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Power Electronics
Dr. Imtiaz Hussain
Associate Professor
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/
Lecture-11
Inverters
1
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.
2
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.
TYPICAL APPLICATIONS
– Un-interruptible power supply (UPS)
4
TYPICAL APPLICATIONS
– Traction
5
TYPICAL APPLICATIONS
– HVDC (High Voltage Direct Current)
6
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
7
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%).
8
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,
9
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.
10
Simple square-wave inverter
• To illustrate the concept of AC waveform generation
AC Waveform Generation
AC Waveforms
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.
14
𝑇𝐻𝐷𝑣 =
𝑛=2
∞
𝑉𝑛,𝑅𝑀𝑆
2
𝑉1,𝑅𝑀𝑆
𝑇𝐻𝐷𝑖 =
𝑛=2
∞
𝐼𝑛,𝑅𝑀𝑆
2
𝐼1,𝑅𝑀𝑆
Fourier Series
• Study of harmonics requires understanding of wave shapes.
• Fourier Series is a tool to analyse wave shapes.
• Where,
15
𝑣 𝑡 = 𝑎𝑜 +
𝑛=1
∞
𝑎𝑛 cos 𝑛𝜃 + 𝑏𝑛 sin 𝑛𝜃
𝑎𝑛 =
1
𝜋
0
2𝜋
𝑣 𝑡 cos 𝑛𝜃 𝑑𝜃
𝑏𝑛 =
1
𝜋
0
2𝜋
𝑣 𝑡 sin 𝑛𝜃 𝑑𝜃
𝑎𝑜 =
1
𝜋
0
2𝜋
𝑣 𝑡 𝑑𝜃
Harmonics of square-wave
16
𝑎𝑜 =
1
𝜋
0
2𝜋
𝑣 𝑡 𝑑𝜃
𝑎𝑜 =
1
𝜋
0
𝜋
𝑉𝑑𝑐 𝑑𝜃 +
𝜋
2𝜋
−𝑉𝑑𝑐 𝑑𝜃
𝑎𝑜 = 0
Harmonics of square-wave
17
𝑎𝑛 =
1
𝜋
0
2𝜋
𝑣 𝑡 cos 𝑛𝜃 𝑑𝜃
𝑎𝑛 =
1
𝜋
0
𝜋
𝑉𝑑𝑐 cos 𝑛𝜃 𝑑𝜃 +
𝜋
2𝜋
−𝑉𝑑𝑐 cos 𝑛𝜃 𝑑𝜃
𝑎𝑛 =
𝑉𝑑𝑐
𝜋
0
𝜋
cos 𝑛𝜃 𝑑𝜃 −
𝜋
2𝜋
cos 𝑛𝜃 𝑑𝜃 = 0
Harmonics of square-wave
18
𝑏𝑛 =
1
𝜋
0
2𝜋
𝑣 𝑡 sin 𝑛𝜃 𝑑𝜃
𝑏𝑛 =
1
𝜋
0
𝜋
𝑉𝑑𝑐 sin 𝑛𝜃 𝑑𝜃 +
𝜋
2𝜋
−𝑉𝑑𝑐 sin 𝑛𝜃 𝑑𝜃
𝑏𝑛 = 0
• When n is even
𝑏𝑛 =
𝑉𝑑𝑐
𝜋
0
𝜋
sin 𝑛𝜃 𝑑𝜃 −
𝜋
2𝜋
sin 𝑛𝜃 𝑑𝜃 =
2𝑉𝑑𝑐
𝑛𝜋
1 − cos(𝑛𝜋)
𝑏𝑛 =
4𝑉𝑑𝑐
𝑛𝜋
• When n is odd
Harmonics of square-wave
19
𝑣 𝑡 = 𝑎𝑜 +
𝑛=1
∞
𝑎𝑛 cos 𝑛𝜃 + 𝑏𝑛 sin 𝑛𝜃
𝑣 𝑡 =
𝑛=1
∞
𝑏𝑛 sin 𝑛𝜃
𝑣 𝑡 =
4𝑉𝑑𝑐
𝜋
𝑛=1,3,5…
∞
1
𝑛
sin 𝑛𝜃
Where, 𝑏𝑛 =
4𝑉𝑑𝑐
𝑛𝜋
0 𝑛 even
1 𝑛 odd
Harmonics of square-wave
20
• 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.
Harmonics of square-wave
21
Filtering
• Low-pass filter is normally fitted at the inverter output to
reduce the high frequency harmonics.
22
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.
23
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
24
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.
25
• 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.
1-∅ Half Bridge VSI
• Figure shows the ideal waveforms associated with the half-
bridge inverter.
26
1-∅ Half Bridge VSI
• The gating signals for thyristors and resulting output voltage
waveforms are shown below.
