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Power Electronics
Dr. Imtiaz HussainAssociate Professor
email: [email protected] :http://imtiazhussainkalwar.weebly.com/
Lecture-11Inverters
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Introduction
• 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.
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Methods of Inversion
• 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
• Electrical inverters use a combination of ‘chopping’ circuits and transformers to change DC power into AC.
• They are much more widely used and are far more efficient and practical.
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TYPICAL APPLICATIONS
– Un-interruptible power supply (UPS)
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TYPICAL APPLICATIONS
– Traction
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TYPICAL APPLICATIONS
– HVDC (High Voltage Direct Current)
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Types of Inverters
• There are three basic types of dc-ac converters depending on their AC output waveform: – Square wave Inverters– Modified sine wave Inverters– Pure sine wave Inverters
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Square Wave Inverters
– The square wave is the simplest and cheapest type, but nowadays it is practically not used commercially because of low power quality (THD≈45%).
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Modified Sine wave Inverters
• The modified sine wave topologies provide rectangular pulses with some dead spots between positive and negative half-cycles.
• They are suitable for most electronic loads, although their THD is almost 24%.
• They are the most popular low-cost inverters on the consumer market today,
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Pure Sine Wave Inverters
– A true sine wave inverter produces output with the lowest total harmonic distortion (normally below 3%).
– 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.
– This type is also used in grid-connected applications.
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Simple square-wave inverter• To illustrate the concept of AC waveform generation
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AC Waveform Generation
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AC Waveforms
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Output voltage harmonics
• Harmonics may cause degradation of equipment (Equipment need to be “de-rated”).
• Total Harmonic Distortion (THD) is a measure to determine the “quality” of a given waveform.
𝑇𝐻𝐷𝑣=√∑𝑛=2
∞
(𝑉 𝑛 ,𝑅𝑀𝑆 )2
𝑉 1 ,𝑅𝑀𝑆
𝑇𝐻𝐷𝑖=√∑𝑛=2
∞
( 𝐼𝑛 ,𝑅𝑀𝑆 )2
𝐼 1 ,𝑅𝑀𝑆
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Fourier Series• Study of harmonics requires understanding of wave shapes. • Fourier Series is a tool to analyse wave shapes.
• Where,𝑣 (𝑡 )=𝑎𝑜+∑
𝑛=1
∞
[𝑎𝑛cos (𝑛𝜃 )+𝑏𝑛 sin (𝑛𝜃 ) ]
𝑎𝑛=1𝜋 ∫
0
2 𝜋
𝑣 (𝑡 ) cos (𝑛𝜃 )𝑑𝜃
𝑏𝑛=1𝜋 ∫
0
2𝜋
𝑣 (𝑡 )sin (𝑛𝜃 )𝑑𝜃
𝑎𝑜=1𝜋 ∫
0
2𝜋
𝑣 (𝑡 )𝑑𝜃
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Harmonics of square-wave 𝑎𝑜=
1𝜋 ∫
0
2𝜋
𝑣 (𝑡 )𝑑𝜃
𝑎𝑜=1𝜋 [∫
0
𝜋
𝑉 𝑑𝑐𝑑𝜃+∫𝜋
2𝜋
−𝑉 𝑑𝑐𝑑𝜃 ]𝑎𝑜=0
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Harmonics of square-wave
𝑎𝑛=1𝜋 ∫
0
2 𝜋
𝑣 (𝑡 ) cos (𝑛𝜃 )𝑑𝜃
𝑎𝑛=1𝜋 [∫
0
𝜋
𝑉 𝑑𝑐 cos (𝑛𝜃 )𝑑𝜃+∫𝜋
2𝜋
−𝑉 𝑑𝑐 cos (𝑛𝜃 )𝑑𝜃 ]𝑎𝑛=
𝑉 𝑑𝑐
𝜋 [∫0
𝜋
cos (𝑛𝜃 )𝑑𝜃−∫𝜋
2𝜋
cos (𝑛𝜃 )𝑑𝜃 ]=0
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Harmonics of square-wave 𝑏𝑛=
1𝜋 ∫
0
2𝜋
𝑣 (𝑡 )sin (𝑛𝜃 )𝑑𝜃
𝑏𝑛=1𝜋 [∫
0
𝜋
𝑉 𝑑𝑐 s∈(𝑛𝜃 )𝑑𝜃+∫𝜋
2𝜋
−𝑉 𝑑𝑐 s∈(𝑛𝜃 )𝑑𝜃 ]
𝑏𝑛=0• When n is even
𝑏𝑛=𝑉 𝑑𝑐
𝜋 [∫0
𝜋
sin (𝑛𝜃 ) 𝑑𝜃−∫𝜋
2𝜋
sin (𝑛𝜃 )𝑑𝜃 ]=2𝑉 𝑑𝑐
𝑛𝜋[1−cos (𝑛𝜋 )]
𝑏𝑛=4𝑉 𝑑𝑐
𝑛𝜋
• When n is odd
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Harmonics of square-wave 𝑣 (𝑡 )=𝑎𝑜+∑
𝑛=1
∞
[𝑎𝑛cos (𝑛𝜃 )+𝑏𝑛 sin (𝑛𝜃 ) ]
𝑣 (𝑡 )=∑𝑛=1
∞
[𝑏𝑛sin (𝑛𝜃 ) ]
𝑣 (𝑡 )=4𝑉 𝑑𝑐
𝜋 ∑𝑛=1,3,5…
∞1𝑛sin (𝑛𝜃 )
Where,
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Harmonics of square-wave
• 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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Filtering• Low-pass filter is normally fitted at the inverter output to
reduce the high frequency harmonics.
