If you perceive something new, if you learn something new, it's such a joy. But why schooling such a pain? Somewhere, the way we're delivering it, we have not looked at it carefully enough. We have not put ourselves sufficiently into it. When I was supposed to go to school, I did everything possible not to go there. So when I thought of creating schools, I thought we must create schools in such a way that children want to go there. What is the point creating a school that a child doesn't want to go with? So if we have to create schools that a child wants to go, I think the adults need to go to school first. So, in a way these conferences are just that effort to put the adults to school so that we, all of you can create schools where the child wants to go. Of all the different manifestations of human beings, the categories are getting more and more. A child is the simplest and the easiest to make them happy and joyful. With such a segment of population, which is naturally joyful, I don't see why it is so difficult to create a joyful way of delivering education. There is substantial scientific and medical evidence today that if you remain in a pleasant state of experience, that is when your body and your brain works at its best. Some studies are trying to establish that if you remain just one day, twenty-four hour segment in your entire life, one twenty-four hour segment, if you remain without a moment of unpleasantness within you, not a moment of agitation, anxiety, irritation, anger, nothing, just joyful.  When we've always known this by experience, but today there is data, because without data nothing is true anymore. If there is no data about you, that means you're not here. Yes, if you did not register your data into the computer, all of us will think you're not here. So without data, nothing exists. So today there is data that a joyful way of existence, it will enable and empower a human being to a very high level of perception and performance. So this is a must. How is the question? There may be many methods, but one fundamental is if you are not joyful, you're not going to inspire anybody to be joyful. That's for sure. Why? I should be telling you this, because is there anybody here that you have not known a single moment of joy in your entire life, is there any such person here? You cannot be, because if you did not have a single moment of joy, you would be dead. You would have no reason to live. So you know how to be joyful. I'm sure right now if I ask you, would you like to be joyful or miserable in your life, every one of you will say, I want to be joyful. So you know how to be joyful, you want to be joyful. So why do you need a guru to tell you? Ah, yes, Sadhguru, I want to be joyful, but to kick that butt, you need a guru. Yes, I want to be joyful, but you know what he did. Yes, I want to be joyful, you know what she did. Yes, I want to be joyful, you know the weather is not good. So there is butt, butt, butt. If we just kick this one butt out of our life, creating a joyful education will be a natural consequence. If we are joyful, whatever we do, whatever we create, we will make it that way, isn't it? Without we being that, now if I say let's create a joyful education, it looks like a very complicated problem. If you are joyful, is it not natural that you will cook joyfully, you will serve joyfully, you will talk to people joyfully, you will do everything joyfully? Is it not natural consequence? So right now, one of the biggest mistakes, I would say a crime that we are committing is we have taken to this western mode of doing things that is goal-orientedness. That is, we want mangoes, but we are not interested in the tree. Soil is not even our concern. We've gotten into this mindset. Goal is the thing, we want this big mango. You have this one, we want this big mango, but we are not so much interested in the tree. Soil is not even in our perspective. In yoga we say, if one eye is on the goal, you have only one eye to find your way. It's very inefficient. If you have both the eyes to find your way, you will find your way. This too much goal-orientedness has made human beings not only in schooling, in life, in the way we create our businesses, the way we run the nations, the way we conduct the affairs of the world, everything has become so skewed and not about life. It is about some product in the end. At the end of our life, there is only one product which is grave. Either we are investing too much in the graveyards of the past or futuristic. If it's futuristic it's grave-oriented because that's our only future that's going to happen to you and me. Yes or no? Hello? That is the only future. So the thing is whether it is schooling or business or anything that we are doing is about how we conduct it now, what will come out of it, whatever comes out of it. Are we doing it in the most beautiful way possible? If this one thing is embedded in us, methods, various tools, all these things will come from experts. There are many, many experts. There are a whole lot of people who invested their life thinking through these things. We can use that but to use these methods in a way that it works, we need people who are first of all joyful, who know how to make their life beautiful. If you do not know how to make your life beautiful, this aspiration of making everybody's life beautiful is not going to work. So many of you who are as teachers, as educators, as experts in the field, variety of expertise that is here, the evolution of the teacher is the most important thing. Everything else is secondary. Who delivers this is the most important thing. We have not invested enough in that. Evolution of the teacher does not mean just more and more tricks about education. As a human being, if the teacher blossoms, becomes a joyful, loving, compassionate human being, above all conscious human being, everybody has to strive for this. Otherwise this will not happen. This is not something that will happen because you got a PhD. This is something that you have to work upon yourself. If this process is brought into the teachers, you must see the tradition of this country is this. Usually parents gave away their children to a teacher. An acharya or a guru took the children totally in his fold because people saw him as a evolved human being. Not just a knowledgeable human being, a evolved human being. So they saw that if we leave our children in their in his hands, in his or her hands, naturally our children will blossom. So this has to come. This may look like, oh, is this possible to train all the teachers? Why not? I'm asking. For the first time, for the very first time in the history of humanity, we have capabilities that no other generation could ever dream of, isn't it? For the first time we can speak to the entire world, it is time we make truth mainstream. If truth becomes mainstream, truth means what really works if I sit here, what will leave me in the highest state of experience is the truth of my existence, isn't it? If every human being knows what is the truth of their existence that you can sit here in the best possible way a human being can be right now, you don't have to worry about tricks of education. In those hands, every little tool will become a powerful process. In this effort to make educators, teachers and others who are involved in children's education in this country or anywhere else, we want to offer these things to you, simple tools for transformation, something that you can do with yourself, which will leave you in a better place than you are right now. This must happen. It's my wish and my blessing that this must happen.


So, today we are going to cover in this lecture junction field effect transistors and I should make sort of generic command before we begin and that is this like these lectures are actually not sort of full-fledged you know classroom lectures and that is because you already know many of these things. It is essentially to present a way maybe a different way to cover some of this material and in a way it sort of explains it also illustrates the philosophy of teaching that we follow here in IIT Bombay and we thought everybody could benefit from that a little bit. So, teaching is we always feel that teaching is not really about you know getting some giving some formulas and say here is a formula and then mug it up and write in the exam you will get marks that is exactly the opposite of teaching that is what we believe and therefore, we want to do things in a different way and hopefully at least some of you will be able to follow this in your teaching. So, teaching we think is like more about you know thinking and not just thinking yourself, but also making your students think and for that to happen the student should feel that there is a link between what he or she has already learnt in the past and what he is going to learn he or she is going to learn today. So, that is something that we always keep in mind and that gives the students some confidence that he or she also can you know understand the development of a topic and not just sort of mug it up and write and get marks. That is the last thing that is of importance. So, let us I will talk about JFETs in this particular lecture and of course, let us begin. So, before we even get down to the structure of junction field effect transistor the students should really know what is a transistor I mean what is the basic job of a transistor and that is illustrated in this figure. A transistor is to differentiate it from a diode or a resistor it is essentially a first of all is a three terminal device. Second there is a conduction between source and drain and that conduction is controlled by the gate terminal. So, that is the sense of transistor action and then of course, you can have various transistors there are field effect transistors there are bipolar junction transistors the field effect transistors are of various categories and so on. But, essentially the idea is this there is a resistance not of course, it is not a constant resistance between the source and drain and that is controllable by a voltage that you apply on the gate terminal. So, this is something that we feel must be step number one in sort of illustrating transistor any kind of transistor that there is a the current is going to vary between source and drain on two counts one is a drain voltage itself and the resistance itself is getting controlled by the gate. So, the drain current is going to depend both on the gate voltage as well as the drain voltage. So, just like I said a field effect transistor as a gate terminal which controls the current flow between the other two terminals that is the source and the drain. So, in simple terms and that is why that is what must appear before any equation appears in simple terms of an FET can be thought of as a resistance connected between the source and drain which is a function of the gate voltage that is that is the basic operation of a field effect transistor and what do you do with it etcetera these are questions that will come later in applications digital circuits analog circuits whatever you all right the pointer. Now, as I said the mechanism of the gate control would be different in different devices for example, in junction field effect transistors as we know it is p plus n p n junction that controls the resistance. In MESFET it is a short key barrier that controls the resistance between source and drain MOSFET it is again controlled by gate, but there is an oxide and there is an immersion charge and so on high electron mobility transistor similar to MOSFET all right. So, FETs can be used for both analog and digital applications in each case the fact that the gate is used to control current flow between source and drain plays an extremely crucial role and that is why you must have these transistors and a diode for example, will not be able to do the same job no matter what you do because it is a two terminal device and it has its intrinsic limitations ok. So, now we proceed to the structure of a junction field