﻿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