1
00:00:17,060 --> 00:00:23,529
Here is a Cascode amplifier which is similar
to a normal amplifier this is your driver
2
00:00:23,529 --> 00:00:33,489
amplifier and one which are all standard value
of W/L VT etc. series to that M1 there is
3
00:00:33,489 --> 00:00:43,530
another transistor M2, please look at it this
is very relevant and good concept to understand
4
00:00:43,530 --> 00:00:47,530
there is a another transistor in M2 please
remember this is common source M1 is a common
5
00:00:47,530 --> 00:00:54,730
source system inputted at the gate source
is grounded wherever M2.
6
00:00:54,730 --> 00:01:05,740
Actually has a DC potential at the gate what
AC what is it ground for AC it is ground so
7
00:01:05,740 --> 00:01:12,880
what kind of this amplifier or circuit will
be common gate it is a common gate we already
8
00:01:12,880 --> 00:01:18,320
done common source common gate common drain.
so now I say this is a common source amplifier
9
00:01:18,320 --> 00:01:26,840
over which we are sitting a common gate amplifier
and it is biased by a constant current source
10
00:01:26,840 --> 00:01:32,000
I this is I told you the other day there are
different way of biasing circuits.
11
00:01:32,000 --> 00:01:38,399
One is my resistance network the other of
course is by active load like a diode connected
12
00:01:38,399 --> 00:01:43,469
loads or a current sources as shown here they
can be also taken from Miller sources ok,
13
00:01:43,469 --> 00:01:50,880
Miller that mirrors a normally see mirrors
current mirrors. so it is biased in such a
14
00:01:50,880 --> 00:01:58,299
way that both M1 and M2 are in saturation
okay is that point clear so, the first problem
15
00:01:58,299 --> 00:02:00,859
one can see from here.
16
00:02:00,859 --> 00:02:04,219
If you are drawing the circuit rest you do
not write just read listen and then think
17
00:02:04,219 --> 00:02:12,760
if M1 and M2 are in series as shown here in
normal case if M2 would have been absent only
18
00:02:12,760 --> 00:02:19,720
M1 would have been there so how much would
have been supply this is our VDD. Let us say
19
00:02:19,720 --> 00:02:25,610
source requires some drop but the most of
the VDS would have come in M1 is that layer
20
00:02:25,610 --> 00:02:33,790
most of the drop would have come in M1, so
VDS would have been larger to make it in saturation.
21
00:02:33,790 --> 00:02:40,799
Now I want 2 transistors M1 M2 also to be
in saturation if they are to be in saturation
22
00:02:40,799 --> 00:02:47,239
this voltage should be sufficient enough that
VDS2+VDS1 should be such that both makes it
23
00:02:47,239 --> 00:02:55,510
insanity is that clear that means power supply
needs to be slightly boosted because otherwise
24
00:02:55,510 --> 00:03:00,209
I most of the voltage goes here then you have
nothing left but to make them both them in
25
00:03:00,209 --> 00:03:06,620
saturation, so one of the major problem in
Cascode Sharpe you may require larger power
26
00:03:06,620 --> 00:03:07,720
supply voltages okay.
27
00:03:07,720 --> 00:03:14,170
It is not very true what I am making but roughly
larger than the single stage amplifiers I
28
00:03:14,170 --> 00:03:19,590
will not say large or something because also
made in the chip so it cannot be more than
29
00:03:19,590 --> 00:03:25,590
1.5 also we can adjust VDS we can adjust VT
and therefore Vgs-VT still can be smaller
30
00:03:25,590 --> 00:03:33,819
than VDS but irrespective of other number
I chose I need larger drops for M1 M2 to be
31
00:03:33,819 --> 00:03:34,819
saturation.
32
00:03:34,819 --> 00:03:41,599
This is one is problem one you should understand
now right now assume this current source is
33
00:03:41,599 --> 00:03:50,500
ideal and we say it is output resistance is
infinite, infinite is that okay this is a
34
00:03:50,500 --> 00:03:57,190
Cascode amplifier is that clear this sentences
have written just for the sake of it you need
35
00:03:57,190 --> 00:04:01,510
not because I already explained you what I
am talking these are some okay.
36
00:04:01,510 --> 00:04:07,090
Basically I am saying this is driver this
is the gate common gate transistor this is
37
00:04:07,090 --> 00:04:12,890
the essentially current shows biased.
38
00:04:12,890 --> 00:04:19,820
Now I want to analyze what is the condition
what is the criteria I said last time to make
39
00:04:19,820 --> 00:04:27,600
cascode better than cascade that normally
I say ft or unity gain amplifier frequency
40
00:04:27,600 --> 00:04:34,169
or the gain bandwidth product as we called
GBW that is constant for a normal common source
41
00:04:34,169 --> 00:04:35,440
or any amplifier.
42
00:04:35,440 --> 00:04:43,710
So if I increase gain bandwidth goes down
if I increase bandwidth n goes is that clear
43
00:04:43,710 --> 00:04:48,591
and the maximum frequency is you need again
frequency is that correct you cannot go beyond
44
00:04:48,591 --> 00:04:56,259
that because why because beyond again is -DB
means >1, so we have been now worried that
45
00:04:56,259 --> 00:05:02,069
how can I increase gain without actually losing
on the bandwidth or which out getting that
46
00:05:02,069 --> 00:05:07,569
cutoff point to be any different if I remain
there and still boost the gain.
47
00:05:07,569 --> 00:05:14,820
Then I achieved something which was not possible
in a single stage that first case gain bandwidth
48
00:05:14,820 --> 00:05:20,729
constant or UFT is constant is essentially
called the figure of merit you cannot go beyond
49
00:05:20,729 --> 00:05:28,150
that now I want to say even at that frequency
gain is higher if I do that and I have achieved
50
00:05:28,150 --> 00:05:34,270
something is that clear and this is exactly
what Cascode is trying to do in this circuit.
51
00:05:34,270 --> 00:05:41,800
If you see carefully the gain there is no
additional load resistance I have put here
52
00:05:41,800 --> 00:05:47,819
the load resistance is essentially due to
M2 M1 okay series some combination of this
53
00:05:47,819 --> 00:05:57,870
will lead to R,Ro and this R is also parallel
to this R of this source is that clear why
54
00:05:57,870 --> 00:06:04,319
because this is ground for AC so this R is
also parallel to RO of these combinations
55
00:06:04,319 --> 00:06:06,630
but that we treated as how much infinite.
56
00:06:06,630 --> 00:06:12,979
So, essentially I said this arrow equal and
what I am going to see here is the only load
57
00:06:12,979 --> 00:06:20,449
resistance I have and therefore gain is gm
x R0 and this R0 will call RO effective because
58
00:06:20,449 --> 00:06:27,229
without this whatever is out over this with
2 of them. Whatever we will call it say we
59
00:06:27,229 --> 00:06:37,020
will call it RO effective so if I can make
gm RO effective larger than gm RO with and
60
00:06:37,020 --> 00:06:44,650
what is the bandwidth frequency and just show
you gm/C say gm/C is not touched and I increase
61
00:06:44,650 --> 00:06:45,650
R0.
62
00:06:45,650 --> 00:06:51,639
Then, I am increasing the gain with not losing
bandwidths is that clear so essentially what
63
00:06:51,639 --> 00:06:58,090
Cascode is trying to do is to book output
boost output resistance without losing anywhere
64
00:06:58,090 --> 00:07:04,230
in gm factors is that that is exactly what
we are trying and this is an achievement which
65
00:07:04,230 --> 00:07:09,509
is actually not the device is not able to
tell us directly it is we say device is always
66
00:07:09,509 --> 00:07:11,060
cut off cut or you can go beyond.
67
00:07:11,060 --> 00:07:16,960
Now I am circuit where I am showing you I
can beat the system this is the trick in analog
68
00:07:16,960 --> 00:07:22,090
how to beat the system some cost will play
somewhere which we are not actually telling
69
00:07:22,090 --> 00:07:27,190
but at that cause I will boost this gain without
losing the bad news is that clear this is
70
00:07:27,190 --> 00:07:33,250
why Cascode are used where do you use goals
you want gain boosting okay without losing
71
00:07:33,250 --> 00:07:34,310
the bandwidth okay.
