Harmonic Fundamentals, Application and Mitigation Methods for AC Drives.pdf

Harmonic
Fundamentals Applicatio
Fundamentals, Applicatio
ns and Mitigation Methods
for AC Drives
for AC Drives
Copyright © 2011 Rockwell Automation, Inc. All rights reserved.

What are Harmonics?
Copyright © 2011 Rockwell Automation, Inc. All rights reserved. 2

What are Harmonics? What are waveforms?
• A sinusoidal waveform Rfund.V =...
0 40.00m
10.00m 20.00m 30.00m
50.0
• A sinusoidal waveform has no harmonics
does not contain any harmonics
00.0 100.0
A sinusoidal waveform has no harmonics
0 0
50.0 50.0
50.0 -50.0
• This is an example of a linear load
150.0 -150.0
100.0 -100.0
p
0 40.00m
10.00m 20.00m 30.00m

What are Harmonics and Waveforms
• This non-sinusoidal waveform contains harmonics
Rtotal.V =...
0 40.00m
10.00m 20.00m 30.00m
50.0
00.0 100.0
50.0 50.0
0 0
50.0 -50.0
100.0 -100.0
0 40.00m
10.00m 20.00m 30.00m
150.0 -150.0

Let’s create a distorted waveform
Fundamental (1st harmonic) Only
• fundamental at 60Hz

Fundamental and 5th Harmonic
• Some 5th harmonic , 153deg

1st, 5th and 7th Harmonics
• A little of 7th harmonic , 282deg

1st, 5th, 7th and 11th Harmonics
• A bit of 11th harmonic , 0deg

1st and Sum of the 5th, 7th and 11th
• Sum the 5th, 7th and 11th harmonic currents

Fundamental, Harmonics, Total
• Sum the harmonics with the fundamental

Electrical loads and current harmonics
Power Source Load Type ?
yp
Line Current Harmonics?

What are Loads that do not have Current
Harmonics?
• A sinusoidal waveform has no harmonics
p

Examples of Linear Loads
• Induction motors
• Incandescent lights
• Incandescent lights
• Resistance heaters
• Power Factor Correction Caps
• Electromagnetic devices
g
– Transformers
• non-linear
– During energization
– Over-voltage

What are Loads that have Current
Harmonics?
• A non-sinusoidal waveform contains harmonics
0 40.00m
10.00m 20.00m 30.00m
Rfund.V =...
Rtotal.V =...
0 40.00m
10.00m 20.00m 30.00m
150.0
100.0 100.0
0 0
50.0 50.0
0 0
50.0 -50.0
0 40.00m
10.00m 20.00m 30.00m
150.0 -150.0
100.0 -100.0
p
0 40.00m
10.00m 20.00m 30.00m

Examples of Non-Linear Loads
• Single Phase
– Fluorescent lights (ballast)
• Three Phase
– Welders
g ( )
– Incandescent lights with
light dimmers
– Arc furnaces
– UPS
– Anything with an ac-dc
power supply
– DC power supplies
– DC Drives
• Computers (ac-dc PS)
• Monitors (ac-dc PS)
TV ( d PS)
• Phase control
• PWM
• TVs (ac-dc PS)
• Fax machines (ac-dc
PS)
– AC Drives
• 6-Step
PS)
• PWM

Frequency, Amplitude, Phase Angle
• Harmonics are simply integer multiples of the
fundamental frequency
fundamental frequency
– for example, if 60Hz is the fundamental (sometimes
referred to as the 1st harmonic) then the 2nd harmonic is
referred to as the 1st harmonic), then the 2nd harmonic is
120Hz, the 3rd harmonic is 180Hz, etc.
• Any non-sinusoidal waveform can be created by the
addition of harmonics at various amplitudes and
addition of harmonics at various amplitudes and
phase angles

FFT and How are Harmonics measured?
Power Source
Harmonic
Power Meter
P f i FFT
AC Drive
Performing FFT
A1 Waveform
178.67 Arms, 34.40 %THD
Motor
M
20
40
60
80
100
0
1 5 10 15 20 25 30 35 40 45 50
10/28/2010 - 1:46:18.297 PM

