 you can follow along with this presentation using printed slides from the nano hub visit www.nanohub.org and download the PDF file containing the slides for this presentation print them out and turn each page when you hear the following sound enjoy the show so we'll get started now this is lecture 33 on MOS capacitor electrostatics we are still thinking about applying bias to a MOSFET to a capacitor structure this is a metal insulator semiconductor capacitor and it is in some way a little bit more complicated than the electrostatics of a p-n junction the reason is that here when you apply a bias charges may accumulate next to the insulator region and there is a build-up of charge that wouldn't have happened if this oxide was not like a dam sort of was not preventing the flow of the electrons and therefore the electrostatics is a little bit more complicated but then electron transport is significantly simpler because you do not have any current flow so that simplifies it so in some way that you spend more time on electrostatics so let's get started we'll talk about the the review we'll start with a review of the problems that we discussed last in last class and then we discuss the depletion and inversion charges this is one of those problems where the exact solution of the electrostatic problem is possible what do I mean by exact I mean had you not been doing exact things for a while now actually we hadn't been because we are looking at depletion approximation we assume that when you have a junction then most of the mobile carriers are negligible you only have a certain amount of depletion charge and that gave us a rectangular charge structure a triangular electric field and a corresponding potential so it was not exact in the sense because we didn't consider the electrons and holes but this time you will see that an exact solution is possible but then I immediately will show you that exact is not really exact so let's get started with the DC equilibrium equilibrium solution so you remember that we are looking into this particular problem where on the y-axis we look into the amount of electrons that have been accumulated or amount of charge that have been accumulated on the semiconductor side as a function of surface potential phi sub s and there are these three regions that if the phi sub s is negative that means with corresponding to a negative gate voltage we have the blue region where the accumulation charge increases exponentially with with the applied voltage on the other side we have this square root of phi sub dependence up to a point and beyond that point if you apply higher gate voltage then the charge the surface charge again begins to increase exponentially with surface potential remember this word this is not with gate voltage gate voltage is a completely different dependence this is with surface potential it increases exponentially however you will see it becomes very difficult beyond 2 phi sub f to change phi sub s at all as a result although in principle it will increase exponentially if you could change phi sub s but you will see it's very difficult to do so and that's a important thing to to understand so you remember that this is depletion region where you have applied a positive bias and as a result you have essentially pushed the holes away you know positive bias in the blue and you have pushed the holes away and therefore you have exposed red region in the in the bottom which are this simply the acceptors exactly balanced like a parallel plate capacitor and then we also talked about this region having this particular square root of phi sub s dependency as a function of vg do you remember that from this expression you can actually calculate phi sub s for any gate voltage how so you say let's say you have 2 volts you know all the doping density oxide thickness x naught and every other all other things and it will have a form of equation x square plus ax equals b and therefore you can calculate x and that gives you phi sub s now something very strange happens when this phi sub s or surface inversion potential when that becomes equal to 2 phi sub f what is phi sub f by the way phi sub f is the equilibrium distance between the Fermi level and the intrinsic level right so that that's phi sub s so if you have more doping then phi sub s would be larger right that will be larger because the the the the the donor levels will be closer to the valence band and therefore the Fermi level will be closer to the valence band in this particular case and therefore you will have larger phi sub s so phi sub f I'm sorry so in this case I want to tell you something about this phi sub f what something very strange happens at 2 phi sub f and I want to explain to you what that is and that is a very important point because that's called a threshold voltage a key parameter in MOSFET okay now you can see by the way the phi sub s is where this square root of dependency changes into the exponential dependence so we need to know why that happens so let's go slow very slow so what happens in the beginning is a that when you have begin to apply a positive bias then gradually the holes going to move away are pushed back you know they don't like positive charge so from the gate and so they gradually push back and as they push back this red acceptor charges are exposed these are space charge those are exposed they cannot move anywhere so they are sort of left behind but at the same time you realize that as the positive holes are moving away because of the band bending now there is a possibility that this is now becoming a depleted regenerate and there will be a generation of carriers so now the electrons will begin to generate by this trap assisted process as a result this green electrons will gradually build up and you can see at some point that greening electron the magnitude of it not the total amount but the height of the grain will equal to the rate so you can see at that point the rate is positive the electron is beginning to take over and that is exactly the point where what it be related to the threshold voltage and that is exactly when the surface