 Now, let us quickly see some application depending on the time. Let us see some actual circuits and also how some of these parameters improve. Now, one of the most important thing to remember about negative feedback is by applying negative feedback, your ordinary amplifier or the basic amplifier goes from its non ideal situation towards the ideal situation. Now, we know that when you talk about a voltage amplifier what are the ideal conditions as far as the input resistance and the output resistance is concerned. I know that in a voltage amplifier ideally my input resistance should be infinite, my output resistance should be ideally 0. Now, we know that we can never get that with a voltage amplifier, but by applying negative feedback we can go from the non ideal stage towards the ideal stage. We may not reach there, but we will reach very close to it. Now, how is it done? We will not time does not permit us to do the derivation, but what will happen is the way it is the how you would analyze it shown here. Now, you could also intuitively say that since now the input voltage is getting reduced here, effectively the current gets reduced therefore, that is equivalent to the input resistance increasing. So, that is a intuitive way of looking at it. So, we see that by applying feedback what I have done is I have essentially reduced the current flowing because I have essentially reduced the V i here. So, by reducing V i effectively I have increased my input resistance. So, this is so we are going from the non ideal state to ideal. So, if our input resistance without feedback was R i, after applying negative feedback your input resistance would become R i times 1 plus a beta. That would be a substantial increase if you plug in numbers especially in an op amp. We know that a is 2 into 10 to the power 5. Now, if you take a beta of typically let us say 1 10, you would get a beta of 10 power 4 approximately. So, you are going to jack up the input resistance by a factor of 10 power 4 huge increase tremendous increase. So, this is why we said the by applying negative feedback we are taking the amplifier from is almost non ideal state towards the ideal state. What is the other advantage we get here? Now, again the derivation we will skip because of lack of time, but the way you do the derivation shown here, what you do is to measure the output resistance we connect a source in our analysis. You would never should never do this in an actual amplifier, but in analysis this is the way we do it. You connect a source fictitious source say v x and if you name the current i x in this scenario. Now, when we measure if you remember the what we wrote in our expressions when we talked about common amplifier for measuring the output resistance the input side should be shorted. So, you short circuit the input there. Now, you find the current here you would find that when you derive you would find that the output resistance r naught which is here reduces by 1 plus a beta which means if it was r naught earlier. Now, it is r naught divided by 1 plus a beta that is an extremely significant decrease. Now, take the example of an op-amp if you plug in the numbers you would be surprised 741 op-amp the input resistance without feedback non open loop input feedback input resistance is approximately 1 mega the output resistance of the open loop 741 is 75 ohms very high, but if you use is it in the non inverting amplifier mode if you use it as a non inverting amplifier and you substitute these numbers 1 plus a beta will just see quickly how you calculate beta. Now, if you substitute it you would see that these numbers drastically change that depending on the beta you have there the input resistance can jump from 1 mega ohm to maybe let us say at this theoretically 100 mega ohm you may not get that because 100 mega ohm getting 100 mega ohm in a circuit is very difficult because of leakage resistance your PCB you can never get that in a micro board because of the leakage currents, but at this theoretically you have a possibility therefore, you would see that very high in a very precision application this is why amplifiers are soldered and printer circuit boards are made to avoid leakage currents. Now, so we see that in a voltage amplifier the input resistance increases by 1 plus a beta the output resistance decreases by 1 plus a beta. Now, another way of arguing is we can say that whenever we have a series connection at the input the input resistance goes up by 1 plus a beta we know that we get a series connection only when we have a voltage there. So, whenever you have a series connection at the input the input resistance would go by 1 plus a beta whenever you have a shunt connection at the input the input resistance would go down by 1 plus a beta. So, the same thing you could apply to the output also. Now, let us quickly look at the others other amplifiers also. Now, for the current amplifier by applying feedback this is the detailed diagram the equivalent circuit. Now, as we said we have a shunt connection here and we have a series sampling here we know that in a current amplifier the in an ideal current amplifier my input resistance here should be 0 and we know that in any current amplifier you can never get that number, but