 You can follow along with this presentation using printed slides from the nanohub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. Okay, so let's get started. Today I will be talking about MOSFET current voltage characteristics and some advanced topic of it. This is lecture 36 of solid state devices. And today we'll be talking about the DC characteristics of MOSFET as you realize. This is IV characteristics or current voltage characteristics. I will quickly review the square law simplified bulk chance theory that we have been talking about. Now this is called a square law because the current increases with the gate voltage as a square of VG minus V threshold squared. So as you let's say you have one voltage of one volt overdrive. They call it overdrive VG minus VTH. If you have one volt versus if you have two volts overdrive then the current would increase four times. Three volt overdrive nine times current will increase. That's why it's called a square law. So we'll quickly review that to see why for last 20 years people haven't really seen square law being experimentally correct or experimentally observable. So this is a very simple theory for that. We'll talk about velocity saturation as being one of the reasons. And then we'll make some comments about very small transistors and the additional physics that happens here not captured by the square law. I will not even capture by many semi-classical or more advanced models. So we'll talk about that a little bit. Now if you remember that if we operate a transistor a MOSFET with a certain drain bias VD greater than zero and a fixed gate bias then we realize that close to the source region which is marked at zero let's say close to the source region there's a lot of inversion because the essentially the full inversion occurs there and the effect of drain bias is not as significant. So you have a lot of charge there in the beginning of the channel. As you go down in the channel along from source to drain the charge gradually decreases the amount of charge gradually decreases because you do not have as much inversion because of the drain bias effect of the drain bias. As a result in order to keep the current continuous you realize that the velocity of each of the section electron velocity in each of the section average velocity must increase right. That is the only way current can remain constant. And the only way the velocity might increase is if you allow a larger electric feed. If you allow a larger electric field in that region in the ith section let's say so therefore the velocity V shown here in red is given by mu times the electric field E and larger the electric field larger the velocity so that it can compensate the reduction in the charge and keep the current continuous. That was an assumption we made. And using that assumption we were able to sum up over the channel require that in each section without recombination current must be continuous. So we pulled it out as something called JD and since we are summing over the dy its small section from the beginning to the end of the channel we call that. We can add the channel length and we integrated it out and got this very simple formula. Now that's fine except you realize that there is something wrong here or at least potentially that could be wrong here. Assume like in a modern transistor you apply 1 volt in the drain and let's say you have 20 nanometer source to drain separation channel channel length that would be a typical latest Pentium type transistors. In that case the electric field you will have will be huge will be huge very large electric field much larger than the saturation velocity we talked about before. Do you remember that at some point if you push the electric field too much the velocity doesn't increase anymore it's going to saturate. So then for that physics of it is not shown in here. Here it says put on an electric field the velocity will always keep rising linearly that is the first assumption in the top line where the rate V velocity has been replaced by mu times electric field E directly. That is not generally true and that's something we need to fix in order to make sure that our expressions are correct. Now you remember that this is essentially the IV characteristics before we go on and the square law simply says that at the current essentially peaks at the saturation velocity and the saturation velocity we took the derivative last time remember this is like a parabola that comes up and down and then at the peak point we said the theory is valid only up to that point we calculated by taking derivative of the current we got the saturation drain voltage and from the drain voltage once we inserted that in the full current expression we got this square law. So that's something you know what I'm trying to tell you now is that if you go into an lab and made a measurement of a MOSFET characteristics plotted in in this way you will never find it that it goes as Vg minus Vt square and that's because of that wrong assumption that velocity is directly proportional to the electric field no matter the magnitude of the electric field that's wrong and that's what we need to correct. So let me explain how this saturation velocity is accounted for in the MOSFET current voltage characteristics first of all these are very simple so there's no reason we shouldn't be able to follow it very easily. So the main problem is this that if I were to look at the velocity of the electrons as a function of electric field you remember that in MOSFET in silicon or even gallium arsenide in other materials by the time you sort of exceed 10 kilo volts per centimeter at that point the velocity field characteristics rolls over beyond that point you may have let's say 10 mega volts per centimeter but in that case the velocity cannot increase anymore so you have the velocity saturates to a particular value why does it do so because at very high field the scattering is such this is chapter 6 in the first part of the thesis or first part of the of the course so if you remember so the reason what why it happens is because at very high field the scattering back scattering of electrons is exactly proportional to the number of electron that is coming in so it immediately begins to have two stream going against each other if you push this one more the other one goes back up also and the difference remains more or less constant and that's why the velocity saturates and an expression for the