potential will become 2 phi sub f so the crossover point but remember the total area under the green the total amount of charges far smaller yet compared to the rate but this is at least in the magnitude on the height side they become comparable let me prove you where I prove this to you so for example I want to know what happens at 2 phi sub f I'm focusing on the electron charge shown here in group you know this is sort of getting closer to the Fermi level and therefore they are gradually filling up that pocket with electrons so let's think about how many electrons I have right so how will I calculate that I have n1 sub n1 s s meaning surface so I'm asking the questions how many electrons do I have next to the surface well the formula is obvious you know the formula from before that ni which is the intrinsic concentration multiplied by e f minus e is right e is generally I would have written it an e I the intrinsic level that you have seen many times but this time the intrinsic level as you can see from the right hand picture that I is not a constant everywhere and I is sort of bending over as the bands are bending Fermi level is not changing why not because there's no current flow so there's no gradient of the quasi Fermi level whatever value you hold it the right hand side remember is grounded the semiconductor side is grounded so therefore the red line corresponding to the Fermi level is not moving as a result the difference between e I and e f are changing so therefore the carrier concentration next to the surface will be e f minus e is and beta is 1 over kt so from that you can easily calculate that now the problem is that I still do not know it's let's say what e is is but that wouldn't be a problem as well show you now what I have done the exponent factor e f minus e is I have split it into two pieces you can see in the magenta I have taken out e I bulk whatever value was in the bulk you know far to the right hand side where no band bending has occurred and I have put it back in in the second exponent otherwise exactly the same now e f minus e I at bulk that is something I know because that difference is exactly what phi sub f is by definition because I know the doping therefore I know where the e f is and I know of course what you are e I is in a bulk semiconductor so I know phi sub f and correspondingly the quantity on the red e I bulk minus e is that's something I still do not know but do you see that that's equal to phi sub s because that's the amount of band bending e I bulk is whatever was the intrinsic level far to the right and you can see it has bent gradually but do you see that the bending is equal to phi sub s q phi sub s because that's all three bands e c e v and e I they have been together so therefore you see that that will be equal to phi sub s the surface surface band bending and now therefore let's consider the case when the band bending is twice by phi sub f so let's say phi sub f was 0.3 milli electron sorry 0.3 electron volts and therefore 2 phi sub f 0.6 electron volts let's say I'm thinking about the surface potential when it's equal to 2 phi sub f now what is that value that value because you have 2 in one place and minus phi phi sub f so that will give you one but that one is exactly equal to n a because that's by definition is equal to n a so you can see what happens when you bend the band twice the amount corresponding to the phi sub f then the majority carrier charge instead of holes remaining majority carrier first they deplete and then the electron becomes sort of equal to the original hold density and that's when the inversion will begin right inversion what is the word mean that initially you had the holes as a majority carrier now electrons will become sort of the majority carrier and there is an inversion of the charge and that's what happens at 2 phi sub f now this is a very important quantity for many reasons and let me just take a take a minute to explain to you that why this 2 phi sub f understanding 2 phi sub f is very important because what will happen that most of the time when you will discuss transistor physics a little bit later but you realize that we are looking into a vertical cut through that MOSFET structure right through the gate oxide and the substrate and we are looking at the amount of electrons this green electrons which are forming in a very thin sliver very close to the oxide we'll apply an electric field in the drain and then this green electrons will flow right so eventually you want them to accumulate easily and in large quantity as soon as possible so therefore what do you want generally you want that threshold voltage to be low now how would you reduce the threshold voltage look at look at that expression it's already telling you how to design modern modern MOSFET first of all well you know if you wanted to reduce the threshold voltage what you do you will decrease the oxide thickness oxide thickness is x naught is the oxide thickness so therefore in 1970s it was let's say 1000 angstrom oxide thickness these days the oxide thickness is about 10 angstrom sort of same thickness of this as in a DNA and therefore there is a lot of interesting consequence of that that I'll come and tell you a little bit later the other thing is that well you say that oxide thickness I can only make it so 10 you know 10 angstrom is say 5 atoms 5 silicon dioxide atoms and I always tell this story that these days you have 12 inch wafers you know big wafers 12 inch you're trying to put five atoms on top of that it is like covering the whole united states with six inch of snow not a inch variation anywhere because if you had a little bit low there'll be too much current flow and that IC is gone so therefore this level of control that this is possible technologically it's simply it's impossible to believe unless you have actually seen the degree of precision and control people can have so what I'm showing you on these pieces of papers and this math and everything this is not the exciting part once you see how this technology actually plays out then you really get the sense that what a remarkable wonder this