Bismillah ar-Rahman ar-Rahim. My name is Dr. Syed Abdi. I'm from Liverpool. I'm working as a doctor in Liverpool. I've been there in the United Kingdom for the last 23 years now. Although originally I'm from Pakistan, born in Karachi, brought up and did my basic education in Pakistan. I'm fortunate I belong to one of the Shia family and I was raised in the Mahabad of Alibath, love and affection of Alibath, peace be upon him, and I always wanted to come here. So this is my first visit to Karbala, inshallah, and I hope it will be first of many. I think, as I said, we're fortunate that we're born in Shia families who were raised by our mothers, our father and our brother and sister to love and affection of Alibath. And when we're hearing from the very early age about what happened to Imam Alayhi Salam, Maslume Karbala, what happened to them, so I think people think about going for holidays here or there, but for every Shia, we all want to, that's the biggest holiday of our life to come to Karbala. And so obviously, this was from the day one and Allah gave me tafeeq now, so I came this time. And we arrived from Najaf airport, so as soon as we saw Karbala sign, and you can feel like sinking that you can feel the sorrow, you can feel what happened to Maslume Imam, and immediately you want to cry, so that exactly happened to me. We did not see the sign, but I thought about that I think it's a Karbala now, you can feel internal feeling, that it is a land of sorrow. I think, obviously when you don't see anything you want to see, but generally what happened if you go somewhere, which you've never seen, you said, okay, I've seen, let's move on. But the feeling of Karbala, once we're here, I'm already planning what to do next, when I'm coming next, and inshaAllah Arba'een Arfa, so it's a more, you want more and more visits here. So this is the feeling different from any other I've been to all over the world, from Disneyland to here and there, but the feeling which you, when you come here, you see there, you see the affection, it looks like your Imam is inviting you, and you want to be here all the time. As you said, you're from Karbala, so I don't know, I'm already planning when I'm retired, or whether I saw the sign of Imam Hussain, as-salam-Safire Imam Hussain Hospital, I was thinking about going there and find out whether I can come once in a blue moon or once a year to come and do some charity work. So that's a feeling, that's a different feeling from anywhere else. Even before coming, when I was coming, people were asking, why are you going to Iraq? Because when our Western friends find out about Iraq, they know about ISIS. So they worried about why I'm going on this time. So I was explaining to them what is Karbala, what is Iraq, where I'm going, and I explained to them, I said to them, this is very difficult to say why we want to go, we know we may be killed there. And I'm quite happy to go there in Karbala, likely because if you have someone else say that you go there where people will be killed, you try to go to the white, but this is the feeling that you want to, I think it will be honour for, when they said you may be, I said that will be honour for me if I can be killed in Karbala. That will be biggest honour of my life if I can be killed. So I think that's I'm trying to explain to them what happened in Karbala, what was the history of Karbala, who was Imam Hussain, as-salam, who was his immediate family who were killed, and how I think the history of the human being never been like this story, like a 72 companion and immediate family, especially six months old, has been murdered by these coward people. And I don't think so, we have any history of the inhuman being, the way our Imam went through this tragedy. I came here on last Thursday, Wednesday, Thursday morning, we came here, and for last, today is like a Sunday, so it's a six, five, six day. As I said, I've been to Mashallah Hajj, Umrah, and where you see the people selfish. When you go there, you say they're performing something which is like an actor performing some act. Here you see people are doing things from their heart, and you can see everybody's trying to help each other. It's not everybody trying to do their self. That's a different feeling. That's a different things, what I can see. And I think the internal feeling is very, very difficult to explain how I feel. I think being a Shia, Rais and Mahabata alaihi ba'at, as-salam, so it's automatic, and as soon as you come and go inside, you just feel like your heart is sinking, a feeling of what went through the Imam alaihi salam. So it's very difficult to express. I don't think anybody can express what they feel. It's purely internal. So that's how I feel. And the memory is basically, yes, I think every memory is good because, as I explained to you, I think looks like Imam calling you to come again and again. So I think that's my feeling that, I think I convey the same message to everyone that, obviously, for our religion and our belief that Ziyarat-e-Imam-e-Hussain alaihi salam is, wajib those who can afford. Even I think the belief is like a Hajj you can afford, but Ziyarat is a wajib even if you can't afford, even if you have to borrow money, you should do it. So my message is, I think you're missing something. If you haven't been here, I think you're missing biggest things in your life. I don't think so you can experience any of these feelings anywhere in the world. So it's not that Imam is going to lose if somebody is not coming. I think they are missing. They should come. They should experience. I think it's a life-changing experience, especially if we think that Imam has accepted our Ibadat, then we should feel chained to ourself after this Ziyarat. Bismillah ar-Rahman ar-Raheem, Allahumma salli ala Muhammad wa ala Muhammad wa ajil farajam. My name is Mamina Abbas. I'm 41 years old and was a Montessori teacher in Austria, Vienna. Since five years, I'm settled in Pakistan now and I'm just giving my duties to my kids, home and my Ahlebaat, Allahumma salli ala Muhammad. During past five years, I think I started my journey. I changed when my life was totally changed. I was having a very different life before. But once there was an incident that happens with me that I just saw a dream in which somebody told me that show me something which changed my life. And now, Alhamdulillah, day by day, I'm walking on the path which is near to Muhammad wa ala Muhammad. And before, it's my third visit to Karbala, Aba Abdelil Hussain, Allahumma salli ala. Two years before, I was here in Urbain and this changed my life. Urbain walk and Urbain, it changed my life more than anything else. This visit, when I come to my Imam, to visit my Imam, I just think that I was just, one day I was just sitting and thinking about to come again on Urbain. Then I thought that why shouldn't I take my kids with me this time so that they can also feel Karbala, to feel Karbala so that they can apply Karbala in their life. So I decided and Alhamdulillah, I came here with my kids and it is a very, very, I think it was a very good decision to bring kids with me and like it's easier for me to, for the Tarbiyah of my kids. And Alhamdulillah, I'm just thinking that I'm in heaven because to Karbala, I never think that it's a place on earth or it's a piece of land or a country, something like that. I always think it's a piece of heaven and I'm just walking in heaven and paying my regards to my Imams, my lords. The thing which made me to bring my kids, it was a very good, I said this before, it's a very, I'm thinking that it's a very good decision on this age, I bring them here. So that's, it's a very good decision because it's a very, it's a very strong step to take for their Tarbiyah, first of all. Now they can think about from where they belong and about Karbala because Karbala make, make what you say a personality. I think so, Karbala is always, if you visit Karbala, it is, it is a strong bridge to make a man, a strong man or a strong person, to build up a strong personality. It's, it's, it's always been like that. So that