 So, we start our third session on the carrier transport in nano MOSFET devices and MOSFET non-classical MOSFETs. So, we continue on some effects of short channel lengths on carrier velocity. We just began on this last time. The three aspects we will be discussing will be one is the velocity saturation effect on the drain current and then the a quick look at what is the implication of ballistic transport and velocity overshoot effect, whether we should worry about that that also we will discuss. For silicon, we will see how good is important is the velocity overshoot is there. Then we will also discuss injection velocity, which is ultimately which is the one which controls the drain current in short channel devices. Now, before we go into that velocity saturation, we take a quick look at the velocity versus electric field characteristics of the carriers. I use the word carriers because this sort of curves hold good both for electrons and holes. What I am showing is mainly for electrons right now. Similar argument can be extended to the holes also. Now, in thermal equilibrium when you do not apply an electric field, you know the free electrons are moving in random directions because they have the thermal energy. They move with the thermal velocity which is about 10 to power 7 centimeter per second. Now, even though they are moving there is no current flow because they are in random motion and each of them cancel in all the directions. So, there is no net current flow. Now, when you apply an electric field at the electrons have an additional component in the direction of the electric field. That is, it moves towards the plus thermal opposite to the electric field direction. So, it has a component of velocity. So, the carriers move in that particular direction directed by the electric field. If it is holes, it will move in the direction of the electric field plus to minus. If it is electrons, they move from minus to the plus. So, now this component is superimposed on the thermal velocity. So, you get the current due to the velocity of these electrons which can be expressed in terms of electric field and collision time. What is collision time? When the electrons move in the particular direction, they get scattered by the lattice atoms. They get scattered by various effects. So, in a time tau m, there will be collision. So, the velocity or I am sorry the electrons get accelerated by this force q into e. q into e is the force on the electrons. e is the electric field divided by m star. m star is effective mass of electrons because you talk of moment of electrons in the crystal not in the free space. So, this is the acceleration. Acceleration into time of collision gives you the velocity. Acceleration into time gives you the velocity of the carriers at the time of collision. So, that is the maximum velocity that it gets. We can say there are different electrons which will suffer different accelerations or different collision lengths and collision times. So, what we talk of this velocity will be an average value for all the electrons put together which gives rise to current. Now, this thing whatever we write here, we usually express as mu into electric field where mu is actually q tau q tau m divided by m from here. I just pull out this electric field term out and call this term as the mu. That is the mobility. Now, you can see here there was an assumption that when you change the electric field, the collision time was not changing. So, that is why this mobility is constant. So, v is proportional to electric field and velocity field characteristic will be linear when the electric field is low. At higher electric fields, when the acceleration is more, the time of collision will reduce because the distance between collisions is same, the time of collision or collision time will get reduced. So, that means the mobility actually gets reduced, but electric field is increasing. So, the velocity field characteristic does not increase linearly, but it becomes nonlinear. This is because of a collision time reduction. So, even though electric field is increased, it does not increase, velocity does not increase as much as you think it should. When you go still to high electric fields, you can see that the velocity will become higher and higher and it will become comparable to the thermal velocity which is there even due to thermal energy. So, total energy of the electrons now becomes thermal energy plus this additional energy which is absorbed from the electric field. Now, the kinetic energy which you call it as half m v squared is related to the temperature or k T. So, once you say that the kinetic energy has gone up because of the thermal energy and the energy absorbed by the electric field. In fact, finally, the kinetic energy, net energy is this one due to the energy absorbed from the electric field. So, you can say that it looks as if the temperature of the crystal has gone up or temperature of the electrons has gone up. Even though crystal temperature is room temperature, the electron energy is higher by this amount. So, you say that they are hot electrons. Electrons are not really hot. It looks as if they are hot because their energy is much more than that of thermal electrons. So, when they collide. So, you can see now these hot electrons have addition and energy over and above the regular thermal energy. So, when they collide with the lattice atoms, they excite the lattice into a new vibrational mode which has large cross section for intercepting the more