 Now, this is the last module of our course, so far we have shown how you can model the MOSFET in the various steps, namely square bus tip. Now in this module, we would like to show some of the effects that we have neglected in our model so far, how they can be included, specifically we will discuss the inclusion of series resistance, non-uniform doping and small geometry effects. So at the end of this module, you should be able to show that the effect of series source or drain resistance is equivalent to a reduced mobility due to transverse field. Then you should be able to derive an expression for the effective channel length including channel length modulation. You should be able to derive an expression for the threshold voltage including vertical and lateral non-uniform substrate doping, short channel effect abbreviated as SE and narrow width effect abbreviated as NWE. You should also derive an expression for drain induced barrier lowering also called DIBL. Let us start with the effect of RS and RD, here is a diagram which shows the current flow contributing to RS or RD, this is the source or drain N plus region, this is the contact window from where the drain or source current is taken. There is your gate and these lines show the current path, suppose this is the source then your current is flowing like this, since your gate is positive for an inch channel device and this substrate is N plus, a positive voltage on N type substrate will result in accumulation. So this region is an accumulation region, therefore the current flow through this accumulation region contributes to this resistance R accumulation. From this point onwards the current spreads ok and in this region the current flow paths are parallel ok then again the current flow paths curve right and finally they pass through this contact. Therefore this region contributes to the spreading resistance RSP, this region contributes to the resistance because of the sheet resistance ok due to this N plus region that is why it is called RSH. So this will be contribute, this resistance will be calculated using the sheet resistance of the N plus region and this is the contact resistance which will include resistance of this current path as well as the resistance of this contact. These are the components of the series resistance, however we are not going to extract these components separately, if you want to model the series resistance theoretically you will have to model these components ok. Our goal is to see how if there is a series resistance present the current voltage characteristics are affected. This is a symbolic representation of a device, now this is the MOSFET part without any source or drain resistance. This is usually called the intrinsic device and including the series source and drain resistance the device is called the extrinsic device, so this is an extrinsic device. Let us assume that RS is equal to RD and is equal to R, further we will assume linear region of operation and we will assume that the gate to source voltage is much more than R into IDS. Now what is R into IDS? So here you can see that R into IDS is this voltage drop, so this is RS into IDS where IDS is the current through this, we are neglecting the gate leakage and bulk leakage that is why our current is only between drain and source and this RS we are assuming it to be R, this is really the voltage drop. So we are saying VGS which is the voltage drop between these two terminals is much more than R into IDS, so in other words this is small compared to VGS, we are assuming these conditions under which we will derive a simple formula for the effect of RS and RD on the current. Now let me give you the rationale for considering the linear region of operation, how the effect modeled under these approximations can be used to model the characteristics over the entire region of operation from linear to saturation. For that purpose let me draw the ID versus VDS without any source or drain resistances, suppose this is a curve corresponding to some VGS. Now on the same graph let us plot the curve when you include RS and RD, so for any given ID if I include RS and RD I have to add these voltage drops, IDS RD and IDS RS or 2 ID into R, so that is this, if I add that voltage drop here the corresponding drain source voltage point will be this it will shift to the right, like this I can do for every ID and then I will get this curve. Now if I assume that the R into IDS is much less than VGS then I can assume that the saturation current is not affected right, so the saturation voltage is affected it is shifting from say this point to this point but saturation current is not affected because our VGS is not affected by this particular voltage drop. Now if I want to model this which includes RS, RD this one is RS, RD equal to 0 then a simple approach would be that I model this region this change in slope in the linear region this is the slope with RS, RD equal to 0 this is the slope with RS and RD non-zero. Then I can model this entire curve using our square law MOSFET model that we have derived okay how so that model you know contains the critical field for velocity saturation and that critical field is given by VSAT saturation velocity by mobility that mobility is nothing but the slope of the ID versus VDS curve for small VDS okay that gives you the mobility the mobility depends on this slope. So if I change this slope