effect transistor and this again it is better to show three dimensional view first. So, that the students get an idea of what a real structure looks like and then you can go to a cross section directly jump to a cross section students often get confused they do not know what it looks like in 3D all right. So, here is a figure that shows the transistor in three dimensions there is a it is like basically a rectangular it is it is a cuboid you can say there is a source terminal at one end drain terminal at another end and the gate terminal is actually the both the top and the bottom these are connected together and that is the gate once the student gets this picture clear it is much easier for him or hard to understand the cross sectional view all right. So, next we go to the cross sectional view and that is here in the cross sectional view it is important to show the P plus regions. So, what we have here is the channel or the conduction channel is n type silicon that is marked over there and the device I you can see if you should explain to students that it is actually a symmetric device there is a there is an axis and think the symmetric around that axis. The channel width or the this would not be called width this would be called the channel thickness is 2 times a where a is each one of these dimensions and is 2 a because of a grain symmetry all right. So, what happens essentially and that will be shown in the next slide is there is a depletion region which will reduce the channel thickness and so on. So, even this is the little little more complicated than what the basic the very very basic device structure should have and that is because it has got these neutral regions near the source and drain and it is it is best not to confuse matters with these neutral regions in the beginning. So, let us represent the same cross sectional view with a simplified structure and that is here. So, here we do not worry about this source this regions near the source and drain but just truncate that and of course we can add that later fine. So, this is the structure and that is as I was saying this is a simplified structure in which we do not worry about the source and drain regions. The purple region here is the P plus region and the channel region itself is n. So, there is a P plus n junction at the top as well as a P plus n junction at the bottom. So, that is the basic structure the more the important things here is the doping of this n region or the channel region the lengths from source to drain and the channel thickness which is called 2a here where a is half of this that is and it is 2a because of the symmetry alright. So, let us go to the next slide ok. So, the important things to take from this slide is the entire region between the top and bottom P plus region offers a resistance to current flow and the resistance will depend on Vg and that of course we want to see. So, again make things as simple as possible first and then make them complicated later. So, that the students get a feel for you know the basics first and then get an idea of how things are in real life right. So, the very simple case that we can first consider is when the drain voltage as well as the source voltage are both 0 ok. So, that is the first case we will consider ok and that shows this particular situation and maybe I should expand this. So, that you can see it better ok. So, let us concentrate first on the gate voltage equal to 0 volts and as as somebody was pointing out I should go slow now. So, let us take Vg equal to 0 volts and in all of these figures note that the source is grounded the drain is grounded in this particular case gate voltage is also 0 volts right. So, what is what is the situation at 0 volts we have this P plus region the purple one next to that there is a blue region and that is the depletion region right and this is very important because some textbooks actually do not point this out ok. So, there is this purple region which is the P plus region and next to that there is this blue region which is the depletion region and even note that even if your gate voltage is 0 volts there is still a depletion region and that is because of the built in potential between the P plus of the P plus n junction. So, there is it is not as if you have a gate voltage of 0 volts the depletion region is not there it is not at all like that and very often some of the textbooks I have seen do not make this very clear and that can be confusing ok. Next to that and of course, they both the top and bottom are symmetric. So, that is one thing that must be noticed as well next to that there is a light blue region and that region is essentially the n type silicon which has got electrons ok. So, the depletion region does not have any electrons or holes or extremely small number which is insignificant. So, they do not participate in conduction. The light blue region is actually of importance also is of interest as far as conduction from source to drain is concerned because that is essentially neutral that means it has got as many ions as there are electrons and electrons of e to move. So, that is really the region of interest and the dimensions this A of course, is the half channel thickness that is fixed this W is the depletion region width that will obviously depend on the gate voltage H is the width of the other thickness of the neutral region again half thickness of the neutral region. So, this and of course, H and W are both dependent on the gate voltage such that H plus W will always be A ok. Let us go to the next next figure the only difference between this and this is that the gate voltage is now made more negative. So, here is 0 volts here is let us say minus 1 volt. So, as a result what has happened is the depletion width. So, the junction is now more reverse biased and the depletion width has increased and therefore, it has eaten up into the channel. So, the thickness of the channel that is available to you has now decreased. So, that is very clear in this figure and so on. So, once the student understands this point it is easy to go forward and if you increase this gate voltage or rather make it more negative. Obviously, you are going to get more of the depletion region and less of the channel region and of course, at some point what is going to happen is that this whole channel is going to disappear because the depletion region the top and the bottom depletion regions will just simply merge with each other and that situation of course, is called as the pinch off situation ok. So, essentially the point is that as the reverse bias across the junction is increased by making V g more negative the depletion region widens as we have seen here and the resistance offered by the n region increases because now you are the width of the n region becomes smaller and smaller and therefore, the resistance increases and that is essentially the operation of a junction field effect transistor. So, that is the mechanism by which the channel resistance depends on the gate voltage in a junction field effect transistor. Finally, when the reverse bias becomes large enough the depletion region consumes the entire n region and that is little bit away from this one when all of this depletion region will just be one and the channel becomes pinched off and the voltage that corresponds to that particular situation is called the pinch off voltage and that of course, is a very important parameter for a GFET alright. So, now let us get to this pinch off voltage let us see how we can calculate that and again when you are teaching in the class you can say the formula for the pinch off voltage is so and so it is not a very satisfactory thing to do because the students will always have a lingering doubt in their mind where did this come from and they will think that it has just sort of fallen from the heavens and you have to just take it for granted or what and in fact, it is not like that at all it is very simple to derive and therefore, it is always a good idea to derive it in front of the students so that then they develop some confidence that things can be derived and not just mocked up very important ok. So, how do you derive this let us in this slide. So, what is the pinch off voltage basically it is the gate voltage for which this H becomes 0 or in other words this W becomes equal to A that is the pinch off voltage. So, if you know the dependence of W on V G just equate that to A and you get your pinch off voltage that is simple as that. There is no reason why this should not be derived in class in which it must be derived in class so that students develop a better understanding ok. So, what is W for a P plus n junction? So, the depletion region actually extends both into the P plus region as well as the n region and if it is P plus that means is much the P region is much more heavily doped than the n region and therefore, it is a fair assumption to say that all of this all of the depletion region basically is in the n region and not at all in the P plus region that will be extremely small. So, we just ignore that part and we say then that the W depletion width here is given by this P plus n junction expression and what does it have? It has got epsilon for silicon which is 11.7 times epsilon 0 then it has got the built in voltage the built in voltage will actually depend on the doping of the P plus n region the doping of the n region. This is something like 0.8 volts 0.9 volts 0.95 volts and so on of that order V is the difference is the potential difference between the P plus and n. In this case your P plus is sitting at V G and n is sitting at 0 volts because your source and drain everything is at 0 volts. So, that is what V is. So, V is simply V G in this case Q is the electronic charge and D is the doping density of in the channel typically 10 is to 15 10 is to 16 per centimeter cube. So, it is a simple matter now to put this W here and then find out what V G is and that V G will be the same as the pinch off voltage. So, then we substitute this W here and that is it and then you get the pinch off voltage equal to V B I minus let me just zoom on this. Then we get this V P equal to built in voltage minus Q and D A squared by 2 epsilon. So, problems with some of the text books they say that assume V B I equal to 0. Why? Nobody knows I certainly I do not know why because it is not really negligible compared to that term as we will see. So, that is a silly thing to say that ignore V B I and you should not say that Q N D A squared by 2 epsilon. So, of these Q and epsilon are just physical constants you really cannot do anything about that, but N D and A squared are in the control of the fabrication person. So, when the device is being fabricated you can control N D or you can control and you can control A to get the desired value of pinch off voltage and note that this dependence on A is rather strong because of the square here whereas N D comes with the linear dependence. Let us go to the next slide. So, now in the next slide we have a sample calculation of the pinch off voltage. This structure is the same as before we do not need to zoom that. So, here is the formula is just reproduced. So, an example and this is always I am showing this because it is always a good practice. Whenever you give a formula in class it is always good to put some numbers so that it helps the students to get these things fixed in their minds. Then they can connect these formulas to actual numbers and it really helps a lot. Not only the numbers it also helps them to think about the units of each of these quantities and that is very important. So, let us say in this example we compute the pinch off voltage for an N D of 2 10 is to 15 per centimeter cube A the half channel thickness of 1.5 micron and the built in voltage given built in voltage of 0.8 volts. So, let us go further. Now the reason this is actually a simple calculation but the reason I put this in this slide is to illustrate the units of each quantity and it is very important that the units are used consistently. So, that you