 Good morning to all. Today, talk about multistage amplifiers. When we talk about multistage amplifiers, let us see why should we have multistage amplifiers in the first place? Do we really need multistage amplifiers or can I somehow design a single stage amplifier that has all the needed specifications? Is it possible? Let us see whether we can manage with a single stage amplifier. In that case, we did not study this particular topic. Let us consider a voltage amplifier. As we said yesterday, most of the time we are actually looking for a voltage amplifier. Now, what are the parameters we are looking for in a voltage amplifier? Again, from what we studied yesterday, you might be remembering, we talked about four types of amplifiers. We said the most common one is the voltage amplifier and then we have current amplifiers. We also have transconductance amplifiers and also transresistance amplifiers. We also said that the input requirements or rather the ideal conditions for all these four amplifiers are very different. When we talk about a voltage amplifier, ideally we are looking for very high input resistance. We said why yesterday we discussed why we saw that the equivalent circuit of a voltage amplifier, if you see at the input side, we have a resistance r i and only if that r i is much higher than the source resistance, we can get the entire voltage across the input of the amplifier. Therefore, in a voltage amplifier, we must have very high input resistance. We also saw that a voltage amplifier must have very low output resistance. We also saw the reason yesterday. This is because of the fact that the voltage amplifier we said is a voltage controlled voltage source. In a voltage source, if we use the Thevenin equivalent circuit, the Thevenin representation, we have an open circuit voltage in series with a resistance. Now, any current which is that is flowing into the load has to flow through this series resistance. Therefore, if we do not have a low output resistance, lot of voltage will get dropped across that particular resistance and the load will not get the full, the amplified output. Therefore, in a voltage amplifier, necessarily we must have very high input resistance, very low output resistance. Also, we require high gain and let us say high bandwidth. The last requirement is not very often required. We also saw yesterday why. We said the 741 op-amp and other op-amps have become extremely popular, even though they do not have high bandwidth. We said it is for the simple reason that for most of the applications, especially sensor applications, the frequency we are talking about will be most of the time typically less than 100 hertz and definitely not more than 1 kilohertz. Therefore, we are not worried about high bandwidth, but there are applications like video applications or even audio applications, where an op-amp bandwidth may not be sufficient. Now looking at these parameters of the voltage amplifier, we know from our study yesterday that all the above parameters cannot come out of a single stage amplifier. We know that when we based on our discussion yesterday, we saw that most of the amplifiers have maybe one of the properties, not all. Therefore, we cannot manage a practical application with a single stage amplifier. We can almost certainly say that for any practical application, we have to necessarily use more than one stage. Now once again, let us have a take another block which let us consider the 741 operational amplifier. Now we know that we can make, you can build circuits around say the 741 op-amp, a voltage amplifier which is almost ideal. We know that if we build a non-inverting amplifier using the 741, the input resistance will be very high. It can be made several mega ohms. The output resistance can be made very low, say typically within say a few ohms. So, we see that a 741 op-amp when we use it in the negative feedback circuit, we see that it possess the kind of ideal parameters. Now what about this op-amp, 741? Is it a single stage or is it a multi-stage circuit in its construction? Now op-amp is very much a multi-stage amplifier. Some of it was discussed by Professor Sharma and if you see the inside circuit of an op-amp, we see that it is a very complicated circuit, but you can divide the op-amp into about 3 stages. This is 741. Most of the op-amps which you get in the market has typically, let us say about 3 stages. All of them would have an input stage which is a differential stage with high input resistance, high CMRR and high gain. Now that way the input stage can be designed to ensure that you have high input resistance and op-amp being a high gain differential amplifier, the input is a differential stage. This may not be the case for some other amplifiers and in a 741 op-amp, the input stage also has gain or the order of about 1000, slightly less than 1000. Now in a 741 op-amp, you have a second stage which is called a gain stage. Now in 741, this is a compensated op-amp, it is a compensated gain stage meaning internally a capacitor has been introduced between the, let us say the collector terminal and the base terminal. And from our discussion yesterday, we saw that in a common emitter kind of configuration, if you connect a