suitably if I reduce the slope right to include the effect of RS and RD then my critical electric field will increase because I am not changing the velocity saturation and if I increase the critical electric field automatically the saturation point will also shift to the right so that is how by properly modeling the linear region close to VDS equal to 0 okay that is here I can also model the increase in the saturation voltage because the critical electric field for saturation will increase. So under the conditions namely VGS is much more than R into IDS and linear region of operation I can remove the square law VDS term and I will get this particular formula where I replaced VDS by VDS minus 2 into R into IDS. So this is this quantity is the voltage drop between these points here I have neglected R into IDS when I am writing the VGS term right that is why here I have not included any effect of resistance. Now you can reorganize this equation and show that the same equation can be written in this form so all that I have done is this quantity into 2R into IDS I have shifted to the left and then I have taken IDS as a common term out of the two terms on the left hand side and moved the multiplicant of the IDS to the right and you get this from here you can see that the effect of R is to reduce the average surface mobility by this factor okay so we can say our new average surface mobility is this when R is equal to 0 you will get back just the mu NS average. So this is how you can show that the effect of RS and RD can be approximated as reduction in mobility due to transfers electric field why are we saying it is due to transfers electric field because you can see that this reduction factor depends on VGS minus VT so this is the gate to source voltage okay and therefore this effect is termed as effectively reduction of mobility due to field from gate to bulk or transfers field. Now let us discuss a model for the channel length modulation with channel length modulation the drain to source current IDS is given by IDS without channel length modulation into 1 plus delta by L let us show the channel length modulation region delta this is delta and this is L so the effective channel length is equal to L minus delta. So how do you get this expression well IDS is equal to W by L into a function of VDS and VGS when you include channel length modulation this L is replaced by the effective length L minus delta. Now if I assume if I assume delta to be much less than L that is delta by L to be much less than 1 I can write W by L minus delta which is equal to W by L into 1 minus delta by L as W by L into 1 plus delta by L approximately right I am using the fact that 1 by 1 minus x is approximately equal to 1 plus x therefore if I replace this by W by L into 1 plus delta by L I can recognize this term in square brackets as IDS without channel length modulation and this is my 1 plus delta by L ok. So how do you express delta in terms of the bias conditions that is our problem here is the field from source to drain in the x direction at the surface that is why you see here E X S further I have put a modulus because the field is directed from drain to source whereas x direction is from source to drain this is your delta region in which the field starts rising rapidly this is the velocity saturation region and at this point the field is equal to the critical electric field so whenever your E X S exceeds critical electric field that is this region your velocity saturates. So to model the delta as a function of the voltage all we need to do is we need to express this area which is nothing but V DS minus V DS sat in terms of delta and other parameters right like doping and things like that. So once you do that you can solve for delta in terms of V DS minus V DS sat. So to do that we write an expression for the slope D E X S by DX ok now what is that? So the slope D O E X S by D O X according to Gauss law is given by minus Q into N A plus N by Epsilon S here we are not considering D O E Y by D O Y ok we are neglecting that part of the Gauss law. What is N? N is a inversion layer electrons ok. So let me illustrate this point beneath this let me sketch the device structure and the charge conditions so this is your inversion layer this is a depletion layer ok. Now this is the region we are talking about. So here you have in electrons and beneath that you have depletion charge this is dependent on the doping N A. So if I take this delta L region I have to construct both electrons due to inversion layer as well as the doping that is what is being shown here. So while writing this we are punctured only the x directed field that is the field from drain to source right there is a y directed field from gate to bulk that we are neglecting because we are saying this is really the significant field here. Now how do you get the electron concentration? The electron concentration has to obey this equation that is the current density in the x direction in this particular channeling modulation region is given by minus of Q into N electron concentration into saturation velocity because electrons are moving at saturation velocity. Now these are the equations to be used in our modelling so we have used the Gauss law and we are using the current density expression corresponding to velocity saturation ok. First step what we do is we link this current density expression to the terminal current