27
Note: Turn off circuitry for thyristor is not
shown for simplicity
𝑣𝑜 =
𝑉
𝑠
2
0 < 𝑡 < 𝑇/2
−
𝑉
𝑠
2
𝑇/2 < 𝑡 < 𝑇
1-∅ Full Bridge VSI
• This inverter is similar to the half-bridge inverter; however, a
second leg provides the neutral point to the load.
28
• 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.
1-∅ Full Bridge VSI
• Figure shows the ideal waveforms associated with the half-
bridge inverter.
29
𝑣𝑖
𝑣𝑖
1-∅ Full Bridge VSI
• The gating signals for thyristors and resulting output voltage
waveforms are shown below.
30
𝑣𝑜 =
𝑉
𝑠 0 < 𝑡 < 𝑇/2
−𝑉
𝑠 𝑇/2 < 𝑡 < 𝑇
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.
31
1-∅ VSI using transistors
• Single-phase half bridge and full bridge voltage source
inverters using transistors are shown below.
32
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
33
100V
-100V
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 𝑎𝑜 and 𝑎𝑛 will be zero.
34
100V
-100V
𝑣𝑜 = 𝑎𝑜 +
𝑛=1
∞
𝑎𝑛 cos 𝑛𝜃 + 𝑏𝑛 sin 𝑛𝜃
Example-1
• Where,
35
100V
-100V
𝑣𝑜 =
𝑛=1
∞
𝑏𝑛 sin 𝑛𝜃
𝑏𝑛 =
1
𝜋
0
2𝜋
𝑣𝑜 𝑡 sin 𝑛𝜃 𝑑𝜃
𝑏𝑛 =
4𝑉
𝑜
𝑛𝜋
0 𝑛 even
1 𝑛 odd
𝑣 𝑡 =
4𝑉
𝑜
𝜋
𝑛=1,3,5…
∞
1
𝑛
sin 𝑛𝜃
Example-1
• THDv can be calculated as
• Fourier series can be further expanded as
36
𝑇𝐻𝐷𝑣 =
𝑛=2
∞
𝑉𝑛,𝑅𝑀𝑆
2
𝑉1,𝑅𝑀𝑆
𝑣 𝑡 =
4𝑉
𝑜
𝜋
𝑛=1,3,5…
∞
1
𝑛
sin 𝑛𝜃
𝑣 𝑡 =
400
𝜋
sin 𝜃 +
400
3𝜋
sin(3𝜃) +
400
5𝜋
sin(5𝜃) +
400
7𝜋
sin 7𝜃 +
400
9𝜋
sin 9𝜃 + ⋯
Example-1
37
𝑇𝐻𝐷𝑣 =
𝑉3,𝑅𝑀𝑆
2
+ 𝑉5,𝑅𝑀𝑆
2
+ 𝑉7,𝑅𝑀𝑆
2
+ 𝑉9,𝑅𝑀𝑆
2
+ ⋯
𝑉1,𝑅𝑀𝑆
𝑣 𝑡 =
400
𝜋
sin 𝜃 +
400
3𝜋
sin(3𝜃) +
400
5𝜋
sin(5𝜃) +
400
7𝜋
sin 7𝜃 +
400
9𝜋
sin 9𝜃 + ⋯
𝑇𝐻𝐷𝑣 =
0.707 × 400
3𝜋
2
+
0.707 × 400
5𝜋
2
+
0.707 × 400
7𝜋
2
+
0.707 × 400
9𝜋
2
+ ⋯
0.707 × 400
𝜋
𝑇𝐻𝐷𝑣 =
1
3
2
+
1
5
2
+
1
7
2
+
1
9
2
+
1
11
2
+
1
13
2
+ ⋯
𝑇𝐻𝐷𝑣 = 0.45 𝑇𝐻𝐷𝑣 = 45%
Example-1
38
𝑇𝐻𝐷𝑣 =
𝑉3,𝑅𝑀𝑆
2
+ 𝑉5,𝑅𝑀𝑆
2
+ 𝑉7,𝑅𝑀𝑆
2
𝑉1,𝑅𝑀𝑆
𝑇𝐻𝐷𝑣 =
0.707 × 400
3𝜋
2
+
0.707 × 400
5𝜋
2
+
0.707 × 400
7𝜋
2
0.707 × 400
𝜋
𝑇𝐻𝐷𝑣 =
1
3
2
+
1
5
2
+
1
7
2
𝑇𝐻𝐷𝑣 = 0.41 𝑇𝐻𝐷𝑣 = 41%
• THDv by first three nonzero harmonics
END OF LECTURE-11
To download this lecture visit
http://imtiazhussainkalwar.weebly.com/
39

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  • 1. Power Electronics Dr. Imtiaz Hussain Associate Professor email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/ Lecture-11 Inverters 1
  • 2. 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. 2
  • 3. 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.