FFT, RSS, THD
Iharm = 30.83A
Harmonic
Number
Frequency
Hz
Amplitude
RMS
Amplitude
RMS^2
Phase
Angle
0 DC 0.00 0.00 0
Ifund 1 60.00 70.71 5000.00 357.00
Ifund = 70.71A
3 180 0.00 0.00 0
5 300 27.97 782.50 153
7 420 10.85 117.67 282
9 540 0.00 0.00 0
11 660 5 54 30 74 0
Itotal = 77.14A
I(THD) 43 6%
11 660 5.54 30.74 0
13 780 2.79 7.79 81
15 900 0.00 0.00 0
17 1020 2.54 6.44 189
19 1140 1.45 2.09 246
I(THD) = 43.6%
= Iharm / Ifund
21 1260 0.00 0.00 0
23 1380 1.37 1.88 10
25 1500 0.91 0.84 58
27 1620 0.00 0.00 0
29 1740 0 78 0 61 185
29 1740 0.78 0.61 185
Sum of squares 3rd to 29th 950.56
Square Root of Sum 30.83 Iharm
S f 1 t t 29th 5950 56
Sum of squares 1st to 29th 5950.56
Square Root of Sum 77.14 Itotal

What Is I(THD)?
• I(THD) = Iharm / Ifund
–So, Iharm = I(THD) * Ifund
• I(THD) is a ratio between two numbers it does
• I(THD) is a ratio between two numbers, it does
not stand alone!
W d I(THD)
We can decrease I(THD)
by either decreasing Iharm or increasing Ifund

Why do drives create harmonics?
A Typical Drive System and Harmonics
M
P S AC Drive Motor
M
Power Source AC Drive Motor
Line Current Harmonics –
Covered by IEEE 519

What does an Adjustable Speed Drive do on
the inside?
AC Drive
Input =
Fi d V
Output =
V V
Bus = Fixed Vdc
AC Drive
AC
ut
Fixed V
Fixed freq
Var V
Var freq
C
Motor
Outpu
AC
Line
Inpu
Converter Inverter
DC
ut
Converter
AC to DC
Inverter
DC to AC
DC
Bus
Filter
480Vac 0-460Vac
480Vac
60Hz
0 460Vac
0-60Hz
650Vdc

Let’s look at some voltages and current.
Ia
Vab
Vac Vbus

First Current Pulse
Into A
Into A
Out of B
1

Second Current Pulse
Into A
Into A
Out of C
Out of C
2

Typical current waveform for 6 diode (pulse)
Rtotal.V =...
0 40.00m
10.00m 20.00m 30.00m
50.0
00.0 100.
0 0
50.0 50.0
50.0 -50.0
150 0 150
100.0 -100.
0 40.00m
10.00m 20.00m 30.00m
150.0 -150.

Spectrum – 3ph Diode Bridge
70
80
90
100
40
50
60
70
Amplitude
10
20
30
40
%
A
0
1 3 5 7 9 11 13 15 17 19 21 23 25
Harmonic Number

Why Be Concerned?
SO WHAT !……what do I
care if my drive draws
current harmonics from
current harmonics from
the transformer?

Root Cause of Problems with Other Equipment
Current Harmonics
Current Harmonics
create
Voltage Distortion

What other problems do they cause?
• Increased Utility current requirement
– Inability to expand or utilize
equipment
Larger wire size needed = increased
PLC
EQUIPMENT
– Larger wire size needed = increased
installation costs
• Component overheating
– Distribution transformers, generators
& i
TELEPHONE
EQUIPMENT
PCC
& wires
• Reduced Utility power factor
– Increase in utility costs
• Equipment malfunction
DATA
PROCESSING
CENTER
IH
PFC PFC
Equipment malfunction
– Due to voltage distortion with multiple
or loss of zero crossing
– Due to voltage distortion such as flat
topping
HARMONIC SOURCE
H
topping
• Excitation of Power System
Resonance's creating over-voltage’s
– If PFCC in system

