 Let us start a fresh module, this is on testing improvement and parameter extraction. So far we have covered the following steps of MOSFET modeling. Let me put here the abbreviation for the procedure for MOSFET modeling. In fact this is an abbreviation which applies to any device right. So for the MOSFET we have covered the structure and characteristics, the qualitative model, then equations, boundary conditions, approximations, solutions. In this module we are going to cover testing improvement and parameter extraction, TIP. At the end of this module you should be able to explain how the threshold based model is improved with the help of smoothing functions to get the Berkeley simulation model, BCM model. Then you should be able to explain how threshold voltage VT, the parameter phi T, flat band voltage VFB, body effect parameter gamma and the effective mobility parameters are extracted from measurements. This is phi suffix T. Then you should be able to describe the evolution of the MOSFET model or MOSFET models for circuit simulation. So once again here this is phi suffix T. Let us start with the testing step. We recall from the module on the procedure for device modeling that the solution is tested for the following criteria namely generality, continuity, accuracy, physical basis and simplicity over the entire range of interest and we abbreviate this criteria as G caps. Let us apply this criteria to our models on surface potential based model and threshold based model and see how they fare. Here is a surface potential based model summary. As we can see from here that although we have derived this model for a bulk MOSFET, in principle we could extend this MOSFET model to even smaller geometry devices. The central feature of this model is the calculation of the surface potential and then substitution of the surface potential into the expression for drift and diffusion currents ok. So if we can modify this expression to include effects of small geometry then I could calculate the surface potential using this modified expression and then apply it to these currents and then get the currents ok similarly for the mobility expressions here. So in principle this approach is very general ok with small modifications you can extend it to small geometry devices even though it has been derived for a bulk MOSFET. Let us look at continuity. This model is continuous as we have remarked the surface potential as a function of channel voltage as well as the gate to bulk voltage is a continuous function ok. It is a saturating function for psi s as a function of channel voltage V as well as psi s as a function of VGV. So we can get continuous expressions across all regions weak inversion, strong inversion ok moderate inversion and so on and also saturation non-saturation right. So this model is continuous. Next comes accuracy this model is accurate as we have discussed the Pausa model is indeed an accurate model and then its approximation namely the Bruce model charge sheet model is also a fairly accurate model. So then comes physical basis you can see here there you can see here that there are really no empirical parameters in this model except let us say this parameter n ok and maybe to an extent this electric field E0 critical electric field E0 is empirical but otherwise if you run through these equations and see each of these terms all these terms have physical basis ok. Even the term E0 here has a physical basis right but its value is not derived from very fundamental theory similarly the value for the parameter n here ok. So there are a few empirical constants and parameters here but then by and large this model has a strong physical basis. So final thing is simplicity s the problem with this model is that its expressions look simple but when you try to evaluate the surface potential expression right then you run into difficulties because this is an implicit expression and you have to do a lot of computation to derive the surface potential size for any given BGB and channel voltage V ok or surface or source to bulk voltage VSV and drain to bulk voltage VDB ok. So that is the problem with this model it is computationally intensive and therefore we cannot say it is computationally simple ok so that is its problem. As we have remarked in the previous module that if you want to have a model for circuit simulation purposes it should be computationally simple because in circuit simulation we are trying to solve equations for hundreds of transistors simultaneously and unless the equation for each transistor is simple the computation involved in simulating a circuit would be enormous. Let us look at the threshold based model how does this fair on the GCAPs criteria. Now is this model general well it is not general to the extent the surface potential based model is that is because you know this is threshold based model as we will see in a later module the threshold voltage for small geometry devices is affected by many phenomena. So you cannot use the simple expression to predict the threshold voltage that is one of the issues ok. So this model cannot be easily applied to small geometry and other devices as it is ok. It requires more modifications similarly if you want to apply this model to let us say SOI based MOSFET ok and so on it will not be that simple ok. Surface potential based model on the other hand with less number of modifications can be extended to other device structures like SOI MOSFETs and things like that. So this model is not that general so it does not fair well on the generality criteria. It fair miserably in the next criterion namely the continuity criterion because as we have seen here as we have seen in a earlier module that the drift current expression of this model is valid for VGS greater than VH while the diffusion current expression is valid for VGS less than V dash t. So what happens between V dash t and VH will do not have an expression this is really the moderate inversion region ok so we do not have an expression there. So there is a gap it is a serious gap ok in this model so it does not work for all regimes of operation. Let us look at other criteria let me put down the abbreviation for the criteria here so that we do not forget the next criterion we should look at is accuracy. Now definitely if this model cannot work for moderate inversion region right then it is not accurate because in that region it cannot really predict anything ok. In other regimes of operation it predicts currents reasonably well ok physical basis well this model is more or less physical right there are not many empirical there are not many empirical parameters here as compared to the surface potential base model. So N and E0 are used even in this model as in surface potential base model so there is some amount of empiricism here but if you consider the other expressions right here you go