 So, we were discussing last time the silicon on insulator analysis of that we are discussing and I will just go little bit back one or two slides. So, that because I had thrust a bit during my last lecture on that portion. So, we will just start from there one or two slides compared to the bulk MOSFET. There are different modes of operation in the case of SOI. So, that we have said you can have the front channel, front channel inverted all the time, but back channel can be either accumulated, depleted or inverted. So, that is two channels are there. Now, if I make this thickness very thin thickness of the SOI layer T silicon very thin to nano meter scale, 10 nanometer, 20 nanometer of that order, then a phenomena called as a volume inversion takes place entire channel will be both the inversion layer will be merging that also we will discuss. Now, I just wanted to bring in some aspect of this depletion and accumulation. To make things clear, if you take a look at this bulk MOSFET, you can see the gate red is oxide, the silicon will be depleted if I apply plus voltage to the gate if this is P type. Depletion charge is negative in the P type of substrate. So, this is the depletion plus voltage. If I keep on increasing, you know that you can invert the surface. When the surface potential is twice FOF, then you say it is inverted. So, I have shown the plot last time itself I have discussed this. The voltage drop across oxide leads to the electric field which is constant. Then the electric field at the oxide and at the interface will be different, there will be discontinuity because the flux is continuous. Epsilon oxide into E oxide is epsilon silicon into E silicon. So, this field will be about 3 times smaller than that because the permittivity of silicon is about 12, that of oxide is about 4. So, 12 by 4 that is the ratio is 3, 3 times smaller. And if the doping is uniform as it happens in most cases, the electric field is linearly falling. The electric field is linear. I do not have to derive that because you already know it. That you recall that. And the slope of this depends upon the doping. Higher the doping, there will be steeper. Now, if you integrate this, you get the voltage. X, if you integrate it is x square. So, if it is linear, that will be parabolic, the voltage. And if it is constant, it will integrate and it becomes a straight line. So, you have got the voltage drop. Field actually is V G minus V silicon divided by T oxide. That is the field there. So, V silicon is like this. Now, if it is inverted, it is inverted when V silicon becomes twice 5. So, that is depletion or inversion. Usually, you operate in this region. But if I apply negative voltage to the gate, bulk MOSFET, they are still on bulk MOSFET. Black is a gate. Then you have got this red oxide. Then the silicon. Now, you can see when I apply minus here, the plus charges can be supplied very easily by this majority carriers in the P type region. So, they accumulate plus charges. They are mobile charges. Remember what you see on the left hand side here when it is depleted. I put it inside a circle to show that it is not mobile. It is the dopant which has been depleted of the holes. Whereas, this is the plus charges which have accumulated there. And this is a charge sheet on that. So, whatever voltage you apply appear completely across the oxide. So, the field is actually equal to or the voltage is minus V G here and it is 0 just at this point. So, the voltage is linearly falling across the oxide minus to plus. But that is 0, 0 to minus V G. And the electric field of course is constant. No charge within the oxide. So, V G divided by T oxide will give you the field to be. So, this is a charge sheet. Now, when you go to the SOE MOSFET, you have got the front gate which almost looks like this. I have shown the MOSFET source drain also here. Here, I can put a source and drain here if I want to invert and collect carriers. Here, I have plus here minus here. The schematic diagram, front gate metal or polysilicon doped. Red color is the front gate oxide. This is the SOE layer P of thickness T silicon and this is the back oxide. So, here you have got front gate and you also have got the back gate. So, front gate oxide, back gate oxide. Now, you can invert the front gate by applying sufficient gate voltage. At the same time, you can have the back channel accumulation, depletion or inversion. You can accumulate it by applying a BGP negative like what you saw here. If I apply negative voltage here, it will be accumulated. If I apply plus voltage I can deplete it from the back side. If I apply large voltage, I can invert it. That portion we will see now. So, this is the structure which I did not show in the previous lecture for analysis. So, you can see now here whatever we have discussed yesterday is here or previous class. That is this diagram is written on here. Voltage, that is the bulk MOSFET. Now, let us see this back channel is accumulated case. So, this we have discussed yesterday. What you do is apply a negative voltage to the back gate. So, there will be plus charges here. Now, if I apply plus voltage to the gate with respect to the source or with respect to the substrate