72
00:07:34,310 --> 00:07:43,140
Here is that, we wish to find gm effective
and R0 I have a tendency to earlier because
73
00:07:43,140 --> 00:07:50,300
this is my old papers and this since I used
to use all instead of R0 conductance GO is
74
00:07:50,300 --> 00:07:57,229
nothing but 1 upon RO okay GO is nothing but
1 upon or so I want to calculate for this
75
00:07:57,229 --> 00:08:00,909
Cascode stage a gm effective.
76
00:08:00,909 --> 00:08:08,260
What is gm effective if the M2 would not have
been there would have been only gm1 is that
77
00:08:08,260 --> 00:08:16,490
correct with empty present the if a few gm
is we call gm effective we divide that left
78
00:08:16,490 --> 00:08:23,560
gm and gmo1 are the trans conductance and
output conductance of transistor into and
79
00:08:23,560 --> 00:08:33,470
gm2 and gm2 and go2 are similar numbers for
M2 means gm1 1 is for M1 2 is for M2 is that
80
00:08:33,470 --> 00:08:34,470
ok.
81
00:08:34,470 --> 00:08:42,570
So if I define that gm1 and g01 andgmo2 are
the patterns for M1 and M2 by a network Theory
82
00:08:42,570 --> 00:08:52,370
the small signal current flowing in this transistor
is I which is equal to gm effective x VL+gm
83
00:08:52,370 --> 00:08:58,510
x VO this is a simple 2 port network.
84
00:08:58,510 --> 00:09:03,720
If you are not doing done so far in your network
course do it again otherwise the easiest way
85
00:09:03,720 --> 00:09:15,150
to represent there to put outputs okay I=gm,
l+ gm effective V0 okay, if you cannot I will
86
00:09:15,150 --> 00:09:19,540
show you how but other I should be able to
give it I to any 2 port Network in and out
87
00:09:19,540 --> 00:09:24,890
Vin and Vin and you will get these expressions
ok.
88
00:09:24,890 --> 00:09:36,029
Now in this figure I call this voltage asVO1
this of course I still call it V0 output of
89
00:09:36,029 --> 00:09:43,700
we are not taking it but the drain voltage
of M1 we defined as output 1 VO1 is that okay
90
00:09:43,700 --> 00:09:52,610
different definition nothing great about however,
if you see carefully the VO1 which is the
91
00:09:52,610 --> 00:10:02,269
drain voltage of M1 is also acting like a
source voltage of M2 is that point clear.
92
00:10:02,269 --> 00:10:12,019
VO1 is the drain voltage of M1 but it also
source voltage of M2 now do you get the point,
93
00:10:12,019 --> 00:10:21,190
so Vgs is still available if this is your
0 for example for AC also 0-VO1 which is good
94
00:10:21,190 --> 00:10:26,700
enough because then Vgs-VT still will be possible
is that clear, so this is a trick which I
95
00:10:26,700 --> 00:10:32,110
have played I said okay I put a series there
the drain voltage of this is same as source
96
00:10:32,110 --> 00:10:40,620
voltage of burst and therefore I can say my
Vgs is still plus VO1 and if that is larger
97
00:10:40,620 --> 00:10:43,779
than this it is still conduct ok.
98
00:10:43,779 --> 00:10:55,910
So, this is what I did I took on that for
AC signal that is what I say Vgs2 is 0-VO1
99
00:10:55,910 --> 00:11:05,029
is-VO1 my now from the current equation which
I wrote I is equal to gm effective o u VN+GO
100
00:11:05,029 --> 00:11:15,130
effective times VO if I make V of 0 do you
get the point what I am saying if I make VO
101
00:11:15,130 --> 00:11:24,870
0 I upon VN is gm effective is that clear
I upon V is-FB 0 0 knots ok if V0 is tends
102
00:11:24,870 --> 00:11:28,210
to 0 I by V in is gm effective.
103
00:11:28,210 --> 00:11:37,390
So, I write gm effective is I/Vin when V0
is 0 by same argument effective is I/V0 when
104
00:11:37,390 --> 00:11:49,170
Vin it is in equation do you understand you
make one codes output 0 next time you put
105
00:11:49,170 --> 00:11:54,720
input for 0 is that correct and then you get
gm effective and go effective I/V is both
106
00:11:54,720 --> 00:12:05,730
cases is I/VN is gm effective I/V0 is G effective
is that okay as shown we apply if now there
107
00:12:05,730 --> 00:12:18,360
is a game we are playing ok why I say it is
this is DC.
108
00:12:18,360 --> 00:12:32,870
So, not easy I add this node, I apply V fixed
DC1 at the output where or V0 its I apply
109
00:12:32,870 --> 00:12:42,850
fixed DC value V fixed so what is the AC value
applying 0 so VO is 0 is that point clear
110
00:12:42,850 --> 00:12:52,269
VO is 0 when I say I applied fixed DC bias
cannot change DV DV/DZ is 0, so we say it
111
00:12:52,269 --> 00:12:59,699
is V0 there is 0, so we say assume a apply
effects we fix at the m20 terminal therefore
112
00:12:59,699 --> 00:13:07,650
bias that is the condition what is the condition
I was asking for gm effective that I buy VN
113
00:13:07,650 --> 00:13:10,310
is GM effective and V0 is 0.
114
00:13:10,310 --> 00:13:18,230
So, I made that condition true and then I
write VO-VS which is nothing but variation
115
00:13:18,230 --> 00:13:27,990
of DVD s 2 is nothing but 0-please remember
this is 0 this is VO1 so how much is change
116
00:13:27,990 --> 00:13:44,630
in VO1 AC value 0-VO1 which is-VO1 so VDS
is how small vds that is AC is-VO1, now look
117
00:13:44,630 --> 00:13:50,639
at the 2 for the 2 transistors I like to current
though I same in both transistors is that
118
00:13:50,639 --> 00:13:56,769
clear I cannot have 2 in a circuit there can
be only one current which is flowing through
119
00:13:56,769 --> 00:14:00,209
M2 as well as through M1.
120
00:14:00,209 --> 00:14:09,630
So for a transistor 1 I=gm1Vin say 2 port
Network so I is equal
121
00:14:09,630 --> 00:14:15,380
to this is what we say intrinsic values external
resistances will put light on if whenever
122
00:14:15,380 --> 00:14:20,330
we need in actual circuit we want to know
intrinsically whether we can break that gm
123
00:14:20,330 --> 00:14:28,540
that is gain bandwidth limits okay, we can
know so I=gmVin+go1VO1.
124
00:14:28,540 --> 00:14:40,320
What is VO1 VDS RM1 VDS FM1 is VO1 so VO1
for the transistor 2 gm2-VO1 because that
125
00:14:40,320 --> 00:14:52,010
is the input now 0-VO1 so this is VDS+go2
again VDS-VO1 is that okay just substitute
126
00:14:52,010 --> 00:14:58,040
values for both transistors M1 and M2 and
currents are same please take it in a circuit
127
00:14:58,040 --> 00:15:04,660
same current can only flow okay, so the at
the drain of the M2.
128
00:15:04,660 --> 00:15:11,670
I am only still looking at this that is why
it is 0-VO1 is Vo1 for the VDS for this is
129
00:15:11,670 --> 00:15:16,399
that correct that is what I put-view is that
correct okay.
130
00:15:16,399 --> 00:15:27,970
If I do this I can now find out from the equation
3 VO1 3 or 4 I do not know which one where
131
00:15:27,970 --> 00:15:36,120
3 VO1 is I upon go1+go2 simple math okay.
132
00:15:36,120 --> 00:15:45,500
VO1 is-I upon gm2+go2 please remember G is
R conductance 1 upon G R whenever you wish
133
00:15:45,500 --> 00:15:56,500
it you can convert to back to R now this VO1
value which I got from equation 3 can be substituted
134
00:15:56,500 --> 00:16:07,350
where in equation 2 in that case I=gm1Vin
please remember what was there -go1 VO1 so
135
00:16:07,350 --> 00:16:23,040
it is 0 1 x I upon gm –go2 so if I collect
items I=1+0.1 upon gm go2=gm Vin under what
136
00:16:23,040 --> 00:16:24,040
condition.