through all these expressions you will find most of the terms here right have physical basis they have physical meaning and they are not purely empirical parameters they are not purely empirical parameters which are adjusted for achieving match with the measured data. What about simplicity criterion well this model fares quite well on the simplicity criterion you look at this that all its expressions this expression this expression here the expression for saturation voltage right and so on all these expressions even the expressions for mobility these are all closed form solutions right. In other words the amount of computation involved in this model is much smaller than in surface potential base model that is what is the advantage of this model we have highlighted this fact in our previous module that is why circuit simulation based models have tended to be threshold based right. Now if we can somehow improve this model so that we cover the moderate inversion regime of operation between VT dash and VH okay so between this VT dash and VH for gate source voltages between VT dash and VH then this model would be really good and we should ensure that when we bridge this gap our model remains continuous. Another thing if we can do here we find that you have separate segments for VDS less than VDS sat and VDS greater than VDS sat so even here if we can somehow remove this partitioning okay then we will get a continuous model. So this model this threshold based model which suffers on the G and C criteria and to an extent on the A criteria right so GCA it suffers on these three criteria if somehow we can overcome this limitation by doing some modification then this model would be really good okay for circuit simulation purposes. Let us look at some other criteria on which models are tested so criteria for physical basis of the solution we said that a model should have a physical basis. So how do you test the physical basis of a model first thing you do is dimensional correctness check right and then you also try to see whether for large voltages and small voltages and so on the model reduces to some simple expression that we already know right this is how you check limiting cases. Let us apply this let us apply these checks okay to our threshold based model dimensional correctness okay let us choose any one expression here and then try to see the dimensional correctness let us take the drain to source current expression for a diffusion. I am choosing this particular expression for illustration because most students are familiar with the square law expression right so when they are asked to write the square law expression they will not have a problem right but the diffusion current expression students are not familiar because first level course normally does not discuss diffusion current so how do you you know remember the diffusion current expression based on dimensions. So let us use the square law expressions as a guideline then we know that the right hand side of this IEDS expression consists of the W by L ratio aspect ratio of the MOSFET this is dimensionless then a mobility term then a capacitance term and square of voltage. So we know that when you multiply mobility by capacitance and square of voltage you get current okay so let us quickly check this mobility is centimeter square per volt second capacitance is farad per centimeter and voltage square so one voltage term cancels here one centimeter term cancels here farad into volt is charge coulombs divided by second so coulomb per second is amperes okay now we are left with one centimeter term okay now where is the problem so what we have done is we have written the capacitance term as farad per centimeter which is wrong actually it should be farad per centimeter square okay so now the centimeter square cancels with the centimeter square and we are left with amperes. So this is what we need to remember so you have an expression for current in terms of product of mobility a capacitance and voltage square let us look at the diffusion current expression in this slide so you have the W by L aspect parameter you have the mobility parameter and the capacitance term is the depletion capacitance this capacitance term is analogous to C ox here and then you have some exponential terms here which people might forget but they can remember that these terms are dimensionless so what are in the exponents people might forget but they should remember that these exponential terms have no dimensions and therefore you are left with only the square of the voltage term analogous to this particular term in the drift current expression and that square of the voltage term is the thermal voltage square term okay okay after that one can remember that in the exponential you should get the gate source bias and the drain source bias because otherwise there are no biases right in the terms that have appeared here so far and when you get those terms since they should be dimensionless because they are in the exponent you know that the normalizing voltage would be thermal voltage so that is how you are sure that this is dimensionless because you have thermal voltage here similarly you have thermal voltage here so this is how one can check for dimensional correctness of expressions now prediction of limiting cases let us take a somewhat complicated expression such as this for saturation voltage as an example we have discussed this point in our previous module how to check an expression like this with for asymptotic situations or limiting cases right so we have remarked for example that one limiting case you can consider is when velocity saturation is or the saturation velocity is very high in which case this term this expression should reduce to that you obtain for the pinch of a MOSFET model based on pinch of condition right in other words it should reduce to only VGS minus VT by alpha so you can check that it does indeed happen saturation velocity very high is equivalent to saying the EC or critical field for velocity critical field for velocity saturation is very high right and when that happens this entire term goes to 0 so you are left with just one this one plus this one becomes two and there is a reciprocal here so 2 divided by 2 this is VGS minus VT by alpha so you see that we have checked using a limiting case that this expression seems to be fine at least on that particular checking criterion similarly you can have another check right if you know that the term alpha is basically introduced to model the effect of increase of the depletion charge from source to drain then you know that if you do not take into account this increase of depletion charge from source to drain and assume the depletion charge to be constant from source to drain your saturation voltage expression will be just VGS minus VT so if you set alpha equal to 1 you indeed get VGS minus VT