it's, it's a thing that you can, you can to be near to your Imam. If you, if you feel Karbala, if you think Karbala, if you don't take Karbala like a story, it's not a story. It's a, it's a whole life. You can say, it's a, it's a meaning that what, how to, how to lay your life, lead your life. Because if we are belonging to Ahle Bad, so if you don't come to Karbala or you don't think Karbala, then you, then you can't be near to them. I always think that if you want to be near to your lords, you always have to visit Karbala and then you can shade it in your whole life. When the star, when they saw the, the tomb first time, I think what can be a, a moment or a, or it's a, first of all, this, this thing was a dream for them to visit Karbala. It was a big dream for them that is going to be true one day. So this was the day that their dreams come, came true. And when I think, so what should be there when your dream come, came true. So that was the feelings to them and always for the kids, the, they always think that Imam Hussein as love for their family, especially for Hazrat Ali Akbar. This is the, for my elder one, he's always fond of this part of Karbala and Imam Hussein's life that he always want to be, walk on the path of Janab Ali Akbar, alaihi salam. And the younger one, his, once he, he told me when we, when we decided to come to, came to Karbala, he, he told me that it says, it was my dream to visit Karbala and he was just thanking to me that you have fulfilled my dream and my dreams came true. I would like to say to my family that do visit to Karbala because Karbala, if you, you came to Karbala, you know the meaning of life. And I'm so glad to see here the, the people of Karbala are so cooperative, loving, caring. And I don't think so there is any security issues, because security is taken, the security is under, I think, Aba Fajlil Abbas, alaihi salam. I never think that there is any security issues nowadays. And it's a very, I think so it's fine. They have to come and even there are any issues and something like that. Everyone have to visit to Karbala to, to build their lives and their according to alaihi bayat and to be more honest to their lords and to be more near to them, to walk on the right path because it's, it's always said by Prophet Muhammad sallallahu alaihi wa sallam always say that Hussainu minu ana minal Hussain. So if we came to Hussain, it means we came near to our Prophet Muhammad sallallahu alaihi wa sallam and if we came to them, we are near to God more than anything else. Yeah, it is right. One visit is never enough to come to Karbala because how can a man live without water? I think how water, how water is necessary for your life, Karbala is like that. So I only think Karbala is breath, Karbala is blood running in your veins. This is Karbala, Karbala is inside you. It's not only outside, it's inner. Your inner is Karbala. So you always have to be, to come to Karbala. It's, it's more than necessary to anything else. There is a hadith about one Imam. I just forget the name. I think Imam Jafarisadek or Imam Ali Raza. I don't know, just forget about that.  You have to come to Karbala. You have to visit Karbala. So Karbala is always, Karbala is as necessary as what you can say breath for your life. It's a breath. Our breath is necessary for, for our life. Karbala is also necessary for our life. So everyone have to visit to Karbala. Should, I will, I will say should come to Karbala and fulfill your desires because these are the ones who give you everything with the wish of Allah. They are all the one to give you everything. I, I believe on this thing.


At 9, 50, and 45 seconds, coordinated universal time, on the 14th of September 2015, a signal was detected by the LIGO detector in Livingston and 6.9 milliseconds later in Hanford. It was a chirp signal that lasted just over two-tenths of a second. When we route the wave into a sound generator, here's what it sounds like. This plot combines the data from both sites. The waveform is consistent with coalescing masses, with a 10-cycle 200 millisecond in spiral that gives us the frequency, the rate of change of the frequency, and a peak wave amplitude. A merger that takes around two milliseconds, and a ring down as the coalesced objects cease to radiate gravitational energy. Detector noise introduces errors into all the calculations based on these figures. That's why we'll provide a range for each item. The amplitude and frequency data points give us the luminosity distance. It is important to note that gravitational waves experience redshifting as they travel across the cosmos, just like light does. Having traveled around a billion light years, this wave would have experienced a redshift near 0.1. So the frequency we see here is a bit smaller than the frequency at the start of the wave's journey here. The frequency data also gives us the chirp mass. Taking the redshift information gleamed from the merger and ring down portions of the waveform, we get the binary system masses. These masses are too large for neutron stars that are only a few times the mass of the sun. So we must be witnessing the merger of two large stellar black holes. During the last 200 milliseconds of their in spiral, the orbiting velocity of the black holes increased from 30% the speed of light to 60% of the speed of light. Over the same period, the distance between the two black holes went from around a thousand kilometers to just under 200 kilometers when their event horizons made contact. Modeling the final ring down shows that the mass of the resulting curve black hole is around 62 solar masses. That's three solar masses less than the sum of the masses of the two inspiring black holes. This mass was converted to the radiated gravitational energy. In other words, during the final 20 milliseconds of the merger, the power of the radiated gravitational waves peaked at about 3.6 times 10 to the 49th watts. Let's take a second to get a feel for how large this number is. In our How Far Away Is It segment on nearby stars, we found that the sun converts 4.26 metric tons of matter into energy every second. The resulting power output is equal to 4 billion hydrogen bombs exploding every second. The sun is an average star, so we can use this as an average stellar power output. From our segment on local superclusters, we saw that there are 250,000 trillion stars within 1 billion light years. This represented around 7% of the total number of stars in the universe. We get the total power emitted by all the stars in the visible universe by multiplying the average watts per star times the number of stars. The power generated by this merger of the two stellar mass black holes is 26,000 times greater than the combined power of all the light radiated by all the stars in the universe. That's the signal we saw in September 2015, a billion years after it happened.


Hi and welcome to the session, I am Shashik and I am going to help you to solve the following question. Question is if y is equal to 5 multiplied by cos x minus 3 multiplied by sin x, prove that d square y upon dx square plus y is equal to 0. Let us find the solution now, we have given y is equal to 5 cos x minus 3 sin x, now differentiating both sides with respect to x, we get dy upon dx is equal to 5 multiplied by minus sin x minus 3 multiplied by cos x, so we get dy upon dx is equal to minus 5 sin x minus 3 cos x, now again differentiating both sides with respect to x, we get d square y upon dx square is equal to minus 5 cos x minus 3 multiplied by minus sin x or it is equal to minus 5 cos x plus 3 sin x, now it can be further written as minus multiplied by 5 cos x minus 3 sin x, here we can see clearly that 5 cos x minus 3 sin x is equal to y, so we can write d square y upon dx square is equal to minus y or we can say this implies d square y upon dx square plus y is equal to 0, so d square y upon dx square plus y is equal to 0, this is our required answer, hence proved we were required to prove this equation only in this question, so hence proved this completes the session, hope you understood the session, good bye.