by giving negative feedback you can make that r i very close to 0 or you could reduce it substantially and you can see that how it happens here you have a shunt connection. So, therefore, the previous r i gets reduced by r i by 1 plus a beta. So, you see that in a current amplifier by applying negative feedback you are able to reduce the input resistance by a factor of 1 plus a beta. Now, we know that in a current amplifier the output is a current therefore, ideally this r naught should be infinite and again we know that we cannot get infinite resistance, but because of the series connection we said whenever we have a series connection we get a multiplication. Now, in this case also because of the series connection the output resistance here gets multiplied. So, again it is going from its non ideal state towards the ideal state. Now, let us quickly look at the other two we have the trans conductance amplifier where we have a voltage mixing here. So, therefore, immediately we can say the input resistance increases by 1 plus a beta. Now, since we have again a series connection here at the output the output resistance increases by 1 plus a beta and we know that this is a trans conductance amplifier which means you input is a voltage input therefore, ideally the input resistance should be infinite. So, by connecting the series connection we are going towards that direction similarly, the output is a current and we know that ideally r naught should be infinite and by connecting this negative feedback what would happen is r naught would go towards that side. So, it would increase drastically. So, the output resistance increases by 1 plus a beta and the input resistance increases by 1 plus a beta. Coming to the last one we have the detailed diagram here the trans resistance amplifier now here we know that the input is a current and the output is a voltage and as we said ideally the input being a current this r i should have been 0. Now, by connecting a shunt connection here we are going towards that. So, if the previous input resistance is r i the one width feedback would be r i divided by 1 plus a beta. Similarly, the output side we have a voltage here and we are sampling a voltage and we said whenever we do a shunt connection here the output resistance go down and this being a voltage here we know the ideal situation the r naught should be 0. So, it would go towards that direction we would get. So, in summary in a shunt amplifier feedback amplifier the input resistance decreases by 1 plus a beta and the output resistance also decreases by 1 plus a beta. Now, what I will do is may be another 10 minutes I will spend on the feedback topologies before going to the second topic which is oscillators. Let us see some examples of what we studied let us see a few examples we have 4 topologies. So, in each case I have chosen a ideal situation using op-amp and most of some of the circuits would be very familiar to you some of the may not. Now, in a good a very good example of series shunt feedback is the non-inverting amplifier which we are very familiar with and now we see what is beta. Now, V f is nothing but V naught by V naught times r 1 by r 1 plus r 2. So, we know that V f is equal to beta times V naught therefore, beta is nothing but r 1 by 1 by r 1 plus r 2. So, we said in the case of a series shunt feedback or voltage mixing voltage sampling feedback the input resistance increases and intuitively we could see that here the connection there is no other resistance connected to ground or anything here. So, intuitively we can see that the voltage which is fed back will decrease this V i will try to reduce it. So, effectively the current flowing into the amplifier gets reduced drastically and because of the shunt connection here the output resistance would decrease drastically and that again would make it towards the ideal situation. So, this is a circuit which is an example of series shunt feedback. So, we could remember at least this particular example we could remember that a non-inverting amplifier is the case of a series shunt feedback and that is an example of a voltage amplifier therefore, it is a series shunt feedback circuit. Now, what we have here is another very very familiar circuit again another example of series shunt feedback which is nothing but an emitter follower. So, emitter follower is again an example where you see that the because of the resistance here unbypass resistance here in the emitter this is subtracted and we said when we talked about biasing we said that by putting a resistance unbypass resistance in the emitter we said that any changes especially we said that V B is very sensitive to temperature, but that change is countered by the resistance unbypass resistance put on the emitter. So, we see this as a very good example of voltage I mean sampling voltage sampling mixing or series shunt feedback and we know that the input resistance will increase and we know it very well from the example of an emitter follower the input resistance is much higher and again from the example of an emitter follower we know the output resistance is very small. So, again a very good example a discrete circuit example of series shunt feedback. Now, let us look at the next topology. So, let