which is somewhere, it is a cross section. On the surface, you will have a contact you can put outside this gate region. Otherwise, you can apply voltage between these two also. So, if I apply voltage, then this will start depleting, completely depleted and once it depletes, the additional field lines will cross this and go into the other side. That is what we discussed already. So, you get this is the situation when it has just got depleted dotted line. I have got actually plus charges here on this side. On the surface, you have got plus charges exactly same way you have got here. You have got the plus charges here. That is giving rise to the field here. So, after this depletion, if I increase the voltage, the field lines will cross this and the entire field will go up. This portion I think I have discussed this. This is not a very important need for the SOE MOSFET, but I just showed you. You can get a inversion here and you can because you have applied a plus minus voltage here, the voltage is 0 here because if you recall here, the potential where this is accumulated is 0. So, you have got 0 potential here. Surface is inverted here and the electric field will be will depends upon twice y of minus 0 divided by T silicon. That is the average field. If this were completely insulated layer, then twice y of minus 0 divided by T silicon would have been flat. Now, what happens is you have got this particular field present here already. So, you superimpose that whatever field was there, you superimpose on that. Original field was like this varying. Now, beyond that, a constant field from here, here, they both are same. This gets lifted up by the amount that it is an additional field. So, you can actually find out how much this field is. How do you find out? Twice y of divided by T silicon is an average field. So, it is an average field. For example, if I just draw this like this, I have at when this is the silicon, I have the oxide on either side. Now, it is just depleted. It is like this. Lengths are not straight actually. It is not coming less straight. So, if I draw it here like this, now, if I take twice y of minus 0 and take the average field, it is like this. This is ESF. This is the backside field. So, average field is like that. Now, what I want to find is, so, I know this is actually equal to twice y divided by T silicon, average field. So, what I drop is that. But now, I want to find out how much this quantity is. That quantity is because this whole thing has got shifted up. So, whatever I have here is half of that. If the whole thing is shifted up, I get that like this. Now, this quantity is the same as half of that. Average is between these two midline. Here, it is maximum ES 1 and 0. So, what is this quantity? What is the field here? When it is just depleted, whatever charge is there in silicon divided by epsilon s. That is the electric field here. So, half of that is that. So, if I have this quantity as twice of the T ox T s, that is the field here. This quantity is, that is that. So, this is the ESF. So, I know how much is the field here is. That is ESF. I am writing that surface, front surface. So, that is the ESF from here to here, total. So, that is how you find out. Average field add to that. So, whatever distribution you have, all that you do is supposing I have a potential difference like this. This is phi f and this is phi s, phi back. If there is phi b back side. Now, this minus that divided by T f s is the average field. This may not be equal. Let us just do it once again. I remove that and I go back to this. So, you have got, let us say this is phi b back channel. This minus that divided by T f s is the average field. To that, you know that there, if I take the electric field, it will be like this. So, the average field is whatever I get, this minus that divided by T f s. In this case, it was twice phi f only, because this voltage was 0. Here it will be, some phi b is there. So, I find out the average phi f minus phi b divided by T f s is that average field. And what is that quantity? That is half of this. That is same as this. So, this is a general case. So, if you understand that, there is nothing more to, no more difficulties will be there. So, I will just go back to this. So, here you have got that electric field like that. This field is actually twice phi f by T f s plus whatever field is there, half of that you add to that that you get that. Now, let us go to more realistic situations. So, you can see, I change the whole situation now. This was back channel was biased with negatively. So, you have got accumulation layer here. Another practical situation. I have the back gate grounded. This is where I think we started last time and some understanding needed for a bit more analysis. So, I apply plus voltage. We are finding a situation where the front channel is inverted, back channel is, you do not apply to the back gate. So, it will be depleted. What happens? When I apply both voltage, plus voltage to this front gate, this is grounded. This is with respect to that you are applying. This is connected to the ground somewhere, substrate somewhere else. We are not able to see in the cross section. So, I am applying