 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 we'll get started now this is lecture 34 on MOS capacitor and today we'll be talking about the capacitors frequency response now we'll start with a background that why it's an important problem and we'll primarily discuss about small signal capacitances and you can immediately realize that if there is no current flow per se physical current flow electrons moving from one side to another then there will be certain capacitances which will be present and certain capacitances which wouldn't be present so that's an important thing I will talk about large signal capacitance for a MOS capacitor large signal capacitance have some special features because the way charge gets redistributed among fixed charge and mobile charges we'll see that how it works and then we'll conclude so just to remind you where we are we have of course talked about equilibrium band diagram you know and we are assuming this idealized case where everything is flat we have assumed at zero bias all the vacuum level is essentially flat in reality that doesn't happen we'll take care of this proper band bending and other things a little bit later if you class down we also talked about DC operation and you realize and DC there wasn't too much to talk about because generally the current flow is not very much right because there is this big oxide barrier if you had to calculate current then you would use thermionic emission by essentially looking at fluxes from both sides because there is a discontinuity in the conduction band but of course there is tunneling current these days that one has to calculate self quantum mechanically today let's talk about a small signal and what I always want you to do when you read this things that follow the vertical vertical column and try to see a common pattern of how small signal is handled among all the devices so that any new device you can handle them easily now this is a configuration you will see on the left hand side I have this blue gate gate region it could be a metal it could be a polysilicon but here we are assuming it's a metal the red is the oxide and P is the bulk region now in this case we have just the gate and the body or bulk or substrate contact over there you can see a DC bias which puts the capacitor in a certain degree of depletion even in the absence of that little small signal which is the signal coming from the antenna let's say then you have a given DC condition certain amount of depletion accumulation or inversion already present in the system and the small signal is trying to modulate around that DC or equilibrium point and that has very interesting characteristics as we will see now in general anytime you have a diode you might want to represent them through this parallel combination of junction depletion conduct capacitances and as well as of course the conductance G0 now you realize that because there is no current flow so I doesn't depend on V directly so DI DV which is the small signal change of the current in response to a small change in the voltage doesn't exist so G0 in principle if you do not consider tunneling current it will be essentially zero no conductance right now of course these days when you have very thin oxides there are certain component to it but that we will not get into in this course now the only capacitance that is relevant is that junction capacitance why because you can see if you look at the picture and diagram on the left that there are majority carriers of course in the blue region the electrons and holes will come in and out very quickly and also in the bulk region the holes can come in and out very quickly so the majority carriers are of course present so in that case junction capacitance must always be there so if you look at the capacitance this has this very strange form and let's think about that for a second because this is a important figure that we will be discussing many times later you'll see when the gate voltage is negative remember 0 is a flat band no voltage applied gate voltage then flat band on the left side when you have negative voltage what do you have there you have accumulation of holes right so you have accumulation of holes you see a flat region regardless of your accumulation bias you have a fixed amount of capacitance now as you go in the positive bias you can see the junction capacitance dropping rapidly and that rapid drop happens when this is in depletion region and then what something happens beyond that point that as soon as the inversion happens inversion happens the capacitance immediately goes back to whatever it was in the accumulation region that's something strange but the more important thing is this this only happens at low frequency if your small signal over there remember that signal coming from the from from your antenna if that signal is low how low on the order of kilohertz or less so very low you really have to work at it then it will go back as soon as the inversion occurs it will go back to the original capacitance of CO CO is the oxide capacitance epsilon not a over D D being the oxide thickness you see now the transition point where it goes up back that's a very important point what should be that voltage called that's the threshold voltage because beyond that point the inversion the depletion changes into accumulation right depletion is charged have been pushed back an accumulation where minority carriers in this case electrons have just gotten accumulated next to the rate region remember these are mobile charges through shock to read hall generation has gone back there but the more interesting thing for most of the relevant frequencies you know megahertz gigahertz this type of thing the capacitances you will see in