 So, we discuss these aspects in the series of these particular lectures which will be in this 14 to 15 sessions. Now, in this I have just summarized what it is going to be like in this series of non-classical MOSFETs. I have put 14 to 15s having a one buffer lecture if necessary. Module 1 which we start today will have transport in nano MOSFETs and of course, when we do not get down into the transport straightaway, we will first discuss the first order MOSFET model and go through some of the effects of interface states in sub-threshold slope. Highlight some of the short channel effects which have already been discussed hot electron, LDD, light due to drain, extended source in all that has been discussed. So, I will not I will just rush through them. Velocity saturation, ballistic transport, injection velocity, these are the things which one has to see particularly because you will see that ultimately the performance of the classical MOSFETs is decided by the injection velocity rather than any other parameters. And the injection velocity you will see that it is controlled by the low field electric low field mobility of electrons or holes. So, the goal in all these devices non-classical devices will be to realize devices which will result in high carrier mobility, which automatically will give you high injection velocity. You can also switch to devices like LM arsenide, where you can get velocity overshoot effects. You can have velocities in addition over and above the velocity saturation that is the module 1. Once we see what are the problems and what is required for high performance, we get down to these will cover about 2, 3 lectures. Then we go down to module 2, which is a special device which is very popular today even in industry like IBM they regularly use SOI MOSFETs for their integrated circuits as and when it is required. Silicon on insulator MOSFET, SOI MOSFET and here we discuss very different versions of this type of MOSFETs including the FinFET and also discuss the quantization effects on threshold voltage mobility and ultimately we also get down to discuss the latest version of these devices called junctionless transistor. Then we have the module 3 in which the focus is on metal source, source strain junction MOSFETs. Instead of PN junction, we will have metal junction, Schottky barrier junctions or ohmic contacts that type of thing. Here we will have to focus quite a bit on the properties of Schottky junctions on silicon, germanium, compound semiconductors etcetera, because we will be discussing about the non-silicon based MOSFETs, FETs in the next section that is the module 4. So, there we take on germanium. So, in order to improve the mobility of the carriers, one way of doing is use silicon itself and ensure that doping levels are very low. You know in conventional devices you have to go to very high doping levels 10 to power 15, 17 percentimeter cube, so that the Schottky channel effects are reduced. Now that is achieved by the SOI MOSFET. Now further if you want to improve the mobility, you have to change the material. Materials like germanium, materials like gallium arsenide, those are the two major contenders for high performance devices, germanium devices PMOS, NMOS. That is what we discussed, germanium MOSFETs, then compounds semiconductors based on gallium arsenide, aluminium gallium arsenide and hetero junction devices. We will try to cover all that. So, this is where I kept option of going 4 or 5 sessions. So, depending upon that, depending upon the availability of slots, we will put 14 or 15 lectures. So, now we take the module 1. The module 1, we begin with some sort of a review, so that you are focused on this topic. So, we will talk, take the conventional MOSFET and conventional analysis long channel model and substrate short swing and the effect of interface state density on the substrate short swing. I was informed that DIT effect on substrate short slope was not covered in the previous presentation, though the DIT is discussed in the great detail. Then short channel effects mainly on substrate short swing and current, very big field touch up on this. Then we will go on to the velocity saturation, ballistic transport and velocity overshoot effect, injection velocity and need for high value of low field mobility, those are the things which we will cover. So, you see that there is bit of overlap in this particular module, but when we go to next, these are required. So, I just quickly running through this thing. Today it will be mostly some of these things. I will not be able to cover the entire thing today. This will be covered in two, three lectures. So, this is the diagram which you will see everywhere in text books, all text books. We can see that you have got the substrate which is made up of silicon in the conventional MOSFET and you have got the red color is the oxide. Then this is the polysilicon gate. If it is N channel MOSFET, it is N plus stable doped and you have got the source drain, N plus source, N plus drain. And when you analyze this, you will actually not bother about the entire region. Your