voltage across that. So, you see at a particular voltage, as I keep on increasing, the depletion layer keeps on moving. I will just go back to this now. So, let us see where I can do this. So, what we are trying to see is the, if I have the layer like the silicon, T silicon. Now, if I apply plus voltage to the gate, plus here, I have grounded this. This is the back oxide. This is oxide. So, what happens is the moment you apply voltage, there is a depletion layer formed. That is depleted. They are all negative charges immobile. Keep on applying, increasing the voltage. This keeps on moving. Ultimately, this merges with that. So, when it just reaches this point, that is called punched through. That depletion, the silicon layer is punched through. So, then what you will get will be, I will just show that at that punch through, you will have the, I will just remove this particular thing. So, I will go back to this, back to this. Just when it is punched through, electric field will be something like this. That is 0. 0 electric field here. This is the maximum field. What happened? So, you have got the electric field like that. And you know, at this point, the electric field is q s by epsilon s. So, to charge, what is q of s? q n a, T of s by T of s. That is a charge. That divided by epsilon s is the field here, like before. Now, if I increase the field voltage further, beyond punched through, what happens? There is no charge here. There is no charge here. There is no charge in the oxide. This is oxide. That is oxide. I hope you can read that. There is an oxide there. So, if I increase further, additional field lines will cross here. Additional field lines will cross there. And whatever additional field lines there, there is no charges here. It will be down by factor 3. The oxide, field oxide divided by 3 is a charge here, a field. So, here, if this has gone up, this will go up by the same amount, but by factor 3. Now, whatever electric field has increased here, everywhere it will increase, because there is no charge there. Or whatever field line originates from there, terminates there. And once it reaches here, there are no charges in oxide. So, what will happen? It will cross the oxide. But, the electric field here in the oxide will be higher, like the electric field here. The field here will be higher by factor 3. So, you can see the electric field distribution will be like this now. Now, I can find out. This is the electric field at the surface, ESF. Why are we finding electric field at surface? If I know the electric field at the surface, I can find out what is the voltage drop across oxide is. Because of, voltage drop across oxide is ESF into epsilon s. That is the total charge beyond that point divided by, see oxide f, charge by see oxide f. Charge is ESF into epsilon s divided by see oxide f will voltage drop. Now, when you go into this situation, go back to the diagram now. So, I hope it is clear enough now. So, what we are trying to find out is, how much is this field? The ultimate goal is to find out how much is the surface field? When the surface potential is, if you take a look at the surface potential, I think I have that here. If I take a look at surface potential, see this is the dotted line is the electric field when I just depleted it. That is what is the thing. Now, when I increase the voltage, the field has gone up. The entire thing has gone up. The diagram which I have drawn, I put ESF d to tell you that the back channel is depleted. Front field, surface field and the back surface is depleted. So, how much is the field here? How do you find out? What is the potential here? Psi s f. Because the field lines are crossing this oxide and because this is grounded, this will drop across oxide. So, the potential here is plus channel potential on the back surface is plus with respect to back gate. So, that is psi s b positive. So, the voltage difference between the front channel and the back channel is psi s f minus psi s b. I am trying to find out how much is the threshold voltage of the front gate. At threshold voltage psi s f will be twice f. I have kept on increasing the voltage till this has reached twice f. There is some voltage here with respect to ground psi s b. So, what is the ESF? How do you find out? Psi s f minus psi s b divided by T of s. You can see that that is smaller than what you got in the previous case. Previous case, it was psi s f minus 0 divided by T s, average field. To that average field, you add half of this field, T s b twice epsilon s. So, here twice f minus some voltage divided by T of s will be smaller average field compared to back channel accumulated. So, if you want to find out how much is this average field plus half of this quantity. So, I will get in this case also I will get in this case ESF will be pi psi of minus psi of s b divided by T of s that is average field. Now, let me just remove that. So, plus this is average field plus what will be the thing? Half of this. This is just when it had, once you through Q s divided by epsilon s is total field. So, this is the key situation. So, L T field is that. So, once you know this multiplied by epsilon s that is charge beyond that point. We do not