 So, this set of nodes was indeed made for exactly this course and put up on our Moodle side for this course. So, essentially in this you have a generic amplifier, then bipolar amplifier first take the transistor characteristics, introduce early voltage which determines the G 0, take the pi model of the transistor which you can then use in designing the thing, then bipolar common emitter common base and common collector amplifier. After that there are single ended amplifiers, differential amplifiers, some definitions of differential and common mode quantity, operational amplifier and then various configurations of that and some of the things that you wanted me to discuss like a virtual shot, where does it come from etcetera. And then the non-inverting configuration and a few applications like inverting adder differentiator integrator and non-linear uses of op amp like a logarithmic amplifier. And finally feedback and oscillator, so first negative feedback then positive feedback, amplifier with positive feedback and finally oscillator and as an example of that phase shift oscillator. So, this is my old notes which I had used for this course itself, I think it more or less mimics the kind of things that we have discussed that they and which are there in your course. So, essentially and I think they have some problem chasing down what I am writing on the board. So, for figures we will use the projection, but for discussion we will use the board. So, that way they would not have too much difficulty filming and we can discuss at a comfortable pace here. So, let us start right from the top, not make any assumptions about amplifier. Now, the first thing is that an amplifier takes something which is small amplitude and makes its amplitude larger linearly, that is the basic idea of an amplifier. Now, because the input is of small scale, essentially lots of linearization approximations can be made. Now, you do not have to use this high polluting technology, you know terminology with your students. Essentially what it means is that therefore, we have two separate discussions whenever we talk of the amplifier. First is the DC conditions which determines at what bias point will be operating and the second are small fluctuations from this bias point. So, that means we separately discuss the DC analysis of the circuit and then small fluctuations which is the small signal analysis of that circuit. And in case of amplifiers it makes all the more sense because the inputs are guaranteed to be very small. And therefore, we can assume that the deviation from the operating point of a circuit is always small and therefore, we can use first order approximations there. So, let us assume that we are going to use all of this. Remember that for bias considerations you cannot use this linearization because the value of the bias is not small. You may have 1.1 volt, 1.2 volts, 1.3 volts whatever. So, therefore, that is not small. Given that background, let us take a generic amplifier in which we have not said whether it is common emitter, common base, JFET, MOSFET, bipolar, whatever. First of all, let us consider the general properties of an amplifier. So, this black box is our amplifier. I have the input here and I have some magic inside here which I am not going to talk about right now. Obviously, some amplifying device will be required, but our initial discussion does not drag you through the nitty-gritty of what this device is. Let us say that it is some device whose current voltage characteristics are known to us. We will later see what kind of current voltage characteristics should this device have so that we can get good amplification. So, right now I have connected some device here. It could be a bipolar transistor, FET, MOS, could be several bipolar transistors connected in a particular way. Right now I do not care. Now, this load I have taken as a simplified in a current source and before we talk of this, we should therefore talk of current source, what current sources are. So, we will look at the practical implementation of current sources a little later, but right now we just introduce a current source as a source which supplies a fixed amount of current irrespective of the voltage at its terminal. Just like a voltage source provides a fixed amount of voltage irrespective of how much current you draw from it. A current source is a source which will provide you with a fixed amount of current irrespective of how much voltage is across it. So, these are two parallel approximations. While practical examples of constant voltage sources or at least approximations to constant voltage sources are easier to imagine and see. For example, a battery cell is pretty close to being a voltage source, it has some resistance. However, current sources are not so common and therefore, it requires the practical implementation of current sources takes a little bit of discussion, but that we will postpone for later. Right now let us assume that I do have an ideal current source. So, this essentially what it means is that it will always give you that much current, will not allow any deviation from this current whatever you do to the voltage at this particular point. So, we have put such a device here. Alright, I do not know whether you can read it through. Actually, I have a presentation version of this. This is the printable version of that course. I also have a presentation version, but that you know I never trust these windows things. So, I have not brought that along. This PDF file is somewhat better. I do not use power point. So, the current through the device, let us say it is a function of two voltages. By the way, this might be the point to introduce common emitter, common base, etc. amplifier. Because if you look at it, there are two input terminals and two output terminals, ground and input, ground and output. So, overall there are in principle four terminals for this. However, most devices that we normally use for amplification have three terminals. For example, a bipolar transistor has a collector base and emitter. A JFET has a source, drain and gate. So, there are only three terminals in an amplifying device normally. Therefore, at least one of them has to be common to the input and output because you need four. And whatever is common then defines what kind of amplifier it is. So, to get four, one of the terminals has to be seen