focus will be only on this portion because that is the portion which gives you the MOSFET action. And these are the thick oxides which allow you to isolate one device from the other device in the integrated circuit. So, if it is N channel device, you use beta substrates. And while analyzing, you have some simplifications done. The most important simplification which is done in the Shockley analysis is GCA. That is, gradual channel approximation. What is the meaning of that? The electric field GCA, gradual gradual channel approximation says or considers or assumes that the electric field along the channel, I marked it as y direction, along the channel is much smaller compared to the electric field along the vertical direction. That is, x is marked as vertical which would mean that you can ignore the two dimensional effects while you solve for the carrier concentration in this direction. You solve the Poisson's equation along the x direction, that is top to bottom. I am marking it non-conventional, y is in the horizontal direction, x is in the vertical direction. So, you can solve Poisson's equation in one dimensional direction because you consider the electric field only in the E x because E y field is small. This assumption will fail when you go to high fields in the y direction. That is where you have to think other methods of solving it. Then also you assume that Ohm's law holds good. When you say Ohm's law holds good, what is the meaning? Immediately you say v is equal to i into r. When you say v is equal to i into r, what you assume is that the velocity of electrons or carriers is proportional to the electric field. That is the meaning of Ohm's law holds good. So, if you know the mobility and the electric field, I can straight away say the velocity is equal to mobility into electric field. Velocity is proportional to electric field. That is Ohm's law. Now, other assumptions in this analysis is channel doping is uniform. There are minor modifications on this if you have the non-informed doping concentration. That I do not get down to discussion. I just took this up for continuity sake. So, here these portion I have shown that is where the channel is present. You can see that when the voltage is applied here, the or when you apply voltage to the gate, the p channel p region gets inverted and becomes n type. So, I would say electrons are collected here. That is the inversion layer is present here. If there is a voltage drop along this direction by means of applied voltage, you will see that or you know that the potential available for inverting is less than p g s. So, we will see what it is. Now, quickly notice we have taken only this portion of that because that is where the transistor action is. And what does this portion do? The bottom portion is only mechanical support. When you have a after all this region will be as thin as point 1 point 2 microns. But this whole thing is a wafer which will be 300 microns, 400 microns depending upon the wafer thickness. If you have a 4 inch wafer, if I can say so, that is 100 millimeter diameter wafer that will be about 400 micrometers thickness. So, that is to give an idea. So, that is why this portion is a mechanical support. Since the entire action is on the top portion, I put only that layer. Source, drain that is the region between is p type, the gate oxide which is very thin, whole transistor side about 100 nanometers. Today you go down to 1 nanometer to avoid the short channel effects. And this is a gate. The gate can be metal, but usually it is polysilicon doped very heavily. Now, you can see if there is no voltage applied between the source and drain, V d s equal to 0, the voltage applied between the gate and the substrate. I have shown the contact here, but it is far away from that. Then plus voltage will apply, the field lines terminate on this semiconductor here. It will initially deplete and then finally, when the voltage is large enough, there will be electron accumulated. This has been elucidated in detail in the last series of lectures, your electron which is known as the inversion layer. So, you say that it is inverted when the gate voltage is threshold voltage that is v threshold voltage. So, once see till you till the channel is inverted, the voltage applied voltage to the gate is shared between this oxide and the top depletion layer. V oxide V g is equal to V oxide plus V silicon, V silicon is depletion layer voltage, but once this there is there are enough electrons accumulated in the channel device. Then the plus charges which are generated here due to the applied plus voltage require negative charges, they come from this channel itself. That means the voltage over and above the threshold voltage appears totally across the oxide. So, the charge in the inversion layer here will be V g minus V threshold that is the voltage that appears across this oxide. So, V g minus V threshold into the capacitance of the oxide. So, that is what is written here. V g minus V threshold into C oxide, C oxide is the oxide capacitance per unit area. So, what we say if this is unit area, the q n that