switch-mode dc/ac...
TRANSCRIPT
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 1
page 1
Switch-Mode DC/AC Inverters
2005年6月7日
鄒 應 嶼 教 授
國立交通大學 電機與控制工程研究所
Filename: \電力電子 (研究所)\PE-08.PWM DC-AC Inverters.ppt
國立交通大學電力電子晶片設計與DSP控制實驗室Power Electronics IC Design & DSP Control Lab., NCTU, Taiwan
http://powerlab.cn.nctu.edu.tw/
POWERLABNCTU
電力電子晶片設計與DSP控制實驗室Power Electronics IC Design & DSP Control Lab.
台灣新竹交通大學 • 電機與控制工程研究所
page 2
Switch-Mode DC/AC Inverters
IntroductionBasic Concepts of Switch-Mode InverterSquare Wave InverterSinusoidal PWM InverterPWM Switching SchemesSingle-Phase InvertersThree-Phase InvertersEffect of Blanking TimeAdvanced PWM Switching SchemesRectifier Mode of Operation
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 2
page 3
Introduction
What is an Inverter ?
AC/DC: RectifierDC/DC: ChopperDC/AC: InverterAC/AC: Cycloconverter
~=
Inverter
AC OUTPUTDC INPUT
page 4
Applications of DC/AC Inverters
PWM Inverters for AC Motor DrivesPWM Inverters for AC Power SourceUPS & AVRInduction HeatingPower Supply for Fluorescent Lamp
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 3
page 5
Inverter in AC Motor Drive
+
−
vdAC
motor60 Hz
AC
diode-rectifier
filter capacitor
switch-mode inverter
Unidirectional Power Flow
+
−
vdAC
motor60 Hz
AC
filter capacitor
switch-mode converter
Switch-mode converter for motoring and regenerative braking in AC motor drive.
switch-mode converter
Bidirectional Power Flow
page 6
Circuit Architecture of Advanced AC Motor Drives
uεud
• PWM Control• Inverter Control• DTC Vector Control• Sensorless Control• Servo Control• Auto-Tuning
• Power Factor Control
• Regenerative Braking Control
• DC-Link Voltage Regulation
• DC-Link Capacitor Minimization
Cd
to sw itches
Inputconverter
Outputconverter
ud
to sw itches
u’1u’2u’3
N S
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 4
page 7
Classifications of Inverters
Voltage Source Inverter (VCI)
Current Source Inverter (VCI)Square Wave CSI
PWM CSI
PWM InverterCurrent-Controlled PWM Inverter
Variable Voltage Inverter
Processed Energy Source and Load (Voltage Source, Current Source)Topology (Single-Phase, Three-Phase, etc.)PWM Strategy (Square, PWM, Sine PWM, Regular PWM, Space Vector PWM, etc.)Switching Devices (SCR, Power Transistor, Power MOSFET, IGBT, etc.)Switching Schemes (PWM, Resonant, Quasi-Resonant, Soft PWM, etc)Control Schemes (Hysteresis, PID, Dead-beat, Variable Structure, Fuzzy, etc.)Controller Implementation (Analog, Microprocessor, DSP, etc.)
page 8
Three-Phase Inverter Drives
b1
b2
b3
b4
b5
b6
Voltage (Line to Neutral)
Current (Line)
3-PhasePowerSupply
N
S
S
N
b1
b2
b3
b4
b5
b6N
S
S
N
3-PhasePowerSupply
Voltage (Line to Neutral)
Current (Line)
VSI: Voltage Source Inverter
CSI: Current Source Inverter
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 5
page 9
Three-Phase Inverter for AC Power Source
3-phaseload
3-PhaseMotor
3-PhasePowerSupply
page 10
Voltage Source Current-Controlled PWM Inverter
Current
Controller
ias*
i as
b1
b2
b3
b4
b5
b6
3-PhasePowerSupply
--
-
ibs*
ics*
i bsi cs
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 6
page 11
Applications of Inverters in UPS
整 流 器負 載
市電
充電器
升壓器 換流器濾波器
蓄電池
控制功能:
• 開機程序控制• 蓄電池充電控制• 直流鏈電壓調節• 功率因數補償
• 輸出交流電壓調節• 自我保護機制• 電源監測• 緊急事件處理
控 制 器
功率級
輸 出
濾波器
輸 出
轉換開關
page 12
Basic Concepts of Switch-Mode Inverter
+
−
vd
1-phase switch-mode
inverter + filter
ioid
+
−vo
iovo
4 1 2 3
1 inverter
3 inverter
2 rectifier
4 rectifier
io
vo0
Switch-Phase Switch-Mode Inverter
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 7
page 13
One-Leg Switch-Mode Inverter
DA+TA+
TA- DA-
io
+
−vAN
N
o
+
−2dV
+
−2dV
+
−
vd A
( $ ) . ( )V V VAo d d14
21 273
2= =
π
( $ ) ($ )V VhAo hAo= 1
1.41.21.00.80.60.40.2
00 1 3 5 7 9 11 13 15
x
xx x x x x x
( $ ) / ( )V VAo d1 2
(h: harmonics of f1)
t
Vd2
−Vd2
v Ao
11f
0
Spectrum of Square-Wave PWM
page 14
Analysis of a Step Approximated Inverter for UPS
π ω t
V
2πu21
ππ u
D−
=
)(tv
一個由責任比(D)控制的方波如下圖所示:
1. 計算此方波的諧波。2. 計算此責任比控制方波之基本波(fundamental)與諧波的均方根值。3. 計算此責任比控制方波的均方根值 Vrms = ?4. 當此方波之均方根值與峰值為V的正弦波相同時,責任比D = ?5. 此方波的總諧波失真(THD)為何?