When Should You be Concerned
About Harmonics ?
• If Service Transformer is Operating at about 60%
of its Rated current.
of its Rated current.
20 % f T t l L d i N Li El t i L d
• 20 % of Total Load is Non-Linear Electronic Load
• When PowerFactor Correction Capacitors Used or
Planned
• When Voltage Distortion Exceeds 8%
• When Voltage Distortion Exceeds 8%

Current Harmonics Cause Voltage Distortion
Example: 1500kVA transformer, 75hp ASD
Ithd = 37% Vthd = 0.9%

Current Harmonics Cause Voltage Distortion
Example: 75kVA Transformer, 75hp ASD
Ithd = 29% Vthd = 9.3%

Flat-Topping

Voltages with High Peaks
Event Details/Waveforms
750
250
500
477VAC RMS
720 volts peak
0
Volts
-250
-750
-500
D V A-B V B-C V C-A V
Event #397 at 12/23/2009 22:19:59.800
Timed
22:19:59.86
12/23/2009
Wednesday
22:19:59.87 22:19:59.88 22:19:59.89
C C

IEEE Std 519-1992

What are the IEEE 519-1992 standards?
Application Maximum THD (%)
Harmonic Voltage Limits
Low-Voltage Systems
Table 10.2
Application Maximum THD (%)
Special Applications - hospitals and airports 3.0%
General System 5.0%
Dedicated System - exclusively converter load 10.0%
y y
Application Max Notch Depth
Low-Voltage Systems
Special Applications - hospitals and airports 10%
General System 20%
Dedicated System - exclusively converter load 50%
y y

Table 10.2
A li ti M N t h D th
Low-Voltage Systems
Special Applications - hospitals and airports 10%
General System 20%
Dedicated System - exclusively converter load 50%
Dedicated System exclusively converter load 50%

Current distortion Limits for General Distribution Systems (120V through 69,000V)
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
Maximum Harmonic Current Distortion in Percent of Iload
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
Table 10.3
>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 PCC
Iload=maximum demand load current (fundamental frequency component) at PCC

What is Ithd vs Itdd?
• Ithd = Iharm / Ifund at any speed or load level
• Itdd (IEEE519) = Iharm / Ifund at max load
• Itdd (xfmr) = Iharm / Ifund at rated transformer current

How does motor load affect I(THD)?
6-Pulse Buffered Drive Currents
90
100
60
70
80
und.
Curren
Iharm
Ifund
I(THD)
30
40
50
%
of
Full
Load
Fu
I(THD)
I(TDD)
0
10
20
%
Ifund decreases as load decreases
NOTES:
0 20 40 60 80 100
% Load
Iharm decreases as load decreases
(drive is at full speed)
I(THD) = Iharm / Ifund
I(THD) increases as load decreases

Vthd vs Load
45
50
30
35
40
I(TDD)
20
25
Ithd, %
Iharm, A
5
10
15 Vthd, %
Max
0
5
0 10 20 30 40 50 60 70 80 90 100
% Load
Max
Vthd
% Load
100hp drive on 250kVA xfmr, 6%

Why Itdd on Table 10.3?
• Itdd is called for because that is a worst case condition.
• Please note:
Please note:
– This is not where Ithd is maximum
– But, it is where Vthd is maximum because Iharm is maximum
– When Vthd is a maximum, then greatest likelihood of problems exist

The Goal of IEEE 519
Thou
Shalt Not
Thou Shalt Not
Mess Up
Thy
Mess Up
Thy
Neighbor’s
Line Voltage
Thy
Neighbor’s
Line Voltage

Who is your neighbor?
utility
utility I(TDD) is measured at each metering point
transformer
transformer
PCC1 Other
Customer
Customer
A
Iharm A
Iharm
( ) g p
2500kVA
5.75%Z
480Vsec
2500kVA
5.75%Z
480Vsec Other
Customer
Iharm B
Ifund A
Ifund
480Vsec
480Vsec Ot e
Customer
B
Ih C
Ifund B
Goal is to
keep the V(THD) at
Other
Customer
Customer
C
Iharm C
Ifund C
keep the V(THD) at
PCC1 <= 5%,