electrons. In other words, what I am trying to point out is these electrons when they collide with these lattice atoms because they have a large energy. They create extra vibrational energy in the lattice. So, if there is extra vibrational energy, the chance of collision of these electrons becomes more. For example, if the atom is moving like that, you know the cross section of that movement is much larger. So, it what happens is almost all the electrons collide and lose the energy. So, the maximum energy that is acquired by that is equal to is decided by that velocity that is achieved by that. That velocity is equal to the thermal velocity or close to that. See, when the velocity has become equal to thermal velocity or close to that, the energy is almost much large compared to thermal energy. So, at that point what is the velocity acquired that is close to thermal velocity. It is not exactly equal to thermal velocity, but very close to that slightly less than that. Still we can say V is actually that velocity at which the maximum velocity is reached is the saturation velocity. So, what happens is if I keep on increasing the electric field initially linear, non-linear because of the collision time decreases. Finally, velocity saturates because all the energy that is acquired within that collision time is lost to the lattice. So, within that time it is whatever velocity is acquired is very close to the thermal velocity, which is the limiting velocity of the electrons. It cannot go beyond that because the entire thing is absorbed by the lattice. So, this is the principle of velocity saturation is basic semiconductor devices. This discussion comes for completeness sake. I just brought this into. So, what happens in the case of MOSFET? So, remember that linear, non-linear saturation is the field and that saturation velocity is about 10 to the power of 7 centimeter per second for electrons or whose it is about 6 times 10 to the power of 6 centimeter per second. Now, if you take a look at the velocity saturation effects on the short channel MOSFET, what will be the electric field? The electric field along the y direction that is from the source to the drain in that region there is a voltage drop equal to gate voltage minus threshold voltage that is the channel potential, which we have discussed earlier. So, voltage drops across the channel from between the drain end and the source end once saturation current has reached is this VGS minus V threshold. So, you can you can call or you can think of how much will be the electric field that is whatever voltage drop is there divided by channel length that is average electric field. So, now you will see as the short channel length becomes shorter and shorter electric field becomes larger and larger. That means, if the electric field is larger than about 30 kilo volts per centimeter for electrons the velocity saturates to 10 to the power of 7. So, what you are telling is this 30 kV per centimeter can be reached very easily. See for example, even if you take a number like 1 micron channel length if the volt VGS minus V threshold is 3 volts 3 volts divided by 10 to the power of minus 4 that is 30 kV per centimeter. So, even there you can have velocity saturation you can see if the channel length is smaller even if the threshold VG minus V threshold is 2 volts 1 volt you will have velocity saturation. So, when you go down to 0.1 to 0.2 micron channel length definitely there will be velocity saturation and that value will be 10 to the power of 7 per centimeter for electrons and whole velocity saturates at slightly smaller value because of its higher effective mass. Now, let us see how this will change the drain current. If we are operating in this region where the velocity is proposed to the electric field the Ohm's law holds good and there we said it is a square law. IDS is equal to VGS minus V threshold whole square into some constant which depends upon mu C oxide W by L. Here let us see that. So, due to high electric field I just take a look at the drain current in the scaled down MOSFETs that is short channel length everything reduced in size due to the high value of electric field in the y direction that is along the channel length the velocity of carriers will be saturation velocity. Now, we have assumed that electric field is very high everywhere that means velocity saturation will be there everywhere. So, if you take the current the source end itself whatever current enters here will flow through the thing. So, we have written this formula the drain current is width into charge into velocity W Q into V W is the width of the channel Q is the electron charge here 0 is implying that x or y equal to 0 here. So, Q n so W Q into V sat is actually the drain current this is the velocity saturation model very simple. Now, if you write substitute for the inversion charge inversion charge is C oxide into V G minus V threshold that we have put here. So, the drain current now you becomes W C oxide into V G minus V threshold into V sat. Now, T is at this point the drain current does not depend upon the channel length. In the in the low field case you remember that the drain current is W C oxide mu divided by L into V G minus V threshold whole square. Now, here it is not square law it is I D increase a linearly with V G minus V threshold and it is independent of the channel length. Because the current is the because the velocity is saturated velocity no longer depends upon the channel length. So, it is independent trans conductance