10 ≤≤ D
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 8
page 15
Harmonics of Odd-Quad Symmetric Square Waveform
) cos sin(
3cos 2cos cos 3sin 2sin sin) (
10
321
3210
∑∞
=
++=
+++
++++=
nnn tnBtnAA
tBtBtBtAtAtAAtf
ωω
ωωωωωωω
L
L
L+++= tAtAtAtv ωωω 5sin3sinsin)( 531
1. 由於此方波具有奇函數對稱 f(ωt) = -f(-ωt) 且半波對稱 f(ωt) = -f(ωt + π) 的特性,因此稱之為奇四分對稱(odd quad symmetric),其傅立葉係數(Fourier Series)僅含有正弦函數的奇次項。
∫=π
ωωπ
2
0 cos)(1 tdtntfAn
[ ]
2sin
142
cos14
coscos
)( sin)( sin1
5.025.0
5.05.0
5.0
5.0
5.02
5.0
πππ
ωωπ
ωωωωπ
ππ
π
π π
π
nDVn
unV
n
tntnnV
tdtnVtdtnVA
uu
uu
u
u
u
un
==
+−=
⎥⎦⎤
⎢⎣⎡ −+=
−+
−
− −
+∫ ∫
page 16
Harmonics of Duty-Ratio Controlled Square Wave
1.0
31
71
91
111
131
1VV n
0 n1 3 5 7 9 11 13
51
⎩⎨⎧=
n/nVn
Vn of values oddfor of valueseven for 0
1
) cos991+7sin
715sin
513sin
31(sin
2sin4 L++++= tttttnDVVn ωωωωω
ππ
2sin14 π
πnDV
nVn =
2. 基本波的均方根值:
tDVtV ωππ
sin2
sin4)(1 =
2sin
24
1π
πDVV rms =
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 9
page 17
RMS Value of Harmonics
第n次諧波的均方根值:
tnnDVn
tVn ωπ
πsin
2sin14)( =
2sin
214 π
πnDV
nVnrms =
諧波的均方根值:
21
227
25
23 rmsrmshrms VVVVVV −=+++= L
2cos8)
2(sin8 22
22
uDVDDVVhrms ππ
π−=−= π
π uD −=
page 18
RMS Value of the Square Waveform
ππ
ωπ
ωπ
ππ
uVDV
tdVtdvVD
rms
−==
⎥⎦⎤
⎢⎣⎡== ∫∫
2/1
0
22
0
2 )(1)(21
DVVrms =
3. 由於週期波之平移,不會改變其均方根值,因此:
[ ]2/1
,..5,3,1
22/12
52
32
1 2sin14
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛=+++= ∑
∞ ππ
nDVn
VVVVrms L
2VDVVrms ==
4. 當此方波之均方根值與峰值為V的正弦波相同時:
%50 =∴ D
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 10
page 19
THD of a Square Waveform
rms
rmsrms
rms VVV
VVVV
THD1
21
2
1
27
25
23 −=
+++=
L
5. 此方波之總諧波失真:
2sin
214
)2
(sin8
2sin
24
)2
sin2
4()( 22
22
1
27
25
23
ππ
ππ
ππ
ππ
D
DD
DV
DVDV
VVVV
THDrms
−=
−=
+++=
L
當責任比 D=1 時:
%3.48
214
81
2sin
24
)2
(sin8 22
2=
−=
−=
π
ππ
π
ππ
DV
DDTHD
page 20
Sinusoidal PWM
b1b2b3b4
PWMModulator
L
R
C
v irc
v c
v o
ic
ioiLrL+
−−−
+
+b1
Vdcb2
b3
b4
AB
vcontrol
vcarrier
0 t
vcontrol
vcarrier vcontrolvcarrier
+_
b1 b4
b2 b3
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 11
page 21
Pulse-Width Modulator
+
-Amp comparator
repetitive waveform
switch control signal
vcontrolvo (desired)
vo (actual)
ton toffTs
on on
off off
switch control signal
stV̂vst = sawtooth voltage vcontrol (amplified error)
vcontrol > vst
vcontrol < vst
(switch frequency fs = 1/Ts)
sts Vv
Tt
D ˆcontrolon ==
page 22
Natural Sinusoidal PWM
⎭⎬⎫
⎩⎨⎧ <
+− off : ,on :tricontro
AA
l
TTvv
⎭⎬⎫
⎩⎨⎧ <
−+ off : ,on :tricontrol
AA TTvv
Vd2
−Vd2
1 lfundamenta , )( AoAo vv =uAo
controlV triV
( )1f s
0 t
t0
t = 0
tri
controlˆ
ˆ
VVma =
1ffm sf =
Amplitude Modulation Ratio
Frequency Modulation Ratio
dAoA VvTvv 21 on, istricontrol => +
dAoA VvTvv 21 on, istricontrol −=< −
DA+TA+
TA- DA-
io
+
−vAN
N
o
+
−2
dV
+
−2
dV
+
−
Vd
dAoAN Vvv 21
+= hAohAN VV )ˆ()ˆ( =
A
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 12
page 23
Spectrum of Sinusoidal PWM
x
xx
x
x x x x x x x xx x x xxxx
( $ )/
VV
Ao h
d 2
15 ,8.0 == fa mm
( )m f + 2m f
( )2 1m f +2m f 3m f
( )3 2m f +Harmonics h of f1
00.20.40.60.81.01.2
Vd2
−Vd2
1 lfundamenta , )( AoAo vv =uAo
controlV triV
( )1f s
0 t
t0
page 24
Sinusoidal PWM
controlvtriv
0
Vd2
( )− Vd2
vAo
controlvtriV̂
0 t
triv
t
tricontroltri
control ˆ2ˆ
VvVV
vV dAo ≤=
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 13
page 25
Linear Control of the Sinusoidal PWM Output
tricontrol1controlcontrolˆˆsinˆ VVtVv ≤= ω
0.1for 2
sin2
sinˆˆ
)( 11tri
control1 ≤== adadAo m
VtmVtV
Vv ωω
0.1 2
)ˆ( 1 ≤= adaAo mVmV
which shows that in a sinusoidal PWM, the amplitude of the fundamental-frequency component of the output voltage varies linearly with ma (provided ma ≤ 1.0). Therefore, the range of ma from 0 to 1 is referred to as the linear range.