ideas by writing it as the current is given by the current density into the area of current flow that is W into Tx where the Tx is the thickness of the inversion layer at any x. Substituting Jnx given here into this relation ideas becomes Q into N into Vsat into W into Tx. So now I can use this expression to get a formula for N. I will find N is a function of the drain to source current ideas and the inversion layer thickness Tx ok. W is a constant saturation velocity is also a constant Q is also a constant. Now let us see what is this Tx so Tx is given by T inversion at x dash equal to 0 and by xj at x dash equal to delta. Let us see what is this x dash this is your point x dash is equal to 0. So instead of x we are using x dash to simplify our math. So this x dash is given by x minus L minus delta because you can see that this distance is L minus delta. So x dash equal to delta is this end. So variable x dash varies from this from x dash equal to 0 to x dash equal to delta ok. Now this means that inversion layer thickness is varying ok from T inversion to xj. Now let us show that what it means. So this is your device structure I am approximating the drain contact by rectangle to show things clearly your inversion layer is varying like this. At this point the thickness of the inversion layer is T inversion and what we are saying is that the thickness of the inversion layer at x dash equal to delta is xj this is xj junction depth. Now you might ask but how is it that the inversion layer thickness is increasing like this right we thought the inversion layer thickness is constant in the channel length modulation region. Well numerical simulations have showed that actually the current in the channel length modulation region or velocity saturation region spreads as shown here. So an approximation for that is used in this analytical model and we assume that the thickness of this current flow at the drain contact is equal to xj that is what is being assumed here. Further we approximate Tx to vary linearly ok. So Tx is the thickness at any point x dash that is this formula here. So you can see that if I put x dash equal to 0 I will get T inversion and if I put x dash equal to delta then this T inversion will cancel with this T inversion and I will get Tx equal to xj. So this is really the picture that is being used. So if I use this variation of Tx in this formula I can then write an expression for n in terms of ids. I can substitute that n into this formula and then I can integrate this expression to get Exs as a function of x that is this line. And once I get this line I can easily find out the area under this curve right or line and that would be Vds minus Vds sat and I can use that equation to solve for delta that is the approach. So here is a solution neglecting the effect of inversion charge that is if I neglect n then I get this particular formula for the delta in terms of Vds minus Vds sat. Now here Vds sat has been replaced by Vds effective. We will explain why this is done but first let us see how do you get a formula like this if you neglect the inversion charge that means I should not take n here into account. Now this can be seen very easily. If I do not take n into account the slope of this line is nothing but q na by epsilon s according to this formula okay. Note that I am taking modulus of Exs okay so this slope will be positive. Now if I know the slope I also know how do I know this height so this height is equal to the distance delta into the slope. So I can write this as q na into delta by epsilon s. So once I know this height I already know this is EC and this is now a trapezoid. So under the approximation that inversion charge is neglected this is a straight line okay of constant slope q na by epsilon s. Then this is a trapezoid and area under the trapezoid should be equated to Vds minus Vds sat. So I can write half the sum of the parallel sides that is EC plus EC plus q na delta by epsilon s into the distance between them that is delta is equal to Vds minus Vds sat. So if I remove this half then this one EC will go away and here I will get 2. This is my formula. Now this is a quadratic in delta. I leave it to you as an assignment to show that when you solve this quadratic for delta you will end up getting this expression okay. Without of course this with this Vds effective replaced by Vds sat okay this is straight forward. It is just that this product epsilon s EC square by 2 q na has been replaced by a variable 5D because this is this has the units of a potential. Now let us explain why do we replace Vds sat by Vds effective. Now as we have discussed in our BC model that we should not have regional models right. We should have a model that works continuously from a linear to saturation from sub threshold to super threshold and so on. So if I use Vds sat here this formula would work only for Vds greater than Vds sat. If Vds is less than Vds sat okay this quantity will become negative and actually no channeling modulation exists. So this formula is not applicable for Vds less than Vds sat physically speaking but mathematically we do not want to say that use this formula for Vds greater than Vds sat alone and for Vds less than Vds sat do not use this formula. This is going to introduce some discontinuity problems in the slope of the curve and so on. So I should use this same formula even for Vds less than Vds sat but it