 My interest of course is in gain and if I see a equivalent circuit of a normal amplifier using a MOS transistor, very very low frequency circuit nothing very great happens and let us say it is a common source and I figure it out that this is what is going to be seen by me. This current source and this are 0 somewhere related to the device, this is a small signal low frequency equivalent circuit of an amplifier common source, we will come back to it again I just trying to say you. So, my worry is if in a circuit if I have a current source which essentially I am showing you something like this, now this current source is a function of g m essentially means it has something to do with the device as well as the amount of biasing I am going to do with it and since g m is what is most worrying me, I would like to have a model of a transistor which will replace or which will give me equivalent of g m expression because at the end of the day I can only solve something numerical. So, I must create a equivalent circuit for a current source in terms of device parameters and once I know what those device parameters I have to govern which will control my g m and similarly r 0, then I will say I have designed an amplifier for a gain of v 0 by v n which you can see very clearly is the if the gain is simple like this it is minus g m r 0 parallel r l divided by there is no input impedance right now. So, it is infinite there. So, 1 upon infinite. So, there is nothing 1 only appear 1 plus that. So, obviously if I want to control gains I must control g m and r 0, if r l is very very small compared to r 0 I am not very keen about r 0 either, but in normal case I may be interested to know what are the values of r 0 and what are the values of g m so that I can control v 0 by v n that is my ultimate aim. Of course, this is a simple circuit we will actually modify the circuit more complex way and we will see whether we can have some equivalent model of a transistor which represents equivalently of this circuit because at the end of the day when I design a chip I have processes and I have sizes you have done a MOS transistor theory earlier. So, we know we can only control sizes w by l we can control partially mobility which is not very much in my hand which is technology dependent. So, all that I will control is threshold and w by l and of course, power supply is also not in my hands by technology node someone say 1.2 volts supply that is the end of it. Some extent therefore, control is only on the sizes and also on the threshold I adjust. So, I want to create a model for g m r 0 many other things so that I will be able to control g m and r 0 so that I can attain a gain of my choice. That is the design if I analyze analysis is so straight forward, I want to know I want to gain of 1004 10000 and above above means I do not mean how much but 10000 minimum I want. So, what should I choose the values in a device so that when I interconnect on the chip that all circuit it will give me a gain at least 10000 or may be lower or higher whichever value you choose that is the difference between I keep saying between analysis and design I am now want to know this value now I figure it out that that is not so straight forward if I control something I will figure out something I may lose. So, how to get optimal use or optimal parameters so that I get whatever everyone is asking I can attain to a great extent that values and that is what the design is all about. So, if I had to create a equivalent circuit or equivalent model for this I must do something we are done in second year but I will do little more detail here that time you know you probably were not very receptive at least many of you I should I know about so now at least I have word gain so that you know that why these values are relevant for for example threshold how do you control a threshold sometimes that threshold control is beyond you sometimes it is well within you the same technique which I am using for analog threshold control can be used even in the case of digital circuit where we will may probably do what called low power circuits. So, threshold control is a major area where we actually look for design at times technology forces you to do only this much then circuit wise what do I do if everything is fixed by technology then what do I design. So, there must be some circuit way of fooling a device or vice versa may say so that we still have some leeway to design. So, that is the aim of this course we were looking for MOS transistors already for this is all n channel device and we say that if VGS is positive and substantially higher one can see from here it is going to say let us say VDS is 0 and source is also grounded substrate is also grounded it is like a MOS capacitor which has the oxide thickness of T ox which has a dielectric or permittivity of that dielectric is epsilon ox and because we are applying VGS as we did last time initially there will be a depletion charge because we want to balance charge put on the metal plate or gate plate and this is positive gate voltage is applied you need negative charges to appear. So, first thing the holes move away and you created depletion there and I said last time that this can continue infinitum because that depletion thickness can keep on increasing, but it does not occur because one figures out that at certain value of threshold VGS equal to VT the electric field in the depletion layer is large enough. So, that the whole electrons which are constantly generated everywhere including depletion layer they get separated in depletion layer because there is a electric field please remember Poisson's equation says d by dx is rho by epsilon if there is no rho. So, there is no P is equal to n means no rho. So, E is constant or 0 in most cases the neutral regions have electric field assume 0 whereas, in the depletion layer you have an electric field and that can separate electrons and holes. At the point where it occurs we say it is VGS is actually defined as threshold. Now, this is what we said last time and we said depletion layer width is proportional to the surface potential that is drop in the semiconductor surface and one can see it is inversely proportional to the doping in the substrate whereas, psi s itself is a function of any psi s which is called surface potential twice of Fermi potential is given this expression. So, we say if I had 2 phi f I already termed in any in x d I have another term in any. So, there are there is a transcendental equation. So, for a given any you can get a good charge equivalent of this you have to solve a quadratic kind of nonlinear terms. Now, we defined further I mean this is what we said last time if we continue to increase VGS further above VT we need more minus negative charges in the semiconductor and now we say the depletion layer has already reached its maximum and we said this upwards because now free electrons can be created holes will still move down, but electrons can come to the surface which is called inversion layer because it is opposite of the substrate which we started with not going to 2 details of mass transistor mass capacitor theory. One can say VGS which is now equal to VT when threshold starts the turn on starts it is 2 phi f which is the maximum surface potential which we reach which is twice the Fermi level minus Fermi energy Q phi minus Q s which is semiconductor charge divided by C ox and this as I said it seems from the very simple relationship if you wish one can derive very fast not very accurately, but psi s plus V ox is equal to VGS. Now, we know D is epsilon s E s D is assuming the electric vector and we assume right now which is not very bad assumption across the insulator and silicon D is continuous this is slightly very some essentially I am saying epsilon E s is epsilon ox E ox the E E for both is same that is the continuity of D vector along the semiconductor to oxide. Once we assume this we also know the charge in the semiconductor can be written as epsilon s capital E is the field and small e is the dielectric permittivity. So, the