page 26
Harmonics of the Sinusoidal Modulated PWM Waveforms
1)( fkjmf fh ±=
kmjh f ±= )(
x
xx
x
x x x x x x xx x x x xxxx
( $ )/
VV
Ao h
d 2
15 ,8.0 == fa mm
( )m f + 2m f
( )2 1mf +2m f 3m f
( )3 2m f +
Harmonics h of f1
00.20.40.60.81.01.2
The frequencies at which voltage harmonics occur can be indicated as:
The harmonic order h corresponds to the kth sideband of j times the frequency modulation ratio mf:
where the fundamental frequency correspond to h = 1.
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 14
page 27
Generalized Harmonics of vAo for a Large mf
0.2 0.4 0.6 0.8 1.0
fundamental1
mah
0.2 0.4 0.6 0.8 1.0
mf
mf ± 2
mf ± 4
1.242
0.016
1.15 1.006 0.818 0.601
0.061 0.131 0.220 0.318
0.018
2mf ± 3
2mf ± 5
0.190 0.326 0.370 0.314 0.181
0.024 0.071 0.139 0.212
0.044
2mf ± 1
0.016
3mf ± 2
3mf ± 4
0.335 0.123 0.083 0.171 0.113
0.139 0.203 0.176 0.062
0.157
3mf
0.104
3mf ± 6 0.0440.016
0.044
0.012 0.047
4mf ± 3
4mf ± 5
0.163 0.157 0.008 0.105 0.068
0.070 0.132 0.115 0.009
0.119
4mf ± 1
0.084
4mf ± 7 0.0500.017
0.012
0.034
Generalized Harmonics of vAo for a Large mf.
Note: is tabulated as a function of ma.])2/1/()ˆ([)2/1/()ˆ( dhANdhAo VVVV =
page 28
Example 8.1: Harmonic Analysis
In the following circuit, Vd = 300 V, ma = 0.8, mf = 39, and the fundamental frequency is 47 Hz. Calculate the rms values of the fundamental-frequency voltage and some of the dominant harmonics in vAo using Table 8-1.
Example 8.1: Harmonics Analysis of a One-Leg Switching-Mode Inverter
DA+TA+
TA- DA-
io
+
−vAN
N
o
+
−2dV
+
−2dV
+
−
Vd A
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 15
page 29
Solution of Example 8.1
2/)ˆ(
07.1062/)ˆ(
221
)(d
hAo
d
hAodhAo V
VVVV
V ==
Fundamental: (VAo)1 = 106.07 × 0.8 = 84.86 V at 47 Hz(VAo)37 = 106.07 × 0.22 = 23.33 V at 1739 Hz(VAo)39 = 106.07 × 0.818 = 86.76 V at 1833 Hz(VAo)41 = 106.07 × 0.22 = 23.33 V at 1927 Hz(VAo)77 = 106.07 × 0.314 = 33.31 V at 3619 Hz(VAo)79 = 106.07 × 0.314 = 33.31 V at 3713 Hzetc.
Solution:The rms voltage at any value of h is given as
From Table 8-1 the rms voltages are as follows:
page 30
Selection of Frequency Modulation Ratio
Small mf (mf 21)
1. Synchronous PWM
2. mf should be an odd integer
1. Subharmonics due to asynchronous PWM becomes small as mf is increased.
2. In applications of inverter motor drives, asynchronous PWM should be avoid when operating in low frequency range.
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PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 16
page 31
Harmonics Due to Overmodulation
1harmonics h
3 5 7 9 11 13 150
)2
/()ˆ( dhAoVV
0.5
1.0
17 19 21 23 25 27
x
x
x x xx x
xx x x x x x
ma = 2.5mf = 15
(mf)
Harmonics due to overmodulation; drawn for ma = 2.5 and mf = 15.
page 32
Voltage Control by Varying ma
0
)2
/()ˆ( 1 dAoVV
1.0
(= 1.278)
ma0 1.0 3.24
linearover-
modulationSquare-wave
π4
(for mf = 15)
24)ˆ(
2 1d
Aod VVV
π
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 17
page 33
Square-Wave Switching Scheme
t0
1
x
h (harmonics of f1)3 5 7 9 11 13 150
00.20.40.60.81.01.21.4
xx x x x x x
V d2
2dV−
11f
v Ao
( $ ) . ( )V V VAo d d14
21 273
2= =
π
( $ ) ($ )V VhAo hAo= 1
( $ ) / ( )V VAo d1 2
page 34
Single-Phase Half-Bridge Inverter
D+T+
T− D−
io
+
−vo
N
o
+
−2dV
+
−2dV
+
−
vd
Half-bridge inverter.
VT = Vd IT = io, peak
The input capacitor C+ and C- act as dc blocking capacitor, there will no dc component flows through io.The peak voltage and current ratings of the switches are:
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 18
page 35
Single-Phase Full-Bridge Inverter
io
+
−vo
o
+
−2dV
+
−2dV
+
−
vd
Single-phase full-bridge inverter.
