1
00:00:17,850 --> 00:00:23,440
So, this is the BUCK converter. We are familiar
with the circuit by now, we need to rate the
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00:00:23,440 --> 00:00:32,410
components, the switch, the diode, the inductance
and the capacitance. These are the four elements
3
00:00:32,410 --> 00:00:44,059
that we need to design this. To rate the components
properly we need to have an understanding
4
00:00:44,059 --> 00:00:52,380
of the various waveforms that are pretty critical.
Now, the pole voltage VP is a very important
5
00:00:52,380 --> 00:00:59,890
waveform that you should understand because
it will help in rating both voltage rating
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00:00:59,890 --> 00:01:07,659
of both these diode and the transistor. The
current waveform of the inductor is very critical
7
00:01:07,659 --> 00:01:11,600
because this is an important part in this
component, in this particular topology see
8
00:01:11,600 --> 00:01:17,930
that is current through the inductor gives
information about the current flow in the
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00:01:17,930 --> 00:01:19,670
input and therefore, the current flow in the
transistor, the current flow in the diode
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00:01:19,670 --> 00:01:21,909
because the freewheeling current also flows
through the inductor, the current flowing
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00:01:21,909 --> 00:01:35,100
through the inductor also divides into the
capacitor current and the current through
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00:01:35,100 --> 00:01:40,869
the load.
As we had discussed earlier, the current through
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00:01:40,869 --> 00:01:48,329
the capacitor will have zero average value
in steady state and the current through the
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00:01:48,329 --> 00:01:55,290
output load will be at pure DC. So, keeping
these things in mind, let us construct the
15
00:01:55,290 --> 00:02:02,979
waveforms and based on the waveforms we can
just read off the rating values of the various
16
00:02:02,979 --> 00:02:14,660
components. Let us now start by placing the
x and y-axis. Now, I have to here x-axis which
17
00:02:14,660 --> 00:02:23,920
is the time axis and on the y-axis, we have
three important things Vg; Vg is nothing,
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00:02:23,920 --> 00:02:29,280
but the gate signal which we will be applying
at this point. I did not want to put it as
19
00:02:29,280 --> 00:02:33,860
Vb because it may be confused with battery
voltage. Therefore, I am calling this one
20
00:02:33,860 --> 00:02:41,420
as Vg as we had indicated in the simulation
and this is the signal to drive this on or
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00:02:41,420 --> 00:02:47,220
off.
So, what is that signal? And before that the
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00:02:47,220 --> 00:02:53,640
other important variable voltage that we need
to see is the pole voltage; pole voltage the
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00:02:53,640 --> 00:03:00,010
voltage at this point P with respect to the
ground and then the current that is the inductor
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00:03:00,010 --> 00:03:09,040
current. Now, the pole voltage will give information
on this diode rating in terms of voltage and
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00:03:09,040 --> 00:03:15,100
the rating of the transistor and also the
voltage across the inductor to check for the
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00:03:15,100 --> 00:03:20,130
volt second balance. The current of the inductor
will give lot of information on the input
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00:03:20,130 --> 00:03:24,850
current, the current through the diode the
current through the capacitance current through
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00:03:24,850 --> 00:03:27,810
load.
So, let us look at them one by one. For the
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00:03:27,810 --> 00:03:41,710
voltage, let me install this here. So, this
is the waveform, the voltage waveform that
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00:03:41,710 --> 00:03:56,860
we would see at the base or the gate of this
power semiconductor switch. The time, when
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00:03:56,860 --> 00:04:03,090
the pulse is high then the switch is on the
time when the pulse is low, this switch is
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00:04:03,090 --> 00:04:10,070
off and during that time the diode is freewheeling.
So, this time would be the dTs time and this
33
00:04:10,070 --> 00:04:33,140
time would be the 1 minus dTs time, this is
the on-time and this is the off-time. So,
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00:04:33,140 --> 00:04:49,480
we shall mark them accordingly. So, let me
place the marks here. So, this is the dTs
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00:04:49,480 --> 00:04:55,690
time and this is the 1 minus dTs time.