should give 0 delta. Now that is what is achieved by using this Vds effective expression right. This Vds effective expression works for all Vds and we know that the Vds effective expression in BCM model had this particular behavior okay. So for small Vds this is Vds effective is Vds itself. This is line with slope 1 and for large Vds it saturates at Vds sat. Now that being the case you will substitute Vds for Vds effective when Vds is small and therefore this will go to 0 and when Vds is large automatically this will change to Vds sat. So you will get a continuous expression right. Now if you include n then the formula is a little bit more complex. So including the effect of inversion charge the formula looks something like this. Here I have not written the formula for delta because this is a quadratic expression which rather long coefficients right and so if I write in terms of delta it would be unwieldy. So I have left it in the form that Vds minus Vds effective which is the area under this curve is given by the right hand side formula when n is included. Let us just check that this formula reduces to our previous formula which gave you this delta when n is neglected. So you can see that when n is neglected this term will not come about okay. This is the term that has come from n and the result will reduce to this formula okay. Let us further try to get some idea of this expression you know. Let us familiarize with this expression because as we have discussed we must understand an expression physically right and by limiting cases we should check whether you know it predicts the results properly for limiting cases. So that we can be sure that this unavailability expression actually makes sense. So in fact that is given to you as an assignment. Let me give you some hint as to why this expression is reasonable. Why are you getting a logarithm term here and why are you getting these terms in the denominator of an expression where numerator is ids. Just look at this n if I want to include n here the n is obtained from this formula okay. So what is n according to this formula n is ids divided by q vsat wtx. Now tx varies linearly with x dash. So when I substitute this n here in the Gauss law this q will go away and I will get the epsilon s term in the denominator okay. So this is the kind of term that you will get additional term. Let us come back here and I find ids in a numerator vsat in the denominator w is there okay but I do not seem to have the epsilon s but that is because I have taken out q na into delta square by 2 epsilon s here. If I were to take this inside the q na and this q na will cancel and this 2 will cancel and I will get delta square by epsilon s. So I will get epsilon s w vsat into xj. So epsilon s w into vsat is here. Now let us understand why you are getting this xj into logarithm term. Now this tx has a linear variation with x prime. So when you integrate this expression with respect to x right you have something like 1 by a plus bx integrated with respect to x that will give you a logarithm logarithmic expression. So this will have a ln a plus bx as a result right plus some constant and so on. So that is how you are getting this logarithm expression here and when you put the limits you will get the constant and so you will get basically this form of the expression where this is xj by t inversion right. Your limits of integration would be from x is equal to 0 sorry x dash equal to 0 your t is t inversion and x dash equal to delta your tx is xj. So when you put the limits you get this particular term okay the more details you will have to work out yourself in your assignment. Let me show you the current voltage curve the measured current voltage curve and where the channel line modulation phenomenon is effective. So this is effective in this region just beyond the saturation point just beyond the saturation point. You go to higher values of VDS then it turns out that another phenomenon drednities barrier lowering becomes stronger than channel line modulation and for the VDS region near breakdown yet another phenomenon called substrate controlled body effect starts dominating over Dibble and CLM. So these 2 things we will consider later. As a prelude to the Dibble let us talk about the change in threshold voltage right because of various phenomena. Let us start with VT including the effect of non-uniform doping. Now this is your device we will start with the VT for uniform doping and long channel and then see what modifications we can make in the expression to include the non-informed doping. Of a long channel device is 1 with channel length 3.5 microns here your VT is given by VT for 0 bias plus gamma into square root of phi T minus VBS minus square root of phi T where phi T is twice phi of plus 6 VT. Now how do you get this expression? You write the expression for threshold voltage which consists of the flat band voltage and then potential drop in the silicon for 0 flat band voltage and the potential drop across the oxide for 0 flat band voltage that is this. Now you can write this as VFB plus phi T plus gamma square root phi T plus this term and minus gamma square root phi T. So I have added and subtracted gamma square root phi T which is the potential drop across the oxide for 0 VBS right this term for 0 VBS. I can then identify this part as VT0 okay and then get this formula okay. Now let us modify this formula to include