DA+TA+
TA- DA-
TB+ DB+
TB- DB-
A
B
VT = Vd IT = io, peak
With the same dc voltage, the maximum output voltage of the full-bridge inverter is twice that of the half-bridge inverter. The peak voltage and current ratings of the switches are:
vo = vAo -vBo
page 36
PWM with Bipolar Voltage Switching
0
0
( )1f s
controv triv
vo1
V d
−Vd
vo
t
t
io
+−vo
o
+
−2dV
+
−2dV
+
−
vd
DA+TA+
TA- DA-
TB+ DB+
TB- DB-
A
B
)()( tvtv AoBo −=
)(2)()()( tvtvtvtv AoBoAoo =−=
)0.1( ˆ 1 ≤= adao mVmV
)0.1(4ˆ 1 >
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 19
page 37
Example 8.2: Harmonics of Bipolar PWM
In the full-bridge converter circuit of Fig. 8-11, Vd = 300 V, ma = 0.8, mf = 39, and the fundamental frequency is 47 Hz. Calculate the rms values of the fundamental-frequency voltage and some of the dominant harmonics in the output voltage can be obtained by using Table 8-1, as illustrated by the following example.
2/)ˆ(13.212
2/)ˆ(
22/)ˆ(
22
21)(
d
hAo
d
hAod
d
hAodhAo V
VVVV
VVVV ==⋅⋅=
Fundamental: Vo1 = 212.13 × 0.8 = 169.7 V at 47 Hz(Vo)37 = 212.13 × 0.22 = 46.67 V at 1739 Hz(Vo)39 = 212.13 × 0.818 = 173.52 V at 1833 Hz(Vo)41 = 212.13 × 0.22 = 46.67 V at 1927 Hz(Vo)77 = 212.13 × 0.314 = 66.60 V at 3619 Hz(Vo)79 = 212.13 × 0.314 = 66.60 V at 3713 Hzetc.
Example 8.2: Harmonics of Full-Bridge Bipolar PWM Inverter
Solution:
page 38
DC-Side Current id
filter
+
−eo
+
−
vd
*di
Lf1
Cf1
idfilter
Lf2
Cf2+
−
vo
ioload
Lload
switch-mode inverter
fs → ∞Lf1, Cf1 → 0
fs → ∞Lf2, Cf2 → 0
fs
Inverter with “fictitious” filters at the input dc side and the output ac side.
tVvv ooo 11 sin2 ω==
)sin(2 1 φω −= tIi oo
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 20
page 39
Analysis of DC-Side Current id
)sin(2sin2)()()( 11* φωω −== tItVtitvtiV oooodd
)2cos(2
)2cos(cos)(
12
21*
φω
φωφ
−−=
+=−−=
tII
iItV
IVV
IVti
dd
ddd
oo
d
ood
φcosd
ood V
IVI =
d
ood V
IVI2
12 =
On the dc side, the LC filter will filter the high-frequency components in id, and idwould only consist of the low-frequency and dc components.
*
Therefore,
where
page 40
Waveforms of DC-Side Current id
0
0
ω1t
ω1t
vo (filtered)
idioid2
Vd
-Vd
Id
φ
The DC-side current in a single-phase inverter with PWM bipolar voltage switching.
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 21
page 41
Voltage Ripples at DC Link
The voltage ripples at the dc link are caused by two reasons:Rectification of the line voltageSecond harmonic current component occurs (of the fundamental frequency at the inverter output) at the dc link due to the sinusoidal output.
page 42
Unipolar PWM
io
+−vo
o
+
−2dV
+
−2dV
+
−
vd
DA+TA+
TA- DA-
TB+ DB+
TB- DB-
A
B
vcontrol > vtri : TA+ on and vAN = Vdvcontrol < vtri : TA- on and vAN = 0
(-vcontrol) > vtri : TB+ on and vBN = Vd(-vcontrol) < vtri : TB- on and vBN = 0
leg A switches:
leg B switches:
vtri vcontrol
t0
(-vcontrol)TA+on
TB+on
-vcontrol > vtrivcontrol > vtri
Vd
Vd
t
t
0
0
vAN
vBN
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 22
page 43
PWM with Unipolar Voltage Switching (Single Phase)
x
d
hAo
VV )ˆ(
00.20.40.60.81.0
vtri vcontrol
t0
(-vcontrol)TA+on
TB+on
-vcontrol > vtri vcontrol > vtri
xx x
x
Vd
Vd
t
t
0
0
vAN
vBN
Vd
-Vd
0
vo(= vAN - vBN)
vo1
t
1x x x x x x
(2mf - 1) (2mf + 1)2mfmf 3mf 4mf
h
(harmonics of f1)
page 44
PWM with Unipolar Voltage Switching
TA+, TB- on : vAN = Vd, vBN= 0; vo= Vd
TA-, TB+ on : vAN = 0, vBN= Vd ; vo= -Vd
TA+, TB+ on : vAN = Vd, vBN= Vd ; vo= 0
TA-, TB- on : vAN = 0, vBN= 0 ; vo= 0
Vd
-Vd
0
vo(= vAN - vBN)
vo1
t
3-level switching
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 23
page 45
Unipolar vs. Bipolar PWM
Compared with the bipolar PWM, the unipolar PWM has the following advantages:
“effectively” doubling the switching frequency Reduced voltage jumps in the output voltage
Lower current ripples and harmonic distortion!
page 46
Harmonics of the Unipolar PWM
)0.1(ˆ 1 ≤= adao mVmV
)0.1(4ˆ 1 >
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 24
page 47
Example 8.3: Analysis of Unipolar PWM
In Example 8-2, suppose that a PWM with unipolar voltage-switching scheme is used, with mf = 38. Calculate the rms values of the fundamental-frequency voltage and some of the dominant harmonics in the output voltage.