137
00:16:24,040 --> 00:16:34,069
I derived this VO is external VO big DC value
therefore ACVO it is 0 and under that case
138
00:16:34,069 --> 00:16:42,949
III/VN is gm, gm effective so gm effective
is I/gm1 upon this quantity is that correct
139
00:16:42,949 --> 00:16:49,389
so I are now derived expression for GM effective
is that unconditional I have made what was
140
00:16:49,389 --> 00:16:53,310
the condition they wanted me to say that VO
is 0.
141
00:16:53,310 --> 00:17:00,579
So I have made it V0 and I solve the equation
and I got my gm effective now I want to see
142
00:17:00,579 --> 00:17:09,670
all these values which I put if I do little
extensions of this I can bring this denominator
143
00:17:09,670 --> 00:17:21,500
back to numerator so the gm effectively gm1
x gm2+go2 upon gm1 gm2+0.1 2 just remove that
144
00:17:21,500 --> 00:17:31,570
one part and just put the denominator to this
now typically lambdas of the transistor is
145
00:17:31,570 --> 00:17:39,780
how much close to 0.00100.2 or+2.0 as people
is saying.
146
00:17:39,780 --> 00:17:47,660
So R0 are normally infinite or even if it
is higher tens of mega ohms okay or at least
147
00:17:47,660 --> 00:17:53,030
a mega ohm is that correct unless your current
is very high R0 cannot be reduced to kilo
148
00:17:53,030 --> 00:17:58,210
ohms if it is very high current and the power
will actually take care of your device failures
149
00:17:58,210 --> 00:18:01,580
you do not worry on that okay.
150
00:18:01,580 --> 00:18:11,310
So if I R0 very large what does conductance
is smaller or larger very small normally gm
151
00:18:11,310 --> 00:18:18,070
are the order of ten to millions per volt
gm are typically of the order of millions
152
00:18:18,070 --> 00:18:26,590
this go the order of microns go are the order
of microns is that clear, so I say gm2 is
153
00:18:26,590 --> 00:18:35,060
much larger than either 0 1 or go so I need
love go1 and go2 come the denominator same
154
00:18:35,060 --> 00:18:40,260
way I neglect go from the top is that correct.
155
00:18:40,260 --> 00:18:47,410
What is the condition I said since R0 are
very hard gm go are very small compared to
156
00:18:47,410 --> 00:18:56,270
trans conductance s which is typically one
order or 3 to 3 order higher therefore equivalently
157
00:18:56,270 --> 00:19:06,450
saying it is gm gm2 upon gm2, which is equal
to how much gm1 is that correct, so even by
158
00:19:06,450 --> 00:19:09,350
using your cost code how much trans conductance.
159
00:19:09,350 --> 00:19:19,410
I obtain same as equivalent of their normal
single stage amplifier transistor M1 if I
160
00:19:19,410 --> 00:19:26,210
were used only among Rgm1 with 2 in this also
I got equivalent of gm, so what is that if
161
00:19:26,210 --> 00:19:32,260
gm months are equal you can say gm is gm1=gm2
also if you wish and still it is valid for
162
00:19:32,260 --> 00:19:35,350
it if the transistor identically can always
use.
163
00:19:35,350 --> 00:19:41,200
Otherwise you can always get accurate value
please remember this is the net value I have,
164
00:19:41,200 --> 00:19:46,770
but in your life number y gm will be very
close to or sorry gm effective will be very
165
00:19:46,770 --> 00:19:57,020
close to gm1 so if you see this expression
again okay, I will come back first let us
166
00:19:57,020 --> 00:20:02,700
keep this and we will use it on bandwidth
part later, so what value we have calculated
167
00:20:02,700 --> 00:20:03,700
gm effective.
168
00:20:03,700 --> 00:20:09,310
What is the next value I must calculate Q
effective what is the condition in go effective
169
00:20:09,310 --> 00:20:20,309
I said range should be 0, so if I say Vin
it 0 that is the condition you wondered GN
170
00:20:20,309 --> 00:20:32,110
should be 0, then I said you are effective
is 1/V0 when Vin is 0 is that circuit okay
171
00:20:32,110 --> 00:20:35,630
input its grounded that is what you said being
a 0.
172
00:20:35,630 --> 00:20:42,040
So I grounded it I use the same equations
of current again for this as well as for this
173
00:20:42,040 --> 00:20:54,870
for the second transistor I is equal to -gm
to VO1 okay what is videos for this 0-VO1
174
00:20:54,870 --> 00:21:07,210
please remember Vgs is how much 0-VO1 that
is this –VO1 is appearing here, so the first
175
00:21:07,210 --> 00:21:25,020
is –gm to VO1+gm2 x VDS how much is VDS
VO-VO-VO1 so go2 is okay I made a mistake
176
00:21:25,020 --> 00:21:27,590
so I rewrote VO is VO2.
177
00:21:27,590 --> 00:21:41,840
Which is actually so VO-VO1 this is another
equation now if I write for M 1 I had it on
178
00:21:41,840 --> 00:21:57,730
it for M2 now I write for M1 current are in
this is gm1 x 0YVin is grounded so gm1 x 0+go1
179
00:21:57,730 --> 00:22:10,870
x VDS,which is VO1, ok yes input signal ground
small signal every single parameter I am using
180
00:22:10,870 --> 00:22:13,530
right now a small signal.
181
00:22:13,530 --> 00:22:23,100
So from this equation VO1 is R/y go1 from
this equation pi VO1 is I/go1 substitute this
182
00:22:23,100 --> 00:22:36,100
VO1 in this expression like yeah from five
this time I by VO1 Rgo1 is VO one use this
183
00:22:36,100 --> 00:22:42,800
VO one value here and give new equations.
Now is that okay all that I note from the
184
00:22:42,800 --> 00:22:50,070
circuit five equation I calculate VO1 and
sorry substitute back in equal s the biasing
185
00:22:50,070 --> 00:22:57,140
DC bias is coming from I source and nothing
to do with AC signals is that clear to you
186
00:22:57,140 --> 00:23:00,460
M1 M2 are biased.
187
00:23:00,460 --> 00:23:06,650
I am not saying this is Vgs is biasing greater
than VT small which have nothing to do with
188
00:23:06,650 --> 00:23:14,260
bias is that correct this is signal what is
input to this signal is for him to 0-VO1 is
189
00:23:14,260 --> 00:23:20,740
the Vgs power that is the signal it is seen
it is that clear it is a AC signal is that
190
00:23:20,740 --> 00:23:32,670
clear 10 month AC signal is grounded but bias
is not removed bias is coming from I.
191
00:23:32,670 --> 00:23:43,270
I have a little told you if I=beta x beta/2Vgs-VT
square if I keep fixed I am keeping Vgs-VT
192
00:23:43,270 --> 00:23:50,730
fixed which is biasing it for saturation is
that clear I repeat you are not that my issue
193
00:23:50,730 --> 00:24:02,059
anytime I say I is fixed please look at it
the way I say it knowing once for all capital
194
00:24:02,059 --> 00:24:10,160
I is if the device is in saturation assuming
lambda 0 right.
195
00:24:10,160 --> 00:24:26,590
Now so if I is fixed Vgs-VT is fixed we make
i sufficiently such that Vrs-VT is always
196
00:24:26,590 --> 00:24:37,650
less than VDS choice of i is such that Vgs-VT
is smaller than VD s so when I fixed active
197
00:24:37,650 --> 00:24:45,250
load which is current source I am actually
biasing the device in saturation the choice
198
00:24:45,250 --> 00:24:54,080
of I is mine is that correct choice of I is
mine which will fix my bias for any transistor
199
00:24:54,080 --> 00:24:59,370
is that correct same current is passing in
and one same current is passing in M2.
200
00:24:59,370 --> 00:25:06,420
So both will be up till saturation is that
correct Vgs-VT for both should be smaller
201
00:25:06,420 --> 00:25:13,500
than videos for each transistor is that clear
that I am adjusting externally which is called
202
00:25:13,500 --> 00:25:19,360
bias current okay so all that analysis is
clear all that we are talking is AC analysis
203
00:25:19,360 --> 00:25:24,610
and not DC and all DC we are assuming that
I am keeping both b of the first one.