charge is minus epsilon s E s D is continuous so I have one relation here and another relation here now I can say oh I can now start playing games I say from here this expression you can choose from here I can write epsilon s E s is equal to epsilon ox by T ox now this term from T ox is coming from here because this E ox oxide voltage whatever you are seeing the field across this is nothing but voltage divided by the thickness. So, this V ox by T ox is epsilon ox not epsilon E ox now this epsilon ox by T ox by capacitor theory is the oxide capacitance per unit area. So, this C ox is defined as oxide capacitance per unit area. So, C ox into V ox is minus Q s is that clear Q s is by Gauss's law is minus epsilon s E s. So, I just substitute here so from here I get V ox is equal to minus Q s by C ox. So, what did I do that this expression in which I know at threshold how much is psi s to phi f that I know because it is only function of n A n A by n temperature n A by n i ln of ln A by n i into k T by q twice of that is twice k T by q. So, I know psi s I know now here because if I know Q s which is the charge in the semiconductor charge density please remember this is all density per unit area that is why this is per unit area this is per unit area. So, it is a voltage. So, if I now have a term which is I know Q s directly otherwise I know C ox anyway because the oxide thickness has started with T ox. So, I know how much oxide thickness I have and therefore, I know the oxide capacitance is that clear. So, I know my V ox once I know my V ox at V T one can therefore, say psi s is to phi f minus Q s by C ox at V g s equal to V T this Q s at that time. So, at how much is Q s at the V when the inversion starts Q n A x d max is that correct Q n A with a minus sign because the negative charge. So, we say this ok. So, at threshold voltage therefore, I have a 2 phi f term which is known to me I can calculate and this is minus for n channel minus Q n A x d max phi c ox. So, one can see from here every term is known from the device doping concentration is known from where you can calculate phi f I know phi f. So, I can calculate x d max because it is a function of n A and x phi 2 phi f. So, I can calculate the Q s by C ox term I know 2 phi f. So, I know my threshold ok. So, that is how I know threshold since this quantity is minus and the sign here is minus and 2 phi f as I say band bending down phi is positive. So, both terms are positive adding to each other which means threshold for n channel device is always positive. However, this statement is can be modified little later, but that is what we want threshold of a n channel device should be always positive. If it is 0 or negative what does that mean that the inversion has certain even at 0 bias because the threshold is lower. So, by the time you reach 0 volt already inversion has is already available and that was one of the major worries in earlier technologies. We started with as I said earlier we started with p circuit p device circuits for simple reason because getting the other 2 term which I will show you now we are so strongly negative that the V t is to become net negative and once that becomes then that means device is always on in terms of device called depletion modes already in the depleted situation ok which means always on ok. So, you have to go minus higher voltage to switch it off in a circuit we never wanted that to occur we want on a particularly for digital for example. So, 0 to something is switch on now that was not possible. So, we said look for p channels in p channel this is negative this value will become plus q n d x d max. So, this will minus and we will hold this will be minus the other 2 terms are always minus. So, all the terms which you will get is minus V t, but p channel I want a minus V t. So, if there are worse situation those 2 terms which are minus only may add to minus value more. So, to turn on a device you will require much higher minus voltage. So, that it crosses V t that was possible because it is a 3 volt I may put 5 volt supply 5 it does not I will put 10 supply minus of that. However, if I have a minus V t for a p n channel then I have a problem that I cannot then turn it on at normal by that is why I say we started with. So, what are those 2 terms which cause so much worry for most of us there are 2 assumptions we made so far one is we said there are no charges in the oxide which in reality they are there are charges very close to the interface. They are normally always positive normally I will not I will not go to detail on that they are called fixed charges and normally within 100 amps longs from the interface they are situated. Now, take a situation in which you have a charge at the interface what was the problem with us. If I have a device and this is my oxide and this is my semiconductor let us say positive charge occurs very close to the this is oxide and this is semiconductor what it will do I have not applied bias, but this positive charges itself will now expect negative charges to appear in semiconductor to balance the charge neutrality that means anyway by V g s what we are going to create negative charge. So, even without V g s I have some charge already available is that clear to you already there are charges available to you similar thing happened if you have a metal this is your metal our assumption is between metal and silicon the work functions are same work functions means the energy taken for electron to leave the material is called its work function. We assume initially phi m is same as phi s, but in reality the material will have different phi m s for example, if you put aluminium as the gate aluminium has a 4.2 electron volts as phi m where semiconductor is typically 5.2 minus phi s phi f values will around 5 e b for example. So, it means phi m s phi m minus phi s is negative quantity for aluminium even with the silicon gate which we use poly gates as we say the polarity of poly and the substrate will be opposite. So, there will be again difference between phi m and phi s which essentially means it is normally and normally depends on the metal use there is something is used phi m s is always negative that means if phi m s is negative it also wants something additional already available to you semiconductor is already having some negative charges already sitting there we also have positive charge which also is already created this. So, normally how do we measure equivalent level? So, we say apply additional as if this voltage so that the equilibrium is that in Fermi level on both sides is are equal this is called flood band and from there we now say inversion can be set in, but this additional voltage was already charges were available that much you would not need now actually this is already present with you without worry. Now, the problem is this is giving you negative both values on negative charges and you are going to create some negative charges due to VGS, but these charges may be sufficient for the inversion to be there which essentially means that the device is pre at 0 itself at is an inversion this essentially is the worry which means the VT of a n channel transistor could be fully negative two terms were positive, but these two next two terms which I will now show you may be strongly negative compared to these two positive terms and here is that expression I am talking about this is my two phi f q s C ox is what normal VT would have been, but two assumptions I made q ox is 0 which is not 0 and phi m s is 0 which is also not 0. So if I add those two terms without going to details of band diagrams because some other day specific someone wants come to me I will explain you mass theory even much more in detail. So why we say that because that is our bread and a mass transistor we have been working for 25 years or 30 years or maybe more 35 years. So for us it is very trivial to understand what is going on, but for you it may be interesting some of you not all of you VT is equal to phi m s plus two phi f minus q ox by C ox this relation comes from the fact you can see from