chapter4 DC to AC Converter.ppt

DC to AC Converter

1. CHAPTER 4
INVERTERS:
Converting DC to AC

2. CONTENTS
۩Introduction
۩Basic Principles of Inverter
۩Single-phase Half-Bridge Square-Wave Inverter
۩Single-phase Full-Bridge Square-Wave Inverter
۩Quasi Inverter
۩Three-phase inverter
۩Fourier Series and Harmonics Analysis
۩Pulse-Width Modulation (PWM)

3. Definition:
Converts DC to AC power by switching the DC input
voltage in a pre-determined sequence so as to
generate AC voltage.
Applications:
Induction motor drives, traction, standby power
supplies, and uninterruptible ac power supplies (UPS).
INTRODUCTION

4. INTRODUCTION
V
dc V
ac
+
-
+
-
General block diagram

5. Three types of inverter:
INTRODUCTION
Vdc Vac
+
-
+
-
AC
Load
Iac
Inverter
switching
control
“DC LINK”
Vdc


+
-
AC
Load
ILOAD
Inverter
switching
control
L
IDC
(a) Voltage source inverter (VSI)(b) Current source inverter (CSI)

6. Three types of inverter: (cont.)
INTRODUCTION
Vdc


+
-
AC
Load
Iac
Inverter
switching
control
Comparison
circuit
Reference
Waveform
Output current
sensing circuit
(c) Current regulated inverter

7. BASIC PRINCIPLES
The schematic of single-phase full-bridge square-wave
inverter circuit D1
D2
D3
D4
T1 T3
T4 T2
+ V0 -
I0
VDC

8. BASIC PRINCIPLESS1
S2
S3
+ V0 -
VDC
S4
V0
t1
VDC
t2
t S1
S2
S3
+ V0 -
VDC
S4
V0
t3
-VDC
t2
t

9. Single-phase Half-bridge
Square-wave Inverter
V
0
1
2
V
0
G.
V
DC
+
-
The basic single-phase half-bridge inverter circuit

10. The total RMS value of the load output voltage,
The instantaneous output voltage is: (refer to slide 30/pp:88)
The fundamental rms output voltage (n=1)is
Single-phase Half-bridge
Square-wave Inverter22
2 2/
0
2
DC
T
DC
O
V
dt
V
T
V 
















 2,4,.... n for 0
sin
2
,...5,3,1


n
DC
O tn
n
V
v 
 DC
DC
O
V
V
V 45.0
2
12
1









11. In the case of RL load, the instantaneous load current i
o,
where
The fundamental output power is
Single-phase Half-bridge
Square-wave Inverter





,..5,3,1
22
)sin(
)(
2
n
n
DC
o tn
LnRn
V
i 
 )/(tan
1
RLn
n 

 











22
2
1
1111
)(2
2
cos
LR
V
RI
IVP
DC
o
ooo



12. The total harmonic distortion (THD),
Single-phase Half-bridge
Square-wave Inverter





 

 ,....7,5,3
2
1
1
n
n
O
V
V
THD  
2
1
2
1
1
oo
O
VV
V
THD 

13. Example 3.1
The single-phase half-bridge inverter has a resistive load of R =
2.4Ω and the DC input voltage is 48V. Determine:
(a) the rms output voltage at the fundamental frequency
(b) the output power
(c) the average and peak current of each transistor.
(d) the THD
Single-phase Half-bridge
Square-wave Inverter

14. Solution
VDC = 48V and R = 2.4Ω
(a)The fundamental rms output voltage,
V
o1= 0.45V
DC= 0.45x48 = 21.6V
(b)For single-phase half-bridge inverter, the output voltage
V
o= V
DC/2
Thus, the output power,
Single-phase Half-bridge
Square-wave InverterW
RVP
oo
240
4.2
)2/48(
/
2
2




