 So, let us continue our derivation of the surface potential base model recapitulation in the previous lecture we have discussed how we can derive the x dimensional current equation and y dimensional voltage equation or surface potential equation. Further we showed how the x dimensional current equation can be reduced to an equation of this form where drain to source current is equal to W by L into an average surface mobility multiplied by a term which involves integration of the inversion charge with respect to the surface potential from source to drain and another term which is difference of the inversion charge at the drain and the source okay multiplied by the thermal voltage where the average mobility, average surface mobility is defined as reciprocal of the average of the reciprocal of the surface mobility. Now it is clear from here that the next step is to derive the Q i as a function of psi s okay and also expression of psi s in terms of the various biases namely the gate bias, source bias and bulk bias. So, if you have these two pieces of information then this equation can be converted into an equation for drain current in terms of the various biases. Let us look at derivation of Q i as a function of psi s. Now this we have to do from y dimensional analysis. So you can see that in the MOSFET we are doing an analysis in the x direction and we are doing an analysis in the y direction okay. Now it is important to note that this solution cannot be obtained using the following integral which defines Q i. So basic definition of Q i is this it is integral of the distribution of the electrons, the volume distribution of the electrons. Let me sketch the diagram to show this. So this is your source and drain. In this direction if you go your inversion charge is going to vary something like this okay. This is your n as a function of x and y. So this is actually y direction. If I go to another place I will have more inversion charge right. So in other words if this is the height of the charge, height of n of x y at this x, at this x I will get say something like this. Evidently this is not to scale because inversion charge, inversely I will not be that thick if this is your depth of the source and depth of the drain okay. So maybe if I have to show it to scale I should show the drain something like this sorry the source depth something like this and the drain depth also something like this right okay. But those are the other details. So now what we are saying is it is the area under this curve. So now let me remove this. So this was to show that n is a function of x as well as y. Supposing I take this x okay I take electron concentration as a function of y at this x say it may look something like this. So this is my n of x comma y. If I take here I will get more electron concentration so it may be something like this okay at this x. So this is how n is varying along y and also it is varying along x. Now qi is nothing but the area under this that is what this is saying okay you are integrating from 0 to bulk that is actually qi by q with a negative sign because this is area is positive whereas qi is negative okay. Now so what we are saying is you cannot do this integration analytically because the n of y is a complicated function right it cannot be integrated with respect to y analytically. So now this is the very important step in MOSFET modelling. So what we do is the following we write qi as difference between qs and qb. Qs is a total silicon charge and qb is the depletion charge depletion component of the total silicon charge. Now it is interesting to note that we can express qs in terms of psi s very easily and qb in terms of psi s let us see how. So this is achieved using the surface potential equation gradual channel approximation charge sheet approximation and depletion approximation and the result is the following. So here we are showing the milestone and then we will discuss the derivation. So first we are showing the result and then we will discuss the derivation. So using this approach you get qi as a function of psi s as minus of C ox into the gate to bulk voltage minus flat band voltage minus the surface potential minus gamma into square root of the surface potential. Now let us see how we can do this. Now let us take the surface potential equation. So any y dimensional analysis will be starting from the surface potential equation. Any x dimensional analysis will be starting from the current density equation right. So vgb is equal to psi ox plus psi s plus phi ms. Now here we can write psi ox as qs plus qf upon C ox with a negative sign. This is nothing but the parallel plate capacitor formula. You want to know the potential drop across the insulator in a parallel plate capacitor it is simply given by the charge on the electrode divided by the capacitance right. Now the charge on the electrode is silicon charge and the fixed charge. Let me show that here recapitulate. If you take a simple MOS or let me draw this in the same direction as the MOSFET. So this is your oxide of the MOSFET right and this is your gate and this is your substrate. Now what we are saying is that any x so this is the x direction and this is your oxide thickness. So you have a depletion layer and an inversion layer together this charge is qs the silicon charge ok. If I show that at any in y direction it is something like this. So this is inversion charge. So let me and the depletion charge ok this is the inversion charge. So this is really this whole area is really qs this is y. You also have a fixed charge in the oxide so that should also be taken into account that fixed charge is positive. This is qf. So this qf and qs together can be regarded as a charge on one electrode of this capacitor between the gate