Now, such a waveform we will apply at the
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00:04:55,690 --> 00:05:18,990
gate now. So, what happens, during the time
when this is on Vin comes to the pole. So,
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00:05:18,990 --> 00:05:36,010
the pole voltage will be Vin value and during
the time when this is off the diode is freewheeling
38
00:05:36,010 --> 00:05:39,120
this path and the pole voltage is connected
to the ground. So, it will be 0, let us mark
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00:05:39,120 --> 00:05:46,560
that. So, I am going to place this waveform
here. So, see here that the blue colored waveform
40
00:05:46,560 --> 00:05:50,260
is the pole voltage waveform that appears
here during the time, when the Q is on and
41
00:05:50,260 --> 00:05:51,260
the pole voltage is high and it is Vin value
and the value there let me indicate by this
42
00:05:51,260 --> 00:05:52,260
symbol Vin.
Now, this has an average value. So, let me
43
00:05:52,260 --> 00:05:53,260
indicate the average value by a line like
this. So, this is the DC value that you would
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00:05:53,260 --> 00:05:58,790
see at the output because this is the average
of this, what is the average of this Vin-this
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00:05:58,790 --> 00:06:09,139
area and this portion will get average down
this will fill up here and then you will get
46
00:06:09,139 --> 00:06:20,600
a smooth and that is our Vo, Vo which is Vin
into d is the average of the waveform
47
00:06:20,600 --> 00:06:36,280
Now, if I take the average at Vo whose amplitude
I am indicating by the arrow will be Vin minus
48
00:06:36,280 --> 00:06:42,310
Vo and this amplitude from the average value
to 0 would be Vo. So, this portion in fact,
49
00:06:42,310 --> 00:06:57,220
would be the waveform that you would be seeing
across the inductor. So, you see that the
50
00:06:57,220 --> 00:07:03,080
portion above the average and one below the
average. So, that yellow portion is the one
51
00:07:03,080 --> 00:07:09,170
which will be appearing across the inductor
and there should be volt second balance doing
52
00:07:09,170 --> 00:07:18,091
the top portion of the yellow and the bottom
portion of the yellow portion of the waveform
53
00:07:18,091 --> 00:07:20,710
above and below the average value.
So, the yellow portion of the waveform is
54
00:07:20,710 --> 00:07:36,020
what is called the VL or the voltage across
the inductor. So, this wave the voltage the
55
00:07:36,020 --> 00:07:42,830
pole voltage here has so much information
in it. It gives you what is the average output.
56
00:07:42,830 --> 00:07:49,610
It gives you the voltage across the inductor
and the volt second balance that is happening
57
00:07:49,610 --> 00:08:01,169
and with this we can rate these two components
Now, let us see during this period the dTs
58
00:08:01,169 --> 00:08:08,960
period, the transistor is on. If the transistor
is on the pole voltage here is Vin then what
59
00:08:08,960 --> 00:08:10,760
should be the peak inverse voltage of the
diode. The peak inverse voltage of the diode
60
00:08:10,760 --> 00:08:23,449
should be Vin or at least greater than Vin
because this diode has to withstand a value
61
00:08:23,449 --> 00:08:33,290
of the potential equivalent to Vin.
Now, let us say this is off and this is freewheeling.
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00:08:33,290 --> 00:08:37,460
So, when this is freewheeling this is pulled
to the ground. So, the emitter is pulled to
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00:08:37,460 --> 00:08:43,710
the ground and this is at Vin. So, the voltage
withstanding capability of this BJT or any
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00:08:43,710 --> 00:08:51,339
other power semiconductor switch which is
placed here should be Vin which is this value.
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00:08:51,339 --> 00:08:55,339
So, you see that this by looking at the waveform
and the pole voltage waveform, we have found
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00:08:55,339 --> 00:09:00,589
two important ratings of these two devices,
the voltage withstanding rating of the switch
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00:09:00,589 --> 00:09:07,060
and the voltage withstanding rating or the
peak inverse voltage of the diode.
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00:09:07,060 --> 00:09:21,170
Now, let us look at the inductor current,
I have drawn a green line here, the green
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00:09:21,170 --> 00:09:24,759
line here is not the inductor current, but
it is the current that is flowing through
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00:09:24,759 --> 00:09:32,050
the load assuming very negligible ripple across
the output, I have approximated it by a straight
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00:09:32,050 --> 00:09:35,269
line and this is the current that is DC current
flowing through the load, we will call that
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00:09:35,269 --> 00:09:58,860
one as Io. So, this is Io, and now how will
be the current through the inductor when the
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00:09:58,860 --> 00:10:05,740
transistor Q is on the voltage here is Vin
and the voltage here is Vo. So, in the steady
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00:10:05,740 --> 00:10:11,350
state Vin minus Vo is applied across the inductor.
