 Welcome to this session of our course on basic electronics. Today's topic is operational amplifier. I would like to just take up a few topics. First of all, thanks for all the questions that you have been sending to me. I have responded to a few on the Moodle site. So, you will find those answers there. However, some questions have been sent to me by email which I do not find on the Moodle. I do not know how these were sent. And when I try to reply to those, then because it is sent from a no reply address, it does not go through. So, whosoever has sent those questions, please join the discussion group on the Moodle for this course. And then that way my answers will be visible to all the participants of this course. But because some questions were relevant and will be required for the future classes, before I begin with the topic of operational amplifier, I would like to take just a couple of questions which were there on the Moodle site. My answer is there, but I would just like to discuss. So, essentially we are still on the topic of semiconductor devices actually. We will go to op amp. So, one of the questions and I think that doubt is quite natural. Ask about the difference between ions which are positively charged could be donors and holes which are also positively charged. Now, both are caused by locations which are deficient in electrons. So, the question was what is the difference between a hole and an ion. So, I had covered it, but I think I went a little fast over it. So, therefore, this doubt has remained and let us just take it up so that it is clear in your mind. Consider a semiconductor in which there are four nearest neighbors always. Now, we will consider two separate scenarios. One is that we have put in a donor. That means, this is a phosphorous atom and we know that these sites are very scattered. There is only one part per million dopant and so on. So, all its neighbors are actually silicon and the lattice structure, the crystal structure is also that of silicon. Therefore, it has four bonds to be shared with its nearest neighbors in the silicon lattice which is the diamond type lattice. So, it has four bonds to share with its nearest neighbors. So, all these are silicon. However, phosphorous actually has five electrons in its outermost orbit. So, it has one additional electron for charge neutrality purposes. However, nobody wants it. Nobody wants to shake hands with this. So, this one being the outside electron is very loosely bound to this atom and because it is not shared by two neighbors, it is kind of free to detach and go away. The amount of energy required to detach this, it is like a hydrogen atom where the inner electrons and the nucleus together form a pseudo nucleus with charge plus one and this one isolated electron outside very loosely bound to this. Therefore, the amount of energy required to remove this electron from the phosphorous atom is quite small and at ordinary temperatures practically none of these electrons are attached to their parent phosphorous atom. There is enough energy at room temperature or even at reasonably cool temperatures. There is enough energy present to detach this electron from the phosphorous atom. So, now what happens is that this guy moves away from here. It could be because of local electric fields. It could be because of diffusion. Whatever happens, this electron has now moved away leaving this phosphorous atom pretending to be a silicon atom. It has four electrons shared with in four bonds and those four electrons are tightly bound. However, it has a net charge left here because its nucleus is not like that of silicon. Its nucleus is that of phosphorous. So, therefore, it has one additional positive charge in the nucleus and hence this side is positively charged. This is an ion. This is an ion because it is an atom which is positively charged and which is fixed in its place. This positive ion cannot move about in the crystal. It is just fixed there. This positive charge is therefore called an ion. Now, consider a corresponding scenario where instead of phosphorous, we have silicon. This is where happens. This is what happens in most of the lattice. In order to create an electron hole pair, now we need pair amounts of energy. Therefore, this happens quite rarely. Indeed, out of 10 to the power 20 or 22 atoms or so, we know that number of electron hole pairs intrinsically without dopants is of the order of 10 to the power 10. So, this is a very rare event. However, it does happen. One time out of 10 to the power 10, it does happen and when it happens, then this bond is broken. One of the bond is broken. This makes two electrons here disconnected from each other and one of the electrons now moves off because this bond is broken. However, this lack of electron is not localized. All the electron clouds which are forming these bonds, they are connected to each other. Therefore, there is no stamp that this electron belongs to this particular silicon atom. All of these are in an extended wave function and are jointly shared by all these neighbors. Now, if you were to apply a field, let us say that you applied a positive field and there is a net positive charge here, which is somewhat similar to the