1
00:01:05,840 --> 00:01:11,679
Last two classes we were discussing the p
n junction diode and now we move on to the
2
00:01:11,679 --> 00:01:18,679
next device in terms of complication, BJT
the bipolar junction transistor. The model
3
00:01:25,100 --> 00:01:32,100
which we shall discuss for the bipolar junction
transistor is the Ebers Moll model which is
4
00:01:33,310 --> 00:01:40,310
simple yet very useful model and it can be
applied in most cases and we shall be applying
5
00:01:44,509 --> 00:01:51,509
it also in our discussion on integrated circuits.
So
6
00:02:05,159 --> 00:02:10,929
Ebers Moll model of the bipolar junction transistor
considers the bipolar junction transistor
7
00:02:10,929 --> 00:02:17,929
which you know is a three terminal device
consisting of 2 p n junctions as two diodes
8
00:02:18,739 --> 00:02:22,510
connected back to back.
9
00:02:22,510 --> 00:02:29,510
If you have an n p n transistor, you have
one p n junction and the other p n junction
10
00:02:29,989 --> 00:02:36,989
like this, in addition you have two current
sources. These current sources gives us the
11
00:02:53,810 --> 00:03:00,810
coupling between the two junctions. So this
is the emitter terminal say this is the collector
12
00:03:02,959 --> 00:03:09,959
terminal and this is the base terminal. If
we call this current IF which is the forward
13
00:03:14,329 --> 00:03:21,329
current which is given by IES which is the
same diode relation which we are writing now
14
00:03:25,090 --> 00:03:32,090
and
15
00:03:37,269 --> 00:03:44,269
you
16
00:03:47,599 --> 00:03:51,129
have the other current IR.
17
00:03:51,129 --> 00:03:56,659
So this is the base collector junction, this
depends on the base collector voltage IR is
18
00:03:56,659 --> 00:04:03,659
equal to another constant
19
00:04:09,310 --> 00:04:12,989
and these current sources are actually the
coupling between the two junctions. So this
20
00:04:12,989 --> 00:04:19,989
is given by a constant into IF, the constant
is called alphaF and this current source is
21
00:04:20,969 --> 00:04:27,969
given by alphaR IR where alphaR is another
constant. So the idea of this model is that
22
00:04:30,280 --> 00:04:35,700
if you know the applied voltage between the
junctions. For example here you have two voltages
23
00:04:35,700 --> 00:04:41,300
emitter base and base collector, you should
be able to evaluate the different terminal
24
00:04:41,300 --> 00:04:46,960
currents. That is the emitter current which
in normal n p n transistor will be flowing
25
00:04:46,960 --> 00:04:53,960
in this way IE, you have the collector current
which flows this way and you have the base
26
00:04:56,360 --> 00:05:02,389
current which is this way. So we should be
able to evaluate the three terminal currents
27
00:05:02,389 --> 00:05:08,500
in terms of the two voltages that is the emitter
base and the base collector voltages.
28
00:05:08,500 --> 00:05:15,500
In order to do that now we see that we have
4 constants which are to be or 4 parameters
29
00:05:17,450 --> 00:05:24,450
which have to be known beforehand that is
IES, ICS, alphaF and alphaR. If you know these
30
00:05:27,580 --> 00:05:34,580
parameters you will be able to evaluate the
terminal currents. So now let us see what
31
00:05:36,620 --> 00:05:43,620
these parameters depend on, we shall go ahead
in developing this model. Just before we go
32
00:05:48,500 --> 00:05:55,500
ahead, if we take for example VBC =0 in this
transistor, what does it imply? IR is equal
33
00:06:13,560 --> 00:06:20,560
to zero, if vBC is equal to zero IR is equal
to zero. Then what do we have?
34
00:06:25,560 --> 00:06:32,560
We have IC is equal to alphaF IF because this
current is zero, IR is equal to zero so this
35
00:06:55,910 --> 00:07:02,910
current source is zero and IB is equal to
1- alphaF IF which gives us that if you take
36
00:07:15,650 --> 00:07:22,650
the ratio of the collector current to the
emitter current, it is equal to alphaF which
37
00:07:23,270 --> 00:07:30,270
is called the common base current gain and
if you take the ratio of the base current
38
00:07:30,379 --> 00:07:37,379
to the collector current, you have this IB
will be equal to alphaF by 1- alphaF.
