1 00:00:24,000 --> 00:00:35,000 In the previous lecture we developed a model for the transistor for FET, that model is 2 00:00:36,680 --> 00:00:43,680 applicable and will be used for the analysis of amplifiers, which are constructed using 3 00:00:46,110 --> 00:00:53,110 either junction field effect transistors or MOSFETs. So, the analysis, which we shall 4 00:00:53,640 --> 00:01:00,640 be currently carrying, that is applicable for MOSFET circuits, as well as, for JFET 5 00:01:01,690 --> 00:01:08,690 circuits. We will recall, let us recall, that the bipolar transistor can be used in three 6 00:01:13,110 --> 00:01:20,110 configurations and that gives common emitter circuit, common collector circuit and common 7 00:01:20,360 --> 00:01:27,360 base circuit. In the same way, a field effect transistor can be used in three configurations 8 00:01:28,780 --> 00:01:35,780 and we have three kinds of amplifiers; three types of amplifiers. 9 00:01:48,810 --> 00:01:55,810 The common source amplifier, that is, CS, common source, CS, like C amplifier in BJT, 10 00:02:07,560 --> 00:02:14,560 so similarly here, common source, CS amplifier, and then, we have common drain, that is, CD 11 00:02:25,989 --> 00:02:32,989 amplifiers and then, we have common gate, that is, CG amplifiers. We will study all 12 00:02:59,140 --> 00:03:06,140 these three amplifier circuits and by study we mean, that we are going to derive expressions, 13 00:03:07,780 --> 00:03:14,780 analytical expressions for the characteristic parameters for the amplifier. 14 00:03:15,260 --> 00:03:22,260 And you will recall from our previous study, that characteristic impedance for an amplifier, 15 00:03:24,180 --> 00:03:31,180 these are for example, voltage gain, voltage gain, input impedance of the amplifier. These 16 00:03:31,989 --> 00:03:38,989 two are most important parameters for the, which will decide the performance of the circuit. 17 00:03:40,319 --> 00:03:47,319 And then, there are some other considerations, like what is the output impedance, which is 18 00:03:49,709 --> 00:03:56,709 important in some cases. Now, out of these three common source amplifiers, 19 00:03:57,269 --> 00:04:04,269 in junction field effect transistors or in MOSFETs, this is most widely used. This is 20 00:04:05,720 --> 00:04:12,720 like common emitter circuit for BJT in FETs and MOSFETs, the most commonly used, most 21 00:04:14,189 --> 00:04:21,189 widely used amplifier is the common source amplifier. Then, we will study common drain 22 00:04:22,120 --> 00:04:29,120 amplifier, which is like a and that CC, common collector or emitter follower, this is known 23 00:04:32,090 --> 00:04:39,090 as source follower and it has the similar application as the emitter follower for matching 24 00:04:41,380 --> 00:04:48,380 purposes, so it is used as a buffer amplifier. And common gate amplifier is like common base 25 00:04:49,020 --> 00:04:56,020 amplifier and this has very limited applications for the simple reason that the input impedance 26 00:04:57,560 --> 00:05:04,560 of common gate amplifier is very low; it is few hundred ohms only, so it has limited applications. 27 00:05:06,990 --> 00:05:12,810 We go one-by-one for the study of these amplifiers. 28 00:05:12,810 --> 00:05:19,810 First, we take common source amplifier, common source amplifier, which for example, the JFET 29 00:05:36,250 --> 00:05:37,500 can be drawn like this. 30 00:05:37,500 --> 00:05:44,500 This 31 00:06:17,919 --> 00:06:24,919 is the circuit, here we connect the biasing source, the battery, V DD and this is R D, 32 00:06:29,069 --> 00:06:36,069 this is R S, you are familiar with these resistances and this is R G, and this is the input signal 33 00:06:38,669 --> 00:06:45,669 v in, this is to be amplified and the amplified signal is taken at the output here. This is 34 00:06:55,210 --> 00:07:02,210 the common source amplifier and here, the input is given between gate and source and 35 00:07:05,610 --> 00:07:12,610 output is taken from, from drain and source. Now, this is the coupling capacitor and here, 36 00:07:16,900 --> 00:07:23,900 this is, C 1 is coupling capacitor, coupling capacitor and C 2 is by-pass capacitor. If 37 