the capacitor this type of capacitor structure is that it will not go back it will stay essentially flat at that depletion value and that's happens at high frequency now I want to emphasize strongly this happens only when it's diode if you have a MOSFET which we'll talk about later that put those two additional contacts near the source and rain this is not going to happen the only characteristics you are going to see is a low frequency regardless of what the frequency is you see so this is a very special case about diode that the high frequency behaves this way so we'll start with by considering the junction capacitance only and then not worry about the diffusion capacitance I mean where is the diffusion so therefore we'll not talk about that at all so let's talk about the total gate capacitance now you can see the total gate capacitance here of course there is this red region is oxide and you can see the bulk p-silicon I have drawn it here also but the deep blue is sort of the inversion region and the green is a depletion that how far the charges have been pushed back and exposing space charges so this would be NA right acceptor essentially sort of being exposed in the green region now how would you calculate the capacitance well the simplest way to calculate capacitance is to take a derivative of the gate charge with respect to the gate voltage because q divided by v essentially is the capacitance but we are talking about small signal because remember that capacitance is not really constant it's changing at every point so therefore it will be a gate voltage dependent thing so you can easily do that but of course I don't know the gate charge that's the amount of charge I'm supposed to have in that blue top part I do not know that however one thing I can say that whatever that charges I do not know what it is but that must be equal to the charge that I have in the semiconductor what type of charge I have in semiconductor green region we have fixed charge blue deep blue that I have mobile charges so this sum of these two pieces now the VG this I have already discussed in the last class that this equals to the band bending price of s and the oxide voltage drop q sub s divided by c c o c sub o do you realize how it comes about do you sort of remember this is actually e ox the oxide field multiplied by x naught the oxide thickness now the oxide field you will always have to remember that whatever charge you have remember this Poisson peel box whatever charge you have the electric field that is coming out that will be equal to charge divided by the epsilon of wherever the electric field is coming out so electric field is coming out in the oxide so therefore you will have a kappa oxide and epsilon naught and then you have on one x x naught on the top that gives you the c naught you can look it up it's not complicated so you have that and using this relationship you can say well I cannot directly calculate it cg but let me try to calculate it the inverse of it so I flip it so now dvg dqg is what I am looking for and vg depends on price of s the amount of band bending and the amount of the charge that I already have if I take a derivative you will agree perhaps that the red qs as soon as you take a derivative with respect to qs that will become one so you will have one plus c naught on the second term now what about the first term do I know the first term I already know the first term remember in the last class we talked about that as soon as you have a certain amount of band bending then you know what the w is how far the depression with this and from the w you can calculate how much charge you have very very simple so you'll just need to connect it up in the last lecture but if you give me that let me let me assume that you remember then in that case I will call that the semiconductor capacitor semiconductor capacitor is that when I change the band in a semiconductor up and down the amount of charge that comes in and out in response to that that's a semiconductor capacitor oxide capacitance you understand and so the sum of these two will be the gate capacitance so essentially that that too right so I have CO CO which is the oxide capacitance and the C semiconductor I have the semiconductor capacitance and those two are in series they are they are one one after another and one thing you see that I have drawn a arrow on that because this capacitance is not fixed right why is it not fixed because as soon as they apply larger gate bias then the depletion region changes and you remember right that the depletion region is a denominator for the capacitance calculation depends on where you are operating it and as a result that capacitance in general is not fixed however you can still if you wanted to know the value for it you could easily calculate it because these values are already known okay from the last class so go ahead try to see that the Q sub s how it depends on phi service how should it depend on it do you remember square root of phi service remember this is just a junction depletion didn't you have VBI minus VA square root of that when you had talked about diode so it's that VA being replaced by phi service that gives you the depletion so that's what the degree of the capacitance change and charge change you will have here now I want to use this information a little bit later so I want to define this quantity m the called the body factor right here m is by definition that m is equal to 1 plus the semiconductor capacitance divided by oxide capacitances well why why should I do that you can immediately see in a second first of all do you agree that CS is 1 over kappa 0 into WT and and then correspondingly CO that is in the denominator so the X naught gets flipped back so