1
00:00:15,320 --> 00:00:20,040
welcome back to this ah lecture series on
pulse width modulation for power electronic
2
00:00:20,040 --> 00:00:28,710
converters so the first seven or eight lectures
we focused on various power electronic converters
3
00:00:28,710 --> 00:00:33,840
dc to dc converters and dc to ac converters
voltage source and current source and also
4
00:00:33,840 --> 00:00:38,350
multilevel converters now we have started
to focus on pulse width modulation firstly
5
00:00:38,350 --> 00:00:42,690
we have been looking at the purpose of pulse
width modulation our focuses will be on you
6
00:00:42,690 --> 00:00:47,539
know among all the power electronic converters
we will focus on a three phase voltage source
7
00:00:47,539 --> 00:00:52,899
inverter that would where you know our focus
will be on and we are trying to look at certain
8
00:00:52,899 --> 00:00:56,789
modulation methods for that
so to start with we are looking at the purpose
9
00:00:56,789 --> 00:01:02,780
of pulse width modulation so in the last lecture
we saw saw few things ah in terms of why we
10
00:01:02,780 --> 00:01:08,520
need such kind of pulse width modulation and
this is going to be a continuation on that
11
00:01:08,520 --> 00:01:14,979
so in pulse width modulation what we do is
basically you have certain you know you have
12
00:01:14,979 --> 00:01:20,700
a dc bus voltage you have an inverter lets
say let me put an inverter like this you have
13
00:01:20,700 --> 00:01:26,450
certain inverter and this inverter has certain
dc voltage it has certain dc voltage excuse
14
00:01:26,450 --> 00:01:36,460
me it has certain dc voltage and what you
have is you need certain ac voltage here now
15
00:01:36,460 --> 00:01:40,450
how do you get the required amount of ac that
is through pulse width modulation that is
16
00:01:40,450 --> 00:01:44,130
by gating all these signals
so what needs to be given by the pulse width
17
00:01:44,130 --> 00:01:48,119
modulator is basically the gating signals
for the various devices in the inverter so
18
00:01:48,119 --> 00:01:54,710
the gating signals should be such that your
ac output voltages as this side with the fixed
19
00:01:54,710 --> 00:01:59,520
dc bias voltage so the first purpose of pulse
width modulation is to control the ac side
20
00:01:59,520 --> 00:02:05,170
fundamental voltage and the dc bias voltage
being fixed or it could sometimes be unregulated
21
00:02:05,170 --> 00:02:10,730
when it has been derived from let us say a
diode bridge rectifier you know or some rectifier
22
00:02:10,730 --> 00:02:14,070
and a open loop
now the second purpose is what you are going
23
00:02:14,070 --> 00:02:19,350
to get here on the ac side is going to be
a pulsed waveform is not going to be a sinusoidal
24
00:02:19,350 --> 00:02:24,690
waveform its a waveform is going to have certain
non sinusoidal components so those are harmonics
25
00:02:24,690 --> 00:02:29,410
other than the fundamental you are going to
have certain harmonic voltages these harmonic
26
00:02:29,410 --> 00:02:34,079
voltages are going to cause certain harmonic
currents these harmonic currents again could
27
00:02:34,079 --> 00:02:38,770
increase the copper losses in the various
elements through which they flow and the you
28
00:02:38,770 --> 00:02:43,260
know the harmonic fluxes and the harmonic
currents interacting with the you know steady
29
00:02:43,260 --> 00:02:48,160
current or steady flux ah can produce pulsating
torque if you are talking about a motor drive
30
00:02:48,160 --> 00:02:52,540
so they have lot of effects you know which
are undesirable effects of harmonic voltages
31
00:02:52,540 --> 00:02:57,720
the second purpose of pulse width modulation
is to reduce or mitigate these harmonic voltages
32
00:02:57,720 --> 00:03:02,230
are at least one or more of their harmful
effects so what we are going to look at these
33
00:03:02,230 --> 00:03:07,250
are the two broad purposes now and we are
going to quickly review ah whatever we saw
34
00:03:07,250 --> 00:03:12,010
in the previous weeks now so how are you going
to analyze the pwm waveform thats what here
35
00:03:12,010 --> 00:03:17,350
we are seeing some very very simple pwm waveforms
the actual waveforms could be much more complicated
36
00:03:17,350 --> 00:03:22,280
but these such simple pwm waveforms are of
great instructional value and you know and
37
00:03:22,280 --> 00:03:26,900
also in very very high power converters you
know you still have pwm waveforms which are
38
00:03:26,900 --> 00:03:31,971
reasonably simple because the switching frequencies
are low in hypo converters well now this is
39
00:03:31,971 --> 00:03:36,260
an example is what i have given is a square
wave and as i have said before the square
40
00:03:36,260 --> 00:03:42,100
wave has certain amplitude like this so if
v it is plus v and minus v this amplitude
41
00:03:42,100 --> 00:03:50,470
is going to be something like ah four v by
pi that is the amplitude that did you are
42
00:03:50,470 --> 00:03:55,970
going to have here and what you can do is
this is another example what is being done
43
00:03:55,970 --> 00:04:00,190
is its not a complete square wave but you
have pulses and these pulses have certain
44
00:04:00,190 --> 00:04:03,560
duration all of them have equal durations
you have positive pulses in the so called
45
00:04:03,560 --> 00:04:07,151
positive half cycle you have negative pulses
in the so called negative half cycle of the
46
00:04:07,151 --> 00:04:11,570
voltage of the fundamental voltage so you
can vary this bits if you increase the widths
47
00:04:11,570 --> 00:04:15,320
the fundamental voltage you get is going to
increase otherwise its going to decrease so
48
00:04:15,320 --> 00:04:21,280
its an example of a periodic signal and its
an example of a pwm waveform and you know
49
00:04:21,280 --> 00:04:25,650
and ah this is how you control the fundamental
voltage here and this is a another example
50
00:04:25,650 --> 00:04:29,410
where you can intuitively see that the fundamental
voltage is controlled and also the harmonics
51
00:04:29,410 --> 00:04:33,560
are probably better controlled because your
fundamental component is going to be something
52
00:04:33,560 --> 00:04:40,400
like this when the fundamental component is
high those pulses durations are higher and
53
00:04:40,400 --> 00:04:43,700
then when the fundamental is going through
zero the pulses duration are lower
54
00:04:43,700 --> 00:04:48,490
so you do certain things like you vary the
pulse widths width of pulses in certain fashion
55
00:04:48,490 --> 00:04:53,550
so that the the you get the desired fundamental
voltage but the harmonics are suppressed now
56
00:04:53,550 --> 00:04:57,690
so you i can one way i can say is you would
look at these widths there are pulses and
57
00:04:57,690 --> 00:05:01,180
they have certain widths and those widths
are modulated in certain fashion and that
58
00:05:01,180 --> 00:05:06,980
is why we call it as pwm modulating the widths
of these pulses in certain fashion such that
59
00:05:06,980 --> 00:05:11,230
you get the desired fundamental voltage and
you get you know the harmonics are reduced
60
00:05:11,230 --> 00:05:14,640
now
so all these signals that we see are essentially
61
00:05:14,640 --> 00:05:19,400
periodic signals all the signals we see are
essentially periodic signals so you we want
62
00:05:19,400 --> 00:05:23,210
to analyze we want to understand how much
is the fundamental voltage quantitatively
63
00:05:23,210 --> 00:05:26,850
and how much are the harmonics what are the
various harmonics how much are there how do
64
00:05:26,850 --> 00:05:32,240
we do that being periodic signals they can
be expanded into a fourier series they might
65
00:05:32,240 --> 00:05:36,550
have a dc component in general a periodic
waveform might have a dc component and might
66
00:05:36,550 --> 00:05:43,800
have a fundamental component zero omega might
have two omega three omega four omega etcetera
67
00:05:43,800 --> 00:05:49,790
where omega is the fundamental angular frequency
two omega three omega four omega correspond
68
00:05:49,790 --> 00:05:54,030
to the harmonics it goes on like this now
so in general it can have all these harmonic
69
00:05:54,030 --> 00:06:01,370
components but the waveforms we saw previously
ah these waveforms they have zero dc value
70
00:06:01,370 --> 00:06:06,590
they have no average value so the d the dc
is zero here so they only have a fundamental
71
00:06:06,590 --> 00:06:11,200
component and then they have the other harmonics
they are fundamental and they are the other
72
00:06:11,200 --> 00:06:16,720
harmonics now so what is the amplitude of
the fundamental i once again as i told you
73
00:06:16,720 --> 00:06:21,480
this amplitude of the fundamental is for such
a waveform it is going to be the peak value
74
00:06:21,480 --> 00:06:27,370
is going to be four v by pi
so this is the amplitude now you cannot vary
75
00:06:27,370 --> 00:06:31,380