here why did I say total charge initially was 0, but if there are oxide charges this is the new equation charge neutrality if that occurs we can say q m is minus q s plus q ox and this I call since this is fixed I call it q s dash okay and then I substitute here q s dash okay because that is a constant value q ox is always fixed from the technology positive always known to me how much if that is so I will get additional term of minus q ox by C ox here okay now this is negative please remember in the expression this term is coming negative this is positive this is negative this is positive because minus q n a so this but these two term may offset these two positive terms this has happened because initially when technology started in 60s the q ox was extremely high in the process we made okay it was of the order of what we say 10 to power 12 per centimeter square as the density multiplied by charge it will be coulomb per centimeter square so it was a very very high charge density was available in the interface so whatever we do for n channel it will always be on in respect to what we do now we figured out in 20 or 15 year down then that I can control q ox by technology details I can do something which will reduce my q ox now I can give you a q ox of 10 to power 10 per centimeter square into q of course which means this term will be very small and in which case this may not be strongly negative of course this I cannot play much this is a inbuilt gate whatever I create but this stronger term may become smaller and then the net value may become positive now one catch from all this if I want to increase VT with a given q ox and fine as what is the term I should improve on so that the VT goes higher that is the circuit requirement from here expression can you check what should I increase in a the doping in the substrate if I enhance I will be able to enhance the threshold that is the control that word which I say control I can do some technology control now I say okay boost the any it may have something else a problem so I will okay selectively boost the in a wherever I want that is in the channel only I have higher doping substrate I do not have which I can selectively dope the device okay so the way thresholds could be adjusted also you can have very interesting thing you can have two areas or two separate transfer may have a different doping and we have a multiple VTs on the circuit itself so tomorrow you design a good analog block or a digital block and you expect variable VTs technologically it is called additional mask okay extra one mask is typically cost around 1 million dollars on process okay so you say one mask 1 million dollars another VT is 31 another mask for just for that another million dollars okay typically 24 masks are required for making a good chip at CMOS chip or an equivalent of a chip now you add this as now ISO new technologies are coming and many controls are being given you may require 34 mask so now remember how much money per mask you are adding that adds to the chip cost okay so what Intel Pentium 4 initially was sold may be 140 dollars and now may may sell it 200 dollars because now I improved something which is multiple something I am giving you which can control your power requirements so all technologies cost you on of course if you make billions and trillions that cost may actually go down that is what the whole game in the circuits are so is that clear so the basic idea of a mask transistor VT control is to control alternatively you can also control it from C ox or T ox but T ox is not very much in mind is that technology dependence I say okay 90 nanometers should have so much thickness of oxide 0.25 should have so much I mean that is the technology so I am not very much within my this but if I see see there is a epsilon term going on oh so I can need not work on silicon dioxide I may look for other dielectric with higher abstinence okay and in that case high k dielectrics actually came up okay of course they are their own problems so in general a typical capacitance voltage characteristics of a mask capacitor look something like this is for N channel device what does please remember P substrate always gives you N channel device N substrate will give you P channel device so this is N channel device means P substrate device so we apply what is the threshold for P channel N channel device positive we assume right now that Q ox and these are small enough so they still shown VT so if I have VGS negative since there is a accumulation you can see in accumulation there are plus charges and there is a metal charge plus plus so it is a good MIM capacitor metal insulator larger charges means equivalent metal so you have a metal insulator metal okay so fantastic capacitance came okay so you say C ox is the only capacitance available when you are in accumulation however if I start increasing VGS towards positive somewhere down I said as you cross 0 VGS becomes positive depletion starts the depletion layer starts and you start getting negative charges due to depletion now what can you see you have already seen that figure which is a capacitive figure you can see from here this is metal this is silly sorry this is metal this is oxide this is the inversion layer and also there are acceptors okay now this has two capacitance one due to the oxide and one due to the semiconductor semiconductor charge means it is equivalent of a capacitance there and you see they are in series so what we say therefore it is like saying you have a C ox in series with yes which is semiconductor capacitance so after VGS become positive there is a semiconductor capacitance which is essentially how much epsilon X by XT this is KS epsilon not KS is the dielectric constant epsilon 0 is free space permittivity so this is die please remember epsilon is K silicon N2 epsilon 0 silicon dielectric constant of silicon dioxide is 3.9 okay so epsilon is known to me because this is 8.854 10 to power minus 14 for arts and per centimeter so I know now this is series in compared to see ox what will happen if you have a series capacitance the net capacitance seen between these two node will what will be it will decrease you are in series as you increase VGS XD increases CS decreases so what will happen to the net series see further goes down okay see further goes down because CS is started going towards smaller value as this reaches maximum you will reach CS minimum okay now you cannot have further depletion layers XD max has reached now the net capacitance is see ox in series with CS minimum where this will occur this XD max when the threshold occurs so if I keep increasing VGS somewhere at VGS equal to VT I reach CS minimum so the net capacitance then becomes constant now this is slightly catch I showed you two curves one shows constant the other shows it goes back to higher value of its original see ox this is essentially because of the frequency of measure how do I measure a capacitance I apply a current source or a voltage source through a resistor and pass current through a capacitor so what is the current through a capacitor CD V by DT is the current RJ omega C times V is the current in the capacitance if I know my omega and I measure the impedance on a impedance bridge and I know what currents I am voltage or current I am pushing then I will be able to evaluate the capacitance okay assuming ours are practically 0 once I know see but then that omega term appeared that means the frequency at which I am monitoring will give me the impedance if omega is very small 2 pi F then I am in a low frequency zone if I omega is megahertz tens of megahertz of one megahertz and above I say I am in high frequency now why this at high frequency what happens when I apply higher frequency basically what I am doing I have a DC over which I am superposing AC if the frequency of AC is this whatever this capacitance here I was monitoring I must say that the depletion charge must vary with the frequency because you are going plus minus okay at much higher frequency depletion charge cannot follow that is that clear to you so