capacitor between the collector and the base terminal, we said that particular capacitor gets multiplied by the gain. In a 741, internally they have introduced a 20 pf capacitor, approximately 20 pf capacitor which gets multiplied by a gain of about 1000. So you get a highly compensated meaning you have an extremely low frequency cutoff, rather cutoff frequency, high frequency cutoff frequency. In an op-amp, like 741, the cutoff frequency is 5 hertz. Now in some other op-amps, you might not have compensation, internal compensation. In a 741, this compensation has been done to ensure that when you use it as a negative feedback amplifier under no circumstances, it should go into kind of a positive feedback. Now in a op-amp, once again all op-amps would have a third stage, sometimes let us say the final stage which is called the output stage. Now this particular stage is meant to drive an output load. Now 741 can drive a maximum current of about 20 milliamps. If you have to drive more current, you need to add some extra discrete components, transistors at the end. So we see that an op-amp is also a multi-stage and you get this kind of ideal situation precisely by combining three very different stages. So in summary, we can think of multi-stage amplifiers as very good options to get the desired amplifier specifications. Now typically, they would have two or more stages, definitely two, but most of the time it would have three or more. So another very important thing to remember, just like from the example of 741 op-amp is also that in a multi-stage amplifier, the individual stages are not identical. In fact, these stages are been chosen with a purpose, each one with a very specific purpose. In the case of a 741 op-amp, we saw that the input stage is designed specifically to provide very high input resistance and differential, it is a high gain differential amplifier with very high CMRR also. So that is a very purpose of the first stage, whereas the second stage does not have high gain, sorry, it does not have high input resistance, but it has high gain. Now its output resistance is also not that low, whereas the third stage in an op-amp is very, very different from the first three stages. So we can say, most of the time when you talk about a multi-stage amplifier, it would definitely have more than two or more stages and almost always the first stage would require large input resistance. This is to ensure that when you interface this particular multi-stage amplifier with a sensor, the finite rather the high resistance of the sensor need not upset the performance. So if the input stage has very high input resistance, almost entire voltage would get developed across the input terminals. Now at the same time when you talk about the first stage, the first stage need not have high voltage gain, it need not be high, if you get it is okay, but that is not the major requirement. The main requirement is to have very high input resistance. When you talk about the second stage in a any almost all multi-stage amplifiers, we see that there is no requirement to have a high input resistance, but that stage must provide very high voltage gain. So the main requirement as far as the second stage of a multi-stage amplifier in a practical situation, the main requirement is to have very high gain and it need not have high input resistance. Now again coming to the final stage, the main purpose of that final stage is to interface with the output load. Therefore, it need not have any voltage gain, it can be just a unity gain, but it must have low output resistance. Now before we consider a numerical example to illustrate what we said, let us look at another example which we discussed yesterday. Let us look at the public address system which we are very familiar with, right now I am speaking into a microphone and the output of that microphone is extremely small, definitely not more than 100 millivolt. Typically it would be about 50 millivolt, but unfortunately if you consider this microphone as a sensor, you would see that the source resistance is extremely high, typically the order of 100 kilo ohms or more. Therefore, if I directly take the output of the microphone and connect it, let us say to a common emitter amplifier, you would not get any input to amplify. Therefore, in a public address system, we see the same like what we saw in op-amp, typically a public address system would have three or more stages and in a public address system, the scenario is slightly different. Now the first stage must take care of interfacing to the sensor which is the microphone and as I said, the microphone has very high source resistance and it gives you extremely small outputs. So, what is done in a public address system is, the first stage would be a low noise pre-amplifier. Now that is stage, the main purpose of that particular stage is to ensure that there is some amount of amplification and the fairly high amplification, but extremely with low noise. So, therefore, the pre-amplifier devices are very carefully chosen, they are very special devices which are low noise devices and then the second stage in a public address system would be what is called let us say a post gain stage or a gain stage and you might have