 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 let's get started and today we'll be talking about non-ideal effects in MOSFET and presumably so far we have been talking about an idealized version of MOSFET in which several things were simplified but in real life there'll be a few things more things that we would like to add and in some ways best discussed separately so that you can clearly distinguish between the essential operation of a MOSFET and this additional considerations that are present in every MOSFET and are very important for the operation but are not fundamental to it. So let's the first topic we'll be talking about is the so-called flat band voltage. I want to define what it is how to calculate it and how to measure it from experiments. Then we'll talk about the threshold voltage shift due to trap charges. Now we hadn't been talking about charges within the oxide. The oxide we assumed was charge free the electric field in the oxide was constant but in many times the oxides are amorphous material remember silicon dioxide these are random material lots of defects in there and how to once charge often gets trapped in there and that shifts the threshold characteristics and we'll see how that affects transistor performance. You can see that if you have traps in the oxide then or charges in the oxide then the charges that you need to balance the inversion charge on the gate would be slightly different so you can begin to see that there will be some some differences. We'll see how to handle that. I will talk about physics of interface traps as well a very important consideration and finally make some come to some conclusions. Now just like in bipolar transistor we use the gain of the transistor remember beta was done as a ratio of the doping of the emitter and the collector emitter and the base and ni squared for the base and the emitter and we used to hang our ideas of improving the bipolar transistor performance so to understand various features using the gain as sort of the central point through which we relate various ideas. So here I will use this current expression for the current ID is a drain current obviously above threshold using this law to essentially bring together various ideas that we want to want to think about. Remember that although I have written it as Vg minus Vth squared Vth of course the threshold this square is not always correct right the exponent is close to 1 rather than to 2 for modern transistor. So the first thing we'll talk about is the non-ideality of the threshold voltage. Next class we'll talk about the mobility mu naught and then we will talk about the physics of C ox and then that will come to the conclusion of the MOSFET analysis. Now the threshold voltage is more than what I have told you so far. What I have told you so far is this Vth threshold ideal which is on the first term that's that what I have told you before but there are these additional things that you haven't seen before. One is this phi ms that there's a work function difference between metal and semiconductor that comes in and we'll see where that comes in. Then there is this Qm is the amount of charge within the oxide divided by C ox there is this factor gamma there are fixed charges divided by C ox that also changes the threshold voltage we'll see how it works and there's also a finally an interface trap you can see it stands for interface and the amount of charge that you have on that interfaces depends on the band bending phi service. So we will discuss each one of them as you will see separately and then we can show that you can get a feel for what these things are. So let's get started first thinking about just briefly reminding you about what this Vth ideal was you know we have done it for last few classes and then we'll start by talking about this the phi ms the metal semiconductor work function difference. So now you remember this picture this was this one side is metal insulator and the semiconductor is P one thing I should remind you that in modern transistors this is how transistors used to be in 1970s that the first Intel transistor had the metal as aluminum as a metal in 2004 so in 1970s that has a metal in it and interconnects and others these days most of the transistors have for the gate polysilicon highly doped polysilicon it's a semiconductor not a metal not a metal anymore but for such a this is a strange twist of life in some way that we have actually gone back to metal now so the latest Pentium would again have metal and this is a different type of metal but again it will have a metal related compound over there oxide and of course semiconductor and we assumed a hypothetical metal and a hypothetical oxide such that the chi's eyes and phi sub m and chi s all essentially gave you a flat vacuum level but you know that that's not generally the case right for practical material that's generally not always the case so this was a spatial case and from that we were able to calculate above threshold above threshold the inward inward amount of charge induced mobile charges these are and that was c ox vd minus vth ideal and what is that vth ideal well the vth ideal was phi service and the phi service is 2 phi service the twice the bulk Fermi level that's the first term and qb divided by c ox that's the voltage drop across the oxide and you remember that you have to manipulate them a little bit to find that the electric field is related to the bulk charge and the c ox has in it the x0 the oxide thickness so you get that right that's the amount of inversion charge that you need if if the bands were flat and in that case life was very simple it was flat at zero voltage so the voltage was flat everywhere electric field was flat everywhere essentially zero