  • 6. TYPICAL APPLICATIONS – HVDC (High Voltage Direct Current) 6
  • 7. 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 7
  • 8. 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%). 8
  • 9. 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, 9
  • 10. 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. 10
  • 11. Simple square-wave inverter • To illustrate the concept of AC waveform generation
  • 14. 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. 14 𝑇𝐻𝐷𝑣 = 𝑛=2 ∞ 𝑉𝑛,𝑅𝑀𝑆 2 𝑉1,𝑅𝑀𝑆 𝑇𝐻𝐷𝑖 = 𝑛=2 ∞ 𝐼𝑛,𝑅𝑀𝑆 2 𝐼1,𝑅𝑀𝑆
  • 15. Fourier Series • Study of harmonics requires understanding of wave shapes. • Fourier Series is a tool to analyse wave shapes. • Where, 15 𝑣 𝑡 = 𝑎𝑜 + 𝑛=1 ∞ 𝑎𝑛 cos 𝑛𝜃 + 𝑏𝑛 sin 𝑛𝜃 𝑎𝑛 = 1 𝜋 0 2𝜋 𝑣 𝑡 cos 𝑛𝜃 𝑑𝜃 𝑏𝑛 = 1 𝜋 0 2𝜋 𝑣 𝑡 sin 𝑛𝜃 𝑑𝜃 𝑎𝑜 = 1 𝜋 0 2𝜋 𝑣 𝑡 𝑑𝜃
  • 16. Harmonics of square-wave 16 𝑎𝑜 = 1 𝜋 0 2𝜋 𝑣 𝑡 𝑑𝜃 𝑎𝑜 = 1 𝜋 0 𝜋 𝑉𝑑𝑐 𝑑𝜃 + 𝜋 2𝜋 −𝑉𝑑𝑐 𝑑𝜃 𝑎𝑜 = 0
  • 17. Harmonics of square-wave 17 𝑎𝑛 = 1 𝜋 0 2𝜋 𝑣 𝑡 cos 𝑛𝜃 𝑑𝜃 𝑎𝑛 = 1 𝜋 0 𝜋 𝑉𝑑𝑐 cos 𝑛𝜃 𝑑𝜃 + 𝜋 2𝜋 −𝑉𝑑𝑐 cos 𝑛𝜃 𝑑𝜃 𝑎𝑛 = 𝑉𝑑𝑐 𝜋 0 𝜋 cos 𝑛𝜃 𝑑𝜃 − 𝜋 2𝜋 cos 𝑛𝜃 𝑑𝜃 = 0
  • 18. Harmonics of square-wave 18 𝑏𝑛 = 1 𝜋 0 2𝜋 𝑣 𝑡 sin 𝑛𝜃 𝑑𝜃 𝑏𝑛 = 1 𝜋 0 𝜋 𝑉𝑑𝑐 sin 𝑛𝜃 𝑑𝜃 + 𝜋 2𝜋 −𝑉𝑑𝑐 sin 𝑛𝜃 𝑑𝜃 𝑏𝑛 = 0 • When n is even 𝑏𝑛 = 𝑉𝑑𝑐 𝜋 0 𝜋 sin 𝑛𝜃 𝑑𝜃 − 𝜋 2𝜋 sin 𝑛𝜃 𝑑𝜃 = 2𝑉𝑑𝑐 𝑛𝜋 1 − cos(𝑛𝜋) 𝑏𝑛 = 4𝑉𝑑𝑐 𝑛𝜋 • When n is odd
  • 19. Harmonics of square-wave 19 𝑣 𝑡 = 𝑎𝑜 + 𝑛=1 ∞ 𝑎𝑛 cos 𝑛𝜃 + 𝑏𝑛 sin 𝑛𝜃 𝑣 𝑡 = 𝑛=1 ∞ 𝑏𝑛 sin 𝑛𝜃 𝑣 𝑡 = 4𝑉𝑑𝑐 𝜋 𝑛=1,3,5… ∞ 1 𝑛 sin 𝑛𝜃 Where, 𝑏𝑛 = 4𝑉𝑑𝑐 𝑛𝜋 0 𝑛 even 1 𝑛 odd
  • 20. Harmonics of square-wave 20 • 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.
  • 22. Filtering • Low-pass filter is normally fitted at the inverter output to reduce the high frequency harmonics. 22
  • 23. 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. 23
  • 24. 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 24
  • 25. 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. 25 • 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.