and the bulk. And from one dimensional analysis in the y direction you can use the capacitor formula qs plus qf by cox should give you cox. You are putting negative sign because the cox is positive in gate with respect to negative in the substrate and gate voltage will be of opposite polarity to the polarity of charge here. So that is why there is a negative sign. Now when we have written this equation using a one dimensional analysis what we have done is we have ignored the variation of this charge in the x direction. In other words we have used the gradual channel approximation. So this is fine when the situation is entirely one dimensional in a MOS capacitor but in a MOSFET things are varying. Now what is the starting point for this equation? It is the Gauss law. So in a MOSFET since the conditions are two dimensional the Gauss law contains dou e x by dou x as well as dou e y by dou y. What we have used is we have only used the dou e y by dou y term here to get this okay and we have neglected the dou e x by dou x from gradual channel from gradual channel approximation and Gauss law this is obtained. Now let us substitute this and then what do we get? So we get VGB is equal to minus QS by C ox plus psi s plus phi ms minus QF by C ox. Now this quantity can be recognized as flat band voltage okay. Therefore we can rearrange all the terms here and we can write this as QS is equal to C ox into VGB minus VFB minus psi s with a negative sign. Now note here the QS is a function of x and psi s is also a function of x. Now let us apply the charge sheet approximation to this result. Then what happens? So charge sheet approximation means that no part of psi s drops across QI because QI is in QI is like a sheet here. So the surface potential psi s part of it which is dropping across a inversion charge of zero thickness will be zero if thickness of a charge region is zero there is no potential drop right across that region. So no part of psi s falls across QI. So therefore entire psi s falls across QB because if nothing falls across inversion charge then the psi s is the potential drop across the depletion charge now if that is so then I can use the gradual channel approximation and write QB in terms of psi s based on our knowledge of the depletion charge in a diode in terms of the potential drop right across the depletion charge and we can write this simply as so using gradual channel approximation that is based on y dimensional 1D analysis in y direction. We can write QB is equal to minus of gamma into square root psi s I am sorry here gamma C ox into square root psi s ok where what is gamma? So gamma is square root 2 Q epsilon s Na upon C ox that is we are writing so this quantity is same as square root 2 Q epsilon s Na into psi s with a negative sign. So this relation square root relation between the psi s and QB we all know this we can get from 1 dimensional analysis of the potential drop across the depletion charge and using the terminology or parameters used in the MOSFET that is the gamma with the body effect parameter we can write this as gamma into C ox into square root psi s with a negative sign. So I want to repeat just as we wrote psi ox in terms of these charges on the electrode of the capacitor using 1 dimensional analysis because we use the gradual channel approximation and ignored the variation of the charge in the y direction or ignored the change in the field in the x direction right because of the charge. So we are assuming that the entire charge is controlled by the y directional field alone that is another way of looking at the gradual channel approximation. So that has been used here also to write QB in terms of psi s. So now our expression for QI as a function of x is equal to QS as a function of x minus QB as a function of x where QS is given by this formula and QB is minus gamma C ox square root psi s. So if you now substitute this here clearly you will get this is equal to so QI as a function of x is equal to VGB minus VFB minus psi s minus gamma square root psi s. So this is your formula for QI in terms of psi s. So I can write this instead of x I can write this as QI as a function of psi s. Each of this term psi s here psi s is a function of x. So let us go back to the slide. So this is how we have obtained this equation for QI. Now substitution of the above QI as a function of psi s in the x dimensional ideas solution leads to an equation for ideas in this form ideas drift plus ideas diffusion where ideas drift is this term minus W by L into average surface mobility into integral of QI with respect to psi s from psi s 0 to psi s L. So you substitute this QI as a function of psi s here and clearly you can get the following equation. So VGB minus VFB is a constant therefore it gets multiplied by psi s on integration and when you put the limits on psi s you will get psi s L minus psi s 0. Then this term psi s when you integrate you will get a square lot term psi s square by 2. So that is what you are getting here half of psi s L square minus psi s naught square and this one will lead to a 3 by 2 power right because square root of psi s when you integrate you will get psi s power 3 by 2. So this is how you are getting the 3 by 2 terms power terms psi s L power 3 by 2 minus psi s naught to the power 3 by 2 into 2 by 3 into gamma. Now the ideas equation contains another term and that is W by L into average surface mobility into thermal voltage into QIL minus QI 0. This is the diffusion current. So again substituting this QI here what will happen is this VGB minus VFB will get cancelled when you take the difference between the QIs at drain and source and you will get psi s L plus gamma into square root psi s L as the term coming from here QIL and from QI naught you will get this