15. Solution
(c)The transistor current I
p= 24/2.4 = 10 A
Because each of the transistor conducts for a 50% duty cycle,
the average current of each transistor is I
Q= 10/2 = 5 A.
(d)
Single-phase Half-bridge
Square-wave Inverter2
1
2
1
1
oo
O
VV
V
THD   
%34.48
45.0
245.0
1 2
2








DC
DC
DC
V
V
xV

16. The switching in the second leg is delayed by 180
degrees from the first leg.
The maximum output voltage of this inverter is twice
that of half-bridge inverter.
Single-phase Full-bridge
Square-wave Inverter

17. The output RMS voltage
And the instantaneous output voltage in a Fourier series is
The fundamental RMS output voltage
In the case of RL load, the instantaneous load current
Single-phase Full-bridge
Square-wave InverterDCDCO VdtV
T
V 







22 


,...5,3,1
sin
4
n
DC
O tn
n
V
v 
 DC
DC
V
V
V 9.0
2
4
1

 
 
n
n
DC
o
tn
LnRn
V
i 






sin
4
,...5,3,1
22

18. Example 3.2
A single-phase full-bridge inverter with V
DC= 230 and
consist of RLC in series. If R = 1.2Ω, ωL = 8 Ω and
1/ωC = 7 Ω, find:
(a) The amplitude of fundamental rms output current,
i
o1
(b) The fundamental component of output current in
function of time.
(c) The power delivered to the load due to the
fundamental component.
Single-phase Full-bridge
Square-wave Inverter

19. Example 3.3
A single-phase full-bridge inverter has an RLC load with R
= 10Ω, L = 31.5mH and C = 112μF. The inverter frequency
is 60Hz and the DC input voltage is 220V. Determine:
(a) Express the instantaneous load current in Fourier series.
(b) Calculate the rms load current at the fundamental
frequency.
(c) the THD of load current
(d) Power absorbed by the load and fundamental power.
(e) The average DC supply current and
(f) the rms and peak supply current of each transistor
Single-phase Full-bridge
Square-wave Inverter

20. Three-Phase Inverter

21. Viewed as extensions of the single-phase bridge circuit.
The switching signals for each switches of an inverter leg are
displaced or delayed by 120
o
.
With 120
o
conduction, the switching pattern is T6T1 –T1T2 –
T2T3 –T3T4 –T4T5 –T5T6 –T6T1 for the positive A-B-C
sequence.
When an upper switch in an inverter leg connected with the
positive DC rail is turned ON, the output terminal of the leg
(phase voltage) goes to potential +V
DC/2.
When a lower switch in an inverter leg connected with the
negative DC rail is turned ON, the output terminal of that leg
(phase voltage) goes to potential -V
DC/2.
Three-Phase Inverter

23. The line-to-neutral voltage can be expressed in Fourier series
The line voltage is v
ab= √3v
anwith phase advance of 30
o
Three-Phase Inverter





























,..5,3,1
,..5,3,1
,..5,3,1
6
7
sin
3
sin
2
2
sin
3
sin
2
6
sin
3
sin
2
n
dc
cn
n
dc
bn
n
dc
an
tn
n
n
V
v
tn
n
n
V
v
tn
n
n
V
v











  























,..5,3,1
,..5,3,1
,..5,3,1
sin
3
sin
32
3
sin
3
sin
32
3
sin
3
sin
32
n
dc
ca
n
dc
bc
n
dc
ab
tn
n
n
V
v
tn
n
n
V
v
tn
n
n
V
v












24. Fourier series is a tool to analyze the wave shapes of the
output voltage and current in terms of Fourier series.
Fourier Series and
Harmonics Analysis




2
0
1
dvfa
o 




2
0
cos
1
dnvfa
n 




2
0
sin
1
dnvfb
n
Inverse Fourier  



1
0
sincos
2
1
n
n
nbnnaavf 
Where t

25. If no DC component in the output, the output voltage and
current are
The rms current of the load can be determined by
Fourier Series and
Harmonics Analysis 