one or two and as we are all aware in a public address system, occasionally you might take input from some other place, let us say from a cassette recorder, you might take output from a cassette recorder. Now the output from a cassette recorder is fairly high, much higher compared to the output from a microphone. So, therefore, in most of the public address system, what is done is the output of the microphone is brought after the pre-amplifier and the first gain stage and then it comes to a mixer. Right now in the studio also, we have a very professional mixer at the rear of the room and is managed by professionals, where you could mix different types of signals in the way you want. Now in a public address system, you would see right in the front panel a console by which you could choose either from a cassette recorder or from microphone or you could even mix things. Now therefore, in a public address system, the third stage would be another kind of a gain stage, where the gain need not be very high, may be somewhere between 10 to 100 and that gain also as we all know can be controlled from outside and again as we know the mic gain also can be controlled from outside. And then finally, you have a output stage which is a power amplifier. My last lecture will be on power amplifier, where we will talk about the specific requirements of an output stage. Now in the case of a public address system as compared to an op-amp, the scenario is very different. In an op-amp, the load you are talking about is only about 20 milliamps which is very very small. Whereas in a public address system, the load have to cater to extremely low resistances or in fact very high currents. Now the requirement may be anywhere from 10 watts to 100 watts, very high power requirement. Therefore, as we all might have noticed, if you look behind a public address system, you would see power transistors put there with heatsink. Anyway, coming back to our topic of multi-stage amplifier. So, we see we saw in a actual practical application like a public address system also, we have the case of a multi-stage amplifier where we have 3 or 4 or more stages. Each one is chosen with a particular purpose. Now let us take an example, a numerical example to illustrate what we are talking about. Now what we have here is basically the equivalent circuit of a multi-stage amplifier. In fact, stage by stage, now this particular multi-stage amplifier has an input which is fed by a source V s and the source resistance is fairly high 50 kilo ohms and that feeds to the first stage. Now the first stage amplifier input resistance is 1 mega ohm and the gain the open loop gain rather the without load the gain is 10 and it has a source rather a internal resistance or output resistance of 2 kilo ohms. Now the output of the first stage is connected to the second stage which has an input resistance of 50 kilo ohms. Now that particular second stage has an open voltage gain of 100 and the output resistance is 1 kilo ohm. Now this particular output of this second stage is fed to the third stage which is essentially has an input resistance of 5 kilo ohms and an output resistance of 20 ohms, but the gain of that particular stage, voltage gain of that particular stage is just unity and then the output of the third stage is connected to the load. Now let us try to analyze this how this particular amplifier works and how it has been chosen. As I just now said about the examples of both op amp and a publicator system, we see an extremely we see a lot of similarity between what we said and also in this particular example. We see that the input is from a sensor which has very high output the source resistance which is about 50 kilo ohms it may be higher 100 kilo ohms it could be even higher. Now as we said the first stage we saw that in the both the previous examples that the first stage must have very high input resistance. This particular stage has an input resistance of 1 mega ohm so that satisfies that condition, but the gain is not that high the voltage gain is only about 10 without load. Now the second stage if you see the input resistance is not high compare the input resistance of the second stage with the input resistance of the first stage. First stage had 1 mega ohm whereas the second stage has only 50 kilo ohms, but this stage has 10 times the gain of the first stage first stage only had a gain of 10, but this stage has a voltage gain of 100. Now the output resistance of this stage is roughly let us say almost comparable with that of the first stage. Now when we come to the third stage which is the output stage we see that the input resistance is still smaller it is only 5 kilo ohms in fact the smallest of all the three, but look at the output resistance of this stage. This stage has 20 ohms which is the smallest again of all the three stages and this then drives a load which is 50 ohms. Let us analyze the circuit and let us see it stage by stage let us see the voltage gain stage by stage. Now what the way we would do is we would look at the gain of the first stage and then as the V i 2 by V i 1 and then we would see V i 3 by V i 2 and then finally we will see V L by V i 3. So, let us let us do this step by step. Now in the first