it shows constant but at very low frequencies it has time constants available enough that it also modifies so the average value of capacitance now starts increasing increase Vgs and at maximum Vgs it shows as if it can follow anything like 1 hertz if I do it it will go back to its original value more details other books so the idea is I can measure for a given technology a mass capacitance at high frequencies or low frequencies together individually otherwise that is what all that all that my other colleagues in the so called device area only thing major thing they measure is a CV measurement and keep telling we did great this is all that we do okay keep measuring CV at the end for the circuit what I am really looking for equivalent circuit I am not interested as I say yeah I this theory I did fine I why it satisfied my ego oh I understood but for a circuit how does it matter it only wants equivalent put it what you want to equivalent of that so we figured out in a mass transistor shown here just for the sake of forget about this figure right now you can write down this expression there are three capacitances of interest at the input side which are the three one is gate to the bulk which is called CGB gate to the bulk bulk means up there is a capacitance between gate to the source so we call that is CG now you say from where this is coming if you see the expressions this mass transistor just a minute before I will come back you can see from here there is a depletion layer here so there is a diode sitting there is a diode sitting here diode has a capacitance in the reverse by both are reverse pass anyway so diodes have capacitances source to bulk okay which side is a higher capacitance no no drain side has a larger depletion layer now so smaller capacity but smaller capacitance have more difficulty than a larger capacitance if they come in parallel or in series depends how do they come okay now there is a capacitance of from the gate to the bulk okay but if there is a channel bulk screen because then there only small r in series to that is that correct so the CGB is 0 if there is a channel existing once channel exists the bulk is screen that means there is no connection with the capacitance is that clear so these otherwise if there is a short channel here there will be a capacitance here oxide capacitance and also bulk capacity so these are the capacitances associated with now what we do here which is what the trick we are saying in a mass transistor in fact if you see all positions have different capacitances we will see later the channel does not have same thicknesses okay so the voltages across every point is different so essentially we are saying we are n capacitance going from source to drain all in parallel okay so what is the model we can do for a circuit we say lump it out half this side and remove distributed something solving is a difficult task is like a transmission line theory RCRC network solving here I say okay use lump so there is the catch word I am using analog circuit design essentially uses lump models is that clear as all the word derivations which we will do later are essentially assuming that lump models are valid once you say lump models are not valid we say this is the highest frequency you can use because for the design you can only do this much that frequency we want to know up to where my circuit will function okay you know and then I have to find now my analog circuit will function at 1 megahertz 100 megahertz gigahertz where is that cutoff which I have okay now that frequency is essentially decided about up to which frequency by the device you can have lump model model equivalently fitting the experiments this is the basic understanding we use so beyond that not that one cannot solve but we will have to do something transmission line theory is like we do in microwaves and probably one can solve a more microwave involved problem we using fields but field theory is not so trivial third year people who have heard it must be knowing about it is not very easy to pick up the fundage unless you know very well what is divergence in curl in a better fashion and to get that feel has to be there so we will say okay we will use only up to lump circuits and we will be safe in all our analysis is that clear to you is that point clear how do we increase that frequency maybe we will see the cannot increase this cutoff to a higher level higher how much higher I can do can I go to 100 megahertz possibly yes possibly no so we will see that so my issue is that I must know my capacitance is because if you see my earlier circuit which is that equivalent circuit I drew for you first day first point where is that maybe okay if I have now a capacitor here a capacitor here and many other capacitances then I have a problem because circuit has a feedback circuit is connect some input cannot be going beyond certain impedances shown here that means the limitation of a frequency will appear as soon as I get actual capacitance in the circuit and that is what I want to know how much is the maximum frequency I can have so that I can operate my amplifier whatever circuit I want and those now this capacitances is what we are trying to figure right out how much are they of course what I am showing here can be useful for you in the digital course who are taking VLS design course there this is same it does not matter here or there okay so we have three possibilities CGS CGD and CGB now CGS which is gate to source capacitance there are three regions of operation what is sub VT it is 5 is still there but not 0 but assuming right now that inversion is so small practically we are not assuming as if they are okay so we say sub VT capacitance is 0 which capacitance will if there is no channel what is the capacitance type between gate and the bulk so CGB exists okay but CGS and CGD does not exist because there is no inversion at the either ends okay so we say CGS is 0 CGD is 0 this additional term which I shown here is that partial depletion layer which was coming there even at 0 bias is taken care and this is our oxide capacitance is that correct now if you see the linear what does that linear we are not just talk but in which that VGS minus VT is less than VDS assume right now that is called linear mode in which case we can now lump it half CG half oxide capacitance to the source and half to the drain and then we say there is no CGB because channel exists both so cleaned out however when the device is active mode or what we are interested in all theory which we call saturated mode we are just will come to it soon now at that time it is a trapezoid of the channel and therefore we say it is two third area into oxide capacitance since the channel is up to corner to this the bulk is screen drain there is no charge left there so this is also screen so only capacitance available is so in general what we do is for a general purpose in respect to where the device is operating we add all these capacitors at Z in a DC case okay if that occurs you can see if you add any term which is very close to Seox this is two thirds so maybe one third plus add so Seox is upper value of that this also will give up sorry this is I think minus sign you just check so essentially if I use only Seox I may not be over I may be over estimating that means my frequency actually will go down than what I expected really may happen but I am safe okay so many a times the first design goes with Seox itself okay fine thank you very much okay otherwise if you are solving on a computer how does it matter if the circuit is given all the specifications and all the values so it will calculate and substitute every voltage whatever is happening and use that capacitance at that point okay so in numerical analysis I do not do any assumptions I say this is the solution you solve in the case of analytical I have worried because every time then I have to remember or correctly okay so I want to make some simpler assumptions not that I may use this also but I am just trying to so many books you will say they are using every Seox the reason why