know where it is all. We know this field, how to calculate? Multiplied by epsilon s gives me the charge divided by C oxide of Q s by voltage drop. So, pressure voltage at this point is, go back to that pressure voltage at this point is actually equal to twice psi f. We know that that inversion condition twice psi f voltage drop across that is twice psi f minus psi s b divided by T of s plus half of this field into epsilon s divided by C oxide. So, whatever we are telling Q substrate that has got a bigger expression like this. This whole thing is there. So, it should have Q. Only this charge is there, but there is charge coming on the other cat also. Whole thing is taken into account by this. That is the beauty of this using this substrate potential. That into epsilon s divided by C oxide with voltage drop. So, that is about the bag gate depleted. Now, let us go and see the situation when I see here. Here the bag surface potential is psi s b because this is grounded. There is a drop across oxide. Now, I will take a situation where I deliberately apply voltage to the bag gate plus. That is, if it is making 0, I apply plus here plus there. So, I keep the voltage plus here and this green color that I have shown is the depleted region due to the bag gate. See, I am applying that with respect to that. That is with respect to the substrate, which is connected to this. So, you have got, this is the depletion layer here up to this point. Now, I start applying front gate voltage with respect to this. That is substrate. What happens? As I keep on applying the voltage, I am not disturbing this. I fix that at VGB. This may be thick oxide. It is not inverted. You have to invert depletion layer which is much more than the thickness. It is a fully depleted case. So, I start applying the voltage. Just like in the previous case, you will have the similar situation where that situation is slightly different now. So, the situation now is, if you see, it is like this. This is the back channel. This is the oxide and this is the oxide and part of it, that portion is already depleted because I have applied plus voltage here. So, this field lines are like this in this direction here. Now, I come from this side. As I keep increasing VGF, what happens? This oxide straight away, there is some voltage across, voltage of across oxide. That is I will repeat. To increase the voltage, what happens? That goes up. This goes up. At a particular point, I have this coming and that is coming later. Now, at this point, you see this is a T-silicon. Part of this T-silicon has been depleted and has been occupied by the back gate VGB. This has been taken over by that. You have fixed that. Now, when I have applied up to this point, you have got 0 field here because if you plot this electric field lines from the left hand side and right hand side, if you draw, that was the depletion layer, which is already occupied by the back channel. Now, if I take the electric field now, how is it? Only here in the channel region, 3 of us. Over here, up to this point, depletion electric field is from right hand side. From here, left hand side. So, electric field comes down like this. I think I better take a different color for that. That is like that, electric field. If you take the electric field from here, that is like this because of the back gate bias. That is plus higher here, 0 here, so you have got a region where the electric field is 0. So, if you take the potential here, how will that be? Here it is actually potential here will be 0. It will come from here, change that. So, the voltage will come like this. Corresponding that point, there will be 0 and from this side, it will come like this, but this potential is smaller. Depending upon what was the gate oxide thickness, what was the back gate bias, so this is the point at which 0. So, that is the situation you have got here. Now, at this point, what is the, it is not inverted here. It is the surface potential is not twice 5. So, I have to increase the voltage further. So, what I do at this point is, I will go back to that slide. It contains all the things. So, that is the situation that you have drawn there. What I have drawn there is this situation, the bottom line. So, you have got it like this. Now, I increase, see this is Vgf 1 and this is Vgp 1. So, this portion belongs to the back gate. This portion belongs to the front gate. Electric field lines you see as I went on doing like this. Here, I have drawn it like this. See, instead of drawing it like this, instead of drawing it up like that, this is in this direction, that is in that direction. So, what we do is, we take it like this, draw the 0 line and draw like this. You take it up, you draw it like this, negative. So, that is what is done there. So, you have got that continuously there. In plying that, this is 0 here, this is negative here. Your field is in that direction. So, this is the situation where what you have got here. Now, from here onwards, what I do is, I increase the front gate voltage by some voltage V, back gate by voltage V. See that there is no charge in between, completely depleted. The entire layer is depleted. All that you have done is now, you increase this, you