204
00:25:24,610 --> 00:25:32,480
I showed you is so just hope that M1 and M2
remains in saturation that I externally I
205
00:25:32,480 --> 00:25:41,150
did it is that correct I am NOT saying that
if the voltage goes to a fixed value it does
206
00:25:41,150 --> 00:25:46,550
not pass a current who said you will do-the
effect still I can be adjusted all let us
207
00:25:46,550 --> 00:25:52,980
say as if pure AC is made I am only looking
for VOA see now your question is if I make
208
00:25:52,980 --> 00:25:58,360
ground itself there then the whole of all
current will actually go to the ground.
209
00:25:58,360 --> 00:26:03,300
So I must put some potential bear so that
current can keep flowing is that correct otherwise
210
00:26:03,300 --> 00:26:09,670
all of current will be grounded I want a she
ground but I do not want DC ground there is
211
00:26:09,670 --> 00:26:14,610
that correct and that is why I said I put
some voltage there so that the whole current
212
00:26:14,610 --> 00:26:20,110
does not short-circuit out it should it will
not go through M1 M2 then is that clear that
213
00:26:20,110 --> 00:26:24,980
is why that condition was me is that clear
to you why that voltages are good otherwise.
214
00:26:24,980 --> 00:26:31,280
If I physically ground that then all of I
will actually go to the ground physical ground
215
00:26:31,280 --> 00:26:38,160
so no m1 m2 will be on is that correct so
these conditions have to be made so assumptions
216
00:26:38,160 --> 00:26:45,760
enthralled we are saying which arrow I do
M1 M2 will remain in saturation this is DC
217
00:26:45,760 --> 00:27:00,470
biasing nothing to do with AC signals okay
AC I am calculating but that is DC that decides
218
00:27:00,470 --> 00:27:05,670
gm to beta I under root of that decides gm.
219
00:27:05,670 --> 00:27:14,050
So whenever I fix I I am fixing GM is that
clear to you I am as far as please get confused
220
00:27:14,050 --> 00:27:21,280
on this AC signals theory is based on values
of DC you fix is that correct that is what
221
00:27:21,280 --> 00:27:30,640
we keep saying RA=1 upon lambda x I whatever
I fix gm=under root beta x I so once fixed
222
00:27:30,640 --> 00:27:41,720
I I fix both gm and R0S but that is a DC current
yeah why but that is what is he what is there
223
00:27:41,720 --> 00:27:46,150
is he at any point is the slope if we calculate
what is the area of small signal.
224
00:27:46,150 --> 00:27:51,760
We discussed all these days that at that point
the variation is V near enough that we do
225
00:27:51,760 --> 00:27:59,200
not have to say major value changes DC value
does not change every does IDC+ small idc
226
00:27:59,200 --> 00:28:06,360
will come but that current is how much so
it is I am putting 5 milliamps of 1 million
227
00:28:06,360 --> 00:28:14,720
and this is microns thousand times less or
even lower is that clear that on any small
228
00:28:14,720 --> 00:28:15,720
signal analysis.
229
00:28:15,720 --> 00:28:20,660
This value is always going to change you cannot
say star but how do I calculate slopes if
230
00:28:20,660 --> 00:28:27,280
anything I hope this value and I get a slope
here and I say yeah within the small change
231
00:28:27,280 --> 00:28:35,120
I hope yes 2 values are actually being changing
there is nothing I can do about but this value
232
00:28:35,120 --> 00:28:41,480
being what are the change here and here is
negligible that is what a Cs are all about
233
00:28:41,480 --> 00:28:49,450
correct it is always possible power supply
will adjust to that small AC variation will
234
00:28:49,450 --> 00:28:56,110
over it all the time on DC okay always that
is what where and when we want to get output.
235
00:28:56,110 --> 00:29:02,480
We remove that DC part by putting a capacitor
so that the DC does not go to the output is
236
00:29:02,480 --> 00:29:09,110
that clear that is what we have been doing
all through so far is that okay okay okay
237
00:29:09,110 --> 00:29:14,960
sorry he is right and I think I must thank
him for getting clarifications otherwise I
238
00:29:14,960 --> 00:29:22,040
will be doing and doing and then everyone
will confuse myself and yourself okay.
239
00:29:22,040 --> 00:29:29,560
So if I substitute VO one into our second
whatever number of that equation I have 3
240
00:29:29,560 --> 00:29:40,040
or 2 equation 2 then I rewrite is gm2 I/y01
0 2 V0 0 2-I by is that clear mam that just
241
00:29:40,040 --> 00:29:46,900
substitute VO 1 in upper equation and you
get this equation I forgot to I forgot this
242
00:29:46,900 --> 00:29:49,090
term so I do you wrote again.
243
00:29:49,090 --> 00:29:57,790
So I is equal to collect items collect items
on the left the other terms is 0 to V0 so
244
00:29:57,790 --> 00:30:10,420
I some of this is 1+0 2+01 a +gm2 I is go2
V0 and then what is the definition of go effective
245
00:30:10,420 --> 00:30:20,240
I said I/V01 v NH this equations were derived
based on VNH 0 so go effective is I/V0 which
246
00:30:20,240 --> 00:30:32,330
is 0 2 upon 1 upon G o2/0 1+gm2/gm1 again
if I linear it j 1 so I get 0 1 0 2 upon 0
247
00:30:32,330 --> 00:30:35,410
1+0 2+GM–ok.
248
00:30:35,410 --> 00:30:42,220
Then if you wish I can write our effective
is go1+go2+gm2 upon go1u-R can be written
249
00:30:42,220 --> 00:30:53,960
as 1 upon goH-upon upon Gou+gm2 upon g1 0.2is
that ok what did I get I get auto effective
250
00:30:53,960 --> 00:30:59,390
is that correct I got Ro goH effect is 1 upon
ROh effective so I just got other effective
251
00:30:59,390 --> 00:31:07,990
which is 1 upon goH-what is 1 upon 0-Ro 2
1 upon goH 1 is Ro1.
252
00:31:07,990 --> 00:31:25,570
Now if I do this you can see from here our
effective is ro1+ro2 this 1 upon go1 I made
253
00:31:25,570 --> 00:31:39,960
it ro1 x gm2/go2or gm2 ro2 also you can write
okay but what is gm2/0 to r gm-ro-a v2 so
254
00:31:39,960 --> 00:31:50,920
I write our effect it is ro1+ro2 a VT x ra1
is that correct if AV is large enough.
255
00:31:50,920 --> 00:32:00,780
How I make a V launch by increasing the W
pile of those transistors or by increasing
256
00:32:00,780 --> 00:32:10,130
the currents okay if I increase sufficiently
a V which I can then ro effective eights how
257
00:32:10,130 --> 00:32:21,560
much 1+AV times R 1+ro2 if the gain is say
50 or 100 how much is other effective 100
258
00:32:21,560 --> 00:32:32,010
times ra1 if I do not put M1 M2 how much would
have been output resistance ro1 if I put M-how
259
00:32:32,010 --> 00:32:38,460
much I got gain tines ro1 got is that correct
+ro of course this plus term is there but
260
00:32:38,460 --> 00:32:40,510
this is much higher.
261
00:32:40,510 --> 00:32:50,880
So what I have boosted the output resistance
got boosted gain times in this cos code then
262
00:32:50,880 --> 00:33:02,080
we say a V effective is gm effective ro effect
essentially so I get gm1 x ro2+1+AV1 if I
263
00:33:02,080 --> 00:33:10,890
expand this well please note down this AV
effect in is gm effective upon go effective
264
00:33:10,890 --> 00:33:18,150
is-gm1 I wrote this I put here ro2+1+AV2R1.
265
00:33:18,150 --> 00:33:28,830
Then I say this can be written as gm1 ro1
is AV1 gm this is our route assuming gm1=gm-everything
266
00:33:28,830 --> 00:33:37,200
is same then the effect will be equal to AVS
square if G ends are equal that is sizes are
267
00:33:37,200 --> 00:33:44,960
equal same current is flowing will be equal
so any effect you will be equal to AV1 square
268
00:33:44,960 --> 00:33:54,300
so what did I move gain is that clear gain
I boosted how much square AV square.
269
00:33:54,300 --> 00:34:04,020
How can I have boosted my gain up any amplifier
let us say I have 2 such amplifiers A1 and
270
00:34:04,020 --> 00:34:13,290
A2 then what would have been if this is my
way in this would have been also output of
271
00:34:13,290 --> 00:34:22,659
this VO1 would I become input for this and
next gain times it would have come like this
272
00:34:22,659 --> 00:34:23,659
okay.