saturation velocity is given by this particular formula with electric field now do you see that this formula might work how this five formula might work one is if the electric field is small e is small so you can see in the denominator the e plus divided by ec ec is the critical field this 10 kilo volts per centimeter that's that's the centimeter per second that I'm sorry 10 kilo volt per centimeter that's ec so if the field is lower than that then the second time on the denominator will be zero and then the velocity will be just mu times e the minus sign is because these are electron in the direction of the electric field will be the electrons will be going opposite to the direction of the electric field no problem so now if the field is very large then what would happen then the denominator the second time in the denominator that will be the dominant one and the square root and the square will cancel each other and you can see that the electric field will cancel top and bottom and the ec will go up so ec is a number so the velocity will saturate at mu times ec so that's why this actually sort of describes the saturation characteristics of this velocity field current now there are various forms that people use sometimes people use directly under the square root especially for the holes and sometimes they will use instead of the square and the square root in the denominator they'll just use it proportional to this regardless that's simply makes the transition region slightly different other than that you can see both expressions in the low low field and high field regime will be exactly be the same in between they might be slightly different now we'd like to then use that information and rederive rederive the current voltage characters let's see how it's done very simple you see so again we go exactly the same way did before take each segment one two three four here for example I have divided it into four segments let's say it's 1000 angstrom I have divided it into 250 angstroms four segments and again I have written exactly the same expression as before q charge amount of charge in that section new time see electric field but this time this mu is not a constant this mu will be a field dependent quantity just as we saw in the last last slide since this will not be a constant anymore it will depend the mu will depend on the local electric field and that local electric field is varying at every point right there therefore this time when we sum them all up we will be able to pick out the ji the current because that current is continuous without recombination current from one section to next is same however the mu this time we cannot pick it up outside the sum because at every section the mobility depends on the local electric field this is the only difference nothing else this is done so that is the only difference that will give us the right answer so again we'll all you have to do is to insert this value of qi at each section so let's do that you remember that the qi is in the simplified bulk charge theory we have c ox vg minus vt minus mv right m is a body coefficient you remember that's the effect of the drain voltage coming in the charge that's why the charge goes smaller as you go from source to drain right that's minus mv factor now you can integrate the right hand side i will not even worry about it but do you see on the left hand side what i have done for the mobility the mobility is no longer just mu not because in that case it would be linearly proportional to the electric field the velocity rather i have that one plus e divided by ec that extra factor that was giving us the saturation i've inserted in there now this looks like a horrible integral but you see in two seconds it will work out this is how the right hand side you get the integration you get the conventional looking square law type expression on the right hand side no problem but do you see on the right left hand side how this works so the mu not you pull out no problem j d you pull out current is continuously pull it out but then this is if you flip the numerator of the mobility field characteristics over then you see there'll be a one that's fine and instead of the electric field you will write what should you write electric field is dv dx or dv dy in this case because y is along the channel direction so you have dv dy and you can see that ec e sub c sitting there that's the saturation field that's 10 kilo volt per centimeter approximately right now you can do this integral right there's nothing in nothing in there because the first term first term with one you pull it out if you pull it out then the integration of dy well that gives you the channel length the first term is the channel length what about the second term you can see the dy will go away and you are left with dv and if you integrate over dv then correspondingly the only thing you have is you will pick up a gd over ec that you can pull out and you will have a vd left pdf and that once you take the you can then divide out throughout and you will see that you have exactly the same expression as before except that little term vd over e sub c sitting there apart from that this is exactly what we had in the square law so apart from that we shouldn't have to worry about anything else but that of course has a important meaning see what happens if you have one volt and ec is how much 10 to the power 4 right so that has a dimension of a length if that length turns out to be larger than the channel length and that will happen if your voltage is high right if your voltage is high then that might happen and in that case the lc will actually disappear lc will disappear because the second term on the numerator is actually larger and if lc disappears what it says that you will not get more current by making the channel length smaller you see because there is not no lc over there anymore it's a very strange thing but do you realize why why why that might happen the reason is that as you increase the voltage the electric field is going up right the electric field going up at some point the electrons cannot go any faster so if you make your channel length smaller and therefore at every point the electric field goes up well in that case the velocity cannot go up so therefore current remains constant independent of voltage right so therefore you will see there's another consequence of that also because the velocity cannot go up you will see the charge in q3 and q4 they also