increase this by same amount. The difference between the two voltages is not changing. That means, the electric field will remain same. The electric field distribution here will not change. Whereas, that extra voltage has gone. That is the thing. When I apply like that, see here, once we have depleted completely and this was also biased positively, beyond that point, just when it is fully depleted, beyond that point, I increase both of them by the same amount V. That extra V does not appear here. The difference is here same, but that extra V appears across with respect to ground that is here. So, that this point becomes plus with respect to that. That is the source junction gets forward biased by that extent. So, if this is forward biased, from that point on to the surface, whatever drop was there is present. See, from here to the midpoint, there is some drop. When this situation, see from here, surface to this point, there is a drop. Now, if I increase this by V, this potential here also has increased by V. That means, the source junction has got forward biased by that V extra here. But here, what is the forward biased voltage? Whatever V is there plus whatever was there, there has a potential here. See, when the point just when it has depleted here, potential here is here. What we are plotting is just outside surface here. That was 0. It was racing to that phi phi S 1. But now, it has gone up here. This has gone up here. So, you can keep on increasing it so that this is inverted. This becomes phi phi F. But this should not have been inverted because that that differential layer is narrower here. Supposing you have applied voltage so that the entire thing is at the middle, both would be the same potential. It will be both will be inverted. So, that is the situation that you have got. Now, let us see. So, here in this case, what will be the field now? Psi S F minus Psi S B divided by T silicon. That is the average field. That is the field there. Psi S F minus Psi S B divided by that. That is the average field. Now, if I want to find out how much is this field, what do I do? Psi S F minus Psi S B divided by T silicon. And how do I find out this one? U S B divided by epsilon and subtract. That is all. So, it is you can get the whole thing by solving Poisson's equation. But you can see this very easily. You can see for any bias condition. So, this will be inverted. This is depleted. Now, you have got a situation where I have applied a let us say this is a situation where need not be oxide, need not be the same thickness. But I can apply a large enough voltage, so that the depletion layer from backside comes up to the center. So, now I can come from the front gate till the depletion layer merge. That is the situation. So, if the depletion layer were coming up to this point, the potential will be symmetric. Even for thicker back gate, I can do that. But the difference will be the back gate voltage will be more, where more voltage drop will be oxide. So, here the drop here from the midpoint to the surface and from midpoint to surface both are equal. But for the same field, the voltage drop here will be something here will be more. Now, beyond this point, I increase both the voltages by the amount equal to v till that goes twice y f because it was symmetric originally when it is depleted. When you started applying voltage, this also will be twice y f. So, what is average field here? 0. This minus that is 0. So, average field is 0. So, average field is 0, but the field here will be how much? The average field is 0. The field here will be this minus this by twice t s by 0 plus q s by twice epsilon. That is the field. So, you can see that q s is actually, what is q s? Charge in the depleted region q n A t of s. So, if I use lower and lower doping, q s by twice epsilon that is this field will be smaller. If this field is smaller, this field also will be smaller. You can see now, we are seeing it like this, but in the device, you are able to reduce, if you reduce the doping here and also reduce the thickness here, the field here will be reduced. If the field there is reduced, delta field there also is reduced. That means the voltage drop across the oxide will be reduced. That means, your threshold voltage will be lower in this case. When I got into twice five here, the voltage drop was less here. Correspondingly, voltage drop here will be less, because field is less. If this field is less, this is less, drop is less. So, what we are telling is you can have low electric fields in this direction, which we are looking forward to for increasing the mobility. So, electric field in the vertical direction, which transfer direction will be reduced by reducing the doping, by reducing the thickness. Now, you may question, if I reduce the doping, will I have short channel effect? That aspect will address later. The difference between the bulk MOSFET and this is, you can make it thinner and thinner. You can make it thinner and thinner, so that these two have more tighter control on the channel. If the thickness is small, if I have the drain here, if this is long compared to this thickness, this has a better control, gate has a better control. It is like two