ion, but there is a crucial difference here. There is a net positive charge here. If you apply a field with the positive end to the left here and a negative end to the right, then this entire electron cloud will have an attraction in this direction and a repulsion from this direction, all the electrons. As a result, electron cloud will shift towards this deficient silicon leaving this guy deficient in electrons and then eventually this guy deficient in electrons and eventually this guy deficient in electrons because the entire electron cloud feels this force of the field and shifts to the left towards the positive. As a result, the net deficient silicon location is not fixed in space. That is the difference. In the previous case, the electron deficient place was in phosphorus. Once this electron went away, it is slightly bound by four bonds and nobody is willing to move electrons even if you apply a field. The electron cloud may be distorted, but there will be no net movement of charge. Whereas here, all the electrons are connected. They are all equivalent. All are silicon and you apply a field. There is no reason why this electron, which was shared and more towards this silicon atom will not move here and will be more towards this silicon atom making this atom now electron deficient and then eventually this atom more electron deficient and so on. So, as a result, the site of electron deficiency will move when you apply a field. This is the major difference between a hole and an ionic charge. So, in the previous case, this is an ionic charge. In the current case, you have a hole. In the previous case, when the electron was plugged away from the atom, the net output apart from the crystal was that there was a free electron and a fixed site positive charge, which is fine. In the new case, you have a free electron. This bond was broken and one electron moved off and a hole, which is mobile and it is mobile because the entire everything is equivalent. This is the silicon atom nearest neighbors are also silicon. There is nothing peculiar about it and if it is electron deficient, the entire community of silicon atom is electron deficient and when you apply a field, the location of major electron deficiency moves in a direction as if a positive charge was moving and that virtual particle we call a hole and it is mobile that is the main difference. So, I hope now the difference between an ion and a hole when ionization occurs. In both cases, an electron is wrenched away from its rightful place. So, that part is common. However, what follows is quite different and the major difference to remember is that in one case, the charge is localized. In the other case, the net charge due to electron deficiency is free to move in the lattice if you apply a field. Indeed, it is also free to move if there is a diffusion. So, it follows as if there is a concentration of particles, which are positive in charge whereas, actually it is actually deficiency of electrons. However, the major point is that they are mobile. So, I hope the difference between the two cases is now clear to you. There were a few other questions, but I think we will end up doing a devices class here and we need to move on with op-amp. So, I will answer some other questions on Moodle. They had to do with very interesting points like for example, why should n times p remain n i squared even when the semiconductor has been changed from intrinsic to expensive. We have doped it. So, why should the constant that we had assumed for that chemical reaction? Why should it remain the same when the properties of the material have changed? It is a very interesting question. It has a good answer. Unfortunately, we do not have the time to discuss it right now. I will perhaps take it up in a similar little insert pocket in some other course if you have time. Otherwise, I will answer it on Moodle. So, finally, let us begin with Seymour's op-amp. So, what are op-amps? This term is very widely used and the motivation for having an op-amp should be understood first and then we will go on to see its properties and see what it actually does. So, we have all kinds of analog circuits and each circuit is unique in its own way. You may have a filter, you may have a differentiator, you may have an integrator, you may have an analog adder, sub tractor, what have you. All sorts of circuits exist and these circuits have to be individually designed. The analog designers then look at the digital designers. Life is very simple in digital design. The circuits can be taken to be black box. As long as it performs that logic, then you do not care what is inside that black box. And the choice of operational blocks and gate and gate etcetera is quite limited. So, you have standardized devices and you just learn tricks of putting these standardized devices to accomplish the function that you want. You can make now a counter, an adder, what have you using this logic, using combinational logic or storage device. So, it makes us analog designers quite envious because here you have anything you have to start taking a transistor and its resistor load and its operating point and its bias and it is a place in the neck to design it right from the beginning. So, the