this amplitude if you want to vary this amplitude
we should very v that means the dc bus voltage
76
00:06:31,380 --> 00:06:37,410
is being varied it is not the case that we
are considering we want you know v to be fixed
77
00:06:37,410 --> 00:06:41,889
so what we have to do here is as we will see
we have to introduce certain notches this
78
00:06:41,889 --> 00:06:47,139
is the maximum fundamental voltage that can
be obtained for certain four v by pi so what
79
00:06:47,139 --> 00:06:51,930
we can do is we have to introduce make certain
values zero or certain negative values
80
00:06:51,930 --> 00:07:03,880
now for example if a change is waveform such
that it is zero here and zero here this waveform
81
00:07:03,880 --> 00:07:10,060
has certain amplitude fundamental amplitude
this fundamental amplitude is lower than four
82
00:07:10,060 --> 00:07:14,390
v by pi this fundamental amplitude is lower
than what we had in the previous case that
83
00:07:14,390 --> 00:07:20,280
is four v by pi so how much lower it depends
on the notch how big is the notch now so if
84
00:07:20,280 --> 00:07:23,941
the notch is zero it will be four v by pi
as the notch goes on increasing it will come
85
00:07:23,941 --> 00:07:30,640
down now so you control your you know amplitude
of the fundamental ah doing like this now
86
00:07:30,640 --> 00:07:35,260
so this is another part you know how do you
want to control the amplitude of fundamental
87
00:07:35,260 --> 00:07:39,150
voltage now so there are various ways that
in which you can do that that there i had
88
00:07:39,150 --> 00:07:43,450
introduced a single notch now whereas you
can also introduce several notches like what
89
00:07:43,450 --> 00:07:53,210
we said we can introduce several notches like
this so this is an example which is very similar
90
00:07:53,210 --> 00:07:57,040
to the previous examples that we saw here
so what you can do is you know you introduce
91
00:07:57,040 --> 00:08:01,460
some notches few notches more number of notches
and more number of notches means the device
92
00:08:01,460 --> 00:08:03,960
is inside the inverters which is more number
of times
93
00:08:03,960 --> 00:08:08,480
so that places an upper limit so in the olden
days the device is switched at very low frequencies
94
00:08:08,480 --> 00:08:13,470
so in a fifty hertz cycle one could have very
few notches or so nowadays in modern power
95
00:08:13,470 --> 00:08:18,120
power electronic devices ah they can switch
at much higher frequency particularly the
96
00:08:18,120 --> 00:08:22,080
power levels are not too high i mean the power
levels are not like tens of megawatt or something
97
00:08:22,080 --> 00:08:27,270
like that you can certainly do that now so
what we do is basically we you know this widths
98
00:08:27,270 --> 00:08:31,960
of the pulse is basically determine what you
want to do so you come you control the amplitude
99
00:08:31,960 --> 00:08:36,289
of the fundamental voltage by doing this now
so this is about the dc value we are looking
100
00:08:36,289 --> 00:08:42,899
at waveforms which have no dc value these
are the ac side waveforms in a pwm inverter
101
00:08:42,899 --> 00:08:46,920
so they dont have an you know they dont have
dc value and the the fundamental voltage can
102
00:08:46,920 --> 00:08:52,680
be controlled as given here now and further
they have no even harmonics or they have half
103
00:08:52,680 --> 00:08:58,110
wave symmetry this is what we saw in the last
class now assuming now let us say you you
104
00:08:58,110 --> 00:09:03,190
want to check whether it has a second harmonic
what you do you multiply it by a second harmonic
105
00:09:03,190 --> 00:09:11,740
as shown here now right now take some arbitrary
instant say here and the same instant one
106
00:09:11,740 --> 00:09:23,800
eighty degree is later on the same instant
let me just change the color of ink right
107
00:09:23,800 --> 00:09:30,279
this is certain instant theta let me take
some other instant one eighty plus theta at
108
00:09:30,279 --> 00:09:37,180
one eighty plus theta the second harmonic
has the same value but the waveform in question
109
00:09:37,180 --> 00:09:43,970
has plus v here and it has minus v here therefore
if you take the product at theta and if you
110
00:09:43,970 --> 00:09:50,100
take the product of the two at one eighty
plus theta the sum is zero so the same is
111
00:09:50,100 --> 00:09:53,800
valid so when you integrate this over the
over entire like you know this is valid for
112
00:09:53,800 --> 00:09:57,830
any theta start from zero go all the way up
to one eighty degree it is zero so it does
113
00:09:57,830 --> 00:10:04,339
not have any even harmonic
now this is what we say half wave symmetry
114
00:10:04,339 --> 00:10:08,310
and this is the definition for waveform with
half wave symmetry if v is the voltage waveform
115
00:10:08,310 --> 00:10:14,820
and theta is a measure of time it is in terms
of the fundamental angle it is omega t where
116
00:10:14,820 --> 00:10:20,890
omega is equal to two pi f where if f is your
fundamental frequency two pi f is your fundamental
117
00:10:20,890 --> 00:10:27,120
angular frequency and theta what we mean here
is omega t so you take it at any theta if
118
00:10:27,120 --> 00:10:30,440
you take it as theta if you take it as one
eighty plus theta you see that at one eighty
119
00:10:30,440 --> 00:10:34,270
plus theta its value is the negative of whatever
it was at theta
120
00:10:34,270 --> 00:10:40,380
so such waveforms satisfy half wave symmetry
and they do not have even harmonics so of
121
00:10:40,380 --> 00:10:46,350
the various ah you know um ah harmonics or
frequency components we saw they have no dc
122
00:10:46,350 --> 00:10:51,070
they are fundamental they have no even harmonics
so the waveforms in question may have only
123
00:10:51,070 --> 00:10:57,500
odd harmonics the frequency components are
omega three omega five omega seven omega nine
124
00:10:57,500 --> 00:11:01,700
omega etcetera these are all the frequency
components that you will find here now so
125
00:11:01,700 --> 00:11:05,700
in addition to half wave symmetry you see
that these two halves are similar ah you know
126
00:11:05,700 --> 00:11:10,710
symmetric that is you take this half and you
invert it by one eighty degree you will get
127
00:11:10,710 --> 00:11:14,510
this half now
further to this if you look at about these
128
00:11:14,510 --> 00:11:19,710
points it is pi by two or ninety degrees you
move certain distance here in the same distance
129
00:11:19,710 --> 00:11:24,339
this side to the left or to the right you
find that the values are equal so this is
130
00:11:24,339 --> 00:11:29,580
what you call as quarter wave symmetry this
is again what we discussed last class now
131
00:11:29,580 --> 00:11:36,240
so its not exactly at ninety degrees so you
can consider any theta p theta p is an instant
132
00:11:36,240 --> 00:11:41,270
where the fundamental component of the waveform
v has a positive peak or a negative peak so
133
00:11:41,270 --> 00:11:44,720
it could be theta is equal to ninety degrees
or two hundred and seventy degrees going by
134
00:11:44,720 --> 00:11:49,740
our convention now
so about this so you can say that v of theta
135
00:11:49,740 --> 00:11:54,940
p plus theta is equal to v of theta p minus
theta this is what we have for quarter wave
136
00:11:54,940 --> 00:11:59,200
symmetry so now on top of half wave symmetry
if you have this quarter wave symmetry your
137
00:11:59,200 --> 00:12:04,440
calculations become simpler you can just use
one quarter of the waveform to do your calculations
138
00:12:04,440 --> 00:12:09,279
it also means that whenever your fundamental
has a zero crossing the fundamental component
139
00:12:09,279 --> 00:12:14,940
of the waveform has a zero crossing the harmonic
components also have their zero crossings
140
00:12:14,940 --> 00:12:20,420
so now ah up to this we have seen you know
its about fourier series and waveform symmetries
141
00:12:20,420 --> 00:12:25,220
and how you calculate the various harmonic
components so that you know we have we now
142
00:12:25,220 --> 00:12:29,610
have an ability to calculate given a pwm waveform
which is actually a periodic waveform we are
143
00:12:29,610 --> 00:12:34,330
able to calculate the fundamental and the
various harmonic voltages now now when harmonic
144
00:12:34,330 --> 00:12:40,800
voltages are fed to certain loads which we
normally come across you know how would the
145
00:12:40,800 --> 00:12:42,580
harmonic currents would be is what we are
looking at now
146
00:12:42,580 --> 00:12:46,830
so is the next thing let us say you apply
a harmonic voltage you know non sinusoidal
147
00:12:46,830 --> 00:12:51,760
voltage to in a induction motor this is what
happens in an inverter feeding an induction
148
00:12:51,760 --> 00:12:56,910
motor you dont get to apply pure sinusoidal
waveforms you apply waveforms which have sinusoids
149
00:12:56,910 --> 00:13:02,890
