resulting in circulating currents is to open up the loop at
one junction point and calculate voltage across the break:
Starting with the right winding (120 V ∠
120o) and progressing counter-clockwise, our KVL equation
looks like this:
Sure enough, there will be zero voltage
across the break, telling us that no current will circulate within the
triangular loop of windings when that connection is made complete.
Having established that a Δ-connected
three-phase voltage source will not burn itself to a crisp due to
circulating currents, we turn to its practical use as a source of power
in three-phase circuits. Because each pair of line conductors is
connected directly across a single winding in a Δ circuit, the line
voltage will be equal to the phase voltage. Conversely, because each
line conductor attaches at a node between two windings, the line current
will be the vector sum of the two joining phase currents. Not
surprisingly, the resulting equations for a Δ configuration are as
follows:
Let's see how this works in an example
circuit:
With each load resistance receiving 120
volts from its respective phase winding at the source, the current in
each phase of this circuit will be 83.33 amps:
So, the each line current in this
three-phase power system is equal to 144.34 amps, substantially more
than the line currents in the Y-connected system we looked at earlier.
One might wonder if we've lost all the advantages of three-phase power
here, given the fact that we have such greater conductor currents,
necessitating thicker, more costly wire. The answer is no. Although this
circuit would require three number 1 gage copper conductors (at 1000
feet of distance between source and load this equates to a little over
750 pounds of copper for the whole system), it is still less than the
1000+ pounds of copper required for a single-phase system delivering the
same power (30 kW) at the same voltage (120 volts
conductor-to-conductor).
One distinct advantage of a Δ-connected
system is its lack of a neutral wire. With a Y-connected system, a
neutral wire was needed in case one of the phase loads were to fail open
(or be turned off), in order to keep the phase voltages at the load from
changing. This is not necessary (or even possible!) in a Δ-connected
circuit. With each load phase element directly connected across a
respective source phase winding, the phase voltage will be constant
regardless of open failures in the load elements.
Perhaps the greatest advantage of the
Δ-connected source is its fault tolerance. It is possible for one of the
windings in a Δ-connected three-phase source to fail open without
affecting load voltage or current!
The only consequence of a source winding
failing open for a Δ-connected source is increased phase current in the
remaining windings. Compare this fault tolerance with a Y-connected
system suffering an open source winding:
With a Δ-connected load, two of the
resistances suffer reduced voltage while one remains at the original
line voltage, 208. A Y-connected load suffers an even worse fate with
the same winding failure in a Y-connected source:
In this case, two load resistances suffer
reduced voltage while the third loses supply voltage completely! For
this reason, Δ-connected sources are preferred for reliability. However,
if dual voltages are needed (e.g. 120/208) or preferred for lower line
currents, Y-connected systems are the configuration of choice.
- REVIEW:
- The conductors connected to the three
points of a three-phase source or load are called lines.
- The three components comprising a
three-phase source or load are called phases.
- Line voltage
is the voltage measured between any two lines in a three-phase
circuit.
- Phase voltage
is the voltage measured across a single component in a three-phase
source or load.
- Line current
is the current through any one line between a three-phase source and
load.
- Phase current
is the current through any one component comprising a three-phase
source or load.
- In balanced "Y" circuits, line voltage
is equal to phase voltage times the square root of 3, while line
current is equal to phase current.
-
- In balanced Δ circuits, line voltage
is equal to phase voltage, while line current is equal to phase
current times the square root of 3.
-
- Δ-connected three-phase voltage
sources give greater reliability in the event of winding failure than
Y-connected sources. However, Y-connected sources can deliver the same
amount of power with less line current than Δ-connected sources.
Three-phase transformer circuits
Since three-phase is used so often for
power distribution systems, it makes sense that we would need
three-phase transformers to be able to step voltages up or down. This is
only partially true, as regular single-phase transformers can be ganged
together to transform power between two three-phase systems in a variety
of configurations, eliminating the requirement for a special three-phase
transformer. However, special three-phase transformers are built for
those tasks, and are able to perform with less material requirement,
less size, and less weight from their modular counterparts.
A three-phase transformer is made of
three sets of primary and secondary windings, each set wound around one
leg of an iron core assembly. Essentially it looks like three
single-phase transformers sharing a joined core:
Those sets of primary and secondary
windings will be connected in either Δ or Y configurations to form a
complete unit. The various combinations of ways that these windings can
be connected together in will be the focus of this section.