question we ask ourselves is could I could I please have a standard circuit like the digital guys have, why cannot I have it. So, please give me a standard circuit which I can then use as a black box that my circuit will be free of the details of the implementation of this circuit as is the case with logic gates. I can perform the same function with a CMOS gate with a TTL gate with any gate as long as it is an AND gate it is an AND gate. So, could you please give me something standard which I can use in a variety of applications without bothering about the innards of this circuit. So, because I am a sincere kind of a guy and I have made such an impassioned plea for this component. So, there is suddenly Akashwani and it says could you be please more specific could you decide what this component will be this standard component. So, as always whenever you ask for something you are given a homework. So, let us do that homework first what would be this standard device with which I could perform many, many analog functions without worrying about its innards. So, let us say and because I am a simple kind of a guy I will choose a simple kind of a symbol for this device which I do not know what it is right now. Now, first of all because nothing should depend on this on the innards of this circuit it should not load the previous driver at all that means it should look at its input, but not draw any current draw not draw any charge or in anything that means the previous stage should not even know that there is something connected to it. Now, obviously I am asking for the moon, but when Akashwani do not occur every day. So, I will ask for the moon. So, could I please have a device which can be connected to something and that something is not disturbed at all by it. So, what I am saying is that here is some complicated circuit as analog circuits often are all sorts of R L C transistors God knows what mess is there and what I am saying is that I will take this place and connect little magic component of mine actually it has more input than 1 we will come to that later and it should not disturb the circuit at all that means there should be no current flowing in this direction if there is current flowing in this direction then the circuit will know that something is connected to me the voltage drops across these resistors etcetera will change the net voltage at this point will change and I have disturbed the original circuit I do not want to do that that means the input impedance of this should be infinite. So, therefore, this is one of the boons that I am asking from the Akashwani that I want an infinite input input at the output I should be generous that means if the output is suppose to have some voltage you draw any amount of current from me and my voltage will still be that predictable voltage. So, you load me with 10 ohms resistor or 10 mega ohms resistor I do not care my output voltage will be what it is right. So, I want to make that promise as well that means if this circuit is connected to something may be connected to a similar mess here God knows what. So, it is connected to all sorts of complicated things and they might not be as well behaved as I am they may actually draw current and they may sometimes draw less current sometimes they might draw more current. So, then my output voltage should be predictable it should not change what does it mean what does it mean for my output impedance. So, I can think of this as a Thevenin source with some output impedance right. So, my model of the output of my op amp is some voltage source with series impedance and what I am saying is that this is the visible output voltage and what I am saying is you draw any amount of current from here could be less could be more could be heavy could be light I do not care draw any amount of current and this voltage will not change. So, what does it say about this resistance it should be 0 if it is non-zero then whatever voltage I intended here there will be a drop across this resistance and the amount of that drop will depend on how much current I am drawing. So, if I have to be totally independent of the amount of current drawn from me. So, essentially what it says is I am an ideal citizen I shall not ask for anything at my input, but I will give everything at my output and never complain never let the world know my lips are sealed. So, take what you want from me and my out there will be no change in my behavior and that is only possible if the output impedance is 0 finally, I do not care at which frequency I am operating. So, because this has to be a catch all device should not its behavior should not be dependent on which frequency I am operating on. So, my behavior should be the same at all frequency that means it should be frequency independent all right. Now, we have to figure out how much gain this guy should have after all it is an amplifier. So, it has some input which is probably small and you probably want an output which is much more. So, how much gain it should have right. So, here we now have to do some more homework before we go with our next demand from the source of that Akash one. So, the question that we are asking ourselves is what should be the gain. So, let us say