defined as delta ideas divided by delta V G s. When you differentiate that this term goes off and you will W C oxide into V sat. So, you can see couple of things here to be noted your drain current depends on the channel width of course, but you do not want to make this large the maximum that you can get will be decided by the saturation velocity. And of course, larger the C oxide larger will be the current that means thinner the oxide more will be the current. So, it will be the trans conductance with all the integrated circuits you would like to have best trans conductance. That means, the velocity should be highest and you get why do you need best trans conductance because the driving capability for a given change in gate voltage is maximum if trans conductance is maximum. So, I D and G m are independent of the channel length that is one thing to be noted out noted. The transit time that is the time required for the electrons to move from the source to the drain that is length divided by velocity. And the transit time will control how much will be the cutoff frequency that is 1 by tau t will be the omega t f t is 1 by 2 pi tau t. So, high frequency transistors will have of course, the shorter channel length and because of shorter channel length you will get faster field of operation not because of improved velocity. Now, let us see whether you get the drain current independent of channel length when you go to smaller channel length. There is some experimental result which is available as I reported in one of the thesis in the transferred university PHD thesis. I think that this results are given by Intel very interesting. You see you keep on reducing the channel length, excessive use of the channel length, y axis gives you the saturated drain current that is after the drain current is large enough. So, as you keep on reducing the channel length this is for different MOSFETs shorter channel, shorter channel, shorter channel. You can see the drain current keeps on increasing because you are not in the velocity saturation region. The drain current is in this region is controlled by mu C oxide W by L into VGS minus with threshold whole square. So, because it is inversely proportional to channel length the drain current keeps on increasing where you reduce the channel length. This is for the P channel and channel device the bottom curve is for the P channel devices. But, once you go down to channel length which are something like 0.2 microns and below these are all experimental points they are all scattered, but on an average you can see that it is reaching constant. It is independent of the drain channel length which almost shows that this is correct. Drain current is independent of the channel length. You can see that it is independent of channel length. Both N channel MOSFET and P channel MOSFETs show the same characteristics. There is a difference in these two numbers because of the difference in the mobility and the difference in the saturation velocity you can see. But, if you calculate this I think you have to watch very carefully. These devices have an oxide which is about 1.75 nanometers around that for these channel lengths or 1.5 and I have taken this number 1.75 and corresponding oxide capacitance per centimeter square is 2000 nanofirons per centimeter square. I am trying to find out if I substitute this oxide and if I substitute with this V velocity saturation do I get this number for electrons. See what we are telling is if there is velocity saturation I can substitute let me take VGS minus V threshold is 1 volt. C oxide is 2000 nanofirons per centimeter square which is 2 into 10 to the power of minus 6 farads per centimeter square and V sat is 10 to the power of 7. So, 10 to the power of minus 6 into 10 to the power of 7 that is 10. 10 the 2 is there. So, this is this product on this side is 20. Usually they define the current as drain current per micrometer width of the channel. I have taken what they have given there in the experiment is per micrometer W is per micrometer. So, when you substitute this as 2000 nanofirons this is 10 to the power of 7 product is 20, 20 into 10 to the power of minus 4, 1 micrometer is 10 to the power of minus 4 centimeter. So, that 20 into 10 to the power of minus 4 is actually 2000 micro ampere per micrometer 20 into 10 to the power of minus 4 is 2000 into 10 to the power of minus 6 that is 2000 micro amperes or 2 milli amperes per micrometer. So, what we are telling is if there were velocity saturation you would have got this current as 2000 micro amperes per micrometer what you get here is less than that which would mean that there are 2 things one is what you say is correct that channel the drain current is independent of the channel length, but the velocity is not velocity saturation. There is a different velocity which is smaller than the velocity saturation. So, that is what we get from here. Now, let us see what the model that we have taken here what we have done is you have taken the charge at the source end and taken the velocity is velocity saturation. Now, that would mean the current here is whatever charge is present in the velocity saturation. If you go down along the channel the velocity cannot increase beyond that point velocity is maximum is 10 to the power of 7. So, if you say this is velocity saturation here also there is velocity saturation here also is velocity saturation. What happens to charge in that would imply that charge also is constant because I d if it is 1000 micro amperes per