therefore you have that particular expression the X naught value and the WT value the WT will depend on the voltage at which you are operating it why am I interested in that ratio what does it give me this random definition well not completely random you can see that if I wanted to know what phi services in a series connected capacitor right series connected capacitor do you agree with the statement below that it will be given by C naught divided by C naught plus CS multiplied by VG do you agree with this because capacitances when they are in series how do you how do we add them up it is the inverse of capacitances we add them up right so therefore you will have to see that that's how this one is but then you can immediately realize look at the definition the definition is one m is 1 plus CS divided by C naught so you can see if I divide C naught top and bottom on the denominator then I will have I will pick up this factor called m now this m in general is voltage dependent so in principle I shouldn't really it's not a very good quantity in practice it turns out that the m value of m is on the order of 1.1 to 1.4 in most modern transistors as a result you can see the phi service will be like a 40 percent or 60 percent less than or 40 percent or 60 yeah less than the gate applied voltage you apply 1 volt surface voltage well it may be 0.5 volt that's coming across the surface so that's what this is trying to tell you and we'll use this later on you'll see so let's talk about small signal capacitances with this basic definitions in place now this is in accumulation right accumulation I have what type of voltage I have applied a negative voltage negative voltage brings in negative charge and so there's negative charge on the gate and so it will pull in positive charges and positive charges are holes so the holes will pull in close to the surface you can see the formula level is flat but the band has bent so that the difference has been reduced so I have more holes and that's the red holes you can see close to the surface looks like a parallel plate capacitor right doesn't it so it will essentially the majority carrier charges will come there and so the accumulation capacitance for the junction I'll just have the oxide capacitance and be done with it now it really doesn't matter what voltage you are in right of course the magnitude of charge is very different depending on the voltage but the charges are same distance apart and change of charge as a function of change in a voltage is the same you may have a huge amount of DC charge sitting there but if you are just making a small change let's say 0.1 millivolt 0.1 volt change in the by gate bias moving up and down so the extra amount of charge you bring in is exactly the same therefore the capacitance in accumulation is essentially independent of bias accumulation bias now in principle the accumulation region if you were born 30 years ago then perhaps you could ignore it all together it turns out that these days oxides are only 10 angstrom right very thin oxides and the red region well the red region remember W a accumulation inversion we calculated the other day we just looked at there was this exponential green region we compressed it into a rectangular region you remember and that had a certain amount of depletion so I cannot really these days completely ignore the extent of that red region so then what do I do how do I calculate capacitance one way you could say I will just take the centroid of the red charge and centroid of the blue charge and make that one my new x naught and be done with it well that wouldn't really work because the dielectric are the the dielectric constants of the oxide and semiconductor are not the same so therefore there is a trick anytime you have let's say you have a parallel pet capacitor you have air let's say and then some other dielectric in series two materials what do you do in that case so of course you can do a complicated calculation but there is a very simple trick you see you can put some virtual charges there the negative and you put some negative charge there and immediately take it out in the next point with that red dotted line right because these are two parallel capacitors in that and then you can write it this way because then the two blues will make one capacitance that has only one dielectric the two reds will make another capacitor that has only one dielectric in between and these two are essentially two capacitors in series and so you just put down the formula for two capacitance in series do you remember that right capacitance in series is the inverse you have to add and then flip back so that's that that's it and you can easily calculate it so the bottom line I want to make that only if only if the oxide thickness is reasonably thick you know reason even even maybe five years ago you could say that if it is reasonably thick then you would only about think about the capacitance being the junction capacitance being exactly equal to C naught but these days if you go in the lab and make a measurement it will be a little bit less because of the charge that you have in the accumulation region right so this is something very easy to understand and you should be able to do it that trick that trick these days is often very useful because the gate oxide these days have multiple layers among themselves they start with a thin layer of silicon dioxide then put a they put a hafnium silicate or hafnium oxide then they have a polysilicon so when you have a stack get stacked they call when you have a get stacked if you use that virtual charge that in every interface you put that plus and minus up and down you just can cascade them easily and you have the entire capacitance in one shot okay so