right so alpha is equal to 1 when the depletion width is constant from source to drain and when alpha is more than 1 your saturation voltage is less than VGS minus VT this also you can intuitively satisfy yourself based on the observation that when your depletion width varies from source to drain your depletion charge is more and current would be less right so if the saturation voltage is less your saturation current would be less so when alpha is more than 1 your saturation voltage reduces because alpha more than 1 is the representation for depletion width increasing from source to drain okay so this is how you can check for dimensional correctness and you can check the expression for limiting cases for checking the physical basis of the model. Now yet another method of ascertaining the physical basis of the solution is to check whether the solution is consistent with the approximations we have made let us illustrate this for the MOSFET so this table represents the all the approximations that we have made for deriving the MOSFET model so we can take each approximation and check whether the solution we have obtained is consistent with the approximation or does it violate the approximation that we made okay now this is a fairly detailed exercise I am going to illustrate only one particular approximation how we can check its validity okay and I am going to take a very important approximation that is the gradual channel approximation for illustration you can find your own ways of checking other approximations now even this I am not going to go through the detailed validation or checking but I am going to lay out the steps in which you can check so this is an assignment for you estimate the fraction delta L by L of the channel length L over which the gradual channel approximation GCA is invalid in an MOSFET biased at VGS greater than VT and VDS equal to VDS SAT equal to VGS minus VT by alpha so how do you check this you can check as follows now before I tell you the steps in which the approximation is checked let me recall what we are talking about here so delta L is the channel length modulation region okay so as we have brought out in a previous module what is the interpretation of the channel length L in the threshold based model channel length L is the region over which GCA or gradual channel approximation is valid so delta L is the region where it is not valid okay so that is the delta L we are talking about so you check the approximation in the following steps assume that the GCA is valid at a channel position X if at that position D square psi S by DX square is less than 0.1 times space charge rho by epsilon S okay where we take the modulus of the space charge so we normally work with the P type substrate or N channel MOSFET and in the P type substrate the space charge is negative right so we want to take only the positive part of it to check this particular relation note here that psi S is the surface potential so this talks about the rate of change of the surface potential with distance X along the channel. What we are saying is that if D square psi S by DX square is less than one-tenth of this term rho by epsilon S then we can assume the rho by epsilon S to be equal to D square psi S by dy square right so please recall this gradual channel approximation this approximation is applied to the Gauss law where divergence of E is dou by dou X of E X plus dou by dou Y of E Y now when you put this in terms of the potential you get the Poisson's equation so this term is nothing but dou square psi S by dou X square when you are applying this particular law at the surface okay then this is dou square psi S by dou X square and this is dou square psi S by dou Y square okay but of course you have to put a negative sign here because when you convert from field to potential you have the negative sign okay. So now if this term is less than one-tenth of the right hand side that means this term is 90% of the right hand side okay so that is how gradual channel approximation would be valid I can neglect this part so that is the logic of this assumption. Next to start with assume the gradual channel approximation to hold so that the Y dimensional surface potential equation can be written as QI as a function of psi S is given by minus C ox into VGS minus VT minus alpha into psi S minus psi S naught so if you recall this particular expression was derived assuming the gradual channel approximation right so assuming that we can use the one dimensional capacitance formula to relate charge to voltage okay so one dimensional formula along the Y direction that is the vertical direction from gate to bulk. Next assume the current to be due to drift so that X dimensional current equation can be arranged as IDS into DX is approximately equal to W into the surface mobility into minus of the inversion charge QI into d psi S. Now you make these assumptions that is so we are going to assume the gradual channel approximation to be valid and then we are going to look at what are the results and then we see how they contradict the assumption of gradual channel approximation. So derive psi S as a function of X by integrating the current equation after substituting QI psi S from the surface potential equation and assuming the surface mobility to be constant independent of X express the result in a compact form in terms of the normalized variables psi S minus psi S naught by VDS sat X by L and IDS by IDS sat where IDS sat is given by this particular expression okay in terms of VDS sat. In the next step substitute the surface potential expression psi S expression derived above into the condition d square psi S by dx square is less than 0.1 into rho by epsilon S okay using rho is approximately equal to minus QNA and IDS equal to IDS sat since the device is biased at VDS equal to VDS sat and then derive the maximum X by L over which the gradual channel approximation is valid to get an expression for delta L by L express your result in a physically meaningful form in terms of normalized quantities L by LD and VDS sat by VT where LD is the Debye length. Part C estimate delta L by L for L equal to 0.25 micron substitute doping NA equal to 1 into 10 power 17 centimeter cube and the saturation voltage VDS sat equal to 2 volts so that you can see for a short channel device with L equal to 0.25 microns what is the channel length over with the GCA is valid and therefore how much is the delta L. Now let us return to our slide on testing we talked about consistency of the solution with approximations right so far. Now the next thing we should look for is number of empirical parameters so if you have a large number of empirical parameters which have to be adjusted in the model to get a match with measurement then you know that model is not physical on the other hand if the number of empirical parameters is small then the model