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Harmonic Mitigation Selection
This paper is a review of the methods available to reduce line current harmonics seen with AC motordrives. There are several methods that can reduce the Ithd, and by doing so, also reduce the Vthd on
the secondary terminals of the transformer feeding the drive. However, each one impacts the power
distribution system in a different way, and each one impacts the drive operation in a different way.This paper will review the effects along with the advantages and the disadvantages of each method.Included in this paper are test results of several of the harmonic mitigation methods using a 100hp
1336F drive.
1. Introduction to Harmonics
2. The Basic Drive System
3. AC Line Reactor4. DC Link Choke
5. Pseudo 12-Pulse
6. Multi-Pulse Converter
7. Passive Filter8. Active Filter
9. Active Front-End
10. Other Comparisons11. Comparison Checklist
12. Summary Rules-of-Thumb
Rick Hoadley
April 2005
Acknowledgments: Special thanks to Nick Guskov, Gary Woltersdorf and John Streicher for
engineering, technical and marketing assistance.
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Harmonic Mitigation Selection
When faced with the task of reducing line current harmonics caused by AC drives, there are severalalternatives available. How does one pick the best solution? What criteria should be used to weigh
the alternatives? This paper will deal with these topics.
1. Introduction to Harmonics
ConverterAC to DC
DC BusFilter
InverterDC to AC
AC Drive
A C
L i n e
I n p u t
A C
M o t or O u t p u t
Why do drives create current harmonics?
We know that an AC drive creates line currentharmonics because the line current does not look
like a sine wave. Why? Because the converter
within the drive takes the 3 phase AC line voltagesand converts them into DC that feeds a filter and the
inverter section. The inverter section is what then
takes the DC and creates the variable voltage,
variable frequency output for the motor. See Fig 1. Fig 1. Sections of a 6-Pulse AC drive
During the process of 3-phase AC to DC conversion,
the line current can resemble a pair of distinctpulses. VMab.V = f...
La.I = f(t, b...
Cbus.V = f(...
VMac.V = f...
VMan.V [V]...
81.50m
81.50m
93.50m
93.50m
82.50m
82.50m
85.00m
85.00m
87.50m
87.50m
90.00m
90.00m
92.50m
92.50m
100.0 -100.0
00.0
0 0
00.0 200.0
00.0 400.0
00.0 600.0
Vab Vac
Van
Vbus
Ia
VMab.V = f...
Cbus.V = f(...
VMac.V = f...
VMan.V [V]...
81.50
La.I = f(t, b...
m
81.50m
93.50m
93.50m
82.50m
82.50m
85.00m
85.00m
87.50m
87.50m
90.00m
90.00m
92.50m
92.50m
100.0 -100.0
00.0
0 0
00.0 200.0
00.0 400.0
00.0 600.0
Vab Vac
Van
Vbus
Ia
Vab Vac
Van
Vbus
Ia
Why does the current look like that? Remember,
current can only flow whenever the instantaneousline-to-line voltage exceeds the DC bus capacitor
voltage. So for phase A, when can current flow?
Only when the peak voltage between phases A and
B exceed the cap voltage, or only when the peak voltage between phases A and C exceed the cap
voltage. No current flows in phase A due to thepeak voltage between phases B and C. This is
easily seen in Fig 2.
VMab.V = f...The currents in phases B and C are the same, simply
phase shifted with respect to each other.La.I = f(t, b...
Cbus.V = f(...
VMac.V = f...
VMan.V [V]...
181.5m
181.5m
193.5m
193.5m
182.5m
182.5m
185.0m
185.0m
187.5m
187.5m
190.0m
190.0m
192.5m
192.5m
100.0 -100.0
00.0
0 0
00.0 200.0
00.0 400.0
00.0 600.0
Vab Vac
Van
Vbus
Ia
VMab.V = f...
Cbus.V = f(...
VMac.V = f...
VMan.V [V]...
181.5
La.I = f(t, b...
m
181.5m
193.5m
193.5m
182.5m
182.5m
185.0m
185.0m
187.5m
187.5m
190.0m
190.0m
192.5m
192.5m
100.0 -100.0
00.0
0 0
00.0 200.0
00.0 400.0
00.0 600.0
Vab Vac
Van
Vbus
Ia
Fig 3. Current in phase A with an increase inline impedance
Fig 2. Current in phase A with very little lineimpedance
If impedance is added to the circuit, either byadding a DC link choke, or an AC line reactor, the
current does not change as rapidly, causing thewaveform to change shape as seen in Fig 3.
What causes voltage distortion?
Since the current looks like these pulses and do notlook like a sine wave, when we analyze the current
waveform we will see that it is composed of afundamental sine wave along with a combination of
several harmonics. These are the harmonic currents
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that cause voltage distortions on the powerdistribution system.
Rfund.V =...
Rharm.V ...
Rtotal.V =...
0
0
40.00
40.00
10.00
10.00
20.00
20.00
30.00m
30.00m
150.0 -150.
50.0
0
100.0 -100.
50.0 -50.0
0.0 50.0
00.0 100.0
The voltage distortions appear whenever currentharmonics flow through the impedance within the
transformer feeding the drive, and through anyimpedance in the wiring between the transformerand the drive. As you may realize, the voltage
distortion is least at the transformer terminals, and
greatest at the drive terminals. It is this voltage
distortion that causes problems for other pieces of equipment fed by that transformer.
Fig 4. The red line current is made up of thegreen fundamental current plus the blueharmonic currents
The next two figures show how the voltagedistortion will change depending on the size of the
transformer feeding a drive. Shown in Fig 5 is the
voltage waveform at the terminals of a 1500kVAtransformer, and the current waveform of a 75hp
AC drive. There is very little distortion in the line-
to-line voltage. However, Fig 6 shows the
waveforms at the terminals of a 75kVA transformer.There is a significant amount of flat-topping in the
distorted voltage. This amounts to about 10% Vthd.Fig 5. The red line current of a 75hp drivecauses little distortion seen in the blue line-to-line voltage at the 1500kVA xfmr sec terminals
How much is too much?
The question we now ask is, “How much is toomuch?” This is where the recommendations found
in IEEE std 519-1992 come into play. Table 10.2
recommends that the voltage distortion be less than3% for hospitals and airports, less than 5% for all
other facilities, but may be as high as 10% if
converter loads are the only loads on a particulartransformer. This is shown in Fig 7. The location
for measuring the voltage distortion can be
anywhere in the facility where non-linear and linear
loads are connected together, typically at thesecondary terminals of distribution transformers. Fig 6. The red line current of a 75hp drive
causes significant distortion seen in the blueline-to-line voltage at the 75kVA xfmr sec
terminals
Why is there a concern for limiting the voltagedistortion? When motors are fed line voltages that
contain voltage distortion, there will be additional
heating in the motors and a loss of torque. It isdifficult to predict how much the motor needs to be
derated due to the level and content of the voltage
harmonics present. However, in NEMA MG1-1998,Revision 1, Part 30, Section 30.1.2.1, they define an
Harmonic Voltage Factor and a give a derating
chart. This was part of the consideration that IEEE
Application Maximum THD (%)
Special Applications - hospitals and airports 3.0%
General System 5.0%
Dedicated System - exclus ively converte r load 10.0%
Harmonic Voltage LimitsLow-Voltage Systems
Table 10.2
Application Maximum THD (%)
Special Applications - hospitals and airports 3.0%
General System 5.0%
Dedicated System - exclus ively converte r load 10.0%
Harmonic Voltage LimitsLow-Voltage Systems
Table 10.2
Fig 7. Table 10.2 from IEEE 519 showingrecommended Vthd limits for variousapplications
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used when determining the values in Table 10.2. Inaddition to overheating motors, other equipment can
also malfunction due to voltage distortions on the
power lines.