they are using is because of the assumption is that little overestimation but fair enough okay okay let us quickly finish up the device part so that we will actually go on the major issue of our interest but as I say why I am showing a device because I want to show you those parameters which process and design is required I mean with design parameters from where they can be controlled because that is why I know want to know the theory otherwise for me expression is good enough spice may be there are versions okay we will only give you the first version why we are doing that because there are the problems which we may give will be only simulation problems okay and you must run spice okay essentially spice solves a small network what is it solves I is equal to G times V with all three are matrix is that correct that is all that it does now these G's could be voltage dependent current dependence which be sources are same okay so that nonlinearity certainly comes larger the parameter matrix may become bigger larger circuit may have so many nodes to solve so the complexity of matrix solving may increase because of the nonlinearity terms and because of the larger sizes is that clear but basic idea is Kirchhoff law G into V is I is that clear so that is what spice does now what is G you must know device parameters you must give so that it can calculate dependent sources for this so it needs data from the device it needs data from the connectivity from which node to what node what component you are putting once it knows the circuit it just does G is equal to V and you can calculate V or I any position any mesh okay that is essentially spice what is my stand for spice life in both of them similar to hey Abhi to S word again simulation program S spice I C integrated circuit E emphasis okay this program was written by Berkeley group headed by Professor Newton unfortunately is no more he was the provost of not provost he was the president of Berkeley sometime and his group has written this program spice now all industries actually working on spice but they are modified model they are modified few things and then keep telling oh this is our mind the mentor graphic has his own spice some other spectrum now I am doing a basic spice kahimala reason is in if I have a what we used to call a small routine I may block that routine for anyone else so I may still run spice there for outsider you say I do not know what program I ran there you know I just call from somewhere on that okay so everyone is using spice in one way or the other okay H spice she now she came with some other name okay because they say it is it takes care of both high power and highest frequency S spice okay some better models they put but basic idea is GV is equal to I and nothing more is that clear so do not get too much read and it is best thing to happen because then you can listen music and talk people things will be done by someone else for you okay good okay quickly we find out the current because at the end of the day for a circuit person I want to know given a voltage on the device what is the current I am going to get okay because that is what my model is asking what is G okay so I say okay I must find the relationship between currents and voltages if I get that expression some way I will convert into a equivalent source and if I do that I my job is over so I figured out what is the most understood doing if VGS is greater than VT on a N channel device VSB is 0 source is 0 and I am applying positive VDS and applying positive VGS greater than threshold that can be two possibilities of course but first we say VGS is large enough okay VGS is large enough and VDS is relatively smaller please remember once VGS is greater than a channel may exist in the substance as soon as channel exists the one shown here okay this is so this is drain how does it look this is an N plus area this is N plus area this I am applying plus VDS small enough but and this I am grounding what is it look like a semiconductor bar with voltage of my VDS on one side and grounded the other it is like a resistor a small resistor since it is like a resistor which law I am following Ohm's law or drift current only flows okay J sigma E is the equation of drift currents sigma essentially is conductivity which is related to resistance okay since Ohm's law is to be followed now I must know what is R because if V I know and I want to know I so I must know V by R R is what I calculate for this device this is what all that we do and we solve it so the first assumption we make there are three electric fields or other two strongly feel electric fields let us say this direction downward is X across the channel is Y and along the third dimension plan dimension it is a Z okay this is the axis I used so you can see from here if this is X there is the electric field downwards positive VGS which I called is E X across off side because there is a current in the channel there is a voltage means there is a electric field this be divided by lengths equivalently not so there is a E Y and there is E X so the first assumption I make E X is much greater than E Y which is called gradual channel approximation we can remove that what is essentially telling that the electrons available in the channel are essentially governed by the gate and no one is that clear who per se he control nowhere else so gate is controlling the charge is that clear so this essentially says E X is stronger than this so this controls the charge okay that is what gradual channel approximation is all about all to say the inversion charges what is the charge density is capacitance per unit area into voltage so if you see it what is the capacitance here C ox just what is the voltage you can see from any point here I applied a VGS here okay current is flowing in this electrons are going in this so current is going in this direction drain to source since this is a resistor what does that mean at every point of the channel that is from Y is equal to 0 Y is equal up to Y is equal to L there is a voltage drop what is the net voltage drop at L VDS which I applied but at the source 0 from 0 to VDS there is a voltage drop everywhere is that correct so if I take a channel let us say this is my channel and at any point here Y this is my Y is equal to 0 and this is my Y is equal to L at any point on the channel I use an element of the channel DX as such and then I say okay if I can calculate the resistance here by roll by a expression and then integrate it out across the channel length across the X both side I take a piece and I actually integrate on X side and integrate on Y side to get the net equivalent average resistance and I is equal to V by R that is what I am doing so I figure out the charge density is VGS if I apply VGS minus VT because that is required for inversion minus VY because at every point now substrate is equivalently saying at PY in normal case of capacitance substrate was grounded now at the lower side of the voltage is potential there so what is the voltage across the oxide now VGS minus VT minus VY multiplied by capacitance per unit area will give me the charge density available of electrons in the channel below is that clear now this is an approximation more accurate expressions can be done but we just see we also know from the very simple electrostatic law electromagnetic electrostatic laws the current in a resistor essentially is Q times velocity times the width the width comes because as if you have a number of such channels and you have to add all of them so multiply by W okay assuming that is constant okay that means this W is uniform throughout so if I do that I have a Q VW is the this is charge per unit area please remember what is the velocity there mobility times the electric field is that correct I have a drain current which is charge density electron charge density into velocity and velocity is essentially mu EY into width of the channel better expressions can be derived but just to take from this so if you now know ideas if I know my QN which is here I assume now mu is constant but in reality