term okay. So this is straight forward. So you get this as the diffusion current. So we have now obtained the drain to source current in terms of the surface potential at source and drain. This is really the surface potential based model for the drain current. Now the final step is to express the surface potential in terms of the gate bias, drain bias and source bias. The solution of psi s as a function of VGB and channel voltage can be obtained from y-dimensional analysis okay. So we are alternating between analysis based on y-dimension and analysis based on x-dimension. Now note here that the channel voltage V varies from VSB at source to VDB at drain. Therefore if you get this expression then in that expression if you replace channel voltage by VSB you will get psi s at source and if you replace channel voltage by VDB you will get psi s at the drain. Now let us first say how this is obtained and then you go about showing the details of the derivation. So what we are going to do is we will use the gradual channel approximation on the left hand side of the Gauss law and the fact that the y-direction electron current density and hole current density is 0 which will lead to the Boltzmann relation. We will use that on the right hand side of the Gauss law. Now the form of the Gauss law used here is the Poisson s equation okay wherein divergence of E is written as dou square psi by del square okay divergence of E is written as del square psi. So we get the y-dimensional Poisson Boltzmann relation. So this is our milestone which we have to now show how to achieve. This is called the Poisson Boltzmann relation because you are taking the Poisson s equation version of the Gauss law and in that you are replacing the carrier concentrations in terms of the Boltzmann relation okay. So here let me familiarize you with the various terms dou square psi by dou y square psi is a potential in the channel region is given by minus cube epsilon s into a constant k into pp0 is the whole concentration in the substrate p type substrate under equilibrium into exponential of minus psi by vt minus the electron concentration in the equilibrium in the substrate into exponential of minus v by vt where v is the channel voltage the exponential of psi by vt where psi is the potential at that point minus the acceptor doping in the substrate okay. So where k times pp0 that is this quantity and np0 into exponential of minus v by vt that is this term here are values at the edge of the space charge layer okay. So what we are saying is the following if I take the situation in the MOSFET in any direction like this let me show you the depletion region now this is the edge of the space charge layer okay this is the space charge layer in the channel. Now what we are saying is at this point the value of the whole concentration is k times pp0 will show what is k shortly okay and at this point the electron concentration is np0 into exponential of minus v by vt where v is the channel voltage. Now based on body effect this is this is the consequence of body effect. So let me show you the electron concentration and whole concentration here in this direction. So if I take this and draw it here separately this is y direction and let us say this is the depletion edge here. So this is the so called yd so this is yd how will the whole and electron distributions look like when I move in this direction. So the electron distribution would look something like this in the log scale I am plotting okay. So this is np0 in the bulk somewhere here and the whole concentration would look something like this this is your depletion edge this is pp0 now the value here is not exactly pp0 okay less than pp0 this is k times pp0 and the value here of electron concentration this is the surface concentration of electrons this is ns and this is surface concentration of holes this is not exactly to scale the ps can be less than this. So this value you can see that this is not the same as np0. Now this value is np0 exponential of minus v by vt where v is the channel voltage here you will recall that in a previous lecture we have remarked that you can regard this region in analogy to the situation in a p-injunction okay and you can regard this as a reversed biased p-injunction this is the inversion layer this is n plus and this is p. So this is like a reversed bias n plus p-junction where the applied voltage between this point and this point the applied voltage between this point and this point is v. So if this is a n plus p-junction across which there is a reverse bias of v you know that at the edge of the space charge layer or depletion layer on the p side the carrier concentration will be given by np0 equilibrium carrier constant into exponential of the negative of the applied bias by thermal voltage okay from analogy to the reverse bias p-injunction. So that is what this is so this quantity here is this carrier concentration here and this quantity is a carrier concentration of holes at the edge okay depletion edge. Now let us explain how this is obtained using the gradual channel approximation because what we have said we are using gradual channel approximation on the left hand side and we are using this on the right hand side. So when I write this relation here for electron concentration at the depletion edge okay in terms of the applied voltage I am using the Boltzmann relation right I am using the Boltzmann relation you recall the p-injunction similarly if I want to express ns here in terms of the potential drop across the depletion layer that is a psi s okay so this potential drop is psi s. Then what will I do I will say this ns is equal to this carrier concentration into exponential of the psi s by vt because psi s is the potential drop between these points