plus a lot of harmonics the fundamental component
of that of the inverter output voltage sees
150
00:13:02,890 --> 00:13:07,600
this induction motor as its fundamental equivalent
circuit this is the well known equivalent
151
00:13:07,600 --> 00:13:11,890
circuit that you would have studied in your
undergrad courses on basic ah electric machines
152
00:13:11,890 --> 00:13:15,930
or the basic electrical technology kind of
courses you would have studied this
153
00:13:15,930 --> 00:13:21,430
so the machine you know the fundamental component
is seen as something like this its basically
154
00:13:21,430 --> 00:13:26,520
it is seen as certain impedance it is seen
as certain impedance and this impedance is
155
00:13:26,520 --> 00:13:32,270
a function of the slip as i said at no load
the slip is zero as the motor gets loaded
156
00:13:32,270 --> 00:13:36,420
and loaded the slip increases and the rated
slip could be something small like about five
157
00:13:36,420 --> 00:13:41,000
percent or so it is a typical value so what
you have is you know you we have the zn as
158
00:13:41,000 --> 00:13:46,660
a function of s on the other hand instead
of fundamental if you considered the harmonics
159
00:13:46,660 --> 00:13:51,290
what happens is the harmonics see a much simpler
equivalent circuit which is only the leakage
160
00:13:51,290 --> 00:13:56,070
inductances because for the harmonics if you
look at this the slip is very high and therefore
161
00:13:56,070 --> 00:13:59,540
rr is very small therefore this is basically
a shot
162
00:13:59,540 --> 00:14:03,920
now when this leakage inductance of the rotor
comes in parallel with the magnetizing inductance
163
00:14:03,920 --> 00:14:08,870
this is small and therefore leakage inductance
is small therefore this dominates and the
164
00:14:08,870 --> 00:14:13,070
reactance is dominate over resistances at
high frequencies considering these factors
165
00:14:13,070 --> 00:14:17,350
the whole thing simplifies to this now if
you apply certain harmonic voltage then the
166
00:14:17,350 --> 00:14:21,900
harmonic voltage sees the motor as such an
equivalent circuit which is simply its total
167
00:14:21,900 --> 00:14:27,779
leakage inductance so if you apply as this
n th harmonic you can calculate the n th harmonic
168
00:14:27,779 --> 00:14:31,830
current using this now
this is these are the models for the fundamental
169
00:14:31,830 --> 00:14:37,830
and the harmonic ah components for a induction
motor now in inverter you can consider a grid
170
00:14:37,830 --> 00:14:42,339
connected converter that is an inverter is
connected to a grid through line inductors
171
00:14:42,339 --> 00:14:48,160
and this is the line inductor l now the invert
this is the grid voltage we can assume it
172
00:14:48,160 --> 00:14:52,490
to be sinusoidal though it is not always so
grid voltages are distorted and they could
173
00:14:52,490 --> 00:14:57,120
be unbalanced but here lets say for the time
being we assumes balanced and sinusoidal grid
174
00:14:57,120 --> 00:15:00,820
voltages now because vg is the grid voltage
which is sinusoidal
175
00:15:00,820 --> 00:15:05,270
since they are balanced we can consider aper
phase equivalent circuit so now what you do
176
00:15:05,270 --> 00:15:09,250
is this is the sinusoidal voltage now how
about the inverter terminal voltage the inverter
177
00:15:09,250 --> 00:15:15,960
terminal voltage has sinusoidal as a sinusoidal
component it has a fundamental component plus
178
00:15:15,960 --> 00:15:21,510
also a lot of harmonics so let me say the
sinusoidal component or the fundamental component
179
00:15:21,510 --> 00:15:26,890
there let me give this name v inverter and
all the harmonics added up constitute this
180
00:15:26,890 --> 00:15:30,399
v ripple
so this is sinusoidal and this is sum of several
181
00:15:30,399 --> 00:15:35,450
harmonics and it is non sinusoidal now this
is the per phase equivalent circuit for a
182
00:15:35,450 --> 00:15:39,960
grid connected converter now so if you are
interested in the fundamental current you
183
00:15:39,960 --> 00:15:44,130
want to see how much fundamental current is
drawn the ripple voltage produced by the inverter
184
00:15:44,130 --> 00:15:48,830
makes no sense here they are at various frequencies
so what you have is the grid voltage and this
185
00:15:48,830 --> 00:15:51,690
side what you have is the inverter output
fundamental voltage the fundamental voltage
186
00:15:51,690 --> 00:15:58,060
in the inverter line side and then the leakage
inductance comes in between this circuit helps
187
00:15:58,060 --> 00:16:04,180
you calculate the fundamental component
if you are interested to calculate the harmonic
188
00:16:04,180 --> 00:16:08,500
components now the grid voltage is a sinusoidal
voltage at fifty hertz or sixty hertz and
189
00:16:08,500 --> 00:16:13,860
whatever the line frequency is in that country
now that is not going to affect any n th harmonic
190
00:16:13,860 --> 00:16:18,089
component so if you want to calculate some
n th harmonic component how much it flows
191
00:16:18,089 --> 00:16:23,330
the grid voltage is zero there it has no effect
and the fundamental component of the inverter
192
00:16:23,330 --> 00:16:27,200
output voltage has no effect so it is just
a specific harmonic voltage produced by the
193
00:16:27,200 --> 00:16:32,790
inverter and the line inductance so this gives
you basically vn upon omega l is going to
194
00:16:32,790 --> 00:16:36,130
be the amplitude of the current that was flow
here right
195
00:16:36,130 --> 00:16:44,200
so this is how we can calculate harmonic currents
that is first we are able to calculate harmonic
196
00:16:44,200 --> 00:16:48,650
voltages or we are considering a voltage source
inverter which applies certain voltage waveforms
197
00:16:48,650 --> 00:16:53,460
these voltage waveforms are periodic waveforms
to the extent we have considered but ah they
198
00:16:53,460 --> 00:16:59,160
are not sinusoidal so we know how to calculate
the fundamental and harmonic components you
199
00:16:59,160 --> 00:17:03,930
know thats a fourier series helped us and
once the harmonic voltages are known the harmonic
200
00:17:03,930 --> 00:17:09,169
currents can be obtained through the models
we just discussed now so to repeat we have
201
00:17:09,169 --> 00:17:15,289
done two things one is given the fundamental
voltage i mean given a pwm waveform we know
202
00:17:15,289 --> 00:17:19,279
how to calculate the fundamental voltage and
harmonic voltage and given the fundamental
203
00:17:19,279 --> 00:17:23,089
voltage and harmonic voltage we know how to
calculate the fundamental current and harmonic
204
00:17:23,089 --> 00:17:25,799
current
we have considered two situations here one
205
00:17:25,799 --> 00:17:30,259
is the inverter feeding an induction motor
the other is the induct the you know it is
206
00:17:30,259 --> 00:17:35,940
connected through grid the line side is connected
through the grid through inductors so both
207
00:17:35,940 --> 00:17:39,999
cases have been one is a motor drive case
and the other one is a grid connected converter
208
00:17:39,999 --> 00:17:43,739
both the cases we have considered and we know
how to calculate the fundamental current and
209
00:17:43,739 --> 00:17:48,489
harmonic current based on the fundamental
voltage and harmonic voltages so with these
210
00:17:48,489 --> 00:17:53,309
two capabilities we now move on to the more
serious business of controlling the fundamental
211
00:17:53,309 --> 00:17:59,489
voltage first and to see if we can reduce
the harmonics to certain extent
212
00:17:59,489 --> 00:18:04,019
now this is the three phase inverter that
we we we have been looking at or we will be
213
00:18:04,019 --> 00:18:08,470
looking at for a few more lectures to come
now so as you can see that you know there
214
00:18:08,470 --> 00:18:15,159
is dc applied here and these are the load
terminals r y and b the three phase loads
215
00:18:15,159 --> 00:18:20,309
are connected here and as you can see that
every load terminal is basically the pole
216
00:18:20,309 --> 00:18:25,610
of the leg you have a every leg is a single
pole double throw switch and the load is connected
217
00:18:25,610 --> 00:18:29,499
to the pole and so we will commonly use the
term pole henceforth
218
00:18:29,499 --> 00:18:37,240
so we will define certain terminology certain
voltages here we will or you know we will
219
00:18:37,240 --> 00:18:40,580
identify certain nodes if you look at what
are the various nodes that you have in this
220
00:18:40,580 --> 00:18:46,340
circuit the nodes are r y and b which are
basically the load terminals or the midpoints
221
00:18:46,340 --> 00:18:53,169
of the three legs or the poles the other terminals
that we have are the dc terminals so these
222
00:18:53,169 --> 00:19:00,270
are the various nodes and you also have the
dc bus midpoint o so this dc bus midpoint
223
00:19:00,270 --> 00:19:04,480
o need not have to be a physical point i mean