technology is but for the time being on more mundane things so let's talk about induced charges in various places mainly in inversion so I have gone to inversion region meaning the green has taken over the red right rain is mobile remember these are electrons which can move and these are generated through shockly redoll type generation right because it's a depleted region now of course if I keep pushing more and more voltage what will happen well I'll get we'll be getting more and more green electrons but how much more that's the question I want to discuss discuss next on that top I have reread in the equation because now I might be going above 5 service above 2 5 sub f 5 sub s being greater than 2 5 sub f which is that I'm going beyond the beyond the inversion point and well if you want to do that that's fine you will put the same expression but this time you realize that will not stop at 2 5 sub f so the second factor on the first line of the equation e i bulk minus e is I will not put 2 5 sub f anymore I'll put whatever the 5 sub s is and you can immediately see the carrier concentration now will increase exponentially with 5 sub s if I could change 5 sub s it will increase exponentially but what I am going to show you now you cannot and as a result it will only increase linearly now this is a sort of a puzzling statement I want you to understand this clearly at least through the lecture or later on when you go home because this is one of the most important concept here so first let me and we will get to that in a second but let me tell you a little bit about that green electrons and how they are spread out because I had been writing this or drawing the green region as a little rectangle now you should have complained in the sense that that doesn't really make sense look at the Fermi level the e f flat and look at the e i e i is bending so therefore if I am to somebody tells you that how many electrons you have as a function of position you will say I'll take e f minus e i at every point put it in an exponential and get the electron concentration and you can see that therefore it will go exponentially down it will spread out like this and q i is my induced charge so that's the i stands for and the green will gradually go down there's no square sitting there right so where is that square coming from that's a very sorry the rectangle coming from actually that's a very important concept and that's I want to show you that this area under this green region you can represent it with an equivalent rectangle and that is a very useful concept so let me show you how that works so what is the amount of induced charge well you say well I will take it from nx dx what was the extra whatever the extra electron is this is the area under this green curve and so whatever that is I will just integrate over now do you realize that that expression nx is ni square nb nb is a bulk charge it's not base anymore there's no base base is gone so here it is a bulk ni square nb what is that charge that's the charge far out to the right that's the minority carrier in the bulk region right in the bulk region as you are bending the bands then at every point there is a phi sub phi of x whatever the distance x and then as you are bending it more and more you increase exponentially get larger and larger amount of charge beta is 1 over kt you remember now you can integrate this right you can integrate that and if you integrate that you will see that first of all that in a state of the variable dx you could because the integration variable is phi so you therefore you can multiply and divide d phi dx and multiply with d phi so you see that that's equal to dx and d phi dx what is that that is the electric field so I place the electric field over there now in general of course the electric field is not constant right electric field is varying in this region but you know if you take a derivative of this it will look like a triangle region right this is a highly doped one side and low doped another side you remember that will be like a triangle but let me assume just let me assume that this is a constant on the average whatever the average of the triangle is let me pull that out and then you can see that the phi sub x you can integrate this one now easily and if you do you get a very simple and wonderful expression because you see why does this kt over q come from in the blue in the bottom line well that's the q and beta beta is 1 over kt remember so when you integrate you pull out that factor this is the exponential of ax so when you integrate you have exponential of ax divided by a so that is that factor whether you have you can see the average electric field and what do you what else do you see you see ni square nb and q phi sub s what is that ni square nb q phi sub s well that's equal to the electron concentration that is the green curve and kt over q what is the dimension what is that that's a voltage right kt is energy divide by q right and then therefore if you divide by the electric field what dimension does it have this has a dimension of a distance and so w inversion is essentially kt over q divide by the electric field that is the width of that rectangle you see that is the width of the rectangle because instead of doing that complicated integral you know the amount of surface charge n of s n sub s because as a function of bias you are easily calculate that one quantity you just multiply with w inversion on the order of maybe 30 or 40 angstrom and then that gives you the equivalent amount of charge of the more complicated green curve right now what this is trying to tell you that the electrons flow through about 20 30 40 angstrom next to the oxide and for the rest of the things first it is depletion then it's a completely charged neutral region so thin sliver of region next to the oxide where all the action happens in a MOSFET so this is what the purpose of this calculation okay now what will happen above threshold as I said that that's a very important and interesting thing so above threshold you say okay the green electrons is moving up with the width of w inversion that you