 So, we will continue our discussion on the non classical MOSFET, which I explained what it is last in my last lecture. So, this is module 1 continuing on the carrier transport in nano MOSFET. So, we discussed last time just this is the repetition of what I said last time because I am continuing on that. So, the sub threshold region has an exponential characteristics which already you have heard number of times and on the log scale it will be linear. And we also saw that a sub threshold slope or sub threshold swing is 60 millivolts into n 60 millivolts is k T by q along 10. And we also saw that this n is nothing but an indication of how good is the coupling of the gate with respect to the channel. If delta V g is exactly equal to delta psi of s the coupling is excellent, but because of the sharing of the voltage between the gate oxide and the depletion layer the coupling is not that good. So, what you get will be V g s will be shared between these two and we also wrote since psi s is less than V g s we wrote V g s by n is equal to psi of s. So, then delta V g s by delta psi of s will be n. So, that n is actually the coupling factor ratio of these two psi of s and V g s. And from these two capacitances you can see that the voltage or any change in the gate voltage will result in a change in psi of s decided by the capacitance ratios and that is delta psi of s is equal to delta V g s divided by sum of the two capacitors multiplied by c oxide that is the standard basic relationship. So, from here delta V g s by delta psi of s is 1 plus c d by c oxide where c d is the depletion layer capacitance per centimeter square I call it centimeter square, but per unit area. And c oxide is oxide capacitance per unit area. So, you can see that if the depletion layer with this smaller c d will be larger and then n will become larger that one indication of this sub threshold slope will be much more that is you want n as small as possible ideally one you will never get one, but you will see you can have some special structures where the n can be brought very close to 1 1.01 like that of that order. Now, we want to see is this is the ideal situation where there are no surface is excellent, but in that is you have got interface states interface state density is there between the oxide and silicon particularly if you are talking of I k dielectric it will be unbearable. So, you have to see what will be the effect of that. So, just before going to that I thought I will give a quick glance of what is implied by this surface state density and how it arises. I know that Professor Naukan, but has already discussed some aspects of this may be some other things which I would like to elucidate I will just quickly go through. So, if you take a semiconductor energy band diagram has no states in this between the conduction band and valence band that is a forbidden band. Now, any loss of periodicity or presence of any impurity will give rise to levels between the conduction band and valence band. So, to give a very good example of that donors, donor is a fifth group element which has one valence electron which I have shown here and core is plus charge and this one valence electron is a fifth electron of the fifth group which will be revolving around that which is very loosely bound. So, you can give even if you give very very small energy even at room temperature this electron can be raised removed away from this parent atom and it can go to the conduction band. So, when you put the energy level here corresponding to donor level a shallow level implication is you can give you can raise the electrons from this donor level to the conduction band with very small energy from 50 milliolectron volts for that is called shallow donor. That is why I put it near the conduction band I have a why I am going through this is because we want to see what happens in interface state density. If you take acceptor see here donor the core is plus if you remove the electrons from there is positively charged acceptor core is negative there is a site where plus charges can it can you can look into as vacancy you can assume that there is a plus charge. So, that whole thing as it is neutral. So, if this plus charge is removed from there that is how can it be removed by accepting electrons from the valence band that plus charge is neutralized. So, you will get negative charge. So, here you donate electrons here you donate plus charge and you donate the plus charge by taking electrons from the valence band. So, when it accepts an electron this is plus charge is gone its negative charge. So, that is what you to put it once more if the electron is occupying the donor level it is neutral. If the electron is occupying the acceptor level it is negatively charged because the plus charge is not there that is the thing to understand to understand the interface state density and its impact. Now, let us see what is the. So, these are the shallow donor and shallow acceptor implying very small energies are required for interaction between the donor level and conduction band and the acceptor level and the valence band. Now, if you take a surface I take this surface this is the bulk this is on the surface if I take the atoms are in surface of semiconductor atom does not have neighboring atoms. Usually in the bulk of the silicon the atom is surrounded by 4 neighboring atoms. So, all the bonds are continuous it is stable is staying there. The atom on the surface has got 2 on the side 2 at below it is a very qualitative description, but there is no atom on the top surface. So, above the surface there are no