1
sin)(
n
nno
tnVtv   



1
sin)(
n
nno
tnIti  2
11
2
,
2












n
n
n
rmsnrms
I
II
Where n
n
n
Z
V
I

26. The total power absorbed in the load resistor can be
determined by
Fourier Series and
Harmonics Analysis





1
2
,
1 n
rmsn
n
n
RIPP

27. Since the objective of the inverter is to use a
DC voltage source to supply a load that
requiring AC voltage, hence the quality of
the non-sinusoidal AC output voltage or
current can be expressed in terms of THD.
The harmonics is considered to ensure that
the quality of the waveform must match to
the utility supply which means of power
quality issues.
This is due to the harmonics may cause
degradation of the equipments and needs to
be de-rated.
Total Harmonics
Distortion

28. Total Harmonics
Distortion

29. The THD of the load voltage is expressed as,
The current THD can be obtained by replacing the
harmonic voltage with harmonic current,
Total Harmonics
Distortionrms
rmsrms
rms
n
rmsn
v
V
VV
V
V
THD
,1
2
,1
2
,1
2
2
,)( 



 rms
n
rmsn
i
I
I
THD
,1
2
2
,)(




30. Harmonics of Square-
wave Waveform2  t DC
V DC
V 0
1
0
2
0 





 
 



DCDC VdVa 0)cos()cos(
0
2






 
 



dndn
V
a
DC
n  
 
  











 

n
n
nnn
n
V
nn
n
V
dndn
V
b
DC
DC
DC
n
cos1
2
)cos2(cos)cos0(cos
)cos()cos(
)sin()sin(
2
0
0
2









 

31. When the harmonics number, n of a waveform is
even number, the resultant of
Therefore,
When n is odd number,
Hence,
Harmonics of Square-wave
Waveform1cosn 0
nb 1cosn n
V
b
DC
n
4


32. Spectrum of Square-
waveNormalised
Fundamental
1st
3rd
5th
7th
9th
11th
n
(0.33)
(0.2)
(0.14)
(0.11)
(0.09)
•Harmonic 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.
•Due to the small
separation between the
fundamental and 3rd
harmonics, output low-
pass filter design can be
very difficult.

33. Quasi-square wave
a
n= 0, due to half-wave symmetry

34. Quasi-square wave
Therefore,
If n is even, b
n= 0;
If n is odd, 

cos
4
n
V
b
dc
n

35. Example 3.4
The full-bridge inverter with DC input voltage of 100V,
load resistor and inductor of 10Ω and 25mH
respectively and operated at 60 Hz frequency.
Determine:
(a) The amplitude of the Fourier series terms for the
square-wave load voltage.
(b) The amplitude of the Fourier series terms for load
current.
(c) Power absorbed by the load.
(d) The THD of the load voltage and load current for
square-wave inverter.

36. Amplitude and
Harmonics Control 2 t VDC
-VDC   
S2
S4
S1
S2
S1
S3
S3
S4
S2
S4
0 VDC 0 0-VDC
S1 Closed Opened
S2
S3
S4
Vo
The output voltage of the full-
bridge inverter can be controlled
by adjusting the interval of on
each side of the pulse as zero .

37. The rms value of the voltage waveform is
The Fourier series of the waveform is expressed as
The amplitude of half-wave symmetry is
Amplitude and
Harmonics Control





2
1)(
1
2


DCDCrms VtdVV 
oddn
nO tnVtv
,
)sin()(  )cos(
4
)()sin(
2






n
n
V
tdtnVV
DC
DCn 








38. The amplitude of the fundamental frequency is controllable by
adjusting the angle of α.
The nth harmonic can be eliminated by proper choice of
displacement angle αif
Amplitude and
Harmonics Control

cos
4
1 






DC
V
V 0cosn
ORn

90


39. Pulse-width modulation provides a way to decrease
the total harmonics distortion (THD).
Types of PWM scheme
Natural or sinusoidal sampling
Regular sampling
Optimize PWM
Harmonic elimination/minimization PWM
SVM
Pulse-Width Modulation
(PWM)