both by the input and by the output. And that is the common terminal. And this is the way to give at least the terminology of common emitter, common base and common collector amplifiers. We will see the analysis of these a little bit. However, let us forget about bipolar transistors or anything that you have to consider and look only at this black box picture of an amplifier. Now, the output current of that device, this device, the output current of this device I out that can depend only on these two voltages because it cannot see anything else. This voltage and this voltage. Now, notice that following standard convention capital letters will represent the DC value and small letters will represent flux phase of those values. So, then I out is some function right now we do not know what function it is, but it is some function of V in and V out right because V in is this voltage and V out is this voltage and nothing else is visible to this. So, the current has to be determined by these two along with the internal properties of this device right. Therefore, just from simple calculus if I was to write the total differential of D I out, that will be partial differential D I out by D V in times the amount of change at the input plus the partial differential of the second with respect to the second variable times D V out. Now, this does not involve any electronics. It is just that if you have a function of two variables, its total variance of this is given by the partial differential with respect to first times the amount of change in the first plus the partial differential with respect to the second times the amount of change in the second. This is just mathematics that all of your students would have done in first year. Now, what we do is that I am going I design I call this gm this variable as gm ok. What is gm? The rate of change of output current with the input voltage right assuming that this term is 0 right. So, when will this term be 0? If the output voltage is held constant right. So, gm is the rate in rate of change of current of this device if you keep the output voltage constant right rate of change with the change in input. So, if I make D V out equal to 0 then this term is nothing but D I out upon D V in right. This partial derivative becomes equal to the total derivative because the it is a function of only two variables and one variable I have made constant. Therefore, partial derivative and the total derivative are the same that is what we are saying. So, let us call that gm similarly I define this to be g 0 and what is g 0? g 0 is D I out upon D V out while D V in is 0. That means, if I hold the input voltage constant and change the output voltage how much is the change in the current of this device ok. So, that I call g 0 right. So, this is just a matter of definition of terms. Now, I can write I out notice what is this D I out? This is the small signal change in the output current right small I out. So, I out equal to gm times this remember is gm what is D V in is the small v input small small change right that is a signal. So, time small v in plus this is g 0 into V out right. Now, notice that I out what is I out fluctuation in the output current right it is a lower case I not the upper case I, but the fluctuation in the current is 0. Why is it 0? Because it is fed by a current source which does not allow this current to change right. So, therefore, the variation in the current is 0 because it is fed by a current source. Therefore, we must have 0 equal to gm into V input plus g 0 into V out right this left hand side will now become 0 and that tells us that the ratio of the output voltage to the input voltage is nothing, but minus gm by g 0 or minus gm r 0 right. So, even without getting into any transistor theory anything at all what we have got is that the amplification can still be defined in terms of these circuit variables gm and g 0. Now, if you want a high value of gain what do you want? You want gm to be large and g 0 to be small right. So, what does large gm mean? What is gm? The I out upon rate of change of the output current with the input voltage. So, what you are saying is that the current change in the device should be large with respect to changes in the input voltage right that is what large gm is and what is g 0? g 0 says that with respect to the output voltage the current should not change very much right. So, in short what it means is that give me a device which I shall now bias and with respect to the output voltage the current should remain reasonably constant with respect to the output voltage ok. Now, we know devices like this a bipolar transistor has characteristics like this. So, if I use a bipolar transistor and this is the plot of collector current versus collector voltage. So, if I make collector the output terminal then I am fine right then I will be able to get high gain because the current will not change a lot with voltage and which is what I want for high gain right. At the same time its rate of change with input voltage in short a high beta transistor right or a high gm MOS that means the rate of change of current with the input voltage should be large and rate of change of current with the output voltage should be small. This we can derive without getting into any device physics just from amplifier theory right. So, this is the first point to be established. Now, not surprising that most of the most of the devices that we use for amplification have IB characteristics like this. If you look at a transistor the characteristics are like this. If you look at a JFET the characteristics are like this and if you look at a MOSFET the characteristics are like this. So, it is not accidental we choose those devices for amplification which have characteristics like this ok. So, only those devices become popular those are the devices that we use for amplification ok. This can be derived from a totally general consideration. Now, suppose I do not have a current source alright. So, you can say I went to Leamington road and I asked for a transistor and they have a transistor, but I asked for a current source and they did not have any current source cannot buy a current source from Leamington road. So, what do I do? So, let us say we will put a resistor here let us see what happens. If I put a resistor here then can I make an amplifier alright. So, let us put a resistor here