because that's the derivative of the potential and the charge derivative of the electric field so take as a derivative of zero that remains as zero so life was good at zero voltage that is where the transition point was between accumulation remember if you put negative voltage accumulation if you got positive voltage that's our depletion first and that transition happened at one zero volt idealized case but in reality things are going to be slightly different as you can as you can realize and also by the way the vbi the built-in voltage was equal to zero because remember the vbi is essentially difference of the vacuum level on minus infinity and plus infinity though those two last two materials matter flat everywhere zero vbi in reality of course what you find that the metal work function depending on what metal it is if it's aluminum copper or any other material for that matter you will have a different a work function five seven now in general depending on the semiconductor you have if you have silicon then you have one chi-sub-s or if it is germanium you know many transistors these days are germanium transistors you'll have a different chi-sub-s the oxide inside the oxide could be silicon dioxide many times these days hafnium oxide so that chi could also be different so in general these numbers can float around so therefore you have to follow the rule about drawing the band diagram you know that how to draw band diagrams in presence of this right and generally i'm not going through the details you will have a band diagram something like this you can see the blue line flat quasi Fermi level do you see on the right hand side i have redrawn the bulk region and the continuity of the vacuum level eventually you see a diagram like this i'm sure all of you know by now how to draw a band diagram now you see in this particular case the band is no longer flat band is no longer flat and this has banded in a particular way because phi sub m is less than this phi sub s and i have defined what those quantities are because this is one is bigger on this side you can see it has sort of tilted towards the metal side had it been some other metal and some other semiconductor combination it could have tilted the other way so we cannot a priori say that which way it will tilt but in general let's let's start with one and then we'll get the basic idea for this spatial case now for this spatial case you realize that there is something very interesting here first of all even at zero bias even when i haven't applied any gate bias body is grounded haven't applied anything on the gate so first of all you can see that phi sub s the surface band bending is no longer zero right previously it was all flat it is has sort of banded down towards the towards the oxide already therefore if you look close to the surface you will realize that not only the holes have been pushed back holes have been pushed back even at zero bias and the whole thing has been depleted near to the surface oxide semiconductor surface but in addition i have electron build up on that on that corner that green corner the electron has built up why more than the equilibrium one equilibrium one is ni square over na that's the equilibrium one on the bulk side but i have electron build up of course through generation and on that corner now do you realize that this doesn't happen in a diode why not why what is in diode if you have these two are metal semiconductor junction right and if you have banded like this what will happen it will happen there'll be depletion on one side there'll be depletion on the other side and then essentially there'll be a depletion region we can calculate that we know how to calculate it there will not be a pocket of electrons piling up that happens because we have an oxide here an oxide is acts like a dam it doesn't let the electron go in when you had a diode we didn't have a sort of a oxide or a dam sitting in between the p and n side preventing the electrons to go in that case the green electrons will go to the other side and mingle with the electrons on the metal side so this extra build up is sort of unique for MOS configuration or MOSFET configuration do you realize why there's extra electron because you can see the conduction band has gotten closer to the Fermi level and therefore you expect more electrons to be there and less holes okay now let's ask ourselves this question if I now start applying a gate bias right I want to invert this transistor do I need more or less gate voltage in order to invert it now you can almost clearly see that you might need a little less right in this particular configuration it's already sort of bended in the right way of how I want to bend it I put a little bit more I have 2 phi sub f surface where band bending I have my inversion point so you can see immediately that my total amount of voltage needed to reach threshold will be a little less because of this band bending in this particular way now you realize had it been the other way around phi sub m being greater than phi sub s in that case I might have needed more because it have gone bended the opposite direction and therefore I might have needed more to bring it back and invert it so this is then extra factor the phi m s that sort of brings in comes in in the threshold voltage considerations now let's talk about what is this flat band voltage issue now physically what flat band voltage means is the amount of gate voltage amount of gate voltage you have to apply in order to make the bands flat you know that's flat band voltage so which direction do you think we should go in order to make the bands flat vg is zero now now do you remember that when I apply a negative bias to a contact it generally goes up apply a positive