us see again two examples of shunt series feedback or essentially the feedback for a current amplifier. So, what we have is a opamp based current amplifier. So, you have the current source here and you can see the a certain amount of the current flowing into the opamp and you have the feedback current here flowing into the output. So, you can see whatever is I s here I s minus I f is what is flowing into the input which is almost 0 and the current I naught is shown here. Now, this being a current source the current flowing through the load in this case the load here is R 1 here. Now, in a current source the same current should flow irrespective of whatever resistance you put here. So, effectively we by analysis we can see that the output resistance of this particular amplifier would be very high. So, that the current I mean irrespective of I mean the you would get the same current. Now, let us look at the another discrete circuit example a quick example of a shunt series feedback which is a current amplifier and you might remember when we talked about common emitter amplifier and I mentioned that a common base amplifier is a example of a current amplifier. So, what we have is a conceptual diagram of a current amplifier. Now, again the same thing you have the way is connected here is to show the way a feedback topology. Now, you have the input current here and you have the feedback current this is the output current and you have a that feedback here and you have the input current shown here. So, this is conceptually a diagram. Now, what I have here is an actual diagram which looks quite complex, but let us have a quick look at it. So, what you essentially have here two stages you have a common emitter stage here and another common emitter stage here. The second common emitter stage you have an unbypassed emitter resistance here and this is a current amplifier. Therefore, what I need to do is I need to take a shunt connection. So, you can see a shunt connection taken here and you have a feedback resistor here R of here and output we need to sample current. So, we know that approximately the current flowing there would be approximately a rather a proportion of that would be flowing in the emitter here and that is sampled here. Now, one and what we have here is an equivalent circuit. Now, I would upload this in modal. So, you could look at these circuits leisurely, but one very important thing need to be kept. Now, when we talk about feedback you would see that when you use an op-amp you would get almost close to ideal situation, but discrete amplifiers like the ones which you made make with transistors are very often very difficult to analyze and things may not be as straightforward as it is. Now, the way you would identify is to look at the way a feedback is done like you need to figure out whether the voltage I mean the feedback at the input is there a shunt connection and again of the output side. So, sometimes things may not be very straightforward may be difficult to figure it out. Now, this is an example of a an op-amp base trans conduction amplifier again you have a voltage as the input a current as the output and the voltage which is fed back here is V f here which is a fraction of the output current. Now, you have another example of a trans conduction amplifier using a discrete example. So, this is nothing but the common ampere amplifier which we are familiar with unbiased. So, the unbiased common ampere amplifier is an example you can think of that as an example of a trans conductance amplifier with negative feedback and we know the reason input side you have a voltage here. So, you have a series mixing here because of which the input resistance increases and the output is a current and if you analyze if you analyze a common ampere amplifier with unbiased resistor you would see that the output resistance increases. So, this is why this is a series feedback. Now, coming to the last one again this is a circuit which is very familiar with this is the circuit of a inverting amplifier. So, an inverting amplifier is a very good example of a shunt feedback now where now generally we do not appreciate this as an example of a trans resistance amplifier, but I assume that you would have definitely noticed something very interesting in a inverting amplifier in a inverting amplifier because of the negative feedback the because of virtual ground property these two terminals will be the same potential. So, essentially if you use this as a voltage amplifier the input resistance is nothing, but R 1. So, therefore, inverting amplifier is a very very bad choice as a voltage amplifier. So, if you need a voltage amplifier you always need to use non-inverting amplifier. So, strictly speaking an inverting amplifier is essentially an example of a trans resistance amplifier and the current which is flowing here. So, you can think of the input as nothing, but a current flowing here and one thing very very interesting here is we have the ideal situation we have the short circuit. So, this is beautiful op-amp gives you the ideal situation. So, if you have a short here and we know because of virtual ground property this also will be a short here. So, the entire current from the current source will flow here and