mu is also not constant so we have to take care of mu variations EY how much is EY 0 to L it goes from 0 to VDS how much is the voltage across channel as the length goes from 0 to L voltage goes from 0 to VDS or VDS is that correct so if I integrate these two terms which I am going to then I will get net average I is equal to ideas equal to average resistance multiply equal to voltage across that this is what I do the second assumption which I make which I said V is equal to mu is constant is always available but that is not true because if E increases too much the velocity starts becoming constant so right now my assumption is V is not reaching V sat which may actually okay so there is another assumption I made first assumption what did I made EX is much stronger than EY second I made assumption velocity is still not saturated which what does when this can occur what is electric field is VDS divided by so when EY will be smaller either the VDS is smaller or L is larger larger channel channel link devices called long channel devices so in a long channel devices this assumption is good enough we say velocity does not saturate what is the new problem will now come as you go for a smaller and smaller dimensions the velocity will become saturated day one okay and that is our worrisome part as you come to the models. Third assumption I made which is also an assumption that net current in the channel is essentially drift current that means that is the only current I have which is the other possible current in a semiconductor diffusion current but I said there is no diffusion there is no gradient though there is but that current is a million times smaller than the other I said there is nothing there so the third assumption only drift current occurs this is also an assumption in real devices if you are doing a device simulation projects or something you can get rid of all such assumptions and get exact values but to a great surprise you will find it does not give the exact experimental result in spite of great modeling then finally what you do you try to fit some parameters which fits into experiment once you start fitting then whole physics is lost okay so physics and fitting do not go together but what do I do it I still want to retain my physics so I add constants to it may be less than one or exponential some term which is say constant so it is called fit constant fit functions so key physics with some fit functions okay weightage and then you say oh it fitted my physics also it is good my you are just trying to fool yourself by why choose the same physics you could have written alpha 0 plus alpha 1x plus alpha 2x polynomial 100 kebyl Kushner Kushner alpha we should be adjusted to the same physics with no scope constant very good spice what does it does in model as you scale down technologies it changes its constants fitting functions because experimental given technology is known to you okay so you try to come closer to it and you say look I have got exact physics and exact this is game all device simulation people including replay as I say I already said new structures and new smaller devices diffusion current may not be as small though we do tricks so that it becomes smaller there are another assumption I mean we assume as I said mu is constant when in real life will it varies with electric fields okay so even mobility is not very much constant function of y it is a function of x also okay x it comes from what we call interface state density right now we will not going to so mu is not even a constant but all these assumptions when you make and get ideas videos characters what do we call textbook model which probably and then the actual model what will be give it k1 k2 something so that it fits to the reality okay so I say look this expression I am still using that is the game all spice models people do okay including us so if I now substitute using first order assumptions I write E y is dv by by d y electric field is slope of voltage okay V y I substitute in this expression which I wrote I write ideas d y is W mu C ox V gs minus V t minus V y into dv by y changes from where 0 to L V y varies from at source 0 voltage a drain V ds I integrate from 0 to L 0 to V ds and I get this expression ideas is mu C ox W by L bracketed V gs minus V t V ds minus half this is the simple integral of this what is it trying to say mu is constant that is why it has come out before integral I also assume as I say V t constant which may not be V t is a constant value so I just in difference integral I did not use that I said V t in real life that may not be now this in my analysis analytical and see I use mu C ox term I call it beta dash is that correct and this beta dash into W by L I call beta which is like telling you know bipolar transistor used to have a gain factor so this is equivalent of a gain factor okay so I may use in my final expressions beta dash is mu C ox and beta is beta dash into W by L this is my writing it is not necessary you can keep writing mu C ox W by L everywhere and fair enough this is just to reduce the size of the expressions alright is that okay I repeat how did I do it I wrote Q V W as my current I know V is mu V so and E I substitute a dv by dy then I figure it out I can integrate this from 0 to length because that is what the resistance blocks are okay so I just summed it them all and once I sum them all I get this relationship now the trick worry some part is now after you write down this expression you will find if I really plot this ideas versus Vds characteristics and I am some funny characteristics I suddenly see now the fun starts here if I plot this expression ideas this expression okay for all values of videos okay I figure out initially you can see from here if videos is smaller initially this term is small and ideas is proportional to videos if videos is smaller please remember half videos square term is smaller neglected so ideas is beta times Vgs minus Vt and if Vgs is fixed by me ideas is proportional to videos that is why it is called linear mode ideas is proportional to videos what is the condition Vgs minus Vt is much larger compared to videos and therefore this term is neglected however if this assumption does not go and I keep increasing videos that expression now says as videos starts increasing the current start decreasing because minus half videos square term start increasing is that correct and at some days videos actually ideas is 0 it is going down so it looks parabola because that expression is half square I mean parabolic so it shows a parabola but in real life no one has seen that beyond certain voltage current suddenly dropping down that means current remains constant people have seen that expressions okay so what has happened so we say okay in real life if there is something has to happen like this the this point I must know where this can start so I said okay what is the maximum value up to which current is still one directional so I differentiate this equate it to 0 and I get Vds equal to Vgs minus Vt at the point beyond which the current will fall so there must be something related to this point Vgs minus Vt equal to Vds is the issue at this point onwards something else is happening certainly linearity is not there and no other term going higher negative so that current goes down now what is beyond then this value beyond this Vgs minus Vt if I further increase Vds I do not get this so the theory is something which is what device is very interesting about is that plan clear in reality I never see going down but the expression shows it should go down it does not so what could have happened at this voltage beyond now this voltage is very crucial as if you see when the threshold occurs in a capacitor if you see you have a capacitor this is your metal this is your substrate what is happening Vg minus Vs is the voltage across oxide this should be greater than Vt to have inversion or plus or minus whatever if it is plus then I want inversion to occur is that clear this is always defined it essentially now saying if Vg minus Vs is less than Vt