Table 10.3
Current distortion Limits for General Distribution Systems (120V through 69,000V)
Isc/ Iload <11 11<=h<17 17<=h<23 23<=h<35 35<=h TDD (%)
<20 4.0 2.0 1.5 0.6 0.3 5.0
20<50 7.0 3.5 2.5 1.0 0.5 8.0
50<100 10.0 4.5 4.0 1.5 0.7 12.0
100<1000 12.0 5.5 5.0 2.0 1.0 15.0
>1000 15.0 7.0 6.0 2.5 1.4 20.0
Even harmonics are limited to 25% of the odd harmonic limits above
Isc=maximum short circuit current at PCCIload=maximum demand load current (fundamental frequency component) at PCC
Maximum Harmonic Current Distortion in Percent of Iload
Table 10.3
Current distortion Limits for General Distribution Systems (120V through 69,000V)
Isc/ Iload <11 11<=h<17 17<=h<23 23<=h<35 35<=h TDD (%)
<20 4.0 2.0 1.5 0.6 0.3 5.0
20<50 7.0 3.5 2.5 1.0 0.5 8.0
50<100 10.0 4.5 4.0 1.5 0.7 12.0
100<1000 12.0 5.5 5.0 2.0 1.0 15.0
>1000 15.0 7.0 6.0 2.5 1.4 20.0
Even harmonics are limited to 25% of the odd harmonic limits above
Isc=maximum short circuit current at PCCIload=maximum demand load current (fundamental frequency component) at PCC
Maximum Harmonic Current Distortion in Percent of Iload
Table 10.3 puts limits on the current harmonics atthe place where multiple customers are connected tothe utility, usually at the metering point. See Fig 8.
This is the primary PCC (Point of Common
Coupling). Table 10.3 allows each customer
connected to the utility to draw some harmoniccurrent and thus create some voltage distortion
which everyone else needs to live with. However,
no one customer should draw so much harmoniccurrent so as to cause the voltage distortion at this
common connection point to exceed the
recommended limits at this point and causeproblems for the other customers.
Fig 8. Table 10.3 from IEEE 519 showingrecommended Ithd limits for various Isc/Iloadcategories
Within a customer’s plant, though, Table 10.3 does
not apply. Instead, only Table 10.2 applies –voltage distortion limits. The user should plan his
equipment type and placement and wiring to make
sure that the voltage distortion at any transformersecondary and at any point where linear and non-
linear loads are connected together are within thelimits in Table 10.2.
Fig 9. Diagram showing overview of whatvoltage distortion limits should be followed atvarious locations in multiple systems
If drives or other converters are the only load on atransformer, this table allows the voltage distortion
to be as high as 10%! For most users, the voltage
distortion should be less than 5%. For hospitals andairports (life-critical applications), the voltage
distortion should not exceed 3%. This is
diagrammed in Fig 9.
If, after a survey and harmonic estimation is made
of the plant, it is determined that there is a need to
reduce the harmonic current drawn by a drive or agroup of drives, several methods are available. We
will review these methods, compare their ability to
reduce the line current harmonics, list the pros andcons of each one, and discuss the effect each
method has on the operation of the drive and on the
power distribution system.
We will start with the characteristics of a basic,
bare-bones drive and draw comparisons from that.
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2. The Basic Drive System Transformer
xfmr% Z
Drive
DC
AC
AC
DCM
The basic drive system, in Fig 10, consists of a
transformer as the power source, a 6-pulse diodebridge converter, a DC bus filter capacitor, an IGBT
inverter bridge, and a motor connected to amechanical load. Fig 1 shows the details of thedrive. The user needs to realize that the transformer
is as much a part of the system, and has an impact
on the harmonics produced by the drive.
Fig 10. The basic drive system, with atransformer feeding an unbuffered drive (a drive
without a DC link choke)
La.I = f( ...
150.0
150.0
200.0
200.0
162.5
162.5
175.0
175.0
187.5
187.5
00.0 -400.
00.0
0 0
00.0 -200.
00.0 200.0
The line current will typically look like what is
shown in Fig 11.
Analyzing this current waveform, we see that it
contains the harmonics shown in Fig 12. The %
Ithd is typically 80 to 120%, meaning that for a100hp drive, with a fundamental current of about
100Arms, the harmonic current content could be
between 80 and 120Arms! Please note that the onlyimpedance limiting the rate of rise of the current
pulses is the transformer impedance and any other
impedance that exists in the cabling between the
transformer and the drive.
Fig 11. Typical line current for this driveconfiguration
0
10
20
30
40
50
60
70
80
90
100
0 6 12 1 8 24 30 36 42 4 8 54 60 66 72 78
Harmonic Number
% H a r m o n i c C u r r e n
Most drives that are rated 5hp or less are like thisbasic drive. One or two drives in a plant usually
don’t cause any problems, but if you have 200*5hp
drives, this is the equivalent to a 1000hp drive. Thepeak currents can be around 3000Apk – causing
heating problems in the transformer and wiring,
causing circuit breakers to open or fuses to blow,and greatly distorting the voltage on the secondary
of the distribution transformer that could cause
other equipment connected to the transformer tomalfunction.
Fig 12. Harmonic spectrum showing themagnitudes of the individual harmonicscontained in the line current
30.00
40.00
50.00
60.00
70.00
80.00
90.00
100.00
0 10 20 30 40 50 60 70 80 90 100
PF disp
PF dist
PF total
Simply adding an AC line reactor, as discussed in
the next section, will eliminate a lot of the problemsassociated with this drive configuration.
Fig 13 shows how the power factor is affected bythe harmonics as the load is varied from 0 to 100%.
The green trace is the displacement power factor,
the blue trace is the distortion power factor, and thepink trace is the total power factor (the product of
the other two).
Fig 13. Power factor vs load - green isdisplacement PF, blue is distortion PF, pink istotal PF
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3. AC Line Reactor Transformer
xfmr% Z
Line Reactor
Drive
DC
AC
AC
DCM
Adding an AC line reactor to the drive, as shown in
Fig 14 and 15, has a dramatic effect in reducing theline current harmonics. The waveform in Fig 16
shows peak currents that are about half of what wasseen in the previous configuration. The % Ithd isnow typically 30 to 60%. Fig 17 shows the
harmonic spectrum. Notice how the 5th and 7th
harmonic currents have dropped.
Fig 14. Drive configuration with the addition ofAC line reactors
ConverterAC to DC
InverterDC to AC
DCBusFilter
AC Drive
A C
M o t or O u t p u t
A C
L i n e
I n p u t
Not only does the AC line reactor reduce the line
current harmonics, but it also helps to prevent the
drive from tripping as a result of power line voltagedips and the voltage transients that occur when
power factor correction caps are connected to the
power system. Fig 15. Details of drive configuration with ACline reactors
Fig 18 shows what typically happens during a
power dip. When the line voltage drops to half of
nominal for only a portion of a cycle, the DC busvoltage starts to drop quickly because the drive is
providing power to the motor during this time and
the energy from the capacitors is not beingreplenished from the power line. When the line
voltage returns to normal, there is a large surge of current “recharging” the DC bus caps. The DC bus
then quickly shoots up. This recharging current
surge can damage the diode bridge. Simplyinserting a 3% line reactor in the circuit will greatly
reduce the peak recharging current as seen in Fig 19.
This can prevent nuisance fuse blowing.
La.I = f(t...
150.0m
150.0m
200.0m
200.0m
162.5m
162.5m
175.0
175.0
187.5m
187.5m
00.0 -400.0
00.0
0 0
200.0 -200.0
00.0 200.0
Fig 16. Typical line current for this driveconfiguration
0
10
20
30
40
50
60
70
80
90
100
0 6 12 18 24 30 36 42 48 54 60 66 72 78
Harmonic Number
% H a r m o n i c C u r r e n
An AC line reactor also helps prevent OverVoltage
trips whenever power factor correction caps areswitched onto the power lines. The voltage
waveform that normally results during this time is
seen in Fig 20, showing how the DC bus voltage in
the drive quickly tries to charge up to the peak of the ringing waveform. Again, a large surge of
current can also result. In addition, the DC bus
voltage can be taken to a level that will cause anOverVoltage trip. Inserting a 3% AC line reactor
limits the peak charging current and limits the effect
on the DC bus voltage as seen in Fig 21. If youhave a drive that trips on OverVoltage every
morning at 7:00am, this could be the reason.
Fig 17. Harmonic spectrum showing themagnitudes of the individual harmonicscontained in the line current
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The only drawback in the use of AC line reactors isthat it creates a drop in the DC bus voltage as full
load, full speed is approached. Fig 22 shows how
the DC bus voltage decreases as the load increases,comparing a basic drive to a drive with a 3% line
reactor, a drive with a 5% line reactor, and a drivewith a DC link choke. A 3% line reactor typicallycauses a 3% drop in bus voltage. A 5% line reactor
causes a 5% drop in bus voltage. The effect this has
on the drive system is that full power would not be
available from the motor because rated V/Hz couldnot be achieved from the drive. If full load, full
speed operation is not required, then this is not an
issue. However the user needs to be aware of thiseffect.