.
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00:10:11,350 --> 00:10:30,759
So, let me take the writing pad to give a
view of what is happening. So, consider this
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00:10:30,759 --> 00:10:45,029
inductor and it is receiving the voltage across
this VL, which is equal to Vin minus Vo at
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00:10:45,029 --> 00:10:56,430
some index. So, we know that VL which is equal
to L di by dt and di by dt where i is the
78
00:10:56,430 --> 00:11:13,309
current flowing through the inductor is given
by VL by L ratio or it is Vin minus Vo by
79
00:11:13,309 --> 00:11:22,819
L ratio.
So, this is the di by dt or the slope of the
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00:11:22,819 --> 00:11:31,079
current waveform. So, at this slope the current
waveform should increase. So, if I draw this
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00:11:31,079 --> 00:11:38,829
slope, during the time when the voltage across
the inductor is Vin minus Vo it should increase
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00:11:38,829 --> 00:11:45,670
like this because Vin is a constant Vo is
a constant, L is a known constant. So, the
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00:11:45,670 --> 00:12:04,079
rate has to be a linear one and this is Vin
minus Vo by L and during the time when the
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00:12:04,079 --> 00:12:20,850
switch is off. We saw earlier that this side
of the inductor becomes 0 and here it is Vo
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00:12:20,850 --> 00:12:27,920
and what is applied is minus Vo.
So, what is the rate di under such conditions
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00:12:27,920 --> 00:12:42,119
di by dt would be, a change in color some,
VL by L, becomes minus Vo by L. Now, minus
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00:12:42,119 --> 00:12:53,350
Vo by L indicates negative slope. So, it should
start falling at this state. So, this would
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00:12:53,350 --> 00:13:04,749
be minus Vo by L and negative slope falling
and then the subsequent cycle again should
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00:13:04,749 --> 00:13:11,269
go in this fashion and start falling. So,
this would be the shape of the current through
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00:13:11,269 --> 00:13:21,399
the inductor
Now, what will be the average value? The average
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00:13:21,399 --> 00:13:34,360
value flows the average value flows through
the output and that is nothing, but Io and
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00:13:34,360 --> 00:13:41,519
the ripple value the one with that component
with 0 average will flow through the capacitor.
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00:13:41,519 --> 00:13:48,569
So, through the capacitor the one marked like
this is what is going to flow. So, this is
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00:13:48,569 --> 00:13:57,019
what we can expect and let us go back and
have a look at that slide. So, the average
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00:13:57,019 --> 00:14:03,699
value of the inductor current is Io and that
is what will go through. The ripple component
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00:14:03,699 --> 00:14:10,570
of the inductance will go through the capacitance
and there would not be the ripple component
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00:14:10,570 --> 00:14:15,120
here. So, let us place the ripple component
here and see.
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00:14:15,120 --> 00:14:24,970
After placing the ripple component here, you
see that the wave shape is just like as we
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00:14:24,970 --> 00:14:32,070
had discussed during the time, when Q is on
and that was charging up and this charging
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00:14:32,070 --> 00:14:38,920
up rate is Vin minus Vo by L and then the
inductor is discharging during the time when
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00:14:38,920 --> 00:14:45,910
it is off when it is discharging it is freewheeling
like this and it is falling down at the rate
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00:14:45,910 --> 00:14:56,649
of minus Vo by L and it keeps on repeating
under steady state condition like a triangle
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00:14:56,649 --> 00:15:03,309
and the one that is filled up here the colored
filled up portion is the 0 average portion
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00:15:03,309 --> 00:15:09,660
and that is what will flow through the capacitance
the filled up portion is the capacitance current
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00:15:09,660 --> 00:15:13,763
the average value of that is current flowing
through Ro and this entire wave shape is the
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00:15:13,763 --> 00:15:19,899
inductor current.
So, you see that you get again here, lot of
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00:15:19,899 --> 00:15:28,170
information in rating the components, if you
look at the current flowing through the BJT;
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00:15:28,170 --> 00:15:29,769
the BJT the current flows only during this
portion something like this. So, let me. So,
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00:15:29,769 --> 00:15:45,430
only during this portion the current through
the BJT flows. So, you can rate your BJT for
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00:15:45,430 --> 00:15:57,819
this it has a peak current of Io plus this
ripple peak. So, if I say that the ripple
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00:15:57,819 --> 00:16:11,939
amplitude is like this delta IL, then the
rating of the current for BJT will be Io plus
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00:16:11,939 --> 00:16:16,862
delta IL by 2, which would be this peak that
will be the peak current.
