 you can follow along with this presentation using printed slides from the nano hub visit www.nano hub.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 so this is lecture 32 on MOS electrostatics here is written as short-hand for metal oxide semiconductor and it could be a capacitor or it could be a transistor the capacitor would be two-terminal structure like a diode and or it could be a transistor so in that case it will be called MOSFET and we'll discuss this very important device which is at the heart of the microprocessors and all sorts of electronics today I would probably say 90% of the market today for electronics is essentially based on MOSFET so that's a very important important new cloud class of devices we'll start with a brief background you know this is a very important topic and most than likely that when you go for any job or any if you join any company or if you have to teach a class let's say like this then you certainly want to understand MOSFET without understand it perfectly so that would be good now like any new device we'll begin with band diagram that's work number one with and without bias and then we'll talk about the capacitor aspect of it the two-terminal aspect of it and it's very interesting the way the charge moves with voltage in a MOSFET because MOSFET has an insulator and I'm coming to that that has the oxide therefore it is in some way in details different from bipolar bipolar did not have a oxide in inside the device oxide meaning insulator so there was in MOSFET in bipolar current was flowing through emitter base and collector we had to balance the current but in MOSFET that's slightly different so I will then conclude now let me bring back that original picture that we had in perhaps the first or second lecture of the semester and you can see that we have covered a lot of things by now yeah I said that the 606 involves you know quantum mechanics statistical mechanics and transport equation that was the first five weeks now you know right the effective masses the band gaps this various crystal orientation so on so forth and we have talked about diodes and bipolar and in the last three weeks which will about be nine nine lectures or so we will talk about MOSFET and then the semester will end I wanted to bring this back one more time to give you the historical perspective of course the diode on the very left for the vacuum tubes we haven't discussed that that's really a device that went away but in some modern form like a MEMS device MEMS transistors metal metal no what is it MEMS is microelectromechanical microelectromechanical switches in some form it is coming back again as a small thermionic emitter so for so that diode everything is dead rises one more time it appears now you understand that bipolar transistor right that was a short key barrier bipolar transistor that's why it has a wage that was the extra credit problem that you had for the homework and finally we will be talking about this MOSFET and the MOSFET you can look that it looks like a very strange structure first of all there's the lots of line on the 1960s until now that one the lots of lines going in different ways you can see hopefully by the end of next four lectures you will understand exactly why this device looks like the way it does and why this is of coming to an end even the scaling of a MOSFET is coming to an end very quickly so that's where that's the discussion we'll have and the reason it's coming to an end very quickly is because these days you have how many maybe 30 million transistors that you can fit in the head of a pin this pin that you have you can fit 30 million of them so five pins is all you need to give every person in United States a transistor that's the five heads of pin and that's it is fantastic the degree of sophistication and integration that has been been possible for bipolar for MOSFET a key question is why is it that we can do it for MOSFET and couldn't do it for bipolar and the main thing that I'll ask you to see as I go through this lecture that in MOSFET when you turn the MOSFET off the amount of leakage current is minuscule so you can put them a lot of them together and still or not have that much power dissipation you couldn't do it with bipolar bipolar simply consumes too much power and so this is something I haven't explained yet but this is something you should look out for as I go through the lecture that why this statement could be true now a MOSFET looks like a lateral bipolar transistor you remember that bipolar transistor is a vertical device emitter on the top base in the bottom in the middle and the collector on the bottom right and this was a double diffuse structure do you remember we had a uniformly dope material diffuse the base once and then diffuse the emitter once the current was coming down vertically and then laterally flowing through the sub collector and coming back up through the collector basically a vertical device for a MOSFET you will notice that this is a lateral device the two green fingers that you see on the left hand figure those two green fingers are let's say n-dope region and the magenta that you see is a p-dope region so you have essentially an n-p-n transistor and if you contact it somehow let's say the ash colored region in the middle if you contacted directly in some way it would look like a bipolar one question you immediately ask yourself that if it is looks like a bipolar couldn't it act like a bipolar meaning why don't people simply go ahead and contact it on the top and I think the answer you should recognize by now that that would be too far out emitter and collector would be too far out if you put a base contact there all the current will flow out through the base contact nothing will go to the collector as a result here you put a thin layer of insulator that's more clearly shown on the right hand figure more schematic but more clearly shown here the blue hatcheted region that's the insulator so in the bottom this is all silicon at least in the beginning you know nn plus p and n plus these regions are all silicon let's say or germanium one material and the blue is silicon dioxide and the n plus region you see vertically close to the gate it could either be polysilicon or it could be