do not end up with you know 3.48 million volts or 3.48 micro volts and so on and that can happen I have seen that in answer papers. So, it is very important that you bring this point out very clearly in class. So, q 1.6 10 is to minus 19 coulomb and you should actually ask the students to tell you what q is and if they do not know the value of q they should leave the class and know the value of q and come back may be. So, q is something that everybody must know epsilon 0 8.85 10 is to minus 14 farad per centimeter that is something that everybody should know 11.7 is the epsilon for silicon something that everybody must know and these are all constants basically and. So, what is what are the variables 2 10 is to 15 per centimeter cube that is the doping density and 1.5 micron now this is very important we need to convert this of a micron into centimeter. So, that the units work out well and that conversion is 10 is to minus 4 and then you should actually show this in class you have centimeter raise to minus 3 there centimeter squared here and this is there is a per centimeter in the denominator. So, all the centimeters will actually cancel you are left with coulomb divided by farad coulomb divided by farad is coulomb divided by coulomb by volt and that is volt. So, if you do this in class in this way or an equivalent manner then students actually begin to develop an appreciation for all the units and do not make mistakes later on. So, all this comes to 3.48 volts and the built-in voltage is 0.8 volts and so the net pinch of voltage therefore is minus 2.7 volts in for this device what does it mean? It means that if you apply gate voltage of minus 2.7 volts this channel will get pinched off and you will see a very large resistance between source and drain that is what happens at pinch off. Things are of course different if you apply a drain voltage and we will come to that later. So, let us continue we will continue with I think some there may be some questions, but we will take them at the end all right. So, the pinch off voltage in this particular case is minus 2.7 volts and as I was saying the built-in voltage as a built-in voltage of 0.8 here is certainly not negligible compared to this 3.4 3.5 volts right. So, therefore, there is no particular reason there is no good reason to ignore it although some textbooks do that and that is not proper. Let us go ahead and this is exactly what we said if a gate voltage of minus 2.7 volts is applied the channel gets pinched off that is the resistance between the source and drain becomes very large. So, that is that was with the gate drain voltage equal to 0 and now we need to consider the more general case of drain voltage not equal to 0 right and we took this vd equal to 0 case earlier because it is much simpler to handle and also gives the students a feel already of the pinch off voltage. So, the pinch off voltage is something that you should that can be taught even before one talks about a non-zero drain voltage because it is completely independent of the drain voltage. So, let us go to this slide let me zoom a little bit and just explain these figures first. So, now here what we have is there is a certain gate voltage which is constant and we are not going to vary that from this figure to that figure to the next figure, but what we are going to vary is the drain voltage. So, in this particular example in this particular figure the drain voltage is 0 volts and source voltage is also 0 volts. So, there is nothing there is all points in the channel are sitting at 0 volts and there is no change as you go from x equal to 0 to x equal to l it is all the same. You notice that this depletion width as well as the channel thickness h they are also constant that is irrespective that is independent of x that is because the drain voltage is at 0 volts and this is the case we have already seen in the previous slide.  So, what happens now is the reverse bias between the gate and the source is the gate voltage is some value let us say minus 1. So, this will be minus 1 minus 0 minus 1 here between the gate and the drain the reverse bias would be minus 1 minus 0.05. So, it will be minus 1.05. So, here we have a reverse bias of 1 here you have a reverse bias of 1.05. So, the drain essentially it is sitting at a larger reverse bias as compared to the drain as compared to the source. As a result what happens is the depletion region near the drain end is wider as compared to the source and that is basically what a drain voltage will do. If you keep increasing the drain voltage the depletion with the depletion region width at the drain will increase and of course, the source width is not going to change because the source we are not changing the source voltage. The gate voltage remember is a constant let us say minus 1 volt. So, this keeps on increasing and therefore, the channel thickness near the drain will keep on decreasing and if you plot the potential between from source to drain along let us say this line that would look something like this. So, it is at 0 volts here it is at 1 volt here that is because you applied a 1 volt drain voltage and it will vary in some manner we do not know what that form is, but it will vary in some continuous manner between 0 and 1. So, it is always good to show these pictures to students explain to them qualitatively what happens and then come to equations. If you jump to equations it basically just becomes overhead transmission for them alright. So, let us go to the next consider an n. So, this we have already discussed in this. So, it is important to note that w and h are now functions of x in this figure let me expand it again. So, what we are saying is this w and h are both. So, the w here is that much w here is that much and similarly h will change. So, w and h are both functions of x now depends on where you are in the channel, but w plus h will always be equal to a that is what we are saying here. So, the important point is that since the p n junction bias at a given x is v g minus the potential at that particular point that is shown in this figure approximately the drain end of the channel is sitting at sitting with a larger reverse bias than the source end and therefore the replacement region is larger there and therefore the channel is thinner. And now let us derive the I.D. equation as I said it is always good to derive this and this derivation is a little more involved, but certainly something that you can do in class and you should do in class. So, that the students have some confidence in how things are done. So, let me expand this figure I hope after expansion is showing up already at your end. So, it is the same as same figure as before I am in addition we have shown here a cross section of the device. So, that the equation derivation becomes a little more clear to the students. So, this is w that is h and if I just take this line over there and look at the cross section of this device this light blue region actually is the region of conduction. The region next to that the blue region is actually useless as far as conduction is concerned because it is just a depletion region with no carriers and of course, next to that is the p plus n p plus gate region. So, the important area for the drain current calculation is just this light blue area and what is the area the cross sectional area of that area there it is 2 h that is this dimension times z where z is the device dimension in the direction perpendicular. So, let us begin. So, consider a slice of this device that we have done here the current density at any point in the neutral region is assumed to be in the x direction that is very important and that is a crucial assumption that must be made and this is of course, a very good assumption and that is because normally this length is much larger than this a here the channel thickness and that is why this assumption is quite good. So, what is the current given by it is given by the usual drift term and the diffusion term and in this device in the junction field epic transistor the doping density everywhere is the same and the electron density is roughly equal to the doping density and therefore, the electron density everywhere in the channel is also the same and therefore, there is no derivative this d and d x is just 0.  So, e actually is minus d v by d x, but I have just put d v by d x there finally, we will introduce all these signs correctly. So, we have neglected the diffusion current since n the electron density everywhere in the channel is roughly the same that is equal to the doping density and therefore, d n d x is 0 when you differentiate this constant electron density you will get a 0. So, what is important in this figure note that only the neutral part of the n silicon conducts that is the light blue region since there are no carriers in the depletion region. So, the depletion region depletion regions are essentially not contributing to any current at all. So, at a given x the current i d is obtained by integrating j n over the area of the neutral channel region that is this region here and in this device v this j n is actually the same everywhere in this area and therefore, the integral simply becomes the current density times this area.  So, that is what it is. So, there is this j n which is integrated over this light blue area and that area is simply 2 h times z and then what you see in this bracket is nothing but j n and that is this. So, this j n has been substituted right here that is it. So, i d at any given x at any given x let us say at this point is given by 2 times q times z times mu n times n d all of these are constants q z mu n n d all constants times a times the electric field there p v d x times h, h is nothing but a times 1 minus w by a you can show that by from the fact that h is basically a is h plus w and this will follow from there. So, let us look at this equation somewhat. So, what it says is the current at any point x in the x direction is some constant times d v by d x which depends on that particular x and multiplied by 1 minus w by a. Now, this factor is also dependent on x because w itself is a function of x and we know that is i d at this point is the same as i d at this point. So, what happens in this device is that this d v d x and this 1 minus w by a these 2 terms basically go on adjusting as x is changed such that their product remains the same and that must be the case because we have the same current throughout the device. So, all these things are good to know for the student. Next so what do we do next now of course we do not like this d v by d x here and we do not like this 1 minus w by a because w is also a function of x. We want an expression for i d which is independent of d v by d x etcetera. So, let us see how to do that and that is done by using the fact that i d is actually constant from the source to drain right. So, i d is just a constant and therefore, if I integrate i d as a function i d d x from 0 to l I will simply get i d times l. So, I can do that on this left side and I can integrate this also with respect to x and when I do that this d v d x d x will give me integral with respect to d v with respect to v and I can replace the limits instead of putting limits on from 0 to l on x I can put limits on v the channel potential from 0 to v d and that is what this is. And notice that we have used for this w we have used this expression which we have seen before where this v b i minus v g minus v is the net reverse bias across the p plus n junction. So, the next step is to just carry out this integral. If you notice this integral has got 1 here can be easily integrated 2 epsilon q and d i square all of these are constants this will simply come out v b i is a constant v g is a constant and then there is this minus v. So, the integral will only involve this term can be easily integrated square root of v. So, you will get v raise to 3 by 2 divided by 3 by 2 and so on and with of course, appropriate substitutions. So, when you do all that evaluating the integral and using this