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1
Power Electronics
Dr. Imtiaz HussainAssociate Professor
email: [email protected] :http://imtiazhussainkalwar.weebly.com/
Lecture-11Inverters
2
Introduction
β’ Converts DC to AC power by switching the DC input voltage (or current) in a pre-determined sequence so as to generate AC voltage (or current) output.
Methods of Inversion
β’ Rotary inverters use a DC motor to turn an AC Power generator, the provide a true sine wave output, but are inefficient, and have a low surge capacity rating
β’ Electrical inverters use a combination of βchoppingβ circuits and transformers to change DC power into AC.
β’ They are much more widely used and are far more efficient and practical.
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TYPICAL APPLICATIONS
β Un-interruptible power supply (UPS)
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TYPICAL APPLICATIONS
β Traction
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TYPICAL APPLICATIONS
β HVDC (High Voltage Direct Current)
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Types of Inverters
β’ There are three basic types of dc-ac converters depending on their AC output waveform: β Square wave Invertersβ Modified sine wave Invertersβ Pure sine wave Inverters
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Square Wave Inverters
β The square wave is the simplest and cheapest type, but nowadays it is practically not used commercially because of low power quality (THDβ45%).
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Modified Sine wave Inverters
β’ The modified sine wave topologies provide rectangular pulses with some dead spots between positive and negative half-cycles.
β’ They are suitable for most electronic loads, although their THD is almost 24%.
β’ They are the most popular low-cost inverters on the consumer market today,
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Pure Sine Wave Inverters
β A true sine wave inverter produces output with the lowest total harmonic distortion (normally below 3%).
β It is the most expensive type of AC source, which is used when there is a need for a sinusoidal output for certain devices, such as medical equipment, laser printers, stereos, etc.
β This type is also used in grid-connected applications.
Simple square-wave inverterβ’ To illustrate the concept of AC waveform generation
AC Waveform Generation
AC Waveforms
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Output voltage harmonics
β’ Harmonics may cause degradation of equipment (Equipment need to be βde-ratedβ).
β’ Total Harmonic Distortion (THD) is a measure to determine the βqualityβ of a given waveform.
ππ»π·π£=ββπ=2
β
(π π ,π ππ )2
π 1 ,π ππ
ππ»π·π=ββπ=2
β
( πΌπ ,π ππ )2
πΌ 1 ,π ππ
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Fourier Seriesβ’ Study of harmonics requires understanding of wave shapes. β’ Fourier Series is a tool to analyse wave shapes.
β’ Where,π£ (π‘ )=ππ+β
π=1
β
[ππcos (ππ )+ππ sin (ππ ) ]
ππ=1π β«
0
2 π
π£ (π‘ ) cos (ππ )ππ
ππ=1π β«
0
2π
π£ (π‘ )sin (ππ )ππ
ππ=1π β«
0
2π
π£ (π‘ )ππ
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Harmonics of square-wave ππ=
1π β«
0
2π
π£ (π‘ )ππ
ππ=1π [β«
0
π
π ππππ+β«π
2π
βπ ππππ ]ππ=0
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Harmonics of square-wave
ππ=1π β«
0
2 π
π£ (π‘ ) cos (ππ )ππ
ππ=1π [β«
0
π
π ππ cos (ππ )ππ+β«π
2π
βπ ππ cos (ππ )ππ ]ππ=
π ππ
π [β«0
π
cos (ππ )ππββ«π
2π
cos (ππ )ππ ]=0
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Harmonics of square-wave ππ=
1π β«
0
2π
π£ (π‘ )sin (ππ )ππ
ππ=1π [β«
0
π
π ππ sβ(ππ )ππ+β«π
2π
βπ ππ sβ(ππ )ππ ]
ππ=0β’ When n is even
ππ=π ππ
π [β«0
π
sin (ππ ) ππββ«π
2π
sin (ππ )ππ ]=2π ππ
ππ[1βcos (ππ )]
ππ=4π ππ
ππ
β’ When n is odd
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Harmonics of square-wave π£ (π‘ )=ππ+β
π=1
β
[ππcos (ππ )+ππ sin (ππ ) ]
π£ (π‘ )=βπ=1
β
[ππsin (ππ ) ]
π£ (π‘ )=4π ππ
π βπ=1,3,5β¦
β1πsin (ππ )
Where,
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Harmonics of square-wave
β’ Spectra characteristics
Harmonic decreases as n increases.
It decreases with a factor of (1/n).
Even harmonics are absent.
Nearest harmonics is the 3rd.
If fundamental is 50Hz, then nearest harmonic is 150Hz.
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Harmonics of square-wave
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Filteringβ’ Low-pass filter is normally fitted at the inverter output to
reduce the high frequency harmonics.
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Topologies of Invertersβ’ Voltage Source Inverter (VSI)β Where the independently controlled ac output is a voltage
waveform.β In industrial markets, the VSI design has proven to be more
efficient, have higher reliability and faster dynamic response, and be capable of running motors without de-rating.