that I have some way of feeding back my output to the input to stabilize the behavior of the circuit after all we want this device to be very stable all right. So, let us say that this is some voltage that I am looking at and notice that I shall take no current in. So, let us say I have put some resistance here I am taking just one particular configuration here the results. However, are general here is my black box I am not drawing the triangle because I will start drawing the triangle only when I have completed my list of demand. I take my output and feed it back now this is my input this is my output and let us see what happens. Now, if the input increases and the output also increases then I have a problem on my hand why all right the input increased the output increased as a result the potential at this point will increase right because output is feeding back to this input and if the output is now higher the potential at this point will become higher if this output increases if this point this input increases the output will increase even further that will raise this even further and this will be a situation which will go run out of control. So, we do not want that happen what we would like is that the direction of change of these two should be opposite. So, that on the net R 1 and R 2 together will be able to hold this voltage at some steady constant value this part is clear I want the input to remain stable. Therefore, it is clear that the gain should be negative that means when the input increases the output should decrease and let us say that this gain is g all right. So, let us call this v in call this v out and will call this v x now notice that there is no current flowing input impedance is infinite and therefore, there is zero current flowing that gives us gives us a very simple circuit to analyze you have v in connected to R 1 and of course, no current is flowing here so it we can ignore it connected to R 2 and connected to v out which is minus g times v in right it has a gain of g. So, v out is negative and the absolute value of the gain is g times v in right. So, this potential at this point is minus g v in volts and the potential here is v in volts and current can only flow this there is no flow here all right and that current is in fact provided by the output which has a zero input impedance it holds the output irrespective of what else you are doing to it it holds the output at minus sorry it is minus g times v x right it does not know what v in is the op amp can look only at it is input pin. So, the voltage there is v x right so the output is minus g times v x so this is v x. So, how much current is flowing through these two resistors you have v in minus v out divided by R 1 plus R 2 this is the total current which is flowing. So, what is v x remember we cannot find out v out unless we can find out v x right. So, what is v x it is v in minus ok I am actually taking you through the longer path, but that way you will understand why op amps are what they are all right. So, v x is v in minus R 1 times this current you agree with me this is the current flowing in this direction and the drop across R 1 is R 1 times current we found out the current current is v 1 minus v out divided by R 1 plus R 2. We multiplied by R 1 that gives me the drop I subtracted from v in and that gives me v x. So, then now I can rather than writing v out as g times v x I can write v x as v out divided by g right. So, from here it is v x is minus v out divided by g right we have defined the gain to be g and to be negative right. So, g is the absolute value of the gain the actual gain is negative g is its absolute value and therefore, v out was minus g times v x and therefore, I can write v x equal to minus v out divided by g. So, now let us put all of this together you had v in here coming to R 1 this is v x this is R 2 and this is v out which was equal to minus g times v x and therefore, v x equal to minus v out divided by g. We had written that I is equal to v in minus v out divided by the total resistance in the arm which is R 1 plus R 2 remember no current flows here 0 current flow into the black box. This is just for reference do not worry if you can see this is v x and no current flows into the open into the device. So, now we are evaluating v x and substituting for v x as v out by g. So, v x is equal to minus v out by g from here and that is equal to v in minus R 1 times the current which is v in minus v out upon R 1 plus R 2. So, now we can collect terms in v in and v out on two sides and can find out v out as a function of v in. So, let us collect all the terms in v in on one side and v out on the other side. So, this term so minus v out by g that is equal to v in minus R 1 by R 1 plus R 2 into v in plus R 1 upon R 1 plus R 2 times v out. So, this is equal to v in R 2 upon R 1 plus R 2 if I combine this two I will get R 1 plus R 2 minus R 1 divided by R 1 plus R 2. So, this is just R 2 upon R 1 plus R 2 and plus R 1 upon R 1 plus R 2 v out and this is minus v out by g. Now, we are lazy we will just multiply both sides by R 1 plus R 2 we do not like this R 1 plus R 2 occurring in the denominator. Just multiply both sides by R 1 plus R 1 plus R 2 and collect terms in