available for making a connection it could
224
00:19:04,480 --> 00:19:09,950
just be a point you know which is for a ah
theoretical purposes we will use that as a
225
00:19:09,950 --> 00:19:14,619
reference point with respect to which we can
express the other voltages
226
00:19:14,619 --> 00:19:18,470
now let us say we want to say what is the
voltage at this terminal r phase terminal
227
00:19:18,470 --> 00:19:27,989
or this pole r so we call that as pole voltage
the voltage at pole are measured with respect
228
00:19:27,989 --> 00:19:36,700
to the dc bus neutral o this is what we would
call as pole voltage vro now this pole voltage
229
00:19:36,700 --> 00:19:44,049
if the pole is connected to the top throw
as indicated by the solid line here then the
230
00:19:44,049 --> 00:19:52,470
potential will be plus vdc by two the entire
dc bus voltage is vdc and so you have vdc
231
00:19:52,470 --> 00:19:59,780
by two here and vdc by two here so when the
pole is connected to the top throw vro is
232
00:19:59,780 --> 00:20:05,340
plus vdc by two and when the pole is connected
to the bottom throw shown through the dashed
233
00:20:05,340 --> 00:20:11,519
lines here you get minus vdc by two
so vro can be either plus vdc by two or minus
234
00:20:11,519 --> 00:20:19,729
vdc by two similarly you can define vyo this
is another pole midpoint of the y phase leg
235
00:20:19,729 --> 00:20:28,450
and vyo can also take the values of either
plus vdc by two or minus vdc by two similarly
236
00:20:28,450 --> 00:20:37,250
you have v bo voltage at the pole b measured
with respect to the dc bus neutral o this
237
00:20:37,250 --> 00:20:45,129
is also equal to plus or minus vdc by two
so these are pole voltages now what are line
238
00:20:45,129 --> 00:20:54,489
voltages is basically vry if you want to get
vry what you have is vry is essentially vro
239
00:20:54,489 --> 00:21:05,640
minus vyo similarly vyb is vyo minus vbo and
vbr is vbo minus vro you can you like that
240
00:21:05,640 --> 00:21:10,909
things similarly
now if r is connected to the top throw and
241
00:21:10,909 --> 00:21:20,129
y is connected to the bottom throw then you
have vry is equal to plus vdc so vri can take
242
00:21:20,129 --> 00:21:26,779
the value of plus vdc and vro is plus vdc
by two and vyo is minus vdc by two or this
243
00:21:26,779 --> 00:21:32,830
is connected to the bottom throw the other
way if r is connected to the bottom throw
244
00:21:32,830 --> 00:21:37,549
as indicated by the dashed lines here and
y is connected to the top throw as indicated
245
00:21:37,549 --> 00:21:46,690
by the dashed lines there then what you have
is minus vdc vry will be equal to minus vdc
246
00:21:46,690 --> 00:21:50,759
on the other hand you know these are only
two combinations that is r can be either connected
247
00:21:50,759 --> 00:21:54,950
to the top or bottom y can be either connected
to the top or bottom you should take the two
248
00:21:54,950 --> 00:21:59,529
phases together you have two multiplied by
two four possibilities are there these are
249
00:21:59,529 --> 00:22:03,350
only two possibilities
what are the two other possibilities yes r
250
00:22:03,350 --> 00:22:08,019
can be connected to the top throw and y can
also be connected to the top throw what would
251
00:22:08,019 --> 00:22:14,100
be ry then both are connected to the same
point so vry is going to be is zero not just
252
00:22:14,100 --> 00:22:20,309
that when r and y both can also be connected
to the bottom throw in that case also vry
253
00:22:20,309 --> 00:22:28,240
is going to be zero thus the line voltage
vry takes three different values plus vdc
254
00:22:28,240 --> 00:22:36,149
minus vdc or zero so these are the possible
value ah line voltages now
255
00:22:36,149 --> 00:22:40,210
you have a pole voltage if the pole is connected
to the top throw it is plus vdc by two if
256
00:22:40,210 --> 00:22:43,940
it is connected to the bottom throw it is
minus vdc by two there is nothing you dont
257
00:22:43,940 --> 00:22:48,070
have any redundancy there whereas if you start
seeing the line voltage you start finding
258
00:22:48,070 --> 00:22:53,649
one redundancy plus vdc there is no redundancy
r top has to be connected and y bottom has
259
00:22:53,649 --> 00:22:59,009
to be connected similarly for minus vdc but
when you want vry is equal to zero you are
260
00:22:59,009 --> 00:23:03,679
able to achieve it in two different ways one
is by connecting both the poles to the top
261
00:23:03,679 --> 00:23:09,059
throws are both the poles to the bottom throws
so this redundancy will be later on exploited
262
00:23:09,059 --> 00:23:15,470
as we will see later the similar definitions
are possible for vyb and vbr as i mentioned
263
00:23:15,470 --> 00:23:25,269
before now so these are the pole voltages
and the line to line voltages to be specific
264
00:23:25,269 --> 00:23:30,570
now so all of them will take values the line
voltages will take either plus vdc minus vdc
265
00:23:30,570 --> 00:23:36,289
are zero as the possible values now then you
can have further voltages defined here that
266
00:23:36,289 --> 00:23:40,869
is let us say you assume there is a three
phase load let us say what we have is a three
267
00:23:40,869 --> 00:23:47,009
phase load and the three phase load is balanced
and star connected load we let us assume a
268
00:23:47,009 --> 00:23:58,730
balanced star connected load in that case
its possible for us to express vrn vrn as
269
00:23:58,730 --> 00:24:09,610
one by three times of v r y minus vbr
the same way vyn and vbn can also be expressed
270
00:24:09,610 --> 00:24:21,649
now so we we have different things here so
one of them is let me just we have the voltages
271
00:24:21,649 --> 00:24:27,390
vro vyo and vbo what are these these are the
pole voltages these are the voltages at the
272
00:24:27,390 --> 00:24:34,299
poles measured with respect to the dc bus
neutral and we have these voltages vry vyb
273
00:24:34,299 --> 00:24:42,090
and vbr what are these voltages these are
the line to line voltages these are the line
274
00:24:42,090 --> 00:24:49,100
to line voltages so if you ah you consider
a balanced star connected load we will further
275
00:24:49,100 --> 00:24:56,889
have vrn vyn and vbn and how will those be
vrn is as given by this equation one third
276
00:24:56,889 --> 00:25:02,020
of vry minus vbr
similarly if you want vyn that is going to
277
00:25:02,020 --> 00:25:14,309
be one third of v yb minus v ry similarly
you will have vbn also so these are the line
278
00:25:14,309 --> 00:25:21,220
to neutral voltages applied on the load in
case of balanced star connected load now so
279
00:25:21,220 --> 00:25:24,830
these are the various voltages we just needed
to define them and we have been able to define
280
00:25:24,830 --> 00:25:28,679
the pole voltages line to line voltages and
line to neutral voltages applied on the load
281
00:25:28,679 --> 00:25:35,299
now with this let us move on to see how we
can control the fundamental voltage now
282
00:25:35,299 --> 00:25:49,559
firstly if you just take a square wave lets
take a square wave if your inverter is is
283
00:25:49,559 --> 00:25:57,580
switched like a square wave let me take the
horizontal axis let me draw it with black
284
00:25:57,580 --> 00:26:08,380
ink this is the time axis now this is the
voltage axis so this is equal to point five
285
00:26:08,380 --> 00:26:16,149
v or point five vdc and the lower level is
equal to minus point five v if i switch the
286
00:26:16,149 --> 00:26:23,570
inverter in a square wave fashion what i get
is you know i get of the pole voltage to be
287
00:26:23,570 --> 00:26:29,269
point five vdc for half the time and minus
point five vdc for the remaining half the
288
00:26:29,269 --> 00:26:35,260
time and this goes on like that
so i have shown this in red color for the
289
00:26:35,260 --> 00:26:45,509
red phase if this is my vro how will my vyo
be this is vro red phase vro how will my vyo
290
00:26:45,509 --> 00:26:52,159
be that is the y phase let me try and choose
the yellow color here ok so vyo has to be
291
00:26:52,159 --> 00:26:56,970
exactly phase shifted by hundred and twenty
degrees now so this is one eighty degree and
292
00:26:56,970 --> 00:27:02,249
hundred and twenty degree is here vyo has
to be phase shifted the yellow color might
293
00:27:02,249 --> 00:27:06,649
not be very clearly visible but i i mean i
do hope that you are able to see this waveform
294
00:27:06,649 --> 00:27:11,940
that i am drawing now right
so vyo is going to be phase shifted by one
295
00:27:11,940 --> 00:27:21,419
hundred and twenty degrees it goes on like
this now how will vbo be let me choose blue
296
00:27:21,419 --> 00:27:27,679
color here that will be phase shifted by further
hundred and twenty degrees now so if r had
297
00:27:27,679 --> 00:27:34,509
a zero crossing at zero degree y had a zero
crossing at one twenty degree y b will have
298