atoms. So, there is an electron which is not forming in the bond completely. So, there is a possibility that you can knock out that electron and take it into the semiconductor make it available for conduction if possible. So, that means that silicon atom on the surface can act as a donor if it donates an electron. Now, what will where we will locate its energy level is the question. See the shallow donor the electrons can be knocked out with very small energy. So, we locate it very close to conduction band. Now, let me ask you if you have if you knock out electrons in the balance band that is the electrons from the bond you have to apply about energy equal to 1.1 electron volts that is why the conduction band balance band difference, but if the silicon which is on surface from there if you want to knock out electrons you may not required as much energy as this required for the electron band gap, but it will not be as small as what you require for this pentavalent material. So, it is actually this level corresponding to this donor will be very close to balance band it is not there in the near the conduction band implying it will require large energy much more large energy than the shallow donors. They are called deep donors with the far which are far away from the conduction band. Similarly, you can argue that for the balance the acceptor type deep levels can be on the upper band from the same argument. So, why I am bringing out donor and acceptor is the surface atom can either donate an electron or it can take an electron to complete another bond. So, it can act as a donor or acceptor if it is acting as a donor that donor level can be located near close to balance band somewhere in between the mid gap and the balance band. The acceptor level on the same argument will be on the upper half of the band gap or closer to conduction band. Now, one more thing that I want to point out is there are about 10 to power of 14 to 10 to power of 15 atoms per centimeter square on silicon that means large number of atoms are there. Now, let us go back to the shallow donor or shallow acceptor theory. If there is shallow donor you do not put that donor level as continuous level you put it by dotted lines what is the implication of that? The implication is very simple it is very dilute in silicon the number of atoms per centimeter cubed in silicon silicon atoms are about 5 times 10 to power of 22. Now, the donors which are there in dopants when you dope them are 10 to power of 18, 19, 20 about if it is 10 to power of 16, 17 etcetera it is very very dilute out of 10 to 22 atoms of silicon there are 10 to power of 16 or 17. So, these two atoms do not see each other. See a conduction band was a band of levels why? It was interaction of the energy levels of the silicon atom. How the interaction comes up? The spacing between the atoms is very very small from 5 electrons. So, that is why the from Pauli's exclusion principle you can say that no two energy levels can be at the same level corresponding that. So, it splits up. Similarly, if the donor level now you put it dotted line saying that it is not interacting with each other they are isolated from each other. Now, if you raise that donor levels donor concentration to about 10 to power of 20, 10 to power of 18, 19, 20 and all that then number of atoms are very large and number of distance between the atoms donor atoms is reduced there can be interaction between the wave function of these two atoms. That means the two electrons trying to occupy the same energy level, but they are not able to occupy the same energy level they split up. So, this donor level which you put it as a donor level one level with no longer be one level it will split up into a band of energy. It will split up band of energy and it can even merge with conduction band. Similarly, the p acceptor levels also can split up into a band. So, now going back to this theory of surface states you would have seen already. Now, I have argued out that if it is one atom there will be one acceptor level and one donor level. It can go either way it is an importer it. For example, if you put gold into silicon it has one single electron it can either donate or accept. So, in the same way surface states electrons can go and occupy or it can donate that electron that is available there. So, now that is one atom has got one acceptor one level and one donor level. We said donor level will be below and acceptor level will be up because of we argued it out. Here donor level is up acceptor level is below. There deep levels will have donor levels down acceptor level is up. Now, if there are large number of the surface atoms like this each of those donor levels will sit up into a band. See donor levels are below each of those acceptor states is split up into a band. So, that is why you put these acceptor states closer to the in the upper half or nearer the or above and donor levels below and all of them are not one levels several levels split up a band of energy levels because large number of surface atoms are present. Each one of those surface atom corresponds to one level here one level here. Now, other thing that you would see here is these acceptor levels will be actually occupying from top to bottom completely spread out. But when you come down here there are large number of donor atoms which are compensating that. So, you can say that predominantly there are acceptor levels up to a point predominantly there are donor levels in this region between the level called E naught and E v they are all donors between level E naught and E c they are all acceptors. I hope