273
00:34:23,659 --> 00:34:32,760
If they are equal I still could have got the
NEB gain is a square this is called cascade
274
00:34:32,760 --> 00:34:40,880
output of first is given to the input of next
stage so gain boosting is not very difficult
275
00:34:40,880 --> 00:34:47,080
I could have put doing series like this cascade
them I have got the boost of the grids okay.
276
00:34:47,080 --> 00:34:51,590
So what did I achieve out of that that is
what the cost code is different what is that
277
00:34:51,590 --> 00:35:00,440
we ask you please remember gain boosting is
not the only thing we achieved one day to
278
00:35:00,440 --> 00:35:06,730
be held constant in that gm we even with cascade
a cascade we maintain same gm gm1 gm effective
279
00:35:06,730 --> 00:35:08,220
okay.
280
00:35:08,220 --> 00:35:18,360
Because of that thing one can see the gain-bandwidth
product is gn effective/CL gm/C because gm
281
00:35:18,360 --> 00:35:30,310
is same gm1 is same is that correct, so since
if all gm are equal one can say gain of a
282
00:35:30,310 --> 00:35:40,280
cascade is gain single stage square but the
bandwidth remains same is this one impossible
283
00:35:40,280 --> 00:35:46,030
in cascade no because full system would have
gain-bandwidth constant.
284
00:35:46,030 --> 00:35:52,540
So if I increase gain what could I lost in
that the bandwidth so cascade stages actually
285
00:35:52,540 --> 00:36:02,050
improves gain but lose on okay, so this has
have you got the point advantage of cascade
286
00:36:02,050 --> 00:36:09,650
or cascade that the cascade amplifiers boost
gains at the cost of bandits COS coves boost
287
00:36:09,650 --> 00:36:17,450
the gain without losing the bandwidth is that
but what is the penalty I am paying out of
288
00:36:17,450 --> 00:36:25,650
all of it I may require larger power supply
voltages is that correct at the cost of VDD.
289
00:36:25,650 --> 00:36:32,630
I achieved this is that clear so is that point
clear so what is the advantage of COS forward
290
00:36:32,630 --> 00:36:40,640
cross coal is a very interesting device that
it allows you to boost output resistance without
291
00:36:40,640 --> 00:36:47,290
changing the trans conductance okay and that
is something, which no one else can achieve
292
00:36:47,290 --> 00:36:53,680
this is normally not done so much in discrete
amplifiers this is always done in integrated
293
00:36:53,680 --> 00:36:58,530
circuits because that is much easier to adjust
that gm1 um 2 identical others everything
294
00:36:58,530 --> 00:37:02,920
is possible on single chip to different devices.
295
00:37:02,920 --> 00:37:07,860
If you take on a breadboard they will never
be identical try any number take thousand
296
00:37:07,860 --> 00:37:15,390
of such devices whether RVs and series of
them nothing will be equal okay, so any discrete
297
00:37:15,390 --> 00:37:21,710
such experiments are very difficult to through
sometimes but in case of ICS since the areas
298
00:37:21,710 --> 00:37:27,580
are very small everything is almost identical
therefore divided circuit used almost all
299
00:37:27,580 --> 00:37:28,810
such tricks more.
300
00:37:28,810 --> 00:37:36,350
So in analog I see the on mixed signal ICS
since future is only on I see it is not on
301
00:37:36,350 --> 00:37:41,710
discrete so why do we discrete because discrete
tell you which is the variable part IC okay
302
00:37:41,710 --> 00:37:48,560
unless you know discrete you can do I see
today I showed you same methods can be extended
303
00:37:48,560 --> 00:37:52,180
for IC designs is that here okay.
304
00:37:52,180 --> 00:38:00,740
The next chapter for our course is very important
frequency response of amplifiers now R assumption,
305
00:38:00,740 --> 00:38:13,750
so far that there is a frequency of signal
which device does not bother I had amplified
306
00:38:13,750 --> 00:38:21,450
output at the same frequency please take my
word what I am saying let us take an amplifier
307
00:38:21,450 --> 00:38:26,990
and I have an input signal at 20 kilohertz
okay and again off 300.
308
00:38:26,990 --> 00:38:35,430
So I know one minute of 20 kilohertz will
become 100 milli volts at 20 kilohertz at
309
00:38:35,430 --> 00:38:43,440
the output let us say I change that 20 kilo
Hertz to 50 kilo Hertz I still believe that
310
00:38:43,440 --> 00:38:50,470
it may still come, so many milli volts at
50 kilohertz if I increase to mega Hertz,
311
00:38:50,470 --> 00:38:57,540
then I am not very sure whether the same hundred
gain will remain for that on amplifier even
312
00:38:57,540 --> 00:39:05,590
at 100 one mega order about which means the
gain is also not only a function of v-0 of
313
00:39:05,590 --> 00:39:10,300
a normal magnitude wise but also is a function
of frequency.
314
00:39:10,300 --> 00:39:17,850
So I must for an amplifier what is the condition
I must have I must know the region of frequency
315
00:39:17,850 --> 00:39:27,010
band as we call say f1 to f2 in which gain
is constant so that I know any frequency in
316
00:39:27,010 --> 00:39:35,990
this range will give amplified normally but
beyond that the gain may not be same or before
317
00:39:35,990 --> 00:39:41,590
that f1 also cannot be safe, what do you want
is that clear this is what we want to know
318
00:39:41,590 --> 00:39:46,520
this is essentially called the bandwidth or
mid band frequencies of an amplifier.
319
00:39:46,520 --> 00:39:51,960
I want to know where is the mid band what
is mid band where the gain remains constant,
320
00:39:51,960 --> 00:40:00,260
so first thing to notice I must now first
do a mathematics to some extent and see how
321
00:40:00,260 --> 00:40:07,440
do I evaluate the frequency response of any
okay amplifier is a system which has some
322
00:40:07,440 --> 00:40:13,790
input some output any system has input and
an output why I am saying this generalized
323
00:40:13,790 --> 00:40:14,790
because this theory.
324
00:40:14,790 --> 00:40:19,560
Which we are talking has nothing to do with
electrical networks it can be applied to any
325
00:40:19,560 --> 00:40:25,230
energy mechanical or any kind okay this is
what control system is all about is that correct
326
00:40:25,230 --> 00:40:30,910
we are only looking electrical signals but
this is true for any other signals as well
327
00:40:30,910 --> 00:40:36,930
okay so let us start with our electrical for
example I hope that in your math course if
328
00:40:36,930 --> 00:40:42,300
not so far must have done Laplace transforms
if you are not done it do it.
329
00:40:42,300 --> 00:40:51,640
So we do J Omega is what we call s which is
called s domain analysis and if you are a
330
00:40:51,640 --> 00:40:56,440
system which is shown here any typical system
which can be elliptical in our case then it
331
00:40:56,440 --> 00:41:08,950
has then which is a function of frequency
that means frequency can se J Omega so Omega
332
00:41:08,950 --> 00:41:14,840
changes, please remember anything which is
return is only right.
333
00:41:14,840 --> 00:41:20,700
Now imaginary quantity but there will be also
real quantity s is equal to Sigma plus J is
334
00:41:20,700 --> 00:41:25,752
that clear there will be a real quantity plus
imaginary right now I am assuming everything
335
00:41:25,752 --> 00:41:33,120
is imagined but in future in their log we
will use that real quantities also there will
336
00:41:33,120 --> 00:41:39,270
be some output and please remember there can
it many possibilities with voltage input voltage
337
00:41:39,270 --> 00:41:47,120
output voltage input current output current
input current output current input voltage
338
00:41:47,120 --> 00:41:53,390
for possible outputs for possible inputs.
339
00:41:53,390 --> 00:42:03,030
This is called ratio of output to input function
is called transfer function ratio of output
340
00:42:03,030 --> 00:42:08,820
to input is called transfer function so how
many kinds of transfer function in normal
341
00:42:08,820 --> 00:42:16,340
this system could be four kinds voltage by
voltage will be voltage gain voltage gain
342
00:42:16,340 --> 00:42:23,160
and current outdistance conduct gain current
is input and output is current gains current
343
00:42:23,160 --> 00:42:27,410
is input a output is trans resistance gains
is that correct.
