 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. Today we will be talking about the IV characteristics of a MOSFET. This is lecture 35. We have five more lectures to go before the end of the semester, so it's almost over. But at the same time, towards the very end of course, we talk about slightly more difficult things, more complicated things. But given the background that you have based on previous classes, I hope these will not be as difficult. So we'll be talking today about MOSFET, MOS capacitor. MOS capacitor is a two terminal device where you have a gate that could be a metal, oxide and a substrate without the source and drain contacts. But today we'll put those source and drain contacts in there and then try to see the current that flows through a MOS capacitor or MOS transistor structure, metal, oxide, semiconductor transistor structure. I will not talk about equilibrium anymore. You know how to do band diagram by now. But this band diagram or this potential profile is slightly different, more complicated than the MOS capacitor or a bipolar junction transistor in the sense that this is a two dimensional device. Because you know you apply gate bias in a vertical direction. It controls the potential in the vertical direction. But you will see the current flows from source to drain in the lateral direction. So it is intrinsically a two dimensional transistor, a two dimensional current flow. That is something which is a little bit more difficult to think about than the one dimensional structures before. So in principle, although MOSFET is a simpler device, it's much simpler than bipolar you will see. But it is treated at the very end because of the complexity of the geometry. But it's not too bad you'll see. So let me begin by talking about the so called sub threshold current. And this sub threshold current you have already seen. You have already seen you will immediately recognize after a few minutes that you have done every piece of this calculation when you had been thinking about bipolar junction transistor. So therefore the math will not be a problem. The way the currents are handled will not be a problem. But you will see how it's slightly different also compared to bipolar transistors. Now let's get our orientation right. It's very important because I'll be drawing a series of two dimensional pictures. So it's important that we'll get our orientation right. You can see on the left hand side, it is a side view where we have taken a cut across the device. This is an NPN configuration. Therefore starting from the sources N plus, drain is N plus, and the body or the bulk or the substrate is P. And you can see his gate is on the top. And look at that blue bottom on the substrate where I have contacted it on the bottom. Do you see that with a little circular knob in the bottom. Now the same picture I have rotated it. It sort of you flip it towards you and then rotate it a little bit and look at the picture onto the right. And look at the position of the blue line with the knob which is the substrate contact, the grounded substrate contact on the top. You can see the gate is now facing towards me. The source and drain you see that those N plus regions and the body is P. So therefore anytime we think about an electron coming from the source to drain through the inversion layer. Do you remember that very thin layer W inversion, 10, 20, 30 angstrom, that region. So the electrons flows from source to drain over there in correspondingly in this other figure on the right, the electrons will flow in the y direction in the same way. Exactly the same way underneath the gate from the source to drain. Very simple. But now let's think about the potential the electron sees, the red electron sees as it goes from source to drain. Even before drawing the picture if I didn't have the gate then do you realize that this looks like a bipolar transition transistor. I have an N plus P and an N plus region. So therefore we expect two diodes back to back right with a built-in voltage of VBI in each side. We expect that right and we'll see exactly that's what's happening. You see that that if you take a cut and draw the conduction band and a valence band diagrams you see the flat quasi Fermi level which is the dotted line and then you also see that the N plus region is close to the conduction band and the P plus region is the valence band is close to the Fermi level. You can see that and it's exactly like a bipolar junction transistor except that this is in the lateral direction in the plane bipolar junction transistor is a vertical device where current flows vertically from emitter through the base to the collector. Okay now comes the difference. Now if you now take the gate so there's no voltage applied anywhere all four voltages are zero in this in this particular band diagram. Now if you apply a gate bias if you apply let's say a certain amount of positive bias so okay before that the electrons of course will flow from source to drain from just going over the barrier. This would be like your emitter current going to the collector without base recombination so that's very similar very similar. Now if you apply a gate bias okay even before that I have to getting ahead of myself. Now if you take a cut one dimensional clock just below the gate oxide just below it and then they take that blue line and plot it over there then you can correspondingly draw the conduction and valence band and you can calculate that height of the barrier from the doping right. If you have the same material then the chi will be the same so simply from the doping you can calculate VBI. Had it been two different material and these days indeed there are transistors in which source and drain material are silicon germanium whereas the latest pentium so exactly that source and drain have a different material compared to the bulk and as a result you'll have a hetero junction