Example
utility
utility I(TDD) limits are met at each metering point
transformer
transformer
PCC1
y
y
Other
Customer
Customer
A
113Arms
241Arms
I(TDD) limits are met at each metering point
300hp 6-p drives
600hp linear load
2500kVA
5.75%Z
480V
2500kVA
5.75%Z
480V
Customer
A
Other
Customer
101Arms
981Arms
2960Arms
80hp unbuf drives
480Vsec
480Vsec Other
Customer
Cus o e
B
2A
926Arms
at PCC1:
p
700kW linear load
Other
Customer
Customer
C
72Arms
1053Arms
V(THD) = 3.6% 1000hp 12-p drives

What about within Customer A or B or C?
V(THD) <= 10%
300hp 6-P drives
600hp linear loads
PCC1
113Arms
V(THD) <= 10%
V(THD) <= 5%
PCC1
Isc/Iload = 53 3
981Arms
V(THD) <= 5%
Isc/Iload 53.3
V(THD) = 2.0%
I(TDD) = 11.5%
V(THD) <= 5%
Meets IEEE 519 at PCC1 and within plant
Meets IEEE 519 at PCC1 and within plant

What about back-up generator?
V(THD) 10%
Customer A 300hp 6-P drives
600hp linear loads
PCC1
V(THD) <= 10%
V(THD) <= 5%
PCC1
Isc/Iload = 53.3
( )
G
V(THD) 5%
Isc/Iload 53.3
V(THD) = 2.0%
I(TDD) = 11.5%
V(THD) <= 5%
Meets IEEE 519 within the plant if the
Meets IEEE 519 within the plant if the
generator is sized properly

Information Needed for Generator
Applications
Generator Issue Information Required
kW Rating
Prime Mover / Engine Specifications
Generator Reactive Capability Curve
G C C
kVAR Lagging Generator Reactive Capability Curve
kVAR Leading Generator Reactive Capability Curve
G t I d X ”
Voltage Distortion Generator Impedance, Xd”
Voltage Notching Generator Impedance, Xd”
Harmonic Current Regulator Control
Drive Precharge Regulator Control

What about the Power Factor?
What is the Power Factor
of a
of a
Non-Linear load?
Ireact
Itotal
Iharm
REACTIVE
Current
Q
S
D
y-axis
2
2
2
D
Q
P
S +
+
=
PF = Watts/VA
or phase angle between
voltage and current
HARMONIC
Current
Ireal
Ifund
(in phase with
line-to-neutral
voltage, VLN)
P
S1
S
x-axis
z-axis
REAL
Current
Linear Load Power Factor Non-Linear Load Power Factor
No Current Distortion Includes the Effect of Current Distortion

Power Factor
Total PF = PF(disp) * PF(dist)
• Displacement power factor - PF(disp)
– PF(disp) = Ireal / Ifund = a number between .01 and 1.0
– involves only the fundamental quantities
i l d th l d ti t
– includes the real and reactive currents
• Distortion power factor - PF(dist)
PF(dist) = Ifund / Itotal = a number between 01 and 1 0
– PF(dist) = Ifund / Itotal = a number between .01 and 1.0
– includes the fundamental and harmonic (distorted) currents
It t l f d t l and h i t
– Itotal = fundamental and harmonic currents

Displacement Power Factor - PF(disp)
If d
Ifund
Ireact
Ireal
Ireact
θ Vac
Ireal
PF(disp) = Ireal / Ifund
(In phase with line voltage)
PF(disp) = Ireal / Ifund
Ifund = sqrt(Ireal2 + Ireact2)

Distortion Power Factor - PF(dist)
It t l
Itotal
Iharm
Ifund
PF(dist) = Ifund / Itotal
Itotal = sqrt(Ifund
2
+ Iharm2
)
I(THD) = Iharm / Ifund
Itotal sqrt(Ifund + Iharm )
( )
Note: Itotal = Ifund if there is no harmonic current