centimeter square it is the same thing everywhere because current continuity. So, what we are trying to point out is if you assume this model which you have written based on velocity saturation here that would mean that throughout there is velocity saturation that would mean the charge is throughout same which implies there is no field in this region. If there is field in this region plus here minus here the charge will be keeping on decreasing. If charge is keeping on decreasing velocity will be keeping on increasing. If the charge is not decreasing if charge is constant there is no voltage drop. So, the argument that we have put forward is not quite correct. So, what we are telling is velocity saturation may not take place here because charge is maximum. So, velocity is less than velocity saturation. As you move towards the train end charge keeps on decreasing because there is a voltage drop here because charge keeps on decreasing velocity keeps on increasing you may have velocity saturation towards the train end not at the source end. So, what we are trying to point out is at the source end the velocity is less than the velocity saturation. So, I should calculate the current taking this same formula, but V is less than that velocity saturation. What is that velocity we do not know right now we will find out that. So, whatever I have mentioned I have summed up and put it here. The train current formula in the velocity saturation model assumes that the electron velocity is V s throughout the channel. This assumes that charge does not vary throughout the channel which is not correct. So, from what we have discussed we you know the charge does not vary means this is possible there is no voltage drop, but voltage drop across the channel if there is current flow. So, this will lead to the conclusion that velocity saturation is absent near the source end of the channel where the electric field is low. Lower than the such lower than saturation that we give you or so and velocity saturation can exist only near the train end. Now, let us examine that quickly go through some of these things. So, this is the basic thing that is involved. So, experimental results ideas is indeed independent of L. We are very happy, but this formula also holds good except the charge when you take at the source end the velocity is not value saturation, but some effective velocity which is lower than saturation velocity, but it is a function of saturation velocity and some other factor comes into picture. So, W c oxide into V s Q and I am substituting here. So, all that we replace is this by V f A t. Now, in V silicon from the result that you have seen this V f A t is less than the saturation velocity. I will not go through this at this stage, but I will just point out that if you take gallium arsenide devices that effective velocity will be even more than the saturation velocity because of what is known as the velocity overshoot effect. If the electrons are injected to the channel suddenly to a high field it goes through a transient phase where velocity gets much more than the saturation velocity then comes back to the saturation velocity. So, that effect is dominating in gallium arsenide whereas in silicon that is very marginal. So, it is always less than this. Now, let us examine this. There is one more term that I bring in here that is a ballistic transport. Can there be ballistic transport of carriers in silicon? In gallium arsenide question mark I will not discuss that now in silicon. What is ballistic transport? Normally, when the electrons get accelerated they get scattered. Due to collisions with lattice atom or impurity ions everything whatever will be the scattering centers. Now, the ballistic transport is referred to the motion of electrons through the semiconductor material when collisions are completely absent. When will the collisions be absent? If the collision centers are far away. So, ballistic motion is in the uniform electric field will be governed by the equation. The velocity there is acceleration into time. So, what we are implying is if there is no collision, if the electron moves without collision the acceleration will be forced by the mass. There are Q into electric field divided by mass. So, for example, if the electron is moving in vacuum there is no collision that is ballistic transport. The velocity will keep on increasing. That is why vacuum tubes the velocity of electrons is much more than that in semiconductors. That is why people are thinking of bringing back the vacuum tube for high power high frequency operations. There is one activity going on when to bring in vacuum tubes into very high frequencies. There are frequencies. So, but if the time is T if you take the velocity at that time is acceleration into time T. Now, ballistic transport will take this so long as the time is less than the collision time. That is the collision. Now, let us see what it is in silicon. This is something people have simulated from Monte Carlo simulation etcetera. They have seen that for silicon the average collision time is about 3 into 10 to power of minus 13 seconds. That is 0.3 picosecond very small within that time the collision takes place. So, you will have ballistic transport in a time less than that collision time. There is no collision. You can use this equation to compute the that is what we actually do. So, what is that length let us see. Let us say the electron is moving with the velocity of 10 to power of 7 due to ballistic action. Average velocity 10 to