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00:16:16,862 --> 00:16:27,740
The average current rating will be Io into
this duty cycle d and for the diode it will
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00:16:27,740 --> 00:16:36,519
appear like this during this time, whenever
the switch is off the diode is freewheeling
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00:16:36,519 --> 00:16:41,929
and during this time. There will be current
flowing through the diode and it is the same
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00:16:41,929 --> 00:16:50,490
inductor current which has split into it therefore,
the diode also has to have the same peak current
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00:16:50,490 --> 00:17:02,139
value Io plus delta IL by 2 and the average
value would be value of Io one minus d.
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00:17:02,139 --> 00:17:13,020
So, this way you get the value of the current
that would flow through the input and the
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00:17:13,020 --> 00:17:21,959
value of the current that flows through the
diode 2, now these are the critical waveforms
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00:17:21,959 --> 00:17:28,459
that you have to be clear about and when you
do the simulation, you should actually try
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00:17:28,459 --> 00:17:37,450
to observe these effects in the simulation
and see that this is what you are getting
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00:17:37,450 --> 00:17:47,950
at this waveform is the waveform of the capacitor
current, which flows through the capacitor
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00:17:47,950 --> 00:17:54,120
this would actually be the 0 line in case
the inductor current this line would be the
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00:17:54,120 --> 00:17:58,990
Io line.
Now, let us consider the current flowing through
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00:17:58,990 --> 00:18:07,880
the capacitor. Assuming that this green line
is 0, there is no average Io component has
126
00:18:07,880 --> 00:18:18,760
flown to the load side, now what is important
in the capacitor is that for charged balance
127
00:18:18,760 --> 00:18:28,279
and second balance this area should be equal
to this area. So, the amount of charge put
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00:18:28,279 --> 00:18:35,330
into the capacitor should be equal to the
amount of charge taken out. So, this will
129
00:18:35,330 --> 00:18:36,510
be delta Q
.
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00:18:36,510 --> 00:18:51,779
Now, looking at the capacitor waveforms we
know that this is the time that inductor current
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00:18:51,779 --> 00:19:08,409
has magnetically charged up. So, this is dt
and this is the time when the inductor current
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00:19:08,409 --> 00:19:18,169
is falling with a slope of minus Vo by L and
therefore, this has to be the 1 minus d Ts;
133
00:19:18,169 --> 00:19:30,370
however, if you consider the triangle this
upper triangle and the bottom inverted triangle
134
00:19:30,370 --> 00:19:37,190
are similar triangles because you see that
the total height is delta IL therefore, the
135
00:19:37,190 --> 00:19:52,490
height of this triangle is delta IL by 2,
the height of this part of the triangle is
136
00:19:52,490 --> 00:20:06,260
also delta IL by 2 and therefore, the bases
are also equal because they are similar triangles
137
00:20:06,260 --> 00:20:21,360
the height is the same and you can easily
say that the base is Ts by 2.
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00:20:21,360 --> 00:20:32,220
So, going from minus delta IL by 2 to plus
delta IL by 2 and this is a straight line
139
00:20:32,220 --> 00:20:39,759
the midpoint which is at 0 in the case of
the capacitor will be half the distance likewise
140
00:20:39,759 --> 00:20:44,520
coming down the midpoint will be half the
distance in time and therefore, this will
141
00:20:44,520 --> 00:20:50,050
be Ts by 2, the other base will also be Ts
by 2.
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00:20:50,050 --> 00:20:58,710
So, now let us look at this triangle, the
area of the triangle will give you the charge.
143
00:20:58,710 --> 00:21:10,700
So, the charge will be half base into height
of the triangle, which is half the base is
144
00:21:10,700 --> 00:21:28,610
Ts by 2 into height of the triangle delta
IL by 2which is Ts delta IL by 8 or further
145
00:21:28,610 --> 00:21:37,570
delta IL by 8 times, fs which is the switching
frequency, Ts is the switching period fs is
146
00:21:37,570 --> 00:21:43,380
the switching frequency.