bias it goes down right you remember that so in that case in order to make this band flat which direction I have to go I have to apply a negative bias exactly right and that negative bias will make the whole thing flat and that is called a flat band voltage fb is for flat band that's the small part of it and this is telling me that the flat band voltage is equal to the difference of work function between metal and semiconductor but also equal to this minus vbi this is the same built-in potential that we have seen before do you remember you know diode we have built-in potential same so let's calculate this because once we calculate it we'll know how far away from the idealized case are we so that we can calculate the actual threshold voltage so let's do that let's calculate what the vbi is now at this point I shouldn't really have to even do anything you should be able to calculate vbi for the left hand side structure immediately right you know how to do it so what do I do I always have followed one single rule and the rule to calculate vbi is essentially some of the work function till the flat vacuum level part on the top on both sides on the metal and semiconductor side and you know what you have to write do you agree with this on the right hand side I have chi service the semiconductor chi the band gap that's the eg and the delta p is where the Fermi level is with respect to the valence band because that's the extra I have gone in that direction and on the left hand side I should have written phi sub m plus q vbi that's what I should have written and therefore I have flipped it on the other side and you know one second I calculate what vbi is do I know everything here well chi service and phi sub m I can give in two materials I immediately know them from the textbook or any reference book band gap yes material known band gap known and delta p well that's if I know the doping from the doping I know how far away the Fermi level is so I can easily calculate it and that will give me the flat band voltage and you should convince yourself that this flat band voltage is also the difference between the metal and semiconductor work functions these two work functions so I'll leave it as an exercise is a very simple thing you can just see from this from this particular diagram therefore therefore if I know what my flat band voltage is then you realize that my threshold voltage will now be reduced will now be reduced by this idealized threshold voltage remember that's the first two term on the bottom two phi sub f and qb at c ox that's the idealized one that must be reduced by the flat band voltage because I already have that amount of voltage applied to the gate and so therefore I will need to apply a little less in order to invert it now you realize that if the flat band voltage went the other way we had a negative sign then I would have needed a more voltage to invert it so as a result my this expression for the threshold voltage because of this non-ideal effect will be slightly different from what we have calculated before but it's a very simple calculation you can see as soon as I tell you what the two materials are for the metal and the semiconductor then you immediately know what what this new threshold voltage would be right now now that's how you will calculate it but how would you know what you calculate right or not first of all you can punch wrong numbers in your calculator that's one problem the other is that many times you do not know whether there are other problems there or not so many times what you have to do is to go ahead and make a measurement and the measurement would be the capacitance divided by the oxide normalized to the oxide capacitance so you remember that on the left hand side of this plot these are accumulation accumulation charges are next to the oxide it's a parallel plate capacitor majority carriers from two sides so therefore you have a certain capacitance and then the as the oxide as a semiconductor begins to deplete with positive bias remember the holes are being pushed back your capacitance begins to decrease because you're essentially the total capacitance distance the D is larger than the original one so therefore it begins to decrease what type of frequency do you think I am at this is a MOS capacitor this is high frequency absolutely absolutely because in that case it didn't have time for the minority carriers to be generated and get the inversion charge respond to the small signal right so therefore I have that now threshold voltage you all know how to calculate threshold voltage from this car right the point where the curve essentially stops dropping and essentially becomes flat at that point in the dc the accumulation charge has formed only in the ac the accumulation charge cannot respond so therefore it responds to the age of the depletion on two sides that's why it becomes flat so at that point we call that the threshold voltage that's my ideal threshold voltage now again from an experimentally measured curve from the ideal threshold voltage you should be able to calculate back the what the bulk doping was remember 2 phi sub b phi sub yes 2 phi sub b minus qb divided by c ox so from that you should be able to calculate what the bulk charges from experimental curve now if you have a non-ideal device then what's going to happen for the particular one that I just showed you this whole oxide curve or this capacitance curve will shift to the left because I need less voltage to invert it so therefore the whole curve will shift to the left equally important the transition between accumulation and depletion that point will also change right because it's already zero gate bias is already sort of partly inverted or partly depleted and so the whole