now junction capacitance and depletion is the part where this is rapidly plunging this junction capacitances with the positive gate bias let's see how it works out well again it's not very complicated because you apply a positive bias if you apply a positive bias then you induce positive charges in the gate so it pushes back any positive charges in the body or in the bulk and therefore it pushes back and therefore the red region here is these are exposed charges right in a exposed charges and the holes have been pushed back and so that you do not see it in this close to the surface well then you can again do this if you wanted to calculate the capacitance right because again you see that if you wanted to know that how much charge if you change the blue charge a little bit remember first of all this is DC first is DC that I have a bias then I am changing the frequency a little bit moving up and down so first of all you have this blue region and the red charge already present even if before you bring in any charge bring in any small signal voltage now when you bring in the small signal voltage the blue moves up a little bit and the red on the age of it or the far end age of it there it moves out a little bit and so therefore the charges are sort of one end on the gate and the other end on the far edge of that red region therefore when I want to look at the total capacitance I essentially look at the blue arrow and the red arrow on the age of those two regions and then if you see the virtual charges I have put in and again I can easily calculate I can easily calculate the capacitances you remember W how to calculate W but this one is actually very simple has a very simple formula and let's see how it works do you remember this equation sort of I think the which one is the oxide fear or potential drop in the oxide the first one you see first term after BG that is NAW that term that term is the electric field that is coming out in the oxide multiplied by X naught you could you could have cheated and see whichever X naught occurs in whichever terms that must be the oxide the oxide drop and this W squared you remember if the charge is a constant integrate electric field is a triangle and the potential is the W squared part of it so you have that and I told you before right you can solve for W from here is a quadratic equation everybody can solve it and if you solve it you will get an equation like this VG minus V delta 1 plus on the right hand side minus 1 and the V delta is a constant and what you can do therefore remember if you go back in the top top equation where I am trying to calculate the depletion capacitance on the top what is the only unknown I see there I know it's not dielectric capacitances are the constants are known only thing unknown over there is W right and I have just calculated W so therefore this I can put back in what happened to the plus 1 by the way you can see there is a minus 1 sitting on the bottom so that took care of my minus 1 now do you see as you increase the voltage that it will go down is a square root of gate voltage right as you screen is started and that's what you see in the depletion region that as you increase the gate voltage because of 1 plus VG divided by V delta underneath the square root therefore it goes down I mean that it will go down you understand right because the charges the depletion is getting larger and larger so your capacitance have to go down that it should be square root of VG that of course comes from the calculation okay so this I can do up to what point till the point I do not have the inversion charges coming in because then that will change the ball game one more time now in inversion what will happen in version then so we are talking about low frequency in the inversion region where the oxide have capacitance have decided to go back to the original value again do you remember that I have those green charges over there these are inversion charges that were generated through shock read hall because these are majority carriers were depleted and the minority carriers were thermally generated okay now we are talking about this time charges is the small signal bouncing charges back and forth on the top of that blue region right that's why when you apply extra little extra bias the original configuration the deep blue green and the red those are DC DC conditions whatever you are you have biased it in inversion and then you are applying a small bias and so if you do it small enough sort of slowly enough then you can see the way the charges will respond well will be this two yellow regions those charges will respond if they could and in that case again you wouldn't worry about it right because these are parallel plate capacitors so that's what you have see not and you will expect a constant low voltage independent of the applied DC bias if you wanted to account for this width of the green region then how would you do that that's the same as the accumulation one nothing nothing special because you have this inversion capacitance and the inversion capacitance will go as W inversion you know in the last class we calculated that how much it is how much is it very small 20 30 angstrom not very much right very thin sliver of charges that moves just underneath underneath the gate going from one end to another so it's a very very thin region that allows you to do this unless see not is is relatively big you didn't don't really worry about this very much okay now that's now first tell me what frequency when somebody says low frequency what frequency is determining this what is determining that response of these electrons this green electrons forming this is the rate at which they can go back in the conduction band right they have