Now, delta Q are the change in the charge,
147
00:21:43,380 --> 00:21:53,320
in the capacitor is given by c delta V by
V operation of the capacitor. Now, this we
148
00:21:53,320 --> 00:22:06,230
know is delta IL by 8fs and therefore, we
can find out C, C is equal to delta IL by
149
00:22:06,230 --> 00:22:18,900
8 fs deltaV. now this is an important relationship
which you can use for finding the value of
150
00:22:18,900 --> 00:22:24,370
the capacitance that you need to put for the
buck converter.
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00:22:24,370 --> 00:22:34,150
Now, out of these things fs is a design step
which is known delta V what is supposed to
152
00:22:34,150 --> 00:22:42,240
be the ripple across the capacitance is also
obtained as an output ripple spec that the
153
00:22:42,240 --> 00:22:48,730
capacitor is also a design spec the capacitor
should not have ripple beyond this particular
154
00:22:48,730 --> 00:22:54,799
value the value of delta IL is also a divine
spec while designing the inductor because
155
00:22:54,799 --> 00:23:09,389
you see the delta IL value is assumed as around
10 percent of Io max. So, this is I have a
156
00:23:09,389 --> 00:23:16,360
design, and starting designed value which
people use or in some of the specification
157
00:23:16,360 --> 00:23:22,330
minimum value of Io is specified. So, based
on the minimum value of Io, delta IL value
158
00:23:22,330 --> 00:23:24,130
can be obtained.
.
159
00:23:24,130 --> 00:23:32,450
So, let us look at the inductor finding the
value of the inductor, finding the value of
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00:23:32,450 --> 00:23:38,909
delta IL becomes clear.
So, for now you can see that the capacitor
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00:23:38,909 --> 00:23:43,860
value can be calculated knowing the value
of delta IL deltaV and fs; f s is known delta
162
00:23:43,860 --> 00:23:56,480
V coming from design spec delta IL also coming
from design spec. So, let us now look at finding
163
00:23:56,480 --> 00:24:11,100
the value of the inductor L. So, we know that
in the case of the current, it is rising at
164
00:24:11,100 --> 00:24:20,340
the rate of Vin minus Vo by L at that rate
and then it is falling at the rate of minus
165
00:24:20,340 --> 00:24:30,710
Vo by L this is. So, for the case of the buck
converter, where the average value of the
166
00:24:30,710 --> 00:24:36,899
inductor current is Io?
So, this we have seen just now, just for this
167
00:24:36,899 --> 00:24:45,130
the various waveform, when we are studying
the various waveforms the inductor current
168
00:24:45,130 --> 00:24:56,759
goes by the Faraday’s law VL, the voltage
across the inductor is equal to L di by dt
169
00:24:56,759 --> 00:25:03,080
and VL by L is there. So, this is what we
have here now you could use either of the
170
00:25:03,080 --> 00:25:08,269
slope there is one variable in the only single
variable in the output is the falling slope.
171
00:25:08,269 --> 00:25:15,620
So, therefore, we could use that. So, let
us say that this rate, I will just take the
172
00:25:15,620 --> 00:25:31,630
rate absolute value of the rate this slope
finally, is the delta d iL by dt and because
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00:25:31,630 --> 00:25:47,409
they are all linear straight lines people
say delta IL by delta t now in this particular
174
00:25:47,409 --> 00:25:58,379
buck converter waveform case delta IL is this
much. So, there is a change in delta IL from
175
00:25:58,379 --> 00:26:03,509
here to here in terms of time within this
time.
176
00:26:03,509 --> 00:26:12,580
So, this time was one minus dTs for the buck
converter case you could do the same in the
177
00:26:12,580 --> 00:26:21,600
upward slope also and you will get the same
result. So, delta IL that was, is delta IL
178
00:26:21,600 --> 00:26:33,620
itself. So, this is nothing, but delta IL
by 1 minus dTs. So, in this time period we
179
00:26:33,620 --> 00:26:45,210
have this rate Vo by L. So, we are writing
here you will see that Vo by L will be delta
180
00:26:45,210 --> 00:27:01,860
IL by one minus dTs or L equals rearranging
this equation you will see Vo one minus d
181
00:27:01,860 --> 00:27:15,050
by delta IL fs. So, I will replace it by fs
because fs is another design spec which is
182
00:27:15,050 --> 00:27:24,750
normally given d is the duty cycle. So, to
get the maximum value of L we could use the
183
00:27:24,750 --> 00:27:33,910
minimum value of d that is d min.