have flattened out they are not going down remember in the original one at each section the charge was going down look at this one in the q3 and q4 we have higher than the saturation electric field as a result the velocity is constant if velocity is constant current is constant can the charge change of course now so in that case therefore you will have a constant q once the velocity saturation point has been reached so this is very important so therefore you will not have a depletion in the velocity saturated regime the charge will essentially plateau at one point and it will not go down any further again how would you then calculate how the characteristics look like you should take a derivative with respect to the drain voltage to see how the curve looks like and at what point they become maximum so you have that iv characteristics you know just derived it so we can take a derivative of this now the derivative is a little complicated because we have the numerator and the denominator all depending on the vde so it takes a little bit but the simplest is if you take log on both sides and then take the derivative then becomes very simple and then you set the derivative to zero and that gives you this horrendous looking expression 2 vg minus vth divided by m and you have the also a long denominator this is something you should check out it takes five minutes of algebra but very very simple but first thing i want to point out here do you realize that the point where this curve would saturate is a little less than where the square law curve you know the simpler one that we derived in last class and also in the beginning little less than that you realize because look at this if i had channel length equal to infinity channel length equal to infinity then the numerator of that curve that has that mu naught vg minus vth and all those things if my channel length was infinity then the second term on the denominator in the middle expression that would have gone to zero then i would have one plus square root of one that's two then two cancels the two on the numerator that would have been equal to vd minus vth divided by m therefore anytime i have finitel anytime i have finitel definitely the denominator would be larger than the previous case then two as a result the overall expression vd sat will be little lower and that what simply says that the current will saturate at a little lower value compared to the square law so again you can you can do this you can plot out this various quantities but one thing you will notice now look at this this expression for id doesn't have that square anymore it is vg minus vt but it doesn't have any square and also you do not have any mu left in this expression you have simply v sat which is mu naught multiplied by ecs of c the saturation electric field so this curve spaces out linearly so therefore if i give you an experimental curve or if you are reading a journal paper and going through a particular paper in five seconds by just looking at the iv characteristics you can immediately diagnose whether this is a short channel transistor or whether it's a long time channel transistor because if these are going out linearly then you immediately realize that this is a short channel transistors velocity saturation is occurring and immediately from the pre-factor you should be able to calculate the v sat out from there from that expression right because this is telling you that this is only will happen if the velocity saturation occurs and you can take that expression and assuming that vd over ec is actually much larger than channel length then you can essentially put those uh you know i'm trying to evaluate the current at the maximum point because beyond the maximum point the current will be constant and so therefore you put the vg side value in that we just calculated in the last slide and that will give you the expression for the current v sat c ox vg minus vth do you realize that this is a correct expression c ox vg minus vth is what is that that's charge right v sat velocity charge multiplied by velocity that's current so dimensionally it is correct you will notice that there is a w missing between that top and the bottom expression and the w is because in one case i have looking at id which is that total width dependent current total current and the top one is jd is per unit width so therefore you have a w difference but other than that exactly exactly the same expression and uh one way to show that how that current comes about is to i'm sorry i think yeah i think uh the real way to calculate the id sat i just showed you an approximation is to insert that entire vd sat value in the id expression if you do the whole thing rather than the simpler expression i showed then you will get a iv characteristics which looks very complicated but the bottom line is that once you allow the l to be small because we are talking about short channel you will realize that how you get the final expression very easily that goes as vg minus vth linearly rather than as a square you should also try that if you let the l to go to infinity this expression will go back to the square law so therefore both ends come out from this expression so this is a few lines of algebra but you should try it out when you're at home and see that it indeed works out so let's look at some how would you just what i was mentioning a few seconds ago that how one would know that you are in a velocity saturated regime that if you look the iv characteristics if it goes as a square of the vg minus vth that will tell you you are in the long channel regime and although i'm old but i haven't seen any and i have worked many years in industry i haven't really seen any square law dependence in any transistor in last 20 years so this is something historical remnant of how the transistors used to be but all transistors now essentially will be either linear or slightly super linear on this so most of the time it's not vg minus vth per se but raised to the power alpha and the alpha is on the order of 1.1 to 1.3 something like that but not definitely 2 not vg minus vt is square but not exactly linear either but it's somewhere in between but a close to one then close to two so this distinction is extremely important that tells you about transistor characteristics now many times in various exams i set up a problem like this i take something from a journal and put it there and ask you to calculate for example from here i give