You can follow along with this presentation using printed slides from the Nano Hub. 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. All right, so what I would like to do is to give a very concrete example of the application of some of the ideas that Professor Dada talked about. Now I can't do blackboard lectures the way that he does them. I have to do it in PowerPoint. So the pace might be a little quicker, but you'll see concepts that you heard Professor Dada discuss. And for those of you that would like to see more about how you apply these ideas to small transistors, the 2008 summer school, which you can find on the electronics from the bottom-up web page, which you can find by just googling it, you'll see a set of five lectures on this topic. So I'm just going to give you in the next hour or so just an idea of how you apply those ideas to a very practical and very important problem. So it's nanotransistors. And as Professor Dada mentioned, transistors are getting very, very small these days. This is an SEM cross-section of a typical transistor of a few years ago. And today, current day technology, the channel lengths are something on the order, the physical channel length, is something on the order of 30 nanometers. But Ian is telling you, that's roughly correct. So these devices are getting incredibly short, much shorter than we imagined a few years ago that they could even be manufactured. So they're true nanoscale devices. And any talk on transistors should have a plot of Moore's Law. You've probably all seen Moore's Law plot. I like to do my Moore's Law plot a little differently. I do it on a linear scale. Because when you plot on a log scale the number of transistors on a chip per year, it just sort of looks like, well, everything has a positive slope. It gets a little bit bigger every year. What's the big deal? You don't really get a sense as to what's happening with exponential growth unless you plot it on a linear scale. And this is intel data on the number of transistors per microprocessor chip per year. About the year 2000, I remember all of the discussions we had about, boy, this is getting really difficult. How much longer can we continue to shrink transistors and put more transistors on a chip? When you plot the number of transistors per chip versus year, you can see that it looks like almost nothing was happening until the year 2000. And then things started to happen. And of course the big question now is what's going to happen in the future. So right now, manufacturing technology, the latest manufacturing technology is 45 nanometer technology. 32 nanometer technology will be deployed soon by companies like Intel. And 22 nanometer is well underway in development and will come along after that. 15 nanometer has started. And of course the big question marks for a lot of people like you who will graduate and contribute to technology development. Is how do we do the next step and then the step beyond that. If I make this plot 10 years from now, I'm expecting that it'll look like nothing happened until about the year 2010. And then things started to happen. Okay, so how do we understand these very small transistors? Well, a lot of our understanding and a lot of the design that gets done in companies like Intel and others uses very sophisticated simulation programs. This is a Monte Carlo simulation, some results that came from IBM a number of years ago when they were looking ahead. And this is a semi classical simulation. So it's a simulation that treats electrons as particles. So what you're seeing on the left is the conduction band versus a position or a hypothetical. I think this was an SLI double gate transistor with a 30 nanometer channel length. Each dot represents an electron that's being tracked through the transistor from the source to the drain by the computer. As it is accelerated by the electric field and undergoes random scattering processes. So you can see a lot of electrons in the source. You can see them going over a barrier and flowing across the channel. You can see that they're sort of going quasi-ballistically. A few of them scatter and lose energy but most of them go to the drain without losing a lot of energy. And you can see on the right the average velocity of those electrons versus position. So you can see the average velocity is low in the source and then in the channel it gets very high. And then it comes down in the source where they relax and you'll go back near equilibrium and the velocity is low again. Now one of the things that I'm sort of giving this lecture to two audiences. To some of you who know a lot about transistors, what I hope to do is to give you a different way of understanding transistors, especially smaller ones. And some of you who probably don't know very much about transistors and I'm trying to convince you that they're really quite easy to understand. Now if you've seen transistors before and taken courses, one of the things that you'll remember is that the slope of the energy band diagram is the electric field and yet the electric field is very strong near the drain because the slope is very steep. And one of the things that you learn in basic introductory, semi-conductor courses is something called velocity saturation. When the electric field in silicon gets very high, the electric field in bulk, silicon, gets very high, electrons scatter a lot and you approach a saturated velocity that's almost exactly one times ten to the seventh centimeters per second. So if you look at the drain in this simulation, the electric field is very, very high near the drain end. But if you look at the average velocity, you can see that it's very much higher than that saturated velocity of one times ten to the seventh. And that's just a reflection of what can happen in a very short device where there isn't time for the scattering really to come to steady state and to equilibrium. The devices are out the drain before they've had a chance to scatter enough for their velocity to saturate. So this is called off-equilibrium transport. Things get very complicated in these small devices. Now you can ask about quantum mechanics too. In these very small devices, you can ask whether we really should be thinking of these electrons as classical particles or should we be treating transport quantum mechanically. You can do that too. These are some simulations of ten nanometer channel length MOSFETs done by our nanomoss program. What you're seeing here on the left is the energy resolved electron density, energy and position resolved. So lighter means that there's a high density of electrons. Black means there's no electrons. The dashed line here is the conduction band profile, just like in the previous one, but this is a quantum mechanical simulation. So you can see some electrons underneath this potential barrier. There aren't supposed to be states underneath the bottom of the conduction band. Those are just electrons tunneling in from the source. And then you can see them streaming across ballistically. Or you can turn on scattering. And now when you turn on scattering, if you look carefully, you can start to see potential drop series resistance in the source and the drain. And you're starting to see that it's not so easy to see this ballistic stream because the electrons are losing energy and everything is getting mixed up. And it's not so clear to see which contact the electron came in from because it's all been mixed up. So a lot of our understanding of these small devices comes from very detailed simulations like this. And things began to get complicated 15 or 20 years ago. And I started hearing people actually say that these small devices have gotten so complicated, the only thing you can do is to run the simulation and that's the answer. But that's not really what we mean by understanding how a transistor works. It's something that you can understand what those simulations are telling us in some simple way. And I like this quote by Eugene Ligner. Somebody must have brought him a computer simulation and said, ah, we understand the problem. It agrees with experiments. And I think his feeling was that, well, okay, we really don't understand it. Can we understand what those simulations are telling us? And that's really what this lecture is about. Here we spent five years, ten years, I don't know, doing detailed physical simulations trying to boil down what we were understanding and see if we could make sense of it. And what I'm going to give you is a very simple picture which I believe captures the essence of what's happening in these devices. And what's happening, the details are extremely important. This is the difference between making a profit if you're working at Intel or not. But the essence of what's happening is really very easy to understand.  They'll help us understand what these detailed simulations are telling us. They're no replacement. If you're going to design a very small transistor, you're going to need to get into these details. But they'll help us understand what we're seeing. We'll talk about things like ballistic limits. We'll talk about things like velocity saturation and how that occurs in a ballistic MOSFET. We'll briefly compare the simple model to some real data for silicon and for 3.5 FETs. And then we'll find that we're not at the ballistic limit, so we'll have to understand scattering a little bit. And what I do is I'll go through this, I'll use some ideas that you heard from Professor Dada's lectures. And I think, as I said, we should be able to get the gist of the problem. And if you're interested in learning some more about this, I'll point you to my online short course on nanotransistors. Okay, so we're going to take this bottom-up approach and we're really doing it for two reasons. One is it makes what could look like a very complicated problem really look like it's quite a simple device to understand. But the other reason that Professor Dada alluded to is that we may be encountering materials, like we may be trying to make a transistor out of a small organic molecule. There were a lot of people trying to do that a few years ago. And it's very difficult to take the textbook understanding of MOSFETs and predict the IV characteristics of a molecular transistor. But using these bottom-up ideas, you could describe both devices, a silicon MOSFET or a molecular transistor using the same