to be generated now what is the typical range of time constants in which charges can be generated on the order of a mealy to a microsecond right in silicon that's whatever generally recombination time constant or generation time constants are so unless you are doing it slower than that the charges that are building up on the top in the accumulation reimbursion region it cannot respond so the immediately you see that if you have few traps if you have few traps and your surface states are very good then it will take a long time and you have to go slower than that so that the green charges can respond as a result low frequency is really low and very very low most of the time it's very difficult to see this you have to do special things in order to see this now many times people will define an equivalent oxide thickness because they don't want to carry around this inversion capacitance and the oxide capacitance separately and many times these days also they have people have other high K gate dielectric rather than silicon dioxide so it's a very important quantity that we'll see in the literature all the time equivalent oxide capacitance so we have already said that in inversion I have this inversion charges and therefore I have the inversion capacitance because the charges shift it a little bit away from the junction and therefore this is less than c naught right we know that and I have shown that here in the figure by explicitly drawing that red region for high gate voltage lower than minus lower than 1 I have exaggerated it's not that horrible you will not sell any computer if you to add this horrible and then I just told you about the how to calculate the capacitances now many times people don't like to carry around these two capacitances and then calculate them separately because for a given technology the degree of inversion is known x naught is sort of known and so therefore people want to call it an equivalent oxide thickness EOT and that is sort some sort of effective oxide thickness that already accounts for this charge pushback or charge extent of the charge in the accumulation region or in inversion region and you can easily try to if you go home equate the CG on two sides you will see the effective oxide thickness will depend of course on the inversion how big it is and of course on x naught but one little trick is that the oxide dielectric constant and the semiconductor dielectric constants come as a ratio do you know how much is a semiconductor dielectric constant like in silicon 12 or so right so 12 or so and oxide silicon dioxide is 4 so you can see that this is already you get a factor of one-third from here but these days we are replacing it with high K gate dielectric right we want to change it in the high K gate dielectric so therefore the oxide capacitance like in half name oxide will be 20 so that will be larger than the semiconductor semiconductor capacitors the more polar the material is the higher the dielectric constant and so these days you are trying to really build it up and I'll explain that why why you would like to do that okay now what about high frequency low frequency you can see the charges you have let's say on the on the blue region that tip of that charge is moving and at low frequency in the green region you can see the tip of the charge is moving also there is something wrong in this figure what's wrong this figure the charges are upside down or this is I should have an electron concentration is upside down so I should have flipped that one but anyway so I at low frequencies that black regions you can see on the tip of green and the blue that is what should have responded but if I don't give it enough time if I move things up and down very quickly then you can see I do not even give it time to get to the inversion region so therefore what is going to do the only thing it has in option it has to match the charge balance immediately now the only option it has the depletion regions are more compliant because they can move in and out very quickly so instead of burdening the green region the inversion region asking for it to help because it's slow enough cannot help it will ask the red region to push back a few extra holes out because the majority carriers so it will push back the holes and as soon as it push back the holes you can realize that what's going to happen that essentially the inversion region although it's biased in inversion but as far as the small signal response is concerned as if it's not there because it's not useful it's not helping and so therefore the capacitance will look like as if it is still in accumulation you see so the DC bias point is being set by all free charges the green blue and red but the AC response in being governed by the green and the edge of the red region so that is how that's why it gets clamped at high frequency high frequency as I said more than a few kilohertz would be more than high enough frequency and this is why I spent so much time by repeatedly telling you the importance of majority carrier dielectric relaxation how much point one picoseconds or so right so it will come in and out that quickly well if you're trying to push your input frequency faster than that even the majority carriers cannot do it but we are not there yet chances are will not be there and then of course the time constant is this the minimum time constant that you need is the shock read all time constant this distinction is very important these none of these things happen in PN junction right so it's very important to sort of you follow through the arguments in a series linked linked set of ideas otherwise you'll be thoroughly lost hopefully not now okay now let me just graphically just try to illustrate that point one more