Vbus = 677Vdc
Ipk = 32A
Vbus = 587Vdc
Ipk = 120A
Vbus = 677Vdc
Ipk = 32A
Vbus = 587Vdc
Ipk = 120A
Fig 18. Operation during a power dip without anAC line reactor– green is the DC bus voltage,blue is the line-to-line voltage, red is the buscurrent
Vbus = 637Vdc
Ipk = 10A
Vbus = 545Vdc to 692Vdc
Ipk = 37A
Vbus = 637Vdc
Ipk = 10A
Vbus = 545Vdc to 692Vdc
Ipk = 37A
It is possible to place multiple drives on a singleline reactor. A rule of thumb for this is to select a
3% line reactor for 3 drives in parallel, or a 5% line
reactor for 5 drives in parallel. The hp rating of the
line reactor would be the total hp of the drivesconnected to it. Fig 23 diagrams this configuration.
Fig 19. Operation during a power dip with an ACline reactor – green is the DC bus voltage, blueis the line-to-line voltage, red is the bus current
Vbus = 677Vdc
Ipk = 32A
Vbus = 806Vdc
Ipk = 507A
Vbus = 677Vdc
Ipk = 32A
Vbus = 806Vdc
Ipk = 507A
% Vbus vs Load (Isc/Iload = 47)
94.00
95.00
96.00
97.00
98.00
99.00
100.00
101.00
102.00
0 10 20 30 40 50 60 70 80 90 100
unbuffered
buffered
3% line reactor
5% line reactor
1.35*Vac
% Vbus vs Load (Isc/Iload = 47)
94.00
95.00
96.00
97.00
98.00
99.00
100.00
101.00
102.00
0 10 20 30 40 50 60 70 80 90 100
unbuffered
buffered
3% line reactor
5% line reactor
1.35*Vac Fig 20. Operation during a ringing transientwithout an AC line reactor - green is the DC bus
Vbus = 637Vdc
Ipk = 10A
Vbus = 653Vdc
Ipk = 19A
Vbus = 637Vdc
Ipk = 10A
Vbus = 653Vdc
Ipk = 19A
Fig 22. Vbus vs Load – drop in DC bus voltageas the load is increased, showing the effect anAC line reactor has on this parameter
ge, red is thevoltage, blue is the line-to-line voltabus current
Transformer
xfmr% Z
Line Reactor
Drive
DC
AC
AC
DC
DC
AC
AC
DC
DC
AC
AC
DC
M
M
M
Fig 21. Operation during a ringing transient withan AC line reactor – green is the DC busvoltage, blue is the line-to-line voltage, red is thebus current
Fig 23. Diagram of multiple unbufferred driveson a single AC line reactor
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4. DC Link Choke Transformer
xfmr% Z
Drive
DC
AC
AC
DC
DC LinkChoke
M
Most industrial drives larger than 5hp come
equipped with a DC link choke as seen in Fig 24and 25. It usually is made up of two windings on a
common core, with one winding in each of the DCbus legs (one in the positive leg, one in the negativeleg) between the output of the converter bridge and
the bus caps.
Fig 24. Drive configuration with the addition ofDC link chokes (a buffered drive)
ConverterAC to DC
DC BusFilter
InverterDC to AC
AC Drive
A C
L i n e
I n p u t
A C
M o t or O u t p u t
Since this adds inductive impedance to the circuit,the rate of rise of the current into the caps is limited,
just like the effect seen by adding a line reactor to
the input to the converter bridge. There is adifference, though. During operation above about
25% load, the current in the link choke becomes
continuous (seen when the two pulses on the inputmerge into one pulse with two bumps, and the
current does not go to zero between the pairs of
pulses as compared to Fig 11). The line current will
typically look like what is shown in Fig 26. Fig 27shows the harmonic spectrum. Ithd is normally
between 30 to 40%. Since the impedance of a DC
link choke is usually selected to be about 4%, youcan see that this had the effect of further reducing
the 5th and 7th harmonics in the line current.
Fig 25. Details of drive configuration with DC
link chokes
La.I = f(t...
150.0
150.0
200.0
200.0
162.5
162.5
175.0
175.0
187.5
187.5
00.0 -400.
00.0
0 0
00.0 -200.
00.0 200.0
Fig 26. Typical line current for this drive
configurationWhen operating, the voltage drop due to the DC
link choke is due to its DC resistance and not to itsinductive impedance. The result is that the DC bus
voltage does not drop as much as it would when
compared to a line reactor. See Fig 22. This is themain advantage of using a link choke instead of line
reactors. The other advantage is that they are
usually slightly less expensive than AC line reactors
since they only have two windings. An advantageof line reactors over link chokes is that they provide
better current sharing when paralleling diode
bridges.0
10
20
30
40
50
60
70
80
90
100
0 6 12 1 8 24 30 36 42 4 8 54 60 66 72 78
Harmonic Number
% H a r m o n i c C u r r e n
What is the overall effect of a link choke on a drive
and the power grid? Fig 28 to 31 will give yousome overall operational parameters with which to
compare various harmonic mitigation solutions.
Fig 27. Harmonic spectrum showing themagnitudes of the individual harmonicscontained in the line current
Fig 28 shows how the Arms of the total line current
varies with load. It is fairly linear. The current at
100% load is about 108Arms. This current is
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composed of about 101Arms of fundamental currentand 40Arms of harmonic current.
0.0
20.0
40.0
60.0
80.0
100.0
120.0
0 25 50 75 100
6p
Fig 29 shows the total power factor associated withthe line current vs load. At full load, the power
factor is about 0.92 lagging, and slowly drops downto about 0.7 lagging at 10% load.
Fig 30 shows the change in the DC bus voltage as
the load is changed. 100% of nominal DC bus
voltage is taken to be 1.35*Vac line-to-line voltage.For a 480Vac drive, the nominal DC bus voltage
works out to be 648Vdc. At no load, you see that
the bus voltage will charge up to the peak of the linevoltage, or 1.41*Vac. Again, for a 480Vac drive,
the peak would be 679Vac, or 1.05*nominal.
Please notice that at 100% load, the voltage drop isvery slight, due to the voltage drop at the
transformer terminals due to the internal impedance
of the secondary windings, and due to the DC
resistance of the DC link chokes.
Fig 28. Total Line Current vs % Load
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
0 25 50 75 100
6p
0.00
0.20
0.40
0.60
0.80
L a g
L e a d
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
0 25 50 75 100
6p
0.00
0.20
0.40
0.60
0.80
L a g
L e a d
Fig 31 shows how Iharm (the harmonic current)
varies with load. It also shows the % Ithd as adashed line. This is the ratio of the harmonic
current to the fundamental current that is containedin the AC line current.
Fig 29. Total Power Factor vs % Load
90.0
92.0
94.0
96.0
98.0
100.0
102.0
104.0
106.0
108.0
110.0
0 25 50 75 100
6p
Fig 32 is similar to Fig 13, showing how thedisplacement power factor and distortion power
factor vary with load. The displacement power
factor is quite high throughout the load range.
30.00
40.00
50.00
60.00
70.00
80.00
90.00
100.00
0 10 20 30 40 50 60 70 80 90 100
PF disp
PF dist
PF total
Fig 30. % Nominal DC Bus Voltage vs % Load
0.0
20.0
40.0
60.0
80.0
100.0
120.0
0 20 40 60 80 100
Ithd
Iharm
Fig 32. Power factor vs load - green isdisplacement PF, blue is distortion PF, pink istotal PF
Fig 31. % Ithd and Iharm vs % Load
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5. Pseudo 12-PulseDrive
DC
AC
DC LinkChoke
AC
DCM
Drive
DC
AC
DC LinkChoke
AC
DCM
Transformer
xfmr% Z
A Pseudo 12-Pulse system is one that approaches a
true 12-pulse, but where the 5th
and 7th
harmonicsare not totally cancelled. Fig 33. This is
accomplished by placing half of the hp load on adelta-wye transformer, and the other half on a delta-delta transformer (or a simple line reactor could be
used in place of the delta-delta transformer, with the
same impedance as the delta-wye transformer).
Fig 33. Pseudo 12-Pulse configuration with halfof the hp drive load on a delta-wye xfmr, and theother half on a delta-delta xfmr or line reactor
La1.I = f...