there is no channel is that correct that is what capacitor theory says so look at this transistor if you increase Vds beyond Vgs minus Vt or at that point when Vgs minus Vt is equal to Vds what does this value essentially saying this is Vgs this is Vds is that correct this is 0 so all voltages are measured from 0 Vgs as well as Vds so Vg minus Vd is nothing but Vgs minus Vds is that correct because source is grounded Vg minus Vds same as Vgs minus Vds so if I use this expression Vgs minus Vds this is at this point is occurring is that correct if now Vds increases beyond this point what will happen this quantity will become minus so there is no inversion possible so just beyond this point Vgs minus Vt is equal to Vd Vds beyond this value Vds increase there is no inversion at that end so what you see here there is no inversion when Vds reaches Vgs minus Vt value please remember source is grounded is that clear gate voltage you are in fixed very higher than Vt so at the source there always will be channel because Vgs is always greater than Vt that is we started with so source will have a channel all through but at the drain end at the point when Vds becomes Vgs minus Vt channel is not existing at that point is called pinch off so channel is pinched off but since Vy is varying along this point so Vgs minus Vy is decreasing along this line which means the number of electron density available in the channel will keep on changing as you move from source to drain larger here because the maximum potential is given by your Vg minus 0 is that clear in between smaller smaller and then it may become 0 okay which essentially means the channel thickness will be maximum at the source and will keep on decreasing towards drain end normal in case of channel existing throughout even then it will be something called trapezoidal sorry trapezoid but if you further increase it may actually pinch at the other end and it may become triangular so initial trapezoidal will become then triangular is that correct this point we call device now enters saturation at this point that means Vds greater than Vgs minus Vt device is entering saturation so what does that means in the case of bipolar I do not know how many of still collect when I say a IC VC characteristic and I say device is in saturation what does there means actually VC is very small in a saturation the reason we say okay both junctions are forward biased okay IC become maximum so we say this is the saturation in my MOSFET that is called linear region the current when it becomes constant saturated we say you are in saturation that the name change from there so now if you declare this that pinch occur here I further increase Vds what will happen current is increasing so this point may actually shift from the drain end to this because now Vgs minus Vy itself may be small enough to pinch that channel which what does that mean you may have this pinch of point shifting from drain towards source as I increase Vds beyond this value is that correct now the question in physics was that if you have a channel which is pinched here this is my source this is my drain and I am applying a Vds which is greater than Vgs minus Vt electrons were moving in the resistor but what is here the depletion there is no free carriers there so why carrier should go there is no current current should have gone down 0 immediately still did not go this is essentially saying this is a depletion layer this is something like this the large depletion there there is in this depletion there you have a large electric field the direction is positive to negative this large electric field whichever carriers are coming from here are sucked by this electric field and brought down to the drain side this is essentially how bipolar transistor works as the carriers reach base collector junction the electric field there is so high collects same procedure is that clear otherwise the current should have gone to 0 suddenly it did not it actually sucked out okay so larger the field here the current what but how much is the current whatever this resistor can give you only that much carriers can be picked up that is decided by your Vgs Vds values you already decided but once it reaches there at that point whatever carriers are available to me I just pull it up okay is that point here this is a reverse bias PN junction theory whatever is available will fall down okay how much I am not sure whatever you say I will I because that slope is so high electric field is high to fall down okay this is the theory behind the pinch of head if I do this and if I substitute those Vgs- Vt I call this value saturation at which at a given Vg value Vgs- Vt reaches Vds that value is called Vds saturates now why current become constant that is still not answered by us the current is becoming constant because as you increase Vds the net this triangular part whatever it is the charge density is only governed by Vgs so that is not changing is that correct so the available carriers are not really changing for you because as I said gradual channel approach it now up or so many link now I said yeah I said yeah carries with me which essentially with current cannot now change because whatever has happened has is remain there so current becomes constant so it is essentially now trying to tell the following it is telling me if I plot a ideas versus Vgs character what is this curve called ideas is the output current Vgs is the input voltage transfer characteristics this is train this is this for n channel this is direction so input to output relationship is called transfer so I figure out as long as Vgs is small less than Vt how much is the current it is not 0 I said at Vgs equal to 0 is 0 because there are there is nothing there but as soon as you go beyond 0 till Vt what is the situation we are still in inversion but small inversion is that correct Vt was defined as to five value because it started at five itself okay this region is called sub threshold region there is a current okay there is a current and it is exponential we will see that but at Vt which is what you say current then start going heavily okay and as it starts Vgs start increasing Vgs minus Vt takes over in this case Vds is required why though Vgs minus Vt should be much larger than Vds Vds is still required there are current cannot go so Vds is stepped in millivolts 10 sub millivolts 100 sub millivolts and Vgs varies from 0 to Vdd so if you see the current starts shooting heavily Vgs minus Vt times now this is the called transfer characteristics if I do the output characteristics so what do I output characteristics is called ideas against Vds please take it sorry I just draw I will show you the potential between this is Vds which is essentially for output in most cases okay so if I plot ideas versus Vds at different Vgs values I get for a smaller Vgs which is less than Vt a very little ideas which is this current essentially flowing two currents it is which are the two currents I have flowing here one is sub threshold what is the other current in a transistor can flow reverse saturation current of the two diodes whichever is smaller one of them will flow is that correct that reverse saturation current plus sub threshold current is essentially small current the small word is not really small because in newer technologies that is my issue I said you in a one someday that 32 nanometer down the off current may be more than the on current that is why I use mobile from constantly so that power drain come up so that is exactly the reason so off current shown here is very small so analog people are therefore not very happy to go to 32 nanometers or some 0 nanometers kind okay we are worried actually beyond this value initially current rises linearly and for given Vgs minus Vt equal to Vds in a smaller Vgs this point will be faster Vgs minus equal Vg small so Vgs minus Vt small so smaller Vds it becomes saturated larger the Vgs saturation points goes above and above and different characteristics are seen the only problem which I see here is something worrisome and that may be