power of 7 and the time T 3 into 10 to power minus 13 what is the length travelled? That is between 10 to power minus 13 into 10 to power of 7 that is 30 nanometer. That means, if you assume that maximum velocity is 10 to power of 7 is present for the electrons and within that collision time it would have travelled the distance equal to 30 nanometer. That means, there will be ballistic transport if the channel length is less than 30 nanometers. It is very much less than 30 nanometer. If the channel length is 30 nanometers and above it will be governed by the scattering. That is the mobility and the saturation velocity. So, what we are saying is there will not be any ballistic transport that is in silicon till it is about much less than 30 nanometers say 10 nanometers, 50 nanometers. If you go you can say there is ballistic transport will be there. So, you may not have velocity is much greater than the velocity saturation at the force end. There is a meaning of that. So, in silicon MOSFETS pure ballistic transport will not take place until the channel length is well below 30 nanometers. We examine that phi if there is a chance of velocity higher than the saturation velocity. That is not possible. If you take gallium arsenide again the simulation results have shown that the collision time there is 5 picoseconds much larger. So, if the velocity is 10 to power of 7 I am taking the maximum velocity it will not be that much. Then the length travelled by the electrons before the collision has taken place is velocity into this time that is 500 nanometers. What we are telling is in gallium arsenide the collision will be absent or scattering will be absent till it has moved 500 nanometers 0.5 micrometers which that would imply that there will be ballistic transport in channel length which are 0.5 microns. So, if the carriers are injected into that gallium arsenide channel in that short period of time before it gets scattered if it is suddenly injected into the channel where there high field is present it will get accelerated to have velocities higher than that of 10 to power of 7. That means in gallium arsenide you will have velocity over shoot effects like a transient effects in the L C circuits etcetera. There will be over shoot effects will be there because before it comes to studied it the velocity goes over and above the saturation velocity and that may give rise to your injection velocity much more than the saturation velocity. So, that is why in the case of gallium arsenides you may encounter drain currents which indicate that the injection velocity of electron is much more than the saturation velocity that is because of the velocity over shoot effect. I have put it very briefly without going into details of that. The meaning is sudden transient effects in gallium arsenide leads to that over shoot effect because that has at least few picoseconds of time whereas in silicon you do not see that because it is less than much more than the picosecond that is about 0.3 picoseconds. So, you do not really see the over shoot effects. So, we just do not go into the discussion on the gallium arsenide at this moment when I discuss gallium arsenide devices I may bring this back that is focused on silicon because we are happy with silicon devices and we are happy with the use of small devices in silicon. So, the conclusion from this discussion is in gallium arsenide if it is the carrier velocity in the channel may exceed saturation velocity because the collision takes place after few picoseconds due to the one set of ballistic transport and velocity over shoot effects for small even up to 0.5 up to 0.5 micron channel lengths. In silicon if it is ballistic transport cannot take this even when channel length is 30 nanometers or 50 nanometers because that is estimated by taking such velocity 10 over 7 it will be lower than that. So, much smaller channel lengths are required to see ballistic transport or velocity over shoot effects in silicon. So, the drain current is limited by not by the saturation velocity, not by ballistic transport, not by velocity over shoot effect, but the velocity with which it is injected that is something else is controlling the injection velocity. So, at what velocity the carriers are injected? We will see what it is I will not get down into the detailed discussion on that. So, drain current if it is controlled by injection velocity which is smaller than that velocity saturation I can say that I can still write I d s c is w q n at 0 into injection velocity and that is lower than that. Then I will get the drain current which is actually smaller than what you would get with the velocity saturation, but this injection velocity does not depend upon channel length it would depend upon the conditions at this over shoot. Let us see what it is I will not derive it, but I will just show that where the deriving is bit of involved thing there is a detailed paper written by Lundstrom and Ren on the physics of carrier transport in nanoscale MOSFETs published in January 2002 in which what they have said is whether carriers are injected from the source to the channel. They are able to inject from the source to the channel by gaining an energy to raise above that barrier. So, they have got the sub thermal energy in fact they would be injected at a velocity close to 10 to power of 7 there. So, you would think that earlier analysis what we said is injection velocity is the thermal velocity that is what you would get there at the point of injection at the point when