150.0m
150.0m
200.0m
200.0m
162.5m
162.5m
175.0m
175.0m
187.5m
187.5m
-1000.0 -1000.0
1000.0
0 0
-500.0 -500.0
500.0 500.0
The cancellation of the lower order harmonics will
be affected by the amount of load on each set of
drives. Any pre-existing harmonics present on thevoltage feeding these transformers will also affect
the harmonic cancellation.
Fig 34 shows the secondary current from the main
transformer when the drive on the delta-delta
transformer is at 100% load, the drive on the delta-
wye transformer is at 100% load.
Fig 34. Line current with 100% load on delta-delta and 100% load on delta-wye
La1.I = f...
150.0m
150.0m
200.0m
200.0m
162.5m
162.5m
175.0m
175.0m
187.5m
187.5m
-1000.0 -1000.0
1000.0
0 0
-500.0 -500.0
500.0 500.0
Fig 35 shows the secondary current from the main
transformer when the drive on the delta-deltatransformer is at 100% load, the drive on the delta-
wye transformer is at 75% load.
Fig 36 shows the secondary current from the main
transformer when the drive on the delta-deltatransformer is at 100% load, the drive on the delta-
wye transformer is at 50% load.
Fig 35. Line current with 100% load on delta-delta and 75% load on delta-wye
La1.I = f...
150.0m
150.0m
200.0m
200.0m
162.5m
162.5m
175.0m
175.0m
187.5m
187.5m
-1000.0 -1000.0
1000.0
0 0
-500.0 -500.0
500.0 500.0
Fig 37 shows the secondary current from the main
transformer when the drive on the delta-delta
transformer is at 100% load, the drive on the delta-
wye transformer is at 25% load.
Fig 38 shows the secondary current from the main
transformer when the drive on the delta-deltatransformer is at 100% load, the drive on the delta-
wye transformer is at 0% load. This is basically the
same as the current we saw in Fig 26.
Fig 36. Line current with 100% load on delta-delta and 50% load on delta-wye
You can see how the waveform slowly changes
from a 12-pulse shape to a 6-pulse shape.
Fig 39 shows the secondary current from the main
transformer when the drive on the delta-delta
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11
transformer is at 0% load, the drive on the delta-wye transformer is at 100% load. This indicates
how the harmonic current phase angles are shifted
when going through a delta-wye transformer. Thesecondary current of the delta-wye transformer
would look the same as Fig 26 or Fig 37.
La1.I = f...
150.0m
150.0m
200.0m
200.0m
162.5m
162.5m
175.0m
175.0m
187.5m
187.5m
-1000.0 -1000.0
1000.0
0 0
-500.0 -500.0
500.0 500.0
If you were to add the waveform in Fig 38 to thewaveform in Fig 39, you would get the waveform
seen in Fig 34.
Fig 37. Line current with 100% load on delta-delta and 25% load on delta-wye
La1.I = f...
150.0m
150.0m
200.0m
200.0m
162.5m
162.5m
175.0m
175.0m
187.5m
187.5m
-1000.0 -1000.0
1000.0
0 0
-500.0 -500.0
500.0 500.0
There will always be some 5th and 7th harmonics in
these waveforms since the loads will not be closely
balanced, and because any distortion in the voltagefeeding the two isolation transformers will cause an
imbalance in the currents.
Fig 40 shows the harmonic spectrum of the
waveform seen in Fig 34, which is the best you
could hope for. The waveforms in Fig 38 and Fig
39 will be like the spectrum we saw in Fig 27.Notice that the 11th and 12th harmonics have about
the same magnitude in Fig 40 as in Fig 27.
Fig 38. Line current with 100% load on delta-delta and 0% load on delta-wye
La1.I = f...
150.0m
150.0m
200.0m
200.0m
162.5m
162.5m
175.0m
175.0m
187.5m
187.5m
-1000.0 -1000.0
1000.0
0 0
-500.0 -500.0
500.0 500.0
Fig 39. Line current with 0% load on delta-deltaand 100% load on delta-wye
0
10
20
30
40
50
60
70
80
90
100
0 6 12 18 24 30 3 6 42 48 54 60 6 6 72 78
Harmonic Number
% H a r m o n i c C u r r e n
Fig 40. Harmonic spectrum of the waveform inFig 34. Notice the reduction of the 5th and 7th
harmonic currents due to phase shifting.
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12
6. Multi-Pulse Converter Transformer
xfmr% Z
Multi-Phase
Transformer
Drive
DC
AC
AC
DC
DC LinkChoke
3 9M
There are several levels of multi-pulse converter
systems. The most popular are 12-pulse, 18-pulseand 24 pulse. We will concentrate on a well
designed 18-pulse configuration. The heart of amulti-pulse converter is the multi-phase transformeras seen in Fig 41.
Fig 41. Multi-pulse drive configuration utilizing a
multi-phase transformer
Supply Transformer Line Reactor Auto-Transformer 18 Diode Bridge
NOTE: 5 windings per core leg
Supply Transformer Line Reactor Auto-Transformer 18 Diode Bridge
NOTE: 5 windings per core leg
Fig 42. Nine-phase auto-transformer detailalong with the 18-diode bridge creating an 18-pulse converter
Fig 42 shows the details of a 9-phase auto-
transformer used on an 18-pulse drive configuration.Each of the 9 phases from the transformer feed a
pair of diodes in a bridge that contains 18 diodes.
The windings in the transformer are arranged so thatthe taps for the 9 phases are located 40 degrees
apart from each other (9*40 = 360 degrees) and are
located equally distant from a central point. Thisway, the phases are equally spaced in time and have
identical voltage magnitudes.Lsa.I = f...
150.0m
150.0m
200.0m
200.0m
162.5m
162.5m
175.0m
175.0m
187.5m
187.5m
-150.0 -150.0
150.0
0 0
-100.0 -100.0
-50.0 -50.0
50.0 50.0
100.0 100.0
A normal 3-phase transformer has three phases thatare shifted 120 degrees apart from each other
(3*120=360 degrees). Again, with 2 diodes per
phase you get 2 current pulses per phase, and thattotals 6 pulses per cycle. Hence the name 6-pulse
operation. This is clearly seen in Fig 18. Fig 43. Typical line current for an 18-pulse driveconfiguration
The current waveform as seen in Fig 43 shows very
little harmonics. Typically, a well designed 18-pulse converter will have a % Ithd between 4.5 and
6%, creating very little voltage distortion on a
transformer secondary. The harmonic spectrum isseen in Fig 44. Notice that the characteristic or
largest harmonics are the 17th and 19th. This is the
same as the pulse number plus and minus 1 (18
minus 1 = 17, 18 pulse 1 = 19). Similarly, thecharacteristic harmonics for a 6-pulse converter are
the 5th and 7th; the characteristic harmonics for a 12-
pulse converter are the 11th
and 13th
; thecharacteristic harmonics for a 24-pulse converter
are the 23rd and 25th.
0
10
20
30
40
50
60
70
80
90
100
0 6 12 1 8 24 30 36 42 48 54 60 66 72 78
Harmonic Number
% H a r m o n i c C u r r e n
Fig 44. Harmonic spectrum for the 18-pulsedrive configuration
DC
ACM
Another method of creating a 9-phase transformer is
to design one with three isolated sets of secondaries.
Each secondary needs to be phase shifted only 20degrees with respect to each other. See Fig 45.
First, you have a delta primary. Then you have a
wye secondary to give you a +30 degree phase shift.
Fig 45. 18-pulse configuration using an isolationtransformer with diode bridges in series
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Next you have an extended delta for a +10 degreephase shift. And lastly you have another extended
delta for a -10 degree phase shift. Each has to be
wound to provide 1/3 voltage of the rated voltage(for example, 160Vac line-to-line for each
secondary, when used to feed a 480Vac drive).Connect each set to a 6-diode bridge, and connectthe bridges in series with each other. The current
waveform into the delta primary will again be very
close to what is shown in Fig 43. The disadvantage
of this configuration is that each diode needs to besized for the full current rating of the total converter.
The trade-off is that the voltage rating for each
diode is lower. This works well for medium voltageconverter applications and for low hp, low voltage
applications. It becomes too expensive for higher
hp, low voltage drive applications.