particular equation you we get we get and of course, it says do this. So, you must do this on the board or you must ask the students to do this as an assignment or something and not just take it for granted. So, when you do all this you will get when if you want to put it in nice form you get this expression where we have notice that these quantities are now dimensionless there is a voltage in the numerator there is a voltage in the denominator. So, there is a dimensionless ratio here that is raise to 3 by 2 this is another dimensionless ratio raise to 3 by 2 and it can be put in a nice compact form if you replace some of these constants with a quantity called g 0 and that is g 0 here 2 times q times z times mu n times n d times a by l all that is why we do why do we call it g 0 there is a good reason for that, but before you go to that where is the v d dependence in this equation i d equal to all of this the v d dependence is basically here v d and here. So, that is important to point out where is the v g dependence v g dependence is here and here that is the v g dependence all right next. So, let us see what is g 0 is so this g 0 turns out that it is the channel conductance if there was no depletion that means suppose you did not have any depletion at all throughout and if you calculate the resistance of this particular box 1 over that resistance will give you g 0 that is again something that you can very easily show. So, that is the conductance of the channel if there was no depletion that is if h of x was a throughout that means this h is equal to a throughout the channel. So, that is more or less the derivation and often this is the this is not the equation that is shown in textbooks what we show in textbooks is a simplified equation and let us come to that ok next slide. So, before we do that it is good to consider a special case and that is v d equal to 0 volts right. So, now we have this expression for i d and we should see what does it look like if we make v d equal to 0. Obviously, the equation you have derived should apply for should hold for the special case v d equal to 0 and why are we doing this because we already are familiar with this particular situation that we have done before we applied non-zero drain voltage. So, this is your the generic i d expression in terms of v d and v d. So, how does it simplify? And this is something that the students can do as homework. So, there is a v d here this v d of course remains as it is ok.  Then that turns out to be just this simplified format here. So, now it is independent of v d because v d is very small and it depends only on v g and if you use some of these expressions that we have previously derived.  Now this has a very nice interpretation and it is pretty trivial to show what it is. This simply shows that the channel conductance reduces linearly with w as seen before and we have done this before the v s equal to v d equal to 0 volts k this should be v d. So, this is it is good to know good to know that this complicated expression reduces to something reduces to something that we already are familiar with and something we have seen before. Now comes the very important question that is what does the I v characteristic look like so if I plot this equation that we have derived as a function of let us say I keep a v g equal to some constant value of v g for example 0 volts here put v g equal to 0 here vary v d from let us say 0 to 5 volts and plot this. What I get is I get the current which rises and reaches a maximum at some point. What happens beyond this point we do not know right now, but we will come to that later. So, that is the picture if I apply a reverse bias this is 0 volts if I apply a reverse bias of minus 1 volt then this whole I d v g will come down I d v d will come down and but again it will reach a maximum at a certain drain voltage and so on. So, now this is something that you should actually if the students are willing you should ask actually ask them to write a program and generate this plot it is very simple to write a program. So, what is the key here for a given v g I d reaches a maximum at what point at v d equal to v g minus v p. Now it is not obvious from this figure, but you can differentiate this equation with respect to v d and show that the maximum occurs for this particular condition very simple differentiation and so that is the take home point here at this value of v d the bias across the p plus p n junction at the drain is what is the bias across the p n junction it is v g minus v d and that turns out to be v p and what is v p v p has a significance v p has a significance that it is the pinch of voltage right. So, the channel is pinched off at that particular point. So, what it means is at this value of v d the bias across the p p n junction at the drain end is v g minus v d equal to v p in other words the drain end of the channel has just reached pinch off that is what has happened and how do you show this pictorially. So, here is a figure which shows that. So, this is at this particular point at this particular point or this point or this point depending on the v g this is the situation here this is the situation right. So, at drain end the depletion region the top depletion region and the bottom depletion region I just merged with each other and that is the pinch off situation denoted by this circle here. So, let us go to the next slide. So, the important question to ask now is what happens if v d is increased further. So, now we know more or less the story up to this point what happens if I go on increasing this what will happen to this particular region is will the current just collapse of course we know that it does not happen it is important to explain why it does not happen. So, it is good to show this kind of figure right. So, here we have let me just get all these things and then expand.  So, that is the situation at that point at c as we have seen before this drain end has just reached pinch off that means the top depletion region and the bottom depletion region I have just merged with each other that is the situation d. What happens if I increase v d further and of course we know that we reach the saturation region, but why does it saturates not very it is certainly not very obvious for any student who is doing this for the first time. Therefore, it is very important to explain what happens in this saturation region. So, what happens in saturation is that this region this pinch off region expands just a little bit and the excess voltage that we have now beyond v d sat that all of that drops across this particular high field region. Therefore, as you can see that this region here where there is a neutral channel this region has not really changed all that much and therefore, the conditions in the channel are more or less the same therefore, the current is more or less the same and that is what happens in saturation all of these things are actually written over here, but it is very important to actually show this pictures either draw them on the board or just show them just project it or whatever and correlate that with an IV characteristics. Of course, all of this is at a particular vg otherwise the students really we often ask our the candidates who come for m tech examination and we ask them what why this saturation happening they often do not know the answer very important to explain this. Next slide here is an example it is very important to actually ask students to do something like this. So, they develop some they should start believing the equations rather than just mugging the term writing in the exam and getting marks and if they will start believing in equations only if they do something with the equation and that is very important and this is something that they can easily do it is not really asking for much. Here is a here is a problem and n channel silicon j-fat as the following parameters at t equal to 300 k that is the that is room temperature a equal to 1.5 micron the channel that is the half channel thickness l is l is the device length 5 micron z that is the dimension in the direction perpendicular to the paper 50 microns. Nd 210 is to be 10 210 is to be 15 vbi 0.8 and mobility is 300 centimeter square per volt second. What is the pinch of voltage the you can actually this is this example these numbers are the same as before. So, the pinch of voltage is the same as what we saw earlier and you should point out to the students that in this pinch of voltage calculation this mobility will simply not enter because we are not even talking about any conduction. We are just talking about a gate voltage at which the channel is completely pinched off and that is simply coming from electrostatics and there is no there is no conduction coming into that at all. So, where does the mobility actually come into picture that of course, comes into picture in idvd characteristics. So, question b is to write a program to generate idvd characteristics for vg equal to 0 minus 1 y minus 1 minus etcetera. And for each of the above vg values they need to compute vd sat that means the vd value at which the channel gets pinched off or the drain current saturates and show it on the idvd plot the and of course, the part of the part of an idvd characteristic that corresponds to vd less than vd sat is called a linear region and that corresponding to vd greater than vd sat is called the saturation region. So, it is very easy to write this program and students will feel much happier with the equation once they do this exercise. If they plotted they should get something like this. So, this is the idv characteristics of course, the equation is going to give data only up to this point after that it is just constant and then this is the boundary between the linear and saturation regions. The linear and saturation region are also marked over here. So, I think that is more or less all for this topic. I think there are just a couple of comments. So, it is important to point out that in very often we do not use this original equation because we are only interested in the saturation region and there a simplification works quite fine and that is very useful for approximate design of circuits and so on. So, let us just go through this. So, this is our original equation and we are generally especially if you are talking about analog circuits we are only interested in the saturation current. So, saturation happens when vd sat is vg minus vp. If you do that this whole term becomes just one and you are left with this expression. So, that is the actual id sat as it follows from our original equation, but it is a little complex because there is this 3 by 2 here and so on. So, very often this following approximate model is found to be adequate in circuit design and that is id sat at given vg is simply some constant id s s 1 minus vg over vp square that is it where id ss is nothing but the saturation current at vg equal to 0 volts. And in amplifier design we are interested not in the actual current, but in the slope of this id with respect to vg at a constant vd. So, that can be obtained easily from this equation and that is what you get. So, some students might get if you show this equation some students might wonder where this comes from because they would in textbooks they would probably see this equation. So, it is important to point out that this is actually an approximation for what you get theoretically. Now, just one more slide I think. So, now let us bring back the neutral source and drain regions because as we said earlier this part of the device is the so called intrinsic part, but the actual device will also have some neutral regions here which we can represent with some resistances. So, if you want to model this whole thing accurately then this is your intrinsic device which will follow the equation that we derived. Then in addition you have the source resistance here and the drain resistance here attached to it and that becomes a more complete maybe more accurate model. Very important to also talk about the small signal