has a good physical basis. Next how do you check a model for accuracy so you compare the model results with accurate simulation this is one level of checking for accuracy right so what is meant by accurate simulation. If you recall we have talked about the approximation set 1 and 2 so what we said is you have approximation set 1 which are used while deriving the qualitative model okay so even as you start deriving a model you are making approximations right about the carrier transport and structure of the device and so on so these are approximation set 1. Now based on the approximation set 1 you come to the equations which are used for modeling the device these are the 6 equations that we talked about if you are using the drift diffusion model then we are using additional approximations right that is called approximation set 2 where you approximate these equations drift diffusion equations to get an analytical model. Now if I use these equations without using approximation set 2 then I can get a solution now that is a solution we are talking about here inaccurate simulation so you will have approximation set 1 no doubt okay but then you are not going to use approximation set 2 so you will use the 6 basic differential equations and use a computer to solve them simultaneously right using a numerical solution. So this is the so called one level of approximate simulation that we are talking about here so compare the numerical solution with the analytical results right that way you can check the accuracy of the analytical model. You can use a more deeper check if you check the numerical solution by relaxing approximations of this set you will have more complicated expressions to be used there right but you can improve the you know your accuracy of the reference solution with which you are checking the analytical model and you can then see whether the analytical model is accurate even at a more fundamental level. However this is not the ultimate check the ultimate check of a model is comparison of the results with measured data okay since the measured data is difficult to get in many situations we do accept many times in research a comparison with accurate simulation right so you are derived an analytical model you can compare its results with a accurate numerical model right and then you know that will be accepted as some kind of you know a test for accuracy though it will not be regarded as an ultimate test. You see another reason why numerical simulation is so common is because many times supposing you have a new device structure and you are developing an analytical model for it that device structure is not that easy to fabricate right so you have to spend lot of money and effort to fabricate and you want to show that you know your model is actually good in predicting the device characteristics. So even in such cases this particular method of checking the accuracy would be acceptable now let us go to the improvement step the solution is modified to improve one or more of the GCAPs criteria so what is the direction of improvement once we have tested the model on the GCAPs criteria and we know which are the criteria on which the model is weak now we can try to strengthen the model on those criteria by introducing some modifications or improvement okay so that is how we improve a model now let us consider an example supposing I take the threshold based model right I know for example that this model fares very badly on the continuity criterion okay why because you can see here there are many partitioning voltages one is VH another is V dash T then yet another is VT and then you have VDS sat right so can I remove this partitioning voltages and make this model continuous this is what we will try one approach of doing this to gain continuity now while doing so you do not want to sacrifice the advantage of this model and that is all these expressions that you see here are closed form and computationally this is very simple the approach we are going to talk about is what is adopted in B sim or Berkeley simulation model this is not the only approach of improving the threshold based model but we will focus on one approach so the first step in this approach is to replace the field dependent mobility by a constant mobility UN0 okay so this average surface mobility which had a somewhat complicated expression though that expression was closed form and computation was not that much still we want to make it even simpler and just replace it by a constant mobility term now you can ask but then if you remove some physical effects how can you get match of the model with measurements so this is where we are going to adjust the mobility to get match with the measurement and there is going to be an empirical element so what this means is whenever we are trying to improve the model on any of the criteria of the G caps set of criteria your model may have to sacrifice its strength on another criteria right so what we are saying now is I am trying to achieve continuity okay so when I am trying to achieve continuity in other words when I want to strengthen C criterion the criterion C in G caps I may be losing the physical basis somewhat okay so to gain something I lose something but the loss is not much so that is how on all the criteria I can improve right sorry that is how my model can be reasonably good at all criteria right so I am removing the you know some sort of strong weakness of one criterion by sacrificing little bit on another criterion the next step is simplify the saturation voltage expression okay so instead of having this complicated term here I simply remove this complicated term and the factor 2 I end up with this simple expression for saturation voltage next what I do is I say that my model for VGS is valid for greater than VT instead of greater than VH right now let me remind you what is the VH and VT here so on the VGS axis you have the voltage VT dash here VT and VH okay this VT dash to VH is the moderate inversion region so we have said that this model which is for drift current is strictly valid only in the strong inversion region right that is what is this so this is strong inversion this is weak inversion what we are now saying is let us use the model up to VT where we had shown that is inaccurate but now what are we going to do because we are going to fill in this gap by doing some modification that modification will improve the accuracy of our model in this regime where it was inaccurate okay and it will allow you to use the model until VT rather than only beyond VH so we have removed one partitioning voltage the next step is to use a smoothing function and what will this smoothing function do so let us illustrate one example of that to illustrate that smoothing function approach we have repeated the previous equations here but with a gap where we can write some more expressions to