is the inversion layer charge will be charge per unit area. Unit area not looking into the plane of this like this, it is a charge looking from the top per unit area looking from the top. So, we compute the area of the gate region that is the channel length or the gate length into the depth of the gate w that will be the total area. So, this is when a V d equal to 0. Now, you can see when V d equal to V d is equal to 0 there is no drop across the channel. So, whatever voltage you apply V g minus V threshold is available for creating these charges. Suppose you apply V d is positive what happens? Immediately, you have got plus voltage here and this is connected to the drain all through there is inversion layer. Therefore, depending upon the voltage that you apply there is current flow from the drain to the source due to flow of electrons from source to drain. So, source is the supplier of the electrons. When the source is obtained the supplier of electrons is the p type region minority carriers. They will not be able to respond high frequencies, but when you have the source here supplying the carriers they may be minority carrier here, but they are majority carrier here there is a large reservoir of electrons present in the n plus region. So, that can supply this and this drain collects those electrons here. So, this is the source channel and the drain. Now, notice the difference between the previous diagram and this diagram. I have drawn the channel charge maximum here tapering down to close to 0 here. How much close to 0 it is here depends upon the voltage drop across the channel. Why? At this end the voltage is 0. There is a voltage is 0 from here to here if you go V d s is 0 here maximum here. So, if this is 0, but even though this is 0 due to the gate voltage there is a potential which you call it as some psi of s. So, there are some there is some potential due to that, but that potential is used for creating this inversion here. What I am talking of is applied V d s is 0 here. Therefore, voltage available for this inversion is here is equal to V g minus V 3. If you move in this direction from here to here there is a voltage rise equal to V d s. If I take the any point y here there is a potential rise from here to here equal to V y. So, if this was 0 this was V y. At this point the charge the voltage available for creating the inversion layer was equal to V g s minus V threshold, but out of that voltage V y has gone to this. So, the voltage available here is equal to V g s minus V threshold minus V y. So, charge at any point y is equal to C oxide into instead of V g s minus V threshold you have got minus V y because the part of the voltage is gone into this portion. If I go to this end it is equal to V g s minus V threshold minus V d s. So, that is the charge. Now, what is the velocity at any point velocity at any point is equal to mobility into electric field that is Ohm's law. So, I can write electric field magnitude as d V by d y y is in the direction. So, mobility into electric field mobility into d V by d y that is equation number 3. Why are we looking at this q n velocity? The current drain current is given by this expression because q n is a charge per centimeter square here at any y and that that is actually charge sheet. If I multiply by w and d x that is the total charge available in the d of x q n is per unit area into w depth into d of x d of y that gives the area. Now, I have w into q n into d y by d t. What is that? So, q n into w into d y by d t is equal to that velocity w into q n into V is q n into w I am sorry q n into V, V is d y by d t. W into d y is the total charge by d t is the current. So, in time d t charge moves by the density of y. So, whatever was present in the depth d y is collected in the depth d t. So, this is the standard equation. I am sure you have discussed this in the previous presentation. It is a whatever transport phenomena mechanism you talk of. You can write this equation. Suppose in the velocity is saturated velocity, then you can write this as w into charge into saturation velocity. Now, we are talking of a case where velocity is proportional to electric field. Therefore, we find out q n multiplied by this velocity. So, putting these two together, I will not go through the simplification you can just work it out very easily. Substitute these two here that is w into q n is this quantity into V into d V by d y. And integrate that from y is equal to 0 to l and V is equal to 0 to V d s. You get this equation. Let me not go through that spending time. You can sit down and work it out. It is substituted from these two. Integrate from y is equal to 0 to l and V d equals from 0 to V d s. So, this is the equation for the drain current. So long as this path is there right through this region. So, whatever V d s you apply appears between this. I have shown a small opening here. Actually, there is not been opening. It will be continuous there. So, it is a continuous path. It is equivalent of putting a resistor between this point and that. That is implication of this law. You have put a resistor here. If the