that is understood. Now, this argument whatever I have discussed I put here what is the charge state of these they are not showing the intentionally doped shallow donors and the shallow acceptors. This particular material if I put the Fermi level here what does it mean it is a p type material. And I have taken the doping such that the Fermi level actually coincides with the E naught level which separates the acceptor level and donor levels. So, this E naught level is called the neutral level. Why it is called neutral level? In this particular case which I have shown the Fermi level is coinciding with the neutral level with the level E naught. What is the meaning of that? What is the charge state of these acceptors? From the definition of Fermi level if you recall all the levels below the Fermi level are mostly occupied the levels above that are not occupied mostly. In fact, if you go to the 0 degree Kelvin all the levels below the Fermi level are occupied all the levels are empty. But at room temperature if you go slightly deviation but for qualitative understanding you can say almost all the levels are occupied below all the levels are not occupied. So, if you take that abrupt model for the Fermi distribution these are all not occupied. Now, go back to what I have mentioned all the donor levels are occupied because they are below the Fermi level. So, what is the charge state of that? Just go to this and see they are occupied means it is neutral. So, these levels are neutral because they are all occupied and the acceptor levels are not occupied by electrons which means that is neutral. If it is occupied it is negative the charge. So, in this case you call this at neutral level because if the Fermi level is coinciding with that level the levels below that which are donors are occupied. So, it is neutral levels above that E naught which are acceptors are not occupied. So, that neutral. So, charge net charge on the surface is 0. So, energy band diagram should be flat. Now, let us go further down. I am going to be slow because this is very though you had gone through some of these quickly I will see go through that. Now, I will take a situation where see here it was a situation where it is a p type Fermi level doping adjusted bulk doping adjusted such that Fermi level is coinciding with the E naught. Now, I go to a n type material doping is something like 10 to the power 15 or 16 Fermi level is close to the conduction band decided by the shallow donors. Now, surface is free there are no oxides there are very large density of states. So, if it is like this all if I draw the energy band diagram like this like this now all the levels below E f are occupied. What will be the charge state of this? Take a look at these levels below E naught they are donors they are occupied. So, the charge there is 0. Donors charge is occupied if it is donated it is plus charge it has not donated electron is occupying it is neutral. And these are acceptors electron is occupying that means electrons it has accepted electrons that means it is negative charge. So, on the surface of the semi-conductor there will be net negative charge donors are donating acceptor accepted. Where as this electron come from? That is come from semi-conductor that means this n type silicon removed some of this electrons from the nearby region from surface and given it to this acceptors donated which means the surface the region nearer the surface will be depleted of donors. See these are the donors plus charges these are depletion layer the electrons from here have gone to this surface. Surface is negative charge and there are donors which are positively charged. So, once there is a depletion layer like this electric field is in that direction. So, what happens to the energy band diagram? I hope you are familiar about the energy band diagram. So, electric field is in this direction that means the band bending takes place see if there is no electric field band energy band diagram is flat. Wherever there is a depletion layer there is an electric field and the energy band diagram will bend like this. So, up to this point there will be band bending and since electric field is in this direction energy band diagram will bend upwards like that. How do you figure out that? When the electric field is like this you have to spend energy to take electrons from here to here. So, if the electron is here it is at a higher energy level that is a energy band diagram bend here that is why I drawn like this. Now there is some important thing that you have to understand here before we go next into the argument on some of these things is what happens see the because of this band diagram because of the electric field there is a potential drop from here to here. That means this is neutral surface there is a potential equal to psi of s from here to here if you go there is a potential which is negative with respect to this point that is psi of s that is surface potential potential of surface with respect to neutral region. And that potential is actually equal to the voltage drop across the depletion layer which you know is on the first order theory q and d W d square by pi sub square f into 0 that is well known thing. Now charge in this region is n d is number per centimeter cube and computing charge per centimeter square. So, per centimeter cube if it is n d cube multiplied by W d I get total number of charges per centimeter square into cube. Now what will be a charge on the surface now that we will come back little after discussing one particular point here. I mentioned that the Banji band diagram will