illustrate the concept of smoothing functions so the first step is that I replace the partition VDS SAT here okay I compress these two expressions into a single expression by replacing the VDS term by VDS effective term okay now what is the relation between this VDS effective and the VDS here let us look at that this relation is first shown on a plot to get an intuitive feel so the VDS effective as a function of VDS varies as shown here what is the property of the shape so in this region the VDS effective is same as VDS so this has slope 1 VDS effective is approximately equal to VDS this straight line but when you exceed VDS SAT your VDS saturates to VDS SAT this is how this particular term here will automatically saturate once you exceed VDS SAT okay now that is how it will realize this particular segment saturation segment okay so you use the linear segment here and replace the VDS term of the linear segment by an effective drain to source voltage term which has a shape like this okay now that is a trick it is a very intelligent but very simple approach okay to simulate the current from linear to saturation region. Now you can have a single expression to represent this shape you do not have to have one expression in this region another expression to have another expression in this region you do not have to have you do not need one expression in this region and a different expression in this region right so example of this so before I give you the example expression let us put it down on the slide what we have said the VDS effective stitches the saturation and non saturation segments okay it stitches this region and this region there were 2 expressions here this have been compressed into 1 that is what we are saying. So one example of this shape is one example of an expression that gives you the shape that we talked about is VDS effective equal to VDS SAT into 1 plus VDS SAT by VDS power m and this whole thing to the power minus 1 by m I will leave it to you as an assignment to check that this expression indeed gives you the shape this should not be difficult since you recall that such an expression for used such an expression was used to represent the velocity versus field relation yet another expression which is more interesting is this so VDS effective is equal to VDS SAT minus 0.5 into VDS SAT minus VDS minus an empirically adjustable parameter delta and then you add a square root term which contains this particular part of this expression VDS SAT minus VDS minus delta you square it plus 4 delta into VDS SAT okay. Now what is the role of delta the role of delta is to sharpen the transition between the so called linear and saturation regions. So let me illustrate this the sharpening of this corner is what you achieve okay by either changing this m so if your m is high then this corner here is sharp okay. So similarly sharpening the corner can also be achieved by reducing the delta parameter here what kind of delta parameters are used and so on we will shortly discuss okay since this expression is the expression that is used in Berkeley simulation model and it has a property that the computation involved in calculating the VDS effective by this particular expression is much less than the computation involved in calculating VDS effective by this expression okay the power law expression that is why I am going to discuss the derivation of this expression. So the same expression is laid out here in the slide creating some space for the derivation now what we want is this the expression is plotted here how do you get this now before we move on to that we want to show the role of delta. So delta is equal to 0.001, 0.01 and 0.1 this is what is shown here these are the typical values that are going to be used in practice so you can see here as your delta increases the corner becomes less and less sharp smaller the delta sharper the corner. Now the first step in the derivation is to represent this particular part of the expression as a function f of VDS now what is f of VDS like so if this is your VDS effective expression this red line indicates the f of VDS okay all that you are doing is that you are recognizing that f of VDS is the difference between VDS sat and VDS effective so if you subtract this is VDS sat right so this is your VDS sat if you subtract the VDS effective which is a red curve from this VDS sat you get the blue curve. So let us see how we can get the f of VDS consider this particular behavior of modulus of VDS now one point I want to make here is that this is an example of the synthesis level of thinking or creative level of thinking where it will not be so easy to recognize why I should start in a particular manner okay for example when Planck try to explain the black body radiation and try to model the black body radiation he came up with the principle that you can model the black body radiation only if you assume energy to increase in discrete values and not continuously and the smallest energy element right in this discrete discretization of energy is H into the frequency nu right that is called a quantum Planck's quantum now beyond a point it is not so easy to see how did Planck think about it okay so that is why Planck is regarded as a highly creative person ingenious so the point I am trying to make is in every creative product there will be some step that you cannot explain right how the person got it and that is where lies the novelty or the strength of that idea okay so similarly here why do you start with a function modulus of VDS will not be very clear to you right until I do the derivation so the person who proposes this or the set of people who proposes this that is why are well respected in fact that is the strength of the BC model right so this is a creative element in this model okay now modulus of VDS versus VDS the plot looks something like this on the other hand the VDS versus VDS that is y equal to x plot will look something like this now let us put this together this is where you are doing synthesis right let us put this together and let us see the function modulus of VDS minus VDS by 2 how does that look like so when I subtract VDS from modulus of VDS this part is same as this part so when I subtract I will get 0 this is what you are getting here on the other hand when I subtract this from here right I am going to get double of this value and that is why I am doing a division by 2 then I will get back this particular shape for the negative VDS now one thing I want to say is any synthesis is the result of synthesis is not unique there can be multiple solutions when you are using the synthesis approach to arrive at the solution what we are discussing is one possible solution right that is what is important here so this is not the only way to explain how you get this particular expression there may be some other way but this is one very simple