2/)(13.212)(
d
hAoho V
VV =
h = j(2mf) ± k
Fundamental: Vo1 = 0.8 × 212.13 = 169.7 V at fundamental or 47 Hz(Vo)75 = 0.314 × 212.13 = 66.60 V at h = 2mf − 1 = 75 or 3525 Hz(Vo)77 = 0.314 × 212.13 V at h = 2mf + 1 = 77 or 3619 Hzetc.
Example 8.3: Analysis of the Full-Bridge Unipolar PWM Inverter
Solution:
page 48
DC-Side Current id
0 t
id io
(-io)
The DC-side current in a single-phase inverter with PWM unipolar voltage switching.
Vd
-Vd
0
vovo1
t
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 25
page 49
Square Wave Operation
do VV π4ˆ
1 =
t0
1
x
h (harmonics of f1)3 5 7 9 11 13 150
0
0.20.4
0.60.81.01.21.4
xx x x x x x
V d2
2dV−
11f
vAo
( $ ) . ( )V V VAo d d14
21 273
2= =
π
( $ ) ($ )V VhAo hAo= 1
( $ ) / ( )V VAo d1 2
page 50
Single-Phase, Full-Bridge InverterControl by Voltage Cancellation
io
+−vo
+
−
vd
DA+TA+
TA- DA-
TB+ DB+
TB- DB-
A
B
N
id
00
0.2
0.4
0.6
0.8
1.0
60 120 1807th
5th 3rd
total harmonic distortion
fundamental
α
vAN
vBN
vo
0
0
0
α
Vd
Vd
Vd
(180- α )°
(180- α )°
α
β -Vd
180°
180°
°−=°−= )2
90(2
)180( ααβ
(a) power circuit
(b) waveform(c) normalized fundamental and harmonic voltage output and total harmonic distortion as a function of α.
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 26
page 51
Pulsewidth Control
∫∫ −− ==β
β
π
πθθ
πθθ
πdhVdhvV doho )cos(
2)cos(2)ˆ(2/
2/
)sin(4)ˆ( βπ
hVh
V dho =∴
vo
0
Vd
(180- α )°
(180- α )°
α
β-Vd
°−=°−= )2
90(2
)180( ααβ
where β = 90o - 0.5α and h is an odd integer.
page 52
Application in Step Wave UPS
io
+
−vo
o
+
−2dV
+
−2dV
+
−
vd
DA+TA+
TA- DA-
TB+ DB+
TB- DB-
A
B
vo = vAo -vBo
π ω t
V
2πu21
ππ u
D−
=
)(tv
10 ≤≤ D
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 27
page 53
Ripple in Single-Phase Inverter Output
+
−eo(t)
+
−
Vd
induction-motor load
L
switch-mode inverter
(a) circuit
+
−
vo
io+ −
~tEe oo 1sin2 ω=
vL = vL1 + vripple
+
−
L
io1 + −
~ tEe oo 1sin2 ω=
vL1
+
−~vo1
L+ −
+
−vripple
vrippleiripple
(b) fundamental-frequency components
(c) ripple frequency components
vL = vL1 + vripple
iL = iL1 + iripple
Th ripple in a periodic waveform refers to the difference between the instantaneous values of the waveform and its fundamental-frequency component.
page 54
Current Ripple Analysis
Vo1 = Eo + VL1 = Eo +jω1LIo1
vripple(t) = vo - vo1
kvL
tit
+= ∫0 rippleripple )(1)( ζ
Vo1
j(ω1L)Io1 = VL1 Eo
Io1
(d) fundamental-frequency phasor diagram
Th output current ripple can be expressed as
where k is a constant and ζ is a variable of integration.
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 28
page 55
Ripple in the Inverter Output
vo
vripple = vo - vo1
iripple
0
vo1
t
t
t t
t
t
0
0
0
0
0
vo
vripple = vo - vo1
iripple
vo1
(a) square-wave switching (b) PWM bipolar voltage switching
page 56
Push-Pull Inverters
)0.1(ˆ 1 ≤= adao mnVmV
)0.1(4ˆ 1 >
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 29
page 57
Comments on Push-Pull Inverters
In control of the push-pull inverter, it should avoid the dc saturation in the primary side of the transformer.Very good magnetic coupling between the two primary windings is required to reduce the energy associated with the leakage inductance.
D2T2T1D1
+
−
Vd
id
n : 1
+
−vo
Require very good coupling!
page 58
Comments on Push-Pull Inverters
D2T2T1D1
+
−
Vd
id
n : 1
+
−vo
Number of turns will be high for sinusoidal output!