can calculate and you have this particular expression vg equals phi service phi service is amount of band bending right up to the surface now what is this eox multiplied by x0 it is equal to the amount of voltage drop in the oxide eox is the oxide and then multiplied by x0 is the thickness now I can only write it this simply if there is no charges or anything else in the oxide right because if I had charges in the oxide then my oxide electric field will not be unique up to the charge I will have on oxide field away from the charge I will have another oxide field so for the time being I'm saying that that's equal to electric field multiplied by x0 without any charge in the oxide okay now do you agree that the electric field can be written as inversion charge plus fixed charge divided by epsilon now do you agree with this statement do you remember that when you're an undergraduate student then they used to have this uh pill box for calculating charge that if you want if you had a metal plate and you had a charge q if you wanted to know the amount of electric field that is coming out of that charge you'll put a little pill box in here and gauze apply gauze's law and you will see the electric field coming out is the charge included within the pill box divided by epsilon do you remember this if not when you go back just open up any uh any undergraduate book and you will see that this is indeed the total amount of charge divided by epsilon that gives you the electric field coming out of a out of a region and so that's exactly what I have done you see the green electrons I have qi that's the induced electrons and qf what is that that's the acceptors number of acceptors so together is the total amount of charge which will send the electric field into the oxide and so I have I have this particular quantity multiplied by x0 okay that's fine now at threshold voltage at threshold voltage of course my electric field will not be the same right because my voltage is different uh at at 2 phi sub f remember in the top one vg any vg not just threshold voltage so here you will have uh the threshold voltage you will again have it 2 phi sub f remember phi sub s equals equals to 2 phi service is when the inversion begins to work right again the same formula except one little thing do you see the qi in the top place you have evaluated at phi service in the bottom place you evaluated at 2 phi service right there's no complicated thing here but look at this magic in the next line by the way if you look on the other one you will have a green green one which I should have drawn in the right hand figure right hand bottom figure which I haven't drawn so that is what would have happened at this threshold voltage now let's subtract if I subtract initially it looks a horrendous thing that I have phi service 2 phi service there's a difference of that the charges are floating around but you immediately realize many simplifications are possible because what I'm going to show you a little bit later that after inversion if you do a detailed calculation it's very difficult to move phi service beyond 2 phi service so let's say phi service 0.3 milli at 0.3 electron volt 2 phi service 0.6 then you are putting a lot of gate bias now lot of gate bias you will see the phi service will move into 0.65 even whenever you might have changed the voltage by let's say 2 or 3 volts on the gate but the phi service gets clamped pinned to that value of 2 phi service I'll explain that in a second that's why that happens but for the time being I first of all that term can be dropped the phi service minus 2 phi service approximately can be dropped moreover this is exponential dependence on phi service right remember that ni goes up exponentially so compared to that the tiny amount of charge you have in this in this green region on the right side figure right side figure that's a minuscule amount of charge right because this is a tiny region and then the height isn't very much so compared to qi at phi service you can drop that red term red term over there and if you do then you have this simple one liner equation and that tells you how much charge you have when your gate voltage has gone above the threshold voltage so let's say the threshold voltage was something like 0.8 volts you have applied 2 volts and then you will use get this very simple formula by the way do you see where the seahawks came from you see that see this a kappa s and epsilon naught divided by x naught that will give you the oxide capacitance so that is what I have cross multiplied throughout in order to get this charge now look at this formula for a second first of all simplicity is good which is which is very good but more importantly that while the surface charge is moving exponentially the real charge as a function of gate voltage is moving linearly I mean how is that possible the reason that happens is is because of the of the following thing what happens beyond beyond threshold is that when you have a little bit let's say you apply a bias a little bit charge this is I'm trying to explain why it changes qi a lot a little change because it's exponential dependence now think about this for a second you apply a voltage and let's say a little bit more charge has accumulated in the inversion region as soon as the little bit more charge has accumulated the electric field that is coming out of their semiconductor into the oxide all in a sudden that electric field has increased quite a bit right remember the electric field from the peel box is directly proportional to the charge you have within that box so you increase that charge a little bit electric field increases a lot and then that electric field will drop across the oxide and so that will eat up most of the gate voltage that you have applied with a little bit change in the charge it is like a lever so you little bit change in the charge you have a large change in the oxide field and oxide voltage drop and most of the gate voltage therefore the oxide eats up most of the gate voltage above threshold and very little is required to drop across the semiconductor that's why this phi service