a metal so polysilicon historically for I would say in the beginning it was metal so when the first Intel computer came out this was aluminum metal in 1970s then for a while this was polysilicon for last I would say 30 years now the most modern or latest Intel computer that the Pentium that you have this has once back gone back to metal this n plus region along the vertical side in on top of the blue and this is a picture I have shown you before one thing again you notice the silicon is crystalline because I want the electrons to go very fast silicon is crystalline but silicon dioxide is not only an insulator but you will notice that that is amorphous the atoms are not is randomly oriented it's again a very important point to remember or understand why this has to be more or less a amorphous material it cannot be so it has to be insulator of course but at the same time it has to be amorphous because if you didn't have it amorphous there's always possibility that one of the grain boundaries will be present with the oxide and since it is supposed to stop current even a single grain boundary can short your substrate to your gate and the MOSFET will be lost and if you lose one MOSFET that means you lose the whole IC let's say so in that case this is people generally prefer it to be a amorphous material okay the symbols well symbols you may have seen this before there are two types and dominant one is transistors which are normally off which is that n plus and the green region that you see is P and then n plus so that in the off state if I haven't done anything then you realize there'll be a barrier there's a diode between n plus yellow n plus region and green substrate so there is a barrier and as a result current doesn't flow unless you reduce the barrier somehow forward bias the junction somehow current doesn't flow and the corresponding symbol is shown on the right on the right you can see that instead of having three terminals that we had in a bipolar this device has four terminals and it's very important generally MOSFETs have four terminals there are some subclass of devices have three terminals called SOI silicon on the insulator that has three terminals but primarily all MOSFETs have normal MOSFETs have four terminals and the fourth terminal termed here as B is the body contact the four terminal that is sort of grounding the green region you also have gate which is equivalent to the base you have source equivalent to emitter and drain equivalent to collector so you can see more or less very similar electrostatics very similar flow but that's something important to see another thing you will notice that this device is two-dimensional that is to say that if you look at the gate through the oxide which is red when you apply a potential then that potential is perpendicular to the device whereas the current flows from source to drain and so current is flowing perpendicular to the electric field from the gate of course there is a lateral electric field also from the drain but you can see immediately it's a two-dimensional device as a result although this is a simpler device actually this is not something people teach earlier on right earlier on in the in the in the course there is also another type of device these are normally on devices so the green region instead of being P is same so typically the transistor will be on and you will have to shut it off by using a gate voltage so change the potential barrier and thereby change the and you can see the only difference between the two symbols is that the channel region has a little thicker line to indicate that this is a normally on transistor so just very quickly let me show you how if you see a picture of a modern mosfet how you would interpret it in this particular transistor in this particular configuration it looks like the transistor is lost in the crowd of columns like these are looks like all Greek columns right and these columns are essentially the interconnect lines interconnection is a very complicated thing when you have a billion of transistors 1 billion people and you have to connect them all it is 10 layers of highways 10 sort of 10 layers of highways that must connect all these transistors very complicated but at the same time beautiful process by which is done and you can see all those columns these would be the drain contact you see that little transistor sitting there in the bottom that's the transistor and you have the drain contact let's say shown here so there you can see there are two transistors sitting in the middle right you see that if you blow up that region you will see transistors these days looking something like this it's a little tilted you can see the green region is again the body of the transistor where the electrons would flow and these days most of the transistors is strained and I will explain what it means but it means that it's not standard silicon just but it must be lattice constants have been either elongated or compressed compared to normal silicon for which the electron would flow do you see the yellow poly silicon poly silicon gate regions a gate contact and a very thin and tiny oxide region these days oxides are on the order of let's say 10 angstrom or so 10 or 12 angstrom or so and this is the latest this is the latest in you can where you can see that the source and drain has been replaced by silicon germanium it's no longer silicon just beginning to look like a HBT isn't it and HBT is it a single hetero junction HBT or a double hetero junction HBT double hetero junction HBT you can see silicon germanium to silicon I have one hetero junction silicon germanium to silicon another hetero junction so everything that you have learned so far many of the things we could easily apply to this problem you can see again the metal replacing the poly silicon and the oxide these days are being replaced by silicon dioxide is being replaced by some version of hafnium oxide now I'm setting this all up because I want at the end of 8 lectures that you be able to answer that why these have to be the way it is or at least why people have adopted this particular sequence why couldn't they stay with simple silicon all the way through so the purpose of all this derivation and everything is to understand