In a PWM push-pull inverter for producing sinusoidal output (unlike those used in switch-mode dc power supplies), the transformer must be designed for the fundamental output frequency. This requires a high number of turns.A higher number of of turns will result in a high transformer leakage inductance, which is proportional to the square of the number of turns.
i
Nv+
_
L N Al
dBdH
N Alm m
= =2 2µ
lm
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 30
page 59
Switch Utilization in Single-Phase Inverters
TT
oo
IqVIV
max,1 ratio nutilizatio Switch =
22
422 max,max,1max,
max, ==== qnV
Vn
IIVV do
oTdT π
16.021 ratio nutilizatio switch Maximum ≅=∴π
n = turn ratio
where q is the number of switches in an inverter, VT and IT as the peak voltage and current ratings of a switch, Vo1 is the rms value of the output fundament voltage, Io,max is the rms value of the output current. (The output current is assumed as sinusoidal)
The utilization factor of all the switches in an inverter is defined as:
Single-Phase Inverter: Square-wave mode at maximum rated output
Push-Pull Inverter
page 60
Switch Utilization in Single-Phase Inverters
222
422 max,max,1max,max, ==== qV
VIIVV dooTdT π
16.021 ratio nutilizatio switch Maximum ≅=∴π
42
422 max,max,1max,max, ==== qVVIIVV dooTdT π
16.021 ratio nutilizatio switch Maximum ≅=∴π
Half-Bridge Inverter
Full-Bridge Inverter
16.021 ratio nutilizatio switch Maximum ≅=π
Conclusion: all the single-phase inverter has the same switch utilization ratio
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 31
page 61
Comments on Switch Utilization Ratio
aa mm 81
421 ratio nutilizatio switch Maximum ≅= ππ
)0.1 PWM,( ≤am
Switch ratings are chosen conservatively to provide safety margins.In determining the switch current rating in a PWM inverter, one would have to take into account the variations in the input dc voltage available.The output ripple current would influence the switch current rating.
In practice, the switch utilization ratio would be much smaller than 0.16, due to
Single-Phase Inverter: PWM mode
Conclusion: The theoretical maximum switch utilization ratio in a PWM switching is only 0.125 at ma = 1, as compared with 0.16 in a square-wave inverter.
page 62
Example 8.4: Calculation of Switch Utilization Ratio
In a single-phase full-bridge PWM inverter, Vd varies in a range of 295-325V. The output voltage is required to be constant at 200 V (rms), and the switch utilization ratio (under these idealized conditions, not accounting for any overcurrent capabilities).
V325max, == dT VV
14.141022 =×== oT II
4switches of no. ==q
VA200010200max,1 =×=oo IV
11.014.143254
2000 ratio nutilizatio Switch max,1 ≅××
==TT
oo
IqVIV
Example 8.4: Calculation of Switch Utilization Ratio
Solution:
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 32
page 63
Three-Phase Inverter
DA+TA+
TA- DA-
id
A
o
+
−2dV
+
−2dV
+
−
vd
DB+TB+
TB- DB-
B
DC+TC+
TC- DC-
C
N
page 64
Three-Phase PWM InverterDefinition of Voltages and Currents
o : neutral of the invertern : neutral of the loadVdc : voltage of the dc-linkS1 ... S6 : switches of the invertervao, vbo, vco : pole voltage of the invertervan, vbn, vcn : phase voltage of the loadias, vbs, vcs : phase current of the loadvab, vbc, vca : line-to-line voltage
180o180o
0.5Vd
S1S2 0.5Vd
S1
S3S4
S3S4
vac
vba
vca
0ωt
ωt
ωtS6S5
S6S5
(a)
(b)
(c)vab
ωtVd
vbc
ωt
(d)
(e)
vcaωt
(f)
van
ωt
Vd23
ias
120o120o
Vd
Six-Step Voltage and Current Waveforms
(g)
ias
ibsics
S1
S2
S3
S4
S5
S6
o n
vasvbs
vcs
Vdc
A
C
B
vao = van + vnovbo = vbn + vnovco = vcn + vno
vno =(1/3) (vao + vbo + vco)van = vao - vnovbn = vbo - vnovcn = vco - vnovao = 0.5Vdc[ds1 (k)- ds2 (k)]
d k d ks s1 2 1( ) ( )+ ≤
where ds1 (k), ds2 (k) are the on-duty ratio of S1 and S2during the k-th switching interval.
vno
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 33
page 65
Considerations in 3-Phase PWM Inverters
For low values of mf (mf 1.0), regardless of the value of mf, the conditions pertinent to a small mf should be observed.
page 66
PWM in Three-Phase Voltage Source Inverters
vtri vcontrol, A vcontrol, B vcontrol, C
t0
Vdt0
vAN
Vd
vBN
0 t 00.20.40.60.81.0
1
x
(2mf+1)
d
ho
VV )ˆ(
15 ,8.0 == fa mm
mf 2mf 3mf
x x xx
x x x x xx x x x x h
Harmonics of f1
(mf+2) (3mf+2)
Vdt0
fundamental vLL1
Cancel out!
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 34
page 67
Linear Modulation (ma
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 35
page 69
Overmodulation
0
square-wave
d
LL
VV )rms(1
1
linear
overmodulationsquare-wave
78.06 =π
612.0223
=
3.24ma
(for mf = 15)
Three-phase inverter; VLL1(rms)/Vd as a function of ma.
page 70
Square-Wave Operation
00.20.40.60.81.0
d
LL
VV
h
ˆ
TA+0
vAN
ω 1tVd TA
180°
180°vBN
TB-TB+
0 ω 1tvCN
0 ω 1tTC+ TC-
TC+
vAB
Vd0 ω 1t
1LLv
1.2
harmonics of f1
1 3 5 7 9 11 13
DA +
TA+
TA- DA-
id
A
+
−
Vd
DB +
TB+
TB- DB-
B
DC +TC+
TC- DC-
C
N
A B C
x
xx x x
Square-wave inverter (three phase).
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 36
page 71
Harmonics of the Square-Wave Inverter
dLLh VhV 78.0=
ddd
LL VVV
V 78.06
24
23
)rms(1 ≅== ππ
h = 6n ± 1 (n = 1,2,3, ...)