  • 26. 1-∅ Half Bridge VSI • Figure shows the ideal waveforms associated with the half- bridge inverter. 26
  • 27. 1-∅ Half Bridge VSI • The gating signals for thyristors and resulting output voltage waveforms are shown below. 27 Note: Turn off circuitry for thyristor is not shown for simplicity 𝑣𝑜 = 𝑉 𝑠 2 0 < 𝑡 < 𝑇/2 − 𝑉 𝑠 2 𝑇/2 < 𝑡 < 𝑇
  • 28. 1-∅ Full Bridge VSI • This inverter is similar to the half-bridge inverter; however, a second leg provides the neutral point to the load. 28 • 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.
  • 29. 1-∅ Full Bridge VSI • Figure shows the ideal waveforms associated with the half- bridge inverter. 29 𝑣𝑖 𝑣𝑖
  • 30. 1-∅ Full Bridge VSI • The gating signals for thyristors and resulting output voltage waveforms are shown below. 30 𝑣𝑜 = 𝑉 𝑠 0 < 𝑡 < 𝑇/2 −𝑉 𝑠 𝑇/2 < 𝑡 < 𝑇
  • 31. 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. 31
  • 32. 1-∅ VSI using transistors • Single-phase half bridge and full bridge voltage source inverters using transistors are shown below. 32
  • 33. 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 33 100V -100V
  • 34. 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 𝑎𝑜 and 𝑎𝑛 will be zero. 34 100V -100V 𝑣𝑜 = 𝑎𝑜 + 𝑛=1 ∞ 𝑎𝑛 cos 𝑛𝜃 + 𝑏𝑛 sin 𝑛𝜃
  • 35. Example-1 • Where, 35 100V -100V 𝑣𝑜 = 𝑛=1 ∞ 𝑏𝑛 sin 𝑛𝜃 𝑏𝑛 = 1 𝜋 0 2𝜋 𝑣𝑜 𝑡 sin 𝑛𝜃 𝑑𝜃 𝑏𝑛 = 4𝑉 𝑜 𝑛𝜋 0 𝑛 even 1 𝑛 odd 𝑣 𝑡 = 4𝑉 𝑜 𝜋 𝑛=1,3,5… ∞ 1 𝑛 sin 𝑛𝜃
  • 36. Example-1 • THDv can be calculated as • Fourier series can be further expanded as 36 𝑇𝐻𝐷𝑣 = 𝑛=2 ∞ 𝑉𝑛,𝑅𝑀𝑆 2 𝑉1,𝑅𝑀𝑆 𝑣 𝑡 = 4𝑉 𝑜 𝜋 𝑛=1,3,5… ∞ 1 𝑛 sin 𝑛𝜃 𝑣 𝑡 = 400 𝜋 sin 𝜃 + 400 3𝜋 sin(3𝜃) + 400 5𝜋 sin(5𝜃) + 400 7𝜋 sin 7𝜃 + 400 9𝜋 sin 9𝜃 + ⋯
  • 37. Example-1 37 𝑇𝐻𝐷𝑣 = 𝑉3,𝑅𝑀𝑆 2 + 𝑉5,𝑅𝑀𝑆 2 + 𝑉7,𝑅𝑀𝑆 2 + 𝑉9,𝑅𝑀𝑆 2 + ⋯ 𝑉1,𝑅𝑀𝑆 𝑣 𝑡 = 400 𝜋 sin 𝜃 + 400 3𝜋 sin(3𝜃) + 400 5𝜋 sin(5𝜃) + 400 7𝜋 sin 7𝜃 + 400 9𝜋 sin 9𝜃 + ⋯ 𝑇𝐻𝐷𝑣 = 0.707 × 400 3𝜋 2 + 0.707 × 400 5𝜋 2 + 0.707 × 400 7𝜋 2 + 0.707 × 400 9𝜋 2 + ⋯ 0.707 × 400 𝜋 𝑇𝐻𝐷𝑣 = 1 3 2 + 1 5 2 + 1 7 2 + 1 9 2 + 1 11 2 + 1 13 2 + ⋯ 𝑇𝐻𝐷𝑣 = 0.45 𝑇𝐻𝐷𝑣 = 45%
  • 38. Example-1 38 𝑇𝐻𝐷𝑣 = 𝑉3,𝑅𝑀𝑆 2 + 𝑉5,𝑅𝑀𝑆 2 + 𝑉7,𝑅𝑀𝑆 2 𝑉1,𝑅𝑀𝑆 𝑇𝐻𝐷𝑣 = 0.707 × 400 3𝜋 2 + 0.707 × 400 5𝜋 2 + 0.707 × 400 7𝜋 2 0.707 × 400 𝜋 𝑇𝐻𝐷𝑣 = 1 3 2 + 1 5 2 + 1 7 2 𝑇𝐻𝐷𝑣 = 0.41 𝑇𝐻𝐷𝑣 = 41% • THDv by first three nonzero harmonics
  • 39. END OF LECTURE-11 To download this lecture visit http://imtiazhussainkalwar.weebly.com/ 39