39
00:07:43,099 --> 00:07:50,099
So I shall write it here, IC by IB is equal
to alphaF by 1- alphaF which is equal to betaF
40
00:07:56,009 --> 00:08:02,490
which is called the forward current gain.
So in this condition when IR is equal to zero
41
00:08:02,490 --> 00:08:09,490
or vBC is equal to zero all these terms. We
are in the forward motion of the operation
42
00:08:09,909 --> 00:08:16,909
and we have the betaF or the common emitter
current gain as ratio of collector current
43
00:08:18,229 --> 00:08:19,490
to the base current.
44
00:08:19,490 --> 00:08:26,490
You can also have the reverse mode of operation
in which case in reverse mode if vBE is equal
45
00:08:35,120 --> 00:08:42,120
to zero, you have IF is equal to zero and
similarly all IF if you put as equal to zero,
46
00:08:52,800 --> 00:08:59,800
you are operating in the reverse direction
so IE by IB will be equal to betaR. So this
47
00:09:08,579 --> 00:09:11,770
is called the reverse beta of the transistor.
48
00:09:11,770 --> 00:09:18,770
So you can have forward mode of conduction
and the reverse mode and if you consider that
49
00:09:19,300 --> 00:09:26,300
both the forward and the reverse current exists
so the normal mode that is vBE is not equal
50
00:09:28,970 --> 00:09:35,970
to zero, vBC is not equal to zero. What is
IC equal to? AlphaF IF – IR. What is IE
51
00:09:46,850 --> 00:09:53,850
equal to? IF – alphaR IR and IB will of
course be equal to this. So these are the
52
00:10:23,650 --> 00:10:30,650
different currents as given by this model.
53
00:10:31,480 --> 00:10:38,480
Now what we shall do is we
try to evaluate these currents and try to
get a simplified model. Basically what is
54
00:10:42,670 --> 00:10:49,670
done is we shall be able to see that instead
of four parameters which we now have, we shall
55
00:10:49,809 --> 00:10:56,809
be able to eliminate one of them and write
the equations in terms of three parameters.
56
00:10:56,860 --> 00:11:03,860
To do that now let us again take the transistor
57
00:11:26,040 --> 00:11:33,040
so this is the emitter, this is the base and
this is the collector. Suppose this is n p
58
00:11:39,169 --> 00:11:46,169
n transistor so suppose we have forward biased
this emitter base junction, now let us look
59
00:11:48,230 --> 00:11:53,459
at the charge profiles in the device. Now
if you have forward biased this junction so
60
00:11:53,459 --> 00:12:00,230
this is the depletion region. How will the
charge profile look like? This is the excess
61
00:12:00,230 --> 00:12:07,230
charge profile so draw it like this.
62
00:12:21,220 --> 00:12:28,220
We have already seen for a diode, if you have
a forward biased diode how the excess charge
63
00:12:31,890 --> 00:12:37,390
profile looks like. It is basically the same
thing, only thing we have considered the profile
64
00:12:37,390 --> 00:12:44,140
to be of short base diode. That is the profile
changes linearly which is true in all modern
65
00:12:44,140 --> 00:12:50,829
day devices because the actual size of the
device in all modern day integrated circuits
66
00:12:50,829 --> 00:12:57,829
would be much less than 10 microns even much
less in some cases, whereas the diffusion
67
00:13:01,350 --> 00:13:08,350
lengths are much larger. So we can always
assume that these profiles are linear. So
68
00:13:13,270 --> 00:13:20,270
this is the charge profile if you forward
bias the emitter base junction and this is
69
00:13:21,089 --> 00:13:28,089
the emitter base voltage vBE and then suppose
this base collector junction is also forward
70
00:13:33,819 --> 00:13:40,819
biased then what happens? You will have a
similar situation let me draw this like this.
71
00:14:04,030 --> 00:14:11,030
So this charge let me call this QE, this
is QBF the F actually stands for the forward
case that is when you have the emitter base
72
00:14:31,890 --> 00:14:38,890
junction applied voltage, we have seen that
you have two cases when you have the emitter
73
00:14:40,020 --> 00:14:46,789
base voltage applied and the base collector
is zero we call that the forward case, forward
74
00:14:46,789 --> 00:14:53,789
mode of operation and the reverse mode when
you have these charges, this is q c that is
75
00:15:01,059 --> 00:15:08,059
the collector charge and this component this
is the base charge in the reverse direction.