β’ Current Source Inverter (CSI)β Where the independently controlled ac output is a current
waveform. β These structures are still widely used in medium-voltage
industrial applications, where high-quality voltage waveforms are required.
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1- Voltage source Inverters
β’ Single phase voltage source inverters are of two types.
β Single Phase Half Bridge voltage source inverters
β Single Phase full Bridge voltage source inverters
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1- Half Bridge VSIβ’ Figure shows the power topology of a half-bridge VSI, where
two large capacitors are required to provide a neutral point N, such that each capacitor maintains a constant voltage vi /2.
β’ It is clear that both switches S+ and Sβ cannot be on simultaneously because a short circuit across the dc link voltage source vi would be produced.
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1- Half Bridge VSIβ’ Figure shows the ideal waveforms associated with the half-
bridge inverter.
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1- Half Bridge VSIβ’ The gating signals for thyristors and resulting output voltage
waveforms are shown below.
Note: Turn off circuitry for thyristor is not shown for simplicity
π£π={ π π
20<π‘<π /2
βπ π
2π /2<π‘<π
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1- Full Bridge VSIβ’ This inverter is similar to the half-bridge inverter; however, a
second leg provides the neutral point to the load.
β’ It can be observed that the ac output voltage can take values up to the dc link value vi, which is twice that obtained with half-bridge VSI topologies.
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1- Full Bridge VSIβ’ Figure shows the ideal waveforms associated with the half-
bridge inverter.
π£ π
π£ π
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1- Full Bridge VSIβ’ The gating signals for thyristors and resulting output voltage
waveforms are shown below.
π£π={ π π 0<π‘<π /2 βπ π π /2<π‘<π
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3- Full Bridge VSIβ’ Single-phase VSIs cover low-range power applications and
three-phase VSIs cover medium- to high-power applications.
β’ The main purpose of these topologies is to provide a three phase voltage source, where the amplitude, phase, and frequency of the voltages should always be controllable.
1- VSI using transistors
β’ Single-phase half bridge and full bridge voltage source inverters using transistors are shown below.
32
33
Example-1β’ A full bridge single phase voltage source inverter is
feeding a square wave signals of 50 Hz as shown in figure below. The DC link signal is 100V. The load is 10 ohm.
β’ Calculateβ THDv
β THDv by first three nonzero harmonics
100V
-100V
34
Example-1β’ To calculate the harmonic contents we need to expand the
output waveform into Fourier series expansion.
β’ Since output of the inverter is an odd function with zero offset, therefore and will be zero.
100V
-100V
π£π=ππ+βπ=1
β
[ππ cos (ππ )+ππsin (ππ ) ]
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Example-1
β’ Where,
100V
-100V
π£π=βπ=1
β
ππsin (ππ )
ππ=1π β«
0
2π
π£π (π‘ )sin (ππ )ππ
ππ=4π π
ππ {0πeven1π odd
π£ (π‘ )=4π π
π βπ=1,3,5β¦
β1πsin (ππ )
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Example-1β’ THDv can be calculated as
β’ Fourier series can be further expanded as
ππ»π·π£=ββπ=2
β
(π π ,π ππ )2
π 1 ,π ππ
π£ (π‘ )=4π π
π βπ=1,3,5β¦
β1πsin (ππ )
π£ (π‘ )=400πsinπ+
4003π
sin (3 π)+4005π
sin(5 π)+4007π
sin (7 π )+ 4009π
sin (9 π )+β¦
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Example-1
ππ»π·π£=β(π 3 ,π ππ)2+(π 5 ,π ππ )2+(π 7 ,π ππ )2+(π 9 ,π ππ)2+β¦
π 1 ,π ππ
π£ (π‘ )=400πsinπ+
4003π
sin (3 π)+4005π
sin(5 π)+4007π
sin (7 π )+ 4009π
sin (9 π )+β¦
ππ»π·π£=β( 0.707Γ4003π )
2
+( 0.707Γ4005π )2
+( 0.707Γ4007π )2
+( 0.707Γ4009π )2
+β¦
0.707Γ400π
ππ»π·π£=β( 13 )2
+( 15 )2
+( 17 )2
+( 19 )2
+( 111 )2
+( 113 )2
+β¦
ππ»π·π£=0.45 ππ»π·π£=45%
38
Example-1
ππ»π·π£=β(π 3 ,π ππ)2+(π 5 ,π ππ )2+(π 7 ,π ππ )2
π 1 ,π ππ
ππ»π·π£=β( 0.707Γ4003π )
2
+( 0.707Γ4005π )2
+( 0.707Γ4007π )2
0.707Γ400π
ππ»π·π£=β( 13 )2
+( 15 )2
+( 17 )2
ππ»π·π£=0.41 ππ»π·π£=41%
β’ THDv by first three nonzero harmonics
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END OF LECTURE-11
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