over there and that you can also calculate exactly the same way you would have done in an HPT. Now if you now apply a gate bias now if I apply a gate bias then and a positive gate bias for that then therefore I will pull the whole band down towards in that near the gate side you see in both the diagram I have sort of taken away the gate I am not drawing the gate part I'm just looking at what's happening in the substrate and you can see that the whole band has sort of pulled in and immediately you can realize you do realize that the electrons which are trying to grow from source to drain need not climb up over the whole barrier but rather they can go very close to the surface they will go very close now they don't have to go over the barrier so a lot of current can flow right so that is the essence of this now I want you to notice a couple of things first of all you see in the backside in the backside of the device which would be the substrate side the potential has not been changed why not because I have grounded the substrate remember the blue with a little knob on the top because it is held at zero potential therefore I have not allowed the backside the potential on the backside to change had it not been clamped by that blue stick right so then therefore the whole potential would have then come down in that case but in this particular case because of the substrate contact the backside will remain where it is the front side because the gate will be pulled down an electron will flow now do you realize this electron will flow very close to the surface and the amount of band bending do you know how to calculate it you do right that is exactly equal to 5 service that we had been talking about remember if you took a cut a vertical cut through this structure you would see the backside is an infinity in the bulk and gradually it's bending over near the oxide surface exactly that's how much you know that's 5 service so we have we know how to calculate 5 service for a given gate bias you remember and as a result we can calculate how much the barrier has been suppressed once you know how much the barrier has been suppressed you can correspondingly draw take the cut draw it there now is it not beginning to look like a bipolar transistor exactly that right you forward bias the base emitter junction quote unquote and then the current can flow inside so what would be the so there will be a certain amount of band bending and that's band bending is equal to 5 service the surface potential whatever amount you have and the width of this region for over which the electrons would go this narrow region over which electrons will go that is w inversion right that is this thin region over which the electron goes hopefully if you have gotten this pieces right you know the then the rest of the things will come easily and naturally now of course what will happen that just like a bipolar transistor the base emitter junction has been forward biased so you will have an extra amount of electron in that's the green triangle and on the other side the the drain will drain the carriers out and therefore the carrier concentration would go to zero and if you do not have any recombination what do you expect you will have a linear air profile for the electrons right that's that's as simple as that not no no different than a bipolar transistor now let's now think about that let's calculate the current that under this configuration when you have applied a large drain bias or I'm sorry a gate bias which has sort of let's say depleted the device but not yet inverted it not yet inverted it we're still talking about depletion and you can see that's why we say have the gate voltage less than the threshold voltage you see vg minus vg is less than v threshold and we'll call this sub threshold regime right vth is say threshold so it's below threshold so we are talking about gate voltages in that range and we are still talking about here drain voltages which is not too high you see that that dotted lines are a little bit forward bias it's not a huge amount of bias here the drain side so in this configuration what should be the current flow well you should be now be experts in this so the drain current should be q multiplied by dn and this will be small dndx and if you do not have any recombination then you will have qn minus q2 divided by the channel length channel length is in the compared to the bipolar it's like the base width right base width and channel length is exactly the same thing in this configuration now why q1 q2 why don't I have n1 minus n2 is because I'm sort of multiplying with wc I am looking writing current on the left hand side but I am really multiplying with the width density multiplied by width so therefore I have written it as q1 and divided by a channel so you get the idea now do you know what q1 is of course you do it's a forward bias pn junction how difficult can it be it is ni square ever na in equilibrium remember the minority carrier electrons that is that is ni square and do you also see q e to the power q phi service beta minus 1 do you realize that right this is the excess so that's the minus 1 taken out but this time it is not q vg beta beta is 1 over kt by the way it is not vg it is phi service because when you apply a gate voltage the whole thing doesn't come to the surface remember in bipolar your contact base was touching the base semiconductor so whatever voltage you applied on the base the same amount was pulled down by the base contact but here you have a little oxide sitting there so therefore a bunch of voltages or a fraction of the voltage will drop across the oxide and the remaining part will come across the come across the semiconductor that's phi service you get it and w w is the physical width of the transistor remember I'm just looking at a cross section of course there is a certain width that's w w inversion is the thin region