now. Now, if you put a resistor out see earlier what did we do? We had said that D i out be equal to 0 right the output current cannot change, but now I have a simple resistor if the voltage changes the current will change right the current through the load here will now change if this voltage changes correct. So, therefore, I cannot put D i equal to 0 how much is D i in fact D i is V out divided by R right in a negative sign. If the output voltage increases then the voltage difference across R is smaller and therefore, the current will reduce right. If the output voltage decreases then the voltage across this resistor is higher and therefore, the current will be higher right. So, therefore, the rate of change of current through this has a negative sign and it is linear in that right that is what ohm's law. So, let us put that in right. So, what happens when resistor is used instead of D i equal to 0 you have to put it minus output voltage change divided by R L that is the amount of current change. Current change is now no more 0 it is equal to notice this is small V out the fluctuation right and if the voltage increases the current decreases if the voltage decreases the current increases therefore, the sign negative. So, minus V out by R L equal to this these two terms the right hand side remains the same right. So, I can now collect terms in out and output voltage and input voltage and essentially all I will see is that this R L has come in parallel with the output resistance of the transistor 1 by R 0 is same as G 0. So, this tells you that these two resistors have essentially come in parallel otherwise I have the same expression right. So, this is how I can construct an amplifier even using a resistive load. Now, any old device which has characteristics like this and with a resistive load I can get gain what about the gain with respect to this resistor I know now what happens to the gain if g m and g 0 change what about changes with this resistor. Now, notice that this resistor is in parallel with R 0. So, as long as it is much larger than R 0 there will be no problem the smaller of these two resistors will set the gain right the overall value equivalent resistance will be if they are very different then the equivalent resistance will be set by the smaller of the two and that will set the gain. Because I want high gain therefore, I would like to use as high a value of R 0 as I can possibly afford. So, that tells me how to make an amplifier with resistors and amplifying device. At the end of it since you have been doing bipolar transistors and diodes and so on there is a certain amount of mass here which I am going to skip because you have already done. But if you wish you can use this treatment as well otherwise you have done it in your classes. So, essentially this is just the transistor characteristic and this is somewhat exaggerated depiction of the change in output voltage if the output does not saturate exactly and it defines the early voltage. You need early voltage because if you assume that the transistor is perfectly flat then you will get infinite gain because it is g m by g 0 and g 0 will be 0. So, you have to bring in this non-ideality in order to understand the gain of a transistor. Now at this point we can introduce the pi model of the transistor. This is very simple middle frequency pi model in which all capacitors have gone out and then there is this R b, R pi, etcetera, etcetera which can be derived from the transistor characteristics. By the way this is not essential. If you are doing a quantitative treatment this is required but if you are not doing it even the qualitative values will be ok. And now since I know the I v characteristics and I have already defined the g m values I can find out what is the value of g m. So, just by taking partial derivative in this configuration common emitter configuration where the emitter is grounded and the input voltage is applied here output voltage is taken from here. I can define g 0 and now put the load resistor here and now it is easy to see in circuit terms that R 0 and R L are coming in parallel. And then that tells you a practical common emitter amplifier because this guy has to be made forward by to get gain from here to get the appropriate transistor characteristics and that you do by this divider and then the emitter etcetera, etcetera. Now as many of you have said that all of this is very theoretical we need to show them by an example. So, essentially actually in these nodes a specific example is given that you have a V c c of 6 volts I c equal to 1 milliamp and beta is assumed to be greater than 100 need not be a specific value. And then you assume that V B is like 0.633 volts actually calculated from K T by Q L N I by I 0 and so on. And then you 1 by 1 you can calculate the value of all the resistors from that. So, first thing you say that you would like this to be at about half the supply voltage. So, then it has as much room to swing upwards as to swing downwards. So, therefore, I would like to keep it at 3 volts. Now if this is at 6 volts and this is at 3 volts and 1 milliamp is flowing then R c is already decided to be 3 K. So, that gives you the value of R c. So, that tells you that R c is 3 kilo ohms and then emitter resistance we had taken as 300 ohms that gives you the emitter voltage and because you know V B that gives you the base voltage. And because you know the base voltage you can design the voltage divider which will give you from 6 volts this base voltage. So, these are all the values calculated. Indeed this one has been made a little complicated to challenge the students. The first thing that I do in the class I assume that you can ignore the base current and that gives very easy equation. I challenge them saying let us say base current cannot be ignored in that case calculate. So, essentially what happens is that then through this potential divider this guy takes the divider current as well as the base current whereas, this guy gets this guy takes only the divider current. So, you put that in your equation and then the base current is linked to output current divided by beta. But the output current itself depends on what voltage finally results here. So, you have to solve it iteratively and finally, even not neglecting base current you can come you can design the values of this. So, the example which is given