what people do and why they do it okay so after coming back to earth let's talk about the basic band diagram one more time so we are talking about MOS capacitor now two terminal device and the MOS capacitor is essentially a metal gate you know we are just not thinking about poly silicon yet we'll think about that and the red oxide will assume for the time being silicon dioxide amorphous let's say and a substrate which is silicon and then I have the body contact shown in the black on the bottom I do not have in this particular device any source drain contact that's why this is called a MOS capacitor not a MOS transistor because here you can see I have just two terminals as if I have two metal gate and the body contacts are like two metal plates and in between I have like an insulator and the semiconductor that's like the filler inside the inside the capacitor so that's what we'll be going to think about okay now this would be the band diagram now band diagram do you I think by now you should be an expert in drawing band diagram hopefully but first of all you see that this is a heterostructure type band diagram and how does it come about well again you do draw the blue line first and you know this is a n type subs or sorry p type substrate the light blue region so you make sure that your Fermi level is very close to the valence band of that of that material draw the conduction band correspondingly in the insulator you will have a band gap right and then correspondingly you will have the insulator chi for the insulator you can see chi i correspondingly chi for the semiconductor chi sub s and for the metal pi sub m the only thing that is very interesting here is because you do not have any doping in the insulator right so therefore do you see that whatever electric field you have the electric field is continuous how do you know because the potential is linear if I take a derivative of the potential that gives me the electric field so the electric field is a constant and if I take another derivative of the electric field then I get charge and the charge is zero so therefore in the insulator I do not have any band bending right it takes a little bit to draw it but I hope you will be able to draw this draw this particular diagram by the way silicon dioxide has a band gap of 9 ev about 9 ev so it's a huge thing silicon dioxide you know how big 9 ev is if you made the whole earth whole earth made of silicon dioxide which is almost what it is I think you know like not 100 percent silicon dioxide comes from is a sand right sand is silicon dioxide if you made the whole world with sand there wouldn't be one electron there because with 9 ev n c n v e to the power e g over k t that's ni square with 9 ev there you do not have even a single electron in a volume of 10 to the power 27 centimeter cube so it's a very good insulator if for example let's me start with so we'll get back to that original diagram a little bit later it is a little bit difficult to analyze for our purposes we'll start with an idealized situation where I have chosen I'll assume chosen three hypothetical materials three hypothetical materials n with the so I put down an oxide 9 ev gap and then I essentially draw the same diagram but hypothetically this one I have chosen some three combination of metal semiconductor and insulator so that the vacuum level is flat the idea is that we'll understand this one and then any change that there is in actual material we'll just modify the understanding but this is easier to understand so this would be a essentially a idealized MOS capacitor right in general you realize that that's not what's going to happen but let me assume that I have found some metal whose vacuum level exactly coincides in a horizontal fashion with that of silicon and then I have this particular particular combination okay now let's think about this device by the way that's the rate that's the capacitor that's the oxide region how much charge do I have here I do not have any band bending right we see I do not have any band bending so what's my potential zero I don't have any band bending so my ec is flat so and v is flat but I could take that v to be my reference right and then make it flat also so although I shift it a little bit with respect to the x-axis this is actually zero that's what I'm trying to show you if my v is zero what's my electric field zero here and what about the charge zero and you can see that's true right do you see that everywhere I have the whole region is essentially flat ni square is equal to np and it must be charged neutral no depression nothing in here no depression region and as a result I have in this system lots of charge but no excess charge the net charge is zero but of course I have tons of electrons and tons of donors at every point so I'm not saying that there are no charge lots of charge net charge is zero okay we should stop here this is so simple that life may be actually simple for once what is vbi finally at least in one case vbi is zero so because you can see you can go from one side of phi sub m and another side you have to add up this q phi sub s eg minus delta and then when you do that you essentially get vbi is zero you can see it's flat okay so since I have solved my equilibrium problem so simply let's think about the dc dc characteristics okay now let's think about it slowly because this is a very important these are important concept I'm talking about the capacitor two tunnels I have grounded the body which is the p side I have grounded and you can see with the with the rhombus type blue region that's our diamond shape blue region through which a blue symbol I have grounded it so at holding it at zero potential and as a result my corresponding Fermi level remains flat now on the other side of the gate what voltage do you think I have applied I have applied a Fermi level has gone up and I have applied a negative bias when I apply a negative bias then the Fermi level moves up now when the Fermi level moves up then of course in order to compensate the whole bands must bend upward right for just like a p-n junction when you apply a reverse bias let's say then doesn't the whole region inside also follow through so that the same thing happens a very important thing