non-uniform doping. First let us consider vertical non-uniform doping keeping the device as a long channel. So vertical non-uniform doping means the doping variation in this direction okay. We have discussed that you know you have various threshold correction implant then you have the punch through implant and so on okay so that is the non-uniform doping we are talking about. Then of course the doping will go to the uniform substrate doping right. So that is what is shown here these are doping versus distance threshold correction implant and punch through correction implant. We can show that just by modifying the coefficient of this term and subtracting a term that is dependent on VBS we can model the effect of vertical non-uniform doping. Now how do we show this? As we show this you will come to know a approach used to model phenomena in device modeling right and this approach is using pure intuition okay rather than using a rigorous approach of solving the 6 device equations right for the given situation. This is an important approach that is used to derive simple equations one example of which we are considering now. So this equation is developed intuitively using a box type implanted doping profile. So what is this box type implanted doping profile? It is this. So we assume that the non-uniform doping can be represented by a box. So this is some sort of a box right. So doping is ni over a distance d and then it abruptly falls to the substrate doping the suffix i here stands for implantation because this is realized by ion implantation. So we will see how we can derive the formula considering a single step like this or box and then you can have multiple boxes right. You can modify the formula to include multiple boxes okay. The way to develop the formula is as follows. Supposing my doping is uniform and equal to na, threshold voltage variation with respect to VBS can be shown something like this okay. A modulus has been put here because you know for an N channel device VBS is negative. VSB is positive whereas this is a positive x axis. On the same graph let us plot the situation when na is increased to ni. Now your threshold voltage for 0 bias also will rise as will the rate of rise of the threshold voltage with VBS okay that is why this curve seems to rise faster than this curve apart from the fact that the 0 bias value itself is higher because the doping ni is more than na. So how do I model this dashed line? I changed the coefficient here earlier it was gamma corresponding to this doping na corresponding to this doping ni I am making it gamma 1. What is this gamma? You know that this body effect parameter gamma depends on doping as per the formula square root 2q epsilon sn by C ox. So for na the parameter is gamma and if I replace this a by i then this is gamma 1 that is what we are saying here. Now what happens if I now put the step where the doping is not ni all through but changes abruptly to na beyond y equal to d. The curve corresponding to this can be shown to be something like this that is developed by assuming that d is more than the depression with yd at VBS equal to 0 for doping ni. So if I consider 0 bias situation my yd will be somewhere here okay for 0 VBS this is my yd it is below the depth d that is the assumption here if you make that assumption until your depression becomes yd VT versus VBS will follow this dashed curve. But moment the yd goes beyond d okay for higher VBS then it will try to follow this shape it will look something like this where this modulus of VBX voltage is that voltage okay for which yd becomes equal to d. So the depression width is going on increasing for yd equal to 0 it is here you apply VBS it goes on increasing then it becomes equal to d at some VBS okay so that VBS equal to VBX the depression width becomes equal to d and as the depression width goes into the substrate the rate of rise of this threshold voltage falls because the substrate doping ni is less than ni and then this shape tries to follow this shape right. Now to understand this curve a little better you can consider the VT versus VBS variation for what is called the delta implant where the dose of the delta implant is equal to ni minus ni into d what is ni minus ni into d it is this area okay. So what we are saying is assume that this entire charge is compressed as a sheet near the interface as a delta function then for that case how would your VT versus VBS look like once the charge is compressed as a sheet at the interface you can club it with the fixed charge okay that means it is like changing the flat band voltage okay here now if you change the flat band voltage your threshold voltage at VBS equal to 0 will change but the variation of VT with respect to VBS will be the same as corresponding to the doping ni that is why this dashed line is a line parallel to this blue line but shifted up by how much by the implant dose into Q divided by COX okay. So what we are saying is this VFB is equal to phi MS minus QF by COX in a MOSFET where in this implant dose is not there if you include the effect of implant dose it is saying that it is QF plus QI that is the effect of implant dose by COX where if QI is positive then flat band voltage will go negative. Now this is a pre-type substrate and so the implanted dose is acceptor type and when this is ionized it will be negative okay so this QI that is charge due to implanted dose is