stage since we have the source also if you look at the fraction of the voltage which is appearing at the input of the first stage that we can find the fraction as V i 1 which is the voltage at the input of the first stage divide by V s which you would find as 1 mega ohm by 1 mega ohm plus 50 ohm and that you would get as 0.952. So it is less than 1. Now the voltage gain the first voltage gain as I said is measured from the output of the first stage to the input of the first stage. Now one very very important thing to keep it in mind when you talk about a multi-stage amplifier is the load of the first stage is actually the input resistance of the second stage. The load of the second stage is nothing but the input resistance of the third stage and so on. So, if we calculate the voltage gain we see that the open circuit voltage gain was 10 and 10 multiplied by 50 k this is the input resistance of the next stage by 50 k plus 2 k and we get that as 9.615. So, the open circuit gain was 10 now because of this finite output resistance we see that the gain drops to about 9.6. Now let us come to the stage 2 the voltage gain of the second stage can be calculated as the from the output of the second stage to the input. So, at the output this particular voltage V i 3 is developing across the input resistance of the third stage and this is the V i 2 is at the input of the second stage. Now as we said the second stage has fairly high gain a gain of 100 open circuit gain of 100. Now here we the load in this case is the input resistance of the third stage and that is 5 kilo ohms and this stage has a output resistance of 1 k therefore the voltage gain would be 100 into 5 k divided by 5 k plus 1 k that comes to 83.33. Now coming to the third stage we said the third stage is primarily an output stage therefore it does not have any voltage gain the open circuit voltage gain is only unity. Now this particular stage had an output resistance of 20 ohms and the load was 50 therefore the voltage gain of the third stage would be 50 divided by 50 plus 20 that comes to 0.714. Now if we calculate the total gain of the three stages and if we call it as A V this will be nothing from nothing but from V L will be V L divided by V i 1 V L is the voltage developed across the load V i 1 is the voltage at the input to the first stage. So we would see that that is nothing but the product of the three voltage gains we just now computed A V 1 plus into A V 2 into A V 3. Now if you do the calculations you will get that to be approximately 572. Now what is also important is the voltage gain from source to load yesterday we use the term overall voltage gain. Now in this case the overall voltage gain would be the fraction V L by V S where V L is the load voltage across the load V S is the voltage across this voltage the output the open circuit voltage of the source. So that would be nothing but V L by V i 1 which is nothing but the fraction of the voltage developed at the input of the first stage times V i i 1 by V S. Now this we would find as will be nothing but the voltage gain times the fraction of V i 1 by V S. So if we do the computation we see that that drops to 544. So if you see that without that fraction so if we had the input resistance if it were infinite then A V and the overall voltage gain would have been same but here the input resistance being only 1 mega ohm we see it drops. So higher the input resistance of the first stage the closer these two numbers would be. So this is a very good example to illustrate what we are talking. Let us also compute the current gain and also the power gain. Now the current gain can be defined as the current i naught which is the current flowing into the load divided by current i i which is the current flowing into the first stage. Now we know that i naught is nothing but the voltage V L divided by the load which is 50 ohms and the input current is nothing but the voltage across the input first stage divided by that resistance. So this we would see that by if you compute this would come to be 2 into 10 power 4 times the voltage gain and that comes to 11.4410 power 6 fairly high current gain. Now the power gain is again calculated as the power P L divided by P i. Now P L the power delivered to the load is V L into i naught and P i is V i of i 1 into i i which so this would be nothing but the product of the voltage gain and the current gain and this happens to be an extremely large number 6.54 into 10 power 9. Now we need to notice a few points again to reiterate what we saw was that if we want to avoid losing signal strength at the input end the first stage should have fairly high input resistance or the order of 1 mega ohm or higher. This resistance the input resistance of the first stage should be much higher than the source resistance but one very important thing to keep it in mind is that the gain need not be high. Coming to the second stage the second stage need not have high input resistance but it must have large voltage gain. Coming to the third stage the third stage is not required to have any voltage gain but it should work like a buffer amplifier. So it should have reasonably high input resistance and definitely low output resistance and the output resistance