00:07:31,979 --> 00:07:38,979 we do not use this C 2 by-pass capacitor, then there will be drop, ac drop across this 38 00:07:43,280 --> 00:07:50,280 resistance and hence, the gain, the output voltage, which is available, that will fall. 39 00:07:50,729 --> 00:07:57,729 This point we will discuss again little later. Now, first we should draw ac equivalent circuit 40 00:08:01,210 --> 00:08:06,870 for this actual circuit. This is the actual common source amplifier circuit, which makes 41 00:08:06,870 --> 00:08:13,870 use of these three resistances, two capacitances, the input signal and the output is taken here 42 00:08:14,490 --> 00:08:21,490 and we draw the ac equivalent of that. For that, let us remind you we have talked all 43 00:08:23,660 --> 00:08:30,259 these points when we talked about the analysis of bipolar transistor amplifier analysis, 44 00:08:30,259 --> 00:08:37,259 but again, briefly we talk about them, that for drawing ac equivalent, first thing is, 45 00:08:38,699 --> 00:08:45,699 that dc voltage sources are grounded; dc voltage sources are grounded. 46 00:08:59,140 --> 00:09:06,140 Second point is that these coupling and by-pass capacitors, they are taken as short at the 47 00:09:07,209 --> 00:09:13,550 frequency for which this circuit has been designed. We choose these capacitors such 48 00:09:13,550 --> 00:09:20,550 that, that the impedances offered by them at the frequency of interest negligible and 49 00:09:20,820 --> 00:09:27,820 hence, they can be taken as short. So, capacitors C 1 and C 2 are taken as short, short circuited. 50 00:09:37,350 --> 00:09:44,350 So, keeping these points in mind the equivalent ac equivalent circuit is this. 51 00:10:18,870 --> 00:10:25,870 This is the ac equivalent of circuit in figure 1, this is figure 1, this is figure 2. So, 52 00:10:38,070 --> 00:10:45,070 this is ac equivalent of circuit figure 1. Here, once we take this as short, so this 53 00:10:48,550 --> 00:10:55,079 is shorted; once we consider this by-pass capacitor as short, so this resistance is 54 00:10:55,079 --> 00:11:02,079 shorted. So, zero resistance, this is like that and this is grounded, so this is like 55 00:11:02,200 --> 00:11:09,200 this and output is taken against this resistor R d, this is the ac equivalent. Now, we will 56 00:11:11,410 --> 00:11:18,410 replace this transistor, this FET or MOSFET, whatever is, by its, by its model. So, what 57 00:11:22,950 --> 00:11:29,950 we get is this. 58 00:12:14,709 --> 00:12:21,709 This is the gate terminal, this is the input, v in, and this is the transistor in the circuit. 59 00:12:24,690 --> 00:12:29,700 In the ac model we have replaced by its ac equivalent circuit, we have replaced by the 60 00:12:29,700 --> 00:12:36,700 model and this is drain. So, and this is the current source, which is equal to g m into 61 00:12:38,510 --> 00:12:45,510 v in and here, because this resistance is very high, so, v i, v in is equal to v gs. 62 00:12:50,269 --> 00:12:56,529 You will recall, you will recall, that this current source in the model we have taken 63 00:12:56,529 --> 00:13:03,529 as g m into v gs, but here we have shown it as g m v in because of this and this is true, 64 00:13:09,279 --> 00:13:16,279 true because R G is very high. With this, now we can draw these resistances are in parallel, 65 00:13:33,880 --> 00:13:40,880 with this there may be additional resistance, for example, R L, the load resistance or another 66 00:13:42,240 --> 00:13:49,240 stage, which is connected in series with this amplifier, then what will be the input impedance 67 00:13:50,850 --> 00:13:57,850 of this next stage, that will be taken here. So, all these resistances are in parallel. 68 00:13:58,329 --> 00:14:05,329 So, actually, this can be replaced by one resistance r D and which is r DS in parallel 69 00:14:10,540 --> 00:14:17,540 with R D in parallel with R L. One thing, which is very frequently used in this analysis 70 00:14:20,360 --> 00:14:27,360 in, in, at many places in electronic circuits, when two resistances, one high resistance, 71 00:14:28,589 --> 00:14:33,730 one low resistance, when they are connected in parallel, then the effective resistance 72 00:14:33,730 --> 00:14:39,949 of the combination is closer, in fact, smaller than the smaller resistance. 