bend what will be the state of a pair for the Fermi level. Fermi level here was decided by the donuts concentration in the bulk. So, in this region where it is neutral the energy between the conduction band and the Fermi level remains the same as before it does not change. But as you move to the surface because it is depleted it becomes less n time. Let us have been donated that means the Fermi level will be further away from this point. Two points here the Fermi level is decided by the doping concentration. Here the Fermi level difference between the conduction band and Fermi level depends upon how much band bending has taken place. It is less n type here you can think of that less n type means Fermi level distance between these two is larger. Other point that you must note is the Fermi level is horizontal it is not bending. If you recall some of the discussions or earlier models the Fermi level will be continuous there will not be any deflection or deformation in the Fermi level if there is no current flow. It is a thermal equilibrium situation there is no current flow. So, that is why F is constant. So, all that happens is band bending takes place upwards this remains constant. What happens to neutral level? See if you go back to this the neutral level is assigned is fixed for a given material. If you take silicon the neutral level is one third of the band gap here. If this is the band gap energy E g this is E g by 3 it depends upon the particular material. If you take silicon it is about E g by 3 that is E naught minus E f. If you take Gaia-Marsnet it is also almost that if you take germanium it is very, very close to valence band. In fact, 0.66 electron volt is energy band gap of germanium and E naught minus E v will be 0.06 electron volts very close to valence band. Depends upon the material let me not get into the details of that we will opt in that other ways. So, that location of E naught with respect to E v is fixed E v or E c is fixed. So, if there is band bending taking place like that on this band bending at second place here formulae has the gap has changed from here, but the location of E naught with respect to valence band is same on the surface. Which would mean see if it was like this here because of the charge transfer from the semiconductor to the interface the formulae has this has moved up. So, the distance between the formulae and the E naught has reduced on the surface. So, the band bending is such that charge here is equal to the depletion result. So, all this discussion is to go into what happens further. This is n type. Please remember it always moves closer to the neutral level. So, that the charge here is equal to the charge in semiconductor. See the other case of p type where if the formulae is below E naught, what would happen? These are in neutral level is above the formulae. Now, the levels below the neutral level are donors. That means these donor levels what happens to these donor levels above the formulae level here they are formulae level. So, you can see donor levels between E naught and E f are not occupied levels above E f are not occupied. So, these donor levels would have donated electrons to this p type material. So, p type region it is made negative it will be depleted plus charges. See donor electrons have gone and neutralized the p type plus charges. So, there will be depleted layer. In the previous case n type has depleted it is p type has depleted. So, you have got depletion layer once it is depleted because it has donated electrons surface is positive and this is negative electric field is in this direction. You can see band diagram is up like that to the right because field is in that direction. So, ultimately you can see that the band bending takes place and the voltage drop in the depletion layer is such that net charge in the surface there is equal to net charge in the depletion layer. This I will give one more instance which I will not discuss this you at your leisure time you think of that what will happen this is not equilibrium situation. If there were surface states were not there I have a situation where formulae level is above the E naught. That means these are the acceptor states which are occupied there. What way the energy band diagram will bend? What will happen to the semiconductor surface? That I think I will let you to discuss and go back. In fact, what you have to realize is the levels below the formulae level are occupied that means these acceptor levels are occupied that means it has taken electrons from the semiconductor that means it has given more holes to the semiconductor that means it will become more p type in the semiconductor. If it has become more p type in the semiconductor the formulae level will move and the surface will move closer to the balance band. More p type means formulae will be closer. So, here it will be bending like that. You will have an accumulation layer. Formulae will be flat, but this will bend if formulae level and valence electrons are closer together. It is not the formulae level it bends it is the valence electron valence band which moves up. So, you will have accumulation layer that you discuss. I just give a hint on that it will go to accumulation, but please understand that it is not formulae level it bends it is the either it is the conduction band and valence band which bends. If the valence band bends more close to the formulae level it is accumulation. It moves away from the formulae level that is depletion in the p type material. Now, get down to the I think I have taken quite a bit of time on this particular thing, but I in