0.0
20.0
40.0
60.0
80.0
100.0
120.0
0 25 50 75 100
6p
18p
Fig 46. Total Line Current vs % Load
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
0 25 50 75 100
6p
18p
0.00
0.20
0.40
0.60
0.80
L a g
L e a d
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
0 25 50 75 100
6p
18p
0.00
0.20
0.40
0.60
0.80
L a g
L e a d
What is the overall effect of an 18-pulse converter
on a drive and the power grid? Fig 46 to 52 will
give you a comparison to the 6-pulse buffered drivein section 4. The blue traces are for the 6-pulse
drive, the red traces are for the 18-pulse drive. Fig 47. Total Power Factor vs % Load
Fig 46 shows how the Arms of the total line current
varies with load. It is fairly linear. The current at100% load is about 101Arms. This current is
composed of about 101Arms of fundamental current
and 5Arms of harmonic current.
90.0
92.0
94.0
96.0
98.0
100.0
102.0
104.0
106.0
108.0
110.0
0 25 50 75 100
6p
18p
Fig 47 shows the total power factor associated with
the line current vs load. At full load, the power
factor is about 0.99 lagging, and slowly drops downto about 0.92 lagging at 10% load.
Fig 48. % Nominal DC Bus Voltage vs % LoadFig 48 shows the change in the DC bus voltage asthe load is changed. 100% of nominal DC bus
voltage is again taken to be 1.35*Vac line-to-line
voltage. For a 480Vac drive, the DC bus voltagefor this 18-pulse drive is about 667Vdc. The slight
boost is due to the placement of the input
connections to the auto-transformer. This workswell for those applications where the line voltage is
a little soft, or where it is necessary to achieve full
power from the motor at full load since the drivehas full voltage on the bus. 0.0
20.0
40.0
60.0
80.0
100.0
120.0
0 20 40 60 80 100
Ithd 6p
Iharm 6p
Ithd 18p
Iharm 18p
Fig 49. % Ithd and Iharm vs % Load
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Fig 49 shows how Iharm (the harmonic current)varies with load. It also shows the % Ithd as a
dashed line. This is the ratio of the harmonic
current to the fundamental current that is containedin the AC line current.
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
16.0
18.0
20.0
0 25 50 75 100
6p
18p
Fig 50 is a zoomed in detail of % Ithd vs % Load
from Fig 49. As you see, Ithd increases as the loaddecreases due to the different rates of change of the
harmonic current and the fundamental current. Per
IEEE 519, though, the critical point is at full load,which is the minimum point on the % Ithd curve.
Fig 50. Detail of % Ithd vs % Load
0.00
1.00
2.00
3.00
4.005.00
6.00
7.00
8.00
9.00
10.00
0 25 50 75 100
6p
18p
Fig 51 is a zoomed in detail of Iharm vs % Loadfrom Fig 49. It is obvious that as the load decreases,
Iharm also decreases, meaning that the voltage
distortion will be decreasing, too.
Fig 52 is similar to Fig 32, showing how the
displacement power factor and distortion power
factor vary with load. The displacement powerfactor is again quite high throughout the load range.
But since there is so little harmonic current, the
distortion power factor is also very high throughoutthe load range.
Fig 51. Detail of Iharm vs % Load
30.00
40.00
50.00
60.00
70.00
80.00
90.00
100.00
0 10 20 30 40 50 60 70 80 90 100
PF disp
PF dist
PF total
Overall, the 18-pulse converter has very little
impact on the power grid, and is very friendly to the
drive itself. Low harmonics, high power factor,good DC bus voltage at every load point. This is
almost an ideal type of converter.
A 12-pulse converter, on the other hand, would
cause the harmonics to increase 2 to 3 times. Since
there is very little cost difference between and 12-pulse transformer and diode bridge verses an 18-
pulse transformer and diode bridge, and since an
18-pulse converter is able to achieve 5% Ithd at the
input terminals of the drive system, there has been abig drop in demand for 12-pulse converters.
Fig 52. Power factor vs load - green isdisplacement PF, blue is distortion PF, pink istotal PF
NOTE: There are dozens of transformer andbridge configurations available to achieve 18-pulse operation. Each of these 18-pulseconverters have different characteristics withrespect to line current harmonics, power factor,efficiency, DC bus voltage, etc. The test resultsin this paper refer only to the 18-pulseconfiguration as defined in this section.
A 24-pulse or higher converter is able to furtherreduce the current harmonics, but only by a small
amount. The increased costs provide little payback.
However, due to the large currents involved, thiscould be beneficial for 1500hp drive systems and
larger.
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7. Passive Filter Transformer
xfmr% Z
Drive
DC
AC
AC
DC
DC LinkChoke
Passive Filter
M
Since the largest harmonic seen in the line currents
of 6-pulse drives is the 5th
harmonic, a passive 5th
harmonic filter added to the line side of a drive is
often a satisfactory solution. See Fig 53.
Some higher performance passive filters are made
with two filtering sections – a 5th and a 7th. As seen
in Fig 54 and 55, a passive filter will significantly
reduce the 5th
harmonic drawn from the transformer.If we examine a passive filter, we see that it is made
up of line reactors and capacitors. The section with
the reactor in series with the capacitor is the tunedsection of the filter. They are usually tuned to a
frequency just below the 5th harmonic, such as at
the 4.7
th
(282Hz). Please realize that a passivefilter designed for a 60Hz power grid will not work
well on a 50Hz power grid due to the different
harmonic frequencies involved.
Fig 53. Drive configuration with the addition of apassive filter
La.I= f(t...
100.0m
100.0m
150.0m
150.0m
110.0m
110.0m
120.0m
120.0m
130.0m
130.0m
140.0m
140.0m
-200.0 -200.0
200.0
0 0
-100.0 -100.0
100.0 100.0
Fig 54. Typical line current for a drive with apassive filter
The reactor between the power distribution system
and the filter helps reduce the resonance effect the
filter will have on the power grid. This helps toisolate the filter from the rest of the power system
so that the tuned section does not become the lowestimpedance point in the whole plant for 5th
harmonics. There are some problems that are
associated with passive filters, though.
0
10
20
30
40
50
60
70
80
90
100
0 6 12 1 8 24 30 36 42 48 54 60 66 72 78
Harmonic Number
% H a r m o n i c C u r r e n
a) One is that the capacitors can interact with other
passive filters on the power system, even thougha line reactor is part of the filter assembly. This
can cause voltage and current instabilities.Fig 55. Harmonic spectrum for the drive with a5th harmonic passive filter
b) Along with this, the capacitors draw leadingcurrent from the source. In a plant with several
motor running across the line, this will help
improve the overall power factor. However,there is no control for this. Also, when
operating on back-up generators, the leading
power factor current drawn by the caps cancause unstable operation of the generator.
c) Another problem is that the line reactor willcause a voltage drop in the DC bus when
running at full speed, full load.
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d) A fourth problem is that the capacitors willcause the DC bus to rise at no load. The
designer of the passive filter has to make several
compromises in order to balance one problemagainst another in order to obtain a best overall
kind of operation.
0.0
20.0
40.0
60.0
80.0
100.0
120.0
0 25 50 75 100
6p
18p
P1
P2
P3
Some passive filters have the ability to disconnectthe filter capacitors from the rest of the filter. This
would be controlled by a contactor that is turned on
or off depending on the speed or load of the drive,or if the drive is operational or not. This helps with
the high DC bus voltage seen at no load and with
generators.
Fig 56. Total Line Current vs % Load
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
0 25 50 75 100
6p
18p
P1
P2
P3
0.00
0.20
0.40
0.60
0.80
L a g
L e a d
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
0 25 50 75 100
6p
18p
P1
P2
P3
0.00
0.20
0.40
0.60
0.80
L a g
L e a d
In order to reduce the effect one filter may have on
another, it may be possible to connect more thanone drive to a passive filter output. This reduces thenumber of individual filters on the common
transformer. Also, do not have PF corrections caps
ahead of the passive filter.
The only way to counteract the drop in DC bus
voltage at full speed, full load is to either boost thetransformer secondary or not operate at full speed.
Boosting the transformer secondary has thedisadvantage of causing the DC bus voltage to be
too high under no load conditions. Not operating at
full speed is a more likely operating condition witha drive. If the drive is going to operate at 95% of
full speed or less, then the drop in DC bus voltage
will not be noticed.
Fig 57. Total Power Factor vs % Load
90.0
92.0
94.0
96.0
98.0
100.0
102.0
104.0
106.0
108.0
110.0
0 25 50 75 100
6p
18p
P1
P2
P3 What is the overall effect of a passive filter on a
drive and the power grid? We tested three different
manufacturers’ passive filters with a drive. Fig 56to 61 give you a comparison of the three filters (P1,
P2, P3) to the 6-pulse buffered drive from Section 4
and the 18-pulse drive from Section 6.