over which electron is flowing so w minus w inversion that's the cross section over which the thin layer of rectangular layer of current is flowing from source to drain that's a constant so then let's not worry too much about it now if you plot this drain current if you plot this drain current as a function of phi service let's first start with phi service then do you realize that the current will go exponentially up right it will go exponentially up that's exponential current and this is called thus threshold regime what's going to happen if phi service begin to reach sort of the vbi then what will happen if the barrier is gradually decreasing what happens in a diode do you remember the ambipolar region the high injection region do you remember that that's exactly what's happening you can see don't you remember this region 1 2 3 4 and all sorts of region we discussed this is exactly that same curve just plotted in a different way because you can see that the rate curve is rolling over sort of the high injection regime the difference between a bipolar and a MOSFET at least in this context is one of the differences is that in bipolar we generally operate in the linear regime where it's increasing exponentially we don't want to be in the ambipolar regime or high injection regime it turns out in MOSFET we do exactly the opposite we really operate in the high injection gene above threshold and and that's why we'll be operating but let's start by understanding the below threshold part exactly like a pn junction no difference you can see same curve now one thing is phi service I do not know right but I have to calculate but do you remember in the last class I was saying that phi service the amount of surface potential you could take vg and just divide by this body coefficient m the body coefficient I said that is you know 10 percent to 40 percent of more than compared to phi service that will give you vg so m is 1.1 to 1.4 in modern transistors so in fact you can see that if I somehow could calculate phi service in terms of vg the gate voltage which I know then I could easily have written id as a function of vg now this slope is a very important slope it's called a sub threshold slope of course below threshold a slope of course is a sub threshold slope and you do realize that what should this value be sub threshold slope if the y-axis if the y-axis is in log not ln here log you know factor of 10 increase then do you realize that this should be what should it be 60 millivolts per decade why in a lawn every change in one this one volt change in one volt change in phi service essentially increases the current by an order magnitude I'm sorry for beta if the beta is let's say 26 mille electron volts right beta over q is 26 mille electron volt so if your phi service changes by that amount one order current will change because this is exponential so you'll have to multiply with 2.3 because the code word between log and lawn and that will give you 60 millivolts per decade now this is something very important and a big fraud in in one of the biggest fraud in this in probably in science in last 20 years was found out exactly by this because one of the very young scientists named Henrik Sean he was probably when he was in the middle of a few doors down from where I used to work he essentially was producing a series of fantastic results you know science and nature these are very important journals and he was getting one or two every few years it'd be considered a mark of a very important work and very distinctive work now this person was writing two or three or four science papers every year he was 27 years old he was being considered for one of the directorship and he had the offer for max plank institute you know they have five directors I think I don't know exactly how many they have but he was one of the youngest he was supposed to be one of the youngest everything attached was gold and what happened one time there is a paper published in which the slope of this car of course physicists generally don't pay attention to this it looks a nice car transistor is working and one of the electrical engineers noticed that this is not 60 millivolts per decade even in ideally for an organic transistor he made even ideally it should not be below 60 millivolts this is the lowest it can be in this particular case and therefore that started a whole series of investigation so when you go back home today google under henrik shon and you will see the whole thing came crashing down it turns out the 72 or 73 papers he wrote he didn't do a single experiment it was all out of thin air he essentially published it went through the reviews and everything and so there is a documentary that's coming out two three books coming out this year and so that was a very interesting story that how simple observations of simple things can make a huge difference but let's go on to the story of more mundane story I'll tell you the full story later on now these are the experimental car you can see in the y-axis I have it in terms of log and in the x-axis we have it in terms of vg and if you take a slope of this line they call the sub pressure slope then that would be always greater than 60 millivolts per decade one order of increase in the current you need at least 60 millivolt at room temperature now where is that factor aim come from I explained that to you a few days ago that this oxide capacitance and the semiconductor capacitance you have to take the ratio that gives you the factor m I have explained that to you just to remind you that the m would be in the ratio of x0 the oxide thickness and wt the doping so you realize that the body coefficient will depend on the doping through wt and the oxide thickness of course what would have happened if oxide thickness is zero then m becomes one then phi survey is equal to equal to the vg because you know you don't have any oxide so