.
184
00:27:33,910 --> 00:27:54,010
So, if you consider L max you would get Vo
minus d min minimum value of d by delta IL
185
00:27:54,010 --> 00:28:07,350
fs. Now, how to get the value of dmin, now
we know how to get the value of Vo. Vo is
186
00:28:07,350 --> 00:28:20,650
equal to Vin into d. Now, in this Vin is given
and Vo is given normally Vo is the spec Vin
187
00:28:20,650 --> 00:28:28,820
is also an input spec. So, d is actually calculatable.
So, we calculate d which is equal to Vo by
188
00:28:28,820 --> 00:28:39,419
Vin and normally you will see that Vo is supposed
to be regulated, it is supposed to be a constant
189
00:28:39,419 --> 00:28:45,080
value, it should be regulated in spite of
variation in Vin.
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00:28:45,080 --> 00:29:02,919
So, you will see that d could be Vo by Vin
max or Vo by Vin min and any value in between.
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So, this could be the limiting values where
Vo by Vin max will be dmin value and Vo by
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Vin min value will be the d max value. So,
this is how we calculate. Vin max and Vin
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min are found from the input tolerance spec.
So, you can find d min and d max and then
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use this d min value here to find out the
value of L, sorry it was for minimum duty
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cycle. It will also for any other duty cycles
the value of L will be large enough to take
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care of any other duty cycles.
Now, one other issue is here I mentioned that
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Vo is a spec f is also a spec f is, f s is
also a spec and Vo is also known d is known.
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So, everything can be used for calculating
except IL and as a starting value you use
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10 percent of Io max load current use this
and find some value of L.
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.
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There is also another way, sometimes the following
spec is also given sometimes, you will see
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that; let me write down this equation L is
equal to Vo 1 minus d delta IL into fs. So,
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you see that in the case of the inductor this
is you see the inductor current going up and
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down in this fashion and here you have Io
this is Io.
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Now, what would happen when Ro value is increased,
if Ro value the output load resistance value
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is increased you will see Io decreasing. So,
Io will keep decreasing, but you see that
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their delta IL and the rate the slopes here
will not change they will because they are
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determined by Vin minus Vo by L and Vo by
L here and therefore, if they are independent
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of the load resistor value or the load current
value.
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So, the shape of this triangular ripple will
still remain and the height of the ripple
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will also remain. So, in the limit you will
see that it may just touch lines are not straight.
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So, it may just touch the 0 line like this.
So, this would be considered as Io min minimum
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value of Io that value of Io, where the inductor
current ripple just touches the 0 line would
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be the Io min because the average of that
will be Io min.
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00:33:26,179 --> 00:33:35,559
So, in this case, in this limiting case if
this was supposed to be the delta IL. So,
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you will see
that delta IL by two would be equal to Io
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00:33:48,100 --> 00:34:05,360
minimum or delta IL is equal to two times
Io minimum and this is a very important condition
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that you can use for choosing the value of
delta IL; delta IL should be the two times
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00:34:15,620 --> 00:34:23,470
Io minimum where Io minimum is specified.
So, this is specified, if this is specified
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00:34:23,470 --> 00:34:31,310
by the user then you can use this to find
out the value of delta IL. If this is not
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specified then you use delta IL is 10 percent
of Io, this is how you try to find.
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Now, here one more point is that this at Io
minimum, this is just touching the 0 line.
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So, this is at the boundary of continuous
conduction, what it means is if you go any
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further we could have the triangle coming
like this and I am putting this in dotted,
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if there is a bidirectional switch you will
see the inductor current going negative, but
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the series, the switch if it blocks then you
will see that the inductor current goes like
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that. So, this is a sudden change in the slope
during this point.
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So, the inductor current reaches 0, there
is a change in the slope and then again change
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in the slope. So, that is why we say it is
discontinuous or the conduction period itself
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is discontinuous. You have a period where
the inductor current is present period where
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the inductor current is 0 not present again,
it is present. So, this is discontinuous conduction
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zone or DCM for this one, this type of an
inductor current waveform is called DCM and
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all waveforms above the 0 line, this type
of waveform showed in blue or continuous conduction
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or CCM. So, you should ensure that the inductor
current is operating in CCM and not in DCM
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because all the analysis that we have been
talking about is for a continuous conduction
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mode.