40. Several definition in PWM
(i) Amplitude Modulation, M
a
If M
a≤ 1, the amplitude of the fundamental
frequency of the output voltage, V
1is linearly
proportional to M
a.
(ii) Frequency Modulation, M
f
Pulse-Width Modulation
(PWM) tri,
sin ,
carrier ,
,
m
em
m
referencem
a
V
V
V
V
M  e
tricarrier
f
f
f
f
f
M
sinreference


41. Bipolar Switching of PWM
Pulse-Width Modulation
(PWM)vtri (carrier)vsine (reference)
VDC
-VDC
(a)
(b)
S1 and S2 ON when Vsine > Vtri
S3 and S4 ON when Vsine < Vtri

42. Sinusoidal PWM Generator -Bipolar
Pulse-Width Modulation
(PWM)
G1, G2
G3, G4

43. Unipolar Switching
of PWM
Pulse-Width Modulation
(PWM)vtri (carrier)vsine (reference)
(a)
(b)
(c)
S4
S2
Vo
Vdc
0
Vdc
0
S1 is ON when Vsine > Vtri
S2 is ON when –Vsine < Vtri
S3 is ON when –Vsine > Vtri
S4 is ON when Vsine < Vtri

44. Sinusoidal
PWM
Generator-
Unipolar

45. ■Advantages of PWM switching
-provides a way to decrease the THD of load current.
-the amplitude of the o/p voltage can be controlled with the
modulating waveform.
-reduced filter requirements to decrease harmonics.
■Disadvantages of PWM switching
-complex control circuit for the switches
-increase losses due to more frequent switching.
Pulse-Width Modulation
(PWM)

46. Harmonics of Bipolar PWM
Assuming the PWM output is symmetry, the
harmonics of each kth PWM pulsecan be
expressed
Finally, the resultant of the integration is
PWM Harmonics





 



kk
k kk
tdtn-VtdtnV
tdtntvV
DCDC
T
nk








1k
)()sin()( )()sin(
2
)()sin()(
2
0  )(cos2coscos
2
1 kkkk
DC
nk
nnn
n
V
V 




47. PWM Harmonicsvtri vsine
VDC
-VDC k
 kk
 1k
 k

0
Symmetric sampling

48. Harmonics of Bipolar PWM
The Fourier coefficient for the PWM waveform is the
sum of V
nkfor the ppulses over one period.
The normalized frequency spectrum for bipolar
switching for m
a= 1 is shown below
PWM Harmonics


p
k
nkn VV
1
m
a
= 10.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
n=1 1.00 0.90 0.80 0.70 0.60 0.50 o.40 0.30 0.20 0.10
n = m
f
0.60 0.71 0.82 0.92 1.01 1.08 1.15 1.20 1.24 1.27
n=m
f
+2 0.32 0.27 0.22 0.17 0.13 0.09 0.06 0.03 0.02 0.00
Normalized Fourier Coefficients V
n/V
dcfor Bipolar PWM

49. Example 3.5
The inverter has a resistive load of 10Ω and
inductive load of 25mH connected in series with
the fundamental frequency current amplitude of
9.27A. The THD of the inverter is not more than
10%. If at the beginning of designing the inverter,
the THD of the current is 16.7% which is does
not meet the specification, find the voltage
amplitude at the fundamental frequency, the
required DC input supply and the new THD of the
current.

50. Example 3.6
The single-phase full-bridge inverter is used to produce
a 60Hz voltage across a series R-L load using bipolar
PWM. The DC input to the bridge is 100V, the
amplitude modulation ratio is 0.8, and the frequency
modulation ratio is 21. The load has resistance of R =
10Ω and inductance L = 20mH. Determine:
(a) The amplitude of the 60Hz component of the output
voltage and load current.
(b) The power absorbed by the load resistor
(c) The THD of the load current