thing the whole curve will shift to the left and now this becomes your threshold voltage what is the difference between these two threshold voltages that's the flatband voltage or that's the built-in voltage right these two threshold voltage in this particular case the whole thing has shifted and how do you determine the flatband voltage again you can either do it by changing the threshold voltage or you can look from the accumulation to depletion transition the point where it goes from flat region to the depletion reduced capacitance part so that can also tell you how much the flatband voltage you need in order to shift this curve okay so that's one one ideal effect non-ideal effect that I have to take care of but that's very simple I mean nothing could be simpler now the next thing I want to tell you about is something that is really non-ideal and this one depends on device to device this is an issue about trap charges now why is it that I didn't ever discuss any of these things in bipolar doesn't this thing happen in bipolar transistor certainly it does happen over bipolar transistors also the things that I'll be talking about does happen but in many cases bipolar is a vertical device meaning current flows from the top to the bottom and therefore most of the time it doesn't see surfaces and most of the problem that I'm talking about has to do with surfaces where the charge gets injected has to do with amorphous material where charge can be trapped because it's not crystalline defects are present so therefore in many cases these could be more dominant in MOSFET MOSFET is a very sensitive device you realize 20 10 20 30 angstrom close to the surface a thin layer of electrons moving from source to drain that's all there is to it and you can easily see that it may be parted by any imperfection on the surface so therefore this is a topic that people pay a lot of attention when they are designing MOSFETs again let me remind you that this was the ideal one and I'm trying to see what the non-ideality would be if if there are charges if there are charges in the oxide so the first thing I want to remind you that I didn't have any charge in the in the oxide now is that really right is it that I'm not talking about this important to understand not talking about the electrons and holes in a traditional sense yes silicon dioxide 9 ev below the valence band in silicon dioxide you have a lot of electrons because this is a silicon and oxygen of course you have a lot of electron conduction band is so far up you know that the formula ni squared if you wanted to calculate ni squared it will be nc nv e to the power band gap over kt right remember band gap is 9 ev so you have almost no free electron in a silicon dioxide that's why it's insulated so of course in an ideal insulator you don't expect any conduction charges in India and similarly in other places things are charged neutral but the oxide charge I'm talking about is not electrons and holes in the traditional change this is sense this is the charge that sort of come in from outside and sort of start residing within the oxide so that's the extra charge we are talking about nothing to do with the intrinsic semiconductor now as soon as you put the extra charge life becomes a little bit more complicated because first of all I have you can see that even without charge I have shown things as we're having a band bending because you know just what I told you about the flat band voltage and all and therefore because the two work functions are not the same you have a certain amount of band bending you can see that here the threshold voltage will be a little lower right compared to before in addition look at the oxide and look at the oxide band no longer a straight line why not because that if you didn't have a charge whatever electric field came in that same electric field went out because any d e d x the change in the electric field that is equal to charge if you don't have any charge you don't have any change in the electric field a straight line but if you have charge yes now very important notice why I have put the 0 the origin the origin is always in for our calculations it will always be starting from the metal oxide surface don't try to put 0 on the on the oxide semiconductor surface because all reference or everything is referenced to that point so let's say I have some charges in now these charges although I have drawn it within the band gap this is really not within the gap in that sense there are defect levels within the gap just like remember copper aluminum they had defect levels in the silicon similarly there are defect levels and the electrons are sitting electrons and holes are sitting within those defect levels that's where the charges now if you have that charge as a result you can see that your electric field within the oxide will be changing and therefore you have a non-uniform potential within that within that region so I'm going to define a few quantities and then I'm going to go through and calculate how much for the extra effect of that charges so the first thing I want to define is the area under the curve so let's say I have a certain amount of charge the ordinate of that curve is rho sub o ox this is at every point how much charge you have that is like number per centimeter cube right so density you have certain amount of charge the area under the curve I will call that qm q sub m there is a historical reason what does m mean so there is a historical reason why the word m is used because historically in 1960s there is a great drama and I'll come going to talk about that a little later about mobile charges being present in the oxide in fact that almost killed the MOSFET program remember MOSFET was actually patented before