344
00:42:27,410 --> 00:42:32,780
So for possible gains can be possible and
there can be for therefore different we are
345
00:42:32,780 --> 00:42:37,900
right now using voltage transfer function
why because in most cases amplifiers we are
346
00:42:37,900 --> 00:42:44,560
using r voltage amplifiers but later we will
see it can be also any one of them as well
347
00:42:44,560 --> 00:42:52,080
the typical transfer function of a network
shown here is HS or which is a J Omega is
348
00:42:52,080 --> 00:43:02,290
V0S upon Vin s and can be written as arbitrary
in this form K is constant independent of
349
00:43:02,290 --> 00:43:04,220
frequency in numerator.
350
00:43:04,220 --> 00:43:12,060
You have a series of terms which is s-Z 1
x s-Z2 into s-M/ s-P1 into S-V2 x S-PN these
351
00:43:12,060 --> 00:43:19,620
Z is called 0 of the function jets are called
zeros of the function and peas are called
352
00:43:19,620 --> 00:43:30,820
poles of the function
is that correct this is nothing to do it network
353
00:43:30,820 --> 00:43:38,740
per se as I repeat this is true for any transfer
functions why are we looking into this because
354
00:43:38,740 --> 00:43:40,530
let us say okay.
355
00:43:40,530 --> 00:43:48,560
Maybe there is another slide here or just
from here if s is equal to HS how much is
356
00:43:48,560 --> 00:43:59,276
a change if s is equal to HS how much is it
is infinite it just goes to infinity if s
357
00:43:59,276 --> 00:44:05,750
is equal to Z1 or there toward it how much
is a j0 so plane is ascension is saying the
358
00:44:05,750 --> 00:44:11,720
transfer function value goes to infinite and
0 means in both transmission value goes to
359
00:44:11,720 --> 00:44:15,540
0 this is magnitude wise what we are trying
to say is that layer.
360
00:44:15,540 --> 00:44:22,610
So, poles and zeros are those values of frequencies
at which the transfer function may become
361
00:44:22,610 --> 00:44:30,050
0 or may become or turn to infinite is that
clear that is exactly what mathematically
362
00:44:30,050 --> 00:44:35,390
we are saying how does it circuit wise we
see is very important for us because there
363
00:44:35,390 --> 00:44:41,140
is where we are worried about is that general
idea of transfer function here any and please
364
00:44:41,140 --> 00:44:43,000
remember for there in our system.
365
00:44:43,000 --> 00:44:48,510
We may use I either of the four in actual
amplifiers I may use a current amplifier then
366
00:44:48,510 --> 00:44:57,480
what is the output will be I0 and input will
become I n so there will be AI AV r I R and
367
00:44:57,480 --> 00:45:03,470
G both possible in this case is that clear
so we only write now look for voltage but
368
00:45:03,470 --> 00:45:14,260
do not worry it is true for all kinds of transformations
here is a simple circuit okay there is an
369
00:45:14,260 --> 00:45:19,290
input being a see there is a city search series
RS.
370
00:45:19,290 --> 00:45:26,400
Of course, since there is no mass transistor
RS with the source resistance there is a capacitance
371
00:45:26,400 --> 00:45:34,550
of C which has an impedance of 1 upon C s
and there is a load which is RL and let us
372
00:45:34,550 --> 00:45:43,690
say by Kirchhoff's law if the mesh current
is is RS x RM is v-0 is a function of s is
373
00:45:43,690 --> 00:45:58,350
that correct V0 is the current in this mesh
or loop so I x RL is v-0 but how much is there
374
00:45:58,350 --> 00:46:02,600
is only one voltage source then divided by
the net impedance.
375
00:46:02,600 --> 00:46:14,210
This Plus this Plus this are s upon RS/1 upon
CS+RL is the net impedance in the loop series
376
00:46:14,210 --> 00:46:24,990
implements so being upon this is rewrite in
VN x CS upon RS+RL CS+1 knots nothing great
377
00:46:24,990 --> 00:46:34,010
just take CF like this and bring CS about
okay please remember the not seized up as
378
00:46:34,010 --> 00:46:39,720
if something see small as is the Laplace transform
s J Omega term okay.
379
00:46:39,720 --> 00:46:53,450
You can now J Omega C as well if you wish
so V0S upon V in H is RLC h upon RL+RS x C
380
00:46:53,450 --> 00:47:02,330
s+1 this is the transfer function of this
simple what is come where is the capacitor
381
00:47:02,330 --> 00:47:09,140
in this circuit it is in series to the source
and the output where could have been otherwise
382
00:47:09,140 --> 00:47:13,310
across RL also I could have a capacitor it
would be called parallel capacitance.
383
00:47:13,310 --> 00:47:27,430
This is called series capacitance for namesake
okay is that okay obviously if the frequency
384
00:47:27,430 --> 00:47:36,190
0 s is 0 you can see what is the transmission
RC s in the numerator if s is 0 HS it is 0
385
00:47:36,190 --> 00:47:47,140
so there is A0 at frequencies you know okay
soA0 exists at Omega equal to 0 or F=0 however
386
00:47:47,140 --> 00:47:56,720
if you look at the transmission again okay
this to become 0 RL+SC+1.
387
00:47:56,720 --> 00:48:05,600
Then, we can say RL+RS CSE Quilly-1 at that
time HS will become infinite and this-sign
388
00:48:05,600 --> 00:48:10,560
is avoided in character is always on the left
half plane but just for the sake of it is
389
00:48:10,560 --> 00:48:19,360
1 upon RL plus RS x C and this is it is called
is the pole for this transfer function please
390
00:48:19,360 --> 00:48:26,930
take it what is RE called RM to see time constant
is that correct.
391
00:48:26,930 --> 00:48:42,250
So I define RL plus RS times C as the time
constant tau s as the time constant this then
392
00:48:42,250 --> 00:48:51,270
is that okay zero exists at zero and pole
exists at one upon RC of this value of n upon
393
00:48:51,270 --> 00:48:59,930
tau s however we can rewrite the transfer
function again H is zero please look at your
394
00:48:59,930 --> 00:49:09,310
function V zero is upon Venus I use that RL
plus RS I bring it outside.
395
00:49:09,310 --> 00:49:17,410
So I get RL upon RL+RS and I am multiplied
by RL+RS these are symptoms okay I just multiply
396
00:49:17,410 --> 00:49:28,290
and you do not divide so I get RL upon RL+RS
is RL+RS upon 1+ and use RL+RS see as tau
397
00:49:28,290 --> 00:49:42,960
and this RL upon RL+R is SK then HS is K x
tau is S+1+tau s is that correct all that
398
00:49:42,960 --> 00:49:47,890
I did because on the numerator I do not have
a cow getting.
399
00:49:47,890 --> 00:49:57,050
So I multiplied RL plus RS and divide RL by
RL RL+RS so this also becomes tau s this is
400
00:49:57,050 --> 00:50:04,680
a constant why it is called constant it is
independent of frequency so it is called K
401
00:50:04,680 --> 00:50:13,369
tau S upon 1 plus tau S so we now see if it
is a vv voltage gain transfer function is
402
00:50:13,369 --> 00:50:21,490
a voltage gain then abs is our L upon RL+RS
if I write J Omega J Omega tau s upon 1+J
403
00:50:21,490 --> 00:50:24,680
Omega tau s is that correct.
404
00:50:24,680 --> 00:50:34,460
Now one can see now you are 3 terms here what
are the 3 terms you see one is this term the
405
00:50:34,460 --> 00:50:44,430
second is this term and third is 3 terms are
you sure your 3 terms independently can be
406
00:50:44,430 --> 00:50:56,390
seen one 2 and 3 what I am going to do I will
actually see response of each function with
407
00:50:56,390 --> 00:51:04,960
frequency initial and then what I say if they
are all together that is this one how will
408
00:51:04,960 --> 00:51:07,140
they look is that correct.
409
00:51:07,140 --> 00:51:13,960
I want to find variation of gain with frequency
that is my aim frequency response you said
410
00:51:13,960 --> 00:51:20,050
I want frequency response so I want a V as
a function of Omega I want to find so I made
411
00:51:20,050 --> 00:51:26,470
a trick to understand future I may not do
it but initially I say okay I see now 3 functions
412
00:51:26,470 --> 00:51:32,000
one of this the other is this is the numerator
this is in denominator.