of course the 100% of the voltage should appear across the across the gate contact okay so one can calculate it and then you can correspondingly plot that now let's let's talk about the more important part of super threshold characteristics this is above threshold this part we didn't pay as much attention so we'll be calculating current and I'll be proving a series of formula it's not difficult but you'll just have to pay a little attention so the current I'll explain what this is and formula is in a second but the thing is that somehow I must be able to calculate the w the mu effective you know the mobility I'm not explaining the formula I'll do that in a second but the more important thing is that I will have to calculate the inversion charge inversion charge qi that's the blue as a function of voltage integrated over between zero and vds the drain voltage I have applied and that will give me the current I have not explained it I will do that but the thing is the bottom line I want to say is that somehow I need to calculate above threshold the inversion charge and there are four approximations one will be called an square law where you just get this qi as a function of v by this very simple formula now you it sort of looks familiar to you right vg minus vt you have seen this before c ox that the charge above threshold is c ox minus vg minus vt you know that the only extra thing that is coming in is this minus v sitting there this is coming from the drain so when you have a drain bias then you have that extra minus v sitting there now a little bit better theory is called a bulk charge theory and that has the same vg minus vt but in addition it has a factor big factor an exponential of n a and a v two vice of b plus v all these factors built in this extra thing it will be called a bulk charge bulk charge theory it gives a little bit better estimate for q sub i and this one you can see where the doping my would play a role right do you see that in this expression I have an n a sitting there what does it remind you this expression this is associated with the depletion charge do you remember the top side is looking looking like a w the square root of vbi minus va do you remember that so this is really accounting for that bulk charge which in the first one or depletion charge which is not being actually taken care of mean by the square law so I'm not deriving it yet but you get the idea just looking at the equation now many times it's difficult to apply it apply this one so you want something intermediate and what you can show that the last complicated term can actually be simply reduced to that same bulk body coefficient m you know if a few milliseconds ago we talked about that so it's essentially the square law but you account for the effect of doping through that factor m with the simplified bulk charge now you can solve the problem exactly also and of course pao and sha sha was a very famous person he did a lot of work but do you remember do you see the second set of authors this is one of our professors right professor pierre whose book you read and and his student shields so together they have also a very famous famous famous exact solution for this MOSFET characteristics but that is a little difficult and I will not go I will not trouble you with that exact derivation here it's very good but not necessary for this introductory course so let's talk about the effect of gate bias first of all and what does it do to the potential at a given point now again take a second to get oriented with the with the three dimensional plot on the right you see that the red point where I have shown it this is the drain and you see the p region which is in the bulk which is the yellow region and you expect that between drain and a gate there should be a depletion region and that's what is shown on the right and w sub d which is on the top is essentially the depletion do you realize why w sub d is so big is because it is a n plus p region of course most of the depletion should occur within the p region so that's what do you have here and then correspondingly you have a gate on the right hand figure and the gate when you apply a positive bias that will also begin will deplete a certain amount right do you remember w sub t or w dm written here this is the depletion due to gate and you realize that that depletion is fixed why because the gate voltage is above the threshold voltage so it has been fixed and the mobile charges are piling up just underneath the gate oxide now just to just to make sure that the orientation you get the notion of the orientation correct now let's think for a second about the potential it takes a little bit but you'll understand it easily i'm sure going from the green to the yellow region do you realize that this is the p-n junction that is i'm going over the potential and now if i had done the source side it would have come down it would come down on the other side so that is the vbi do you see that vbi because in equilibrium that is the barrier over which you'll have to go now when you apply a gate bias do you see on the right hand side the fourth corner right hand corner bottom corner you have the uh band bending towards the surface do you see that and that is at what point will it clamp to do you remember that no matter what gate voltage i apply the surface band bending how much can it be approximately equal to 2 phi sub b because beyond that point most of the voltage goes across the oxide it doesn't go across the body and if that's the oxide the lecture that we did in the last time so that now do you get the configuration then of course the electron will flow from the source to drain through that narrow region where it is sort of bending over towards the gate okay so these are the basic configurations no problem now the thing is that in this case i have applied a gate bias haven't applied any drain bias yet not significant