that entire current assuming hardly any current flows into the op-amp would develop across this. So, the inverting amplifier is a very good example of a shunt shunt feedback. Coming to a discrete BJT based amplifier this is an example of a shunt shunt feedback. So, you can see that you have a connection from the collector to the base here and again whenever you have a shunt connection at the input we know that the input resistance decreases and again as I said being a BJT amplifier we should remember that it is not as ideal as it is in the op-amp case. So, analysis also may be difficult, but again at the output side also you have a shunt connection. Now, this is a trans resistance amplifier. So, input is a current output is a voltage. So, I hope you have now some appreciation of what negative feedback is and what the feedback topologies are. I have about roughly 15 minutes. Now, what I will do is we will go through the second topic which is on oscillators and depending on the time we would again have a quick recap. Now, the first topic which we talked about negative feedback is an extremely important topic, but unfortunately it is not well done in most of the books and things could be quite confusing. So, I hope you would have got some understanding and the way to understand these concepts is to go over it again and again yourself by looking at the block diagram and trying to argue and things would make sense. Now, let us now look at the second topic. So far we were talking about negative feedback and we said negative feedback when we say feedback generally we mean negative feedback. Now, in a we want to discuss we want to talk about applications of positive feedback. Now, positive feedback generally is thought to be something very undesirable. Yes, it is undesirable in an amplifier. In an amplifier if you have positive feedback you would have instability and you will not get what you want, but positive feedback can be used carefully to our advantage. So, what we have here is again a block schematic, again the same thing which we did at the beginning. The similar exactly similar block except that the mixing which we are doing at the input now is a addition. The first case it was a subtraction now it is an addition that is only difference, but just by doing that the circuit changes drastically. The previous case we had a negative feedback, now we have a positive feedback. I leave it to you to analyze this is extremely simple just three equations just like we wrote there. First equation is x naught is equal to a times x i, second equation is the parameter x f here is nothing but x naught times beta. The third equation is x s rather x i is equal to x s plus x f, three equations from which you could write an expression for x naught by this should be x s or x i, x s should be then you would get this to be a by 1 minus a beta. Now, look at this expression the previous case the expression what we had was a by 1 plus a beta. Now, it is very obvious that we see here that in this case if for any reason the a beta happens to be unity. Then we have a situation where the closed loop gain if we use the same term as we used in the case of amplifiers you get a situation of a closed loop gain being infinite. Now, what do you mean by closed loop gain being infinite? Now, you could argue it this way in an amplifier you gave an input and you got an output just amplified version, but when you generate a signal let us say you want to generate a sine wave you do not give anything all that you are giving is a battery and that battery or the power supply is used to bias the circuit. So, actually as a signal you are not giving anything. So, without giving any signal to the input you are able to get something at the output. So, that way conceptually or intuitively we could understand what you mean by the gain being infinity. Now, in the previous case we use the word loop gain or a beta same thing here also. Now, there is a very big difference between what we did for negative feedback and what we did here. Now, in the context of positive feedback or generating let us say a sine wave what we are interested in mind we know that we are talking when you talked about an amplifier the negative feedback the feedback network was entirely a resistive network you might remember. So, this is something very important to remember the case of a negative feedback the feedback network in the in the context I am talking in the context of an amplifier. Now, in a negative feedback the feedback network has to be resistive. So, essentially you are not bringing any frequency dependent function there. So, the beta which you can think of as the transfer function of the beta network is purely resistive and is real whereas, when we talk about this application here the feedback network is not resistive. In fact, it is frequency selective you could think of this network as a filter. Now, you this filter need to be to have an order of 2 or higher at least should be a second order filter. Again you could think of the say conceptually or intuitively you could think of the working of a an oscillation or oscillator like this you could think of noise being present in the circuit rather in the environment and these frequencies selective circuit which