73 00:14:39,949 --> 00:14:46,949 Take just one example, 100 k resistance and 1 k resistance, if they are connected in parallel, 74 00:14:46,980 --> 00:14:53,300 then what will be, you apply, that parallel resistance rule and find out the effective 75 00:14:53,300 --> 00:15:00,300 value of the resistance. This will be lesser than 1 k, but close to 1 k. So, here this 76 00:15:02,470 --> 00:15:09,279 resistance is highest, which will have least effect on this. So, this will be very close 77 00:15:09,279 --> 00:15:16,279 to R D. So, this model, they, all these resistances we can replace by R D, where R D will be equal 78 00:15:24,480 --> 00:15:31,480 to this. Also, R G is very high resistance, this we have seen, 50 k, 100 k, so this can 79 00:15:33,899 --> 00:15:38,449 also be dropped from this figure. 80 00:15:38,449 --> 00:15:45,449 And then, this model becomes this. 81 00:16:26,970 --> 00:16:33,970 This is the model, which we are going to use and remember, the current source is g m into 82 00:16:34,290 --> 00:16:41,290 v in, v in is the input voltage. Then, the drain current i d, that is, this, this current, 83 00:16:44,250 --> 00:16:51,250 current source and see the direction, so current is flowing i d and this i d, the drain current 84 00:16:53,779 --> 00:17:00,779 in this circuit and this is equal to g m into v in. 85 00:17:01,139 --> 00:17:08,139 And look at the direction, the direction of current here, this is this way and the particular 86 00:17:09,449 --> 00:17:14,909 direction, which we have taken for v out is just opposite, that is, current normally flows 87 00:17:14,909 --> 00:17:21,909 from higher potential to lower potential, but this is flowing this way and hence, it 88 00:17:22,010 --> 00:17:29,010 is in order to write v out equal to minus this resistance r D into i d. We substitute 89 00:17:37,390 --> 00:17:44,390 for i d from this equation v out is minus g m r D v in and from here, the voltage gain, 90 00:17:55,000 --> 00:18:02,000 the voltage gain, A v is equal to v out, output voltage, ac output voltage divided by ac input 91 00:18:10,100 --> 00:18:17,100 voltage. So, v in and this is equal to minus g m into r D. 92 00:18:20,419 --> 00:18:27,419 This negative sign simply says, that there is a phase reversal in this amplifier, meaning, 93 00:18:29,520 --> 00:18:36,520 when input will be maximum positive, this is varying signal, this is varying signal. 94 00:18:37,390 --> 00:18:44,000 When input signal is maximum positive, the output will be maximum negative and so on, 95 00:18:44,000 --> 00:18:51,000 that is the meaning of phase reversal. So, this sign simply indicates phase inversion 96 00:18:51,669 --> 00:18:58,669 or reversal phase inversion, like common emitter BJT amplifier, that also has phase inversion. 97 00:19:04,140 --> 00:19:07,590 So, similarly, it has phase inversion here. 98 00:19:07,590 --> 00:19:14,590 And the magnitude of voltage gain A v is simply, g m into r D in a, just to give an example, 99 00:19:25,860 --> 00:19:32,860 if g m, g m is 4000 micro , that is the unit in which g m is written 4000 micro, and let, 100 00:19:42,240 --> 00:19:49,240 let r D, that is, all these resistances in the amplifier, the parallel combination of 101 00:19:50,080 --> 00:19:57,080 this, for example, gives r D to be equal to 4 kilo ohms, then according to this relation 102 00:20:01,190 --> 00:20:08,190 the voltage gain is 4 into 10 to power minus 3, that is, g m into r D, which is 4 k 4 into 103 00:20:14,530 --> 00:20:21,530 10 to power 3. So, voltage gain is simply 16. 