here is to realize there is no current flow because the insulator is 9 ev barrier it's very difficult for the electrons to go through that region you know if you try to do thermionic emission only the tail above that band gap that's that tail of the electrons would be able to go you do not have any electrons to begin with and so the tail will be miniscued and there will be no current as a result my Fermi level this time is really flat because remember previously it was approximate reflight because the current was small but now the current is zero and what is the expression for the current n mu gradient of the quasi Fermi level that's zero zero so in that case gradient of the quasi Fermi level is also zero that's the distinction from a diode or a bipolar okay now do you realize that if you bend the band then the holes yeah so okay yes just for a second so there'll be after bending the bands there'll be a lot of accumulated holes do you realize that why there are more you can see the Fermi level has gotten closer to the conduction band after band bending and the smaller the separation is then larger is the number right so you have more holes close to the oxide on the semiconductor side and to balance there'll be more electrons now you realize on the metal side I should also have shown a little band bending but you also remember that if you have a metal semiconductor junction metal has so many electrons that we generally don't show the band bending but essentially you'll you might have a little bit and had it been polysilicon you could have quite a bit so that's the situation and we'll call this region accumulation selfish planet right you have accumulated the majority carriers and the holes are the majority carriers and you have accumulated more of them so that's called a accumulation region when that happens when you apply a voltage which essentially builds up the majority carriers so that will depend on whether the substrate is p type or n type in this case if you apply a negative bias then you have accumulation if you apply a positive bias on the gate then what would happen in this case now the electron will be pushed back in the substrate region you see it bends such a way so that the Fermi level remains where it is remember there is no gradient and this is grounded so Fermi level is not going anywhere but when you bend the band then you can see close to the oxide semiconductor surface the band has moved away from the Fermi level as a result you have fewer free electrons so you have more exposed doping so you can see the red region I have shown here these are space charge immobile charges depleted charges right and you have correspondingly the positive charges on the metal side this region will be called a depletion region and finally you have if you keep bending it more and more in of course what would have happened in a in a in a metal semiconductor junction a short key barrier huge amount of current would have started flowing right you have so much and there might have been impact ionization and other things right but because of the insulator we are lucky no current is flowing so you can put quite a bit of voltage until that one breaks down and that we will discuss later and in the process you will gradually see this green electrons building up very close to the surface now in contrast to the red region rates are positive right these are depleted acceptors right now sorry negative these are depleted acceptors green is also negative these are electrons but the green is mobile they can go where they want but the red are fixed accepted charges sitting in the space they cannot go anywhere they want right so this region when that happens when the mobile carriers essentially takes over then that region that high voltage will be called that inversion region why is it called inversion because the majority carriers were holes now close to the surface the majority have become electrons so the carrier type has been inverted as a result these would be called inversion charges and inversion region now in between when it goes from purely red and when the green one sort of comes over and takes over that is called a threshold voltage that at that point actually the inversion has occurred and we will call that vt there is a little prime sitting there just to indicate that these are all idealized devices we will have to correct it for real structures a little later do the same thing for had it been n type n type substrate you can do the whole thing you can see the whole voltage axis simply will flip again you will have accumulation depletion and inversion and you you get the idea now why did those charges come from well why does this holes come from they come from the body contact you can see it is a p region and so these holes will come from there and accumulate over there if you keep bending more it will want more holes it will keep coming from the keep coming from the body contact until the electrostatic requirements are satisfied now what about this one now you have charges coming in here n a where as the holes gone from that region they have gone out if you do not want the holes they can easily flow out through the body contact and be done now the interesting thing is where does electron come from that is the problem holes can come from that one easily from from the right from the body contact but you see electrons if they have to come they will have to jump over this barrier right from come trying to come from the contact and that is a that is the problem so why why do they come from they come from a hole goes out an electron hole pair is generated by shock read hall or thermal generation process do you remember that this is a depleted region just like a reverse bias p n junction depleted region so electrons are generated and as a result the green electron has been generated a hole has been left behind and the hole comes out through the body contact you realize immediately that this will be a slow process because shock read hall generation may be in the order of maybe 10 to the power minus 6 seconds or so right so if you want it done very fast well it is not going very fast if you want it a gigahertz or a 100 megahertz transistor this way you are not going to get it because the