The fundamental-frequency line-to-line rms voltage in the output of the one-leg inverter operating in a square-wave mode:
00.20.40.60.81.0
d
LL
VV
h
ˆ
1.2
harmonics of f11 3 5 7 9 11 13
x
x x x x
The line-to-line voltage waveform does not depend on the load and contains harmonics (6n +/- 1; n = 1, 2, …), whose amplitudes decrease inversely proportional to their harmonic order.
page 72
Switch Utilization in Three-Phase Inverters
Assume a pure sinusoidal output current with an rms value of Io,max (both in the PWM and square-wave modes) at maximum loading.
Each switch has the following peak ratings:
In the square-wave mode, this ratio is 1/2π ≅ 0.16 compared to a maximum of 0.125 for a PWM linear region with ma = 1.0.
axoTdT IIVV m,max, 2==
max,max,max,
max,phase3 11
621
263
6)VA(
ratio nutilizatio Switchd
LL
od
oLL
TT VV
IVIV
IV=== −
)0.1(81
223
621(PWM)ratio nutilizatio switch Maximum ≤== aaa mmm
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 37
page 73
Ripple in the Inverter Output
+
−
VdThree-phase
three-leg inverter
L
(load neutral)
~
id
A
iA
B
C
iB
iC
R = 0
eA(t)+
−n
3-phaseac motor load IA1
VAN1
φ EA j(ω1L)IA1
Three-phase inverter: (a) circuit diagram; (b) phasor diagram (fundamental frequency).
(a) (b)
Back emf
),,( CBAkvvv nNkNkn =−=
),,( CBAkedtdiLv knkkn =+=
Under balanced operating conditions, it is possible to express the inverter phase output voltage vAN in terms of the inverter output voltage.
Each phase voltage can be written as
N
page 74
Phase-to-Neutral Voltage
0=++ CBA iii
0)( =++ CBA iiidtd
0=++ CBA eee
0=++ CnBnAn vvv
)(31
CNBNANnN vvvv ++=
)(31
32
CNBNAnAn vvvv +−=
111 AAAn Lj IEV ω+=
In a three-phase, three-wire load
and
Phase-to-neutral voltage for phase is
+
−
VdThree-phase
three-leg inverter
L
~
id
A
iA
B
C
iB
iC
R = 0
eA(t)+
−n
N
nNv
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 38
page 75
Phase-to-Neutral Variables of a Three-Phase Inverter
vAn1vAn
iripple, A
peak
dV32
dV31
t0
0 t
vAn
dV32
dV31
0vAn1
iripple, A
0 t
peak
Phase-to-neutral variables of a three-phase inverter.
(a) square wave (b) PWM
)(31
32
CNBNAnAn vvvv +−=
page 76
DC-Side Current id
)()()()()()(111
* titvtitvtitviV CnCBBnAAndd ++=
)quantity dca (cos3)]1201cos()1201cos(
]120cos()120cos()cos([cos2 1111*
dd
oo
d
ood
IV
IVtt
ttttV
IVi
==
−°+°++
−°−°−+−=
φ
φωω
φωωφωω
id
Id = id*0 t
Input DC current in a three-phase inverter.
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 39
page 77
Conduction of Switches in 3-Phase Inverters
vAn
0 t
vAn1
vAn
iAiA1 (fundamental)
0 t
DA+ TA+ DA- TA-
Square-wave inverter: phase A waveforms.
Square-Wave Operation:
page 78
Conduction of Switches in 3-Phase InvertersvAN
iA
0 t
φ = 30°vBN
iB
0 t
vCN
iC0 t
(TA- , TB- , DC-) conducting (TA+ , TB + , DC +) conducting
PWM inverter waveforms: load power factor angle = 30° (lag).
PWM Operation:
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 40
page 79
Short-Circuit in a 3-Phase PWM Inverter
iA+
−
vd
id = 0
iB
iCLoad
iA+
−
vd
id = 0
iB
iCLoad
page 80
Effect of Blanking Time t∆
TA+
TA-
+
−
Vd
N
A
DA+
DA-
t
t
t
t
t
t
t
0
0
0
0
0
0
0
vtri vcontrol
)ideal(cont +ATv
)ideal(cont −ATv
+ATvcont
−ATvcont ∆t
∆t
vAN
loss
ideal actual
actualideal
gainTs(iA > 0)
(iA < 0)
(a)
(b)
(c)
(e)
(d)
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 41
page 81
Voltage Drop ∆id Induced by the Blanking Time
actualideal )()( ANAN vvv −=ε
⎪⎪⎩
⎪⎪⎨
⎧
+=∆
∆
∆
0
0
Ads
Ads
ANiV
Tt
iVTt
V
⎪⎪⎩
⎪⎪⎨
⎧
−=∆
∆
∆
0
0
Ads
Ads
BNiV
Tt
iVTt
V
⎪⎪⎩
⎪⎪⎨
⎧
+=∆−∆=∆
∆
∆
02
02
Ads
Ads
BNAN
oiV
Tt
iVTtVV
V
t0vAN ideal actual
Ts
(iA > 0)
t0vBN
ideal actual
Ts
t0
page 82
Effect of t∆ on Vo
TA+
TA-
+
−
VdA
DA+
DA-
(a)
TB+
TB-
B
DB+
DB-
N
iA
io+ −vo
iB
0
Vo
vcontrol
∆Vo
∆Vo
io < 0
io > 0
No blanking time(independent of io)
(b)
Effect of t∆ on Vo, where ∆Vo is defined as a voltage drop if positive.