104 00:20:24,570 --> 00:20:31,570 This is a simple expression, which expresses the voltage gain of the amplifier. Now, in 105 00:20:34,270 --> 00:20:41,270 the circuit we have taken, that R s is bypassed by the by-pass capacitor C 2. In certain circuits 106 00:20:45,330 --> 00:20:52,330 we do not by-pass, we do not by-pass this resistance R s. In that case, as I said, in 107 00:20:53,620 --> 00:21:00,620 the beginning there will be ac drop across this resistance also. So, the drop will fall 108 00:21:00,890 --> 00:21:07,890 or in other words, the gain will fall for the same input. If this is not bypassed we 109 00:21:08,760 --> 00:21:15,760 get lesser output because the gain will fall. So, in that case, if, if resistance R s is 110 00:21:19,530 --> 00:21:26,530 not, is not bypassed, is not bypassed, in that case this gain will be reduced and this 111 00:21:33,850 --> 00:21:40,850 is equal to g m r D by 1 plus g m into R s here. And once we bypass it, then this R s 112 00:21:55,870 --> 00:22:02,870 becomes 0 and we return to this expression. So, this is about the voltage gain. 113 00:22:04,360 --> 00:22:11,360 And then, the input impedance, input impedance of the amplifier; input impedance of the amplifier. 114 00:22:29,090 --> 00:22:36,090 We can see from, for example, from this model that input impedance, if we measure at the 115 00:22:39,299 --> 00:22:46,070 input terminals between gate and source, this will come out to be equal to R G and R G is 116 00:22:46,070 --> 00:22:53,070 very high. So, input impedance Z i, this is equal to R G, this is very high. Normally, 117 00:23:06,220 --> 00:23:13,220 R G will have range, 100 kilo ohms to 1000 or more kilo ohms. So, input impedance is 118 00:23:23,539 --> 00:23:30,539 very high. So, we can summarize, that CS amplifier, it 119 00:23:36,169 --> 00:23:43,169 has high voltage gain, high voltage gain and high, rather, very high, high input impedance 120 00:23:59,330 --> 00:24:06,330 and it has a phase reversal, phase inversion. These are the characteristics of a CS amplifier. 121 00:24:18,460 --> 00:24:25,460 One thing I may mention here, that voltage gain for MOSFET or junction field effect transistor, 122 00:24:27,799 --> 00:24:34,799 if we compare with a bipolar transistor it is much less, but there are other benefits. 123 00:24:39,820 --> 00:24:46,820 So, FETs are very widely used and this is not very high input, not very high gain. This 124 00:24:53,059 --> 00:24:59,950 can always be compensated by adding another stage, amplifying stage, which is not a problem 125 00:24:59,950 --> 00:25:06,640 at all. If two stages, each having gain ten, for example, 126 00:25:06,640 --> 00:25:13,640 if two stages are cascaded, are connected in series where individually each stage has 127 00:25:15,299 --> 00:25:22,299 a gain of ten, then the two stages will have a combined gain of ten into ten; that means, 128 00:25:23,289 --> 00:25:26,880 one hundred. So, this is no problem. 129 00:25:26,880 --> 00:25:33,880 Then, we go for the next amplifier, that is, common drain amplifier, common drain amplifier. 130 00:25:46,279 --> 00:25:53,279 Let us first draw the circuit. 131 00:26:50,970 --> 00:26:57,970 This is the circuit; this is common drain amplifier circuit. Few, few features may be 132 00:27:19,010 --> 00:27:26,010 noted. One is that drain terminal is directly connected to the dc source. When we draw the 133 00:27:30,350 --> 00:27:37,350 ac equivalent, then this is to be grounded. As we have talked, that ac equivalent circuits 134 00:27:38,669 --> 00:27:45,350 require, that dc voltage sources have to be grounded. So, we grounded, so that means, 135 00:27:45,350 --> 00:27:52,350 that drain is at ac ground. This is one point. Second point is that output we are taking 136 00:28:03,539 --> 00:28:10,539 at the load connected at the source. So, the 2nd point to be noted is, that load is connected 137 00:28:16,120 --> 00:28:23,120 at source. So, the output that is, that means, output is taken from source with respect to 138 00:28:44,010 --> 00:28:51,010 ground; here, source with respect to ground. This is, this is the only amplifier in which 139 00:28:54,940 --> 00:29:01,940 the load is connected at the source and output is drawn. Now, we will draw the equivalent 140 00:29:06,830 --> 00:29:13,830 of this, this is, these are the coupling capacitors C 1 and C 2 and we draw the ac equivalent. 