bipolar transistor MOSFET was patented in 1920s but it happened much later might actual MOSFET operational MOSFET is in 1970s bipolar happened before that right 1950s so the reason that happened and how the engineers in Fairchild semiconductor this treacherous eight I always tell you about right who defected from Shockley semiconductor and went and built their own company fair joined Fairchild they in fact that series of five and six years beautifully solved beautifully solved many MOSFET problems and one problem they solved was about mobile charges so although this charges in modern context is no longer mobile the symbol m still persists from those days and I'll explain to that it's a beautiful story but qm is area under the curve area under the curve and disguise of m is a sort of the moment of the charge that if you had to replace all these charges with a delta charge at some point effectively then why would you put it you will put it at x sub m somewhere and I'll show you how to calculate calculate that but the way it's done you can see the numerator has the first moment of the charge and the denominator has the total charge divided by x naught x naught is oxide thickness so I will show you how to calculate chi sub m or x sub m in a second but this is the general framework let's say I can calculate it I mean you know the prescription is clear if I told you row oxide charges you can do this integral right no no problem you can do this and then once you have done it the extra change in the threshold voltage extra change will be the ideal one which is the first two terms right phi sub s and qf divided by c ox and then I'm sorry that should be qb divided by c ox and this this extra charge is what's going to cause a change in the threshold voltage right now why is it physically why does threshold voltage change I could previously see band bending so that's why I have a change in the threshold voltage let's try to understand first and then calculate so in this very colorful plot so I have drawn the yellow depletion region this is the charges depleted of what depleted of holes holes have been pushed back so I have applied a certain amount of gate bias and you can see the positive gate bias in green has resulted in some sort of positive charge and this charge is being balanced by the yellow which is the depletion charge and this what this that color huh this reddish color something and that color the sum of these two is is equal to the green area under the curve right this charge must be balanced now you know how to calculate the electric field in this context of course you just do the area under the curve when you're done you do you agree with this statement that the electric field so you'll have to start from the right let's say go through the yellow yellow is a constant region so the integral is a constant part as soon as you hit the inversion charge if the total amount of electric field goes up jumps up and then in the oxide I have no charge idealized case remains flat not exactly so there has to be a discontinuity by the amount of the two dielectric constants remember the electric field will change there by the dielectric constants and you go to the green side and the greats it sinks all the electric field and it goes back to zero idealized case so we know how this happens okay now it's minus e because when you do the charge integral you remember this is negative charge so you have to do in a particular way okay now what would have happened what would have happened if I now have some bulk charges or all charges within the oxide so you can see something else has to happen so let's say I put some extra positive charges in there first of all do you realize first of all do you realize that in this case if I look at the charge balance the green has to be a little less will need to be a little less why because the green and a blue together is a positive charge yellow and the red together are negative charges if I am at threshold if I want to go at a particular level of inversion above threshold then the yellow and the red must have a particular value now the green will take care of a little bit and therefore I will need a little less green I'm sorry the blue will be take care of a little bit and therefore I will need a little less green as a result I will actually need a little less gate voltage to invert invert this transistor because the sort of the blue the trap charge is helping me to invert right so you can immediately see that there will be my threshold voltage because of the blue charge trapped within the oxide will be a little less physically I mean there's no math here and you can sort of make that argument that if you wanted to look at what voltage you need the voltage is always the area under the electric field by the way do you agree with my drawing of the electric field diagram the yellow part a constant jump in the near the inversion then the electric field continues I should have made it discontinuous by the dielectric constant but apart from that and when it hits the blue then it goes down the total amount of charge goes down and then when it's a green it goes to zero so area under this curve which gives you the voltage area under this car electric field car gives you the voltage do you agree that this is a little less than the idealized case and therefore for the same level of inversion you need a little less voltage and that's why the threshold voltage is a little less for this positive case had this blue blue been negative it would have gone the other way right you realize that now you also should realize that it really matters why the blue charge is because if the blue charge is I'm shown here in red if that is close to the surface do you see the consequence of