drain bias yet therefore in this case you have a certain amount of depletion wt because depletion remember when it happens when you bend the band enough so that the minority carriers begin to become the majority carriers okay now this is the same set of figures we'll just focus on the figure on the very bottom in this very bottom figure you can see in the beginning you have zero bias then you have just the gate bias but in the very end figure figure in the very end what do you see what you see is a you have applied a large amount of drain bias you see that because the positive drain bias the drain has been the drain formula level has been pushed down all the way and that's why the depletion over there is going to be very very different compared to the source side let's let's try to look at it in a slightly different way now this is just i'm trying to set it up because then i'll do a very simple calculation so in the first case looking on the left figure i have no gate bias but i have a large drain bias and the drain bias the amount is vr reverse bias pn junction right you do you see that that's reverse bias pn junction now if you apply a gate bias now then of course you can bend the band but the thing is do you realize that by bending the band you cannot have now with this amount of drain voltage present vr present the minority carriers can no longer be greater than the majority carriers because you see you now have a huge barrier for the electrons to climb up through that region next to the gate and so what you have to do now if you want to invert this region next to the gate you will have to apply a huge amount of extra bias in the gate so that the barrier close to the surface becomes small enough again so that electrons can climb them so in the presence of a drain bias we realize our gate bias has to be significantly more or if we keep the gate bias the same our induced yard will be significantly less compared to what would have been the case if we didn't have a drain bias so if you understood that hopefully then let's let's look at this very simple figure and see whether we can understand it i'm taking a cut for the potential along that red line remember the cut need not be a straight line right cut can be an end line and that is what i have shown here in a two-dimensional plot do you see that when i'm cutting coming through the top first i have gate then i have this blue insulators and i have correspondingly drawn the bond band diagram below then i have a p region do you see the band diagram in the p region starting in the middle and then i turned right and then essentially go to the n plus region and the correspondingly my band diagram follows suit so you can see the band diagram has gone down now what would have happened if i have applied a drain bias okay if i have applied a drain bias then correspondingly the blue the green region would have been pushed down do you see in the middle figure do you see in the middle figure that the v the corresponding the v the whole Fermi levels have been pushed down what is that vb doing there the blue vb blue vb is telling me my body field or the body potential that has not changed so therefore at the p region has exactly the same potential as before but the n region now because i have applied a positive bias the n Fermi level has been pushed down now if i want to if i want to invert this region now you realize that i'll have to make sure that my gate voltage is large enough so that it goes under the gate this red line because the red line is sort of my majority carrier and so therefore i'll have to pull it down so that the electrons can flow as a result what is going to happen is that in addition of the standard formula this qi minus cox vg minus vt minus v that that thing i also have to account for this depletion charge because the amount of region that needs to be inverted sorry depleted wtv that depends on the gate voltage sorry drain voltage i have and as a result this extra factor would come in so let's see how it works on the left hand side i have just the gate voltage no drain bias on the right hand side i have a gate voltage as well as a drain bias let's think about the charge on two sides do you agree the threshold voltage on the left side will be 2 phi sub f minus that factor what is the second factor that is the potential drop across the oxide right eox multiplied by x naught remember that the electric field coming out and dropping across this and the x naught has been absorbed in the cox so we have that on the right hand side i have a new potential because i will now need to put more voltage before i can invert it so i should have to have 2 phi sub s plus v why 2 phi sub s plus v because if i don't invert additional v amount then i cannot be sort of go over the majority carriers on this side but at the same time because of this extra voltage wtv would be the amount of depletion extra depletion that i have and so from here i can write the new threshold voltage to be the old threshold voltage minus this extra thing essentially you'll subtract one from the other and that gives you this particular formula and therefore the new threshold voltage you can simply write it as vg minus v threshold star because that's the new threshold voltage in the presence of a drain bias now that's the whole formula i have and now do you realize where the where the bulk charge theory where that comes from because wtv is essentially will be proportional to 2 phi sub b plus v right this is the amount of depletion and when you the drain bias was zero then correspondingly the depletion charge will not have that extra v in extra potential it now you can calculate it first of all one way to calculate is drop those two terms the last two terms looks complicated drop it now if you do that that will become your square law theory but if you do a little bit more manipulation and that is not i'm doing here