V d s equals to 0, the channel charge is constant. It is equivalent of a resistance which is of uniform depth. Now, when you apply voltage, the charge concentration here keeps on reducing because of voltage drop, which you can look into it as a resistor whose area of cross section decreases. So, as you apply the voltage V d s, the area of cross section of this resistor keeps on falling. What happens to the resistance value? It increases. Rho L by A. Rho is the same. Rho L by A, the area of cross section decreases. So, that means as you increase the V d s, the resistance does not remain constant, but the resistance keeps on increasing. It would mean that if I keep on increasing the V d s, the current will not be linearly increasing the V d s. It is reflected from this equation. In this equation, you can see if the second term is not present, the I d and V d, that is linear. So, that is the point at which the voltage drop is small. V d s is very, very small. So, as the V d s becomes more and more, this portion takes care of the varying cross section to say. It is not really cross section. It is a charge that is varying. So, if you see the characteristic now. So, this is linear. You can see initially it is linear. Then, as I increase the V d s, the current does not increase linearly because the resistance does not remain constant. The resistance keeps on increasing. So, current does not increase correspondingly. Now, at a particular V d s, when the V d s becomes equal to V g s minus V threshold voltage, the current saturates. What happens is, you go back to the slide. What happens is, when that happens, when the V d s is equal to V g s minus V threshold, the voltage drop from here to here is equal to V g s minus V threshold. When V g s is equal to V d s equal to V g s minus V threshold, what is the meaning? The voltage drop across the channel is V g s minus V threshold. What is the voltage drop across the gate and the channel at that point? V threshold itself because this is V g s. Out of that, V g s minus V threshold is there. So, subtract that, you get this voltage V threshold. That means, the channel is just about to be formed there. If I apply more voltage, there is no channel there, but there is a depletion layer. So, you can see now, when the channel has just opened, it is a very interesting phenomenon. I have oversimplifying it and telling you, when the channel is just opened, the voltage drop across the channel is V g s minus V threshold. This is the channel potential. If I increase the V d s beyond that point, this voltage drop, where the channel is opened up remains the same thing, V g s minus V threshold. What about the resistance value between this point? Where it is opened at this point? That is the same because this is C oxide into V g s minus V threshold. This is just equal to V g s minus V threshold is a drop there. So, the charge here remains the same as when it has just opened. If there is a light lead out region on the drain region, depletion layer will move into the drain region, but this opening will be remaining there. How does the current flow from this to this one, when there is opening? That is because this is N plus. There is a depletion layer here. The electric field is actually from the drain to the channel, depletion layer. So, as a result, the electric field is from the N plus to the channel depletion layer. So, electrons reaching this edge of the depletion layer will be collected there, like in the bipolar transistor. See, it is almost like a bipolar transistor. The difference is here the electrons are injected and they flow by drift and they are collected by this collector. Here, we call it as drain. In the bipolar transistor, there is no field in this region. Electrons are injected from the N plus region to the P region. There is no immersion layer because there is no gate. They move through this region by diffusion and they are collected here. That is the difference, the transport mechanism. In fact, you will see if the carrier concentration is very, very low as it happens in the sub threshold region, you will see that the current flow is by diffusion. So, there is a difference in the mechanism of transport when there is inverted and just below the inversion layer. Now, coming back to this point, why does it saturate? When this has just opened and beyond that point, if you consider this as resistance, that resistance value remains the same thing because charge distributions remain the same thing because this voltage at this edge is with threshold, voltage at this edge is V G minus with threshold, which is available for charge. So, total charge is the same thing and the voltage drop from here to here is remaining the same thing. So, what would you say? That means, drop across this region is the same thing, the resistance is the same thing, current is the same thing because rest of the voltage that you apply to the gate does not go into this channel. It goes into the diffusion layer. So, therefore, you get beyond that point current saturates and flat. It is the over simplification, but that is the first order