76
00:15:10,809 --> 00:15:17,809
So you have the base charge for the forward
case given by this, reverse case like this.
77
00:15:17,819 --> 00:15:24,789
Now if vBC is equal to zero then this reverse
charge in the base will not be existing, you
78
00:15:24,789 --> 00:15:31,789
just have the forward component and also if
vBC is negative that is the base collector
79
00:15:34,020 --> 00:15:41,020
is reverse biased this also would not be existing
because you know that the charge at this end
80
00:15:42,850 --> 00:15:48,870
is equal to the minority carrier concentration
thermal equilibrium value into exponential
81
00:15:48,870 --> 00:15:54,850
vBC by vT. vBC is negative this is actually
going to be negative so this is not going
82
00:15:54,850 --> 00:16:01,850
to be existing. So if you have both the cases
that is both the junctions are forward biased,
83
00:16:02,630 --> 00:16:07,079
the total charge will be the sum of these
two charges and basically you have a charge
84
00:16:07,079 --> 00:16:12,610
something like this.
85
00:16:12,610 --> 00:16:19,610
So now just let us take the first case that
is vBE not equal to zero, vBC is equal to
86
00:16:28,860 --> 00:16:35,860
zero. We can write IC is equal to, what is
IC equal to? It is the current which is flowing
87
00:16:46,240 --> 00:16:53,240
here. So we can write
which is electron current, this is diffusion
current and is given by the slope of this.
88
00:16:59,010 --> 00:17:03,860
This is almost a straight line which implies
that there is very little recombination in
89
00:17:03,860 --> 00:17:10,860
the base. So this is the equation for diffusion
current and this is the slope of the electron
90
00:17:15,760 --> 00:17:22,760
concentration in the base. So what is DN DX
here? It is the slope so it is this height
91
00:17:32,370 --> 00:17:39,370
divided by the base width. What is this height
equal to? This height is equal to I can write
92
00:17:45,610 --> 00:17:52,610
ni square by NA exponential vBE by vT -1 divided
by the base width. So let me write, so this
93
00:18:15,740 --> 00:18:22,740
is the collector current which I can write
as IS where IS equal to qDn by ni square by
94
00:18:58,409 --> 00:19:02,460
NA.
95
00:19:02,460 --> 00:19:09,460
So this is the collector current. Now what
is the base current? I will just write here
96
00:19:10,509 --> 00:19:17,509
so this is equal to, what is this equal to?
In this model what is the collector current
97
00:19:18,179 --> 00:19:25,179
equal to when vBC is equal to zero, this is
equal to alphaF IF. What is IB equal to? The
98
00:19:40,860 --> 00:19:47,860
IB or the base current in the bipolar transistor
here in this case would consist of two components
99
00:19:51,059 --> 00:19:58,059
when vBC is equal to zero. What are the two
components? One is the hole current which
100
00:19:58,529 --> 00:20:04,309
is injected from the base into the emitter
and the other is due to the recombination
101
00:20:04,309 --> 00:20:11,309
in the base so holes are flowing into the
emitter. So it has to be flowing in from this
102
00:20:12,000 --> 00:20:18,129
base terminal and also there is recombination.
that is electrons which flow from emitter
103
00:20:18,129 --> 00:20:24,360
to the collector part of it gets recombined
and recombined with holes so to compensate
104
00:20:24,360 --> 00:20:31,360
for that also holes must flow in. So basically
the base current consists of two components.
105
00:20:31,809 --> 00:20:38,809
Now what is the current which flows into the
emitter q Dp this is in the emitter Dp plus
106
00:20:58,049 --> 00:21:05,049
I should say q in the base. That is the charge
divided by the lifetime that is the total
107
00:21:20,350 --> 00:21:27,350
recombination current in the base. So this
q Dp dp dx what is that? It is this charge
108
00:21:33,690 --> 00:21:40,690
so this is the slope here of this line, this
charge divided by the emitter width by we
109
00:21:40,860 --> 00:21:47,860
which we call let us call the emitter width
we. So this is we, this is wb this is wc.