Fig 58. % Nominal DC Bus Voltage vs % Load
0.0
20.0
40.0
60.0
80.0
100.0
120.0
0 20 40 60 80 100
Ithd 6p
Iharm 6p
Ithd 18pIharm 18p
Ithd P1
Iharm P1
Ithd P2
Iharm P2
Ithd P3
Iharm P3
Fig 56 shows how the Arms of the total line current
varies with load. It is not linear. The currents at
100% load are similar to the 18-pulse drive.However, as no load is approached, the currents do
not go to zero. Some level off at almost 50% of full
load current! What is causing this? This effect isdue to the capacitor bank that is part of the passive
filter. The capacitors will draw leading power Fig 59. % Ithd and Iharm vs % Load
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17
factor current, and is most noticeable at no load.This is saying that when the drive is turned off, you
could have 30 to 50% of full load current flowing to
the passive filter.
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
16.0
18.0
20.0
0 25 50 75 100
6p
18p
P1
P2
P3
Fig 57 shows the total power factor associated withthe line current vs load. At full load, the powerfactor is very close to unity, but then approaches 0.2
to 0.4 leading power factor as the load decreases.
Again, this is due to the cap bank in the filter.Fig 60. Detail of % Ithd vs % Load
Fig 58 shows the change in the DC bus voltage as
the load is changed. 100% of nominal DC bus
voltage is again taken to be 1.35*Vac line-to-linevoltage. At no load, all of the filters provide some
boost in the DC bus voltage due to the cap bank. At
full load, there is a wide range of performance asthe DC bus voltage can be anywhere between 93%
and 102% of nominal. The design of the reactors in
the filter have the biggest impact on this parameter.This can cause the motor to operate at less than
rated V/Hz, causing overheating, if running above
93% of full speed.
0.00
1.00
2.00
3.00
4.005.00
6.00
7.00
8.00
9.00
10.00
0 25 50 75 100
6p
18p
P1
P2
P3
Fig 61. Detail of Iharm vs % Load
Fig 59 shows how Iharm and % Ithd vary with load.
They are all quite good. For more detail, we havethe next two figures.
Fig 50 is a zoomed in detail of % Ithd vs % Load
from Fig 59. At full load, they are all able to meet
IEEE 519 (which required 8% Ithd). As the load
decreased, the Ithd of the filters would rise veryslowly due to the capacitor bank current discussed
in Fig 56.
Fig 51 is a zoomed in detail of Iharm vs % Load
from Fig 49. This is the real test since this is what
creates the voltage distortion on the transformer. It
is obvious that as the load decreases, Iharm alsodecreases, meaning that the voltage distortion will
be decreasing, too. In the tests, the passive filters
were just a little worse than the 18-pulse drive.
Overall, the passive filters do a good job reducing
the harmonic currents. The biggest drawbacks arethe leading power factor due to the cap bank in the
tuned section of the filter and the drop in DC bus
voltage at full load.
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18
8. Active FilterDrive
Transformer
DC
AC
DC LinkChokexfmr
% Z AC
DC
I fu nd I fu nd + Ih ar m
Iharm
Active Filter
AC
DC
M
An active filter is a device that is quite remarkable
in its operation. See Fig 62 for a drive systemutilizing one. It is made up of an inverter bridge
section, like the output inverter section of a drive asshown in Fig 63. However, it is connected such thata cap bank is tied to the DC bus terminals, and the
3-phase terminals are connected to the AC line
through small line reactors and a notch filter. It
then has current sensors that monitor the AC currentgoing to the non-linear load, such as the AC drive.
Fig 62. Drive configuration with the addition ofan active filter
InverterDC to AC
DCBus
Active Filter
A C
L i n e
I n p u t
NotchFilter
fThe key to the operation of an active filter is that itwill supply the harmonic currents required by the
non-linear load. Since these currents have an
average power of zero (zero average watts areconsumed by the non-linear load due to the
harmonic currents), then the harmonic current
supplied by the active filter is absorbed by the drive
during one part of the cycle, and then it is returnedto the filter during the next part of the cycle. The
current simply flows back and forth between the
filter and the drive. It acts like a reservoir forharmonic currents. Most active filters are also able
to supply fundamental reactive current to help withthe displacement power factor.
Fig 63. Details of an active filter showing IGBTinverter section
Ia= f( S,...
-25.00m
-25.00m
24.90m
24.90m
0
0
-20.00m
-20.00m
-10.00m
-10.00m
10.00m
10.00m
20.00m
20.00m
-150.0 -150.0
150.0
0 0
-100.0 -100.0
-50.0 -50.0
50.0 50.0
100.0 100.0
The addition of an active filter to a drive circuitmeans that the transformer does not have to supply
the harmonic current s to the drive. The result is a
clean line current, with just a little ripple due to thecarrier frequency switching of the active filter
inverter section. The current waveform and
spectrum are seen in Fig 64 and 65. The notch filterhelps a great deal with removing the carrier from
the line current. A typical carrier frequency used
for active filters is between 10 and 20kHz.
Fig 64. Typical line current for driveconfiguration with an active filter
0
10
20
30
40
50
60
70
80
90
100
0 6 12 1 8 24 30 36 42 4 8 54 60 66 72 78
Harmonic Number
% H a r m o n i c C u r r e n
The effect on the power distribution system is
minimal, similar to an Active Front-End as will be
discussed in Section 9. There is a slight amount of real power consumed due to losses in the filter, but
this is less than ~ 2% of the filter power. The line
current supplied by the transformer will basically be just the fundamental current. The other big
advantage is that one active filter can be associated
with a transformer secondary, supplying whatever
Fig 65. Harmonic spectrum of the line currentfor the drive configuration with an active filter
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reactive and harmonic currents are needed for all of the loads connected to the transformer. See Fig 66.
In order to meet larger power ratings, active filters
can be paralleled.
AC Drive
DC
AC
DC LinkChoke
AC
DCM
AC Drive
DC
AC
DC LinkChoke
AC
DCM
Transformer
xfmr% Z I fu nd I fu nd + I ha rm
Iharm
Active Filter
AC
DC
DC Drive
AC
DCM
One more advantage is that these units can be addedto a system that is already installed, simply byconnecting the filter to the power lines. No series
reactors are needed (unless the drive does not have
a DC link choke nor an AC line reactor already).
The filter needs only be sized for the harmoniccurrents, not full load current, of the drive. Fig 66. Configuration with multiple drives, AC
and DC, where an active filter can supply all ofthe harmonics in the system
To further improve the utilization of an active filter,the user can use the pseudo-12-pulse technique to
reduce the 5th and 7th harmonics. This will allow
the filter to handle more drives. See Fig 67. If thetotal load were 400A (about 400hp of drives), and if
the Ithd were 30%, this would mean that there is
about 30% of 400A = 120A of harmonic current
drawn from the transformer. The size of the activefilter this system would need is 120A rating.
However, by placing half of the drive hp load on a
delta-wye transformer, then the Ithd would bereduced to about 15%. This would mean that there
is only about 15% of 400A = 60A of harmoniccurrent. The size of the active filter could be
reduced to a 60A rating.
Drive
DC
AC
DC LinkChoke
AC
DCM
Drive
DC
AC
DC LinkChoke
AC
DCM
Transformerxfmr% Z I fu nd I fu nd + I ha rm
Iharm
Active Filter
AC
DC
Fig 67. Combination of an active filter with apseudo 12-pulse system, allowing for a reductionin the current rating of the active filter
It should be noted that active filters employ a
current limiting function as protection for overload
conditions. This can be useful if a slightly smallerfilter can be selected to handle the harmonic load
that is seen 95% of the time. The user would then
allow it to go into current limit during the remainder
of the time. This would not harm the filter, and theonly effect is that the current distortion would
increase during that short time period. If the
voltage distortion remains within the limits then thiswould be an acceptable, lower cost solution.
The only disadvantage is the cost of the active filter.The power electronics required is similar to a drive,
but at about 1/3 of the drive current rating.
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9. Active Front-End (AFE) Transformer
xfmr% Z
Drive
DC
AC
AC
DC
Notch Filter
M
An active front-end is very similar in construction
to an active filter. The difference is that IGBTs areadded to the diode bridge in the converter section
for the drive. See Fig 68 and 69. This providesseveral advantages for the drive.