model because if you are talking about an amplifier example like this then it is important to know what the gain of this amplifier is and that is where we need a small signal model. So, a small signal model is required in analyzing an amplifier it is gain or even frequency response and what is. So, what is an appropriate model for a JFET? We notice that between the gate and source we have a P plus injunction between the gate and drain also we have a P plus injunction. It means very large resistances. So, these large resistances are so large that we do not even we can just treat them as open circuits. But apart from that we also have a P plus since it is a P plus injunction apart from that we also have a capacitance and those are these capacitors this between G and S there is a capacitance CGS between G and D there is a capacitance CGD and these capacitance are actually important because these will limit your limit the performance of your circuit at high frequencies. If the frequencies are low then of course these are open circuits and then they are not there. What is the other most important quantity? GM the transconductance in this saturation region that is DIDVG with Pd equal to constant and we already derived a little approximation for that earlier. So, if you apply a VG here the current the small signal current here is GM times VG where VG is the small signal voltage. In addition in our derivation actually this there is no resistance between this between the drain and source because after saturation we said the ID becomes constant. In practice it is not quite constant and there is a small resistance between drain and source and one over that resistance is called GD and this GD can actually limit the performance of an amplifier somewhat. So, that is the small signal equivalent model and of course whenever you talk about small signal equivalent model it is good to tell the students why we want to talk about it even and that is because we want to look at amplifiers. So, let us go to I think the next one is probably the last slide or this is the last slide. So, I can take some questions and there is still some time remaining and I wanted to talk a little bit about some other topic but let us take some questions. If you have any questions you can press the raise hand button and we can there are also some questions in the chat mode so I can take that now. All of these slides are available on the web so you can just download them and look at it in more detail. In what way analog switch using JFET differs from digital switch using JFETs actually are not really used in digital circuits these days or even earlier because they were better alternatives either the BJT or the MOSFET. So, there is no really there is no such thing as JFET as a digital switch it is used in I know that it is used in some of the sample whole kind of circuits and that is where and that is because it can have a very large gate resistance and it is the drain to source resistance you can control with no by appropriately designed a device but there is for low voltage applications at least there is I mean the same job can also be done by MOSFETs these days. If you are shown in N channel then region you are shown as a neutral whereas when I talking about N type so there is a majority carouser electrons. So, you can say that the number of electrons are more than number of holes it is correct or. Number of electrons is of course more than number of holes in the neutral region the light blue region that we showed in the slides the it is N type region right. So, N d of the order of 10 raise to 15 or 10 raise to 16 or 10 raise to 17 whatever. So, and the electron density is roughly the same as the doping density that is the if the doping density is 10 raise to 16 the electron density is also 10 raise to 16 and the. So, the hole density will be much much smaller than that because N times p will be about N i square. So, if N i square is like 10 raise to 20 so if your electron density is 10 raise to 16 the hole density will be 10 raise to 4 much much much smaller that is why you do not even do not even need to talk about it. Does that answer your question over. Hello sir yeah when I talking about P N junction you are using the term barrier potential or cut-in voltage, but you are using here built-in potential. So, as far as my knowledge is concerned when I am going for built-in potential term so you are talking about the work function difference whereas you are in P N junction using the you are using the term built-in potential. Can I use this out. Over. Okay well built-in potential and the cut-in potentials are two entirely different things. What is the cut-in potential if you plot i versus v for a diode you get something like this and of course the current in the reverse direction is very very small and if you well actually is steeper than that and if you extrapolate this, this is called the cut-in voltage or the turn-on voltage. Now the built-in voltage that I talked about is completely different than this. So, if you are familiar with the P N junction diagram let us just take a P N junction in equilibrium you have the P type region here the N type region here this is the boundary between the two. Rosa Sharma must have done this yesterday then you have the potential varying something like this depending of course on the doping densities this is flat right. The built-in potential is basically the difference in potential between these two this is called built-in potential okay. Now this is the situation in a P N junction even if you do not have a reverse bias right and so and this built-in potential is necessarily is responsible for a depletion region here W depletion. So, in a JFET even when you do not have a gate voltage applied or when your gate voltage is 0 you still have a depletion region and that is why we showed those little blue regions know after the P plus regions right. What happens in a JFET is you have this VBI and now you apply in addition a reverse bias what happens is this gets pulled down etcetera and this becomes your net potential dropper as a result of course lot of things will change this picture is not very clear the depletion region will also change the depletion region will expand and many things like that will happen. In a P plus N junction what happens is that this part of the depletion region is negligible compared to that one because if you look at W depletion on the P side to W depletion on the N side you get something like ND sorry ND over NA and if you are NA in a P plus N junction is much larger than NA ND and it is like 0. So, therefore in the JFET figure we are not even bothered about this particular part you know on the P plus side we are only worried about the N side ok is that clear. So, there are cut in voltage is something that comes into picture when the device turns on the PN junction in a JFET we are we are not allowing that to happen at all because we are never going to forward bias the gate to source junction we are always going to operate that under reverse bias. So, this actually not relevant this is what is relevant ok. Sir while showing the quality for the built-in potential it depends upon the it depends upon the work function difference.  So, shall I give the quality why due to the N side who is having work function if one function more than the P side or. In a P plus N junction we actually do not talk about the work function work function we talk about for a metal semiconductor junction not in a P plus N junction.  So, if you have voltage of 0 volts this is what it looks like. So, there is almost no depletion region in the P region and the entire depletion region is on the N side ok. So, there is no question of any work function. So, this EC minus EF will depend on what is this doping density N A here ok.  So, that is the actual difference between this level and that level. And what is this EC minus EF here that again will depend on what is the doping density ok. So, of course the work function will be important if you are talking about a mesh fit where the gate is a metal. So, then the gate the work function of the metal will come into picture. In this case the work function is not really directly coming into coming into picture because these things are determined only by the doping density. So, let us take another. Mofokam college do you have any question? Over. In the different amplifier shown using JFETs you have used a pair of RGs RG1 and RG2. Can you explain the needs for these RGs? Over. I am not sure if I get that question. Let me see if I get your question correctly. Is your question about the source resistance and drain resistance like these? Over. Is your question about R S and R D? Yeah. I will repeat the question sir. In the different amplifier shown. Oh, in the different amplifier. In the different amplifier. Named as RG1, RG2. Okay. Can you explain the needs for using this pair of resistance? Okay. Yeah, well that is actually not really in the context of this particular topic is not very important. But I believe this may be for some protection purpose. I am not 100% sure. Okay. So, there is one question about what happens to, what happens at depletion region when breakdown occurs? Okay. There is a question about what happens to the depletion region when breakdown happens? So, let us look at that part. Alright. So, when breakdown happens, the P plus n junction, the depletion region is not insulating anymore but it actually will conduct large amounts. And as a result your IV characteristics, the current will basically shoot up and you would have seen that in data books and such. So, if you keep on increasing the drain voltage, the P plus n junction will keep getting more and more reverse biased. At some point what will happen is this junction will simply break down and then the current, the ID will start increasing dramatically. And of course you never want to operate in that particular region. It is not useful for amplifiers or any digital circuits. So, you always avoid that region. Then we find the number of electrons passing at one cross section for FET. Find the number of electrons passing at one cross section. Okay. Of course you can. It is very simple. Let us look at the cross section of the device. So, the number of electrons crossing a particular area. What is the electron density? So, this, the light blue area is the one that you are talking about. So, what is the number of, so you can take a small delta x here. Maybe I can just draw here. Okay. So, let us say this is your, this is that light blue region. This dimension is 2 times h. This dimension is z.  So, what is the number of electrons crossing this? Let us say per unit time.  Now, in this particular area at a given time, what is the number of electrons that we have? The number of electrons here is the volume 2 times h times z times delta x multiplied by the electron density. The electron density over here throughout this area is equal to the doping density because that region is a neutral region. So, that is the number of electrons that are in this area at a given time t. Now, we would like to know in how much time will all of these electrons leave this particular region. So, let us calculate that delta t. So, what is delta t equal to? Delta t is equal to the distance divided by velocity. What is the distance? What is the distance is delta x and what is the velocity? Mu n times the electric field. I am just talking about the magnitude not the science. So, these many electrons in this volume leave this particular region in that much time. So, per unit time if you want to know, you need to just divide that by delta t. The unknown here is dv dx and for that you need to actually solve the device equations inside the device and that one can do by numerical solution, not analytically. There are some other questions, but the time is up. All right. So, if your question has not been answered, you are free to send it again on Moodle and I will try to get back to you. Thanks for your attention and I will see you tomorrow again.