110
00:21:52,659 --> 00:21:59,659
So again what do you get ni square by NDE
in the emitter plus this component. What is
111
00:22:18,940 --> 00:22:25,940
the total charge in the base? It is given
by this area of this triangle. So what is
112
00:22:29,509 --> 00:22:36,509
the area of this triangle? Half base into
altitude so here you have we. So half, what
113
00:22:46,649 --> 00:22:53,649
is the base? Wb into again there will be a
charge, this altitude is again of course you
114
00:23:14,149 --> 00:23:21,149
can write it in a better fashion by taking
the exponential vBE by vt -1is common.
115
00:23:23,029 --> 00:23:30,029
Now the interesting thing is this is equal
to 1 – alphaf into If, Ib is equal to 1-
116
00:23:33,320 --> 00:23:40,320
alpha into IF. Now if you take betaF is equal
to IC by IB which is alphaF by 1 – alphaF.
117
00:23:48,230 --> 00:23:55,230
What do you get? IC by IB, so you have to
divide this by this expression. Now obviously
118
00:24:07,049 --> 00:24:12,710
that exponential vBE by vt -1 that gets cancelled
out on both sides, so what you are left with
119
00:24:12,710 --> 00:24:19,710
is if you write it in this fashion one by
q Dn ni square by NA just write down the expression,
120
00:24:32,690 --> 00:24:39,690
you can do that first to get this expression.
Now this expression tells us a lot of things,
121
00:25:06,570 --> 00:25:13,570
so for beta to be high what are the conditions?
What is Dn toun that is Ln square, that is
122
00:25:15,779 --> 00:25:17,679
the square of the diffusion length.
123
00:25:17,679 --> 00:25:24,679
So for high beta wb must be very much less
than Ln then NA must be very much less than
124
00:25:43,419 --> 00:25:50,419
NDE and
the base width must be very much less than
the diffusion length of the carriers that
125
00:26:02,340 --> 00:26:08,490
is quite understandable because if the base
width is quite large compared to the diffusion
126
00:26:08,490 --> 00:26:15,039
length lot of recombination will take place,
the beta will be low. The acceptor impurity
127
00:26:15,039 --> 00:26:22,039
concentration in the base must be less than
the donor concentration in the emitter because
128
00:26:22,649 --> 00:26:29,649
the base current is proportional to the, if
you look at the base current here it is inversely
129
00:26:32,779 --> 00:26:39,779
proportional to the donor concentration in
the emitter whereas the collector current
130
00:26:40,669 --> 00:26:47,100
is inversely proportional to the acceptor
concentration.
131
00:26:47,100 --> 00:26:54,100
For large beta the emitter concentration must
be larger than the base concentration and
132
00:26:54,610 --> 00:26:59,499
of course the base width must be very much
less than the emitter width because that determines
133
00:26:59,499 --> 00:27:06,499
the slope. So if the emitter length is smaller
the slope is going to be higher and you have
134
00:27:08,480 --> 00:27:15,480
larger base current flowing in. So that also
tells us, we always questioned that although
135
00:27:20,299 --> 00:27:27,299
it’s a n p n transistor why we cannot interchange
the collector and emitter terminals because
136
00:27:28,279 --> 00:27:35,279
the collector doping concentration is much
less than the doping concentration. This has
137
00:27:38,230 --> 00:27:42,230
to be made because you are applying a reverse
bias across that.
138
00:27:42,230 --> 00:27:48,570
normally you apply reverse bias across the
base collector junction and also because of
139
00:27:48,570 --> 00:27:55,570
breakdown reasons and all you have to make
the collector doping much less and so that
140
00:27:57,360 --> 00:28:02,369
the depletion region for the base collector
junction goes on the collector side not into
141
00:28:02,369 --> 00:28:09,369
the base side so that you don’t have punch
through condition. For this reason collector
142
00:28:09,730 --> 00:28:13,149
doping concentration is much less than the
base doping concentration.
143
00:28:13,149 --> 00:28:20,149
The collector length is also larger compared
to them, of course that is not a problem.