that the consequence of that is that that charge if the trap charge if it's closer to the surface it has even a stronger impact look at this the yellow region we integrate goes you know straight goes up on the inversion charge but the interface charge that I have immediately brings it down and then of course it remains constant and goes away with the green now the green in this case I will need even less because area under this curve area under the electric field covers that that's the voltage so that voltage is less than much less than the ideal one so the bottom line is that first of all trap charges reduces the threshold if it is positive in this particular case second is that closer it's to the interface more effect it has right these two things will of course come out in the math also as I'll show you in the next slide one thing before I go from here now I'm making everything with respect to phi with respect to a n-type semi a p-type semiconductor a p-type semiconductor now if I inverted it give you a problem where things are in n-type you cannot just copy from here you have to look at the total amount of charge look at the sign of the blue and keep the depletion and the inversion charge is the same that's your boundary condition and then correspondingly adjust things right always start integrating from electric field from the right side not from the left side because the right side is a constant you want a certain amount of current and then you are trying to calculate what gate voltage will give you that current so that certain amount of current is given by this red inversion region so therefore you have to start integrating from the right don't start from the left because that will give you a certain gate voltage but a different amount of current something that we don't want to calculate so let's calculate them a few lines of math very easy the gate voltage is equal to the drop in the oxide voltage and phi service before my life was simple v ox was electric field multiplied by x0 and I went home this time well I'll have to write the Poisson equation so I write the Poisson equation the second derivative of the potential with respect to x that's equal to rho rho ox oxide charge inside previously it was zero so therefore I had a constant electric field but this time it's not zero so therefore I'll need to integrate I'll need to integrate from x0 x0 is the right side boundary of the oxide from any x at any given x and I integrate on both sides this is all happening within the oxide so you can see the kappa ox right that's the dielectric constant we can see that so that's the first integration you put the limits in for the electric field that's the first integration and then if you want to do the second integration because ideally want v ox if I want to do the second integration then I will do integrate one more time what's there is to it there's nothing much e ox at x0 that's a constant value I know that that's coming from the inversion charge and the depletion charge right the e o low and the red the sum of that so I know that that's a constant so I when I integrate the it goes out of the integral and I pick up an x0 so I pick up an x0 and the second one I have a double integral now I again leave it as an exercise to show that the double integral can be reduced to that first second second single integral below there only thing is that you pick up an x multiplying that rho ox right that's that's the extra thing that you pick up now do you realize the first term of v ox is what I had before when I didn't have a charge in the oxide what would have happened if I made my rho ox is zero second term wouldn't exist right first term would be there it was always there so the only extra thing I have picked up is the second term because of this extra blue charge that I have within the oxide and again this one sort of does it make sense to you that you can see now if x is large the charge from the left side interface is further out from the left side the effect is more right you can see that that's why if the charge is close to the semiconductor oxide interface then the effect is actually more and that's already coming through this map what we physically and intuitively already knew so you can calculate this and this is the ideal part that's we have done through this before and this is the extra thing this extra thing if you remember the definition of kai sub m that I had this is a one line of algebra that tells you that if you have picked charge the q sub m within the oxide then your threshold voltage will be shifted by this extra amount it sort of makes sense because the charges either help you to invert a little earlier or hindered to here invert a little later so that that's essentially the physics of it again you can you can do the same CV characteristics and from that can calculate calculate this extra amount of charge to be present now you can ask that what would have happened what would have happened if both there is phi m of a non-ideality from metal semiconductor work function and non-ideality from q sub m both code have caused a shift in similar amount how do you decouple them so I'll ask you to think about it and then in the next class I'll tell you what to think what the answer is but the point is from the CV characteristics by looking at the change in the threshold voltage one should be able to pick up these charges okay now if you have interface charge then I can again use that formula but there is this dreaded delta x minus x not which you have seen also in Schrodinger equation right that if your charge is actually a delta function by the way it cannot be row ox anymore if you have x minus x not then you realize that that should be charged it's no longer per centimeter