141 00:29:16,529 --> 00:29:23,529 AC equivalent of this figure and of common drain amplifier and this is, this is the drain 142 00:29:46,909 --> 00:29:53,909 terminal at AC ground, this is the AC equivalent circuit, this is the input v s and this is 143 00:29:59,929 --> 00:30:06,929 G, this is R g and this is v in, this is g m into v gs and then, R ds R s and R L. This 144 00:30:46,169 --> 00:30:53,169 is the circuit, this is drain, this is source and this we can, as we have done earlier, 145 00:31:00,210 --> 00:31:07,210 we can replace, this resistance is r g of course, these three resistances are in parallel 146 00:31:07,559 --> 00:31:14,559 and that can be replaced by say R L. So, R L will be r ds in parallel with R s in parallel 147 00:31:21,720 --> 00:31:28,720 with R L and normally, this is very high resistance. So, these, this R L effectively, will be very 148 00:31:31,390 --> 00:31:38,390 close to R s in parallel with R L. So, we can replace these all three resistances by 149 00:31:41,240 --> 00:31:48,240 this. And another interesting point is that the 150 00:31:50,260 --> 00:31:57,260 gate terminal with respect to ground, this is at v in, this is the gate, gate at v 1 151 00:32:16,490 --> 00:32:23,490 with respect to ground and source with, with respect to ground is at v out, and source 152 00:32:30,220 --> 00:32:37,220 is at v out with respect to ground. So, what will be v gs? 153 00:32:42,490 --> 00:32:49,490 If gate with respect to ground is at v in and this is, source is at v out, then v gs 154 00:32:57,269 --> 00:33:04,269 will be v in minus v out, v gs is equal to v in minus v out, and let us call this equation 155 00:33:16,279 --> 00:33:23,279 1. Now, what will be v out? I can draw another circuit by just replacing these three resistances 156 00:33:30,980 --> 00:33:37,980 by one. So, in that case it will be, this is R L and here, this is R G, this is g m 157 00:34:03,750 --> 00:34:10,750 v gs and this is i d, this is plus minus v out, and from here we can write v out. From the circuit, obviously, 158 00:34:39,970 --> 00:34:46,970 this current when, when passes through this effective ac impedance R L, that will produce 159 00:34:47,190 --> 00:34:54,190 the output. So, i d into r L. Now, i d is equal to this. 160 00:35:00,170 --> 00:35:07,170 Since i d is equal to g m v gs, so we have v out equal to g m r L into v gs and v gs 161 00:35:28,560 --> 00:35:35,560 from equation 1 we have this. So, we write like that, putting for v gs from equation 162 00:35:44,349 --> 00:35:51,349 1, we have v out g m r L into v in minus v out or v out 1 plus g m into r L equal to g m 163 00:36:23,030 --> 00:36:30,030 into r L into v in. From here, we can write for the voltage gain. The mathematics is very 164 00:36:35,750 --> 00:36:42,750 simple, so simple, that it is almost very straightforward. So, the voltage gain A v, A v by definition is v out by v in and 165 00:37:00,589 --> 00:37:07,589 this is equal to g m r L by 1 plus g m r L. One thing let me make clear that here the 166 00:37:25,859 --> 00:37:32,859 signal is v s, but here we are showing it as v in because there is a small resistance 167 00:37:36,000 --> 00:37:43,000 here R s, source resistance. So, sometimes there is a small drop across this resistance 168 00:37:48,030 --> 00:37:55,030 that makes v s not exactly equal to v gs. So, this is taken as v s and here, this is 169 00:38:01,280 --> 00:38:08,280 going to the circuit v in because there is a, there may be some impedance, which is normally 170 00:38:11,470 --> 00:38:18,470 associated with the ac source at the input, and there may be some drop across this. So, 171 00:38:20,660 --> 00:38:27,320 so, that is why the two have been taken as different. Anyway, so this is the expression 172 00:38:27,320 --> 00:38:30,130 for the voltage gain. 173 00:38:30,130 --> 00:38:37,130 g m into r L, normally is much higher than 1, that is, A v we have obtained as g m r 174 00:38:43,170 --> 00:38:50,170 L 1 plus g m r L. Normally, usually g m into r L is very large as compared to 1, so 1 can 175 00:39:01,710 --> 00:39:08,710 be dropped, then A v will be very close to unity. When A v is 1, that implies, that there 176 00:39:12,790 --> 00:39:19,790 is no voltage gain, there is no voltage gain, it, it is, whatever is the input, same appears 177 00:39:23,210 --> 00:39:30,210 at the output, but there will be some power gain. Now, so this is this. 178 00:39:34,200 --> 00:39:41,200 Now, here, the input impedance we find out, gain we have seen, similarly for the emitter 179 00:39:47,220 --> 00:39:54,220 follower the gain was close to 1. And as I said, that this is used as a buffer between 180 00:39:56,339 --> 00:40:03,339 stages and hence, same is here. The input impedance, input impedance, when we measure 181 00:40:18,070 --> 00:40:25,070 because this resistance is normally very small as compared to r g, so it is the input impedance 182 00:40:28,230 --> 00:40:35,230 is high, z in is equal to R G and R G is very high. Input impedance is very high, gain is 183 00:40:49,760 --> 00:40:56,760 1, but let us see the output impedance. 