v out and v in. So, we will let us bring the term after multiplying by R 1 plus R 2. So, then this becomes minus R 1 plus R 2 by g this term and we will bring that. So, that will become minus R 1 and this whole time whole thing is the coefficient of v out and that is equal to v in R 2. We have multiplied by R 1 plus R 2 to all terms. So, this is v out into R 1 plus R 2 by g with a minus sign. So, that is this term and when we bring this term in since we have multiplied by R 1 plus R 2 we will get minus R 1 times v out and finally, v out is then is minus v in times R 2 upon R 1 plus R 1 plus R 2 by g. So, now, we have found out what is the output voltage as a function of input voltage. Now, what I say is that this particular configuration should be independent of the value of g. That is the theme that we have been following that this is a generic component and the circuit should not bother about the individual properties of this component. So, we are choosing extreme properties for this component. So, that nothing depends on the properties of this component. So, we made the input impedance infinite. So, that it draws no current. We made the output impedance 0. So, that it can supply any amount of current and now we want this gain the input to output relationship to be independent of its property and what property is there? g that is the only property which belongs to the op amp. Everything else is external component. So, what will make it independent of g? What should I choose as the value of g? If g is infinite, this term will just drop up and then we will have this nice relationship that the output is related to the ratio of these two resistors times v in. So, that is very nice. That means, in such configurations whenever I have negative feedback, the output behavior will not depend on the absolute value of g as long as g is very large. So, if g tends to infinity, now I have got a component whose individual values do not matter. Now, I can take an input and just by choosing appropriate values of R 1 and R 2, I can amplify it by any amount that I want and tomorrow if this op amp is not available, I take another op amp and as long as its gain is large, its input impedance is infinite, its output impedance is 0, the behavior of the circuit will be exactly the same. Now, there will be no change in this device. That is very nice. That is the kind of device that we were looking for. So, now, I have done two pages of homework and now the time is to look up towards the Akashwani and say, oh I forgot to tell you one more thing. While you are about it, please make the gain infinite. So, now we have our demand that g should be infinite. But, having done all this, I find that there is a fly in this op amp and there is something which I do not like this smell off and that is the following. Notice that this guy, the black box was not pulling any current, but this configuration is not leaving my V input unchanged. Indeed, there is a non-zero i that we calculated. So, that much current is anyway being supplied by my source. So, why is this current being pulled? When we had this very nice device infinite gain, infinite input impedance, zero output impedance, till it is loading my input, what has happened? And that is because the feedback is pulling this current. No current is going here, but there is current going here. So, the feedback is causing this current to flow. So, I add a last minute rider saying, while you are about it, could you please give me two input terminals. Then, when I can afford to load the input, I will load it, but I should have the option of connecting the feedback to a different terminal and the input to a different terminal. So, now here is my recipe. I have now finished my demand to the guy upstairs. So, what do I want? Now, I can put back my simple triangle. My demands are not very much. I only want a circuit whose input impedance is infinite, whose output impedance is zero, whose gain is infinite, which has two inputs. That is not much, not much for the guy who is able to give Akashwani from the sky. So, while I am asking about it, might as well ask for this. So, now I have this. So, I am a modest guy and this is what I have asked. I have asked that the gain should be infinity, that the output impedance should be zero, that the input impedance should be infinite at both inputs. Now, I have two inputs and it should be frequency independent and while I am about it, why should I have two separate kinds of in many one same kind of input? What I want is that the gain should be minus g with respect to this and plus g with respect to this. In short, my v out should be g times the difference of the two inputs. This kind of an amplifier is called a difference amplifier. It does not amplify individual quantities. It amplifies the difference of two quantities. This has of its own great merit. A differential amplifier is a very convenient device to have. Why? Because very often we have noisy sources and so on. So, suppose you have this mic and there is a wire and this wire is running to an amplifier with a high gain. This wire will pick up noise. However, if I have two wires which are going to a difference amplifier and both wires pick up the same noise, then