gets clamped to to phi service and that's what I wanted to mention that because this electric field is directly proportional to qi qi goes exponentially with phi service and therefore little change in phi service gives you a lot of oxide field and a lot of oxide field then gives you oxide voltage drop and as a result most of the drop goes to oxide not to the semiconductor right and I will show you detailed calculation a little bit later but this is a very important point everybody will ask you down the road that whether you understand this concept precisely or not and at that point when this happens these two becomes almost like a parallel plate capacitor so the blue one is there and the green one responds and the red essentially the effect of red essentially disappears because inversion charge responding to the gate charge that's how a MOS capacitor becomes much beyond inversion but I let me quickly tell you that you know electrostatics I'm almost done but generally you cannot keep making X naught arbitrarily thin for last four generations in semiconductor technology the oxide thickness have been clamped about to 12 to 10 angstrom I mean 12 13 around that range and it has not been scaled the reason there is so a crisis in the field is because you cannot really reduce the oxide thickness anymore it's not technology problem technology could allow you to go even thinner the problem is tunneling current but before that let me first tell you that if you had to calculate the current across this structure what theory would you have used you'd have used thermionic emission theory right why you know discontinuity in the conduction and evalence band what what other theory will you use it's not diffusion theory here right and how will you calculate that again very simple there is current flowing from semiconductor to the gate region and from the gate to the semiconductor region so you have the q i v g the one that you just calculated right the green green electrons just calculated and what fraction of it will be able to make it will you see that exponential of delta ec multiplied by beta right delta ec is this discontinuity discontinuity because that is how much it has to be above in order for the electrons to flow from the other side will you'll have the metallic whatever the metallic electron concentration is n sub m and then you have the delta ec of course but remember for the electrons from the metal side in order to go to the other side it has to additionally go over that q v ox because v ox is the amount extra drop that has happened you can see the seconds are turned will die very quickly and even the first term delta ec for a typical system silicon dioxide silicon 3.1 ev so 3.1 ev over beta well that's 0 very nice this is 0 beyond machine precision in some ways and so therefore this current is in minuscule no problem so generally you do not have any thermionic emission current however of course it's so thin that the tunneling current remember the triangular barrier and other things chapter one we did that will of course will easily get through a 10 angstrom region where have we seen this again in a in another context zener tunneling do you remember then zener tunneling also the reverse bias electron flowing through these are all the same same concepts and therefore in that case there'll be a significant amount of tunneling and this tunneling must be calculated quantum mechanically chapter one the same formula rectangular barrier only difference from chapter one that now you have to use effective mass for the electrons but over there we just use free electron mass that's the only difference but other than that this is exactly the same calculation and what happens that if you have too much too much current flow then essentially you have a lot of tunneling and that is not acceptable because that's why your laptops get hot even when it's off not doing much calculation it leaks out a lot of current and that is not acceptable simply looking more like a bipolar based current you see the sort of current is going out through the base of the current I put the oxide there so that I don't want to worry about the base current but now the oxide has become leaky and it's not no longer doing the function it was supposed to do now very quickly this is something I have to show you one one time in your lifetime just to show how it works out so the electrostatics as I said can be solved exactly so far in every problem that we have done we have looked into just depletion approximation so we have just n d and n a we have worried about and we have a c on p naught and n naught to be zero in the depletion region but in this particular case it's not necessary and let me quickly walk you through that how it works by the way the formula that you saw saw this is on depletion approximation remember when I was calculating the rate region how far it is depleting I was not really thinking about the group this green electrons and I didn't really think about that explicitly I put it back in later on but the exact calculation is possible so let's do that I'll just walk you through that elevation you really have to sit down with a pencil and paper to do this so you generally normalize the variable to save you some writing phi sub x you'll divide by k t over q so that and you know the definition of phi sub s in terms of e i and e i bulk and e i at a given point and the surface potential you can normalize it again with k t over q and rewrite it as u sub s that's just to indicate the normalized surface potential and the Fermi level instead of writing phi sub f you will write u sub f again normalized with k t over q no this is this is no no problem and then once you are done now do realize this statement that once you are you are done the number of holes at any given point is given by n i e i minus e f right and the number of electrons is e f minus e i but other than that is exactly the same and you can see that e f which is sign change I have written as u f in the last one and you don't see any beta there why because the last expression I have normalized it with k t over q that's why you