the electrons have just landed in the channel, but the moment they land in the channel they start moving within a very short length they suffer collisions. They suffer collisions at scattered backward. So, the average velocity of that beyond that point within that short distance is controlled by the scattering mechanism. I am not showing the derivation. So, that is controlled. So, if you go through the analysis what happens is if there is no scattering at all you would have got the thermal energy. Because of scattering the velocity of the electrons is controlled at that point by combination of this injection velocity or thermal velocity and the velocity with which it moves that is mobility into the electric field. Velocity is mobility in the electric field that is the low field. The electric field is not high at the source end. Electric field high only at the drain end. If it is high of course, saturation velocity will get. It is not high. So, whatever electric field at E plus 0 plus means just at the source end that is the velocity that you would get. So, combined effect of that with skipping all the derivation I would suggest that you go through this paper for that because that itself would have taken me about half an hour to discuss that. So, injection velocity is controlled by these equations that is thermal velocity which is 10 to power 7 centimeter per second. If there is no collision at all after injection that would have been 10 to power 7 centimeter per second. But because of the combined effect of the scattering you get injection velocity less than the thermal velocity. You can see that if the mobility is very very large, if the mobility is very very large it could be that in that electric field you will have it is like putting two resistors in parallel. The resistance is governed by the resistance which is lower. So, here if this is larger than this one if the mu into E is larger than the thermal velocity we injection with thermal velocity. If thermal velocity is larger than the second term here denominator injection velocity governed by this mobility it will actually combined effect it will neither this nor that. So, what turns out from that is this is for the conclusion for this our discussion is not ending here. Considering these both these effects together what you would say that you can still write the drain current is equal to W Q into injection velocity. Injection velocity is the combined effect of these two terms. If I have high mobility in the channel you may get thermal velocity itself very close to saturation velocity, but it will only be less than that. But if the mobility is low then it will be far below the thermal velocity saturation velocity. So, you can see that this particular term drain current that you get saturation drain current will be in the very short channel devices are limited by the injection velocity which is controlled by both saturation velocity or thermal velocity and the mobility. And you can see that both the terms which determine the injection velocity are independent of the channel length. So, injection velocity does not depend upon channel length. Therefore, if you are computing this current using the injection velocity the drain current will be independent of the channel length. So, what we say here holds good. So, you get the drain current which is independent of channel length when you go to channel length which are 0.2, 0.1 microns length or below, but the current is lower than what you would get with the saturation velocity because injection velocity is smaller than the saturation velocity. And this is actually you can see the reference is this is a recess caller modeling of nano scale MOSFETs P H T C soft transfer density 2002. There is a time when that model also was given by non-storm etcetera in the particular city. So, what is our aim this from this discussion what we see is you need to have very high mobility for electrons because after all thermal velocity is fixed by temperature 10 to power 7. You need to get higher and higher mobility. What is the way you can get higher mobility? What are the factors which affect the mobility that we will see now. So, low mobility field mobility should be high. The mu is actually the mobility in the low electric field. You go back to this you see mobility is actually and just the slope here low field region at the source and the field is low as you go towards the drain and the field becomes high. So, velocity naturally go up there. So, now going back to this I am skipping all these things which you have discussed now. So, you are looking at either materials which can give high mobility or if you want to stick with silicon what is the way you can increase the mobility. What are the factors which increase the mobility or limit the mobility. If you can plot mobility mu n or this is actually mu it does not show here this is actually the arrow is mu I think because of change in the program. This is the mobility is the y axis and the doping concentration. If you recall to reduce the short channel effects you always increase the mobility in the conventional MOSFET. So, increase the mobility the I am sorry if you increase the doping I am sorry let me restate in the conventional MOSFETs you increase the doping to reduce the short channel effects and also reduce the oxide thickness to compensate for that. So, if you increase the doping the mobility is almost flat. I have plotted the this is actually approximated straight line it will not be that straight it will be