00:27:34,509 --> 00:27:38,539
its zero crossing i mean or rather what i
mean by zero crossing is switching from low
299
00:27:38,539 --> 00:27:43,809
to high r switches from low to high at zero
y switches from low to high at one twenty
300
00:27:43,809 --> 00:27:47,690
so b will be switching from low to high at
two hundred and ten degree two hundred and
301
00:27:47,690 --> 00:27:53,109
forty degrees
so this is how your vbo will be so if it is
302
00:27:53,109 --> 00:27:58,259
switching low to high at two hundred and forty
degrees it will be switching high to low at
303
00:27:58,259 --> 00:28:06,320
sixty degrees now so this is how the three
phase voltages will be what i have drawn now
304
00:28:06,320 --> 00:28:17,249
with blue ink is vbo blue ink is vbo and the
yellow ink is vyo so these are the three phase
305
00:28:17,249 --> 00:28:23,099
pole voltages when you operate with a square
wave mode and what is the fundamental component
306
00:28:23,099 --> 00:28:27,980
it is the same for all the three and how much
is that the fundamental component is simply
307
00:28:27,980 --> 00:28:34,269
equal to two vdc upon pi as we saw earlier
we cannot control this fundamental voltage
308
00:28:34,269 --> 00:28:39,679
so to start controlling the fundamental voltage
what we do is we are going to introduce one
309
00:28:39,679 --> 00:28:46,100
switching angle in a quarter cycle see we
want to have waveforms with you know further
310
00:28:46,100 --> 00:28:50,559
to three phase symmetry we want to have a
half wave symmetry and we also want to have
311
00:28:50,559 --> 00:28:54,700
quarter wave symmetry now if you want to have
a quarter wave symmetry there should be equal
312
00:28:54,700 --> 00:28:59,690
number of switching angles in all the quarters
so we will go for first is one switching angle
313
00:28:59,690 --> 00:29:03,659
in a quarter we will consider a quarter cycle
be quarter cycle if you look at one switching
314
00:29:03,659 --> 00:29:07,899
angle per quarter
so what is that one switching angle per quarter
315
00:29:07,899 --> 00:29:15,809
now this entire waveform i am going to represent
in a simpler fashion now let me just see here
316
00:29:15,809 --> 00:29:23,889
thats ok so here what i am going to do is
this is one switching angle per quarter first
317
00:29:23,889 --> 00:29:29,820
let us look at the square wave this vro i
what i can do is let me change back to the
318
00:29:29,820 --> 00:29:36,659
black color and indicate the voltage in the
time axis let me indicate the voltage and
319
00:29:36,659 --> 00:29:44,169
the time axis along this now so instead of
drawing the entire waveform i can only draw
320
00:29:44,169 --> 00:29:51,779
one quarter of this it is negative just before
zero and at zero it switches from low to high
321
00:29:51,779 --> 00:29:59,029
and then it is high up to ninety degrees it
is high up to ninety degrees if i draw this
322
00:29:59,029 --> 00:30:03,419
much it is enough because the waveform has
quarter wave symmetry
323
00:30:03,419 --> 00:30:09,289
so what does it mean it means i can just simply
extend this if this part of the waveform is
324
00:30:09,289 --> 00:30:15,309
given to me i can just reflect it about the
ninety degrees and i can get the next quarter
325
00:30:15,309 --> 00:30:20,360
and now i have one half if i have one half
then through using half wave symmetry i can
326
00:30:20,360 --> 00:30:27,570
construct the other half ok so what is given
what i have drawn in solid red line is all
327
00:30:27,570 --> 00:30:33,739
that you need to give and once that waveform
is given for one quarter cycle we know what
328
00:30:33,739 --> 00:30:37,990
it is for the another quarter and the rest
of the half cycle
329
00:30:37,990 --> 00:30:44,539
so what we will henceforth do is only to use
one particular quarter now so lets focus only
330
00:30:44,539 --> 00:30:49,710
on ninety degrees so the first ninety degrees
now so what we want to do is we are looking
331
00:30:49,710 --> 00:30:55,749
at a waveform with just one switching angle
per quarter that is when we want to control
332
00:30:55,749 --> 00:31:02,249
the fundamental voltage lets say vro is negative
and switches from negative to positive at
333
00:31:02,249 --> 00:31:09,139
zero degrees like what it happened did here
now then before ninety degrees what it does
334
00:31:09,139 --> 00:31:14,219
is it switches down from high to low and stays
low
335
00:31:14,219 --> 00:31:21,559
so there is one switching angle here and lets
call this angle as alpha so what do you have
336
00:31:21,559 --> 00:31:25,349
i am representing only one quarter i hope
that you can see how it is going to appear
337
00:31:25,349 --> 00:31:29,629
in the next quarter for example if i want
to do that all that i need to do is i need
338
00:31:29,629 --> 00:31:34,899
to use you know mirror symmetry across about
ninety degrees so i can just do it like this
339
00:31:34,899 --> 00:31:39,279
so this is how the waveform will look like
about ninety i can go on till one eighty and
340
00:31:39,279 --> 00:31:42,460
i can use half wave symmetry to constrain
the other half now
341
00:31:42,460 --> 00:31:46,440
so in one switching we have only you know
in one quarter we have only one switching
342
00:31:46,440 --> 00:31:51,559
angle so the wave form will switch from zero
to you know at zero degrees and it will also
343
00:31:51,559 --> 00:31:57,399
switch at one eighty degrees it will also
switch back at one eighty degree from high
344
00:31:57,399 --> 00:32:02,389
to low other than these two there will be
one switching in a quarter so this waveform
345
00:32:02,389 --> 00:32:09,289
is an example of one switching angle per quarter
now so this alpha depending on what the the
346
00:32:09,289 --> 00:32:12,839
value of alpha is you can control the fundamental
voltage now
347
00:32:12,839 --> 00:32:17,879
so what will be the fundamental voltage if
you use fourier series you can say if you
348
00:32:17,879 --> 00:32:23,620
this is v one and you all that you need to
do is you need to integrate this waveform
349
00:32:23,620 --> 00:32:28,620
as we were doing we are considering only up
to pi by ah two and the waveform value is
350
00:32:28,620 --> 00:32:38,889
vdc by pi so what we have is two vdc by pi
and you need to integrate this sin omega t
351
00:32:38,889 --> 00:32:48,269
starting from zero to alpha and minus sin
omega t from this is sin omega t you integrate
352
00:32:48,269 --> 00:32:59,219
this from zero to alpha and again minus sin
omega t from ninety to sorry from alpha to
353
00:32:59,219 --> 00:33:04,119
ninety this is what essentially you are going
to have
354
00:33:04,119 --> 00:33:21,889
so what is this going to be its going to be
two vdc by pi times one minus two cos alpha
355
00:33:21,889 --> 00:33:26,820
if i integrate sin omega t the integral of
sin omega t is cos omega t and i am trying
356
00:33:26,820 --> 00:33:31,759
to put its limits over zero to alpha so what
i will get here is one minus cos alpha if
357
00:33:31,759 --> 00:33:36,889
i consider the second term its again you know
it is ah cos omega t so what i will get here
358
00:33:36,889 --> 00:33:42,440
is cos alpha so cos ninety is anyway zero
so what will happen is this whole thing becomes
359
00:33:42,440 --> 00:33:47,289
one minus two cos alpha so this is what we
have and if you consider the n th harmonic
360
00:33:47,289 --> 00:33:53,119
instead of the fundamental this is going to
give you something like two vdc upon n pi
361
00:33:53,119 --> 00:34:01,649
times one minus two cos n alpha this is just
applying fourier series there is a waveform
362
00:34:01,649 --> 00:34:05,659
has quarter wave symmetry we need to consider
only one quarter of the waveform and we are
363
00:34:05,659 --> 00:34:10,909
able to calculate v one and vn or any n th
harmonic as given like this now
364
00:34:10,909 --> 00:34:16,840
so what do you get here you need it you need
certain fundamental voltage let us say i want
365
00:34:16,840 --> 00:34:21,589
v one is equal to two vdc pi that is the maximum
which is possible with square wave let me
366
00:34:21,589 --> 00:34:28,330
change the ink color to black so i want two
vdc by pi so if i want two vdc by pi what
367
00:34:28,330 --> 00:34:33,160
do i have to do now this is alpha and this
alpha i have to make it as ninety degrees
368
00:34:33,160 --> 00:34:37,291
if i make this alpha as ninety degrees then
this essentially becomes a square wave what
369
00:34:37,291 --> 00:34:41,310
i want you can see that one minus two cos
alpha is simply equal to one
370
00:34:41,310 --> 00:34:46,800
so at alpha is equal to ninety degrees it
gives me a square waveform and as alpha is
371
00:34:46,800 --> 00:34:50,730
reduced from ninety degrees its lets say is
made eighty nine degrees you get some amount
372
00:34:50,730 --> 00:34:54,649
of v one so as alpha is reduced further and
further from ninety degrees your v one goes
373
00:34:54,649 --> 00:35:02,510