you can also show the last two terms essentially at the end you can have it as a fraction appear as a fraction m multiplying that v but that extra derivation i'm not doing but let me just focus on the physics of square law and try to see whether i can explain it to you now what happens along the channel that as the electrons goes from the source to rain then in one side the Fermi levels for fn and fp both are sort of near equilibrium because i have not changed the source and gate biases on the other hand as i have pumped down the drain potential therefore i will have the fn lowered by the amount qvd on the other side and in between the potential will go linearly as from one side to another now what happens if potential goes linearly from one side to another you realize that if my gate voltage is held constant then as i am approaching the drain then this v this effect of the drain bias v that will gradually increase because at every point as it's getting closer to the drain my full effect of vd is coming in and therefore the further along i go along the drain less charge i have over there because in one case i'm almost fully in inversion as i'm going further and further the drain voltage is sort of pulling the point where i need to get in order to go to inversion so therefore the total amount of charge will gradually become smaller and that's what i have shown here the over there in terms of boxes the rectangular boxes that is you go down the channel you have less and less charges just a quick another view because this is sort of very important that you understand this that as you are moving down the channel on the very left side everything is in equilibrium like a pn junction as you are sort of in the middle you can see because of the effect of the drain voltage the fn has been which is a dashed line has been pulled down a little so it's more difficult to invert a further down near closer to the drain the fn has gone down even further so it's getting even more difficult for the inversion to occur and finally very close to the drain the fn is even further down but you see in all these cases fp is exactly where it was throughout because i have not changed the body contact it's exactly where it was before and because this fermi levels are getting split you can see that corresponding to the conduction band the difference is increasing and therefore i will have less and less carriers as i move down the channel so let's make some calculations because i think we are done so the first thing is let's take box four boxes here of the electrons the first box j1 the current in j1 will be q1 mu multiplied by e1 right why this q1 q is the whatever charge you have in the first box mu multiplied by e1 what is that that's the velocity so q multiplied by v is the current what happened to the diffusion current well remember that this is a very strong field this is above threshold below threshold we did the diffusion diffusion was the dominant one above threshold it is mostly drift so that is going through this source drain by drift and e1 you can write is dv dy along the lateral direction in the horizontal direction that's the electric field along y and you can write the other four pieces also now do you realize this current has to be continuous right why does it have to be continuous because if i assume no recombination whatever is coming in box number one that same thing must go to box number two box number three current right not the number but the flux if they are the same then i can easily do this i could sum the left hand side and if i sum the left hand side divide by mu and multiplied by dy then do you also realize the j i could pull it out of the sum is the same it's a constant i don't know what that constant is but that must be the same constant so i pull it out of the integral what is that sum over di dy the sum over dy is the channel length right because dy is the boxes so sum over di is the channel length and qi well you see that that little formula we derived five minutes ago that's what is sitting on the right hand side cx bg minus vth minus mv now can you integrate this well i'm sure you can because this is a simple vg minus vt that's a constant so when you integrate and put the limit that will give you a vd and the second term will be mv squared over 2 and of course when you put the limit that will become mvd squared over 2 left hand side that would have been in l channel and that's what i have cross multiplied up and down and that gives me the current jd and that's the formula this is the current for a MOSFET above threshold not sub threshold above threshold vd now let's look at this formula what is trying to tell us this this particular formula is trying to tell us first of all this formula is trying to tell us that the current will be maximum at some point and then it will go again turn over that's what that formula is trying to tell us because if i took a derivative of the d id as a function of i'm sorry that should be as a function derivative as a function of vd right so then there should be it should i can set it to equal to zero and so i can have a quantity vd sat beyond which this is what's going to happen that if i take a start with the any of the rate curves you can see that the curve coming to a point which is equal to vd sat and you see look at that formula vd sat is vg minus vt divided by m that is exactly the formula that i derived on the top at that point this formula is saying that it should come back down to r which is like a projectile and you can see that's the dotted line of course that's not that's not possible current cannot come down like that you are increasing the gate voltage so of course beyond that point the theory is unphysical but the point is that up to that point up to vd sat that's the correct formula and it turns out beyond that point the current is essentially constant so that is the solid red