concepts. And you could compare the two and decide whether it makes sense to try to build a molecular transistor or not. Okay, so here's the outline. We'll talk about how we think about a nanom MOSFET first of all. Then I'll talk about a ballistic MOSFET, then we'll put scattering in, and then we'll be about done. Okay, so these are the IV characteristics. This is of a MOSFET in 2007 that had a physical channel length of about a... I had a drawn channel length of about 100 nanometers, a physical channel length that was a little shorter than that. So these are typical IV characteristics that you'll see for transistors. And what I find a useful way I think I find to understand these small devices is something that was first introduced by a fellow named Ed Johnson in 1973 who thought about how you think about a MOSFET. Oftentimes, if you take courses, you'll see that there are two main types of transistors that we talk about. Bipolar transistors and MOS. For many years, there was sort of a competition between these two technologies. And as devices got denser and denser, MOSFETs have taken over more and more of the applications. But we still think of these two devices as being two different kinds of devices. And what he pointed out is that, well, they're really the same kind of device. They really operate by the same physical mechanisms. And the way he looked at this was that if you take a MOSFET and we showed a cross-section earlier, we have a source channel drain. And if you draw an energy band diagram across the source, this is the bottom of the conduction band, and then across the channel and then out the drain, there will be a Fermi level somewhere. That tells us where the states are filled up. That's this dash line. So in the n-type source, the Fermi level is way up in the conduction band. We have lots of electrons in the source. The channel is p-type, so the Fermi level is way below the bottom of the conduction band. And then the drain is n-type as well, so the Fermi level is way above. So this is an energy band diagram under low drain-to-source voltage, where the Fermi level in the source and the Fermi level in the drain are just about at the same place because we've applied a small voltage. These are the two contacts that Professor Dada was talking about. And the channel is undoped or intrinsic, so there's a barrier, there's a potential barrier that stops electrons from flowing. Now, you apply a voltage to the gate and it lowers the electron energy. The energy of the electron is minus Q times voltage. So you apply a positive voltage to the gate, it lowers the electron energy. These are electron energy plots. It pushes the barrier down and it allows current to flow. And what you see here in low VDS is more and more current flowing. And when you push it down and get a lot of charge in the channel, you can see a uniform slope, which means a uniform electric field. It just behaves like a voltage-controlled resistor. So that's one regime of operation. Now, if you look at high drain voltage, all we've done now is to lower the energy in the drain. We've applied a significant voltage about a volt, or 1.2 volts. We've lowered the energy in the drain by a significant amount. There's a Fermi level in the source. Now there's a different Fermi level in the drain that's way down here, one volt lower. If we draw an energy band diagram, there is still a potential energy barrier between the source and the channel. So current doesn't flow. Unless we apply a large enough gate voltage, each one of these curves is for an increasing gate voltage. As you apply a larger and larger gate voltage, you push the barrier down and turn the device on and current flows. But you can see now that once electrons get over the barrier, there's very little to stop them from going out the drain. So the drain voltage no longer matters very much once it's big enough. So in this regime, the current is more or less independent of the drain voltage.  But the current is more or less saturated. So the point here is that a MOSFET is a barrier-controlled device. The gate just lowers and raises the barrier between the gate and the channel and that controls the current flow. Same thing happens in a bipolar transistor. There you have a voltage between the emitter and the base. The potential barrier between the emitter and the base determines a collector current. It's really very much the same device. That's why Ed Johnson called the MOSFET, or insulated gate field effect transistor, a bipolar transistor in disguise. So that's the way we want to think about MOSFETs. As barrier-controlled devices, we sketch an energy band diagram like this. This is the source, conduction band. This is the potential energy barrier in the channel and this is the grain region. Now, we have to bring in some things that require a little bit of discussion more than we can get into here if you are not familiar with MOS transistors. One of the things we're going to bring in is the fact that there is this MOS electrostatics. When you put some charge on the gate, the charge is balanced in the channel. And there isn't much charge unless you achieve a critical voltage called the threshold voltage. You've got a significant amount of charge in the channel and charge is always capacitance times voltage. But you don't have any significant charge in the channel until you turn the device on by exceeding the threshold voltage. So this is an expression of charge balance and of MOS electrostatics. The charge on the gate is balanced in the charge at the top of this barrier, which is at the beginning of the channel, and it's just the gate capacitance times the gate voltage minus the threshold voltage. This is MOS electrostatics. This is the Poisson equation. This is exactly true in a long device and what makes device scaling so difficult is to make this relationship hold in very, very short devices. So we would like the charge at the beginning of the channel to be controlled by the gate voltage, not to be influenced by the drain voltage. And as the drain gets closer and closer to the source, that's more and more difficult to do because you've got a potential at the gate that's trying to control the charge here. You have a potential on the drain that isn't much further away that's also going to affect things. But this is what good transistor design is all about, electrostatics. So you've got a region near the beginning of the channel that's strongly under the control of the gate and weakly influenced by the drain voltage. If you've designed the transistor well, that's what you hope to achieve. So there's a region there with some length script L that's some fraction of the entire channel length that's controlled mostly by the gate, not by the drain. Now the rest of the channel is controlled by the drain potential. And as you increase the drain voltage, you mostly just influence the potential in that region. And under high bias, that doesn't increase the current very much because once electrons get over the barrier, they drop down a potential barrier. If they drop down a bigger potential barrier, it doesn't really matter, they're still going to come out the drain. That's why the current saturates. Okay, so that's our picture of a MOSFET. What we want to do is to relate this to the picture that Professor Dada was talking about. We want to calculate the IV characteristics of this device using the ideas we heard about in Professor Dada's lecture. And as Professor Dada talked about, we think about the device as having some density of states. And then we think about having two contacts which are always large, there's a lot of scattering, they're maintained near thermodynamic equilibrium. The current that flows is just a small perturbation and doesn't really disturb equilibrium in these two contacts. Each one is characterized by a Fermi level or electrochemical potential. It's just that we might have different Fermi levels. If we apply a voltage to the drain, we lower the electrochemical potential in the drain. So the second Fermi level is different from the first. But they're equilibrium Fermi levels. And there are some characteristic times here that describe the transit time across the device or how rapidly we can get into or out of the device. Now the starting point for my calculations are going to be a version of what Professor Dada discussed. We're going to write the current this way. Now if you remember, Professor Dada wrote the current like this. So it was an integral, there was an energy resolved conductance that he discussed. And what's important is that current only flows when there's a difference in Fermi level between the source Fermi level and the drain Fermi level. So I'm going to write this expression in just a little bit different way. It's another way that people commonly use. We're going to write this conductance, this energy resolved conductance, as the product of M, which is a dimensionless number. It's a number of conducting channels and transmission, which is another dimensionless number between 0 and 1. And if you relate this to what Professor Dada discussed, when he lumped some of these constants together, M was the density of states times the average velocity in the direction of transport. And there's a constant H that's involved in there. I'll discuss that very briefly and I understand Professor Dada is going to say more about that in the afternoon. And M is a dimensionless number which tells us how many channels are there at an energy E to conduct current. Now T is just a way of writing mean-free path divided by mean-free path plus lambda. You can see that if the device gets very short compared to a mean-free path, then that ratio lambda over lambda plus L approaches 1. T approaches 1. That's the ballistic limit. The way we interpret that is any electron that comes in from the source goes across to the device ballisticly without scattering. In the other regime, where L is much longer than the mean-free path, then T is very small. It's just the ratio of the mean-free path to the channel length. So as the channel length gets longer and longer, the fraction of electrons that come in from the source and are able to get out through the drain just gets smaller and smaller. So same concepts we're just going to rearrange things and emphasize modes and transmission. That's going to be our starting point. The one other thing that we'll need