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Topologies of Inverters• Voltage Source Inverter (VSI)– Where the independently controlled ac output is a voltage
waveform.– 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.
• Current Source Inverter (CSI)– Where the independently controlled ac output is a current
waveform. – 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
• Single phase voltage source inverters are of two types.
– Single Phase Half Bridge voltage source inverters
– Single Phase full Bridge voltage source inverters
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1- Half Bridge VSI• 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.
• It is clear that both switches S+ and S− 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 VSI• Figure shows the ideal waveforms associated with the half-
bridge inverter.
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1- Half Bridge VSI• The gating signals for thyristors and resulting output voltage
waveforms are shown below.
Note: Turn off circuitry for thyristor is not shown for simplicity
𝑣𝑜={ 𝑉 𝑠
20<𝑡<𝑇 /2
−𝑉 𝑠
2𝑇 /2<𝑡<𝑇
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1- Full Bridge VSI• This inverter is similar to the half-bridge inverter; however, a
second leg provides the neutral point to the load.
• 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 VSI• Figure shows the ideal waveforms associated with the half-
bridge inverter.
𝑣 𝑖
𝑣 𝑖
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1- Full Bridge VSI• The gating signals for thyristors and resulting output voltage
waveforms are shown below.
𝑣𝑜={ 𝑉 𝑠0<𝑡<𝑇 /2 −𝑉 𝑠𝑇 /2<𝑡<𝑇
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3- Full Bridge VSI• Single-phase VSIs cover low-range power applications and
three-phase VSIs cover medium- to high-power applications.
• 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.
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1- VSI using transistors
• Single-phase half bridge and full bridge voltage source inverters using transistors are shown below.
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Example-1• 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.
• Calculate– THDv
– THDv by first three nonzero harmonics
100V
-100V
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Example-1• To calculate the harmonic contents we need to expand the
output waveform into Fourier series expansion.
• Since output of the inverter is an odd function with zero offset, therefore and will be zero.
100V
-100V
𝑣𝑜=𝑎𝑜+∑𝑛=1
∞
[𝑎𝑛 cos (𝑛𝜃 )+𝑏𝑛sin (𝑛𝜃 ) ]
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Example-1
• Where,
100V
-100V
𝑣𝑜=∑𝑛=1
∞
𝑏𝑛sin (𝑛𝜃 )
𝑏𝑛=1𝜋 ∫
0
2𝜋
𝑣𝑜 (𝑡 )sin (𝑛𝜃 )𝑑𝜃
𝑏𝑛=4𝑉 𝑜
𝑛𝜋 {0𝑛even1𝑛 odd
𝑣 (𝑡 )=4𝑉 𝑜
𝜋 ∑𝑛=1,3,5…
∞1𝑛sin (𝑛𝜃 )
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Example-1• THDv can be calculated as
• Fourier series can be further expanded as
𝑇𝐻𝐷𝑣=√∑𝑛=2
∞
(𝑉 𝑛 ,𝑅𝑀𝑆 )2
𝑉 1 ,𝑅𝑀𝑆
𝑣 (𝑡 )=4𝑉 𝑜
𝜋 ∑𝑛=1,3,5…
∞1𝑛sin (𝑛𝜃 )
𝑣 (𝑡 )=400𝜋sin𝜃+
4003𝜋
sin (3 𝜃)+4005𝜋
sin(5 𝜃)+4007𝜋
sin (7 𝜃 )+ 4009𝜋
sin (9 𝜃 )+…
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Example-1
𝑇𝐻𝐷𝑣=√(𝑉 3 ,𝑅𝑀𝑆)2+(𝑉 5 ,𝑅𝑀𝑆 )2+(𝑉 7 ,𝑅𝑀𝑆 )2+(𝑉 9 ,𝑅𝑀𝑆)2+…
𝑉 1 ,𝑅𝑀𝑆
𝑣 (𝑡 )=400𝜋sin𝜃+
4003𝜋
sin (3 𝜃)+4005𝜋
sin(5 𝜃)+4007𝜋
sin (7 𝜃 )+ 4009𝜋
sin (9 𝜃 )+…
𝑇𝐻𝐷𝑣=√( 0.707×4003𝜋 )
2
+( 0.707×4005𝜋 )2
+( 0.707×4007𝜋 )2
+( 0.707×4009𝜋 )2
+…
0.707×400𝜋
𝑇𝐻𝐷𝑣=√( 13 )2
+( 15 )2
+( 17 )2
+( 19 )2
+( 111 )2
+( 113 )2
+…
𝑇𝐻𝐷𝑣=0.45 𝑇𝐻𝐷𝑣=45%
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Example-1
𝑇𝐻𝐷𝑣=√(𝑉 3 ,𝑅𝑀𝑆)2+(𝑉 5 ,𝑅𝑀𝑆 )2+(𝑉 7 ,𝑅𝑀𝑆 )2
𝑉 1 ,𝑅𝑀𝑆
𝑇𝐻𝐷𝑣=√( 0.707×4003𝜋 )
2
+( 0.707×4005𝜋 )2
+( 0.707×4007𝜋 )2
0.707×400𝜋
𝑇𝐻𝐷𝑣=√( 13 )2
+( 15 )2
+( 17 )2
𝑇𝐻𝐷𝑣=0.41 𝑇𝐻𝐷𝑣=41%
• THDv by first three nonzero harmonics
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END OF LECTURE-11
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