I have this with some signal riding on it and the same pickup with the opposite kind of signal riding on it. I make it differential. Now, if I have something which amplifies the difference of this, then all this major noise simply will go away, will cancel. So, whatever is common to the two wires will not appear in the output at all. So, that is called common mode cancellation. Common mode cancellation says that whatever signal is common to the positive input and to the negative input will not appear in the output because output is amplifying only the difference. If it is common, then it will cancel out in the difference. This is very good because it makes it possible to make noise immune system. Now, I can bring my feedback to this guy and apply my input to this if I do not want to load my input. If I can afford to load my input and want an inverting output, then I can use this input. So, I have a lot of flexibility that I can use. This element, now I have finished my demands and I hear a so be it with a thunder roll from the skies and I get this device. It turns out of course, this device cannot be made, but very close approximations to it are quite easy to make. So, you can just amble up to Lamington road or whatever the corresponding places in various cities and buy an op-amp for some cool 10 bucks or so. It will not set you off by much more than a couple of cups of tea. So, this particular thing looked like a very immodest set of demands to make from the uparwala, but it is not too bad. Human wings can make it. You do not quite need to die and go to heaven in order to get this device. So, 10 rupees will do may be 15 and you can get this device. So, these are op-amp. Now, having got it, now we need to know what can we do with it. So, first a few definitions. Remember, now I have two inputs. One which is going to the inverting input, the other going to the non-inverting input. This is v in minus, this is v in plus and this is v out. And the same approximations apply. The real device will be somewhat different, but it will have infinite input impedance here, infinite impedance here, infinite gain, 0 output impedance and now because I have added an input, I would say that the differencing is ideal. That means, if the input is 0, then the output should be truly 0. That means, it should not have an offset and its common mode rejection should be infinite. That means, whatever is it should be a true difference amplifier, whatever is common to the two inputs must not appear in the output at all. But, suppose life is not like that. Suppose, I have some arbitrary voltages v 1 and v 2 connected. So, now I define a couple of quantity. I define v 1 plus v 2 by 2, the average of the two inputs as the common mode voltage. And then I define v 1 minus v 2 to be the difference voltage. So, now it is quite simple to see that v 1 and v 2 can be expressed in terms of v c m and v d f. What is v c m plus v d f? v d f by 2. What is v c m plus v d f by 2? The v 2 by 2 term will just cancel and v 1 by 2 plus v 1 by 2. So, this is that means, v 1 is nothing but v c m plus v d f by 2. And similarly, v 2 is nothing but v c m minus v d f by 2. So, any arbitrary voltage v 1 and v 2 can be expressed as a combination of these two. Their average value is the common mode value and their difference is the difference value. Now, if I have a very good op-amp, then given v 1 and v 2, what will it amplify? G times v 1 minus v 2. So, v 1 minus v 2 therefore, only v d f. There is no component of v c m in its output. v 1 minus v 2 will just cancel out v c m. It will just leave v d f. So, essentially only the difference of the two inputs is visible to this amplifier. And therefore, it is called a differential amplifier. So, c mass is a differential amplifier with infinite gain, with infinite input impedance, with 0 output impedance, with infinite common mode rejection ratio and with frequency response from 0 to infinity. An ideal op-amp will have all these properties. Let us say that we have got it. It turns out that you can get components which are very close approximations to it. So, let us say that we have this device now. And let us look at a few configurations of this. By the way, nodes for this lecture will be available. I have already uploaded them. They will be uploaded to your site, but I do not like to use previously prepared nodes. I would like this to evolve. The whole logic of why we are doing things the way we are doing to evolve. That is why I am using it in this mode, but succinct nodes or what we have discussed will be available and will be uploaded to the model site. So, the first thing we do is what we have already done. And that is this amplifier, because we did not need this input, we just ground this. You assume that the power supply to this is bipolar and therefore, because we are taking differences, we will have negative values. So, we need a bipolar supply. So, assume that this power supply is bipolar and the value of the bipolar is much larger than any input or output. So, that you are not limited by the value of the power supply. This is the circuit that we had analyzed. Now, notice that if you have this is this has negative feedback as we have seen. Let us see what happens