is a filter which would pick up only one frequency and only for that particular frequency we would satisfy it will satisfy the condition A beta is equal to 1. Now, as I said in this case A beta is not real it has both it is a complex term. So, when you say A beta is equal to 1 you have two things. So, let me write that. So, the condition for oscillation we saw that is A beta is equal to 1 and we said here definitely beta, but we might assume say A to be real, but definitely beta. So, let us say A beta is a complex term. So, to satisfy this condition what we are essentially saying is modulus of A beta is equal to 1 and angle of the phase of A beta is equal to 0 or say 2 pi or multiples of 2 pi. Now, this is the condition for oscillation. So, if I can satisfy using a frequency selective network these two conditions the condition is that the modulus of A beta must be equal to unity and the phase is equal to 0 essentially saying that. So, what is the meaning of this essentially what we are saying is at the frequency of oscillation A beta must be real and it should be equal to unity. Now, let us look at an actual circuit and see how do we satisfy. So, before we see that let us see how do we actually implement an oscillator. So, the way an oscillator is implemented is this way you would have an amplifier and what you would do is you would connect the amplifier you would have a kind of back to back connection of an amplifier and a filter. So, essentially what we would get is. So, let me draw this and show you. So, what we would we are talking about is we have an amplifier whose gain is A and what we do is we would feed that into a network which let us say is a filter. So, let us talk about this being a filter and the output of the amplifier is fed to this filter and the output of this filter is then fed back to the amplifier. Now, as I said in any circuit we have noise present now the purpose how do we initiate oscillation. Now, the way it happens is you could think of the filter satisfying this condition A beta is equal to 1 only for 1 case and that would be for a particular case. Let us see now an example and see how exactly it is done in actual practice. Now, what we have a very popular circuit which is a very popular sine wave oscillator which you can easily rig it up in the lab and almost always it will work at the first instant. But, let us see how what condition we need to satisfy. So, let us understand this as again as I said you could think of this as two blocks. One block this for first block here if you remove this particular network here the if I assume as some input coming into this v plus here you could think of this block as nothing but a non-inverting amplifier and we know that a non-inverting amplifier the word non-inverting itself means that there is no phase between no phase difference between input and output. So, if I have a signal here x i here my output will be amplified by A times x i without any phase change and the ways connected here we have two resistors here will come to that in a minute and then the output of the amplifier is fed back into a filter here. So, essentially what we have is a network here you have an amplifier and a filter connector connected say back to back. So, the output of the amplifier connected to the filter and that is all if you just connect supply when you switch on power it oscillates how does it oscillate. Now, let us look at the filter characteristics to understand how it works. Now, this is the same circuit. Now, this is a popular second order filter and this can be easily analyzed and if you write the transfer function if you write transfer function v naught by v i you would see that the transfer function would be real only at one frequency and that frequency is 1 by r c at omega is equal to 1 by r c you would see that the v naught by v i will be exactly one third. So, if you plot it you would see that if you plot the modulus of this. So, you would see that this particular second order filter would give you maximum value at omega naught where omega naught is equal to 1 by r c. So, now what is the meaning of this one third here. So, we see that now again going back to our circuit we can see that this non-inverting amplifier. Now, we have 2 r 1 here and an r 1 here therefore, we know the gain of this non-inverting amplifier is 3. So, let us leave this small resistance here for the time being. So, I have an amplifier who has a gain of 3 and here is a network which has a maximum value of one third exactly at a particular frequency. So, in this case the condition a beta will be satisfied easily only at one frequency and that is a frequency omega is equal to 1 by r c at that frequency the whatever is input here will be attenuated by factor of 3 or if input is V i here you will get exactly V i by 3 and the amplifier will amplify exactly by 3 and at the input again back here you got the same V i. So, if you look around the loop the loop gain you see is unity and that is the condition for satisfied to be satisfied repeating let me repeat it once again. So, if I start from here if I break this here apply a V i here and I know that at omega is equal to one by r c my output of my filter will be one third of V i. Now, I amplify that by a gain of 3. So, I