theory. So, that is happening when V D S equal to V G S with threshold. So, current at saturation is, I have just rewritten this equation here and at saturation, you substitute V D S equal to V G S mass with P T N. So, when you substitute that, you get this equation. The state substitution you do not, you can just sit down and work out that. So, substitute V D S equal to V G S mass with threshold, you get that. This standard is parallel. You can see this is independent of V D S because all that V D S does not go to the channel. It goes into the diffusion layer. So, from here you can see that we get better current, better performance if the mobility is high. So, from here you will say, I would choose a material which has got high electron mobility or high hole mobility. Very indicative of this is that. And of course, you want to have larger I D. Why do you want larger I D? Because trans conductance defined as delta I D S by delta V G S is proportional to C oxide and W. You would increase W. You get larger current and larger trans conductance, but you do not like that. Why do not you like it? Because it occupies a lot of space width and it will also give rise to larger capacitance. So, the way you do it is, you will see that you have seen it already that you increase the C oxide and maybe you can reduce W. Increase the C oxide and reduce the channel length. All these increase, give rise to better drain current and better trans conductance. That is delta I D S by delta V G S which is proportional to this quantity. So, the transfer characteristics is at any voltage here. If I go along this direction for a constant V D S, if I take I D S like that, for V D S less than V threshold voltage, current is practically 0. This is oversimplification telling that when V threshold, V G S equal to V threshold, current is 0. Strictly it is not true. It is finite there. That is sub threshold. We will discuss that later. Already some discussion has been done. And this is the curve which will tell you which gives V D S V G S minus V threshold. Because larger the V G S, larger will be the saturation voltage. That is what is projected here. In fact, this also will tell you that maybe I am not discuss that aspect. You can connect either drain and the gate to the together and you can get that characteristic, two dimensional characteristic and a drain and the gate. V D is equal to V G S practically. So, that is what I wanted to mention here. Now, above threshold voltage therefore, you have got in the saturation region. What do you mean by saturation? That region. V D S greater than V G S minus V threshold you get these characteristics. Now, in the same region, if I go down here at a particular point, you have got a threshold voltage. V G S equal to threshold voltage. In the previous diagram, we have put this as 0. But, if you plot it on a logarithmic scale, it is not equal to 0. It will be finite and go down to very low values like this. That is a sub threshold current and that is due to the transport of carriers from the source to drain when the carrier concentration is low. There are not sufficient carriers so that the drip current cannot hold good, cannot support that current. Drip current depends upon the carrier concentration and electric field and that carrier concentration goes down, drip current goes down. In fact, even when you take the previous scale, I will just go back quickly to the next slide here. Even in this case, the carrier concentration is high here and it is 0 there. There is concentration gradient, which would mean that there is diffusion current. But, the drip current is very high compared to the diffusion current because the current drip velocity is high compared to the way the carriers move by diffusion. That depends upon concentration gradient. That velocity is low. So, here the drip current is dominated. We can comfortably neglect diffusion current. But, when you go to this region where there is hardly very, I say I have removed that inversion layer there. I have removed the inversion layer here. But, there are carriers. They are still electrons. If you recall, when you keep on applying plus voltage to the gate, it becomes less p type. That means it is depleted on the surface and then more and more depleted, then becomes n type in electron attracted to the surface. So, when you go just below that threshold voltage, there are not enough electrons. The electron density is less compared to the whole density in the bulk. That is the threshold region. But, those electrons cannot move by drift. That means there is hardly any voltage drop across that. So, it is like a bipolar transistor. Now, it acts like a bipolar transistor. So, you have got the source. You have got the drain. Instead, you can call it as this emitter. You have got the collector and emitter injects electrons to this region. And from here to here, the drip current is negligible. Therefore, current flow is by diffusion. Now, what is the state of affairs as far as this junction is concerned? See, when I apply voltage plus here and ground here or ground here, between the gate and