on increasing till what when your alpha is
sixty degrees two cos alpha is one and one
374
00:35:02,510 --> 00:35:09,020
minus two cos alpha is zero so alpha can be
varied over the range ninety degrees to sixty
375
00:35:09,020 --> 00:35:17,400
degree and your fundamental voltage correspondingly
will go from one per unit to zero what do
376
00:35:17,400 --> 00:35:23,480
you mean by one per unit one per unit is the
fundamental voltage that you get with square
377
00:35:23,480 --> 00:35:29,040
wave operation this two vdc upon pi thus is
what i take as the base and this is one per
378
00:35:29,040 --> 00:35:32,770
unit now
so here you have a capability to vary the
379
00:35:32,770 --> 00:35:37,340
fundamental voltage what we have is a capability
to vary the fundamental voltage we choose
380
00:35:37,340 --> 00:35:42,820
this there is only one control variable namely
alpha we choose this alpha to give the desired
381
00:35:42,820 --> 00:35:46,550
value of fundamental the fundamental may be
a hundred percent one per unit or it may be
382
00:35:46,550 --> 00:35:50,380
point eight per unit it may be point five
per unit whatever it may be we choose a particular
383
00:35:50,380 --> 00:35:54,860
value of alpha which will give us the desired
fundamental voltage once we have chosen this
384
00:35:54,860 --> 00:36:00,780
alpha we need to plug in the same alpha here
and that will give us the harmonics that are
385
00:36:00,780 --> 00:36:04,080
being generated
so we have no control over them we got to
386
00:36:04,080 --> 00:36:13,070
accept whatever harmonic voltages are for
this value of alpha and you see with alpha
387
00:36:13,070 --> 00:36:18,420
this variation is co sinusoidal i mean its
its transcendental function but here alpha
388
00:36:18,420 --> 00:36:23,670
varies only from ninety to sixty degree if
you look at for example fifth harmonic again
389
00:36:23,670 --> 00:36:31,330
for this variation of alpha between ninety
to sixty degree five alpha will vary between
390
00:36:31,330 --> 00:36:36,160
four hundred and fifty degrees to three hundred
degrees so that is a larger amount of variation
391
00:36:36,160 --> 00:36:41,290
so for a small variation in alpha n alpha
will have a substantially larger variation
392
00:36:41,290 --> 00:36:46,440
here so what you will see here in in the fundamental
will be a small cycle you know but whereas
393
00:36:46,440 --> 00:36:52,360
here you will see you know larger cycles you
will be able to see certain cycles of a sinusoidal
394
00:36:52,360 --> 00:36:57,410
co sinusoidal functions here coming up here
so vn etcetera will change much more drastically
395
00:36:57,410 --> 00:37:01,410
than v one would change thats what i would
like to tell you if alpha is changed by one
396
00:37:01,410 --> 00:37:06,540
degree v one will change to certain extent
but if you take the fifth harmonic it is cos
397
00:37:06,540 --> 00:37:12,150
five alpha five alpha changes to by five degrees
if alpha changes by one degree five alpha
398
00:37:12,150 --> 00:37:16,660
changes by five degrees therefore the change
in v five will be much more now
399
00:37:16,660 --> 00:37:20,090
so all this it is possible for you to plot
these are very fairly simple functions for
400
00:37:20,090 --> 00:37:24,520
you to plot and you can try plotting them
now what i am going to see is i am going to
401
00:37:24,520 --> 00:37:29,960
look at the next two thing of looking at an
alternative to get this one switching angle
402
00:37:29,960 --> 00:37:36,510
so what we do in the earlier case was a situation
like this with we i am sorry let me just change
403
00:37:36,510 --> 00:37:42,950
the color to red here so the waveform was
switching at zero degree and it is switching
404
00:37:42,950 --> 00:37:51,490
back at an angle alpha and this is ninety
degrees on the other hand one could also consider
405
00:37:51,490 --> 00:38:01,250
a different kind of a waveform namely something
that switches from high to low and low to
406
00:38:01,250 --> 00:38:07,950
high and still about ninety degrees
first let me indicate the horizontal and the
407
00:38:07,950 --> 00:38:13,860
time axis now so this is the voltage axis
this is the time axis now what i can do is
408
00:38:13,860 --> 00:38:24,170
here i i what i have is vro is positive and
it is switching from high to low and it is
409
00:38:24,170 --> 00:38:29,380
remaining low and then it is going back to
high let me call this as beta and this is
410
00:38:29,380 --> 00:38:35,470
ninety degrees right so in this kind of a
waveform if i were to projected how will it
411
00:38:35,470 --> 00:38:43,750
look in the second quarter i can simply project
it this is just mirror reflection so at one
412
00:38:43,750 --> 00:38:50,080
eighty minus beta it will switch back and
at one eighty degrees it will go higher and
413
00:38:50,080 --> 00:38:59,680
beyond this is and beyond this is you you
can go through half wave symmetry
414
00:38:59,680 --> 00:39:03,850
so this is another kind of waveform you can
see that here if then switches from zero and
415
00:39:03,850 --> 00:39:09,830
stays high till alpha and from alpha to ninety
it is negative whereas in the other case the
416
00:39:09,830 --> 00:39:15,000
waveform switches from high to low at zero
degrees and it remains low for a certain angle
417
00:39:15,000 --> 00:39:20,320
that we call as beta here and then it is up
to ninety degrees now so for this waveform
418
00:39:20,320 --> 00:39:25,400
if you have to write what is the fundamental
component you can see that it is v one is
419
00:39:25,400 --> 00:39:32,260
two vdc upon pi now what you need to do is
you have to integrate this is minus the first
420
00:39:32,260 --> 00:39:39,190
term is negative of sin omega t and this runs
from zero to beta
421
00:39:39,190 --> 00:39:47,260
and the next thing term is you will have plus
two vdc upon pi you will have to integrate
422
00:39:47,260 --> 00:39:52,780
it sin omega t i mean there is a d omega t
here please note that it even if i have missed
423
00:39:52,780 --> 00:39:58,870
out you can take it to that beta to ninety
degrees now so what you have here is basically
424
00:39:58,870 --> 00:40:06,900
two vdc upon pi and this is integral of sin
omega t is minus cos omega t you have a again
425
00:40:06,900 --> 00:40:13,810
a minus ah sign here so this is cos omega
t limits from zero to beta and this is another
426
00:40:13,810 --> 00:40:21,780
you have minus cos omega t evaluated from
the limits beta to ninety degrees now is it
427
00:40:21,780 --> 00:40:26,150
right
so what i am going to get is two vdc upon
428
00:40:26,150 --> 00:40:35,550
pi so this is cos beta minus one and what
you have is another cos beta coming up here
429
00:40:35,550 --> 00:40:43,140
this is two cos beta minus one this is what
i am going to get for such a function so let
430
00:40:43,140 --> 00:40:48,240
us just check if beta is equal to zero so
this waveform becomes a square wave beta is
431
00:40:48,240 --> 00:40:52,540
smaller and smaller becomes close to zero
so there is no switching at all this becomes
432
00:40:52,540 --> 00:40:57,290
square wave and when beta is equal to one
here this is two cos beta is two so two cos
433
00:40:57,290 --> 00:41:02,780
beta minus one is one you get square wave
when beta goes on increasing lets say it goes
434
00:41:02,780 --> 00:41:10,920
to sixty degrees so two cos beta is one one
minus one is zero so beta now can vary from
435
00:41:10,920 --> 00:41:19,560
ah zero to sixty degrees if beta is varied
from zero to sixty degrees you can get a corresponding
436
00:41:19,560 --> 00:41:26,620
variation in v one the fundamental voltage
from one per unit to zero again one per unit
437
00:41:26,620 --> 00:41:31,170
has the meaning of two vdc by pi whatever
it is equal to the square wave now the same
438
00:41:31,170 --> 00:41:35,750
thing if you want the n th harmonic this becomes
you know instead of two vdc by pi there is
439
00:41:35,750 --> 00:41:40,770
an n pi term and you have n omega t you are
going to integrate it with respect to n omega
440
00:41:40,770 --> 00:41:43,160
t
so you will have an additional n coming up
441
00:41:43,160 --> 00:41:51,940
here for any ah vn you may have two vdc upon
n pi and you may have another cos n beta coming
442
00:41:51,940 --> 00:41:56,870
up here minus one this is what would be your
corresponding voltages now so these are two
443
00:41:56,870 --> 00:42:01,840
different ways of controlling ah what we will
do is by enlarge you will use the first option
444
00:42:01,840 --> 00:42:07,450
here so we look at these things a little later
when you are doing selective harmonic elimination
445
00:42:07,450 --> 00:42:10,910
we will use both types for certain number
of switching angles now
446
00:42:10,910 --> 00:42:15,460
so for the time being this is what we are
going to consider now and lets move on from
447
00:42:15,460 --> 00:42:21,780
here to the problem of let us say two two
switching angles per quarter the difficulty