with the proofs of various statements that we have postponed. So, now let us take one by one, this is one of the important results namely, accession theorem for singular homology ok. So, this is central to our theme, singular homology is central to our theme. So, what is accession? If X is union of two subspace is X1 and X2 such that their interiors X1 and X2 interiors of that cover the whole space, then X1 and X2 is an accessive couple. In particular if X1 and X2 are open then this will be automatically satisfied, then it should be an accessive couple is obvious once from this one. Often what you may get is not exactly opens of sets, but their interiors themselves cover is good enough is what this says. So, this is a very useful statement ok. So, what is the meaning of accession? You take the subspace, the subchain complex namely s dot x1 plus s of x2 inclusion map into the whole of s of x ok. Take that map induced by the inclusion on the homology h star of this to h star of x. This h star of x is nothing but h star of s dot x, the full chain complex, singular chain complex. We want to show that this is an isomorphism ok, this is a bijection that is what we want to show. Given any singular and simple x, one of the generators for the chain complex here, a sigma is from mod delta n to x, remember x is written as interior of x1 and interior of x2. Take the inverse image of these open subsets ok. They will give you two open subsets in mod delta n which will cover mod delta n and mod delta n is a compact thing. Whenever you have an open cover by a compact over a of a compact metric space, you have your Lebesgue number and so on. Once you have a Lebesgue number, you can cut down this delta n by various centric subdivisions as often as you want so that the final pieces will be all as small as you want, small means diameter smaller. So putting all these together, I mean I am just recalling the proofs of various statements here in short. What you get is an iterated subdivision, sd remember is the Paris centric subdivision. Power k is it is repeated k times of delta n each star of each vertex, if you take star of each vertex that will be contained in either this open set or in that open set that is meaning of finer than this covering. So sd of k denotes now capital sd denotes remember we have already introduced this. This is the subdivision chain map. This little sd is the operation of cutting down. This is the Paris centric subdivision. This is the subdivision chain map. You take the composite of this chain map with itself k times that is sd capital k. So what we have proved is that capital sd induces the same isomorphism, same identity isomorphism on the homology. So it follows that the singular n chain sd k of sigma belongs to the subgroup s s dot of x1 plus s dot of x2. So that is the final part here. Each piece is contained inside either in x1 or in x2 and then you are taking the sum total of that. So it will be inside this one. The original thing you know part of it may lie in x1 or part of it may lie in n t ref x2. But now when you have cut down each piece will be either in x1 or x2. So it is some total will be the sum of these two chain complexes. Thus if tau is a singular n chain then what is a singular n chain? It is the sum of summation n i sigma i finite sum. For each sigma i i have a k i. Right? So you take the maximum of all these k i's on that k when you take that many subdivisions each of the parts will be either here. There is a sum of these two. Therefore sd of the chain itself now, chain is sum of these two. If tau is a chain sd k of tau itself will be inside this one. Of course this is not this case it is not the same as this one k. There is some other k. Okay? So you can take the maximum of all the k i's where each k i is for corresponding sigma i that does it. Okay? So this sd k of tau is an element of x1, x2. This implies also always this is an obvious thing because of all canonical thing. That is boundary. Boundary is what? Boundary is some summation of various things subjected to restriction of the boundary. The image will be always in x1 or x2. So therefore image is also, this boundary is also inside x1 and x2. So that is not a problem. Further note that sd and capital D was the chain map, chain homotopy. Okay? The chain group map, the chain group group s1, sx2, they are always, once you are inside x1 and x2, x1 or x2, sd will be also inside x1. So it will never change the code domain of those things. So they are always respecting subspaces. So if something is from s1 plus xx2, it will always in sx1 plus sx2. Okay? And sd induces identity homomorphism on homology. Okay? Because it is already there. In particular, for any cycle tau in sx1 plus sx2, sd of k of tau and tau represent the same element in hx. If you already inside sx1 plus sx2 and taking a sd of k, representing the same element because of tau is a cycle. This will be a cycle and homology elements will be the same. This much we have seen for each capital Sd of c. By repeating it k times, this will be true for resource. Therefore, given an n cycle in sx, there is a k in n such that sdk of tau is in sx1 plus sx2. Okay? This much we have seen. Now pass it to homology, phi of this one, phi is just inclusion map. It will be tau because these two represent the same element. So this element represents the same element as this one. So this proves surjectivity of phi. Okay? After cutting down, it is ready. This one is solved. Essentially, all this is built up in the idea when you take a path and write it as composite of two of its parts. You cut a path into two parts, two paths, then you take it as a composite. So the composition of those two, you know, path composition, represents the same chain element up to chain equivalence. This is what we have seen long, long back. Okay? So that theme is usually always here in a much more strong sense. Okay? So these are all technical details here. That is all. Now to prove the injectivity is also similar. Suppose phi, we prove the injectivity, suppose tau is any n cycle in the sx1 plus sx2 itself. And alpha is an n plus one chain in sx such that boundary of alpha is tau. Okay? What I mean by this one? You have started with an element here. Okay? When you go to the larger thing, it represents a zero element. Okay? It is boundary. In the homology, it represents zero element.  That is all I have to do. Right? So this again, for large k, sd of this alpha now, applied to alpha, that will be inside sx1 plus sx2. And boundary of sd of alpha is nothing but sd of the boundary of alpha. And boundary of alpha is tau is sd of k. Okay? Therefore, tau is zero in hx1 plus sx2 because tau, the bracket tau is the same thing as bracket sd of tau, sdk of tau, but that is the boundary. So this proves our injectivity. Okay? So most of these things are very easy and become strategy, but we have to develop these terminologies and concepts. And each concept is very simple. So those simple things we have to verify all the time. So this completes the proof of the big theorem, namely, excision theorem. The people are afraid of that, but this is very simple. Okay? So next thing is not all that simple. Okay? Little more technical details are here. The next thing is, you take a simple shell complex k and a sub complex l, take the relative singular chain complex. Take the relative singular simplicial chain complex. That is a sub complex. Inclusion map is a chain homotopic covalence. That was the statement of theorem 4.1. We are supposed to prove this, but we shall not prove this statement itself, but slightly a weaker statement, which will be the more useful for us, I mean more central to us, namely induced map on the homology is an isomorphism. So we shall be satisfied with the proof of this statement for 10 years. Okay? The actual statement, the proof of actual statement, this one will come in two steps, namely first of all, instead of k comma l, mod k comma l prime, I will just prove it for the pure thing here, right? Absolute thing here. But once you have done this one, it will prove for l also, right? Once you prove for these two things, it is inclusion map, they are both of them. You can put a long homology exact sequence for the chain complex for s dot of k comma l and s of k comma l. Okay? So in the ladder of this long exact homology sequences, each, you know, two of the arrows will be isomorphisms. So third arrow you do not know. Again, two are isomorphisms and so on. So out of the five of them, you will have the middle one you do not know. The other two, two here, two on the other side, they are isomorphisms. By the five lemma, it will follow that the middle one is also an isomorphism. The middle ones are precisely the relative things. Okay? So once you prove it for absolute thing, relative thing follows. So you can just concentrate on this one. Do not worry about the relative part. Okay? After that, the original statement about homology equivalence comes by another general statement on, you know, homological algebra on arbitrary chain complexes, which has free chain complexes. If you have a chain map of free chain complexes, which induces isomorphism homology, okay, then it must be a chain equivalence. So that statement is purely homological algebra. It is not at all difficult, but time consuming. Therefore, I am skipping it. Okay? So in principle, I am only going to prove this part now. Statement 40. So another simplification. You want to prove homology isomorphism. You can do that by first proving it only for the finite case, case finite. Then usually you can take a direct limit argument and then the proof will be true for all k. Okay? Indeed, you do not need direct limit as strong as you want if you do not understand it. It is very simple, just like we did for just now Paris Enter and so on. What you have to do? Suppose you want to prove this one. So you want to prove surjectivity. Take a chain here. Every chain, every actual cycle here, represented by representing a homology element, every chain is supported on a finite simplification complex. So that will be sub-complex here, which will contain that. Okay? So instead of k, you use that sub-complex, say k prime. Then from k prime here, s dot of k prime to that one, you have already proved it is surjective. And go to the inclusion map into this one because everything is inclusion map here, inclusion induced map. So that will be surjective. Similarly, suppose some homology, some cycle here becomes trivial here. Okay? What is my trivial? It is a boundary of something. The boundary is a chain. The chain and boundary of that I have taken. Again that is supported on a finite simplification complex. So take that simplification complex. So there you have already proved that if this is zero here, it is zero here itself. Zero here in this world, there is a boundary here. So that will be the same thing as in the larger one. So even if you don't know, direct limits and so on, as much as, of course we have done it very thoroughly. But direct limit in the homology case can be done by this method, that is all. But it is easy to remember what is happening for finite once you take a direct limit for homology. Homology commutes a direct limit here. So it will go through. For each finite simple complex isomorphism, for every k it will be isomorphism. Now, so how do we prove it for finite simplification complex? Here we use induction. Induction on the number of simplices inside k. If the number is one, what is k? k has to be singleton vertex, that is all. For a singleton vertex, both s of, now ss of singleton and ss of singleton, they are what? They are identical. Every map from delta into single vertex is simply shell already. So they are identical, so this is an isomorphism. So k equal to 1, for r equal to 1, the statement is obvious. Now you assume that it is true for all simple shell complexes, which total number of simplices is less than r and r is some positive number. k is a finite simple shell complex, therefore there is always some maximal simplex. Maximum means what? It is not contained in another larger simplex. So let s be a maximal simplex. Now delete that maximal simplex only. So suppose s is v0, v1, vn, you do not put v0, v1, vn, that is an n simplex. To delete that means rest of them you have to keep, namely all the boundary of this s will be there. So only one simplex you will be deleted.  