this point, there is that V G S. You can see that at inversion, watch this carefully, at inversion, you apply plus voltage. The carrier concentration was equal to the electron concentration was equal to the whole concentration at this end. Now, what is the potential here? The surface potential, you call it as surface potential. That is the potential here at the top surface is twice phi f or twice phi b. You know what it is. When it is inverted, the potential is twice phi b. Phi b is k T by q, logarithm of doping by n i. So, you have got the potential plus minus plus minus. This is plus. So, with respect to this ground, the source here, the potential, there is a depletion layer across the source always. Even when it is inverted case, there is a barrier here. Carriers are injected across the barrier, because that is forward bias. See, when you go to this particular situation here, carriers are injected from the source to the channel across the barrier. And when the channel is not present, what is the barrier? Built-in potential. When a channel is present, if the potential is twice phi f, what is the barrier? Built-in potential minus twice phi f. So, barrier is reduced by twice phi f. So, carriers are injected across that. Same phenomenon is there, bipolar transistor. The difference is instead of that here is drift. When you go to this, the barrier is still present, but barrier is slightly more than b b i minus twice phi f. It is b b i minus phi s, surface potential. There is a barrier is reduced by phi f s. So, still, because it is forward bias by an amount equal to phi s, surface potential, there will be carrier injection. So, that is transported. What is the carrier concentration here? I call it as N p. N p 0. 0 is y equal to 0 here and y equal to l here. Both ends, you can see, they are a depletion layer. This is forward bias junction. This is reverse bias junction. This is the depletion layer width. If it is heavily doped, 0 l. The carrier concentration is N p 0 and this is a boundary condition of a junction. So, N p 0 will be related to the minority carrier concentration in the p region. That is the minority carrier concentration under thermal equilibrium condition into exponential whatever potential forward bias is there across the junction divided by K T by Q. V T is K T by Q thermal voltage. This is a standard boundary equation, boundary condition. Boundary condition for minority carriers here, because the electron transfer minority in the p region, the boundary condition is whatever thermal equilibrium boundary condition is there, that is N p 0, into exponential whatever voltage is there, forward voltage divided by K T. Powered voltage is psi of s. Is it all right? So, you can see that carrier concentration is that. What is the carrier concentration here? 0, because reverse bias is collecting whatever is reaching is collecting. In between, there is no base contact. Here I am showing the base contact, but there is no contact into this particular region there. So, there is no recombination. There is no base current and particularly when this length is small, there is no recombination. So, whatever carriers are injected here will be connected here. Carrier concentration is N p 0. Carrier concentration is 0 here. And the current flow from here to here if it is by diffusion, what is the current flow equation? Current flow is governed by Q d n N p 0 divided by L. What is N p 0 divided by L? That is a slope. So, the diffusion current is Q d n d n by d of x. That is the current density into area. I do not know what the area is. If the whole thing is injecting and current flow is there, then it will be entire depth. But in the case of bulk MOSFET, what we are talking of, it will be inverted only near the top. So, area will be actually w into the depth where the charges are present. So, this is the carrier distribution linear. It is linear because the slope here and slope here will be same thing. Wherever you take, the current is the same thing. There is a current continuity. That is the meaning of that. So, I can write it as Q d n N p 0 by L. That is Q d n d n by d of x everywhere into area. That is the current. I substitute for N p 0 with this quantity. What do you get? You get a term which depends upon the charge diffusion coefficient. What is N p 0? It depends upon doping level here. N i square divided by doping concentration. So, this is a doping independent quantity. So, some constant term into exponential e to the power of psi s by k d. You see that there. So, I have rewritten that equation. I d that is sub threshold current. I d, I am just rewriting that by substituting this there. Then, I naught can be written as I naught e to the power of psi s by s a d, where I naught actually depends upon Q d n by L and N p 0. N p 0 is dependent doping. Now, let us see how this would affect the sub threshold. I am sure this has been discussed in some detail, but I want to bring in that other factors. So, that is why I am going through it. So, this is the sub threshold current psi s by v t. Now, psi s is part of the V g s. V g s gets shared between the