that is the Boltzmann relation. So ns is equal to np0 exponential of minus v by vt which is the value here into exponential of psi s by vt. So when I write it in this form when I relate the carrier concentration as an exponential of the potential drop I have used the Boltzmann relation okay from p-injunction which is a result of assuming that the current density of electrons is 0 please go back and recall that. So when I put current density of electrons is equal to 0 I am saying q dn if I take one dimensional picture dn by dy plus q n mu n into Ey that is 0. So since this is 0 what I could do is I could write it like this and then I can remove the q from both sides and then I can write dn as mu n vt so which means I can remove this dn and simply put vt here and remove the mu n from here and then I can shift dy here and shift n here. So let me remove this dy and put the n here and remove this and then I can integrate both sides then I will get potential this will give me potential and when I integrate d1 by n dn I will get log of n okay and so I will get log of n in terms of potential or I can express this as n in terms of exponential of the potential okay. So this is the so called Boltzmann relation obtained from Jn equal to 0 so Jn by equal to 0 similarly Jpy equal to 0 okay will allow me to express the whole concentration at any point. So if I want for example let me take any point like this I want the electron concentration here I will use instead of psi s I will use the psi, psi is the potential drop between this point and this point sorry not psi s psi. So here I will remove the psi s I will put psi so I will use psi s if I want to express ns so I want I am expressing n so n at any point so it is the function of potential at that point. Similarly I can do the same thing for whole concentration if I want to express whole concentration I will say p is equal to the value at the depletion edge that is k times vp0 into exponential of minus of psi by vt you can see that whole concentration here is less than whole concentration here therefore I have to use minus of psi by vt whereas electron concentration here is more than electron concentration here therefore I am using exponential of plus psi by vt okay. So this is how and this is the consequence of the relation Jpy equal to 0 okay and that is as far as the right hand side is concerned so your space charge equation right hand side so let me write that Gauss law divergence of e is equal to q times p minus n minus na minus by epsilon s. So this is epsilon s here I will remove it so this is that q by epsilon s and then p you write using the Boltzmann relation as k pp0 exponential minus psi by vt that is what we wrote here n we write using again Boltzmann relation that is this and na minus we are simply assuming that all of na is ionized and you get that. Now left hand side we have changing the divergence of e which is dou e x by dou x plus dou e y by dou y gradual channel approximation you are using so you are removing this so you are left with dou e y by dou y and you are writing this in terms of the potential so e y is minus dou psi by dou y so when you substitute that here this quantity will change to dou square psi by dou y square with a negative sign and that negative sign you are pushing to the right okay. So that is how you are getting this particular Poisson Boltzmann relation now we will show that this equation can be integrated to obtain integrate you first integrate and then approximate okay to obtain an expression for qs that is the total silicon charge in psi s as a function of psi s of this form so let me familiarize you with this equation minus of epsilon s into vt by ld ld is called the Debye length into square root of 2 into square root of psi s by vt plus exponential of psi s minus channel voltage v plus twice phi f by vt and the same thing can also be written as minus of C ox into the body effect parameter gamma into square root of psi s that is this vt term has been taken out plus vt into exponential of psi s minus channel voltage v plus twice phi f by vt okay. Now let us see how we can do this so how we can integrate the Poisson Boltzmann relation so first let me write that Poisson Boltzmann relation here which we want to derive okay now the first step to recognize to solve this or integrate this equation is to recognize that this Na I can write as P P 0 minus N P 0 how because in the bulk in the bulk which is neutral I can write the space charge equation as rho equal to 0 that is P minus N minus Na minus equal to 0 which leads you to the equation P minus N is equal to Na minus then I am assuming the entire Na is Na so therefore P minus Na is Na and in the bulk the value of P is nothing but P P 0 and value of N is N P 0 so that is how I get Na is P P 0 minus N P 0 now this is necessary so that this P P 0 from here can be clubbed with this P P 0 term and N P 0 can be clubbed with this N P 0 term. Now to help us simplify matters let us define what is called the normalized potential so psi by vt will write it as psi star v by vt will write as v star you know you are getting v by vt here then we will introduce a normalizing parameter for the dimension y as a Debye length will shortly show you what is a Debye length we will call this as y star the formula for Debye length is following I am giving the formula and I will tell you the derivation later epsilon s vt by q P P 0 and another term we will write N P 0 by P P 0 as which is nothing but exponential of minus twice phi f by vt will write it as exponential of minus twice phi f star ok so we are defining phi f by vt as phi f star. Now here phi f is the Fermi level difference so this is your Fermi level in the P type substrate this is your intrinsic level and this is q times phi