just briefly is the number of electrons in the channel. So the way we get the number of electrons in the channel, if we were doing this in equilibrium, we would just integrate the density of states times the Fermi function over energy. Well, we have two different Fermi functions, one from the source and one from the drain. So the electrons that come in from the source, they all have positive velocity. So they can occupy positive velocity states in the channel, plus k states, say. So they can only occupy half of the states. So it's like Professor Dada's freeway. Half the lanes go one direction, half the lanes go the other direction. The electrons that come in from the source to coming in ballistically, they can only occupy the positive velocity states. So they contribute to the electron density density of states divided by 2 times the Fermi function. The ones that come in from the drain, they can only occupy negative velocity states. So they contribute density of states divided by 2 times the Fermi function of the drain. We'll need that a little bit later, too. But those are the two equations that we need to understand the IV characteristics of this small MOSFET. Okay, so we want to map this MOSFET onto this simple picture of a nano-device, and we do it this way. So we argue that what's really important is the top of the barrier. That's how you make a transistor. The gate voltage, if you put a negative gate voltage on, you have a high barrier, the device is off. You apply a positive gate voltage, you push the barrier down, the device is on. That's what makes a transistor. So it's really what happens at the top of that barrier that makes the transistor. So at the top of that barrier, we have some density of states. So the easiest way to think of it is this is a silicon transistor. I'll just think of that as some silicon band structure. And we're going to move those states up and down. Now, you might wonder, if you get into a really small device, you may not be able to use a bulk band structure. The local density of states there might be modified. Then you need one of those quantum simulations that we discussed earlier. But this really works very, very well, even down to dimensions that are well below 10 nanometers in practice. So we have some density of states, and we're just going to assume silicon density of states at the top of the barrier. Our self-consistent potential that we apply through the gate voltage is just going to push those states up and down. And we have a small, short region there under strong gate control of length L. That's our nanodevice. So the overall channel length might be 30 nanometers, but there might be a region of 5 nanometers or so that's strongly under the control of the gate. That's our little nanodevice. What are these other two regions? Well, we think of them as the contact. The rest of the channel, that just allows electrons to come in from the source or to come in and out through the drain. And as long as they allow electrons to flow freely in and out, they're just ideal contacts and they don't play a significant role. Okay, so that's our model. That's how we take the ideas that Professor Dada used and try to apply them to a transistor. Now, how do you get the shape of that potential energy profile? Well, I mean, to do that, you have to do a full 2 or 3D numerical simulation of the Poisson equation coupled with the transport equations because charge is flowing through there and affecting the potential. But we're trying to understand what goes on in those simulations. So we'll assume we know what the shape looks like. So now we're ready to discuss a ballistic MOSFET. So, first of all, we'll take the case where there's no scattering. And our IV characteristics, one of the things, the IV characteristics of any transistor are going to look similar. It doesn't matter whether we treat them with a classical model or a quantum model or a drift diffusion model. They're mainly controlled by modulating these energy barriers. They have the same shape. The transport model affects the magnitude of the current somewhat, but doesn't change its fundamental shape. They'll all look something like this. They have a region under low voltage between the drain and the source where the current behaves like a resistor, a gate voltage controlled resistor. And they'll have a region at high drain voltage where it behaves more like an ideal current source. The value of the current is given by the gate voltage, the independent of the drain voltage. We're going to assume a ballistic MOSFET. So, this transmission is one. That just means that the mean-free path, or the channel length is much shorter than the mean-free path, so this ratio is one. Electrons just go across ballistically and don't scatter. We have a Fermi function in the source. We have a Fermi function in the drain that has a different Fermi level. The Fermi level is lower by Q times the drain voltage. And we have to figure out what the number of conducting channels are. And then you have to add in MOS electrostatics. One of the things that makes a MOSFET special is that the Poisson equation is so important, this MOS electrostatics. So that has to be part of the picture, too. Okay, so this is a part. I'm not going to be able to do justice to this, but Professor Dada is going to discuss it a little bit more. You get a little bit more comfortable with the number of conducting channels. In his discussion, he pointed out that this number m is proportional to the density of states times the velocity. And if you look a little more carefully in a MOSFET, it'll be proportional. As Professor Dada discussed, in a bulk semiconductor, the number of modes is going to be proportional to the cross-sectional area of the conductor. A MOSFET is a two-dimensional conductor, so it's proportional to the width of the channel, the thing that comes out of the page in the pictures that I've drawn. Then we have some constants. Then the average velocity in the direction of transport, I'm calling it x, the density of states. These are two-dimensional electrons flowing in the channel, so I'm writing this as the two-dimensional density of states. I'm going to assume parabolic bands to make things easy. The velocity, one-half mv squared is equal to the kinetic energy, so we can easily express the ballistic velocity in terms of the kinetic energy and the effect of mass. As Professor Dada mentioned, these electrons are coming in and they have a distribution of angles. What's important is their average velocity in the direction of the channel. So if you do that averaging over angle, you pick up a 2 over pi. And the 2D density of states, the way I'm writing it here now, as Professor Dada mentioned, the density of states in 3D is proportional to the volume, in 2D it's proportional to the area, in 1D it's proportional to the length of the conductor. Now, normally or commonly in semiconductor textbooks when you compute density of states, because you know that the density of states in 2D is proportional to the area, what you usually see quoted is the density of states per unit area per unit energy. So that's what I'm writing here. The 2D density of states per unit area and g sub v is a valley degeneracy. You might remember in silicon in the conduction band, there are six equivalent valleys. When you quantum confine them in 2D, there tend to be two equivalent valleys in the lowest set of sub bands. You always have to ask yourself how many equivalent valleys are there. For that you need to know something about the band structure, the material that you're talking about. Okay, so we can just plug numbers into this expression and we can find out what this quantity m is right here. And we get an expression for it. And that's what it is. So that's all we need to know to work out the IV characteristics. But we can look at it in another way to get a little more feel for what this quantity is. So this is my MOSFET channel. It has a width w. This comes out of the page that I drew of the MOSFET. It has a length l. Another way to think about this is that the electron has to be confined to the width of the channel. So the electron has a wavelength. And the electron wave function has to go to 0 at each edge of the channel. So what I'm showing here is that if I have a de Broglie wavelength, if I have a half wavelength that fits into that channel, that's a conducting channel. One lane on this highway that Professor Donna talks about. If I have one that has a node in the middle, that would be another conducting channel. And it would have a higher energy because it would have a shorter wavelength or a bigger k. So given a dispersion at some particular energy, I can find out what the wave vector is. The wave vector is 2 pi over the wavelength of the electron. And then I can determine how many half wavelengths will fit into the width because at a given energy I get a k, which means I get a wavelength, and then I can find out how many half wavelengths fit into the width of that transistor. Every one of those is a conducting channel. So that's why this quantity m is called modes because people thought in analogies to wave guides. How many modes do you have in a wave guide? We think of each one of these modes as being a separate conducting channel. So, given an energy e, then we can find out how many channels are there at that energy. Each one can conduct current. Okay. All right, so now we're set. Ian? Could you allow, what would the higher modes do with two wavelengths? The higher energy, we will have all of the modes that have an energy less than that will be able to contribute to current flow at that particular energy. So if you're at low energy, you may only have that first one that I sketched. If you're at a higher energy, you may have that one plus you have one mode. If you're at an even higher one, you have those two plus one with two nodes in the middle. So the higher you go in energy in general, the more and more conducting modes you get. If you have a more complicated dispersion, you'll have to go back to your formula, calculate the density of states for that more complicated dispersion. And you can always work out what it is. Okay, so now we have everything we need to calculate the I-V characteristics of this MOSFET. So it gets a little bit more complicated to calculate the I-V characteristics of this MOSFET. So it gets a little bit involved to