way to explain this expression and also this expression is not the only expression that can give you VDS effective versus VDS right so we have seen that power law expression with the power M can be another approach of getting the VDS effective versus VDS okay. Now next let us look at this particular function what is it this is VDS minus VDS sat modulus of that minus VDS minus VDS sat by 2 so what we are doing is we are taking this form and we are replacing VDS by VDS minus VDS sat now you may have difficulty appreciating why am I doing this that is why you look at it pictorially sometimes you can think better using pictures rather than expressions so what we are trying to do here is that we are trying to shift this function rightward to VDS sat why because you go back and see what we want to achieve we want to get this function f of VDS okay and this is where you have VDS sat okay so to resemble this function what we have done is we have shifted this function to the right to VDS sat and when you do this operation intuitively and translate this to an expression what you are doing is effectively you are replacing VDS by VDS minus VDS sat that is all what you are doing here. Now you rewrite this particular function as square root of VDS minus VDS sat square so the modulus of VDS minus VDS sat can be written in one form as square root of VDS minus VDS sat square once again you might ask so why are we doing this well that is where there is this creativity right so we will see where are we going soon so let us rewrite this same expression here to create some space all that I have done here is I have interchanged the location of these two terms okay I have just re-arranged the equation I have just re-arranged the expression. Now if you introduce the parameter delta in this expression you will end up getting this expression okay the role of this parameter delta is to round off this particular corner now let us compare this expression and this expression again there is a jump here this is a creative element okay cannot be explained that easily you have to just appreciate this this kind of jumps right which you cannot explain very easily that is why you get the aha feeling how did someone get it so you can see here that this 0.5 is this half term and VDS sat minus VDS minus delta so you have just introduced a parameter delta within this expression and the same thing is repeated even here in this expression right now when you do that you have to add 4 delta VDS sat okay so that your expression actually has this same shape but a rounded corner again it is not that easy to just explain why are you adding 4 into delta times VDS sat right you can think about it I will leave it to you as an exercise now let us come back to this particular slide on VDS effective which teaches saturation non saturation segments next we have to combine the diffusion current antitriff current into one expression somehow so this way you will be removing another partition now how do you do that so what we are saying is you simply replace the VGS minus VT term here by an effective function VGST effective okay when you do that your saturation voltage will be written as VGST effective by alpha this is straight forward for you to see okay that if you use this expression your saturation voltage will be obtained by putting the condition DIDS by DVDS effective equal to 0 at the saturation point and that will lead you to the condition VGST effective by alpha for saturation voltage now what is this VGST effective term this approach is similar to replacing VDS by VDS effective to combine the saturation non saturation segments so the VGST effective versus VGS has the same shape as the drain to source current versus VGS shape from diffusion to drift so on a semi-lock plot you recall that the drain to source current versus VGS line is straight okay and then it tends to the rate of increase tends to reduce where the partitioning voltage VT is somewhere here so we want an expression which has this particular shape if you get that expression and put it in here then the result will work for drift as well as diffusion this particular square law expression without VGST effective that is with VGS minus VT was applicable in this region okay so we are extending the validity of this expression by replacing VGS minus VT by VGST effective which will give you this shape and also there will be continuous variation from drift to diffusion so the VGST effective stitches sub threshold and super threshold segment this is sub threshold and this is super threshold let me tell you what is the expression that is used it will look somewhat complex but then we are going to discuss how do you get that expression the expression for VGST effective is 2n into VT into logarithm of 1 plus exponential of VGS minus VT by 2n VT divided by 1 plus 2n C ox by Cd into exponential of minus of VGS minus VT by 2n VT now before I discuss the derivation of this expression let me just tell you one small modification but important modification you have to make so that the predictions of this model with VGST effective and VDS effective are accurate and continuous you have to add a 2 VT term to VGST effective in the VDS sat expression okay we will see later why you have to add this 2 VT term let us look at the derivation of this VGST effective now consider the sub threshold and super threshold segments of drain current for VDS much less than VT actually this is IDS okay because we are neglecting gate and bulk currents now what are these expressions expressions are given here for VGS less than or equal to VT this is sub threshold and for VGS greater than VT you have the drift current expression okay this is super threshold now we are considering the situation VDS much less than VT simply because these expressions will become simpler for that condition we will see how the two expressions can be stitched together or you can come up with one expression which will give you these segments for these conditions easily for VDS much less than VT and then we will show that same expression with very little modification can be also used for VDS much greater than VT and therefore for all VDS so what is the simplification for VDS much less than VT here you get only the VDS instead of 1 minus exponential VDS by VT okay and here again you get only VDS instead of this expression for instead of this expression right so let us see how do you get that so for VDS much less than VT this exponential term can be approximated as 1 minus I am sorry is no minus here 1 minus VDS by VT now this is nothing but saying that exponential minus x is approximated equal to 1 minus x for x much less than 1 and therefore if you approximate this exponential like this 1 minus exponential of minus VDS by VT would be 1 minus this quantity so this minus sign will go away and this 1 will go away and you will be only