must be much smaller than the load. So this is what we saw from the amplifier. Now here we have the three stage multistage amplifier we have we considered. Now we could combine the entire thing in by the equivalent circuit of a voltage amplifier which would be more useful for us ultimately. Now that we can do by we know that the equivalent circuit of a voltage amplifier would have a resistance at the input which is the input resistance and then the output side you would have a controlled voltage source which would give you the open circuit gain times the the input voltage and then you would have the source resistance that is the equivalent circuit of a voltage amplifier. In this case the input resistance happens to be the input resistance of the first stage. Now the source and the load we have kept apart so we have we have kept this particular amplifier as one block. Now the gain here coming to the output resistance is nothing but the output resistance of the last stage which is 20 ohms and the gain we have here is the product of the of the two two gains the first and the second stage. So in this case you would see that we can represent what we considered earlier after we do the computation. We can give it show it by a very small very simple voltage amplifier circuit which combines all that which is essentially having a 1 mega ohm input resistance and having a voltage gain of 800 roughly and having an output resistance of 20 ohm. So when we connect this if when we use this for a particular application we can see what kind of gains we will get by just connecting the source and the load. Let us compare this now in summary the purpose of a multi-stage amplifier is to then maximize the advantages of the individual stages and to minimize the shortcomings of the stages. So that overall amplifier specification would be what we want. So we we try to therefore the placement of the the individual stage is very important. What we have here is the equivalent circuit the open loop equivalent circuit of a 741 op amp. Now a 741 op amp the input resistance is 1 mega ohm the open loop gain is 2 into 10 power 5 and the open loop output resistance r naught is 75 ohms. So we see a lot of similarity between what we considered in the previous numerical example and what we saw here except for the the numbers some of the numbers. Now therefore let us see how do we what what kind of a strategy we should be use when we talk about designing or choosing a multi-stage amplifier. Now one of the first things we need to do is to looking at the specifications depending on the kind of gain required or maybe the power the load requirement and so on. We need to decide whether we would have 2 or more in any case we would need 2 stages. Now typically in a multi-stage amplifier you might have as many as 3 or 4. Generally typically you may not have more than 4 stages. Now as I said it is extremely important to choose stages as required and also as they need to be arranged in the correct order. So it is all about maximizing the advantages of each stage which we studied and minimizing the the disadvantages and then we said what we do is you could express the each individual stage by its equivalent circuit. Now again the analysis of a multi-stage amplifier occasionally would be quite involved especially if you look at the the inside circuit of a 740 and op-amp it is extremely tough. It has about 27 transistors and the amplifiers used there are very different from the amplifiers which we studied. The amplifiers which we studied common emitter amplifier and common base and common collector they all were discrete amplifiers. Now discrete amplifiers have very major disadvantage. The major disadvantage being there is a kind of upper limit on the gain you can get. Typically you cannot get more than about say gain of 100 from a discrete amplifier. Typically maybe you could get maximum 200. Beyond 200 is almost impossible whereas when you talk about an IC amplifier you can get gains of the order of 1000 easily. This is done by using active loads and some of these things were covered and will also be covered maybe in some of the lectures. Now in an IC amplifier you would use a current source itself as the load. Now we know that a current source has extremely high output resistance. So a current source if I use as a load then I am getting a very high load which would give me a very high gain. So if you look at a 740 and op-amp inside you would see that you would not find loads being used as resistors and all the loads are active loads and also in an IC amplifier the biasing is entirely done through current sources. This is for the simple reason that if you change the power supply the biasing does not change. This is very different compared to what we are familiar with. We know that in a discrete amplifier if we change and when we design a discrete amplifier we start with the power supply available. Now if you change the power supply value you would see that the biasing will all change and if we let us say change it by 50 percent the operating point would shift drastically whereas in an IC amplifier this is not the scenario. A 740 and op-amp can be used from all the way from plus minus 6 volts to plus minus 18 volts 3 