413
00:51:32,000 --> 00:51:40,490
So I use my tricks and I say ok first I get
the magnitude I want how many quantities I
414
00:51:40,490 --> 00:51:47,500
should really find for any complex function
magnitude and phase so first derive the phase
415
00:51:47,500 --> 00:51:56,080
I want to know the magnitude of the gain as
a function of frequency it is all L upon order
416
00:51:56,080 --> 00:52:05,300
plus RS Omega tau s upon 1+Omega squared this
is just a plus JB the magnitude is under root
417
00:52:05,300 --> 00:52:07,610
A square+d squared is that clear.
418
00:52:07,610 --> 00:52:17,950
So I just substituted the magnitude values
where Omega of course is 2 pi F ok so if I
419
00:52:17,950 --> 00:52:27,730
take a VF I can write RL upon on it plus RS
2 pi F tau s 1 plus 4 PI square F square tau
420
00:52:27,730 --> 00:52:35,530
square to the power half same function rewritten
in the form of frequency is that clear the
421
00:52:35,530 --> 00:52:41,000
function which I wrote here is same as this
only thing now written in 2 pi F forms ok.
422
00:52:41,000 --> 00:52:49,340
Omega is 2 pi now I start like taking this
is gain and gain can be expressed in some
423
00:52:49,340 --> 00:52:59,030
numbers which is called decibels is what is
the why we use decibel as a world because
424
00:52:59,030 --> 00:53:05,590
essentially the decibel was found for method
kind of noise or measurement of audio signals
425
00:53:05,590 --> 00:53:19,369
initially but they define when T log p2/p1
or rather if I put one meter is some function
426
00:53:19,369 --> 00:53:24,830
power in DB.
427
00:53:24,830 --> 00:53:30,660
But since in our case power can be IV and
I being same we can write oh sorry 10 I am
428
00:53:30,660 --> 00:53:40,900
sorry this will be v square by odd this will
be v square bar so it if I say gain voltage
429
00:53:40,900 --> 00:53:53,270
kind of this it is this is 10 square means
20 D log V2/V1 so AV in DB is 20 log V2/V1
430
00:53:53,270 --> 00:54:00,430
which is gain okay is that correct so if I
want to find any DB values all that I do is
431
00:54:00,430 --> 00:54:04,110
20 log 10 of magnitude of this value.
432
00:54:04,110 --> 00:54:15,040
Which is AV J Omega is that clear 20 log V2/V1
is the voltage in voltage gain in DB it is
433
00:54:15,040 --> 00:54:21,260
okay so am but V 2 by V 1 is the gain actually
what we have found so I actually write this
434
00:54:21,260 --> 00:54:28,460
function and star you know in long what is
the log a into B/C will be log a plus log
435
00:54:28,460 --> 00:54:41,000
B here long if lake yep yeah theme Transco
will occur there so the first term is 20 log
436
00:54:41,000 --> 00:54:50,990
RL upon RL+RF the second term will be 20 log
2 pi tau s and the third term is-20 log 1+2
437
00:54:50,990 --> 00:54:55,940
pi F tau a square or 4 pi square tau R square.
438
00:54:55,940 --> 00:55:05,850
So let us start plotting individual terms
please remember if I want this total what
439
00:55:05,850 --> 00:55:12,940
is the theory I am applying when I say like
this like this what is the theorem called
440
00:55:12,940 --> 00:55:19,880
this Plus this Plus this with a-what is it
called so for positions so I actually do individuals
441
00:55:19,880 --> 00:55:24,480
and then superimpose them to get the net AVF
Indy beads.
442
00:55:24,480 --> 00:55:36,150
So the first term which is 20 log RL by RS
I plot as a variation of frequency whether
443
00:55:36,150 --> 00:55:47,700
it will be less than 1 or more than 1 because
this is RL and the denominator is RD+RS so
444
00:55:47,700 --> 00:55:55,720
it will be less than 1 therefore-so I say
from 0 dB it will be some-value which is 20
445
00:55:55,720 --> 00:56:01,200
log RL+RS some value depending on the ratio
of RL and RL+RS.
446
00:56:01,200 --> 00:56:09,900
Please remember why because denominator is
higher than the numerator then this is always
447
00:56:09,900 --> 00:56:16,490
constant because no there is no term in this
which is frequency dependent so this remains
448
00:56:16,490 --> 00:56:22,310
constant is that correct why I do all this
simple things pause because I want to make
449
00:56:22,310 --> 00:56:27,490
it very clear that things are trivial in real
life okay.
450
00:56:27,490 --> 00:56:32,600
Unless I really do not get you know excess
or a analog for both bar here what is the
451
00:56:32,600 --> 00:56:39,850
value I repeat 20 log R l upon RL+RS has a
fixed value throughout any frequency it remains
452
00:56:39,850 --> 00:56:51,720
constant is that okay that okay now use the
second term the second term is 20 log to 5
453
00:56:51,720 --> 00:56:55,119
powers ok 20 log to power towers.
454
00:56:55,119 --> 00:57:08,420
So I say at F=1 upon 2 pi hours please take
the frequency F=2 pi tau s what is the value
455
00:57:08,420 --> 00:57:18,430
of bracketed term 2 pi F tau s is how much
1 how much is value of log 1 0 log 1 is 0
456
00:57:18,430 --> 00:57:28,790
so it is at 0 dB is that clear F is equal-1
upon 2 pi hours if that is equal to 1 so at
457
00:57:28,790 --> 00:57:35,510
F is equal to this frequency the gain is 0
dB this second term only I am saying not the
458
00:57:35,510 --> 00:57:46,000
net one only second term if I multiply it
by ten frequency that is if this becomes 10
459
00:57:46,000 --> 00:57:49,390
upon 2 pi tau s okay.
460
00:57:49,390 --> 00:57:57,920
Then what will be the value 2 pi tau s is
10 how much is locked in drop 10 is 1, so
461
00:57:57,920 --> 00:58:06,090
how many DBs 20 DBS so at 10 upon 2 pi powers
how much is the value of gain dysfunction
462
00:58:06,090 --> 00:58:18,780
20 dB is it okay 20 lakh 10h 20 DB by I take
another other side balance point 1 upon 2
463
00:58:18,780 --> 00:58:27,250
part of it that is 1/10 of that value so what
is the again 1/10 means how much means how
464
00:58:27,250 --> 00:58:40,020
many d beads-20 dbs is that man clear if 0.1
means one more 10 1/10 means log.
465
00:58:40,020 --> 00:58:49,410
So log 10 is 1/4-20 DB will occur at one tenth
of this frequency and 20 DB will occur when
466
00:58:49,410 --> 00:58:59,150
it is plus 10 times the frequency 10 times
this frequency you are 20 DB 10 times less
467
00:58:59,150 --> 00:59:09,010
is-20 DB so if you go from 110 to ten how
many slow how many how much gain DVD I increase
468
00:59:09,010 --> 00:59:17,630
40 DB in how many orders of frequency how
many orders to orders 10 and 10 okay so pearl
469
00:59:17,630 --> 00:59:22,880
10 which is called decayed how much TV we
are increasing 20 DB.
470
00:59:22,880 --> 00:59:31,810
So the slope of this function gain for the
second term is 20 DB per decade slopes are
471
00:59:31,810 --> 00:59:38,760
called 20 DB per decade is that kind of eyes
because they are very 10 times this 20 DB
472
00:59:38,760 --> 00:59:45,650
will come into picture okay 100 if you do
it how much it will be 40 DB if it is thousand
473
00:59:45,650 --> 00:59:52,990
it will be 60 DB is that clear so every times
you increase ten times another 20 DB will
474
00:59:52,990 --> 00:59:54,040
appear is that correct.
475
00:59:54,040 --> 01:00:02,250
So 20 DB per decade is the slope of this is
that correct by same logic if I do the third
476
01:00:02,250 --> 01:00:08,100
function which is 20-20 log under root of
1 plus 2 5 tower square up to when that F
477
01:00:08,100 --> 01:00:19,490
is equal to 2 pi tau s is equal to 1 how much
is the value of that value should be at this
478
01:00:19,490 --> 01:00:34,700
frequency-20 log off no no not 2 root 2 1+1
under root is that correct so under group
479
01:00:34,700 --> 01:00:41,120
which in numbers it is 3 DB-3 dB is that correct
this is-3 dB.