left by you will only be left with VDS by VT and this VT then you can cancel with one of the VTs here that is how you are getting this expression here so VT square is becoming VT and you are having only the VDS term. Now let us look at this expression if VDS is much less than VT this term can be simply removed because when you square a term that is small you get even smaller term right and that is why this will be less than this particular term okay this term will be much less than this term so that is how you are getting this okay. Now let us look at these two expressions I can write this diffusion current expression as shown here so what am I doing I am dividing the CD term by C ox and multiplying by C ox okay why am I doing that then I can see that this term here becomes the same as this term so that is my motivation in doing that now since this VDS term which is multiplying the expression is common and this term here is same as this term now you compare these two terms okay if I can somehow get an expression for VGST effective which will reduce to this expression for VGS less than VT and this expression for VGS greater than VT then I have solved my problem so this is really my VGST effective right that it should give you this segment for VGS less than VT which is less than or equal to VT and it should give you this segment for VGS greater than VT so let us rewrite or summarize what we have written just now so our expression at this point is ideas given by W by L mu N not into C ox into VGST effective into VDS where VGST effective has this property right these are the expressions note that this is valid for VDS much less than VT only but later on we will show that our approach will our approach can be extended to VDS much greater than VT okay how do you get a VGST effective which has this form consider this particular expression for VGST effective here there is an intellectual jump right you will ask no you know why are you considering this expression now this is where the contribution of the researcher right this is something new that gives you a half feeling oh I do not understand right exactly how this has been arrived at that is why this is novel that is why this particular you know contribution is valuable so something that others cannot understand right easily is the contribution of the person who has proposed this equation now let me write the same this particular equation in a somewhat simpler form this is amounts to considering ln of 1 plus e power x okay into some quantity here right so we would like to show that this particular expression will give you this for VGS less than VT minus 3NVT and this for VGS greater than VT plus 3NVT how do we show this so we come back to this particular expression and see its behaviour then this will be clear we would like to show that this VGST effective expression will give you n times VT into exponential of VGS minus VT by NVT for VGS less than VT minus 3NVT and it will give you VGS minus VT for VGS greater than VT plus 3NVT okay now how do you show this so let us consider this expression ln 1 plus e power x when x is less than 0 e power x is less than 1 okay so if x is a large negative number e power x will be actually much less than 1 okay in that case I can write ln 1 plus e power x as approximately equal to 1 plus e power x I am sorry you can write this as simply e power x I am using the approximation ln 1 plus z is approximately equal to z for z much less than 1 now for x greater than 0 and if this is a large positive number then e power x is much greater than 1 then 1 plus e power x is approximately equal to e power x itself and ln of 1 plus e power x is approximately equal to ln of e power x that is approximately equal to x so that is interesting this ln of 1 plus e power x for x large negative number is simply e power x okay and for x large positive number it is x now in this light if I substitute for x VGS minus VT by NVT why am I substituting VGS minus VT by NVT you go back what kind of VGST effective function we wanted it has to have this exponential of VGS minus VT by NVT okay so that is why the exponent x is being replaced by VGS minus VT by NVT then I can write ln of 1 plus e power x is equal to that is if this then is equal to e power VGS minus VT by NVT for x less than 0 or large negative number or VGS less than VT okay and we may be extending it up to VT right if you allow x equal to 0 so we are getting this particular part of the function exponential of VGS minus VT by NVT for VGS less than equal to VT here on the other hand for x greater than or equal to 0 I will get this ln of 1 plus e power x is simply x that means I will get VGS that same thing is approximately equal to VGS minus VT by NVT so that is how for VG x greater than 0 is nothing but VGS greater than VT okay so that is how I have got this particular expression I have not shown you how exactly you had VGS minus VT because here you have got VGS minus VT by NVT right so you have got VGS minus VT in the numerator now since this gives you VGS minus VT by NVT if I want to get this VGS minus VT I have to multiply by NVT now that is the reason why I am multiplying this by NVT okay here to get VGS minus VT okay that is why you are multiplying the log of 1 plus exponential VGS minus VT by NVT by NVT so that you can get VGS minus VT as shown here then this NVT will cancel of course you will get an NVT term here but that is not a problem because you are having a VT term here so we will see how to work with that okay so now actually these expressions are valid when x is a large negative number or x is a large positive number right so what is meaning of large so here you can see what we are saying is if VGS is less than VT minus 3 times NVT so in other words we are putting here x instead of 0 if x is less than x is less than minus 3 and VG and here instead of x greater than 0 x is greater than 3 we know that when x is less than minus 3 this e power x is less than e power minus 3 and then it can be assumed to be much smaller than 1 it will be 1 by 20 e power minus 3 is approximately 1 by 20 so we have used this criterion even when we wanted to approximate the Fermi-Drak function right by the Boltzmann function right Boltzmann approximate of Fermi-Drak function similarly if x is more than 3 this quantity is more than 20 and therefore we can neglect 1 with 5% error right so that is the logic so that is why you have taken here minus 3 NVT and here plus 3 NVT okay so x less than minus 3 where x is VGS minus VT by NVT gives you the condition VGS minus VT by NVT less than minus 3 and if you shift this NVT here it becomes VGS minus VT less than minus 3 NVT and then when you shift VT to the right hand side it becomes VGS less than VT minus 3 NVT so that is how you are getting this and similarly you get this condition VGS greater than VT plus 3 NVT okay so we have explained