Current Amplitudes
• Itotal = 105.6Arms
• Iharm = 33.1Arms
• I5 = 29.3Arms
• I7 = 10.9Arms
• I11 = 7.9Arms
• I13 = 4.5Arms
• …….
• Ifund = 100.3Arms
Ireal = 98 6Arms
• Ireal = 98.6Arms
• Ireact = 18.0Arms

Current Amplitudes
• I5 = 29.3Arms
• I7 = 10.9Arms
• I11 = 7.9Arms
• I13 = 4.5Arms
• …….
• Ifund = 100.3Arms
• Ireal = 98 6Arms

Current Relationships
• I(THD) = Iharm/Ifund =
33.0%
• I5 = 29.3Arms
• I7 = 10.9Arms
• PFdisp = Ireal/Ifund = .98
7
• I11 = 7.9Arms
• I13 = 4.5Arms
• PFdist = Ifund/Itotal = .95
• PFtotal = Ireal/Itotal
PFdi *PFdi t
I13 4.5Arms
• …….
• Ifund = 100 3Arms
• = PFdisp*PFdist
• PFtotal = .93
Ifund = 100.3Arms
• Ireact = 18 0Arms

How can we reduce
(mitigate) the
harmonic current?
harmonic current?

Drive w/o DC Link Choke
D i
• Typical I(THD) of 80
to 120%
Transformer
xfmr
%Z
Drive
DC
AC
AC
DC
• Sensitive to line
voltage transients
• High peak line
AC
DC
Common
fi ti f
• High peak line
currents
M
hp
configuration for
drives < 5hp
Motor
Load
La.I = f( ...
150.0m 200.0m
162.5m 175.0m 187.5m
400.0
200.0 200.0
0 0 NOTE: Ipk about 3x Irms
150.0m 200.0m
162.5m 175.0m 187.5m
400.0 -400.0
200.0 -200.0

Line Reactor, Drive w/o DC Link Choke
D i
to 45%
Transformer
xfmr
%Z
Drive
DC
AC
AC
DC
• Big help for drives
without DC link choke Line Reactor
AC
DC
Typical values are
M
hp
Typical values are
3% and 5%
impedance
Motor
Load
La.I = f(t...
150.0m 200.0m
162.5m 175.0m 187.5m
400.0
200.0 200.0
NOTE: shown is 3% LR
0 0
150.0m 200.0m
162.5m 175.0m 187.5m
400.0 -400.0
200.0 -200.0

Drive with DC Link Choke
D i
• Typical I(THD) of 30 to
40%
• Less sensitive to line
Transformer
xfmr
%Z
Drive
DC
AC
AC
DC
DC Link
Choke
• Less sensitive to line
transients
AC
DC
M
hp
Motor
Load
La.I = f(t...
150.0m 200.0m
162.5m 175.0m 187.5m
400.0
200.0 200.0
NOTE: Ipk about 1.5x Irms
0 0
150.0m 200.0m
162.5m 175.0m 187.5m
400.0 -400.0
200.0 -200.0

Line Reactor in Addition to a DC Link Choke
D i
to 35%
Transformer
xfmr
%Z
Drive
DC
AC
AC
DC
DC Link
Choke
• Big help for drives
w/o DC link choke
• 75 - 95 PF
Line Reactor
AC
DC
Typical values are
• .75 - .95 PF
M
hp
Typical values are
3% and 5%
impedance
Motor
Load
La.I = f(t...
150.0m 200.0m
162.5m 175.0m 187.5m
400.0
200.0 200.0
NOTE: shown is 3% LR
0 0
150.0m 200.0m
162.5m 175.0m 187.5m
400.0 -400.0
200.0 -200.0