don't see any beta but other than that it's exactly the same expression the number of holes well they are also the same and do you see n multiplied by p what is that n i square right so you can see that's what should happen because it's an equilibrium that's what should happen and then n d plus again you can write it and you can see e f minus e i bulk you can write it in terms of phi sub f the Fermi differential and normalize it so that you write it in terms of u sub f n a well what should I say this is this is very simple now one thing you should realize that I in this expression I am assuming either the donor is present or the acceptor is present if they are simultaneously present then the u f that you have of course you cannot insert it that u f in one of them and get the answer so it assumes one of them is present not necessarily both all right so I can write this expression you know just copied it from the last slide and only thing I have done is that normalized the psi I have written as that k t over q multiplied by u remember that one psi is just newly normalized so divided throughout and this whole expression on the right I will call it g u comma u sub f do you see that if you knew somehow somebody told you what u is and somebody told you what u sub f is then you can calculate the right hand side not that anybody will tell you you'll have to find it but if you knew u at any point and use of f the doping then you can do the right hand side left hand side I still don't know now this is the trick why this is people teach you this so you see what I have done in this equation this is a second derivative on the left hand side and the right hand side is a solely function of u and these always have this neat trick of solving this equation exactly that you multiply on both sides with the first derivative that's not a problem that that I can do but the next step is what makes it very interesting because you see the left hand side du dx squared if you take a derivative of that what is that expression that becomes two du dx and then take your that take one more derivative you can immediately see that's exactly the preceding preceding line right but look at what it does to the right hand side also that if you multiply the du dx and then multiply with dx on both sides do you see the dx's will disappear from everywhere right you see this because the dx that will cancel and the dx from the first derivative that will go away so dx's have gone away this is undergraduate math if you remember differential equations from those old days you will remember that this is where where it comes from but you can see I have also done something one over ld squared is whatever that bunch of constant ni kt over epsilon I've called it a ld squared because it's a dimension of a length squared and that has a name called Debye length because that is the distance over which the electrons sort of fall off to the equilibrium value remember the green one the green electrons which is sort of going down exponentially the extent of that could be in this case ld okay I'm not still done but I can calculate this quantity easily you see okay do you agree with this statement because do what is du dx du dx is electric field right du dx normalized electric field and so I have differential of that so when I integrate I will just simply take the square out now you can see on the top side I have put the top limit q ex over kt what happens to the bottom point the 0th point the electric field on the very right side where I'm integrating between very right side next to the oxide right that's the range on the very right side I don't have any electric field bulk region so therefore the second limit doesn't come in and the rest of the thing is again you do this integration and let's say you do this integration you get this factor f u and uf how do you do this integration well you do it numerically it cannot be done analytically right remember the g was a complex expression and so therefore you do it numerically at any point okay now therefore the surface potential if you wanted to know surface potential that this ex is at any position x so if you wanted to know at the surface so on the left hand side you'll put e sub s there's a surface potential on the right hand side what should you put instead of u you should put u sub s because there's a potential band bending on the surface and this is your final expression because look phi sub s multiplied by that e ox multiplied by x0 that from that expression from the expression before I can write it now this is something we have done before in the following form this phi sub s is that phi sub s this simple approximation is that whole thing is over there is the first term over there is the second term over there so this is the depletion approximation the poor man's version of this more complicated expression where all the charges have been accounted for you see now how will you calculate this this is the algorithm and you have already done it for your homework if you have done your homework properly you will begin with the surface potential you'll assume a surface potential let's say I assume 0.5 electron volts then I will correspondingly normalize it divide by kt over q and get a value of u sub s okay so I'll get some number then I'm going to divide it into let's say 10 pieces so every piece will be 0.5 divided by 10.05 so my gaps are 0.05 0.05 0.1 0.15 this list column of numbers that's my u and once I have the u did I not tell you that we can calculate g u sub f I know because that's the doping and u is for every point that I have in the column if you call a matlab function it will in once I can calculate that g the g and once you know g of course you can calculate f right f is a numerical integration of that g so you call another matlab function you get your g you get your f you get your f then you get your f side electric field on the surface remember that's easily related to the f and therefore you can calculate vg and then vg becomes sort of the output now the input becomes phi service so you go the other way around remember in the previous one I said