oxide and the silicon, the junction. So, I can call it as V g s by N times V t, where N is greater than 1 or equal to 1. It is greater than 1 usually, so that there is a, if it is equal to 1, whatever V g s is there goes to be psi of s. Thinner the oxide, more voltage will go to this junction. So, thinner the oxide, N will be closer to 1. That is the meaning of that. So, psi s is psi s by psi s I am writing it as V g s by N. What is N? N is delta V g s by d psi m s from here. I just assumed some factor. I do not know what it is right now. We will see what it is. So, I substitute here. N is equal to, is given by this equation, since it is logarithmic. So, we know that this is exponential. So, on the log scale I can put it as linear curve. That is what is the sub threshold. So, you saw, when you previous curve you saw it was like that after threshold. Below threshold it is linear in the logarithmic scale. Now, I take logarithm on both sides. Logrhythm of this current is equal to log i naught to the base 10, because it is easy to work out in the decades. Log 10 log i naught to the base 10 plus V g s by N V t. Since it is log 10, you have a dividing factor log 10 to the base e mathematics. So, you have got this term coming up, because it you are taking log 10. If you log to the base e, you would have had only V t by V g s by N V t. So, now, you define the term sub threshold slope, which is delta V g s divided by delta of this quantity. So, delta V g s divided by this quantity is, take this whole thing on to that side and bring this side, that is N V t log 10. So, from this equation today we can see that sub threshold slope or sub threshold swing is delta V g s divided by delta of logarithm of I d, that is N V t log 10. That is the sub threshold slope and from here you can see what is V t log 10? V t is k t by Q about 26 millivolts at room temperature, lower temperature of course, it will be reducing. So, and V t log 10, we will see what it is. So, sub threshold swing is N V t log 10, it is actually inverse slope that is delta V g s by that delta log I d. So, sub threshold swing or sub threshold slope is N into V t log 10 and V t log 10 26 millivolts into log 10 that is about 60 millivolts. So, sub threshold swing is 60 millivolts into N, what does it mean? Delta of log I d is equal to 1. When you take this slope log I d, when log I d is for example, delta log I d may be log 100 divided by log 10. So, log 10 that is log 10 that will be equal to 1. So, let us say when it is 1, you get 60 millivolts per decade. If you take this as 1 decade, how much is this? That is the sub threshold swing. So, the best value that you can get, why do you say best value? And N is equal to 1, sub threshold swing is 60 millivolts. That is the change in V g s for 1 decade change in I d, that is the meaning of that. 1 decade change in I d, how much is the V g s required? Now, this is a very important parameter in the MOSFETs, particularly when you go for low voltages. What you need is, when you want to go for low power devices and low voltages, you need to have this threshold voltage as low as possible. Now, if the swing is 60 millivolts per decade, I will give an example to understand this. If the swing is 60 millivolts per decade, that is if N is equal to 1, it is 60 millivolts per decade. What is the meaning of that? The current will for reduce by 1 decade, when the voltage is reduced by 60 millivolts, gate voltage is reduced 60 millivolts below the threshold voltage. So, for example, if the threshold voltage is 300 millivolts here, when you reduce the gate and if let us say current is 1 microampere, I am just for example, I am taking, if the current there is 1 microampere, it is not 0, it is finite, there is a catch here. So, it is at threshold voltage, let us say there was 1 microampere current flowing. Now, when you and if the N is equal to 1, how many decades must the current come down here? This is 300 millivolts. So, I have to reduce the gate voltage from by 300 millivolts. So, if the slope is equal to 60 millivolts per decade, it has to fall by 5 decades. 300 millivolts divided by 60 millivolts, that is when the voltage falls from 300 to 0, the current will fall by 5 decades. So, that is the ideal case. If N is equal to, let us say 2, that means the current will fall by 1 decade if it is 120 millivolts. So, let us say it is 0.362, to make it easy, let us say threshold is 360 millivolts and N is equal to 2. So, when 1 decade current falls, the voltage will fall 120 millivolts. So, if it is 360 millivolts, by the time you reduce the voltage to 0 volts, 360 by 120, current will fall down by 3 decades. So, from 1 microampere, 10 to the power of minus 6, it would have gone to 10 to the power of minus 9. But if the, in the previous case, it has fallen down by 5 decades, that it would have gone down to 10 to the power of minus 6. So, what we are telling is, if you want to reduce this threshold voltage down, it should be steeper. If it is not steeper, if it is falling down like that, current will be at 0 voltage, it will be finite, it will not be 0 actually, then