if there is negative feedback. Why did we use negative feedback to stabilize the output? Because essentially this point is being pulled up by the input, then it will be pulled down by the output. If it is pulled down by the input, then it will be pulled up by the output. So, we have stabilized it. That means our output is defined and finite. What was this value? This value was V out divided by G minus. Correct? The value at V x. However, G tends to infinity. We know that G tends to infinity. Therefore, what is the value at this point? In that limit, it is 0. This is called a virtual ground. This point is not connected to ground, but the rate of change and the direction of change of the output is such that the output changes itself in such a way such that this point is always kept at ground. Therefore, it is not connected to ground. However, its potential is guaranteed to be the same as that of the ground. Therefore, this is called the virtual ground. By assuming that this virtual ground exists, it makes the algebra of calculating the gain and so on of these op-amps quite simple. Now, we do not have to calculate for V x. We painfully calculated the value of V x from input, from output and equated them and then we found out what the V in V out relationship was. But we further know now that if G is infinite, in that case we can make the additional assertion that the value, the voltage at this point is 0 if this is 0. In fact, because it is a difference amplifier, the output is not G times V x. The output is not minus G times V x. The output is minus G times the difference of these two inputs. Therefore, the difference of these two inputs should be 0 and if this is grounded, this will be the virtual ground. If suppose I was to give it some other input, suppose this is V 1 and I give it not ground, I give it V 2, then what will be the voltage here? What should be the voltage here? It should be V 2. If it is anything other than V 2, then the difference of V 2 and this point will be multiplied infinitely and appear at the output and that cannot exist because there is negative feedback. Let us say that V 2 minus this voltage is slightly positive. That means this voltage is slightly larger than this voltage. Then in which direction will the output go? Positive or negative? We are saying that this voltage is slightly positive compared to this summing point. What will the output do? In which direction will it go? The gain is plus G with respect to this and minus G with respect to this. It will go to positive value. The difference is positive and the gain is G. Therefore, the output will go to very large positive value and that large positive value will pull it up making it equal to V 2. This was lower than V 2, but because the output has gone up and quite substantially, it will keep going up and keep pulling this point up till it becomes equal to V 2. Similarly, should this point be a little less than V 2, lower than V 2, then what will happen? That is the point that we had said that V 2 is a bit positive. If V 2 is negative with respect to this point, then the output will go in the negative direction. The output goes in the same direction as this. If there is any mismatch between these, so the output will keep going towards negative value pulling this voltage down till it becomes equal. So, essentially the only stable point in case of this negative feedback is for this voltage to be equal to this voltage. Even a small variation from this, the output tends to bring it at the same value as the other input because the gain is infinite even a small deviation from it will result in a large change in the output which will pull it down in line. Therefore, this point is kept at the same potential as this point provided there is negative feedback. This will not be true if there was no negative feedback, then there will be no way for the output to connect to this point. So, the output must be connected to this point to provide negative feedback. If there is net negative feedback, then these two must be at the same potential and this is called a virtual shot. If that point was connected to ground, it would be virtual ground, but actually the more general term is that this is called a virtual shot. Now, assuming virtual shot or virtual ground in this case, we can quickly redo the somewhat terrible algebra that we had to do earlier and straight away see what the output should be. So, how much current is flowing down here? Now, we know this potential. It is not an unknown anymore. We know that this potential is 0 and therefore, i is nothing but v 1 by R 1. Now, there is no way for this current to flow into this device. The input impedance is infinite. Therefore, the only way this current can flow is through this resistor and if the potential is 0 here, why should it flow that way? Only if the output is negative and how negative should it be? So, that the current flowing through R 2 is i. Therefore, v 0 should be minus R 2 times i. That means it should be minus R 2 by R 1 into v 1. So, we have a linear device. The output is a linear function of input and the gain of this device is negative and it is given by R 2 by R 1. This is called an inverting configuration or inverting amplifier.