480
01:00:41,120 --> 01:00:49,619
So what is it showing that at Ft is equal
to 1 upon 2 pi tau s the gain has fallen from
481
01:00:49,619 --> 01:00:59,240
0 to 3 DB down at this this frequency please
remember initially how much of the game and
482
01:00:59,240 --> 01:01:06,450
f is smaller than 1 what is this spot I am
showing you when f upon 1 upon 2 pi is a bachata
483
01:01:06,450 --> 01:01:14,120
so 1 plus 0 as a for I so I am getting a 0
DB gain as I start increasing the frequency
484
01:01:14,120 --> 01:01:20,600
that 1 upon 2 pi is term coming into picture
and then we may start saying the gain will
485
01:01:20,600 --> 01:01:28,200
start reducing at at that frequency the gain
would have fallen by 3 DB-3 DB.
486
01:01:28,200 --> 01:01:36,160
Essentially 0 to-now this further down if
you say what will happen one may be negligible
487
01:01:36,160 --> 01:01:43,990
or this and again 20 log decade it will start
if you increase F 10 times again it will come
488
01:01:43,990 --> 01:01:54,500
the same way 20 DB per day so now you can
say 3 functions which ones first one is this
489
01:01:54,500 --> 01:01:59,730
second this one is this and this one is this.
490
01:01:59,730 --> 01:02:07,220
So I have 3 functions which I created and
I now plotted 3 different frequency response
491
01:02:07,220 --> 01:02:15,484
term for 3 terms but what is the importance
for us I want addition for this okay so if
492
01:02:15,484 --> 01:02:28,010
I add all of them together and this value
is 0 dB please take it for a sigh but we will
493
01:02:28,010 --> 01:02:40,890
save a function deck tonight for please remember
this is less than zero sum value okay from
494
01:02:40,890 --> 01:02:46,090
this frequency onward the gain is going towards
this is 0.
495
01:02:46,090 --> 01:02:56,790
Please remember if I I am adding now all 3
of them this Plus this and this is 0 okay
496
01:02:56,790 --> 01:03:06,190
so if I add all 3 there then what do I get
initially of course this may be our L upon-20
497
01:03:06,190 --> 01:03:16,150
this and it starts rising up to what value
it was rising towards 0 DB1/2 pi towers but
498
01:03:16,150 --> 01:03:21,760
actually that value is not this is zero but
this value from the third one is how much
499
01:03:21,760 --> 01:03:28,080
3 DB down at this point it is not 0 but us-3
dB.
500
01:03:28,080 --> 01:03:36,430
So it is somewhere-3 DB is that point clear
up to this frequency at this frequency how
501
01:03:36,430 --> 01:03:48,990
much is actual gain-3 dB one is zero the other
is-3 dB so this is-3 dB okay then I if you
502
01:03:48,990 --> 01:03:56,619
see further the gain starts falling and the
other one is gain starts please remember this
503
01:03:56,619 --> 01:04:00,930
is 20 DB per decade this is 20 DB per decade
down.
504
01:04:00,930 --> 01:04:09,760
So what is the sum total of the 3/2 this is
20 DB this is Plus this is-how much is constant
505
01:04:09,760 --> 01:04:22,300
zero so afterwards the function becomes constant
plus 20 you can see in math and geometry whites
506
01:04:22,300 --> 01:04:30,950
plus 20-20 is 0 0 ok per decade 20 debate
or lies 20 degree it will decrease so it essentially
507
01:04:30,950 --> 01:04:39,540
no its opposite yeah you are right and I should
have put instead of 0 but I showing that term
508
01:04:39,540 --> 01:04:40,540
is 0 small ok.
509
01:04:40,540 --> 01:04:46,790
You are right what is saying is correct I
should this term is not 0 what is saying is
510
01:04:46,790 --> 01:04:54,580
this term right now I lifted that you are
right I fully appreciate so what is it trying
511
01:04:54,580 --> 01:05:03,310
to show you this means please remember this
is the crux of it now this is the name which
512
01:05:03,310 --> 01:05:08,150
I am now going to give you this is the frequency
response of the a transfer function I used
513
01:05:08,150 --> 01:05:16,630
this is the 3 DB point and what is this point
is offering at where F=1 upon 2 pi hours.
514
01:05:16,630 --> 01:05:26,960
Beyond this frequency gain is becoming constant
gain is becoming constant if I draw 2 such
515
01:05:26,960 --> 01:05:34,550
lines one line straight array and the other
from the 3 DB point straight way like this
516
01:05:34,550 --> 01:05:42,510
this point essentially represent 1 upon 2
pi tau s this point ok this is called corner
517
01:05:42,510 --> 01:05:49,790
frequency this is called corner frequency
and actual curve is I some critically following
518
01:05:49,790 --> 01:05:52,700
these 2 curves is that correct I some totally.
519
01:05:52,700 --> 01:06:00,070
So this is your real curve this is the curve
which I am now modeling-is that clear this
520
01:06:00,070 --> 01:06:08,160
is the real value is that correct this is
what I am now saying equivalent to lines like
521
01:06:08,160 --> 01:06:16,869
this here it may be so this frequency is also
called 3 DB frequency or-in DB frequency see,
522
01:06:16,869 --> 01:06:31,300
but this is called corner frequency
this thinking that such 3d waypoints can be
523
01:06:31,300 --> 01:06:39,970
put sharply are like this was first suggested
in 1940 by the famous mathematician or other
524
01:06:39,970 --> 01:06:47,330
control system man all control system term
as people okay vice-a-versa all masked people
525
01:06:47,330 --> 01:06:49,030
can do controls.
526
01:06:49,030 --> 01:06:55,021
What is the name of that person beret okay
boy they are bored some people say I do not
527
01:06:55,021 --> 01:07:03,240
know I call it Buddha and yes I will call
first time suggested that to look at the response
528
01:07:03,240 --> 01:07:08,240
I must know the corner frequency because I
am interested in below which frequency gain
529
01:07:08,240 --> 01:07:15,850
is higher or lower or above it and actually
I may calculate by actual function okay 3
530
01:07:15,850 --> 01:07:20,540
DB down but I am NOT interested in the value
of that I am interested which is that frequency.
531
01:07:20,540 --> 01:07:26,330
Which is called the corner frequency therefore
that is substituting that essentially was
532
01:07:26,330 --> 01:07:34,450
first suggested by Boni and therefore all
frequency response plots were called bode
533
01:07:34,450 --> 01:07:43,990
plots of course this is only amplitude but
will also see next time frequency phase part
534
01:07:43,990 --> 01:07:49,420
of the frequency okay is that clear to you
is that clear.
535
01:07:49,420 --> 01:07:54,900
So why it was called Buddha this first part
is not what is people even now called this
536
01:07:54,900 --> 01:07:59,180
Buddha they kill, Buddha only suggested they
should have a corner frequency concept which
537
01:07:59,180 --> 01:08:04,190
was first time suggested by him because in
real life I want to know what is the frequency
538
01:08:04,190 --> 01:08:08,810
up to which gain I should not use or beyond
which I should use or a vice versa I only
539
01:08:08,810 --> 01:08:12,490
want that value the actual magnitude may be
3d.
540
01:08:12,490 --> 01:08:17,750
We know who cared I want that frequency is
that and I will be always away from this corner
541
01:08:17,750 --> 01:08:23,889
frequencies okay to be guaranteed Li on this
side on this side is that clear that is why
542
01:08:23,889 --> 01:08:33,720
these plots were called border so in future
we may not draw the actual cars we made always
543
01:08:33,720 --> 01:08:34,930
border plots okay.
544
01:08:34,930 --> 01:08:41,160
But, when you ask you to real you must do
as some tours on that both side for example
545
01:08:41,160 --> 01:08:49,300
something like this occurs so you know it
will be something like this is that clear
546
01:08:49,300 --> 01:08:55,401
it will be awesome to dig here and showing
there here okay that is actual values but
547
01:08:55,401 --> 01:09:01,180
they will only draw like this is that correct
this is how bad I thought and that is what
548
01:09:01,180 --> 01:09:03,880
the Buddhist diagrams are all about, see you
next time you.