equal to actually minus Q into ni minus na into D okay. So when you take negative of this minus it will become positive so your flat band voltage will shift in positive direction that is why this blue curve is shifting up but the body effect parameter will not change it will remain equal to na and therefore the rate of rise here is same as the rate of rise for the blue curve. Now in the light of this curve for the delta implant you can see that this curve shown by the black line which corresponds to the case when the charge is distributed over distance D this black curve will approach the delta implant curve for large VBS because for large VBS when your depletion width is much more than D this distribution of charge will appear almost like a sheet charge intuitively you can see this if I draw it to scale so let us say I draw the thing like this this is your for large YD suppose your YD is here then if this YD is large compared to D it does not really matter what is this D right over which this charge is present so long as you keep the area same even if I change this D to small value right and then I change take this my field picture would not be very different that is what is being said here so for large VBS the black curve which corresponds to ni distributed over D approaches the curve due to delta implant so now let us come back to our black curve and see how we can model this as this blue curve minus some quantity now this blue curve corresponds to doping ni uniform doping ni and for that the body effect parameter here is gamma 1 so I can write the black curve or rather I can get the black curve from blue curve by subtracting something which is proportional to VBS why do I use proportional well if I take this difference I will find that the difference is increasing linearly though this curve is nonlinear the blue curve is nonlinear and red curve is I am sorry the black curve is also nonlinear the difference between them is increasing linearly ok so that is what I am reflecting here by modifying this equation by including a minus of k2 into modulus of VBS now a problem here is that this term will subtract from this expression even when VBS is less than VBX right whereas for VBS less than VBX that is the condition when the depletion width becomes equal to de the blue curve and black curve are the same so I should not use this term correction term for VBS less than VBX but then again as we have said that this kind of regional expressions are not good in models for circuit simulation because they introduce discontinuity in the slope or other derivatives of the current voltage expression we want to use the expression for the entire VBS range now that is achieved by simply changing this gamma 1 to a higher value k1 this is achieved by simply changing this gamma 1 to a higher value k1 ok now what happens so this part of the function with this higher k1 will be a little higher than blue curve right it will be some it will start something like this but since you are subtracting this k2 times VBS from there it will come back to this black curve ok now that is the trick that is being used here and then you will be able to generate this black curve approximately so various features of this expression are correlated to this curves this difference is this term this difference is being modeled approximately by this term and this initial value that is VT for VBS equal to 0 is this term how do you determine k1 and k2 in practice you know how do you extract k1 and k2 parameter extraction so you can determine this by matching the threshold voltage expression and its derivative with respect to VBS at VBS equal to VBM that is a maximum substrate bias to measure data or rigorous analytical VT formula so you can extract this from measured data by this approach because there are 2 parameters to be extracted you need 2 equations so one equation is match the VT at the maximum VBS calculate account to this model to the measured data and also match dVT by dVBS also match dVT by dVBS derived from this equation to the measured dVT by dVBS at same VBM so when you try to differentiate this with respect to VBS you know you have to differentiate this term and you have to differentiate this term also here ok whereas this term will vanish and also this term will vanish on differentiation supposing you want to derive from a rigorous VT formula how do you do that ok so the rigorous analytical formula for VT is derived from the field distribution for the box doping profile assuming by d is less than d for VBS equal to 0 and by d is greater than d for VBS equal to VBM that is a maximum bulk source bias that is occurring in your circuit ok so what is this field distribution for box doping profile so let us show that here so when VBS equal to 0 your depletion width is less than d so supposing I am increasing VBS so how will the depletion width vary so let us say this is my d so it will go on changing like this the field picture I am plotting field in the y direction when yd is d this is your field picture then your yd becomes more than d then how will it change so here this field will be parallel to these lines but thereafter it will change like this ok this is how it will change so for 0 bias maybe this is your yd but for VBM maybe this is your yd yd for VBS equal to VBM that is at this point so then this is your field picture so what you will have to do is you will have to find