times. This is impossible in the case of a discrete amplifier. Now this is possible in a 741 op-amp because of the way it is biased through current sources. Now coming back to the example of 741 even though it is so complex the way you would analyze the 741 op-amp is to split the entire IC into 3 stages. The first stage would be the input stage which is a differential stage. The second stage is a gain stage and the third stage is an output stage. So the way when we analyze a 741 op-amp what is always done is to isolate these stages and then for each of these stages you would express that stage with its equivalent circuit and it may not be always possible to express it as a voltage amplifier equivalent circuit and very often you would use a transcontactance amplifier model. And once you combine these 3 stages the equivalent circuit of the 3 stages in a 741 op-amp the extremely complex circuit becomes a very simple block as we saw earlier. So the strategy in analyzing a multistage amplifier is to first of all to divide the multistage amplifier into stages the logical stages. And then to express this individual stages by the equivalent circuit and the stages might may be of different amplifier types as I said you may not be able to have a voltage amplifier equivalent circuit. So depending on what kind of is convenient from the analysis point of view we could do this. Now in a in a course like basic electronics this kind of analysis detailed analysis may not be required but what is important is basically the concepts involved in this. So once we understand the concept of a multistage amplifier that it consists of more than one stage each stage having its own different parameters and a multistage amplifier as a cascade of 2 or 3 stages then the analysis becomes much simpler. Now let us come to some actual applications and some realizations. Now when we talk about the BJTs we talked about 3 types of amplifiers and we know that common emitter, common base and common collector have very different performance parameters. Now let us try to compare them before we look at the way how we could pair them. Now a so let us write classify them the type of the amplifier and let us write about the input resistance. Let us talk about the output resistance of these and the voltage gain. Now if you look at a common emitter amplifier we know that the input resistance yesterday we saw that is typically r pi and we said r pi is the order of a few kilo ohms not very high. So let us write let us say low to medium. So this meaning the at best you could get a few kilo ohms not hundreds of kilo ohms. Now coming to the output resistance of a common emitter amplifier we saw that the output resistance is approximately the collector load resistance which is may be definitely in the order of kilo ohms. So from the output resistance point of view it is let us say that it is high. So it is definitely in the kilo ohms range. So let us say it is high. Now the voltage gain we know is much greater than unity. So typically we talk about a common emitter amplifier having a gain of let us say 100 or so. Now when we come when we talk about the common base which is considered yesterday just briefly we said the input resistance is very low. Now the output resistance we said is same as that of the common emitter and the voltage gain also we said is the same as that as the gain typically if you use the same components you would find the gain to be the same as that of the common emitter with the exception that the sign there is no phase inversion. Now when we talk about the common collector which is nothing but the emitter follower we know that the input resistance is high. In fact we can make it as high as possible by choosing different values of the emitter resistance. Now we also know that an emitter follower is nothing but a buffer therefore the output resistance is low that is why we use an emitter follower as a buffer and we know that the voltage gain of a emitter follower is approximately 1. Now we need to keep this in mind this picture in mind when we talk about pairings when we talk about pairing BJT amplifiers. Now before we do that let us look at one very interesting and very useful pairing which is what is called the Darlington configuration. Now you can think of in a Darlington configuration what is done is you are essentially cascading to BJTs by connecting the emitter of the first BJT to the base or the second BJT connecting the collectors of both the BJTs together and then considering taking the output from the second emitter. So if you consider the Darlington configuration as though it were a single BJT then you could think of the base terminal of this compound device being that of the input of the first transistor. The collector they are common and the emitter terminal of this compound device being the emitter terminal of the second device. Now here as we know because of this the way we connected it the base current of the first stage gets multiplied by beta plus 1 let us say approximately equal to beta beta 1. Now the emitter current that would be the emitter current of the first stage and that flows as the base current of the second stage. Now if we come to the emitter current of the second stage we