electron the green electrons will not be able to generate that fast to provide you the inversion electrons right so this is the gating item and of course we will integrate the charges to find potential once we have this then actually we are in good shape now before I get there let me talk about this response time issue just what something that I just alluded to when the holes in the inversion holes in the accumulation region is coming in and out that is the majority carrier right what is the time constant with which majority carriers can come in and out that is the dielectric response time I told this several times right for MOSFETs also I am sorry the diodes and bipolar I have told you several times that the holes don't themselves do not come they just inform the neighbor neighbors and then information is passed along this would be very fast because your sigma is for the majority carriers is very large as a result if you put them in the time constant it will be less than a picosecond with which the holes will come in and out the effect of holes will get in and out very fast so depletion and accumulation happens very fast on the other hand if you want to generate those green electrons in the inversion region you will have to generate them through Shockley Reed Hall process do you realize why I have dropped the N and P so actually reverse bias junction sort of right the whole region has been depleted N and P are approximately equal to 0 in that case you only have ni squared of the numerator and another N on the denominator you have this so you remember this and as a result this is a slower process okay now let's talk about charges so how would the charge look like this is how it's going to look like and I am going to explain to you why it comes from Q service is the amount of charge in the semiconductor I am not thinking about the metal now guess the semiconductor if you put on the negative bias negative bias is accumulation then you will see that the accumulation charge will go with phi service is the surface potential explain in a second and it will go exponentially with that if on the other hand if you apply a positive bias for the surface potential then it will initially have a square root dependence that's up to the depletion region and then you will have another exponential rise that's in the inversion region the green green electrons that's in the inversion region so let me explain where they come from so we will have to solve these equations do I have to solve the electron and whole continuity equation not really I have insulated no currents are flowing so at least I am set that I don't have to follow solve them only thing I have to solve is the Poisson equation and then I'll be done so let's solve Poisson equation now that's the structure where I have applied a positive bias so the system is in depletion right charges have moved back I have a depletion phi service is the degree of band bending in near the surface compared to the bulk that is how much the curve has been turned down compared to infinity so look at the curve on e i e sub i e psi i was in one value if I didn't put a voltage then it would have been a horizontal line after I have applied a voltage it has bent a certain amount that is called a surface potential or surface band bending I want to know how much that is that in next four slides so first of all I can take that one and copy that one here that's my V minus V really right because that was EC so that's minus minus V how do I get the electric field I take a derivative of this I get an electric field and I take a one more derivative I get the charge what is this charge this was that rate chart this is a depleted chart right so I know this now what will be the my total gate potential gate potential is whatever oxide voltage drop that has occurred multiplied by how much surface band bending I have now I obviously know VG because I have applied it with my voltage source so I know 1 volt 1.5 volts I know I do not know V ox and I also don't know if I service somehow I have to find them then my problem is solved so let's do that let's see whether we can do that's very easy actually do you see looking at the electric field curve that the electric field close to the surface close to the surface is simply given by q na multiplied by W which is the charge from the bottom right this is a constant depleted charge depletion approximation divided by that kappa multiplied by epsilon not that gives me the electric field at 0 plus meaning inside the semiconductor that that is my electric field if I know the electric field can I compute the potential of course because potential is essentially area under this blue triangle the electric field triangle right total potential and so what is that I have a half for the triangle on the top I have their electric field e sub e at 0 plus and then I have W W is the base of the triangle so you can see phi service although I don't know what it is yet this is related to the doping it should somehow if I could calculate the W in the magenta if I calculate that I would get phi service okay not yet but let's continue now I can say that somehow if I can calculate that's what I was saying calculate W or calculate phi service either one of them then my other problem will be solved but I haven't done that yet so let's do it now this particular way pay attention to this equation because this equation you will see in various forms many times vg that's fine you see this second term on the right is phi service the one that I just derived w squared you can see that business now the vg v ox the oxide voltage drop do you agree with the fact that e ox 0 minus x not x not is oxide thickness constant electric field right in that oxide region no charge so I have a constant electric field why haven't said e ox 0 minus because the oxide starts with 0 minus not the 0 plus and here it matters because remember the dielectric constant of the oxide is different from the dielectric constant from the semiconductor do you remember this hetero junctions two size different dielectric constants the same thing over here again okay now you can easily then calculate e ox at 0 minus from e semiconductor at 0 plus just by dividing and multiplying with the relative dielectric