regions. See, if I have the gate here and if the drain is here, force here, I have the oxide here, I have the oxide here. Suppose, this is thick, this may have control over here at this point. Suppose, you make it very thin, this is the source drain. Suppose, this is very thin, this has a better say on this region than this region. So, the gate will have more control over this region here than that region. So, you will be able to go to much shorter channel lengths and you will be able to reduce the doping because this region is controlled mostly by this gate. Even if the doping is reduced, the drain depletion layer is not able to encroach into the channel. So, that is the situation we will see more details of that later. But right now, what we are saying is thinner silicon, more lightly doped, fields in that direction will be less. You will be able to invert because of that we had a lower voltage. Now, a most general case. Now, since you understood the whole thing, we will analyze the situation. Vgf is applied, Vgb is applied there. I am showing only front oxide, back oxide and this channel. Gate is here, gate is here. And I have got a situation where electric field is across oxide is there, across the silicon is like this and across oxide. This situation is there when you have got electric field is in that direction, electric field in this direction because this 0 is here. So, general case. So, we already discussed this now in the previous slides. So, what is electric field here? I am just arbitrarily drawn the electric field like that. But how much is the slope here depends upon the doping. This slope depends upon as you start depleting, you can see slope depends upon doping. If it is very lightly doped, it will be flat. If it is very highly doped, it is going like that. So, I have drawn this diagram here to calculate how much will be the threshold or what is the gate voltage applied to this for this situation. General case, the gate voltage at the front gate is actually equal to oxide drop Vxf V oxide front gate plus whatever potential is here, psi sf. See, this is a potential Vgf, psi sf, psi sb, Vgb all with respect to ground point, all with respect to this ground point. So, we are drawing that diagram here. I wish I had this diagram here also, but it is now. So, what is this field? Now, you know how to find out psi sf minus psi sb divided by T silicon is the average field plus q s by twice epsilon c is that field. That is the asf. Exactly the same way we argued out, difference between the potential divided by the silicon thickness gives the average field, the dotted line. I add that half of that q s by twice epsilon s, that is that, q s is that quantity. Now, what is the voltage drop? Voltage drop is whatever field is there into epsilon s that represents the charge beyond that point divided by c oxide. So, I just substitute this here, epsilon s psi oxide into that quantity whole thing. So, that is this quantity. Just a substitution, put this here, substitute that here you get that. Now, using this equation 3 along with that, I will get what is front gate voltage is. We are now not saying whether back gate is accumulated or depleted off, but we know that field distribution like this. So, front gate voltage is actually this quantity plus psi sf. See, this quantity, the potential here is whatever potential here psi sf plus that works here voltage drop. So, add to this quantity this isf. When you add that, this whole, this equation. There is nothing new here that I am adding this, putting it in this form. Only thing that I have made is epsilon s psi t of s is actually the capacitance of this silicon layer when it is fully depleted. It is writing epsilon s psi t of s all the time like epsilon oxide by t oxide. You say c oxide, epsilon silicon by t oxide, t silicon is c c silicon. That is the meaning of that. So, you replace that by c s. So, you will get this equation. Same way, I can write, you know, it becomes very simple now. If you are able to write this equation by trying this diagram, which you can do now, you can write for this back gate also. All that you do is, you have to write field here is that minus this quantity, average minus that quantity. When you do that, you will get vgb is equal to, in terms of the psi sb, you get this. Replace c oxide by c oxide, where you want to go to back of the symmetric equation psi sf by psi sb, c oxide by c oxide, c s remains c s, psi sb by psi sf, all Ulta or reverse, c oxide by c oxide. So, that is, you can now deal with this situation, whichever channel is accumulated, whichever channel is depleted. So, here you can see, from here, how can I get the, I want to see the general equation. I want to find out the threshold voltage when the back channel is accumulated. What do I do here? Front channel is inverted, psi sf will be substituted by psi f. Back channel is 0 when it is accumulated. It is not really 0, slightly negative, but we are putting it very close to 0, 25 millivolts, 50 millivolts of that order. So, I remove these. So, then threshold voltage will be 1 plus c s psi oxide plus twice psi f plus that quantity. When the back channel is depleted, I will have twice psi f and psi sb