smoothening here smoothening here. So, when the doping is up to about 10 to power 16 mobility of electron is written 1500 centimeter square per volt second. But once you reach about 10 to power 16 doping the mobility starts falling mobility starts falling because in the low doping concentration regions this scattering centers are only lattice and atoms. But when you go to higher doping the dopants they in they are in the ionized state the ionized dopants give rise to additional scattering centers. The low doping mobility they are not the there is scattering, but it is not high compared to the lattice scattering. But when you go to higher doping concentrations additional electrostatic scattering due to the ionized dopants takes place mobility falls. You can see the extent of falling mobility what was 1500 centimeter square per volt second can become even as low as 200. So, you can see if you take a look at the doping concentrations this term this is this is not 1500 this will be 200 when you go to 10 to power 18 in 10 to power 17 it may be about 600 or 700. So, moral of the story is you cannot go to such to high doping concentrations in the classical MOSFET you have to have high doping concentrations to reduce that short channel effects. What are the other factors coming into picture the other factor which leads to reduction in the low field and they say that field that field is actually along the y direction. But in the case of MOSFET there is electric field in the vertical direction there is normal direction that is what we stocked here was the I am sorry electric field in that direction that is what we talked of here. Now, what we talk of here is there is also electric field in the particle direction that is normal direction to the channel. So, electric field in the normal direction normal to the channel must be kept also low. But you know when you make the oxide thickness small and the channel doping high the when you deplete the channel there will be very high electric field. Oxide thickness is small the depletion layer charge is high. So, Q D by C oxide is the electric field or whatever charge is there that gives sides electric field Q D by epsilon oxide that is the electric field Q D by C oxide the voltage drop across the thing. So, if the C oxide is large and Q D is large or if Q D is large electric field also becomes large. So, what we are trying to point out is you will have because of the high doping here and thin oxide field will be large in the vertical direction. So, an electric field is large in the vertical direction the mobility of this channels along the y direction gets affected. Why? You can imagine when the electric field is in the direction vertical direction like this normal direction to the channel if the electrons are moving in the direction if the field is like that it tries to keep it hold it for a while in that region. So, there is a restraining force which keeps it back in that position there is a qualitative way of understanding that means actually the mobility of the electrons moving in that direction gets affected if the field is large in that direction. So, you must avoid the field in that large in the x direction normal direction. Now, an empirical formula is given by these people Pecoroni and Wardeman in the year very way back in 1983 by the conventional MOSFETs mobility keeps on decreasing mobility in that direction keeps on decreasing if the E x we call it A effective due to including all other effects if that increases that keeps on decreasing. So, what is the way you can do that you cannot help reducing the oxide thickness you want to keep the capacitance large what you can do will be reduce the doping concentration here. So, if you reduce the doping concentration in the channel two things happen one the ionized impurities scattering gets reduced. So, mobility reduction does not happen you can go to mobility which are as high as 10,500 in silicon if you some other material you can have mobility is decided by the mobility of that electrons that material. And also if you reduce the doping concentration here the depletion layer charge gets reduced. Therefore, the charges in the depletion layer gets reduced mean electric field again is lower in this vertical direction normal direction. So, if you reduce the doping concentration naturally you get improved low field mobility low field in that direction you will get improved mobility and you will get improved velocity you can get very close to the saturation velocity thermal velocity. So, the entire drive in the present day devices is to choose. So, these are the references I am telling. So, to get high performance what should you do just the last slide which I am discussing here today which which has got lot of juice in that. So, high performance nano scale devices therefore, should ensure that doping is low. So, that mobility is high they should also ensure that the vertical normal electric field is low. So, that again the mobility is low field mobility is high. So, ultra short channel mobility devices with improved low field mobility is required for high performance nano scale MOSFETs. You cannot help producing the channel length why should you reduce the channel length you saw channel length divide by velocity use is a transit time higher frequency. To go to higher frequency even the velocity gets saturated because of reduced channel length the time required for transport occurrences reduced the frequency is high. So, to take care of this aspect ultimately to improve the