long run you will see it is worthwhile to analyze any situation. So, now I take a situation where I apply voltage. This particular case that I discussed was for me there was no voltage oxide there was no voltage applied. Now, when I apply when I oxidize it then many of those interface states will be reduced, surface states will be reduced. Now, when you are, but still some surface states will be there depending upon whether it is thermal oxide, high key dielectric. Now, when I apply voltage plus here it is getting shared between the oxide and silicon even when you though there is no interface state. So, v oxide you can see now electric field is in the direction from the surface metal to the interior. So, wherever there is plus to minus if you go plus to minus energy band diagram will be like that because it is difficult for electrons to move to the minus direction. So, you can see the energy band diagram bends up there is voltage drop in the oxide. You will see quietly in textbooks they put the diagram here going up like that imply there is voltage drop across oxide in that direction and semiconductor here also bent like this because plus on the left hand side sorry plus on the left hand side plus minus v oxide and then plus minus v silicon. And what is v silicon? If you are calling v silicon it is history, physics people always call it as psi. We call it as v silicon drop in silicon. So, they call it as psi of a surface potential is nothing but the potential of surface with respect to the neutral region. So, we have this diagram. Now, you can see the E f in the bulk interior here neutral region here is decided by the doping p type material doped. How much doping that decide this distance? Now, on the surface the distance between the E f and the valence band edge is more than this one because it has got depleted. When you say depleted you can very casual way of telling is it is less p type because the depletion is not abrupt it becomes less and less holes are present very few holes are present here. So, less p type. So, the distance between the permeable and this increases, but the neutral level location is always fixed with respect to valence band edge. Silicon this is E g by 3. So, how much charge in this interface density is present depends upon how much is the distance between E f and E naught is. How much is the distance between these two is depends upon how much is this E f and how much is this psi of s total thing. See, I have here only the E f minus E v. Now, here E f minus E v will include this psi of s and that E naught that distance. So, the charge here will be how much will be charge here will it be negative or positive because these is the neutral level. I hope I am not confused to that is there is a band bending and E naught is below the Fermi level. So, these levels which are above E naught and within that Fermi level are acceptors they are occupied there will be negative charge and what will be charge in the depletion layer that is negative. So, the p type depleted negative. So, there will be negative charge here, there will be negative charge in this surface both are negative. If there were no interface state density state density the to all the charge would have been in the depletion layer. Now, part of it goes into this is occupied by this. So, whatever charge in the gate oxide is present is shared by this interface state that is Q D i t. D i t is defined as number of states per centimeter square per electron volt. I hope you do not know about that because I did not mention about that D i t implies how many states are there number per centimeter square per electron volt. So, D i t if it is uniform throughout this distance E f minus C v is the energy per electron volt there are D i t number per centimeter square and energy gap is difference between the two is E f minus C v. So, if you I do not know where I am pointed out there it is left off, but here D i t number into number of levels gives you total number of charges occupied Q gives me the charge. So, Q D i t into that difference that gives me the charge. Now, that is not equal to depletion layer charge because this charge negative charge corresponding plus charge is here on the gate oxide. And there are depletion layer charges which corresponds to this band bending corresponding to how much is the depletion layer width. Q N A into depletion layer width that is the number of charges per centimeter square in the depletion layer all that you are knowing that the charge here beside by the D i t and the gap between the two that negative and the recharge in the depletion layer. Now, if the gate voltage is changed by delta v g that delta v g will be shared between oxide and also psi of x that diagram I have drawn here I have increased that v g by delta v g. So, due to that there will be delta v oxide additional was voltage drop delta psi of s silicon voltage drop changes by delta v silicon or delta psi of s. Fermi level does not change, but this band bending which was originally solid slightly gets bent more by amount equal to delta psi of s corresponding to extra drop across the silicon. Conduction band bends down balance band bends down by additional delta psi of s. Now, watch very carefully Fermi level has not shifted that remains with respect to neutral position same it is flat, but the neutral level what was here is shifted down by an amount equal to how much this conduction band has moved down. Both conduction band and balance band moved together like that it has deflected by an amount equal to delta psi of s. So, that has deflected by delta psi of s the E naught