Passive Harmonic Filter
D i
Transformer
xfmr
%Z
Drive
DC
AC
AC
DC
DC Link
Choke
to 7%
AC
DC
• .3 to 1 PF
M
hp
Passive Filter
Motor
Load
Ia= f( S,...
-25.00m 24.90m
0
-20.00m -10.00m 10.00m 20.00m
150.0
100.0 100.0
0 0
-50.0 -50.0
50.0 50.0
-25.00m 24.90m
0
-20.00m -10.00m 10.00m 20.00m
-150.0 -150.0
-100.0 -100.0

Active Harmonic Filter
to 6%
Drive
Transformer
DC
AC
DC Link
Choke
xfmr
%Z AC
DC
Ifund Ifund + Iharm
• .9 - .99 PF
AC
DC
Iharm
AC
Current from Transformer
M
hp
Motor
Active Filter
DC
Ia= f( S,...
-25.00m 24.90m
0
-20.00m -10.00m 10.00m 20.00m
150.0
100.0 100.0
Motor
Load
0 0
-50.0 -50.0
50.0 50.0
-25.00m 24.90m
0
-20.00m -10.00m 10.00m 20.00m
-150.0 -150.0
-100.0 -100.0

Multi-Pulse
D i
Transformer
xfmr
%Z
Drive
DC
AC
AC
DC
DC Link
Choke
3 9
• 12-Pulse Typical
I(THD) of 9 to 12%
Multi-Phase
Transformer
AC
DC
• 18-Pulse Typical
I(THD) of 4 to 5%
• 90 - 99 PF
M
hp
• .90 - .99 PF
Motor
Load
Ia= f( S,...
-25.00m 24.90m
0
-20.00m -10.00m 10.00m 20.00m
200.0
100.0 100.0
0 0
-25.00m 24.90m
0
-20.00m -10.00m 10.00m 20.00m
-200.0 -200.0
-100.0 -100.0

Active Front-End
D i
Transformer
xfmr
%Z
Drive
DC
AC
AC
DC
to 5%
AC
DC
• Regen
• .8 – 1 PF
M
hp
Notch Filter
Motor
Load
Lx1.I = ...
145.0m 195.0m
150.0m 162.5m 175.0m 187.5m
200.0
100.0 100.0
0 0
145.0m 195.0m
150.0m 162.5m 175.0m 187.5m
200.0 -200.0
100.0 -100.0

Popular Harmonic Mitigation Choices
• 18-Pulse - widely accepted
– Most often specified
W k ll
– Works well
– Has become more expensive
• Passive Filters
– Suited to lower power ratings <100 hp
– Numerous suppliers
• Active Filters
– Showing up on specifications
– Versatile – used for single drives or multiple drives
• AFE
• AFE
– Widely accepted in Europe
– Not viewed as cost effective in North America
Line regenerative feature
– Line regenerative feature

What Did We Learn?
• What are harmonics?
– Distorted waveform, sine wave element that make up
distorted waveform
• How are they measured?
– FFT RSS THD TDD PCC
FFT RSS THD TDD PCC
• Why do drives produce line current harmonics?
– Non Linear Load, 6 pulse rectifiers
• How much is too much?
– Voltage distortion greater than 5%
• How do we apply IEEE-519?
How do we apply IEEE 519?
– Limits guidelines
• How do harmonics vary with load?
– Voltage distortion increase as Iharm increases as a % of
maximum available current
• How can the drive harmonics be reduced? PLC
– DC link choke, Line reactor, Passive filter, active filter, xfmr
config, multipulse converter, active front end
• What about Power Factor?
– Disp pf high, total pf is dist pf * disp pf ( lower than disp pf)
• Total pf proportional to Ithd
PLC
EQUIPMENT
TELEPHONE
EQUIPMENT
PCC
– Be careful with passive filters and leading pf
• What about gensets ( Generators )?
– Increase in Vthd due to greater impedances,
– Voltage regulation issues when applied to drives
HARMONIC SOURCE
DATA
PROCESSING
CENTER
IH
HARMONIC SOURCE

Thank You!
Questions?

Harmonic Fundamentals, Application and Mitigation Methods for AC Drives.pdf

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