get back my V i. So, therefore, if you connect them together at least it is to oscillate. In fact, you can conduct this experiment in the lab break that loop and to you exactly make it one third and then just close it and you will be amazed that will continue to oscillate. So, this is how a V n bridge oscillator works. Now, let me come to this particular small resistor which we have put here what we talked about was all ideal. We said ideally the gain of this amplifier must be 3, but we know that we are using all passive devices here rather passive elements here capacitors and resistors. Now, these resistors and capacitors will have losses. Therefore, if you exactly 3 you will never get oscillation rather if you keep gain very high you would see that the output will not be a sine wave it will be a distorted sine wave. Now, it is very interesting. So, I would if you have not done this experiment yourself I would strongly urge you to do this experiment and you will be thrilled like a child to see this even today when I do this experiment it thrills me to see some of the things which we talk about how you can easily see it in the lab. What you should do is this particular resistor here do not make it 2 R may be make it 1.5 R 1 and put a potentiometer here and you would see that as you increase the potentiometer as soon as it satisfies the condition you will see the sine wave slowly popping up coming into oscillation. And if you just leave it there you would see a sine wave just dancing it might just collapse it might continue and if you increase it further you would see at some stage it is being stabilizing if you go further you will get distorted. So, this particular circuit is an extremely useful circuit the same time it teaches you something very very important very very important that this circuit cannot be used in a as a practical circuit because you need to satisfy that condition exactly which you can never do we cannot do it manually. So, therefore, V and Pitch oscillator when it is implemented what they actually do they would have a non-linear amplitude control very often what is done is you would put a Siner diode across this out of the feedback path and you would choose such a way. So, that the if the by any chance the gain increases the Siner path would decrease a resistance of the overall impedance there and that will keep it in oscillation. So, I have come to the end of my lecture let me spend another 2 minutes to quickly recap on what we have done. Now, we talked about let me maybe spend maybe another minute on positive feedback because I spend lot of time on negative feedback. So, positive feedback we said in this context we are talking about oscillators. So, positive feedback we said we get when if you take a sample from the output and you mix it at the input and add it not subtract it if you do that when you write the transfer function we said you will get an expression which is a by 1 minus a beta and we see immediately that if I can satisfy a beta is equal to 1 then the transfer function will have a value infinite which does not make sense and we said what it essentially means is that you get an output without giving any input and that is exactly what is an oscillator. Now, we talked about the condition for oscillation we said that is a beta is equal to 1 which means because the beta network in this case unlike the amplifier is a frequency sensitivity circuit or a frequency selective circuit and we said in an actual oscillator you would actually use a filter a second order filter or higher order filter. In that case this condition will be satisfied by satisfying two independently two conditions one is the modulus of a beta must be equal to 1 the other one the phase of a beta must be equal to 0 and we said what we actually do is to implement you would choose an amplifier and you would choose a filter and whatever is the attenuation of the filter will be compensated by the gain of the amplifier. So, that your modulus a beta is equal to 1 that is why you would implement. So, in a VN bridge oscillator which we talked about we said in a VN bridge oscillator the gain is chosen as 3 because the this particular the attenuation of this particular network is one third. So, to offset the attenuation you need to multiply by 3. Now, just to give you one quick example you also have another popular RC oscillator you also can talk about LC oscillators. LC oscillators are used at RF frequencies let us say 1 megahertz and high and these are very difficult to let us say analyze and even use them in the lab and RC is very very easy to use to understand concepts. So, this is something you can easily do and VN bridge I encourage all of you to do it yourself. So, that you can teach your students much better and the conceptual understanding will be much much better very simple circuit. Now, you also have an example of a phase shift oscillator where you have you would use a it so happens that you would use a 3 RC network you know kind of networks in series there and the attenuation of that network if you do the transfer function you would see that is our 1 by 29 minus 1 by 29 therefore, the gain of that amplifier should be minus 29 to get unity. So, I think I have come to the end of my lecture.