448
00:42:21,780 --> 00:42:27,250
with one switching angle per quarter is you
cannot control the harmonics you can control
449
00:42:27,250 --> 00:42:33,030
your fundamental voltage but you cannot control
your any harmonic so that is why you require
450
00:42:33,030 --> 00:42:37,760
a minimum of two switching angles now so here
again what i said is this is the complete
451
00:42:37,760 --> 00:42:43,480
pole voltage waveform i have just plotted
it ready electronically so that you know its
452
00:42:43,480 --> 00:42:48,080
its ready for us to see
so you have alpha one and alpha two the the
453
00:42:48,080 --> 00:42:51,720
waveform is initially negative and it is switching
here which is not very evident so this is
454
00:42:51,720 --> 00:42:56,480
switching here then it is switching back at
alpha one from high to low and low to high
455
00:42:56,480 --> 00:43:02,660
at alpha two and this is ninety degrees so
then ninety degrees you have there is a switching
456
00:43:02,660 --> 00:43:06,770
this is pi minus alpha two this is pi minus
alpha one and then you have pi at pi it is
457
00:43:06,770 --> 00:43:11,990
switching back here this is pi this is one
eighty degrees so you have with the half wave
458
00:43:11,990 --> 00:43:15,850
symmetry because the waveform has half wave
symmetry it is enough to consider only one
459
00:43:15,850 --> 00:43:20,280
half of the waveform because the waveform
is quarter wave symmetry it is enough to consider
460
00:43:20,280 --> 00:43:24,910
just one quarter of the waveform
now this waveform has a certain fundamental
461
00:43:24,910 --> 00:43:31,770
component which is like this this has some
fundamental component like this this fundamental
462
00:43:31,770 --> 00:43:38,810
component as you can see is controlled by
certain value of alpha one and alpha two now
463
00:43:38,810 --> 00:43:43,900
lets just ask one question ah ourselves and
you know look at it little qualitatively when
464
00:43:43,900 --> 00:43:48,610
alpha one and alpha two are very close to
one another we will have a fundamental whose
465
00:43:48,610 --> 00:43:54,920
goes like this what happens when alpha one
and alpha two are much farther from one another
466
00:43:54,920 --> 00:44:03,170
what happens when alpha one and alpha two
are much farther from one another so in that
467
00:44:03,170 --> 00:44:06,170
case what may happen is the waveform will
look more negative
468
00:44:06,170 --> 00:44:11,800
now it is clear that it is we are in the positive
half cycle we are looking at one quarter of
469
00:44:11,800 --> 00:44:16,310
the positive half cycle of the voltage waveform
if the gap between alpha one and alpha two
470
00:44:16,310 --> 00:44:20,900
is very large this will look like a quarter
in the negative half cycle and indeed you
471
00:44:20,900 --> 00:44:26,490
may have your voltage waveform like this if
you have you know the alpha one it depends
472
00:44:26,490 --> 00:44:31,470
on the relative values of alpha one and alpha
two so let me just clear this up i just wanted
473
00:44:31,470 --> 00:44:39,060
you to be aware of this in this kind of a
scenario we are going to calculate what is
474
00:44:39,060 --> 00:44:44,950
the fundamental voltage and what are the harmonic
voltages now
475
00:44:44,950 --> 00:44:51,960
so we have the same waveform reproduced here
for just one quarter and what is the fundamental
476
00:44:51,960 --> 00:45:03,120
voltage we can say that v one is going to
be one minus two cos alpha one plus two cos
477
00:45:03,120 --> 00:45:13,170
alpha two in per unit terms in per unit terms
this is what it is and the n th harmonic component
478
00:45:13,170 --> 00:45:26,810
vn is going to be one upon n times one minus
two cos n alpha one plus two cos n alpha two
479
00:45:26,810 --> 00:45:33,960
in per unit terms again one per unit means
two vdc upon pi that is the fundamental voltage
480
00:45:33,960 --> 00:45:37,250
obtained with square wave operation is it
right ok
481
00:45:37,250 --> 00:45:42,110
so now we have these are this we basically
you know coming from fourier series we are
482
00:45:42,110 --> 00:45:46,630
able to evaluate here now and for different
values of alpha one and alpha two you will
483
00:45:46,630 --> 00:45:50,830
get different values of fundamental voltage
conversely for the same fundamental voltage
484
00:45:50,830 --> 00:45:57,160
let me say v one is equal to point eight if
i say v one is equal to point eight there
485
00:45:57,160 --> 00:46:06,830
can be several values of several values of
alpha one comma alpha two if i specify a particular
486
00:46:06,830 --> 00:46:12,230
value of alpha one and alpha two let us say
twenty degrees and forty degrees or sixty
487
00:46:12,230 --> 00:46:18,720
degrees and eighty degrees this will be giving
me a specific v one will give me a specific
488
00:46:18,720 --> 00:46:22,270
v one
on the other hand if i have a specific fundamental
489
00:46:22,270 --> 00:46:27,950
voltage such as v one is equal to point eight
i can realize this not using just one pair
490
00:46:27,950 --> 00:46:33,170
of alpha one and alpha two but i can use it
during several pairs of alpha one and alpha
491
00:46:33,170 --> 00:46:37,740
two so we need to understand what are all
the various possibilities and we need to from
492
00:46:37,740 --> 00:46:42,680
there move on to select which possibility
is better for us so the this gives us multiple
493
00:46:42,680 --> 00:46:49,460
possibilities we will choose one of them which
is better in terms of harmonics that is how
494
00:46:49,460 --> 00:46:54,460
we start exercising some control over harmonics
are their side effects now
495
00:46:54,460 --> 00:47:00,530
so as i have already said the fundamental
and the harmonic voltages we have them as
496
00:47:00,530 --> 00:47:11,690
v one is equal to one minus two cos alpha
one plus two cos alpha two in per unit terms
497
00:47:11,690 --> 00:47:16,210
so you can just plot this you know by varying
alpha one varying alpha two and there are
498
00:47:16,210 --> 00:47:25,640
many many many that you can plot here and
v n is one by n times i am just repeating
499
00:47:25,640 --> 00:47:41,240
it but i think it is all right because i am
sorry
500
00:47:41,240 --> 00:47:52,960
this is this is also in per unit terms now
so as i mentioned before for some value of
501
00:47:52,960 --> 00:47:57,220
alpha one and alpha two you will get v one
if you change those either alpha one slightly
502
00:47:57,220 --> 00:48:02,180
or alpha two slightly you you the corresponding
change in the argument here in the second
503
00:48:02,180 --> 00:48:07,910
one is n alpha one is going to be substantially
higher let us focus on a specific value that
504
00:48:07,910 --> 00:48:15,380
is n is equal to five let us focus on n is
equal to five so let me take this vn of and
505
00:48:15,380 --> 00:48:22,410
let us focus on n is equal to five so i will
say v five is equal to one by five times cos
506
00:48:22,410 --> 00:48:26,650
five alpha one cos five alpha two now why
is it i am focusing on fifth
507
00:48:26,650 --> 00:48:34,040
the waveform as such has a few components
what are the components the waveform has the
508
00:48:34,040 --> 00:48:40,190
order of harmonics are it has fundamental
it has third it has fifth it has seventh it
509
00:48:40,190 --> 00:48:47,190
has ninth eleventh thirteenth fifteenth and
so on it is the waveform on account of its
510
00:48:47,190 --> 00:48:52,380
half wave symmetry has only odd harmonics
it has the fundamental and all the odd harmonics
511
00:48:52,380 --> 00:48:57,200
now what i am going to do is we are going
to apply it to a three phase load we are going
512
00:48:57,200 --> 00:49:05,690
to have your vry is equal to vro minus vyo
recall the definition of the line voltage
513
00:49:05,690 --> 00:49:12,080
now if vro is what we are looking at if vro
has a third harmonic component vyo will also
514
00:49:12,080 --> 00:49:16,780
have the same third harmonic component and
vro and vyo are phase shifted by one twenty
515
00:49:16,780 --> 00:49:21,160
degrees and hundred and twenty degrees for
the fundamental frequency is equal to three
516
00:49:21,160 --> 00:49:25,510
hundred and sixty degrees for the third harmonic
so the third harmonic will be the same third
517
00:49:25,510 --> 00:49:29,920
harmonic will be available at the the same
phase in both the waveforms and when you subtract
518
00:49:29,920 --> 00:49:33,270
vro minus vyo the third harmonics will get
cancelled
519
00:49:33,270 --> 00:49:40,990
so what will happen is the third harmonics
will all go off the third harmonics will all
520
00:49:40,990 --> 00:49:47,320
go off if you look at vry this is in vro you
will have all the harmonics if you look at
521
00:49:47,320 --> 00:49:59,960
vry vry will have only n is equal to one five
seven eleven thirteen then seventeen nineteen
522
00:49:59,960 --> 00:50:06,450
etcetera so you will see pairs of harmonics
this is around six this is around twelve this