Fig 68. Drive configuration with an active front-end as the converter in the drivea) The line current harmonics are greatly reduced,
as long as a filter tuned to the carrier frequency
of the AFE is installed between the powersystem and the input to the AFE. Fig 70 shows
the typical line current, with the harmonic
spectrum shown in Fig 71. If there were nonotch filter, the harmonic spectrum would look
like that shown in Fig 72, where the carrier
frequency of 4kHz appears as side-bands aroundthe 66th harmonic.
A C
L i n e I n p u t
ConverterAC to DC
InverterDC to AC
DCBusFilter
Regenerative AC Drive
A C
M o t or O u t p u t
b) The AFE is able to operate at unity
displacement power factor from no load to fullload.
Fig 69. Details of a drive utilizing an activeconverter
Lx1.I = ...
145.0
145.0
195.0
195.0
150.0
150.0
162.5
162.5
175.0
175.0
187.5
187.5
00.0 -200.
00.0
0
100.0 -100.
00.0 100.0
c) The AFE is also able to boost the DC busvoltage, providing power ride-through during a
brown-out condition of as much as 50%.
d) The AFE can control current flow while
motoring or regenerating, allowing brakingpower from the drive to be returned to the
power distribution system even during a brown-
out.
Fig 70. Typical line current for the driveconfiguration with an active front-end
In many ways it is like a fully regenerative DC
drive, but with distinct advantages and no
disadvantages to speak of. The only drawback isthe same as the drawback of an active filter – its
cost. It can add about 60% to 90% to the cost of a
drive. However, as you review what all it can dofor you, payback can easily be less than a year in
many applications.0
10
20
30
40
50
60
70
80
90
100
0 6 12 1 8 24 30 36 42 4 8 54 60 66 72 78
Harmonic Number
%
H a r m o n i c C u r r e n
Active Front-End are also known as active
converters, active rectifiers, synchronous converters,
synchronous rectifiers, regenerative converters,regenerative front-ends, and so on.
Fig 71. Harmonic spectrum of the line currentfor the drive configuration utilizing an activefront-end with a notch filter
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Conceptually, its operation is very simple. TheAFE is like a voltage source whose voltage
magnitude and phase angle with respect to the
power line can be varied. See Fig 73. By changingthe magnitude and phase angle of the voltage
produced by the PWM waveform at the AFEterminals, the magnitude and power factor of thecurrent through the line reactors of the AFE is
regulated from 100% motoring to 0 and all the way
to 100% regenerating. The result of regulating the
line current, the DC bus voltage is controlled asneeded.
0
10
20
30
40
50
60
70
80
90
100
0 6 12 18 24 30 36 42 4 8 54 60 66 72 78
Harmonic Number
%
H a r m o n i c C u r r e n
Fig 72. Harmonic spectrum of the line currentfor the drive configuration utilizing an activefront-end without a notch filter
Fig 74 shows the line-to-neutral voltage in pink, theAFE line-to-neutral voltage (with its PWM
waveform) in blue, and the resulting line current in
red. Notice how the AFE voltage is slightly laggingthe line voltage. This causes the line current to be
at rated current and in phase with the line voltage.
This is at 100% motoring.
Line Current
AFEVoltageSource
LineVoltageSource
Fig 75 again shows the line-to-neutral voltage in
pink, the AFE line-to-neutral voltage (with its PWM
waveform) in blue, and the resulting line current inred. Notice this time how the AFE voltage is
slightly leading the line voltage. This causes theline current to be at rated current and at 180 degrees
out of phase with the line voltage. This is at 100%
regenerating.
Fig 73. Conceptual operation of an active front-end – a PWM voltage source connected to asinusoidal voltage source through a line reactor
If the AFE voltage magnitude and phase with
respect to the line voltage were identical, then theresulting average line current would be zero.
Fig 74. Voltage and current waveforms duringmotoring operation
What is the overall effect of an AFE on a drive and
the power grid? Fig 76 to81 give you a comparisonof an AFE to the 6-pulse drive from Section 4, the
18-pulse drive from Section 6, and the drive with a
passive filter, P1, from Section 7. Please note thatthe line current and harmonic effects are practically
identical for an active filter as described in Section
8.
Fig 76 shows how the Arms of the total line current
varies with load. It is quite linear. The currents at100% load are similar to the 18-pulse drive.
However, as no load is approached, the AFE current
heads toward 10A. This is due to the capacitor
Fig 75. Voltage and current waveforms duringregenerating operation
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bank that is part of the notch filter. Since the capvalues are so low, there is only a small amount of
current at no load.
0.0
20.0
40.0
60.0
80.0
100.0
120.0
0 25 50 75 100
6p
18pP1
AFE
Fig 77 shows the total power factor associated with
the line current vs load. At full load, the powerfactor is very close to unity, but then approaches 0.8
leading power factor as the load decreases. Again,this is due to the cap bank in the notch filter. As the
carrier frequency is increased, the size of the caps
used in the notch filter will decrease, reducing thiseffect.
Fig 76. Total Line Current vs % Load
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
0 25 50 75 100
6p
18p
P1
AFE
0.00
0.20
0.40
0.60
0.80
L a g
L e a d
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
0 25 50 75 100
6p
18p
P1
AFE
0.00
0.20
0.40
0.60
0.80
L a g
L e a d
Fig 78 shows the change in the DC bus voltage asthe load is changed. 100% of nominal DC bus
voltage is again taken to be 1.35*Vac line-to-line
voltage. This is one area where an AFE shines.The DC bus voltage is controlled by the AFE micro,even if the line voltage were to dip to 50% of
nominal. During regen, the AFE controller will
sense that the DC bus is increasing, causing it tostart regenerating current to the power lines,
keeping the bus voltage under control at all time. Fig 77. Total Power Factor vs % Load
Fig 79 shows how Iharm and % Ithd vary with load.
The results are very good. For more detail, we havethe next two figures.
90.0
92.0
94.0
96.0
98.0
100.0
102.0
104.0
106.0
108.0
110.0
0 25 50 75 100
6p
18p
P1
AFE
Fig 80 is a zoomed in detail of % Ithd vs % Loadfrom Fig 79. At full load, the AFE is able to easily
meet IEEE 519. As the load decreased, the Ithd of
the AFE rises since the harmonic current does not
change as the fundamental current decreases.
Fig 81 is a zoomed in detail of Iharm vs % Load
from Fig 79. This shows something that is uniqueto active front-ends. The harmonic current is
practically constant at all load levels, and whether it
is motoring or regenerating. All other harmonicmitigation methods have a characteristic where the
harmonic current decreases with load. This is due
to the modulating IGBT bridge.
Fig 78. % Nominal DC Bus Voltage vs % Load
0.0
20.0
40.0
60.0
80.0
100.0
120.0
0 20 40 60 80 100
Ithd 6p
Iharm 6p
Ithd 18pIharm 18p
Ithd P1
Iharm P1
Ithd AFE
Iharm AFE
Overall, active front-ends do a great job of keeping
the harmonic currents very low with the use of high
carrier frequencies and a good notch filter. Inaddition, it can handle regenerative load currents
and provide very little impact to the power grid Fig 79. % Ithd and Iharm vs % Load
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system. Its ability to maintain rated DC bus voltageunder brown-out conditions is another plus since it
can keep the process operating during this
disturbance. A drawback is its cost. Compared toan active filter, it is a little less expensive if
integrated with the drive inverter.
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
16.0
18.0
20.0
0 25 50 75 100
6p
18p
P1
AFE
Fig 80. Detail of % Ithd vs % Load
0.00
1.00
2.00
3.00
4.005.00
6.00
7.00
8.00
9.00
10.00
0 25 50 75 100
6p
18p
P1
AFE
Fig 81. Detail of Iharm vs % Load
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10. Other Comparisons
0.00
10.00
20.00
30.00
40.00
50.00
60.00
70.00
80.00
90.00
0 25 50 75 100
6p
18p
P1
P2
P3
AFE
How do the efficiencies compare?
Is there a big difference in overall drive systemefficiency when comparing the above methods of
harmonic mitigation? Surprisingly, no! Thedifferences are at the most about 2% at full speed,full load. Not only are they close to one another,
but the one that is a little better at one load level
may become a little worse at another load level.
There is no clear cut advantage that one mitigationmethod has over another when comparing system
efficiency. The following diagram shows how they
compare.