time that at low frequency the blue and the green charges respond at high frequency only the ages of the green a red and the blue respond right because the majority carriers the holes can move back and forth very easily the green ones cannot in reality the the curve that you see on the left hand side is what I just explained to you in practice if you'd go and make a measurement these are not complicated measurement the students do it in the lab all the time this is standard measurement in electrical engineering and you will see that the capacitances that you will see is more like the one shown in the right and you will see mostly the the the blue line at a high frequency if you don't do anything special there is certain tests like a ramp test that allows you to get to the the low frequency part part as well now do I have I convinced you that that point where the things change is called a threshold voltage right the rate rate region from inversion to from depletion to inversion that to scan your point is the rate point so if I give you a problem to ask you that did give you a CV CV characteristics that what is the threshold voltage of this particular transistor what is the for example the doping of this particular capacitor or particular capacitor structure you should be able to do all that right because as soon as you know deep threshold voltage you know the w how much w is 2 5 of f and related to Vg you can calculate that and from 2 5 of f you can calculate the how much doping you have in the substrate so this is actually telling you everything that you need to know and also what about the blue point that's the flat band voltage point because that is the point where accumulation is going to depletion so that's when the bands become flat and it changes from one side to another so in real characteristics you will also try to see that and that will give you so quote unquote the flat band voltage when the bands are flat now the tricky part is a large signal response and this requires some thought so large signal response how have I treated or have how have we treated large signal response before you just did it in exam we have done it by charge control theory right generally we have done it by charge control theory but in charge control charges were really moving remember I integrated the diffusion equation minority carrier diffusion equation and thereby I got that charge control equation now did I use that when we talking about metal semiconductor junction we didn't because it was all majority carriers majority carries of course there's nothing no minority carriers to integrate and therefore no charge control equation you can see the same is going to happen here I cannot use charge control directly right no minority carrier here no electrons flowing so therefore I'll have to do it in a slightly different way let's see how it's done so I'll be switching things between 0 to 1 and you can see that instead of the small signal tilde I have a sort of a digital signal going back and forth going 0 0 and 0 and 1 I'm sorry that till there shouldn't be there that should be deleted I'll delete that and generally if the signal is large I can see that as soon as the switch switches as soon as it switches the switching is very fast going from 0 to 1 very fast right gigahertz range these days so switching must be even faster gigahertz is one cycle and so it's even faster than the switching is there's no hope that the green inversion charges will be able to respond no hope and so primarily the depletion charges will have to take care and I want to explain to you the most important thing is that when you do high frequency digital then that's the rate curve but the red that rate curve is different from the blue curve both are high frequency but in one case do you see the what the distinction would be in one case the green and I'll show you in the next slide the green even green wouldn't have time to form because in the previous case when you're looking at small signal there was a DC bias the DC bias bought in the blue charges and the green charges everything was sort of ready on top of it when you bring in a small signal then the red and the blue they sort of responded in the back green was there because that is a DC bias would do here however you are getting things from 0 to 1 there was no green to begin with growing electron no inversion charges to begin with so it was all accumulated and so therefore when you bring in this x all electron change the voltage the red one will have to take care of everything it will have to push back out all the way out and therefore this red solid curve in the capacitance is very different from the red the blue one that you see up to the inversion point same beyond the inversion point one case you have the green another case you do not have the help for the green that's make the important distinction between large and small signal and so that's what I wanted to say the small signal you have a green present and that gives you the blue line in the capacitance but a large signal going from 0 to 1 right abruptly switching you can see on the right hand side I do not have any green you didn't have time to form the green and so therefore the entire depletion region essentially pushed back out all the way out and therefore the capacitance is actually much lower again you can calculate this and this is the same capacitance because you do not have the green so as if the depletion is continuing and the depletion is continuing so therefore the same car the red car even beyond the threshold point continues to be square root of 1 plus vg divided by v delta it continues on the same car only for large signal however this is not going to stay right going to stay like this