And the induction hypothesis, what happens now, the inclusion map from s star of l to s star of s l, this is an isomorphism because it has smaller number of simplices. Not only that, the same inclusion map if you restrict it to the s dot, s dot is what? The boundary of whatever you have deleted s, that is already inside l. So this s dot is inside l that also restrict to an isomorphism here. So this is also induction hypothesis that this is also an isomorphism. Here s dot denotes the boundary of simplex s, we also know that both h star of, this is the boundary of simplex s and h star of s of s n. This is simplex itself. This vanishes in positive dimensions. This we have seen as particular argument. The homology of delta n, this is a simplex. This is contractible thing. So this is trivial, we have no, except the 0 dimension it is infinite cyclic. In the positive dimension it is 0. This is also true. This is also contractible. This part we have directly proved in example 2.4. So they are 0 in the positive dimension and identity this h naught to h naught, this is just number of components in both sides. This is an isomorphism always. It is actually identity isomorphism. Since any pair k 1, k 2 of sub complex of a simple complex is excessive. This we have seen for singular and simplexial chains. For both simplexial and singlex chains we have seen these x isocouples. We get, now we can apply meritary sequences. I can write the homologies for these smaller things as you know on k 1 and k 2. So here we are using the, what are we using here tell me. We are using the excision, excision theorem for k 1 and k 2 which we have already proved earlier. We get two corresponding meritary sequences like this. So this is for L union S and singular homology. This is for simplexial singular homology. I have put a hat here. This is not twiddled, this is hat. It is just a notation, temporary notation because this is large and if I put that one it will go out of the slide itself. So this is the first row is here either for a simplexial and the second row is a singular. So this is the union. This is the intersection. This is one of S dot or something. It is h n of S and h n of model is a union here. So this k is a union. This is a direct sum. This is the intersection. This must be something more here which is missing. What should be this one? H n minus 1 of S direct sum S dot here. This may be a sequence after all. Union k, k prior to coming here is no problem. But remember what we have got here. This is this plus this. So this is an isomorphism. This is an isomorphism. This is an isomorphism because induction hypothesis. This is an isomorphism and this is an isomorphism. So only thing that we do not know is this one. So that will be an isomorphism by silent. That is all. So we have one more thing here. Equivalence of singular simplexial and simplexial homology. S dot S double S and C. C is the simplest singular homology, the simplexial homology. So that also you have to see. Here this time we do not have inclusion map but we have quotient map. We shall present a proof for Poincare. For this, just for the proof of that, we would better do it with a total order on the vertices of k. Then for each n, each n simplex sigma, we can display it by writing V i naught V i n in the increasing sequence. V i naught i 1 i n are increasing because I have taken one single order on the vertices. V i naught will occur before V i 1 and so on. That is the meaning of this one. So that will define a unique element of Sn of k. Sn of k includes all the simplexial maps. So this is one simplexial map, 0 going to V i naught, 1 going to E naught, even E n going to corresponding V i naught V i n. So once you have that, this assignment extends linearly, defines a splitting of Sn of k into Sn of k. See this is an element of Sn of k. These are generators here. And this is some element here. This is too large. Actually, we have defined Sn k as a quotient of Sn of k. To define the splitting here, we need to have a order of that map. I am calling it as alpha n. Splitting means what? It is a quotient map here. You come back by quotient map. It is an identity. The quotient map is free from, far free from Sn of k to Sn of k. So if you put all these alpha n's together, that is the totality of this one. That is a chain map. And that chain map follows because of the uniqueness of this, you know, unique simplex and so on. Some seed or sort of a chain map verification is not difficult. Such that it is phi, compute alpha, identity of Sn of k. This is just that it is splitting. That is the meaning of splitting. So we have a splitting which is chain map. So that is the first thing to observe. Now, it remains to define a chain homotopy on the opposite side. See, phi, compute alpha is identity already. Alpha, compute phi is not identity. But I want to say it is chain homotopy to identity of a bigger complex. If we define chain homotopy, then this alpha star will be the inverse of phi star. So we wanted to prove that phi star is an isomorphism. So that will be over. So how to prove, how to define a chain homotopy from alpha, compute phi to identity of Sn? So both phi and alpha preserve sub complexes of k. In other words, if we have sub-complex here, phi of that will go inside Sn of that sub-complex. Similarly, if k is sub-complex of l, say l is sub-complex of k, Sn of l will go into Sn of l. These are all canonical things here, okay. So those things are easy to verify. So sub-complex, namely l is sub-complex of k, p of Sn is c of l, alpha of Sn is s of l. The chain homotopy that we are going to construct will also have this property that it will preserve the sub-complexes. Hence, we can easily pass on to relative chain complexes. So relative version will also come automatically is what I wanted to emphasize here. Okay, so anyway, so we have defined this chain homotopy h. Chain homotopy means what? It is a map from Sn to Sn plus 1 homomorphism so that this boundary composite h plus h composite boundary is the difference alpha composite phi minus identity. This is the way h has to be defined. As usual in many other cases we have seen, we will do this by induction on n, okay. The first thing is alpha naught, alpha naught from c naught of k to Sn of k, okay. He is the identity map. What is Sn of k? 0 delta naught to anything, there is only unique maps, right, where it goes, whichever text it has to take, that is all. It is the c naught of k is also same thing. So c naught of k and Sn of k is identity, the quotient map is identity, so alpha naught is identity. Therefore, there is no problem at this level. You can start with h sigma equal to 0 for all 0 complexes, okay. At the 0th level, that is 0 simplex level, 0 skeleton level, h sigma also can be taken. Now, suppose we have defined h from Sn minus 1 of k to Sn minus Sn of k, so that this is true, okay. Then we have to define it on Sn to have the same property in the next stage, okay. This is induction hypothesis and such that for each n minus 1 singular simplex k, the support of h sigma is contained in the support of sigma. So why I am saying that I want to construct it this way. So that is a part of the induction hypothesis so that it will help me to prove the next step. At the induction h sigma is 0, so here there is induction hypothesis is obvious, okay. So I will assume I have constructed it in this way namely all the singular n simplex sigma, h sigma will be contained in the support of sigma, okay. Now let tau be any singular n simplex in k, okay. So remember we have defined h on only the generators here, okay. Then we take linear combinations, extended linearly. So we are talking only about what is happening on the generator. The both sides here are linear. Therefore, if you verify this identity over the generator, then it will be verified for all of them, okay. So let tau be any singular n simplex in k and s equal to support of tau, okay. We need to find an n plus 1 chain h tau supported on the same s and such that h of daba tau, you see what we are defining here is h in the next stage. So h of this daba of this part, this is already defined. So we take it on the other side. It is alpha composite phi minus tau minus h of daba tau. So daba tau is one dimension lower thing, h of that is already defined. So the left and right hand side here is already defined, okay. And we want to define h here. It is a daba of this one is this one. So this is possible only if this entire thing is daba of something, right. So you have to show that this left hand side, right hand side is actually daba of something. It is in the image of daba. Now that is all one has to show, okay. Obviously, it is necessary that the chain on the RHS of the above equation must be a cycle. If daba of that is 0, then there is a chance. Then it can be daba of something, right. If we should daba of something, daba of that will have to be 0. So it must be a cycle first of all. So, okay, that is if you take daba of this thing, I am taking daba of alpha composite phi minus tau minus h of daba tau. What is this? It is alpha of phi of daba tau, right. This daba is computed. It is no definition. It is alpha phi of daba tau minus daba tau whatever minus daba of h of daba tau, okay. Now why this is 0? This is by induction step. So that we have already this equation on 45 of what we took, okay. So you have to verify that. It is also clear that the support of this cycle is contained inside as we are not going out of a singular simplex x at all completely. Since the homology of h n of s of delta q is 0, the whole thing is seen as single singular simplex. And any cycle which you have shown cycle is what is trivial, it is a boundary. So finally this is where we have used topology. Till then all the time we are using just you know algebra in some sense. So this is an example that we have already done in the case of s of delta q is 0 in positive dimension. So we are already in the high one dimension higher in the positive dimension we are we are argued it in different way. So it follows that this cycle is a boundary and hence there is a chain h tau belonging to s n plus 1 of x, but s is contains an of k satisfying our requirement mainly daba of that is equal to this one, okay. So this because this is a this is first of all it is a cycle but this cycle is inside s therefore it is 0 in the homology which means it is the boundary, okay. So that completes the construction of h and complete the proof of that singular simplex chain complex is equivalent to the simplex chain complex itself, okay. The methods employed here there is a pattern and that pattern is generalized to what is called as methods of acyclic models. Something is free, something is acyclic, this acyclicity these two things will be put together, okay. My idea is to explain these things in the rudiment of this in the in atomic things in this in the simplest case how these things work.  That is why I have explained all these things separately here instead of using the high machinery right in the beginning, okay. So this completes one aspect of whatever you wanted to study namely the homology, okay. Whatever we have aimed up till here that we have completed. So we have completed all the results long back now whatever statements that we have postponed most of them we have completed proofs except things like this homological algebra here that we have that I told you. So except for a few things like this where in clear reference has been also given we have completed all the proofs, okay. So next time we will stop take a different topic namely topology of manifolds. Thank you.