Fig 82. Input kW vs % Load for variousharmonic mitigation solutions
Fig 82 shows the kW into the drive system vs %
Load (measured as torque). Shown are the 6-pulsedrive from Section 4, the 18-pulse drive from
Section 6, three passive filters from Section 7, and
the AFE from Section 9. The drive system is ratedat 100hp. Since the curves are so close to one
another, Fig 83 shows the portion of the chart from
75% load to 100% load only. As you can see, the
AFE takes slightly more watts due to the additionallosses in the converter section and in the notch filter.
The 18-pulse has only about 1.2% more losses dueto the magnetizing current in the multi-phase
transformer. One of the passive filters turned out to
be the most efficient, but not by much.
Fig 84 shows how the system efficiency, including
the motor connected to the drive, varies with % load.For the drive system that is truly the most efficient,
if that is what is absolutely required, the winner is
the basic drive without any filter, without an ACline reactor, without a DC link choke (this is not
shown). However, this gives you the worst
harmonic distortion possible!
50.0
55.0
60.0
65.0
70.0
75.0
80.0
85.0
90.0
95.0
0 25 50 75 100
6p
18p
P1
P2
P3
60.00
65.00
70.00
75.00
80.00
85.00
90.00
75 80 85 90 95 100
6p
18p
P1
P2
P3
AFE
Fig 83. Detail of Input kW vs % Load for variousharmonic mitigation solutions, from Fig 82
Fig 84. % System Efficiency vs % Load forvarious harmonic mitigation solutions
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
0 25 50 75 100
6p
18pP1
P2
P3
This is why the writer feels that it is more important
to select the harmonic mitigation method that
causes the least amount of problems with the powerdistribution system, with the least detrimental
effects on the drive itself, while still achieving the
IEEE 519 recommended limits.
Remember that the harmonic current limits given in
Table 10.3 refer to the Point of Common Coupling Fig 85. Losses, kW vs % Load for variousharmonic mitigation solutions
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(at the power meter to the plant), not at the inputterminals of the AC drive. The IEEE 519 standard
never was intended to be an equipment standard.
The most cost effective solution you will want toachieve is to meet the voltage distortion limits given
in Table 10.2 at every point within your plant.
$0
$10,000
$20,000
$30,000
$40,000
$50,000
$60,000
$70,000
$80,000
$90,000
1 0 H
P
2 5 H
P
5 0 H
P
7 5 H
P
1 0 0 H
P
1 5 0 H
P
2 0 0 H
P
2 5 0 H
P
3 0 0 H
P
4 0 0 H
P
5 0 0 H
P
8 0 0 H
P
Active Filter
Passive Filter
18-Pulse
Active Front-End
Fig 85 shows how the system losses, including the
filter or transformer, the drive, and the motor, vary
with load. Notice that the losses are all within 1 to1.5kW. The P3 curve is the passive filter that
would draw about 50% of rated current at no load.
Fig 86. Cost Comparison of Harmonic MitigationSolutions vs HP
How do the costs of the various methods
compare?
This depends on the hp rating you are looking at.
The chart in Fig 86 gives a rough comparison forvarious hp ratings. .
0
5000
10000
15000
20000
25000
30000
35000
40000
6p Drive Passive
Filter 1
Passive
Filter 2
Passiver
Filter 3
Passive
Filter 4
Passive
Filter 5
Active
Filter 1
Active
Filter 2
18p
Drive
As you can see, the 18-pulse solution is more costeffective today than passive filters, active filters, or
active front-ends for drives larger than about 150hp.
Below that hp level, passive filters are more cost
effective. You can also see that active filters aremade in specific, quantum steps. They can be
paralleled in order to achieve larger units. So theyare not optimized for any particular hp rating. The
same is true today for AFEs. The future will bring
the cost of AFEs down where they will be muchcloser to the 18-pulse solution.
Fig 87. Unit Volume (cu in) Comparison ofHarmonic Mitigation Solutions
Fig 87 shows how the volume of each harmonicmitigation solution compares. The 100hp 6-pulse
drive can be used as a basis and is needed for each
of the solutions. There is a large variability inpassive filter sizes.
What are the trends today?
What we are seeing as the trends for today are:
a) 18-pulse drives are firmly established, and are
much more popular than 12-pulse drives since a5% harmonic current distortion can be easily
achieved. This is not possible with 12-pulse
drives. Also, there is no cost advantage of 12-pulse over 18-pulse.
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b) Passive filters have been improving, with dualstage filters becoming more prevalent.
However, there is still the possibility of
resonance issues among several filters on asingle transformer or on a generator.
c) Active filters are gaining acceptance due to theease of integrating them into an existing system.
Since the cost of iron and copper continue to
increase, and the costs of semiconductors
continue to decrease, this may become morecost effective in the near future.
d) Active front-ends for AC drives is the next stepin the evolution of AC drives. The market will
then have regenerative AC drives, like the
regenerative DC drives that have been sopopular in the workplace. Again, the cost for
the power components will be decreasing.
Integrating the IGBTs into the drive, as opposed
to adding them as an active filter with a drive,will be more cost effective. In addition, as the
switching frequency increases, the notch filter
for the AFE will become smaller and less costly.
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11. Comparison Chart
Harmonic Mitigation
Solutions Check-List
6 - P u l s
e D r i v
e
1 8
- P u l s
e D r i v
e
P a s s i v
e F i l t e
r
A
c t i v
e F i l t e
r
A c t i v
e F r o n t - E n d
Typical Ithd 20 - 45% 4.5 - 6% 5 - 8% 3 - 5% 3 - 5%
Meet IEEE Special Applications No Yes Marginal Yes Yes
Meet IEEE General Applications No Yes Yes Yes Yes
Meet IEEE Dedicated Applications Yes Yes Yes Yes Yes
Effect of 1% Voltage Unbalance Large Moderate Minimal Minimal Minimal
Potential Low DC Bus No No Yes No No
Potential System Resonance No No Yes No No
Typical Total Power Factor, no / full load 0.75 - 0.95 0.90 - 0.99 0.3 - 1 lead 0.90 - 0.99 0.8 - 1 lead
Efficiency 97% 96.5% 96.5% 96% 96 - 97.5%
Cost Effective Good >150hp <150hp Large Sys Regen, MV
Overall Size (relative to 6-P Drive) 1.0 3.3 2 - 6 3.5 - 5 1.5 - 2.5
Reliability High High Medium Medium Medium
Good
Need to confirm application
May not meet IEEE 519
Notes:
1. The 6-Pulse Drive is one with either a DC Link Choke or with AC Line Reactors.
2. For meeting the various IEEE applications, the drive and mitigation assembly would be the only
item at a Point of Common Coupling. It is possible to have a 6-pulse drive meet IEEE general
applications if there are sufficient linear loads in addition to the converter loads on the transformer.
3. The typical total power factor for a drive with an active front-end, at no load, has much less
leading power factor than that seen with a passive filter. This current is due to the notch filter on the
system.
4. Passive filters do have some drawbacks, but if the system engineer takes these issues into account
in his system design, the likelihood of any problem is minimized.
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12. Summary Rules-of-Thumb
The following are some rules-of-thumb to use when mitigating line current harmonics with AC Drives:
1. Identify the required PCC and apply techniques most cost effective for that
location.
2. Add a line reactor (or DC link choke if possible) to all unbuffered 6-pulse
drives.
3. Consider use of an active filter on a multiple drive system or MCC lineup tocorrect for harmonic distortion.
4. For an even number of equally sized drives, consider a pseudo 12-pulsesolution by placing half of the load on a phase shifting delta-wye (delta-star)
transformer.
5. Design the system to isolate linear and non-linear loads and create two systems
with 5% and 10% voltage distortion limits.
6. For passive filters on generator power, select a filter with a dropout contactorterminal block for the filter capacitors. This will limit the leading power factor at
no-load and stand-by operation.
7. Take time to understand the benefits and drawbacks of each type of mitigation
solution to assure you meet the requirements of the application and that you canlive with any negative effects created by the chosen harmonic solution.
8. Consider an active front-end if the application requires regenerative operationand harmonic compliance.
9. Perform a preliminary harmonic analysis on your system and explore theeffects of using various harmonic mitigation methods.
10. Never use power factor correction capacitors at the input (or output) of a
drive, or in parallel with passive filters.
These rules should lead you to a successful AC drive application.