do the whole I-V characteristic. So I'm going to concentrate on the two limits. The low drain voltage limit, where it behaves like a voltage controlled resistor, and then the high voltage, high drain voltage limit. So let's look down here first. We apply a small voltage between the source and the drain. That means that the Fermi levels in the source and the drain are nearly equal. That means that the Fermi function in the source is close to the Fermi function in the drain. This is the regime that Professor Dada was talking about. And this is the regime where we can take, as he discussed, F1 minus F2 is very small. We can expand this in a Taylor series. And this brings in the derivative. Both F1 and F2 are close to equilibrium Fermi functions. That's what F0 is. So we can expand the difference to, in terms of a Taylor series expansion, that brings in this derivative minus dF0, dE, and then times qVds. qVds is the difference between the Fermi level and the source and the Fermi level and the drain. So that's that Taylor series expansion that Professor Dada discussed. So, we see here that now current is proportional to the drain voltage. We have a resistor. What we are interested in calculating is what's the conductance of this resistor. OK. So now there's just some math that we can go through. And I'll leave it as an exercise for you to go through this. It's not too difficult. But I'm going to do it in a different limit than Professor Dada discussed the T equals 0 limit. And he talked about how in general you'll have to compute this average over this broadened distribution minus dF0, dE. I'm going to take the Maxwell-Boltzmann limit. This takes the Fermi function and assumes that all of the states are way above the Fermi level. So that Fermi function reduces to an exponential. Now, that isn't necessarily a good approximation for MOSFETs. It's a good approximation in sub-threshold. But I'm only going to talk about above-threshold. Above-threshold, the device operates somewhere between this limit and the complete degeneracy limit. And that brings in more complicated mathematics, Fermi-Dirac integrals and things. A lot of textbook MOS theory is worked out with Maxwell-Boltzmann statistics. So one of the reasons I'm going to do this is it allows us to connect expressions we see in textbooks and see what's different. So I'm going to do it with Maxwell-Boltzmann statistics. Okay, so here's our expression. Just a slightly different version of what Professor Dada was talking about, but it's the same thing. We know everything in that. We know the number of conducting channels. We know how it varies with energy. We know that the transmission is one because we're considering a ballistic MOSFET. This derivative under Boltzmann statistics turns out to be very easy. It's just f0 divided by kT. And I'm writing T sub e for temperature of electrons only to keep it straight. Because I have a T for transmission and I don't want to get confused. So T sub e just means it's temperature. T means it's transmission. Okay, now I have everything. Now I just have an integral that I can do and it's not too difficult to do it.  But if you look carefully, it'll start to make some sense. Now, I'll remind you I could compute the electron density just by integrating the Fermi function times the density of states. And if you do that for a non-degenerate semiconductor, you'll get an expression that looks like this. It'll be something we call capital N sub 2D, which is called an effective density of states times e to the Fermi level minus bottom of the conduction band over kT. And the effective density of states is just, oops, I have a typo here. It's just the density of states times kT it's an integrated, it's a total number of states times this Fermi function. You'll also, one other thing I want to point out is that there are various ways to define thermal velocities. In equilibrium, the electrons are zipping around in random thermal motion. The average velocity is zero in equilibrium. But if I ask, what's the average velocity of the ones that are just moving in the plus x direction that are crossing a plane? That average velocity is what I'm calling V sub T here. That's square root of 2 kT over pi M. Now, when you do this integral then you can start lumping things into things that you recognize. And when you do that, you'll find that the conductance is proportional to width, that makes sense. If the transistor is twice as wide it should have twice as much conductance. If you start lumping things you'll see it's proportional to the sheet carrier density the number of electrons per square centimeter in the channel. Then we get a thermal velocity divided by 2 kT over Q. Then we can bring in MOS electrostatics. I could do it in a better way but I'm going to assume I'm above threshold and we'll do it in a simple way. Above threshold, if we have a well designed transistor the number of electrons in the channel actually is gate capacitance times VG minus VT and I'm missing a Q there. It's charge. The charge in the channel is gate capacitance times VG minus VT. I know that separately from MOS electrostatics. So that means we have a simple expression for the ballistic conductance of the transistor. And that's the bottom line here. That wasn't too difficult to derive. So I want to discuss in a little bit some of that. Before I do that, I'll just point out we could have, instead of assuming Maxwell Boltzmann statistics I could have assumed Fermi Dirac statistics. And if I had assumed Fermi Dirac statistics I could have assumed that this derivative which Professor Dott had discussed is very sharply peaked around the Fermi function and that it's just a delta function at the Fermi energy. Then the integral is really easy to do. So this is a delta function at the Fermi energy. So transmission is 1. So I just get the conductance is 2Q squared over H times the number of channels. And that's a very common expression that people discuss for the quantized conductance. You can see that at T equals 0. In the previous slide I worked it out for Maxwell Boltzmann statistics. In general in MOSFETs you're somewhere in between depending on what the gate voltage is. Okay. Okay. So we have this characteristic we've computed part of the IV characteristic the part for low VDS. We have an expression if you're used to MOS theory it looks a little different from the textbook discussions that you've seen before. If you're not used to it hopefully it just looks like a reasonable way for low VDS. So for those of you that do know the conventional approach I want to relate it to the conventional approach so you can see the conventional relation. So over here on the right this is the textbook description that we give for the linear region current of the MOSFET. It's proportional to the width of the channel divided by the length of the channel the oxide or gate capacitance the mobility is discussed at all because it's a ballistic device times VG minus VT times VDS and this expression in the red box here is what we've derived. What's the connection between the two? Okay. Well, alright. So we could just do something. We have a W over L out front here so let me divide by L and multiply by L. Okay. Now one of the things I want to bring up this is related to some of the things that Professor Dada discussed. He talked about how diffusion coefficient let me see if I get this right. Diffusion coefficient V sub n is diffusion coefficient and I think the way Professor Dada wrote it was average velocity squared in the transport direction times tau. Now you could write that in a little different way. You know, I could take one of these velocities, multiply it by tau and that velocity times tau would be the average distance between scattering events. That would be a mean free path and then I would have velocity times mean free path and then if you do an averaging over thermal equilibrium I'll just give you the answer it's really a very nice thing to remember and carry around with you. So you can see a derivation of it. Diffusion coefficient is thermal velocity times average mean free path divided by 2 and you can see it has the right dimensions. Thermal velocity is centimeters per second mean free path is centimeters centimeters squared per second that's a diffusion coefficient. So it's a nice way if you know the diffusion coefficient to deduce what the mean free path is. Now remember we also have this Einstein relation mobility is diffusion coefficient divided by kt over q. Now if I look at what I have here we've taken our ballistic expression we divided by L multiplied by L and you can see I have a set of parameters here thermal velocity times length divided by 2 kt over q thermal velocity times length divided by 2 has dimensions of diffusion coefficient except L is the physical length of the transistor not the mean free path. So it looks like a diffusion coefficient where the mean free path has been replaced with a length of the transistor and then it's divided by kt over q so that means that what I have here is mobility. So I could write this ballistic expression just like I write the conventional expression I just replace the mobility by this thing mu B. People call this the ballistic mobility and some people say that's nonsense you shouldn't even talk about a ballistic mobility it's ballistic mobility has to do with scattering but actually you can think about it in a way that it makes some physical sense in these small nano devices there's a lot of scattering in the source that's what maintains thermal equilibrium then the electrons go across the channel without scattering then there's a lot of scattering in the drain that maintains equilibrium in the drain the distance between scattering events is the distance between the source and the drain so in this ballistic device the distance between scattering events is the length of the channel that's what's coming out here so we could take the textbook diffusive expressions and just replace the real mobility by the ballistic mobility and we'll have the ballistic answer now you can see the shorter you make the device the smaller you're making your ballistic mobility and you think that's unphysical why should it depend on the channel length but remember the ballistic mobility is proportional to the channel length we put a W over L here just to make it look like the conventional expression the channel