mobility new approaches are used the new approaches are non classical MOSFETs. Finally, we are getting down to what we wanted to discuss. So, they have emerged these meet some of these things meet this goal of low mobility those devices are silicon on insulation MOSFET insulation MOSFET. If you recall MOSFET that you have got is whole thing is on a bulk region and out of that we discussed right at the beginning only the top layer is the one which is being used below that it is actually mechanical support. So, you can have silicon on an insulating substrate a thin layer of silicon used we will see how that works out Saroo's to improve the low field mobility. So, because of that you have got double gate MOSFET ultra thin SOI MOSFET silicon on insulator MOSFET and fin FET these are the things which we discussed in the coming next few lectures. So, that is actually using silicon itself and this trick that you are using these devices is only the controlling the electric fields controlling the dopants which is enabled by the SOI device. But now you can make use of the SOI device in addition you introduce strain along the channel what do you mean by strain if I have the channel let me go back to one of those slides here this is a channel. Now along this channel if I introduce a compressive strain in that direction if I introduce compress the channel in that direction what happens is the lattice atoms are little bit brought close to closer together. So, if the lattice atoms are bit closer together the collision distance becomes reduced if the collision distance becomes reduced collision time is reduced mobility will be reduced. If I bring in compression the mobility is reduced, but if I stretch it if I introduce tensile stress along the channel then the lattice atoms are slightly pulled it may be delta it may be fraction of an angstrom like that after all distance is about 5 angstroms may be it is fraction of that if you increase the length the spacing between the lattice atoms is increased collision time is increased mobility is increased. So, in n channel MOSFETs if I have tensile stress strain if I can introduce then electron mobility will increase. So, whatever is good for electron is bad for holes. Later on we will again discuss those details in the holes if I stretch it even though electron mobility increases when you talk of this plus charges exactly offset things takes place I will not discuss that at this time frame in fact the hole movement takes place because of the jumping movement of electrons from bond to bond. So, the bond has to break and jump to the neighboring point. So, hole movement takes place by jumping movement of electrons from one bond to the other bond. So, if I stretch it the electron has to jump on a longer path that means the hole movement will be more difficult if I stretch it. So, what we said is the electron mobility increases if I use tensile stress, but because of that the hole movement becomes difficult hole mobility becomes less if I stretch it. How to introduce the tensile stress I can deposit nitride on the top of the gate that nitride has deposited as for tensile stress it will stretch it elongated. So, Intel has made some of the devices where they deposited low pressure nitride on to the on the gate over the gate to stretch the channel. So, that electron mobility can be increased and the hole mobility can be increased by compressing the channel. How do you compress the channel take a look at this by introducing nitride on the top of the gate I can stretch it I can compress it if I introduce germanium into the source and drain region. Source is silicon during diffusion you can add little bit of germanium also. So, this will be silicon germanium small amount of germanium concentration germanium lattice or the atom atomic size is bigger than that of silicon. So, in these regions you have silicon and germanium atom at source end and the drain end if you introduce that volume increases because radius of germanium atom is bigger. So, it is in the given volume you are trying to introduce movements which are bigger. So, it tries to compress it is I am sitting here two fat fellows sit on both my sides they will be compressing me. So, that is what is happening you have germanium atoms added on to the source then compressing the atom. So, if you do it for n channel devices it is mobility will decrease or p channel devices the mobility will increase. So, in the p channel devices they add germanium to the source stream to increase the mobility. So, that is the strain layer. So, other thing is of course, totally use different materials like germanium leave silicon and go to germanium germanium has electron mobility which is about 4000 centimeter square per volt second which is about 3 times source 2 to 3 times than that of electron mobility in silicon. Whole mobility of whole mobility germanium also is about 3000 centimeter square per volt second both electron mobility and whole mobility are high. So, today people are talking of making devices on the germanium MOSFETs and also in addition you can use gallium arsenide as the material for hetero junction devices to sum up what is being done is either go to silicon non-insulator type of materials that is non-classical MOSFET and use strain layers or go to totally new materials like germanium or gallium arsenide to realize conventional type of MOSFET but new materials. So, with that I conclude and next session we will start with the non-classical MOSFET on silicon on insulator. Thank you.