144
00:28:23,629 --> 00:28:30,629
So these are the conditions one must satisfy
anyway. So we shall go ahead, we have derived
145
00:28:34,929 --> 00:28:41,929
a relation for IS here. Now suppose we have
the other conditions that is vBC is equal
146
00:28:46,970 --> 00:28:53,970
to zero, I just remove this. Suppose vBC is
not equal to zero and vBE is equal to zero
147
00:29:10,269 --> 00:29:17,269
then what is it going to be like? The situation
if you calculate IE, which is you can take
148
00:29:26,629 --> 00:29:33,629
the modulus because the signs will be different,
this is given by now you have vBC not equal
149
00:29:37,509 --> 00:29:43,629
to zero, vBE is equal to zero, so this forward
charges are missing so you have a situation
150
00:29:43,629 --> 00:29:50,629
like this and the emitter current is actually
the current injected from the collector into
151
00:29:51,779 --> 00:29:56,149
the base which flows into the emitter.
152
00:29:56,149 --> 00:30:03,149
So it is just like the previous situation
of vBE not equal to zero and vBC equal to
153
00:30:03,440 --> 00:30:10,440
zero but you have just interchanged the emitter
and the collector terminals. So now you have
154
00:30:33,059 --> 00:30:40,059
again q Dn dn dx and this one will give you
the similar relation. What is dn dx? This
155
00:30:48,559 --> 00:30:55,559
is the same ni square by NA by wb exponential
vBC by vt -1 because this voltage now is the
156
00:31:16,110 --> 00:31:23,110
base to the collector voltage and what is
this term? It is the same IS, so we can write
157
00:31:27,509 --> 00:31:34,509
it as IS exponential. Now if you look at these
expressions alphaR IR is equal to IS exponential
158
00:31:58,190 --> 00:32:01,759
vBC by vt -1.
159
00:32:01,759 --> 00:32:08,759
Now what is IR? If you look at the first figure
here if you come back here, see IR is equal
160
00:32:14,720 --> 00:32:21,720
to ICS exponential vBC by vt - 1. Now if you
replace IR there with ICS exponential vBC
161
00:32:25,940 --> 00:32:32,940
by vt - 1 and similarly if you have alphaF
IF is equal to IF exponential vBE by vt - 1.
162
00:32:39,019 --> 00:32:46,019
So if you again in that IF if you replace
IES exponential vBE by vt - 1, what do you
163
00:32:46,840 --> 00:32:53,840
get? You will get alphaF IES is equal to alphaR
ICS.
164
00:33:08,279 --> 00:33:15,279
So basically what you have done is you see
there is a relation between this IES and ICS
165
00:33:18,389 --> 00:33:25,389
through this alphaR and alphaF. So basically
what it means that you can eliminate one of
166
00:33:29,289 --> 00:33:36,289
these four parameters because you can write
one of these parameters in terms of the other
167
00:33:37,649 --> 00:33:44,649
three and that can be eliminated. Basically
you require three parameters. So what we do
168
00:33:45,070 --> 00:33:52,070
is we use this parameter IS and we write the
expressions in terms of alphaF, alphaR and
169
00:33:57,149 --> 00:34:04,149
IS and IS is related through this expression
alphaR IR is equal to IS exponential vBC by
170
00:34:07,249 --> 00:34:14,249
vt -1 and similarly alphaF IF is equal to
IS exponential vBE by vt -1. So that is the
171
00:34:18,940 --> 00:34:19,379
thing.
172
00:34:19,379 --> 00:34:26,379
Now the situation where you have both the
voltages applied that is the most common situation
173
00:34:29,369 --> 00:34:36,369
that is both the emitter base voltage as well
as the base collector voltage is applied then
174
00:34:39,570 --> 00:34:45,690
what happens is the charge in the base or
in the different regions is considered to
175
00:34:45,690 --> 00:34:52,690
be a superposition of the two individual cases.
So you have this charge plus the reverse charge
176
00:34:53,909 --> 00:35:00,909
which is the sum is going to be something
like this. What happens to the slope here?
177
00:35:03,420 --> 00:35:09,640
The slope has decreased so what do you think
that what should happen to the collector current?
178
00:35:09,640 --> 00:35:15,380
The collector current should decrease, isn’t
it? So if you are forward biasing the collector
179
00:35:15,380 --> 00:35:22,380
base junction, this end the charges are going
to go up so the slope is going to reduce.