you let's say the y axis is given drain current is given so i ask you calculate the oxide thickness now in many times you should be able to do that because you can see that from this expression if w is given you know v sat right for silicon 10 to the power 4 centimeter per second volt per centimeter yes or not centimeter per second velocity and the field is volts per centimeter 10 to the power 4 volts per centimeter in that case you know that you know from here the vg minus vth you can calculate c ox and as soon as you know c ox the oxide capacitance you know the oxide thickness so from a curve like this you can pull out a whole set of equal uh a full set of parameters associated with that transistor with that process sure you should think about how to apply the transistor how to apply this expression not just uh how to derive the expressions and as i said that in practice this vg minus vt the alpha is obviously between one and two but close to one these days come not not close to two okay so this is all i needed to say about classical transistors let me make a few comments about the bulk charge theory remember we are talking about simplified bulk charge theory the whole bulk charge theory looked horrendous very complicated i still want you to i want you to understand this a little bit but this is not something i would really place a lot of emphasis on because this is not something that is used widely anymore now you remember that when we looked at the inversion charge then we had the c ox vg minus vth minus v that was the first term and then the whole bulk charge proportional to the accepted density now the second term on the first line that is because of the difference in depletion close to the source you have one depletion close to the drain end you have a different depletion therefore that depletion bulk charge depletion is reflected in the second term of the equation and you realize that this 2 phi sub b plus v that type of expression the depletion which decodes as square root of v do you remember vbi minus va that used to give us the junction with so very similar to that now we actually took care of the second and the the in the second line the last two items we essentially didn't really handle this properly we just took made an equivalent and push it as an m multiplying the v that was the simplified simplified bulk charge theory but if you wanted to keep the whole thing you know it's no rocket science because you can see how to handle this so this was the square law this was a simplified bulk charge so we could handle that no problem very simple integral but the original integral is not impossible either you can see it works very easily all you had to do was do the same sum as before sum them all up segmented segment the channel into n number of sections at every point you sum up the current and you do this integral on the second the square root of 2 phi sub b plus v that integral you do this that's not not complicated at that the end what you get because from the second integral coming from the blue instead of m what you get is a more complicated expression but square root of v you see when you integrate of course you'll have v to the power three halves divided by three halves so that's what you have here and that's why they call it three halves law for this because this vd comes as three halves so this is essentially very similar to the bulk charge simplified bulk charge theory that we talked about but this many many textbook this is shown do you realize the second term in blue or the terms in blue that came as m multiplying vd square divided by 2 in the other case that was m vd square divided by 2 so this whole thing actually only thing it does it explicitly tells you how the doping dependence na the accepted doping dependence how it influences the current other than that this doesn't have any explicit or many more much more inside inside than this so you should be able to take a look and and see how it works now one thing I want to point out that so everything I told you so far are fine it's generally fine but in transistors lots of much more complicated things happen for example here what I'm showing you that let's say you have a field free region up to let's say 0.4 or so micron this is an idealized case then a high field region the field essentially goes up and then then again the field comes down let's say 1.1 micron or so the field comes down and becomes small so a low field very high field and a low field region now here you see something strange happens because we always thought that velocity is given by mu times the electric field therefore in the beginning the electric field should have been the velocity should have been low and let's say and then when the electric field is high the velocity should have jumped up in proportion and then the velocity should have again gone down look at this what actually happens the rate curve says that it not only jumps up it jumps up by 60 percent much higher than the saturation value that's this called overshoot velocity overshoot the velocity versus field characteristics that I showed you before remember that the thing saturating that was actually for bulk semiconductors you take a big chunk of silicon put a uniform electric field and measure the velocity then change the battery or increase the voltage measure the voltage again that's for a bulk semiconductor when in a small transistor the electric field changes very quickly then it is like you know if you are driving a car if you all on a sudden push on the accelerator it doesn't immediately go to the saturation velocity it may overshoot and then it may come down right with the friction it may come down to the final value and this overshoot many times are very important that was not captured in our IV characteristics right so that is something you see it will come down to saturation velocity because the electric field is 10 to the power 5 volts per centimeter which is higher than the critical field so therefore you can see that it has come down to 10 to the power 7 centimeter per second however in between it has saturated it has had a huge velocity overshoot therefore the IV characteristics that we develop is reasonably okay but doesn't really account for these overshoots and other non-equilibrium effects and as a result must be must sort of more be modified then that you will learn how to handle this in more advanced courses and the main point I wanted to make the velocity is