cube when you have the delta so apart from the dimensional change what should be the the whole thing you can pick up and the row ox at the interface that will be the q sub f and multiplied by x not and that x not will cancel the x not outside and you will have this c o the oxide capacitance so if charges are next to the interface then we have this very simple formula where we divide the fixed charge by the oxide capacitance I don't have to calculate any moment or anything I can immediately get the full effect in the threshold voltage so that's many times any interface charge this is done now these charges I'm talking about like trapped holes within the oxide so I'll show you later where the trapped holes come from okay now this is the what I was trying to tell you about 1960s and the time dependent shift of charges so and the experiment is in next slide but the essential point is this you expect that when you make a MOSFET you expect a IV characteristics now let's say in the morning you came to work and you measured IV characteristics you expect that you go to lunch and come back your IV is where it was before it's if you make the second measurement where it was before but after lunch if you see that your IV is completely shifted is completely in a different place that does not mean you did anything wrong with the lunch lunch was fine the MOSFET there is something funny going on in the MOSFET and this was a big problem it's called a bias temperature instability BTI and in 1960s it drove people crazy because what they saw that they would make a measurement in the morning same device a little bit later would have a completely different characteristics of course you cannot make a microprocessor like that right in the morning it's working in the afternoon it's not working not very good microprocessor but so there was a great detective story you can read it but essentially what people found out that the charges from this within the oxide actually was moving back and forth moving back and forth so this charge in the morning you start the charge there now your threshold voltage as you have applied so this is as a function of time in the morning let's say the charge was next to the green and then you went to lunch putting this oxide in the measurement setup and this blue charge would be moving as you are going as you are in during lunchtime and therefore as it goes to the other side you can see that now this x1t where the delta charge is is no longer at the same place therefore your threshold voltage throughout your lunch period is actually your threshold voltage is changing if your threshold voltage is changing then the amount of inversion charge you have the same gate bias right you set it up so therefore your gate same gate bias but your threshold voltage is changing what will happen to your current your current will keep going down because threshold voltage is increasing so your current will keep going down and that would be a big problem right so this turned out to be and also if you come back and apply a negative bias the charges will go to the other side the blue will start going to the other side and the CV will do this oscillation going back and forth so that's a very big problem and it turned out that this is sodium sodium you know many times people in those days didn't know that they have to use gloves in the clean room and all those and in our hand there's this sodium and sodium chloride so anytime you touch this wafers with bare hand right then what happens sodium from the hand gets into the wafer and that resides in the oxide and the sodium is a small atom it's amorphous material so it finds its path going back and forth sodium is a positive atom right so as soon as you apply a positive voltage it doesn't want to stay close and so it's it's pushed back you apply a negative bias the sodium comes back and so this oscillation goes back and forth bias temperature and stability because if you raise the temperature this happens faster and so this was a big problem and eventually it was solved simply by putting on more cleaner condition and putting some globes on these are the experiments so you can see in the morning that's the right hand plots and it can shift as much as 80 volts of course those days oxides were thick also thousand angstrom oxide so huge shift and essentially this would cause a significant problem and so the point I wanted to make this is something you will find in your textbook also that depending on the applied bias if you apply a negative bias the sodium will get close to the metal oxide surface and you can see in the blue and then if you apply a positive bias then the sodium will get to the other side and then therefore depending on the charge where this charges of course you realize that the threshold voltage will be different and as a result you can see the threshold voltage moving back and forth by the way what type of substrate is this can you say from this just looking at this picture all my CV curve I have been drawing in a particular direction right so what type of substrate is this this is a n type substrate this n type substrate right the CV is going the other way do you see for can you calculate from here the flat band voltage the threshold voltage all those things can you do that you should be able to do that can you also calculate how much sodium was there but just by looking at the total amount of threshold shift in the threshold voltage certainly you can do that because when it's stabilized on the other side its sort of charges moved from one side to the other side and so the total amount of shift essentially you can calculate from the shift in the threshold voltage you should be able to do all those things because I just told you how to do it