length and the ballistic mobility and the channel length here cancel and the current is independent of channel length that's the way it should be in a ballistic device there are just two different ways of looking at the same problem ok so now let's go to the high drain voltage a MOSFET person would call this the on current you have the maximum voltage applied to the drain you have the maximum voltage applied to the gate an important figure of merit for a transistor is how much on current do you have so we'd like to compute that on current now under on current conditions we've applied a very large voltage to the drain we've lowered the Fermi level in the drain so F2 is very small so we can just ignore F2 F1 is very much bigger than F2 so the current is proportional to F1 minus F2 but F2 is basically zero so now our current we just have to instead of doing a Taylor series expansion we're just going to integrate an expression that involves the Fermi function of the source ok so again the math is relatively straight forward my Fermi function of the source assuming Boltzmann statistics is just given by this exponential transmission is one number of conducting channels is just what it was before and I can go ahead and I can work out that integral and I can recognize some quantities and then I can rewrite it as width times charge times the density of electrons in the channel per square centimeter times this thermal velocity which again is just the average velocity of electrons in the transport direction under thermal equilibrium conditions now there's only ok I guess the other thing then the final thing I need to do then is to put in what I know about MOS electrostatics the number the charge in the channel is related to the gate voltage by C ox vg minus vt that's all there is to it now I have an expression for the on current there's one kind of curious thing that I've done I should discuss just briefly notice normally when I was doing this for low vds I wrote the electron density as the effective density of states times this exponential factor now I've done it I've divided it by 2 there in order to express the current this way I have to say that the current is half the density of states times that exponential factor you know so either I made a mistake by a factor of 2 somewhere and I just had to put that in in order to make the answer to be what it should be or there's something going on there is something going on remember early on when I told you when we wanted to compute the number of electrons in the channel we integrate half the density of states times the source function because the plus velocity states can be occupied by the source we integrate the other half of the density of states times the drain because the electrons that come in from the drain of negative velocities but the voltage is so high that F2 is almost zero so when the voltage is low F1 is about equal to F2 you just add these two and you get the expression that we had for low BDS when the voltage is high only half the states can be filled the negative velocity states can't be filled because the drain from the level is so low so that's where the factor of 2 comes in but the thing that's really important is in MOS electrostatic it doesn't care the fact that only half the states can be occupied if you put a charge on the gate it has to balance in the channel the same amount of charge has to be accommodated now in only plus velocity states so things are a little bit different inside the channel under low BDS it's very near equilibrium half the electrons have positive velocity half have negative but here it's very far so we can discuss that just a little bit under low BDS electrons come in from the source that gives us a current I plus and I could write I plus as Q times the density of electrons with positive velocity times their average velocity which is this thermal velocity VT some electrons come in from the drain which is low so I negative is almost equal to I positive so that's Q times the density of electrons with negative velocity times the thermal velocity the two currents are almost equal their difference is small and that's what gives us the small linear current the two carrier densities are almost equal so under low drain bias we have both positive velocity states and negative velocity states occupied close to the same and we have MOS electrostatics at the same time there should be a Q times N sub S here under high drain bias it's different we still have electrons coming in from the source and that current is Q times the density of electrons with positive velocity just like it was before but there's almost nothing coming in from the drain so there are virtually no negative velocity states but the total number of positive velocity states has to equal the total charge on the gate that's in order to make the Poisson equation happy charge has to balance so we have twice as many positive velocity states as we had before but the total is still the same now we can use these ideas actually to compute the velocity versus drain bias so I have a positive flux injected I have a negative flux injected I can always think of current as charge times average velocity then I can compute the average velocity so if I want the average velocity I take the net current which is a difference between I plus and I minus and I divide it by the total charge the sum of I plus and I minus okay now I know something about the ratio since I'm assuming maximal Boltzmann statistics then the number of electrons in the conduction band is given by an exponential factor e to the Fermi level minus conduction band over kT the Fermi level in the drain is lower than the Fermi level in the source by an amount qVds so the electron density with negative velocities that come in from the drain is lower than the ones that come in from the source by this exponential factor so I can put that information together and I can develop a very simple little expression for the average velocity and it'll depend on the drain voltage and if you expand that for small drain voltage you'll see that you get a velocity that's proportional to drain voltage bigger the voltage bigger the velocity but if you expand if you look at for high Vds these exponentials become zero and the maximum velocity that you can achieve is this thermal injection velocity so even though there's no scattering the velocity is going to saturate but it's interesting that the velocity we're computing the velocity at the top of the barrier at the top of the barrier the slope of the energy band diagram is zero the electric field is zero in a ballistic MOSFET the velocity is saturating under high Vds where the electric field is zero not at the drain end which is what the textbooks tell us where the electric field is high because it's ballistic but it still saturates and this is something that confused a lot of people myself included for a while if you look at these IV characteristics for this MOSFET one of the signatures of a velocity saturated MOSFET is that the current under high Vds increases linearly as you increase the gate voltage you go in the lab you measure a transistor you see that kind of IV characteristic you say ah, I'm dealing with a velocity saturated MOSFET now that was very difficult for us to understand because the more we understood about transport the more we realized that the velocity does not saturate in high electric fields in short devices it just gets higher and higher but what we've seen here is that we can explain this now the velocity doesn't saturate at the bulk saturation velocity which occurs because of electric fields at the drain end instead it saturates at this thermal velocity which is a little bit bigger and it saturates at the source end instead of at the drain end so that kind of explains what's going on here if you look at these detailed simulations and you see the strong electric field at the drain end and you can see that the velocity is very very high you increase the drain voltage the velocity will get higher and higher but what you'll notice is that the velocity back here has saturated and the velocity there is right at this inflection point that's where the velocity is saturating as you increase the drain voltage the velocity at the drain end gets bigger and bigger but the velocity at the source end saturates at this V sub t so that's how velocity saturation occurs it occurs in these very small devices but it occurs for physics that's very different from the way we understood it 15 or 20 years ago so it's easy I can relate this expression that we've just derived it's almost equal to the textbook expression for a velocity saturated MOSFET but the textbook expressions at least from 10 years ago or so would put a VSAT that was the bulk saturation velocity and it was hard to theoretically justify people started to notice that it's a little bit too small and they would go into their simulation programs and boost it up a little bit to make it clear what they were doing and the reason we understand now is that the appropriate velocity there is this thermal equilibrium velocity that we're injecting the electrons into the channel from the source it's saturating at that velocity okay so you can do it properly we can do the whole IV characteristic it was first done by Kenji Natori several years ago you get transistor characteristics that look just like any transistor you know it's a barrier control device they're all going to look the same just that the current is a little bit different we'll find a ballistic channel resistance no matter how short we make the transistor we're going to have a finite resistance and we'll get a simple expression for the on current and then the next question is okay is this all just an academic exercise or does it have any relevance to modern devices so in order to do that we have to look at some real data so here's some data this is the transistor we've been talking about it had a drawn channel length of 100 nanometers a physical channel length of 60 nanometers the dashed red lines are the measured characteristics and this is some work that my student Chang Wook Jung who's sitting in the back row did a few years ago and what he's showing there is he's superimposing a simple ballistic model for the same device with the same voltages the same oxide thickness onto the measured characteristics and what you'll notice is that if