out the area under this field picture that will give you the potential drop across the silicon for yd equal to 0 similarly you will have to find out the entire area this area that will give you the potential drop in silicon for yd equal to VBM you can then use these potential drops in your threshold voltage expression where you know threshold voltage is equal to flat band voltage plus potential drop in silicon at inversion which are nothing but areas under this field distribution and the silicon charge by C ox now the silicon charge is obtained as ni into yd for VBS equal to 0 and for VBS equal to VBM your depression charge will be ni into d plus yd minus d into na ok so you will have to write it like that you will get the threshold voltage expression at VBS equal to 0 and at VT and at VBS equal to VBM now you can use that threshold voltage expression at VBS equal to VBM and differentiate that ok to determine the K1 and K2. Now this entire exercise is left to you ok I will simply give you the result of that the result of that theoretical exercise is you will get K1 as gamma 2 minus 2 into K2 into square root of phi t minus VBM where VBM is this phi t is twice phi plus 6 VT and what is K2? K2 is given by this formula and what are gamma 1 and gamma 2? Gamma 1 is the body effect parameter corresponding to doping ni ok this doping and gamma 2 here is a parameter corresponding to doping na this doping and VBX that is coming here right this VBX that comes here is actually the potential drop or others this is the bulk to source bias when your depression which becomes equal to d so it is given by phi t minus q ni d square by 2 epsilon s basically the potential drop when your field picture is like this. So yd is d so your field picture is like this the total potential drop is q ni d square by 2 epsilon s this area but that is sum of the applied bulk to source bias and the built in potential phi t ok so that is how it is. Here is an assignment for you you can put my name in scholar.google and search for a publication on the correct equivalent box model so you put this as keywords to predict threshold voltage right well in the title I do not have the term threshold voltage basically what this publication does is it tells you if you have a Gaussian doping profile implanted profile what should be the box type profile that you should use to represent the charge in the Gaussian ok. So this is ni of the box and this is d of the box as far as the Gaussian is concerned it has a peak location rp peak value is n naught and you know that Gaussian has a standard deviation sigma ok the formula for Gaussian doping is nx equal to suppose or this is y direction so let me put ny equal to n naught exponential of x minus rp by sigma root 2 whole square with a negative sign this is a Gaussian doping so how do you replace this Gaussian doping profile by a box profile that will give you the correct threshold voltage ok corresponding to the Gaussian profile a simple formula has been given for d ok as a function of rp and sigma of this Gaussian function and similarly you can also get n naught I am sorry you can get ni as a function of n naught rp and sigma that is not difficult because once you get d you know the area under the Gaussian is equal to the area under the box so you equate the areas you get the ni and this result then has been published ok in a later paper again in my name in general of solid state circuits where it has been shown how you can model a threshold voltage variation with respect to substitute to source bias these are the measured points and this is the model which replaces this particular Gaussian doping and this substrate doping by an effective box profile with ni equal to this value d equal to 0.233 microns and nb same right as that for the Gaussian profile so this is left to you as an assignment now with that we have come to the end of the lecture ok so in this lecture we started a new module wherein we said we will talk about how to incorporate the effect of series source and drain resistances then how to incorporate the effect of vertical and lateral non-uniform doping ok and then how to incorporate the effect of reducing the device size on threshold voltage right in this particular lecture we have covered how source and drain resistances can be incorporated into our current voltage model by simply reducing the mobility right as if the mobility is affected by transfers electric field more than the effect of transfers electric field on mobility is stronger than without rs and rd then we looked at effect of vertical non-uniform doping on the threshold voltage of a MOSFET. So we said that you can derive a simple formula for the effect of vertical non-uniform doping on threshold voltage versus bulk source bias curve wherein the actual non-uniform doping profile is replaced by a step type doping profile now before I close this lecture I just want to tell you how a multiple implanted profile can be replaced by a multiple step profile so what we have shown is I can replace Gaussian profile like this by a box type profile if I have multiple Gaussian I can use a multiple step something like this I can use a multiple step profile like this so each Gaussian I will replace by its step and then I will sum up the steps some of the step functions then I will get multiple step profile and I can use that multiple step profile to simplify my model expressions.