can say that the when we compare the emitter current of the second stage or the emitter current of the compound device we see that the emitter current would be beta 1 times beta 2 times I b 1. So if we talk about a device where the betas are same we are talking about getting a device where the beta gets multiplied. So if you talk about let us say a transistor having a beta of 100 we are talking about by connecting this kind of a device in a Darlington configuration we can think of having a BJT having a beta of 10 power 4 which is extremely high. Now this kind of Darlington configuration is very common and it has very very useful purposes. Now when you talk about the application of a Darlington the one of the first thing is to think about let us say getting high beta from a transistor. Now the major application of a Darlington configuration is to use it as an emitter follower. Now in a emitter follower we said that the output resistance is low and the input resistance is high and in our yesterday's discussion we said that the input resistance of a single stage BJT single stage common collector amplifier is approximately beta times the series resistance of the emitter of the emitter follower. Now in place of the single stage if we use a multi stage that a compound device we get a high performance emitter follower as what is shown here. So in this emitter follower what we have done is we instead of connecting a single BJT we have used that compound device which is a Darlington configuration. With the result the input resistance at the seen by the source would be if with the same device if we have the same device you can say it would be beta square times r e or let us say beta 1 times beta 2 r e. So this can be extremely high if you take a number say r e if you take r e to be let us say 1 kilo ohm then we are talking about getting an input resistance of beta square times 1 kilo ohm which is 10 power 4 times 1 k very high. Now by changing r e we can get much higher values. Now another advantage of the Darlington combination is configuration is because of this the output resistance now become extremely low. In a single stage emitter follower the output resistance may be typically order of a few ohms or may be few tens of ohms. In this case we could get extremely low may be an ohm or less much less and the gain would be in this case much closer to unity compared to the previous case. So Darlington configuration an extremely popular configuration and it is used to get a high performance emitter follower. We need to see this in the context of a voltage follower we know that in an op amp if you consider the non inverting configuration if you directly connect without any resistance if you connect the output through a wire to the negative terminal and if you apply the input to the positive terminal we have a voltage follower. In a op amp voltage follower the input resistance is extremely high it would be hundreds of mega ohms at least theoretically. Now the we need to compare the performance of a high performance emitter follower in the lines of a voltage follower we would see that the we may not get anywhere near there but still the performance of the emitter follower with a Darlington configuration would be much much higher than that of a single stage. Now when we use a Darlington configuration we need to keep a few things in mind one very important thing is to see about the beta of the individual transistors. We assume that if you make a Darlington configuration we assume that the compound device we said would get it would have a beta of beta 1 times beta 2. Now we know from our study of BJTs that the current gain beta is a function of the collector current. Now when a manufacturers makes a transistor and in a data sheet the manufacturer would always specify a kind of recommended current and most of the parameters would be measured at a particular collector current. So, in we know that in a BJT the beta would increase with collector current and especially if you have very low collector currents the beta would be much smaller. Now in a Darlington configuration we can get into trouble what can happen is think about the previous scenario where we had a configuration like what we had a simple cascade. Now here in case we have a large resistance let us say that we want to somehow get let us say tens of mega ohms from a Darlington emitter follower. So, therefore let us say that we put a large resistance at the emitter to get an emitter follower and let us say you we put say a 100 kilo ohms or let us say 1 mega ohm. Now what would happen is the current of the say q 2 the IC of q 2 would be let us say if it is let us say 1 milliamp or let us say 0.5 milliamp. Now the collector current of the first BJT would be less by a factor of 100 and what would happen is at that kind of collector currents which is in the micro amp regime you would see that the beta would drop drastically. Therefore, even though we assumed the compound device to have the same beta assuming the same device, but in actual situation the beta 1 would be much much smaller. So, this is a very troublesome situation and we need to have a solution for this. Now the solution is to introduce a resistor after the first stage and