moves by an amount equal to delta psi of s, because E naught is always defined with respect to the balance band edge or conduction band edge both are same. In the sense if I in order to avoid confusion I will say E naught in case of silicon is one third of E g. So, earlier with respect to balance band it was here one third E g above that. Now, the band bending the balance band has shift moved down by delta psi of s that means this level has moved from here to here by an amount equal to delta psi of s, because when the balance band edge has moved by delta psi of s this level also would have moved down by delta psi of s, because it is written with respect to balance band edge. So, this has moved with respect to this by delta psi of s. What happens to charge on the surface? Originally, but the Fermi level has not moved, because that is fixed with respect to the bulk itself. So, there will be additional charge coming because this neutral level has moved down further down with respect to the Fermi level and how much it has moved is delta psi of s. So, what is the charge now extra charge coming from the interface state density q d i t number of levels multiplied by this shift in that extra charges coming out because due to this shift. So, you will have extra charges coming from the interface state density density equal to q d i t into delta psi of s. And because the delta psi of s is an additional drop in silicon, depletion layer width would have moved by an amount equal to corresponding amount. So, there will be additional charge in the depletion layer width. So, there are two things happening, because delta q in the gate you have increased delta q in the interface state density has increased by this amount q d i t into the shift in this position delta psi of s. And the charge here is multiplied by whatever original charge was there by delta q d. Now, let us see I can define a capacitance corresponding to this. Always you remember delta q by delta v is a capacitance and delta v is the voltage with changes. So, here the voltage changes is the psi of s. So, when psi of s changes by delta psi of s the charge changes by delta q i d which is given by q d i t into delta psi of s. So, you can think of it as a capacitance in the equivalent circuit as c i t which is delta q i t by delta psi of s which is q d i t. It is a constant because we have assumed that the q d i t is same everywhere across the band gap. So, you can write c i t which I think I have already pointed out in some in a different manner may be is delta q i t by delta psi of s. Now, what will happen? Originally, interface state density was equal to 0. We found out the substrate should slope s equal to n v t long 10 and n was equal to 1 plus c d by c coxy oxide. Now, what will happen is when you change delta v g the charge change was only across the c d because of the widening of the depletion layer. But now because of this change in the potential here not only the charge in the depletion layer changes charge in the interface state density also changes that gives rise to an additional capacitance which you put across this. So, delta v g s by delta psi of s now instead of 1 plus c d by c oxide you have got 1 plus that effective capacitance which is sum of c d plus c i t. So, you have got some threshold slope now given by 1 plus 50 millivolts into n that is 1 plus c d by c i t by c oxide. So, what I am trying to point out here is I did not want that to come there. What we are trying to point out here is interface state density will increase the substrate should slope to go back and see when you say substrate should slope this is 1 60 millivolts per decade that is this will be 60 millivolts when the current changes by 1 decade that is n is equal to 1 ideal case. Now, if there is no interface state density it may not be 50 millivolts it may be 70 millivolts, but if there is additional interface state if there is interface state density you will have additional capacitance c i t that will not be 70 millivolts it may be 80 millivolts 90 millivolts 100 120 depending upon how much is your c i t. How much is your c i t depends upon the d i t and d i t depends upon the quality of the oxide in the case of thermal oxide the d i t can be very low people both of even as low as about close to 10 to power 10 per centimeter square per electron volt, but in the case of high k dielectric it can go 10 to power 11, 10 to power 12 unless you use additional passivation techniques. This please remember that this will be very important parameter when you go to non-classical non-silicon FEDs. Now, a quickly I run through some of the things before I go into some more details this is been done thoroughly by professor now counterpart that is short channel effects. I have only one slide which just completed for the sake of completion. When the channel length becomes small you can see the depletion layer of the drain region can encroach encroach into the channel in the conventional structure. So, the width of the depletion layer depends upon how much is the total voltage across the drain is total voltage is V V i plus applied voltage I call it as V d that is equal to q n a w square by this is a standard formula. So, what depending upon voltage you have got depletion layer width. For example, if I have one volt total voltage then doping is 10 to power 15 per centimeter square to give an idea depletion layer width will be one microns. So, if I have a channel length of one micrometer the whole channel will be depleted of carriers which would mean that the entire channel is dominated by the effect of drain and the