here, psi sb here. When the back channel is inverted, what will I get? How do I do? From the same position, back channel is inverted, front channel accumulated. Substitute psi sf here, here, twice psi f. Psi sb also twice psi f. Both channels are inverted. Then both are psi sb. Then you can see this term and this term cancelled out and you will have twice psi f plus q sb twice psi f. Psi sf is twice psi f. I will have written that thing here. See, back channel accumulated, you get that substitute when you substitute in that equation. If you go back to that, you will see that. All that I did is put the 0. Then you get that equation. Middle term is 0. And when the back channel is depleted, more or less it will be that twice psi f plus q sb psi oxide. Some more terms will come, but approximately it will be this. Both channels are inverted. These two terms are cancelled because both are twice psi f. You will have only this term, which is twice psi f plus that term. So, you can see compared to these things, back channel accumulated, back channel depleted, both channels are inverted. Here only front channel is inverted. When I see these symbols are threshold voltage of the front channel with the back channel accumulated. Threshold voltage of the front channel with the back channel depleted. Threshold voltage of the front channel with the back channel inverted. Both are inverted. So, if the back channel is accumulated, you can see threshold voltage is highest. In the case of bulk MOSFET, if you recall, threshold voltage is twice psi f plus q sb psi oxide. So, you can see this twice psi f gets multiplied by a big factor here. So, threshold voltage in the accumulation condition will be much more than even the bulk MOSFET. That will not work out for you if you are looking for reducing the threshold voltage. This one, in the bulk MOSFET, it looks almost like a bulk MOSFET twice psi f plus q d by psi oxide. But in the case of bulk MOSFET, what is the q d? If you see the bulk MOSFET, bulk MOSFET that is the V threshold voltage is twice psi f plus q d by psi oxide. What is the situation where the oxide things are same? Now, in the case of fully depleted also, that is V T f fully depleted, we saw twice psi f plus q s by psi oxide. Is that right? So, you have got that particular situation now. So, here, difference is these two terms. q d in the bulk MOSFET is q n a x t maximum, twice psi f. Here, this is q n a into T of s. T of s is much smaller compared to x t. So, this is much smaller compared to v d. So, this will be V threshold voltage of the bulk MOSFET will be greater than V threshold voltage of that. Now, when you go to inverted case, when you go to inverted case, you have got V T f like channels are inverted. That is equal to twice psi f plus q s. In fact, if you go and I think I made a mistake there, that previous case, it is not there. Two is not there. But, when you go to this now, you have got this twice. This is much smaller compared to the bulk MOSFET. Maybe half one-fourth or even smaller than that, depending upon how thickness you reduce, divided by twice psi f s, twice epsilon s. Am I correct? So, this is a term which actually gives rise to voltage drop across oxide. So, in all the cases, what you are doing is, by taking this situation, what you are doing is, see here. In this case, the drop across oxide is small because of depletion layer charges small. Now, if you reduce the doping, this further can be reduced. Ultimately, you can get the threshold equal to twice of f itself. Very small drop in oxide before inversion, just in this inverting. So, now, if you see the electric fields, psi s be 0. You see this equation, or all the cases, this holds good. Psi s f minus psi s be second plus that half of that. This is average field plus this. So, in psi s be 0, that is that field. And psi s s psi s b is greater than 0. This term is practically same thing, or even if some factor is there, this is there. The difference reduces that term. And I go to back to inverted, both are pi s f s. That is the field. Field in silicon. So, field in oxide is q s divided by whatever is the epsilon s into divided by c oxide f. Whatever charge E s electric field is there into epsilon s divided by c oxide, that is the voltage drop. So, what we are telling is, the field is smallest when the both channels are inverted, higher in this case, highest in that case. In the bulk, it will be even higher, because you have to take care of the replacement layer, which is wide. Now, this is the plots, which you can see now. Electric field is highest there in the case of back channel accumulated. And whole thing comes down. Whole thing comes down to this electric field is over, and the back channel is depleted. And you can see that the average field is here, corresponding to this field is much lower when the both the channels are inverted. If this outer diagram we will see everywhere, it keeps on lower, lower, lowest when it is both are inverted. Now, you have seen the potential distribution twice. If back channel is accumulated, potential is 0 here. The back channel is depleted, potential is positive here. The minimum is somewhere in between, and the both channels are inverted. That