connect it appropriately. Now in an IC kind of a amplifier or in a situation it will be best to put a current source here, but in a discrete amplifier it may be difficult to have a current source. So, we could put a resistor here. Now this resistor need to be chosen very carefully. This resistance is chosen such that you have a certain amount of base current certain amount of emitter current here. Now the current flowing through this particular resistor is chosen such that the beta of the first transistor does not fall drastically. So, that we can still continue with the assumption we had that the total overall beta would be beta 1 times beta 2. So, this is something we need to keep in mind when we talk about a darling when we design an actual darling then and if we try to increase the input resistance by putting a large resistance at the emitter of the second device. So, we need to be very careful about the biasing we need to ensure that both the transistors have reasonable currents. Now let us consider a typical discrete multi stage amplifier. Now what we have here is an interesting multi stage amplifier which is nothing but a cascade of two common emitter stages. What we have here the input the first stage is the standard common emitter amplifier which we considered yesterday and this is an emitter bypassed common emitter amplifier and everything looks very familiar, but the second stage uses a PNP transistor and here also the emitter is bypassed and you are taking the output from the collector. Now it is very common to in multi stage amplifiers to use PNP transistors. Now PNP transistors are you can be cascaded with NPN in a fashion like this. Now one of the things which we need to keep it in mind is the analysis of this kind of multi stage amplifiers. Now this here we have two common emitter amplifiers. So, the AC analysis we could again just like we said the strategy we could split this into two common emitter and we could consider this as a cascade of two common emitter amplifiers and then we would be able to solve this. However the we need to also we concerned before we do the AC analysis we need to get the biasing currents correctly. Now this particular circuit is a bit tricky. Let us see how we would analyze a circuit like this. Now what we have here we know that in the first stage is fairly easy the way we analyze a common emitter amplifier is we convert the input side into we disconnect the source we convert the input side into a Thevenin equivalent and we get a VBB that VBB would be nothing but VCC times R2 by R1 R2 and then the Thevenin resistance would be RB which is nothing but R1 parallel R2. Once we do that in the common emitter amplifier we can write we can apply KVL in the common in this particular loop. So, here once we replace the input side by the Thevenin equivalent a voltage source and the Thevenin resistance there we can apply KVL here and get the base current. Now once we get the base current here now seeming this to be active region we can get the collector current. Now in this particular amplifier circuit the circuit is a bit tricky the collector current IC1 has two components one component is the current flowing from through the first the RC1 resistance here which we we can call as IR1 and the second component is the current flowing from the base towards the collector mind you this is a PNP transistor. Therefore, the direction of current base current would be outward in an NPN the direction of the base current would be inward. So, how do we solve this kind of a situation now here the what we need to do is we need to write a simple equation. Now let us see how we can solve this now we can find the we can write an equation for IE2 the emitter current of the second amplifier as VCC minus VE divided by RE2. Now VE which is the emitter voltage here means nothing but the collector voltage of the first transistor plus VBE mind you this is a PNP transistor. Therefore, the emitter would be more positive than the base. So, we get an equation here for the IE1. Now once we get an equation for IE1 we can divide that by beta plus 1 to get IB2. Now we saw that this particular expression has the term VC1 which is the collector potential here. Now VC1 we know is nothing but or rather we can write IR1 as VCC minus VC1 by RC1. So, now we have expressions for both IR1 and IB2 involving IC1. Now we know the value of IC1 because we calculated IB1. So, from this we would get we can solve easily for VC1 which is the collector voltage. The moment we get collector voltage here we can immediately find out the emitter voltage here by adding VBE voltage to that voltage and then we get the emitter current here. So, that way all the biasing currents can be found out and all the small signal parameters can be found out. So, most of the discrete amplifiers multi-stage amplifiers would have this kind of a connection. Now you can also in this case actually we see that they are cascaded is you can think of this as a direct coupled. Now you can also think of two stages like common emitter or common base or common collector being kind of isolated from each other through a capacitor. Now that way you could consider them much easier because each stage then would be independent.