vg is known vg equals ax plus bx square that is how you get the phi service but here you go the other way around but it's the same result at the end you get so let me tell you just tell you that if you did this calculation at home then this is what you have gotten you would have gotten in the accumulation a strong build up of charge do you see a thin region like a spike moving up in the sky very very high electron concentration I'm sorry hole concentration accumulation right where accumulation next to the surface now in the depletion region do you see that cross hesitated region that is how far the holes have been pushed back this is the red region that there is talking about and you can see that it's not a like a square box moving right because this has a tail because the electron concentration is not sorry the whole concentration doesn't immediately fall to zero at the age of the depletion region it gradually goes to zero so therefore you have that tail look at the surface potential what this is doing the u is the surface potential right how much it is 10 so you have to multiply with kt over q to know exactly what the surface potential is and do you see that that was almost like a square on the top and therefore this has the corresponding potential shape of a potential that we have seen before now you input more gate bias now do you see that the depletion region is sort of is almost clamped to where it was but now extra charges are building up this spike going to minus two let's say the extra charges are building up now these extra charges are electrons now that is the green electrons gradually moving up and you see that how that is exponentially dying that was this rectangular region I showed you right and you will see that is on the order of maybe 20 to 50 angstrom very thin region over which electrons are building up and the strong inversion so you put a lot of electric field and then for you can see it has electron concentration has go to 400 now very thin region and essentially it will increase exponentially but these charges delectri-proportional to the gate field now notice this carefully because I want you to see this see that that's 10 right minus 10 below so you're going from depletion to inversion so when you apply more bias in the depletion region you see you have band bending up to 20 so the band is bending very nicely when you are stealing the depletion region right band is bending quite a bit but look at what happened after you have inverted and beyond that point look at that's about 25 or so and that's about 25 you have gone from 2 to 400 in the charge and look at the surface potential it has only probably moved at 5 5 meaning 5 multiplied by kt so surface potential doesn't want to move anymore and that's the key thing that's how we could neglect that phi sub s minus 2 phi sub f and set it to 0 look at another thing this is the depletion region and that increases a little bit you see point because in the depletion region as you have more charge look at that 0.5 it was almost dead on the charge over there it is now 0.75 or so it has extended a little bit but once the inversion has occurred then look at this this essentially that 0.75 where it has sort of clamped because more new charges will come to the depletion region it will not go to the sorry will go to the inversion charges and will not go to the depletion you see because total amount of charge has to be equal to the gate gate charge so whoever can give you the charge in this case the inversion charge if they can give it the capacitor likes the charge close to as close as possible to the surface the other charge because that minimizes the total electrostatic energy that's why you know in principle charge balance could have told you that this extra charge could be either here or to the right but why is the charge here not over there that is because that when you have a set of charges the charges wants to be as close as possible given the constraints because that minimizes the total electrostatic energy that is the equilibrium concentration okay so that's a very important point that why inversion occurs next to the oxide not anywhere else but finally let me quickly tell you that the exact is not really exact if you try to boast about it you cannot really because first of all you know that it will not that's a triangular barrier in chapter one did I not give you homework airy function and all if you have a triangular barrier where the bound levels are so you will have a viso vest and the electron concentration doesn't increase exponentially like that because of the charge confinement so the wave function needs to be accounted for second is that because of the quantization band gap is no longer just eg because the first place where electrons can sit is not on the bottom of the band but the first quantized level right so your band gap is actually increased a little bit so therefore your exact expression you didn't account for any of this when you are doing the full exact quote unquote exact solution and the non degeneracy assumption remember everything I'm saying exponential of q phi sub s over k t and all when does it does it occur non degenerate assumption where is your fermi level fermi level is now inside a band you are using your non degenerate assumption is that is that a good way of doing business not really so these have to be all corrected for if you really wanted to do a proper simulation especially modern MOSFETs so let me conclude so I just wanted to emphasize on the physics of the induced charge those green electrons and how they change as a functional gate bias now generally current through the oxide is not very important but these days it is this is something why the oxide thickness have not scaled for last four generations and people are trying to change silicon with other substrate so that they can get a little bit extra current previously people could get current simply by reduce the oxide thickness no more and that's why there is all sorts of crisis in the in the industry and finally the exact solution is not necessarily exactly exact okay that's it thank you