here is the tougher part. So, in the nodes what I have given them is the tougher problem. The easier problem I have used in the class to explain for the first time ignoring the base current. Similarly, we are not going to do all of this because we really do want to get to the op-amp. But in these nodes you will find common base amplifier, common collector amplifier. So, all the juicy stuff is here for you to browse through and use if you see it fit for your students, all the algebra is done. So, assume that you have now introduced simple amplifiers and I have some bipolar, you can also do JFET MOSFET like that. All this will change is the basic IV equation and therefore it is derivative. So, the value of gm will be somewhat different, the value of g0 will be somewhat different, but the deriving equations will be the same. Now, a signal ended amplifier is one in which you have v out equal to a times v in. This is what we have done up to now. But the way I motivate differential amplifier is like this. That look you can have a very small signal and you can get very high values of gain you can amplify it. Getting very high values of gain like 10,000 and so on is not a big deal. However, most of the time you cannot use gains as high as this because there is noise on this. And then you will end up amplifying the noise as much as the signal. So, you will not be able to distinguish your signal from noise because you are amplifying the noise as much as you are amplifying the signal if you have a signal ended amplifier. On the other hand, if you have a difference amplifier which amplifies the difference of these two and the two are guaranteed to have the same noise, now you can use gain much more useful. Because even in the presence of noise, you can use large values of gain and the noise will just cancel out. Your signal is the differential of these two, but because the noise is on both of these picked up by both the wires which are put very close together. Therefore, even in the presence of noise like hum like 50 hertz pickup like random noise, if it is picked up simultaneously if it is correlated noise then it can be cancelled using a differential amplifier. So, therefore, we are now looking for amplifier which are not like what we described earlier, but which have this property. So, this is the way we introduce difference amplifiers and the motivation for having a difference amplifier. So, noise picked up by both inputs gets cancelled at the output. Also, if both inputs have the same DC bias, the output is insensitive to changes in the bias. Otherwise if the bias changes then the properties of this device will change quite a lot, but here the bias gets cancelled because the bias is there on both the input. And also later we will use feedback and if you have two input terminals it is easy to decouple the output from the input. The feedback can go to one terminal and the input signal can go to the other. So, these are the motivation points for having a difference amplifier. Then you come up with the standard definition of the differential and common mode quantity. So, then you say that the difference of these two voltages you will call VI differential and the average of the two you will call input common mode. And then output voltage will be equal to a gain with respect to differential and a gain with respect to the common mode. And now if you have an ideal difference amplifier then it amplifies only the difference and therefore ACM should be 0 and V out will be a diff times V diff. In real life of course, life is not so kind to us, it is not possible to reduce the common mode gain to quite to 0 and therefore you measure how close to ideal case you are by defining the common mode rejection ratio by this term. So, you can introduce the CMRR like this. The common mode signal is the average of the two signals. Therefore, whatever is common any two variables x and y can be mapped to two variables u and v where u is x plus y by 2 and v is x minus 1. So, two variables to two variables degrees of freedom remain the same, but the advantage of this is that now the average is defined to be the same for both the terminals whereas the differential always goes in the opposite directions by equal amount. So, that is very useful in this case of differential amplifier. So, you have two voltages at two inputs. You define the average of these to be the common mode voltage and you define the difference of these to be the differential voltage. Now, the idea is that the common mode the average of these should not affect your output at all. So, you are saying that I do not care if one input is one volt the other is 1.1 volt or one input is 3 volts and the other is 3.1 volt my output should not change at all. So, this is the requirement that I should be insensitive to the common mode. My differential is the same 0.1 volt, but my common mode has now changed earlier it was the average of 1 and 1.1 now it is the average of 3 and 3.1. And yet my output does not care. So, the trick now is that let us put the noise in the common mode which will then get rejected. So, that is why for example, you have cables which are twisted together. Why do we use twisted wires? It is actually expensive if you never twisted two wires you will be able to take them much further. Why are you making them spiral like that? The idea is that by twisting them you are making sure that one wire sees exactly the same environment as the other wire on an average. As a result if one wire picks up some signal the other wire will also pick up the same signal from the air from the noise. So, as a result what will arrive at the two inputs will be signal 1 plus n the other one will have signal 2 plus n, but because it is a differential amplifier the output will be proportional to signal 1 minus signal 2 and will cancel. So, you can now make amplifiers which are much more robust with respect to noise. So, that is the basic purpose of resolving it between common mode and differential whatever is common to both terminals get suppressed and we make sure through our design and operating conditions that noise is common that way you will be able to suppress the noise. This is the motivation for difference amplification. So, the ideal difference will have the differential gain to be infinite as high as possible and common mode gain to be 0 it will suppress the common mode completely.