not given by simply the bulk velocity field characteristics it's a very important important consideration the another thing is that something is important to see this is a MOSFET you can see the source side the N and the drain side also and then going between source and drain through the channel region and you can see these black dots at the electron this is called a Monte Carlo simulation a Monte Carlo simulation is where you take a random number and generate an electron with a particular energy and a velocity and then they let the electron go and the electrons are scattered by the lattice by phonons and various impurities and it goes back and forth and finally it gets out either through the drain or through the through the source now had it been a long channel transistor you know like a couple of micron long then you would see all the black dots sort of hugging close to the conduction band because they have lost energy and they would have been very close to the conduction band but these days the transistors are so small that they get over that barrier and many times they did not have enough time to scatter before they get out of the of the drain side therefore all the things that we talked about mobility you know everything sitting in the bottom of the conduction band effective masses all those will not be correct anymore because these electrons are much higher in energy so what I am trying to tell you is the simplified laws that we learned is fine but if you want to apply it to very small modern transistors like the Pentium these days then you have to be careful and do a more sophisticated calculations like this you can also see that the velocity look at the velocity of this of this particular transistor look at 10 to the power 7 that's supposed to be the maximum velocity you are not supposed to go over it right that's the saturation velocity but look that this is getting too close to three times it is seven because of all this non-equilibrium effect because they didn't scatter enough they simply essentially were going from the source to drain without too much scattering so these are very important consideration and has been important consideration for a long time so let me summarize then one thing I wanted to emphasize today in today's lecture was the importance of velocity saturation because if you have a strong electric field at some point the electrons cannot go any faster therefore the charge must remain constant in order to keep the current constant and therefore that is reflected by a linear dependence with vg-vt that I have explained now the Barthes-Chart's theory which is encapsulated and by the number m the body coefficient or in the full Barthes-Chart's theory that had the three halves that we derived that actually explains why doping dependence of the body or doping of the body is so important in controlling the current that na the accepted doping why it's so important and as I mentioned that the velocity overshoot or the non-equilibrium effect in very short channel transistors these actually complicate things very much so therefore if you take a job today in any of the companies you'll see that what I've told you so far about IV characteristics gives you the basics but if you try to make a simulation just based on this and try to make an estimate you will see that that does not immediately go with the experiment but the thing is it seems still important to make simple simple estimate because you wouldn't know otherwise whether a technician let's say when you become a big scientist in a company getting let's say paid a lot of money then most of the time that you request a technician a colleague to make some measurements and they bring you some measurements and you are not yourself doing the measurements so how do you know that at that particular day a knob was not set properly or he may have made a mistake when he was reading off the data or multiplying excelsate multiply with by the wrong constant so the people who do not have simplified understanding about what the result should be in the simple case what the result should be a make estimate most of the time they are not very good engineers because most of the time they cannot see what is normal and what is really abnormal right so therefore although the simple things are looks almost obvious you will find it enormously useful when you actually if you practice in industry and also the same from universities when a student comes back with some measurements how do you know what he did in the morning and the only way to check is to know in the back of your head this simple expression so that you can immediately check now for very short channel transistors I told you very complicated theories but in very short channel transistor turns out most of the electrons go from source to drain almost ballistic meaning that without any scattering and there is you know professor mark lansstrom and in the industry one of the most widely used theory for mosfet about how current goes in a very short transistors he goes by his sort of almost by his name and this is a lansstrom theory of mosfet that encapsulates the ballistic transfer that's considerable simpler when you make things very small electrons don't scatter then the expressions become considerably simple that we'll learn in another course okay now I want to this is essentially the lecture I prepared for this saturation characteristics and others I want to show you a few more slides because in the last lecture after it worked out and after I got a few emails from you I realized that I have not able to been able to explain some things as clearly as I should have so I want to take a moment and explain to you something that that will be helpful to you and this is about this charge getting smaller down the channel and let me try to do it one more time see whether it helps so the point I wanted to make was that in the last lecture I told you about that when you apply a drain bias here for example vr that's the drain the channel is this had hesitated region and that's the depleted the black part is the depleted depletion region the drain is n plus and so most of the depletion goes in the p region this is lower dope so I told you about that and we explained that if we increase vr then gradually there's this voltage potential drop as we go through the channel and then I also explained that in order to invert it invert this channel in a state of 2 phi sub b being sufficient you will have