What's up? What's going on everybody? Welcome back to another video. In today's video we're doing something slightly different from usual. We're sort of varying off the path a little and we're actually looking into marijuana in sports. I am interested in it. I haven't really watched too many videos about it. We've seen guys like Ricky Williams. I've seen Canada legalize it the whole entire country recently which is extremely interesting. I've never been to Canada but that's definitely one reason why I'd like to. Believe it or not I'm actually Dutch. We all know about Amsterdam so with all that being said I've actually got a little bit of greenery right here and so I'm gonna roll something up. Roll up a little wee dooby. We join and we're gonna sit back here and we're actually gonna listen to Snoop Dogg believe it or not rant about marijuana in sports and also Antonio Brown. I have no idea what he's gonna say. No idea at all but before we do that I'm actually gonna go and get my Antonio Brown jersey.  On the Stephen A Smith show which makes it even more interesting. So I'll be back shortly. Okay all good? We all good? Look at that. Still got the tags on. Don't judge me for it. This was a gift a very very special gift and a gift that not a lot of people would have because it is Brown's Raiders jersey and he never actually suited up for the Raiders did he?  Seems a Antonio Brown's year has been similar to Levy on Bells was the year before and I'm very frustrated that because to hear someone you know to see someone take a whole year out it doesn't even matter about the circumstances or why Levy on Bell was because he wasn't getting paid enough and Antonio Brown is all his own doing so regardless you never want to see a player in their prime taking a year off like that so Antonio Brown has been extremely active still on social media and stuff he's he's actually turned a bit of a corner throughout the season and it seems as if he's taken some more responsibility from what happened I guess I yeah I'm pretty sure we'll see him on a team next year it's just who it's gonna be and that's the more exciting part so with all that being said we're back here we're about to watch a video quite a short video Stephen A Smith show probably the first time I've actually seen this it's actually I'm assuming it's pretty old oh no March 8th 2019 rapper Snoop Dogg joins the Stephen A Smith show to discuss the NFL's rules regarding marijuana use and why he feels Antonio Brown is being mistreated by NFL owners and this was in March so obviously Antonio Brown's done a whole lot since then let's get into it guys there's no time to lose but before we do in the name of Snoop Dogg and Stephen A Smith let's have a puff yeah but that they do okay so they must have been talking about marijuana in sports and Snoop Dogg is fucking cooled up on the line and he's gonna chime in with his his thoughts his very experienced thoughts so let's go to me because I want you to remind folks you and I talked to you did you did me the courtesy of coming on first tape you remember that a couple of years ago to talk about the same subject tell my audience what you broke down the folks that day and tell me what you are now well right now I'm in the same place I want to talk to you last time I feel like athletes aren't rappers they aren't entertains we can do whatever we want to do we don't have no boss we don't have no commission we don't have no rules of recognition so we can do whatever we want to and still get into us with deals will still be love for who we are because we don't have rules and regulations and hip-hop but when you sign up for the NFL there was a regulation that is that put in place for the 75 years that you're not gonna change no matter who you are or who you think you are if you sign up for it you know the rules and regulations you must abide by the rules if you violate you out and that's what's happening right now it's a lot of violation going on in this area that was TV and did a video I don't think that was smart at all that's where I was going Snoop that was gonna be my next question to you I want to know what your feelings well when you saw David Irving's video this morning I thought he should have kept it low key and not put his business out there because it's not football season as it is so what David Irving what happened with that it's not even coming up David Irving weed what David Irving quitting the NFL over marijuana okay so that was a live video of him saying he's quitting the league because he can't smoke hot dude David Irving finds post cowboys career stars in his remember look at this shit stars in his own cannabis ads okay that's a story I definitely want to hear about okay let's continue show people that I don't care but you showing a long example cuz a lot of people like yourself look their whole lives to get to that level and it gets that level in the floor behind smoking weed yeah come on man you can smoke weed after the game or the season is over with it's not that serious man this is the habit we smoke already talking to you we're talking about the snoop dog right here with Stephen a ESP radio is Snoop what I went off about and what I've been going off about I take a position like that people going off selling out blah blah blah and I'm like this I'm tired of all the fact of the matter let me stop you're not a seller you bring attention to the situation that need to be talked about you looking out it's a difference between a lookout and sell out you looking out by bringing the attention and telling the truth but putting facts in front of our face that's looking out that ain't selling out selling out is when you don't tell a man he won't you don't tell a man you belong in your position you're born your life and you making a bad example that's selling out looking out is keeping it one file what it is that's what you want that well for sure that's why I'm calling you right now what well I appreciate that you know I do thank you so much new dog right here was even a ESP radio is penis fall at you go here's my thing not just that it's not just we even with an Antonio Brown Antonio Brown is a bad brother you're still a spin I'm gonna steal a spin we know what he's meant to this organization but in the same breath and by the way I love the shop I love what I'm seeing LeBron and Maverick and those boys do I saw you on the show a few months ago you know you know Jamie's my brother I saw him on the show just recently okay they did a hell of a job but the thing about it is this Antonio Brown for our Steelers called out the quarterback called out the coach called out the GM called out the owner this is on separate weeks leading up to that interview and my body yeah okay all right that's that one done I guess what we need to do now is continue with this little story here and just just find out what what happened here you know what I mean I I want to I want to see if we can watch that live video and see what he said David Irving quits 21 seconds it looked a lot longer than that former Dallas Cowboys defensive tackle David Irving rallies to legalize marijuana all right we're gonna watch those two videos to finish off let's let's just let's have a look at this I don't know if it was because he needed to use it for pain or that he just wanted to use it recreationally but if he threw away his career because he wanted to use it recreationally nah that's just stupid basically guys suspension and all this other nonsense I'm out of there I'm not doing it you know it's a lot of reasons why you know I come in this decision you quit football well let's hear why man but first let's hear him a week ago thank you dude we'll see them well there was some action outside of the stadium tonight to former defensive lineman David Irving held a rally to legalize marijuana he left the Cowboys last season after being suspended for violating the NFL's substance abuse policy here's Rebecca Lopez okay why he's now taking on this fight pot over pills is David Irving's message marijuana is healthy I'm not sure how much you all know but I stepped away from my football career cause I was forced to take these pills and I rather choose plants over pills Irving has been on a mission after quitting the doubt respect brah respect that wouldn't have been an easy decision man and if they're trying to give him all these pills that people can easily get addicted to and he's saying no I want to smoke weed instead it's fucked the system's fucked okay it wasn't just to smoke it recreationally it was to use it medicinally which is exactly what it should be used for pain damn man interesting story let's continue Alice Cowboys when he was caught violating the substance abuse policy and suspended indefinitely earlier this year you don't just get suspended so you get suspended you get the fame of a national level he says marijuana is better than the opioids he said the NFL gives players to deal with pain Irving claims he was given six different pain meds by team doctors you know they do a lot to your body start to mess with your mind start to mess with your liver so Irving came to Dallas to tap the medical use for marijuana he held a small rally outside the stadium army veteran Robert head is among those who showed up for support so I was on about 10 pills a day along with about a fifth of whiskey a week head says marijuana helped him get off the pills and alcohol this week the MLB announced it will no longer test for marijuana Irving hopes really will follow and he says he is pushing for laws to legalize marijuana in Arlington I'm Rebecca Lopez well okay so let's say they did stop testing then I guess I assume you know far less athletes would even go to the doctors complaining of pain and saying they needed anything in the first place because that that you know that all be all the ones that would want to would be self-medicating at home and you know doing it on their own sorting out their pain medication themselves basically you know the repercussions really the only negative repercussions and this is what gets brought up a lot is is for the pharmaceutical companies that be selling less painkillers and we'll leave it there anyways guys that is video number two done for the series I'm just giving you guys my opinion on the subject before I go and give it up for 30 days at least from the start of January that's gonna be my New Year's resolution you know wish me luck I'll probably need it and I will say again you know unfortunately I deleted all of my previous videos about the subjects about weed because I wanted more a more clean-cut image I wanted any prospective coaches out there who were gonna recruit me to you know not have anything that could stop them or sway their opinions or stop them from reaching out to me you know so I decided to delete them a few months ago and I guess I'm bringing it back because I do enjoy it but you never know I might have a change of heart and never put any of these videos up for you to see we'll have to see it is Christmas Eve Christmas Day tomorrow I want to say Merry Christmas have a great night or a great day wherever you are in the world whenever you're watching this and have a fantastic festive season and with that being said good night