180
00:35:23,790 --> 00:35:30,790
The collector current reduces, also the total
charge stored in the base is going to increase
181
00:35:30,940 --> 00:35:37,940
which is going to increase the base current,
so the beta is going to fall. So what we have
182
00:35:41,540 --> 00:35:48,540
done now is individually the two cases, you
have calculated the collector current and
183
00:35:53,350 --> 00:35:58,380
the different currents you can do that. IC
is equal to IS exponential vBE by vt -1 which
184
00:35:58,380 --> 00:36:05,380
is equal to alphaF IF. Now alphaR IR is equal
to IS exponential vBC by vt -1, now if both
185
00:36:13,130 --> 00:36:20,130
the voltages are applied the collector current
is given by alphaF IF – IR.
186
00:36:24,590 --> 00:36:31,590
So you can express that now, alphaF IF is
IS exponential vBE by vt -1and IR is IS by
187
00:36:37,690 --> 00:36:44,690
alphaR exponential vBC by vt -1. Similarly
because you can now express the different
188
00:36:50,180 --> 00:36:57,180
currents in terms of IS alphaR and alphaF,
these are now the three parameters nowhere
189
00:37:00,750 --> 00:37:07,750
here do we have IES and ICS occurring, we
have IS alphaR and alphaF, you can write the
190
00:37:08,650 --> 00:37:15,590
terminal currents when both the voltages are
applied in terms of these three parameters
191
00:37:15,590 --> 00:37:22,590
and what you get finally, I shall write that
down finally. Also you can express alpha in
192
00:37:25,940 --> 00:37:32,940
terms of beta because that is the more common
used common emitter configuration, more commonly
193
00:37:38,970 --> 00:37:45,970
rather than the common base configurations.
So beta is the common emitter refers to the
194
00:37:46,530 --> 00:37:51,430
common emitter configuration, current gain
in the common emitter configuration, so we
195
00:37:51,430 --> 00:37:58,430
can write beta in terms of alpha and then
the relations which we get I shall write them
196
00:37:58,980 --> 00:38:05,020
down.
197
00:38:05,020 --> 00:38:12,020
This you can work out for yourself I think.
IC is equal to
198
00:38:37,110 --> 00:38:44,110
IS exponential this is just plain manipulation
basically you have to write IC is equal to
199
00:39:01,180 --> 00:39:08,180
alphaF IF minus IR. So alphaF IF you know
from the forward configuration, when the current
200
00:39:09,330 --> 00:39:16,330
is flowing in the forward direction IR expression
also you have calculated and then you just
201
00:39:16,800 --> 00:39:23,800
write them in terms of IS and then the two
betas betaF and betaR and IB is nothing but
202
00:39:55,660 --> 00:40:02,660
the difference of IE and IC is given by this.
It looks very complicated but we can remember
203
00:40:35,390 --> 00:40:42,390
it very easily. If you look at these expressions
one can look at them as three current sources.
204
00:40:52,310 --> 00:40:59,310
One is given by this IS exponential vBE by
vt - exponential vBC by vt which is occurring
205
00:41:00,320 --> 00:41:07,320
in these two and the other two is IS by betaR
exponential vBC by vt -1 and the third one
206
00:41:07,860 --> 00:41:14,860
is IS by betaF exponential vBE by vt -1. So
207
00:41:25,650 --> 00:41:32,650
you can write them in terms of this is the
collector terminal, this is the emitter terminal,
208
00:41:47,610 --> 00:41:54,610
this is the base terminal so this is IS.
209
00:42:41,870 --> 00:42:48,870
So what is the collector current? This current
source minus this current source, this is
210
00:42:49,940 --> 00:42:56,940
flowing in this way so IS exponential vBE
by vt -1 - IS by betaR exponential vBC by
211
00:42:57,270 --> 00:43:03,090
vt -1. What is emitter current? It is the
sum of these two current sources. This is
212
00:43:03,090 --> 00:43:10,090
the same this current source plus this. This
is given by IS by betaF exponential vBE by
213
00:43:11,490 --> 00:43:18,490
vt -1 and the base current is given by the
sum of these two current sources. Again all
214
00:43:20,510 --> 00:43:27,510
the currents are expressed in terms of the
three parameters IS, betaF and betaR. So now
215
00:43:31,510 --> 00:43:38,510
this is the final Ebers Moll model which is
also available in spice, so you have three
216
00:43:39,570 --> 00:43:46,570
current sources in terms of which you express
the terminal currents.
217
00:43:47,710 --> 00:43:54,710
So you see that when the collector base voltage
say is almost equal to the base emitter voltage
218
00:44:01,510 --> 00:44:06,410
or it is forward biased, collector base junction
if you keep on forward biasing the collector
219
00:44:06,410 --> 00:44:13,410
base voltage these current source value reduces.