184 00:41:00,470 --> 00:41:07,470 The output impedance, look at this circuit, this was the common drain amplifier. Now, 185 00:41:25,599 --> 00:41:32,599 here and its equivalent we draw here. Now, as a rule, whenever output impedance is to 186 00:41:37,810 --> 00:41:44,810 be measured, then we disconnect r L, this is the general practice and always this is 187 00:41:47,119 --> 00:41:54,119 done, that r L is to be taken out. Then, what will be the impedance, which we will get at 188 00:41:55,849 --> 00:42:02,849 the output? That means, between source and drain and this is simply, these two resistances, 189 00:42:03,900 --> 00:42:10,900 this is the drain source resistance and this is the source resistance, so the two in parallel. 190 00:42:13,440 --> 00:42:20,440 Therefore, the output impedance, z out, this is equal to r ds in parallel with R s. And 191 00:42:31,220 --> 00:42:38,220 as I said, as a practice, as a practice, r L is always removed, disconnected from the circuit to measure Z out. So, we are left with these 192 00:43:12,069 --> 00:43:19,069 two resistances in which this resistance is very high. So, Z out is very closely equal 193 00:43:23,430 --> 00:43:29,720 to R s and source resistance normally is kept low. 194 00:43:29,720 --> 00:43:36,720 R s normally may be 200 or 400 ohms, therefore the output impedance only for this circuit 195 00:43:41,150 --> 00:43:48,150 is very low, input impedance is very high, output impedance is low. So, this amplifier 196 00:43:48,890 --> 00:43:55,890 can be used to act as a buffer between very high impedance on one side and low impedance 197 00:43:58,130 --> 00:44:04,510 on the other side. Low impedance will match with the output impedance here and the high 198 00:44:04,510 --> 00:44:11,510 input impedance will match at the other point. So, this is that. 199 00:44:12,359 --> 00:44:19,359 In summary of CD, common drain amplifier, which is also known as source follower, source 200 00:44:30,829 --> 00:44:37,829 follower, here the gain is close to 1, but actually, it is less than 1 and input impedance 201 00:44:45,839 --> 00:44:52,839 very high, R G and output impedance, R s, low and it is used for buffer purposes. 202 00:45:02,910 --> 00:45:09,910 The last one is the common gate amplifier. The common gate amplifier, it is different 203 00:45:19,079 --> 00:45:25,380 from the earlier two, which we have studied, the common source and common drain and the 204 00:45:25,380 --> 00:45:31,160 reasons will be very clear. The input impedance here is extremely low, the circuit we can 205 00:45:31,160 --> 00:45:38,160 draw like this. Just directly I have come to the AC equivalent by grounding it because 206 00:45:41,540 --> 00:45:48,540 we have done it quite few times, so you understand. This is the input signal, this is grounded, 207 00:45:56,069 --> 00:46:03,069 this is source, this is drain and this is gate, this is the common gate, gate is common 208 00:46:04,810 --> 00:46:11,810 between input and output. Output is taken, which means drain and ground and this is the 209 00:46:12,300 --> 00:46:19,300 effective value of r L, which we have talked two times at least in the two circuits. 210 00:46:19,670 --> 00:46:26,670 Here, the v out, v out is i d into r L and i d is, i d is g m into v gs, which is, and 211 00:46:43,150 --> 00:46:50,150 v gs is same, you can see this voltage here, v gs is same as v in, v in. So, this we can 212 00:46:55,430 --> 00:47:02,430 write, v out as equal to g m v in into r L or the voltage gain A v is v out v in and 213 00:47:12,530 --> 00:47:19,530 that is equal to g m into r L, which is high, which is high. And now, we come to the input 214 00:47:29,290 --> 00:47:36,290 impedance. 