dashed line so that will be the formula for a drain current for the from the bulk charge theory a simplified bulk charge theory and in the very early part in the do you realize that there in the very early part the formula gets simplified to what is written on the very left hand corner in the blue part do you realize that why is that because if vd is very small it's a point one volt vd squared is point zero one right so the second term on the full expression that will drop out and once it drops off you see that you have a linear expression so at a very low drain bias we have a linear increase in id right it's a drift current so of course it's a register so therefore you have a linear increase but as you go to higher and higher vd then gradually the curve begins to saturate and then it becomes constant the constant part is not in this theory but the full theory will have that the one that we didn't pausa theory or the peer shield theory those will actually account for the whole rate curve but in this one we just have a piece of it now why does the theory not work why does it not work beyond vd sad because look at that formula what the vd sad point is vd sad is vg minus vt divided by m right now put it back in this formula on the top for the charge so m vd sad if you put it over there instead of v on the top top equation instead of v we'll put vd sad so m vd sad is vg minus vt right what is that at vd sad point this whole charges zero if you put it back in it's zero if you try to put your vd a little bit more this whole thing will become negative right negative in the sense that whole thing inside it but that cannot be this formula is only valid above threshold this is not a below a sub threshold formula so therefore this quantity which is vg minus vt minus mv can never be can never be negative that's unphysical because it should always be above inversion as a result that whole formula which i used to derive that equation that charge formula is incorrect becomes incorrect after vd as a result you cannot use this formula above the vd sad point you see that's why so what happens here the charges in the last boxes they become equal to zero depleted it becomes depleted and this loss of inversion this is called a pinch of region this last region is called pinch of region so if you don't have a charge how is it that electrons are going from one side to another you don't have any charge how this current flowing well you have seen this before many times before when the base current or i'm sorry the emitter current flows over the base base potential and to the collector side doesn't it go over a huge depletion region in the collector mobile charges flowing through the depletion region this is no rocket science in there same thing here yes the last boxes are getting completely depleted but that doesn't mean anything i mean the electrons which are coming from the other side they will essentially simply go over the channel and depleted cross the depletion part and go to the other side current flow did be just fine and beyond that point so at that maximum point where the current become constant if you put that value of vd sad then you will correspondingly get an expression which goes as square of vd minus vt and that's why it's called a square law because it depends on the difference of vd minus vt as a square now for last probably 30 years square law has has not been valid and i'll explain that in the next class in any way one thing that's very important for many characterization is to understand what happens in the linear regime is a low voltage regime if you plot it in a linear plot id and vgs and for small vgs then you can see that you can easily for a given drain bias you can calculate the channel resistance because it's sort of a small drain bias it's like a large amount of gate bias so you have a certain amount of charge and so it's behaving like a register so you can calculate the corresponding channel resistance and in many experimental paper you will see that they report the channel register measurement of the channel resistance that you can easily calculate from that expression sub threshold conduction is of course very small and from the experimental iv characteristics you should always take a slope the red line and wherever that crosses that gives you vt because vt is something you have to measure how else can you get vt you remember from the cv characteristics from a cv characteristics remember where the low frequency sub threshold characteristics or depletion goes to inversion there's a change over there right so if i give you iv characteristics you can also correspondingly take a slope of the right id versus vgs curve and get vt and these two results must be the same so that you know that you have the right analysis and right values in there and once you have the intercept that gives you vt and the slope will give you the mobility so that is how you will calculate the mobility of a MOSFET so let me summarize so the basic point i wanted to make today was that beyond a MOS capacitor beyond a MOS capacitor the way you make it a transistor is by applying a drain bias so in order to collect the current two regimes are there one is sub threshold where the drain bias is small and another is and the gate bias is also small and another is super threshold where the drain bias is very large at the same time the gate bias is also large when you apply a large gate bias then what happens that the inversion at every point along the channel becomes little different and if you look at the electron concentration very close to the source you have a lot of electrons as you go closer to the drain you have less and less and less and when you want this current to be continuous that's how you calculate the bulk charge theory or square law theory that's how you calculate it so we'll pick up more details of this calculation in the next class okay thank you