So you have the current going down inside
220
00:44:19,010 --> 00:44:26,010
in saturation region. In saturation what happens
is both the junctions are forward biased base
221
00:44:26,190 --> 00:44:33,190
emitter as well as base collector. So this
is the difference, the collector current is
222
00:44:33,500 --> 00:44:37,610
mainly made up of this component which is
the difference. If collector base junction
223
00:44:37,610 --> 00:44:44,610
is reverse biased this component is almost
zero because it is very small compared to
224
00:44:45,270 --> 00:44:52,270
this component. In fact the current does not
change much when you go on increasing the
225
00:44:53,180 --> 00:45:00,180
base collector voltage. When you have a constant
vBE and if we go on changing the base collector
226
00:45:02,900 --> 00:45:09,900
voltage so the current will almost be a constant.
So this gives us the relations between the
227
00:45:13,920 --> 00:45:20,920
terminal voltages and the currents and this
model is used in different circuit simulators.
228
00:45:22,820 --> 00:45:28,440
An interesting thing before we wind up because
we shall be coming to that again when circuits,
229
00:45:28,440 --> 00:45:35,440
I think it is the best time to discuss. If
you plot the well known output characteristics
230
00:45:41,610 --> 00:45:48,610
of a transistor for different base currents,
you have this type of characteristics. This
231
00:46:03,080 --> 00:46:10,080
is well known so this is for increasing IB,
this is the collector current versus vCE.
232
00:46:12,170 --> 00:46:17,930
So what is happening when we come to this
region? This is known as the saturation region,
233
00:46:17,930 --> 00:46:24,620
this is the active region. What happens when
we go to the saturation region? What is vCE
234
00:46:24,620 --> 00:46:31,620
given by? vCE is equal to vBE I should say,
so when vCE is going towards zero almost towards
235
00:46:53,430 --> 00:47:00,430
zero vBE and vBC are quite close to one another
and so you can see from this model here when
236
00:47:04,570 --> 00:47:11,570
vBE and vBC are very close to one another
this current is almost going to zero. When
237
00:47:13,180 --> 00:47:20,180
vCE is very large it means vBC is negative
because vBE has a forward voltage cannot be
238
00:47:21,350 --> 00:47:27,570
very large. So this is reverse biased which
means the current becomes almost constant.
239
00:47:27,570 --> 00:47:34,570
The third point I want to make is when vCE
is zero what does it mean? VBE is equal to
240
00:47:39,050 --> 00:47:45,870
vBC so vBE is equal to vBC, what happens to
this current source? This current source is
241
00:47:45,870 --> 00:47:52,870
zero, is the collector current zero? Collector
current is not zero because individually vBE
242
00:47:54,650 --> 00:48:01,650
and vBC are not zeroes, you have a current
flowing here. So IC is actually negative,
243
00:48:05,470 --> 00:48:12,470
there is a negative current collector current
flowing, when vCE is equal to zero.
244
00:48:19,580 --> 00:48:26,580
In fact in order to make the collector current
zero, vCE should be slightly positive. That
245
00:48:29,460 --> 00:48:35,270
is vBE must be slightly greater than vBC to
compensate for this current. These two current
246
00:48:35,270 --> 00:48:42,270
sources must be the same in order to the collector
current is zero. So if you expand this region
247
00:48:44,710 --> 00:48:51,710
for a bipolar transistor, in fact if I draw
it here this is IC, this is vCE it is actually
248
00:48:58,990 --> 00:49:05,990
like this around the zero point. This is called
the offset voltage that is the collector emitter
249
00:49:08,100 --> 00:49:15,100
voltage at which i c is zero, it is slightly
positive and at vCE is equal to zero IC is
250
00:49:20,000 --> 00:49:27,000
slightly negative. We shall see that and this
transistor is actually operated in this point
251
00:49:27,190 --> 00:49:34,020
in when we discuss ttl circuits that is a
particular region, I wanted to introduce this
252
00:49:34,020 --> 00:49:41,020
here itself. So we have seen the Ebers Moll
model of the bipolar transistor which we shall
253
00:49:41,130 --> 00:49:48,130
be using sometimes later on in the course.
Thank you.
254