to use this additional vr phi sub s must be 2 phi sub b plus vr in order to invert the channel and it looks like to some of you or to many of you may not be this might not have been as clear as I'd like and similarly for this one so essentially the same point I want to explain the essence of this slide one more time and let's see whether we can do that and also where that bulk charge term comes from so let's think about it the point I want to make you actually know this you may not have realized this but you actually know these expressions quite well so let's start assume that you have a bulk piece of semiconductor red is a conduction band blue valence band and the dotted lines are electron and hole quasi fermi levels in equilibrium if it is in equilibrium then the fermi level is close to the valence band so here this is you have a huge number of holes that's the p and you have a small tiny amount of electrons that's the red how many electrons ni square divided by na that's how many electrons here tiny amount of electrons now and you know that expressions for n and p you already know that that in terms of ec minus fn and ev minus fp so you know that now let's assume that I have applied a voltage in the gate let's say whereby I have pulled down the conduction band and the valence band without touching the fermi level because fermi levels are sort of independently said by the body contact and the drain contacts so I have just pulled it down what will happen to my electron and hole concentration you know the answer my whole concentration will become smaller do you realize because the fermi level is farther out compared to the valence band my electron concentration will be much larger than before because the fermi level is now close to the conduction band is it already in inversion yes this is already in inversion because I have a huge number of electrons compared to the holes so things have gotten into inversion now let's say somehow and I'm not telling you how let's say somehow the fermi level for the electrons which is the red line red dotted line somehow I have pushed it down compared to where the fermi level for holes are the blue dot that line what will happen you know what will happen this is what's going to happen you see the whole concentration I have not changed because the difference between hole fermi level and the valence band remains the same so that has not changed but look at what has happened to electrons now electrons because I have moved the electron fermi level down how I'm explaining in a second because we have moved it down so now the gap is larger if the gap is larger electron concentration will now go smaller right this is farther out now instead of just having two parallel separation if I instead have a step how would I made this the same time I'll explain in a second but if I made a step for the electron fermi level the red ones then what would have happened this is what would have happened because look the blue fermi level and the valence band they are same separation so I have the same whole whole concentration but because the electron fermi level is moving in a stepwise fashion therefore the difference from this conduction band is changing further out it is more so therefore you have less number of electrons down the channel right this this far should be clear now how do I do that how do I make something step like this and you already know the answer because you already have done a homework like this and that was the purpose of that homework do you remember that if I have a p-n junction and then on the n side if I apply a voltage then the voltage the fermi levels remain flat up to the other side of the junction and then gradually it goes back to the main main region right do you remember that so that is exactly how that step comes about because this is essentially the p-n junction that you have done when you had the recombination problem remember when you did the recombination problem you had this reverse bias p-n junction and you had to calculate the recombination at various points this was exactly that expression and this then is the charge that you see in those expressions you see that along the channel you have the charges actually starting from the source end to the drain end the charges gradually going down this is the same reason the charges go down in the normal p-n junction because as soon as you apply a drain bias the whole thing is reverse biased and that splits the fermi level and makes the electron fermi level lower than the whole fermi level right and then it will eventually come back and come together and that's the being reflected in these steps in the charges and so that's that's all I wanted to emphasize to you because this variation in the charge looks complicated but this is something you have seen before and you should be able to easily understand it without too much effort. Now what will happen regarding the depletion you know there's a second piece now in depletion what happens let's say that's the accepted density the green you apply gate bias as we have done before electron concentration goes up but this time I'm just plotting the depletion the depletion charge the NA and so at some point it will invert the channel will invert at some point now however if your electron fermi level is actually have been pushed down then do you realize that your electron concentration must be going down because the separation between conduction band and the fermi level is now more than before so electron is less now if you want it to invert then what do you have to do you'll have to put more gate bias here right more gate bias so that you again it's like trying to chase the fermi level first time it came close to the fermi level it got a certain amount of charge now it has moving target the fermi level has been pushed down so now you'll have to if you want to again invert it you will have to push it down even more so that the separation again becomes the same as before then again it will invert in order to do that then therefore you'll have to apply more gate bias and that gate bias therefore will result in more depletion right so therefore when you move the fermi level like this with a drain bias then the depletion width also keeps changing so I'll end here let me end here and hopefully we'll go back in the previous lecture and read it along with this particular material or try to understand this material so that everything becomes clear okay thank you