constant so that's what you have because I know e at 0 plus right in the last slide what is unknown here only unknown here is W vg I know I applied the voltage you see everything else is a constant if I know the doping and other things I am set in fact I could have so problem is solved I'm actually done in fact I could have also written it in a slightly different form because w is related to phi service right I told you in five seconds ago in the last slide that phi service is proportional to w squared so from here you can always relate w to phi service and you have the first w first blue w in the term on the top so you have a square root of phi service and anywhere you have a w square you have a phi service what is this this is a vg equals some b square root of phi service plus phi service s stands for surface surface potential that's why now have I solved the problem completely do you see this is a quadratic equation because if I say square root of phi service x then you can say x square plus ax equals b and I know vg I know b is a bunch of constants I know them and therefore actually I have just calculated phi service therefore I have calculated w and my electrostatic problem in depletion is solved this is why in this region on the blue this early part this early part goes as square root of phi service the blue part this is the depletion part right so this is what we have just done just one piece we have to do three pieces I have just done one piece that explain why it goes as a square root of phi service if you wanted to know how much depletion charge you have you know you could calculate that easily because it's q na w and w I know now right I just calculated w so therefore it's related to phi service and if I solve the previous equation then I will get the square root of phi service and that will be my charge density so the depletion region as you apply more and more bias this depletion region expands and it expands as square root of phi service where have you seen this before do you remember in a pn junction when you apply when you apply a reverse bias in a pn junction then that's w expands and that was q vbi vbi minus v that was have the same dependence the same physics of course so of course you will have the very similar similar dependence in the depletion of course right so I will just introduce you to a understanding of this where the other things would come from but take it up take this topic up in the next lecture property so I have just explained to you where the square root of phi service dependence comes in look at the y axis we are talking about amount of charge amount of charge in that inversion region or in the depletion region and then that's what we have just done in the next class what I am going to tell you is how to think about in the inversion region how to calculate the green charge amount of green charge now unfortunately this I cannot do with depletion approximation anymore right this is not depleted huge amount of green electrons mobile electrons moving around so this is no longer this is no longer depleted and this is something I will do in the next class do do this calculation and correspondingly you see those holes generated in the accumulation again it's no longer depletion these are mobile charges so I will have to do something extra to get back those holes and these extra electrons now why is it in a pn junction we didn't have to worry about these green electrons or the magenta like holes over there why is it we didn't discuss this right we just did depletion approximation all the way solved all the problems and everything the reason is that I have an oxide so when I put a lot of voltage electrons can pile up next to it if I didn't have that barrier they would all go down there's nothing to stop them they could simply go flow out through the electron and hole current they would flow it out here because it's like a dam I have a dam here flow getting the outflow of electrons stopping the outflow of electron so the situation here is different I don't have a current but I have the effect of a current which is building up these charges something I do not have in the normal pn junction so let me conclude so as I said MOSFET is a very important electronic device but this is not a very good device in many sense in terms of performance why not do you realize because in a bipolar transistor current flow is vertical so I have a lot of area through which current can flow vertically down for a MOSFET the current flow is lateral and then there is a thin sliver of region that's the green electrons that you saw that flows close to the surface if you looked at the effective cross section of the region in which the electrons are going well MOSFET you have a width no problem right width of the transistors the electrons are going but in terms of depth into the device that's the width of the green region maybe a 100 angstrom maybe 200 angstrom so the effective cross section over which the electron flows is tiny as a result MOSFET has a very small amount of current if you want a lot of current in MOSFET you have to make it very wide that was not the case for bipolar and so intrinsically this is a worse device the one the reason even a worse device sometimes wins up you know the slow and steady business that although it is a worse device compared to bipolar but it has something very important advantage that when you turn this transistor off because of this insulator there is no flow no flow of current in a MOS in a bipolar you can never stop current right this always the base current flowing in always the base current flowing but here you can completely stop the current as a result this is much more power efficient and therefore you can cram more of them into the same IC now this is a two dimensional device as I told you about and so therefore we are looking in the vertical side now and three class down we will talk about the lateral electron flow in response to this but this capacitors we are studying because we want to look at the vertical structure and this is very important that what we are trying to get is how much charges do you get for a given gate voltage and that's different in different regions so today we just got started okay thank you very much