interesting as well I just now made a theory which is said current becomes constant independent of Vgs that is what I say current whatever available I push okay but in real life by monitor I see on all such characteristics there is a slope what does that mean IDS is earlier I said in saturation it is independent of Vds but it seems it is not independent of marginally varying but and another feature which we will see later of course not shown properly curves if these slopes I actually extended in the minus Vds time they all curves meet at one voltage and that voltage we call as early voltage which is taken from early effect in Vgs there is no early effect here but the voltage is still given a name early voltage there I see Vc characteristics at different IV if you extend down at one value of all of them meet which we call as where the base gets punched okay so we say that is early voltage now here also that effect is there now that early voltage are some interesting value because then I can calculate my R0 if I give I am given a early voltage so that is the why I said early voltage now I want to know if I had to represent this few minutes these characteristics as an expression first and then once I know expression I can calculate from expression what if I know IDS function of Vgs and Vds I can differentiate with respect to Vgs so I get gm I differentiate with Vds so I get my R0 okay so I must get a relationship first between IDS and which I first derived but now let us see what can happen if Vgs minus Vt which I defined as please take it Vgs minus Vt I defined as some books defined as Vgt some books like Boyes and Laker's book Baker Boyes and Lee's book defined XS and I defined over voltage my definition why so many years I am talking so I do why I say over over and above which makes sense go of course this is something taken from one of my old colleagues from Sanford so I just use his this Vov so in a book if you are having a different book same names will be a I am using Vov they may use Vx or they may use Vgt whichever name they are giving please remember they are same what is it Vgs minus Vt is that correct now we say if Vov is larger than condition is pds that is vds is small that means channel exists throughout device is in non saturation it is not saturated is that correct device is in non saturation the current can be given as beta which is mu C of W by L Vov times pds minus half Vds is that correct Vgs minus Vt is replaced by Vov this is what mode when the channel exists throughout and that condition can only occur when Vgs minus Vt is larger than Vds once Vgs minus Vt reaches Vds we know it enters saturation prior to this that means channel exists throughout is that okay so in a non saturated mode the current is and why I say it was linear because if Vds is smaller ideas shows linear relationship with so for a smaller Vds you are more linear as you reach towards saturation Vds is not very small so you see curve change slope changing okay you can see from here as you reach here the slope changes and that is the reason it changes so what is the model I am really doing is called two region model one I will say like this the other is sorry like this okay a k region a k region which can a bitch kaka ringa whenever there is an issue two models at that point must have same values okay because that is the fixed value so at the knee point both models should give me same values this is that new value I will say if that I get it I say find two regions it fits now that word is fit it fits and then I use two expressions independently as if I do not know what the other is doing okay that is all modeling people device people do not do it they do not like that if now you say Vov is larger than Vds and ideas is proportional to Vds what is it looks like you can see from this curve again on this region you know slopes are different is that clear a different Vds slopes are different that means it is equivalently saying it is a variable register as long as you are non saturated at any Vgs you have a different registers is that correct linear is the different are so device acts like a variable registers in non saturation as straight as that however if I now increase Vds further I say pinch of me occur at Vov equal to Vds or Vov is become smaller it will further get saturated heavily then one can see from here that expression which I will just show you may I will just repeat again if Vov is equal to Vds my expression writes ideas is equal to beta Vov into Vov minus half at this point this is the expression which is half beta Vov is that correct half beta Vov square so what does this does it have any relationship with Vds no as of now increase that so what is it acting like ideas is independent of Vds what is it looks like output current source so a transistor in saturation acts like a which current source voltage controlled current source Vccs okay it acts like a Vccs okay so Vds versus Vgs I am shown here if Vgs is less than Vt and Vov is you are less than Vds is less than Vov this region is anything below Vov Vgs minus Vt is greater than Vds or Vds is smaller than Vgs minus Vt you are in non saturation which means you have a variable resistor and Vov is smaller than Vds and Vgs is whatever it is for this value you are in saturation which acts like a current source is that clear to you this is what equivalent of a master so now I know if I see a mass transistor depending on the voltages I am using in my circuit I can replace the device by equivalent current source or equivalent resistor depending on the way I am working at if I force the device to always remain in saturation then what will happen the mass transistor will always behave like a current source that is exactly analog region when the device Vovi characteristics show slopes both devices of p channel n channel are in saturation therefore we say the mass transistor analog is always like a current source but at the edges it does not so signal it goes beyond so called non-linear it is may certain okay finally last expression for the day and we will stop here in saturation in real life I say there are slopes ideas beyond saturation point also keep increasing very little way I mean the slope is very low there now if I see it is a very slow low I can fit it this okay I fit the curve and I feared out this is my normal saturation current I multiplied by lambda Vds 1 plus lambda Vds where lambda is called saturation parameter okay which we can derive of course is equal to what we call lambda dash by L where lambda dash is equal to 2 upon Q sub straight under root we can derive all the theory of that when the device is having a flat this from these characteristics can you say if it is very flat what does that mean lambda is 0 0 flat as lambda increases as lambda increases 1 plus lambda Vds factor will start increasing the slope will start building is that clear to you so that technology is more you can see it is substrate dependent it also is depending inversely proportional to length so if I want smaller lambda what should I do you can see I want smaller lambda so I should have larger substrate concentration but what does it will do increase Vt so if I increase Vt what will happen my net currents will go down okay and as I later will say to gm will go down okay however the other possibilities I use longer length devices so is that now clear that in normal good device I want saturation to be very good okay so what should I use device lengths larger than normal technology node for example you are working on 90 nanometers you should not use channel length everywhere use it 180 270 something 2 times 3 times so the analog functions will be better because of lambda will become smaller is that clear to you this is the crux of all analog design how much channel lengths you should use longer channel devices therefore will give you better analog performance any day is that clear to you that is why 0.25 0.35 micron technologies will give you better analog performance and 65 nanometer 45 nanodelecs start killing you how do I get rid of both that is the design