gate has no control. So, you want to make it you increase the doping concentration to do that that is why you go to 10 to power 16, 10 to power 17, but you know that that has other repercussions like reduced mobility and increased threshold voltage. So, you reduce the gate oxide thickness all those things are there. Now, so that is if the channel length is 0.1 micron even if you go to higher doping concentration this can deplete all the way or at least partially. Therefore, I will not get into that discussion due to that charge being taken care of by the drain you will be able to invert the channel with lower gate voltage that is threshold voltage will be lower that is one of the short channel effects which I am not getting down for detailed discussion. So, one is punch you through depletion layer reaching other one is the because of the depletion layer occupying the channel here this charge does not belong to the gate, but to the drain. Therefore, you get you can invert the channel with lower gate voltage that is the threshold voltage gets reduced that is another short channel effect. Both the things can be reduced by by increase the doping of the substrate. Now, I just wanted to bring in this the other effect drain induced barrier lower this also has been discussed. I just drawing what happens here in the substrate shoulder region. In the substrate shoulder region there is no drop across the channel because the there is no drift current. So, potential if you see V B I no drop V B I when you apply voltage to the gate below threshold what happens to the barrier rate it gets reduced because plus voltage apply plus minus plus minus. So, this it gets forward biased. So, V B I gets reduced. If the depletion layer width is only here the first curve here you will have a reduced barrier height. So, V B I by psi of s flat there and going like this. So, this barrier height reduction is completely controlled by the gate. Now, if the depletion layer occupies a good portion of the channel shorter channel instead of saying shorter channel I will say lower doping. This is the depletion layer width encroaching. So, this edge of the depletion layer here. The depletion layer edge is here if increase the voltage. If I increase the voltage it is here. If I increase further if it is reaching very close to the source end its merges it will reduce the barrier height here completely. So, what was getting reduced by psi of s by V G s gate voltage. Now, will be combined if it of the gate voltage and also drain voltage which would mean the reduction in this barrier height which gives rise to your current is controlled by gate and also the drain. So, the gate loses it does not have complete control over this reduction part of it is taken care of by the drain region. So, if that means if I change the gate voltage now it supposing that way it was changing by 60 millivolts per decade. It was changing by 0.1 volt it was changing. When I change the gate voltage now by 0.15 volts the size will not change by 0.15 volts because part of the change takes by drain. So, because of the combined effect gate will not be if you reduce the gate voltage from the threshold voltage. If you gate has got a good control it comes down to 60 millivolts per decade in ideal case. But because of short channel effect threshold voltage has got reduced that is why I have drawn it here gate has lost control because the drain also controls that barrier reduction. So, the reduction in the current is not as much as you get in this case. So, the sub threshold slope will not be 60 millivolts it is worse than that. If it were earlier let us say 90 millivolts per decade in the short channel it may be 100 millivolts per decade or 120 millivolts per decade depending upon how much is the encroachment. So, these are overcome by this encroachment can be practically reduced in present day devices by providing a lightly doped region in the N plus region called extended drain region. So, this is not so important, but I just pointed out this for continuity. Now, for continuity I pointed it out. Now, the last one that I wanted to say here is this sum up short channel effects. I know that I am re doing stating what was done earlier you need to increase substrate doping constructions to reduce the short channel effects that results in carrier mobility degradation and increase in threshold voltage. To reduce you have to reduce the oxide thickness to overcome this effect, but this of course leads to get leakage current when you go to nanometric level, but to overcome that we have discussed already that high k dielectric is used, but when you use high k dielectric you know that there are lot of interface state density effects. So, you encounter those effects. So, you need to have some classification techniques also done there. Now, when you also use high k dielectric you must ensure that this energy between the conduction band of silicon and the conduction band of the dielectric must be high enough. If it is not high enough the electrons which are here can climb up here get into the dielectric. In the case of silicon they do get into that region when you go to large voltages large electric fields, but today you do not have to worry much about that in the case of silicon because you talk of low voltages. So, fields are not that high or your velocity energies are not that high energy is given to be v small energies are not that high. So, but still if this height is not sufficiently large you can have electrons injected into this dielectric. You can consider them as hot electrons and once they are injected into the