there is a drain voltage. So, there is need for having sharp threshold voltage, sharp threshold swing. We will see how that works out here. By taking this example, this has been discussed in detail. So, I will quickly run through that. I will not discuss the, I will not be able to discuss the effect of DIT today, but still I will just see what we can discuss here. So, there is need to reduce the threshold voltage. Therefore, there is need to keep this threshold swing as sharp as possible. There is need to make N as close to 0, as close to 1 as possible. So, now, we will see what are the factors which effect is N. That can be done by doing, making this equivalent circuit. So, you see that below threshold voltage, applied voltage is get shared between the oxide and the depletion layer. And between the gate and the substrate, you can draw an equivalent circuit of C oxide in series of C depletion. Now, that is the, that is the equivalent circuit. Now, from here, what is that N? N was what we defined, if you recall back, we defined it as VGS by N, delta VGS by delta phi of s. N is that. Now, if you have this equivalent circuit like this, you can immediately see for, at a given VGS, if a change is by delta VGS, what we are interested in is in the slope. So, what we are interested in is actually the slope. That means, what we are interested in is delta VGS by delta phi of s, that is N. So, delta VGS, if I apply here, over and above that VGS, the voltage here is delta VGS into C oxide divided by some of these two. This is the basic voltage sharing between the capacitors. VGS divided by C oxide plus Cd into C oxide is the voltage here with respect to substrate. So, you know that delta VGS by delta phi of s is actually equal to delta VGS by delta phi of s is, take this R side, C oxide plus Cd by C oxide, which is that. So, immediately you can see that, the ideality factor, if you can call it N is delta VGS by delta phi of s equal to 1 plus Cd by C oxide. This is a very, very useful information, you know, without going into the device, we think people can understand this using this capacitor equivalent circuit. So, now, what will you say? What is Cd? Cd is the depletionary capacitance. Remember, all these are referred to per unit area, because both areas are same, this is per unit area. Therefore, per unit area Cd is epsilon silicon, epsilon 0 of course, I am assuming, absorbing that into that. Depletionary capacitance is epsilon silicon divided by Xt maximum. I call Xt maximum, because you find out that at this point, where depletionary is maximum. As you go to that side, it will be different. We take it very close to the, you can say Xt at the source end, and C oxide is epsilon oxide divided by T oxide. Remember, epsilon oxide includes is actually epsilon 0 into epsilon R. So, that is the thing. So, what would you do to improve the suppressor slope? That is what we are just want to see here. State of it is clear. Best that you can get is 60 millivolts, it is on 10. I have to make Cd very small compared to C oxide. One way of doing that is reduce the Cd. I can reduce the Cd. How can reduce the Cd? By reducing the doping here. If the doping here is reduced, depletionary width is larger, Cd will be smaller. But you cannot tolerate that in the short channel devices, because you look for reducing the effect of channel encroachment into the channel region. Therefore, you would keep doping concentration higher. So, what is alternate? Increase the C oxide. How can you increase the C oxide? By reducing the oxide. So, this actually goes along with your requirement of improvement of transconductance and the drain current, where you can increase the C oxide and increase the both transconductance and the drain current for a given device geometry. Subpressor slope also can be improved by that. Now, I just want to just begin with this thing, but I cannot go anywhere further. What you will see will be, I will just not go through that now. What you will be seeing will be, this is most ideal case. Only the doping effect is included in this capacitance. When you have interface state density, which you rise to the states at the interface, there you will see that an additional capacitance come into picture, that additional capacitance will actually appear across this. I will not discuss that in detail today, but you can take it as additional capacitance comes across this. And if you say additional capacitance comes across this due to D I T, interface state density, since you have discussed that, I can use the word D I T. If it comes in and across that, what would you say the effect of interface state density on suppressor slope, it will actually increase. Because instead of C D, you will have additional capacitance coming across the C D. So, you will have suppressor slope increasing with the D I T. We will see how it comes about in my next presentation. For today, I will close down with this. We will continue on the threshold effect of D I T on suppressor slope and continue on the transport mechanism in the next two sessions on this. So, thank you. We will see next time.