forever why not because these charges are going to going to relax so you see initially you have this now let's say you go from 0 to 1 and stay at 1 now as you stay at 1 the minority carriers are gradually building up you need you give them time they will gradually build up and as they are gradually building up it doesn't need the help of the depletion charges anymore the depletion charges are gradually coming back and as a result this will capacitances will gradually change from one to another you see there is this little animation let me try to show you one more time just to you understand how this transition takes place because it will start a deep depletion right and then it will gradually if you just hold it to one volt then after a while it will go to the low frequency one at the end how does it go from one to another this is how take a look at the right hand side charge on the right hand side and see how it's working so as you as you hold it at 1 hold it at 1 then you see gradually the green will come up and the red will pull back and the green when the green is fully there well your capacitance is a low frequency capacitance and and eventually the frequency will respond between blue and a red and that's when it will relax back the low frequency value now okay so correspondingly as I said that most of the time if you do this with pulses and other things in the high frequency this is the left hand side is ideal and the right hand side is in reality what you are going to get now the one point I wanted to make was that if you do a pulse then you will have deep depletion because depletion is continuing even above threshold that's called deep depletion if you just do the pulse a little bit ramp then you will have a little bit like this right because you are allowing time for the charges the green charges to build up if you do it even slower I'm sorry if you do it even slower then essentially you will have a DC bias and on top that you can have a capacitance so these three cards you need to understand now let me end with this everything I told you is very good physics but in reality in real devices many of these things may not apply but you understand I want to understand this before you understand the more traditional things most of the time when you talk about a capacitor here is a capacitor charges move in and out of the body contact what charges these are the holes moving in and out through the body contact as you need it and the response time you will have if you want inversion charges to respond the generation time is ni divided by 2 tau let's say and so this time constant will dictate what is low frequency what is high frequency right because you need if you want to do low frequency it means simply means you want the green charges have time to accumulate and so in that case the dividing time is tau which is the generation time kilohertz or so let's say but in real devices in a MOSFET something else is going to happen what's going to happen is that as soon as you bend the bands by applying a positive bias on the gate instead of requiring the electrons to come from the body back contact and then thermally generated in the inversion region remember the bipolar junction transistor npn transistor what happens when you push down the base contact electrons from the emitter immediately get into the base is that right that's right so in that case these electrons will come from here it will come from here in a nanosecond so anything that I have told you about deep depletion about you know this high frequency low frequency response you will not see this only thing you will see is your low frequency response in the sense that here the low frequency is a gigahertz because charge can come in and out very very fast now if this didn't happen your computer wouldn't work because if you needed a millisecond for the inversion charge to form you type you're typing a world document and every time you hit something the transistor is getting into inversion and you say dear sir and then every later it's taking a millisecond well you're not going to write the later because you know every time you hit a key you're inverting a bunch of transistors and if every transistor requires a millisecond you need a million of them then of course nothing your screen is not going to tell you anything at the end of the day this doesn't happen of course because in a real MOSFET structure the majority carriers sort of come from the source and drain and immediately helps you to establish inversion and that is how this time constants play out in actual devices now even in the capacitor structure I'll leave it as a homework meaning that it could be a suggestion for the exam also that if I shine light on that capacitor structure then what would have happened would it help me in some way and how would it help me so try to answer that question let me summarize then although the current flow in the through the oxide is very small but the and therefore the most important thing is junction capacitance all these things I did you know accumulation depletion and all those things all junction capacitance no diffusion capacitance no conductance nothing like that now the high frequency MOS capacitor and low frequency for a diode structure two terminal is very different and as a result we need to really understand how it works and the deep depletion is an important consideration because there are many places where in fact your automotive and many applications people do use this type of diode structures for capacitances high voltage capacitances power electronics people use that so in that case this notions of deep depletion and other things those are also very important that you need to account for okay thank you very much