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 42
page 83
Effect of t∆ on the Sinusoidal Output
0
0
t
t
io
Vo(t)
ideal
actual
page 84
Programmed Harmonic Elimination Switching
-1.0
0
1.0
2/dAo
VV
Notch1 Notch2 Notch3 Fundamental
α2α1 α3 (π - α3) (π - α1)(π - α2)
(π + α1)
2ππ
2π
23π
ω1t
Notch4 Notch5 Notch60 20 40 60 80 1000
10
20
30
40
50
60
α3
α2
α1 22.06°
16.24°
Fundamental as a percentage of maximum fundamental voltage
Not
ch a
ngles
in d
egre
es
Programmed harmonic elimination of fifth and seventh harmonics.
273.142/)ˆ( 1 ==
πdAo
VV 188.1
2/)ˆ( max,1 =
d
Ao
VV
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 43
page 85
Hysteresis Current Controlled PWM Inverter
TA+
TA-
+
−
Vd A
DA+
DA-
N
iA
0 t
TA-: on TA+: on
vAN
actual current iA
reference current iA*nt
iA*
iA
iε ABC
comparatortolerance band
Switch-mode
inverterΣ
Tolerance band current control.
page 86
Fixed-Frequency Current Controlled PWM Inverter
Σ ΣABC
Switch-mode
inverter
comparator
PI controller
0 t
iA*
iA+−
iε vcontrol
Feed forward
+
+
−
Fixed-frequency current control.
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 44
page 87
Rectifier Mode of Operation
eA+
− n
+
−
Vd A
N
iA
~
VAn
j(ω1L)IA
(IA)q(IA)pIA
EAδ
δEA
j(ω1L)IAVAn(IA)q
(IA)p
IA
VAn
j(ω1L)IA
EAIA
(a)
(b)
(c)
(d)
Operation modes: (a) circuit; (b) inverter mode; (c) rectifier mode; (d) constant IA.
page 88
Summary of PWM Inverters
Switch-mode, voltage source dc-to-ac inverters are described that accept dc voltage source as input and produce either single-phase or three-phase sinusoidal output voltages at a low frequency relative to the switching frequency.
These dc-to-ac inverters can make a smooth transition into the rectification mode, where the flow of power reverses to be from the ac side to the dc side. This occurs, for example, during braking of an induction motor supplied through such an inverter.
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 45
page 89
Summary of PWM Inverters
The sinusoidal PWM switching scheme allows control of the magnitude and the frequency of the output voltage. Therefore, the input to the PWM inverters is an uncontrolled, essentially constant dc voltage source. This switching scheme results in harmonic voltages in the range of the switching frequency and higher, which can be easily filtered out.
page 90
Summary of PWM Inverters
The square-wave switching scheme controls only the frequency of the inverter output. Therefore, the output magnitude must be controlled by controlling the magnitude of the input dc voltage source.
The square-wave output voltage contains low-order harmonics. A variation of the square-wave switching scheme, called the voltage cancellation technique, can be used to control both the frequency and the magnitude of the single-phase (but not three-phase) inverter output..
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 46
page 91
Summary of PWM Inverters
Due to the harmonics in the inverter output voltage, the ripple in the output current (current ripples) does not depend on the level of power transfer at the fundamental frequency; instead the ripple depends inversely on the load inductance, which is more effective at higher frequencies.
In practice. If a switch turns off in an inverter leg, the turn-on of the others switch must be delayed by a blanking time, which introduces low-order harmonics in the inverter output.
page 92
Summary of PWM Inverters
There are many other switching schemes (PWM strategies) in addition to the sinusoidal PWM. For example, the programmed harmonic elimination switching technique can be easily implemented with the help of VLSI circuits to eliminate specific harmonics from the inverter output.
The current-regulated (current-mode) modulation allows the inverter output currents to be controlled directly by comparing the measured actual current with the reference current and using the error to control the inverter switched.
-
PWM DC-AC Inverters
NCTU 2005 Power Electronics Course Notes 47
page 93
Control of Inverter Output Voltage
triˆ1
Vma
Vd
k(ma) Vd invertervcontrolvo1 = k(ma)Vd
(rms, line-line)
for ma ≤ 1.0 k(ma) = 0.707 ma 1-phase= 0.621 ma 3-phase
k = 0.9 1-phase= 0.78 3-phase
vo2 = kVd(rms, line-to-line)
(a) (b)
Summary of inverter output voltage: (a) PWM operation (ma ≤ 1); (b) square-wave operation.
The relationship between the control input and the full-bridge inverter output magnitude can be summarized as shown in Fig. (a), assuming a sinusoidal PWM in the linear range of ma ≤ 1.0. For a square-wave switching, the inverter does not control the magnitude of the inverter output, and the relationship between the dc input voltage and the output magnitude is summarized in Fig. (b).
page 94
References
[1] J. Holtz, “Pulsewidth modulation – a survey,” IEEE Trans. on Ind. Electron., vol. 39, no. 5, pp. 410-420, Dec. 1992. [2] J. Holtz, "Pulsewidth modulation for electronic power conversion," Proc. of IEEE, vol. 82, no. 8, pp. 1194-1214,
Aug. 1994. [3] J. W. A. Wilson and J. A. Yeamans, “Intrinsic harmonics of idealized inverter PWM systems,” IEEE IAS Annual
Meeting Conf. Rec., pp. 967-973, 1976.[4] G. S. Buja, “Optimum output waveform in PWM inverters,” IEEE Trans. Ind. Appl., vol. 16, pp. 830-836, 1980. [5] S. R. Bowes and A. Midoun, "Microprocessor implementation of new optimal PWM switching strategies,"
IEE proceedings, vol. 13J, Pt.B., no5, Sep. 1988. [6] I. J. Pitel, S. N. Talukdar, and P. Wood, “Characterization of programmed-waveform pulse width modulation,” IEEE
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