f so you know that I can write N P 0 as N i exponential of minus phi f by vt and P P 0 is equal to N i exponential of phi f by vt ok so that is I will take the ratio P P 0 by N P 0 you will get the N i will cancel and you will get that twice phi f by vt. So now if I substitute this variables here my same equation I can write as dou square psi star by dou y star square is equal to minus q in fact all the q and epsilon s and all will go away and this will be minus of k exponential of minus psi star minus 1 minus exponential of minus twice phi f star and exponential of psi star minus v star minus 1. So we are saying this equation can be written in this form so you can see here what has been done is this equation this part has been clubbed with this P P 0 you get this here ok and then this has been clubbed with N P 0 ok where N P 0 you are again writing in terms of P P 0 ok so let me discuss how I can write this so let me discuss how I can write this Poisson Boltzmann relation in terms of this variables and how the result will be a simple and compact equation. So what we do is you can divide psi by vt ok if I want to do that I should divide by vt on both sides so I will write it like this dou square psi by vt so right hand side also I should divide by vt so q by epsilon s vt now let me keep the denominator as it is in the right hand side let me take the P P 0 out. So P P 0 is taken out then I have in brackets k and exponential of minus psi star psi by vt is psi star minus since I have taken P P 0 out here I have to write N P 0 by P P 0 and this N P 0 by P P 0 we have defined in terms of this term twice phi f star so I will write this as my exponential of minus twice phi f star then multiplied by psi star minus exponential of psi star minus v star ok now minus I have taken P P 0 out so I will get one here and minus N P 0 will be converted to N P 0 by P P 0 which is again written in terms of exponential of minus twice phi f star. Now now look at this term here when I compare the left hand side and right hand side this entire part is dimensionless ok and the left hand side the numerator is dimensionless because I have divided the potential psi by the thermal voltage so I am left with only the so the numerator is dimensionless I have left with a square of the length term in the denominator clearly therefore this quantity here has a dimensions of 1 by length square ok and we write this as 1 by L D square so that length is a characteristic length called the Debye length and that is how the Debye length here has been defined as square root of epsilon is vt by Q P Q into P P 0 what is the physical significance of this Debye length so this Debye length tells you the characteristic length tells you over what distance the space charge will fall significantly or the potential due to space charge will fall significantly ok. So now what I could do is I could shift this 1 by L D square to the right hand side then L D square will go to the numerator or I can put it in the denominator as y by L D so dou square y by L D square so this y by L D is nothing but y star that is what is defined here y by L D is y star ok and this has been removed from here so that is how you are ending up with this compact form of the Poisson Boltzmann equation this psi by vt I will replace by psi star so this is dou not dou square. Now in the next lecture we will look at how this compact form of the equation can be integrated so let us summarize what we have achieved in this lecture so far since we cannot get an analytical expression for Q i using this basic definition of Q i in terms of the distribution of the electrons we use the following trick that we write the Q i as difference between the total silicon charge and the bulk silicon charge and the total silicon charge and the bulk silicon charge each of these can be expressed in terms of psi s very easily using the surface potential equation gradual channel approximation charge sheet approximation and depletion approximation as this relation. So now you can substitute this relation in the equation for drain current as a function of surface potential and you will get two parts the drift part of the current is given by this equation in terms of surface potential and the diffusion part is given by this equation in terms of the surface potential. So then we said that the problem now remains to express the surface potential in terms of the various bias voltages or in terms of the gate to bulk voltage and the channel voltage this channel voltage is equal to the source to bulk voltage at source and drain to bulk voltage at the drain and this equation can be obtained from y dimensional analysis and the equation that we use to solve for the psi s is the Poisson Boltzmann equation which is obtained by writing the Gauss law in the form of Poisson's equation taking the gradual channel approximation on the left hand side so that the potential variation along y alone is considered and on the right hand side we replace the carrier concentration terms in terms of potentials by using the Boltzmann relation and then we showed that this Poisson Boltzmann equation can be written in a compact form using normalized variables where the potentials are normalized with respect to the thermal voltage and so even phi f is here normalized with respect to thermal voltage and the distance y is normalized with respect to a characteristic length called the Debye length which tells you the distance over which the potential in a spatial layer falls significantly it can also be taken to be a metric that can be used to characterize the length of spatial layers so in the compact form it can be written as shown here right this is a dimensionless form in the next lecture we shall look at how to integrate this equation.