523
00:50:06,450 --> 00:50:10,770
is around eighteen and so on so these are
the kind of harmonic voltages which will be
524
00:50:10,770 --> 00:50:15,150
eventually seen into the line to line voltage
so this is these are the ones that get applied
525
00:50:15,150 --> 00:50:22,680
to the motor or the load the triple n harmonics
in vro do not matter they will get cancelled
526
00:50:22,680 --> 00:50:28,200
out so this is why i focus on v five residual
on this is equal to one by five times one
527
00:50:28,200 --> 00:50:33,470
minus two cos five alpha one plus two cos
five alpha two so you can start saying that
528
00:50:33,470 --> 00:50:38,820
you have there are two equations and we have
two unknowns and this is what has led to the
529
00:50:38,820 --> 00:50:43,520
famous problem of selective harmonic elimination
we will deal with selective harmonic elimination
530
00:50:43,520 --> 00:50:46,790
in greater detail in the coming few lectures
now
531
00:50:46,790 --> 00:50:50,400
this is just an introduction so if you want
to say that you can choose your alpha one
532
00:50:50,400 --> 00:50:56,660
and alpha two such that is equal to lets say
some point eight per unit lets say point eight
533
00:50:56,660 --> 00:51:01,330
per unit is what we want its possible for
us to choose not just one value several values
534
00:51:01,330 --> 00:51:05,270
are possible of alpha one and alpha two which
will give point eight per unit at the same
535
00:51:05,270 --> 00:51:10,540
time you can also make sure that v five is
equal to zero so what you have is the you
536
00:51:10,540 --> 00:51:14,280
have two equations and two unknowns alpha
one and alpha two you can solve these equations
537
00:51:14,280 --> 00:51:17,600
of course these are transcendental equations
you may not be able to solve them straight
538
00:51:17,600 --> 00:51:24,190
away it might be better to use you know numerical
iterative procedures to solve these but you
539
00:51:24,190 --> 00:51:26,760
can solve them
its possible for you to come up with some
540
00:51:26,760 --> 00:51:31,701
value of alpha one and alpha two which will
give you a fundamental voltage of point eight
541
00:51:31,701 --> 00:51:38,160
per unit and a fifth harmonic voltage is zero
why it may also give you not just one value
542
00:51:38,160 --> 00:51:43,120
of alpha one and alpha two it may give you
maybe two or even you know possibly three
543
00:51:43,120 --> 00:51:47,880
i dont know but maybe even two values of alpha
one and alpha two might be possible or lets
544
00:51:47,880 --> 00:51:50,790
say for some value of fundamental voltage
it might be possible
545
00:51:50,790 --> 00:51:55,120
so these are the things which we will explore
later when we go on to the specific problem
546
00:51:55,120 --> 00:51:59,360
of selective harmonic elimination this is
a harmonic elimination because we are making
547
00:51:59,360 --> 00:52:03,880
fifth harmonic to be zero so it is harmonic
elimination we are making a specific harmonic
548
00:52:03,880 --> 00:52:08,750
to go to zero so we would call it as selective
harmonic elimination we will deal it with
549
00:52:08,750 --> 00:52:14,390
a little later before that let us try to understand
the various possibilities the same fundamental
550
00:52:14,390 --> 00:52:20,400
voltage can be obtained through various pairs
of alpha one and alpha two thats what we have
551
00:52:20,400 --> 00:52:23,790
now
so here you know it is better to see there
552
00:52:23,790 --> 00:52:28,480
is an alternative you know waveform also with
two switching angles per quarter the alternative
553
00:52:28,480 --> 00:52:34,040
waveform as i was saying could be like this
in the waveform that we considered last time
554
00:52:34,040 --> 00:52:38,660
at zero it was switching from low to high
here in the alternative waveform you can also
555
00:52:38,660 --> 00:52:46,150
it you can have it switch from high to low
you can switch like this and this can be your
556
00:52:46,150 --> 00:52:53,690
alpha one and you can have your alpha two
here and this can be your ninety degrees
557
00:52:53,690 --> 00:53:00,610
this is an alternative waveform from what
we have earlier considered what we earlier
558
00:53:00,610 --> 00:53:15,780
considered was like this um where the switching
was from high to low and you have a not shear
559
00:53:15,780 --> 00:53:23,570
alpha one and alpha two is up to ninety degrees
now so for the time being we will consider
560
00:53:23,570 --> 00:53:31,880
just this waveform here for this also you
can write down v one and vn just as what we
561
00:53:31,880 --> 00:53:35,530
did before this is once again be a function
of alpha one and alpha two but will be a slightly
562
00:53:35,530 --> 00:53:42,770
different function in this case if you want
a fundamental voltage with such a phase then
563
00:53:42,770 --> 00:53:47,690
the spacing between alpha one and alpha two
has to be fairly high only then you will get
564
00:53:47,690 --> 00:53:52,120
a positive voltage otherwise the fundamental
may have an opposite phase here it may be
565
00:53:52,120 --> 00:53:56,540
going negative here
then if if on the other hand if you look at
566
00:53:56,540 --> 00:54:00,240
this waveform alpha one and alpha two have
to be close to one another for you to have
567
00:54:00,240 --> 00:54:05,050
a positive phase otherwise your fundamental
voltage might have a negative phase it might
568
00:54:05,050 --> 00:54:10,380
be going in the negative direction here ok
so we will be considering this for some more
569
00:54:10,380 --> 00:54:16,570
time to come and the equations are as we have
given now so what we will look at is is this
570
00:54:16,570 --> 00:54:24,410
problem of two switching angles per quarter
and once again let me reiterate this is only
571
00:54:24,410 --> 00:54:30,090
one quarter is being represented if i want
to extend this i can extend it like this i
572
00:54:30,090 --> 00:54:34,920
can extend just like this so this will be
one eighty degrees and what is this angle
573
00:54:34,920 --> 00:54:40,620
this angle will be one eighty minus alpha
two and this angle will be one eighty minus
574
00:54:40,620 --> 00:54:44,750
alpha one this is one half
and this will have a fundamental component
575
00:54:44,750 --> 00:54:55,410
like here this will have some fundamental
component like here ok so at zero the fundamental
576
00:54:55,410 --> 00:55:00,570
has its positive zero crossing at one eighty
the fundamental has its negative zero crossing
577
00:55:00,570 --> 00:55:07,369
at ninety degree the fundamental is atits
peak this is how it will be now all right
578
00:55:07,369 --> 00:55:13,350
so we are looking at different values of alpha
one and alpha two first let us see what are
579
00:55:13,350 --> 00:55:17,600
the relationships that we have between alpha
one and alpha two here we have certain things
580
00:55:17,600 --> 00:55:24,290
what is this you have alpha one and alpha
two certainly alpha one is less than alpha
581
00:55:24,290 --> 00:55:29,750
two as a special case alpha one can also be
equal to alpha two
582
00:55:29,750 --> 00:55:34,190
so this is alpha one less than or equal to
alpha two and this once again has got to be
583
00:55:34,190 --> 00:55:40,810
zero degrees alpha one is greater than zero
similarly alpha two is less than ninety degree
584
00:55:40,810 --> 00:55:45,920
now you can consider all this in a switching
angle plane if you look at a switching angle
585
00:55:45,920 --> 00:55:52,510
plane what we have is we can never have any
value of alpha two greater than ninety you
586
00:55:52,510 --> 00:55:59,280
can have no value of alpha one which is greater
than ninety so you have alpha one which is
587
00:55:59,280 --> 00:56:05,760
greater than zero so here and you also have
alpha two you know this is the line alpha
588
00:56:05,760 --> 00:56:11,980
one is equal to alpha two your alpha two has
to be greater than alpha one so you have this
589
00:56:11,980 --> 00:56:16,710
here
so alpha one cannot be less than zero so this
590
00:56:16,710 --> 00:56:23,410
is the line alpha one equals zero this is
the line alpha two is equal to ninety this
591
00:56:23,410 --> 00:56:28,790
is the line alpha one is equal to alpha two
all that you have is these are your pwm waveforms
592
00:56:28,790 --> 00:56:34,570
what we are trying to see here is a point
any point here such as these are pwm waveforms
593
00:56:34,570 --> 00:56:39,840
of the kind we just saw with two switching
angles per quarter so we would like to look
594
00:56:39,840 --> 00:56:44,510
this waveform and this kind of a plane can
make this very clear ok so we will deal with
595
00:56:44,510 --> 00:56:49,200
more of this in the coming lectures we will
use this switching angle plane to discuss
596
00:56:49,200 --> 00:56:53,590
fundamental angle and how they vary in harmonic
voltages how they vary etcetera move on to
597
00:56:53,590 --> 00:56:55,450
selective harmonic elimination
thank you very much