dielectric they can act as additional traps in the dielectric that can change threshold voltage that can change give rise to reliability problems. So, what you have to take care when you go for dielectric materials is not only that it has could have good band gap it must also have this gap between the conduction band and that that is the chi you know this depends on how much is the distance between this conduction band of the dielectric and that is say the 0 level that is say 0 level is there between that and this how much is the gap. How much is the gap between the conduction band and the 0 level silicon it is 4.05 electron volts and you in the case of oxide this gap is 3.3 electron volts the gap is 4.05 that is 0 level and this gap is 4.05 minus 3.3 electron volts. So, this gap is more supposing this is also 4.05 this is 4.05 band gap may be large, but the inverted electrons can get passed through the oxide there will be gate current and or they can get trapped in the oxide. These are some of things one has to remember when one gets into when one thinks of going to high-key dielectrics. Now, the effects of short channel I just come to very close to end of the thing that is there are different effects coming up how many minutes about 7 minutes I will just at least few things I will do now. So, this is the one which I want to discuss in more detail I need one more session on this. So, but I will just get into at least what is this meaning you remember in the beginning itself we said the first order theory of MOSFET we assumed that ohms law holds good. When we said ohms law holds good we said v is equal to i into r which again means that the velocity of carrier is proportional to electric field velocity is equal to mobility into electric field. Now, if you take a look at the silicon MOSFET or any other MOSFET for example, that is the way the charge distribution is charge is more here charge is less there because of voltage drop across the channel. And because of that because the because of this change in charge you also remember you wrote the current is i d is equal to q into v q is charge per centimeter square. When you when you move from the source to the drain we took q is varying we took velocity is equal to mu into e electric field that way you derived the equation. Now, you take a look at this when you move from here to here q is falling what would happen to the velocity current is q charge into q into q n. Into v n is decreasing velocity must decrease. So, if I am saying v is equal to mobility into electric field what happens to the electric field? Mobility is constant if I take electric field is increasing. So, what you say is as I move from here to here evidently the electric field is not constant electric field is increasing. Now, what will be the electric field? If I take an average electric field you can say that what is the drop across the channel when it is opened here the channel potential is v g s minus v th threshold. So, that divide by channel length is the electric field. So, average field depends upon how much is the voltage drop across the channel average voltage drop. So, average voltage drop is always channel drop this v g minus v th threshold we saw after saturation. So, that is the average field. Now, let us see what happens to as we keep on reducing the channel length for a given threshold voltage average electric field actually for a given threshold voltage let us say this is not changing for shorter the channel length more is the electric field average electric field. Even at this end if you take the average electric field everywhere it will be high even though it is varying there will be high electric field. So, what happens when the electric field is high? Can you take mobility is equal to or mobility is constant can we take? What is velocity? The velocity is actually force divided by mass it is acceleration into time. So, acceleration into time. So, as the electron is accelerated it experiences collisions that is it gets scattered. So, at that when the collision takes place it comes back to rest it accelerates collides it accelerates collides. So, the velocity the average velocity you can put it as the force there is the acceleration there is force divided by mass the force is q into e average field between collisions divided by mass into the time of collision. So, that is called mobility to electric field. Now, when the electric field goes up acceleration goes up. So, time between the collisions decreases. So, if the time between the collisions decreases q tau by m decreases. So, when you write nu here q tau by m is the mobility. So, you have got the mobility which you assumed constant is no longer constant it decreases as the electric field increases because the tau falls or to put it in simple terms it is a plot velocity versus electric field. The velocity because of reduction in the collision time for an average electric field the velocity does not increase linearly with electric field it goes on increasing like that then saturates. So, I will not go beyond that this saturation why it comes we will just discuss very briefly next time. So, today I will stop that ultimate velocity of the electrons is limited by the saturation velocity. So, that velocity is about 10 to power 7 per centimeter per second for silicon and also even for gallium arsenide and for many materials and the electric field at which saturates is about 30 kilo volts per centimeter. So, with that I think I will close down today we will discuss more details in my next presentation all these aspects starting from this slide. Thank you.