215 00:47:39,059 --> 00:47:46,059 The input impedance in this circuit, it is not difficult to realize, that input current is the same as the current 216 00:47:58,210 --> 00:48:05,210 between drain and source, that is, i d. If we see the circuit, the, when we discussed 217 00:48:09,829 --> 00:48:16,829 about this current flows between drain and source, so it is not difficult at all to see 218 00:48:18,950 --> 00:48:25,950 that i in is equal to i d and i d, if we replace this by the simple model, then i d will be 219 00:48:30,300 --> 00:48:37,300 equal to g m into v gs and this is equal to v g in g m into v in because gate source voltage is the same as 220 00:48:57,730 --> 00:49:04,730 this, we just discussed this. Therefore, Z in is equal to v in by i in, this is i in 221 00:49:18,349 --> 00:49:25,349 and this is simply equal to 1 by g m, g m. So, this is very low, very low input impedance. 222 00:49:38,809 --> 00:49:45,809 And let g m be equal to, as we have taken earlier, 4000 micro, then Z in will be 1 by 223 00:49:58,010 --> 00:50:05,010 4 into 10 to power minus 3, we have to put in. So, 4000 micro is 1 by, that is, 4 into 224 00:50:10,450 --> 00:50:17,450 10 to power minus 3 and this is equal to 250 ohms, very low impedance. The output impedance 225 00:50:25,440 --> 00:50:32,440 is high, the input impedance is low and so, this amplifier has limited applications. This 226 00:50:39,200 --> 00:50:46,200 is equivalent to common base amplifier in the BJT. So, this way we finish the analysis 227 00:50:50,430 --> 00:50:57,430 of the amplifying circuits. The most, whenever we want to use the amplifying 228 00:51:00,000 --> 00:51:07,000 circuit using a MOSFET or a junction filed effect transistor, then we make use of common 229 00:51:07,900 --> 00:51:14,900 source amplifier and that common drain amplifier is used as a buffer. 230 00:51:18,309 --> 00:51:25,309 Whenever we have here one stage, which is having high output impedance, high z out and 231 00:51:30,410 --> 00:51:37,410 then, here this is some other device, some other circuit where this is having low impedance, 232 00:51:38,869 --> 00:51:45,869 they cannot be connected together because of the mismatch. The most of the power will 233 00:51:48,210 --> 00:51:55,210 be reflected and very little will pass on to this circuit. So, we require a buffer here 234 00:51:55,589 --> 00:52:02,589 and this common drain amplifier, which has high impedance, that will match with this 235 00:52:02,660 --> 00:52:09,660 and low impedance here will match with this. So, this is the buffer amplifier very widely 236 00:52:12,869 --> 00:52:19,869 used in these circuits. So, this was the analysis and we have completed that. 237 00:52:23,030 --> 00:52:30,030 Next thing, which we will be taking, which is very important as far as the circuits are 238 00:52:31,480 --> 00:52:38,480 concerned, many time, the, the many times the MOSFET circuits particularly, because 239 00:52:39,780 --> 00:52:46,780 in integrated circuits very widely used are the MOSFETs. So, MOSFETs or even junction 240 00:52:49,369 --> 00:52:56,369 field effect transistors, but that is normally results for discrete devices. So, MOSFETs 241 00:52:56,819 --> 00:53:03,819 can be connected as resistors as capacitors, that means, whenever a resistance load is 242 00:53:06,710 --> 00:53:13,710 required, that instead of having a resistance there we can use a properly connected MOSFET. 243 00:53:15,510 --> 00:53:22,510 When we connect a simple resistance that is called, for example, this resistance and here 244 00:53:25,140 --> 00:53:32,140 is a MOSFET, this resistance R D for example, this is called passive resistance, passive 245 00:53:47,130 --> 00:53:54,130 load, passive load, and this consumes power, this consumes power. And when this resistance 246 00:54:01,130 --> 00:54:08,130 is replaced by an active device, such as another MOSFET properly connected, then what we have? 247 00:54:14,440 --> 00:54:21,440 Active load, active load, so how to connect the MOSFETs as resistors and capacitors, this 248 00:54:26,660 --> 00:54:33,660 is we are going to take next.