is twice, twice, twice, five. And it is 0 or it is minimum at the center, because half of this belongs to the front gate, half of this belongs to the back gate. So, threshold voltage lowest in this case, higher in this case, higher in this case. Now, what we have said, summing up in this case is that you can go to lower threshold voltage with the fully depleted channel. If both the channels are inverted, and you can make a symmetric gate, symmetric MOSFET. What is symmetric MOSFET? Front oxide, that is the movement like this. Back oxide thickness and front oxide thickness same, everything is same. I can connect them together. I can have a situation where I have the back oxide, back oxide, everything is same thing. And front half go on by that, I can connect, I can connect the back gate and front gate and back gate to the same voltage. Now, what happens? If I do that, when I start applying the voltage from 0, what happens? Depletely starts coming from here, starts coming from here. I think I have to that will come like that. That is how it comes. From here, same way. That is how it comes. Electric field. That is the edge of the deposition here. At particular point, so whole thing is depleted and you have the situation where the potential is like this. That is the voltage across the silicon. So, both are same surface potential. Now, I raise the both, both are connected together. So, both move like that. After if I increase beyond that point, you see difference is not there. Then, afterwards it keeps on moving up like this till I get twice I have all that. That is the double gate, double gate MOSFET symmetric. So, people are looking into devices which are actually realized like this. Both gates connected together and so there is one gate control because after all you do not want to have one supply for the front gate and one couple of back gate. Make it symmetric connected together. There are some issues that we will see when you sort out by some other methods. Now, let us go further down. So, you have seen that if you want to see the threshold voltage versus back gate bias, back gate accumulator, threshold is higher. Back gate inverter, threshold is down. In between somewhere it is falling. This is just for completeness sake. I have got it. You can control that thing. How much is the slope will depend upon the ratio of the oxides in the front and back. If they are equal, it will come steeply down like this. If the back oxide is thicker, it will come like that. You can realize yourself sitting down based on whatever we have discussed now. It will be like this. And difference between the front threshold voltage when the back channel is accumulated and when it is inverted is that. C S by C oxides into pi by f. That is the difference. You can see it by substituting in those equations here. This minus that. This minus that is this quantity. Now, that is you know you can have the transfer characteristic for this MOSFET. I can take the characteristics in the back. I am not showing a symmetric gate. We will come back to that afterwards. This is one of those experiments which some of our students in IIT Madras have done it. Applicated these devices. And you can see back gate is thick. Maybe about 0.4 microns. And T is equal to 87 nanometers. Fully depleted. Something like 10 to the power 14 doping. Front gate oxide is thin. Something like 0.2 micron or so. And W by L ratio is 200 by 15. L is 15 microns. Long channel. We do not want to walk down by the short channel effect. So, transfer characteristics is exactly like in the bulk MOSFET. The only thing is we can change from the accumulation negative. You keep on doing that. You can see threshold property becomes less, less, less, less like that. And if you go, keep the back gate voltage all the time inverted, large voltage will not be able to turn off by the front gate. So, these are just illustrated that. And also you will get. Now, let us go to this particular situation. This is just for showing that you can have a good control of the back gate. You can change the threshold drastically by using that. Unlike in the case of bulk MOSFET, you can bias the substrate, but if you cannot apply forward bias onto that because the source tanks will conduct substrate current flow. Now, what happens to the sub-threshold flow? That is what we want to see. Maybe I think I will take on this particular interesting thing in the next class. So, in summary, what we have said is that you can get complete control of the back gate bias. You can go from accumulation to inversion back channel. And you can have a threshold voltage varying like that. In fact, you will see that you can get also control on sub-threshold voltage by using this quantity. And this is the main advantage of this is threshold voltage can be reduced. The transfer field can be reduced practically so that there is a chance for you to get higher mobility. We will continue on this from here onwards. The life will be much more easier because we understand how it works out now. Not much of mathematics from here, but more of what happens in those short channel devices. We will discuss that in the next one lecture before we wind up. That is why I think.