how you get this particular term from this term right how do you get these approximations these approximate segments from this particular expression. Now compare this and this we have got VGS minus VT here and here however when you compare this and this we have got the exponential term right we have got the thermal voltage right but we are getting N here whereas we should get CD by COX here so that is where we have to now introduce one more trick how do you convert from N to CD by COX to achieve that we want to multiply this function by another function of VGS minus VT by NVT okay which should reduce to 1 when your VGS is greater than VT plus 3 NVT so that this particular segment is unaffected but this function should give you N for VGS less than VT minus 3 NVT that is the nature of this function that is what we have written here for VGS less than VT this function should be CD by NCOX okay so when this function is CD by NCOX that N and this N will cancel and then you will get this CD by COX on the other hand this function should be simply 1 for VGS greater than VT so that this part is unaffected right this is equal to this is unaffected now if you have a function like that for F then you will get instead of N CD by COX here so let us consider derivation of this function F again consider a function F1 okay this is not F yet this is F1 okay where you have the exponential function in numerator as well as in denominator this is again an example of creativity you know you cannot easily see you know why should you do it right that is a novelty here again you get a ha ha feeling however some intuition behind why we should look at this kind of an expression is obtained if you note that it is exponential function which has nice continuity properties so if you compose any function that you want using exponential function that function will have good continuity properties which is essential for a compact model expression right okay now let us look at this you want to get CD by NCOX right ultimately for VGS less than VT so you can see that if VGS is sufficiently less than VT or sufficiently negative then this quantity will be exponential this quantity will be negative and exponential of negative will be a small number this exponential of negative quantity will also be a small number right then what will happen is I can neglect this compared to 1 and then I will end up getting CD by NCOX into exponential of VGS minus VT by NVT okay if VGS is greater than VT then this quantity will be a large positive number if VGS is sufficiently greater than VT this will be a large positive number this also will be large positive number when this is large I can neglect one compared to this number and then I will have this exponential term in numerator as well as denominator which I can cancel and CD by NCOX in the numerator as well as denominator which also I can cancel and therefore this will reduce to 1 so that is what we are saying here. So this function gives you CD by NCOX into exponential VGS minus VT by NVT for VGS less than VT minus 3NVT again we are using the criterion that exponential of minus 3 is much less than 1 and exponential of 3 is much greater than 1 so exponential of minus of a quantity more negative than 3 will also give you the required approximation as you will get the approximation for exponential of a quantity more than 3. Now you put the term 2 here right why do you do it well before I discuss why we do it let us see when you put the term 2 what is going to happen okay so you are replacing NVT by 2VT so here you are getting 2NCOX here you are getting 2NVT right then this particular function will reduce to this particular function. Now why are we doing this because you see that if this function gives you an exponential of VGS minus VT term you are already getting an exponential VGS minus VT term from here okay when this VGS minus VT is very small so I will end up getting exponential of something multiplied by exponential of the same thing in other words I will get exponential of twice that quantity whereas I want only exponential of that quantity in my current expression so anticipating this issue that when I am multiplying this function by F of VGS minus VT by NVT I am going to end up getting product of 2 exponentials okay for one particular condition whereas I should get which is same as saying I will end up getting exponential of twice the quantity but I want exponential of that quantity so therefore I will divide this by factor of 2 so that if I multiply e power x by 2 by e power x by 2 then I will get e power x which is what I want if I multiply e power x by e power x I will get e power 2x but I want only x so anticipating that I divide by 2 that is why you are dividing here by 2 okay so that this leads to an exponential which is suitable for you to derive the ultimate expression now I should evidently divide by 2 here also because only then when I multiply this by this I will get VGS minus VT by NVT and if I am doing that I should multiply this also by 2 okay now you can see that this particular thing that we have derived is same as this VGST effective expression okay that we said the Berkeley simulation model uses okay now with that we have come to the end of our lecture so what is the summary let us summarize our Berkeley simulation model we said that this model is given by IDS is equal to W by L into mu N0 into C ox into VGST effective into VDS effective minus alpha by 2 into VDS effective square this is same as the square law model simple square law model with VDS replaced by VDS effective function and VGS minus VT replaced by VGST effective function right what are these functions so VGST effective has this particular form and VDS effective has this particular form and what is the behavior of VGST effective and VDS effective so VGST effective versus VGS has the same shape as the IDS versus VGS right when you do a semi log plot where this is log scale y axis is on a log scale and similarly VDS effective versus VDS has the same shape as IDS versus VDS behavior for any given VGS right so it rises linearly and then saturates where the VDS SAT term used in VDS effective function is given by VGST effective plus 2 VT by alpha now I have not explained why you add this term 2 VT 2 VGST effective okay I am going to leave that as an exercise for you to figure out you know I have explained to you how do I use the VGS how do I derive the VGST effective and VDS effective right so you can use that experience to figure out why I should add 2 VT here so note that the nice thing about this Berkeley simulation model is it uses simple expressions and these are all continuous across all regimes there are no conditions like an expression is valid for less than some value and another expression is valid for more than some value and so on right that is a nice thing about this particular model.