Whether the winding sets share a common
core assembly or each winding pair is a separate transformer, the
winding connection options are the same:
- Primary - Secondary
- Y - Y
- Y - Δ
- Δ - Y
- Δ - Δ
The reasons for choosing a Y or Δ
configuration for transformer winding connections are the same as for
any other three-phase application: Y connections provide the opportunity
for multiple voltages, while Δ connections enjoy a higher level of
reliability (if one winding fails open, the other two can still maintain
full line voltages to the load).
Probably the most important aspect of
connecting three sets of primary and secondary windings together to form
a three-phase transformer bank is attention to proper winding phasing
(the dots used to denote "polarity" of windings). Remember the proper
phase relationships between the phase windings of Δ and Y:
Getting this phasing correct when the
windings aren't shown in regular Y or Δ configuration can be tricky. Let
me illustrate:
Three individual transformers are to be
connected together to transform power from one three-phase system to
another. First, I'll show the wiring connections for a Y-Y
configuration:
Note how all the winding ends marked with
dots are connected to their respective phases A, B, and C, while the
non-dot ends are connected together to form the centers of each "Y".
Having both primary and secondary winding sets connected in "Y"
formations allows for the use of neutral conductors (N1 and N2)
in each power system.
Now, we'll take a look at a Y-Δ
configuration:
Note how the secondary windings (bottom
set) are connected in a chain, the "dot'" side of one winding connected
to the "non-dot" side of the next, forming the Δ loop. At every
connection point between pairs of windings, a connection is made to a
line of the second power system (A, B, and C).
Now, let's examine a Δ-Y system:
Such a configuration would allow for the
provision of multiple voltages (line-to-line or line-to-neutral) in the
second power system, from a source power system having no neutral.
And finally, we turn to the Δ-Δ
configuration:
When there is no need for a neutral
conductor in the secondary power system, Δ-Δ connection schemes are
preferred because of the inherent reliability of the Δ configuration.
Considering that a Δ configuration can
operate satisfactorily missing one winding, some power system designers
choose to create a three-phase transformer bank with only two
transformers, representing a Δ-Δ configuration with a missing winding in
both the primary and secondary sides:
This configuration is called "V" or
"Open-Δ." Of course, each of the two transformers have to be oversized
to handle the same amount of power as three in a standard Δ
configuration, but the overall size, weight, and cost advantages are
often worth it. Bear in mind, however, that with one winding set missing
from the Δ shape, this system no longer provides the fault tolerance of
a normal Δ-Δ system. If one of the two transformers were to fail, the
load voltage and current would definitely be affected.
The following photograph shows a bank of
step-up transformers at the Grand Coulee hydroelectric dam in Washington
state. Several transformers (green in color) may be seen from this
vantage point, and they are grouped in threes: three transformers per
hydroelectric generator, wired together in some form of three-phase
configuration. The photograph doesn't reveal the primary winding
connections, but it appears the secondaries are connected in a Y
configuration, being that there is only one large high-voltage insulator
protruding from each transformer. This suggests the other side of each
transformer's secondary winding is at or near ground potential, which
could only be true in a Y system. The building to the left is the
powerhouse, where the generators and turbines are housed. On the right,
the sloping concrete wall is the downstream face of the dam:
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.
Harmonic phase sequences
In the last section, we saw how the 3rd
harmonic and all of its integer multiples (collectively called
triplen harmonics) generated by 120o phase-shifted
fundamental waveforms are actually in phase with each other. In a 60 Hz
three-phase power system, where phases A, B, and C
are 120o apart, the third-harmonic multiples of those
frequencies (180 Hz) fall perfectly into phase with each other. This can
be thought of in graphical terms, and/or in mathematical terms:
If we extend the mathematical table to
include higher odd-numbered harmonics, we will notice an interesting
pattern develop with regard to the rotation or sequence of the harmonic
frequencies:
Harmonics such as the 7th, which "rotate"
with the same sequence as the fundamental, are called positive
sequence. Harmonics such as the 5th, which "rotate" in the opposite
sequence as the fundamental, are called negative sequence.
Triplen harmonics (3rd and 9th shown in this table) which don't "rotate"
at all because they're in phase with each other, are called zero
sequence.
This pattern of
positive-zero-negative-positive continues indefinitely for all
odd-numbered harmonics, lending itself to expression in a table like
this:
Sequence especially matters when we're
dealing with AC motors, since the mechanical rotation of the rotor
depends on the torque produced by the sequential "rotation" of the
applied 3-phase power. Positive-sequence frequencies work to push the
rotor in the proper direction, whereas negat | 