you account for the series resistances and subtract those out you'll notice that under low drain voltage the device is operating at 20% of the ballistic limit if you look at the ratio of the currents under high drain voltage you'll see that it's operating at about 60% of the ballistic limit that was actually surprisingly high when we first started computing these numbers that we realized that we were starting to approach some limits that the current couldn't just continue to increase so the message is silicon MOSFETs deliver over half the ballistic limit it's kind of curious we want to discuss this a little bit that under high drain bias the electrons have more energy or giving them higher kinetic energy you would expect them to scatter more but they operate closer to the ballistic limit under high drain bias than they do under low drain bias that's kind of curious okay, so before we discuss that let me let's look quickly at some 3.5 devices so there's a lot of interest these days in the possibility of replacing silicon MOSFETs with 3.5 MOSFETs 3.5 materials have much higher mobilities lower scattering there's a possibility of producing higher performance transistors there's some very nice work being done in MIT similar work being done in Intel but it's harder for me to get their data but if you take a look what you're seeing here are measured characteristics with the blue dots and a simulated ballistic simulation of the same device with the red and you can see here that a 3.5 MOSFET this is actually a 3.5 HEMP high electron mobility transistor actually operates very, very close to the ballistic limit if you were trying to analyze the IV characteristics of a 3.5 transistor it would be much better for you to assume that it's ballistic that would get you closer than to assume a textbook diffusive description okay, but in thinking about silicon for sure we have to understand scattering a little bit so how do we understand scattering in these small devices so we have to understand this transmission T we assume this was one for the ballistic device we assume that the channel length was much shorter than the mean free path so this ratio lambda over lambda plus L was one we need to understand that a little more so let me think of a conductor this is like my channel we inject a flux in from the source it some of it backscatters it undergoes some kind of random walk some of it exits the drain the fraction that exits is lambda over lambda plus L and you can see that that's a number between 0 and 1, that's the transmission the rest of it returns to the source and doesn't contribute to current okay so lambda is this mean free path for backscattering okay, so it seems that all we have to do is to take our ballistic expressions and multiply them by T and we will find out how the device operates in the presence of scattering so actually that's true under low drain bias it's a little more complicated under high drain bias so let's do low drain bias at first so we have our expression for the ballistic MOSFET under low drain bias let's just multiply it by T and in fact we could compare the measured characteristic to this theoretical expression and we would deduce the data I showed you earlier that T is 0.2 20% of the electrons go across ballistic now let's look at the high drain bias that one takes a little more discussion so let me do this here quickly if I current is basically charge times velocity so under ballistic conditions the ballistic current would be proportional to the width of the transistor times this thermal injection velocity times the charge or the charge would be the ballistic current divided by the width of the transistor and the thermal velocity what if there's some back scattering well I'm still injecting the same amount of charge but now some charge is back scattering so a fraction T transmits a fraction 1-T reflects and has negative velocities both of those together so I have to add the forward flux and the reverse flux they both flow at the same velocity VT and the sum of those two will give me the charge now the important point to remember is MOS electrostatic these charges are the same because the charge on the gate is always balanced by the same charge in the channel whether they're scattering or not if I equate these two expressions I'll get a relation between I plus and the ballistic I plus the ballistic I plus has to be different because under ballistic conditions only the plus states are occupied all of the charge on the gate has to be balanced by plus velocity electrons when there's back scattering I don't need to inject as much plus flux to balance the charge on the gate so we get a relation between I plus and the ballistic current that brings in a 2-T and then you can see that the current that comes out is T times the current that goes in so because of this necessity of MOS electrostatics and making sure that the gate charge is always balanced by the channel charge under high drain bias I don't multiply the ballistic current by T to find the current was scattering I multiply it by T divided by 2 minus T so it's a little bit different okay, if you do that and if you just equate that expression to the measured current you'll see that the transmission under high bias is 70% so this is the surprising thing under low bias 20% of them can get across under high bias they're much more kinetic energy they're much more likely to scatter there are many more mechanisms that they can scatter by there are many more states they can scatter to if anything it should be less ballistic under high bias but we find that we get closer to the ballistic limit alright how does that happen? we should talk about that so to do that we have to discuss scattering under maybe yes a little bit this is our expression the way we think about it is the length of the device is the channel length this is the fraction of electrons that come in from the source and transmit across the channel under high drain bias the energy band diagram looks like this solid line and now it's a little bit different all the electrons have to do is to get across this thin short part of the device where the electric field is low where we're strongly controlled by the gate if they get across that short region then they'll see a very high electric field even if they scatter that electric field is going to turn them around and sweep them out to drain so really the only thing that matters in terms of electrons getting from the source to the drain is their probability of transmitting across that bottleneck regime they don't have they can scatter in the drain that'll slow down the dynamic response a little bit but you're still going to be sub-pecosecond and over terror hurts frequencies and DC wise you're not even going to see any difference so what we have to do is to replace the channel length by the length of this bottleneck under high bias and the point is under high drain bias that bottleneck is a small fraction of the channel length that's why you're able to get closer to the ballistic limit you know you have the same mean free paths because you're still under low field near equilibrium there that hasn't changed the mean free path is much shorter here where the carriers are very energetic but it really doesn't matter as much there so this is the explanation it's not the total amount of scattering in the transistor does increase but the probability of getting across the bottleneck region is increased alright so now we did the ballistic case first now we did scattering now I want to put the two together and talk about the connection to the traditional model so here's our traditional low VDS model for MOSFET if you've done semiconductor device courses before you've seen this equation here's the expression that we have we took our ballistic expression assuming Maxwell Boltzmann statistics and we multiplied it by the transmission the two look very different but we're talking about the same physics so at some point they have to be the same thing alright how do we do that well it's actually quite easy we know that transmission is mean free path divided by mean free path it's the channel under low bias so if you do that just do a little bit of algebra and rearrange things what you'll find is that the current is proportional now not to W divided by length but W divided by length plus one mean free path and everything else is the same all you have to do is to take your traditional expression and replace L by L plus lambda and you saw that in Professor Dada's lecture now you can do the algebra a little bit different way if you like just collect terms a little bit differently and you can write the current as W over L not W over L plus lambda but just W over L and I can replace the real mobility by an effective mobility what's the effective mobility well it's the lower of the actual mobility and the ballistic mobility so we've got two different ways that we can think about this we can think about using the traditional expression with the physical bulk mobility and just thinking about the device never being shorter than a mean free path you might ask why when the channel length gets very short is there a mean free path in this expression because it should be ballistic well there's a mean free path in the mobility to cancel out or you can think about it as the traditional expression but I shouldn't use the bulk mobility I should use an effective mobility if the ballistic mobility is lower than the bulk mobility then the ballistic mobility is what controls things if the real mobility is the lower of the two then it's the real mobility so either way you can take the traditional expression and you can smoothly go from the ballistic limit to the diffusive limit depending to see people who do spice circuit models starting to make corrections like this in circuit models because transistor channel lengths are getting very very short you have to start worrying about scaling L to 0 you want to make sure bad things don't happen now how about high VDS how do we make sense of that the textbook expression just says well it's this high field saturated velocity our expression we replace the high field saturated velocity by the thermal ballistic injection velocity and then we multiply by T over 2 minus T so how do we make sense out of this well again we know that T is lambda over lambda plus L so I can do a little bit of algebra and I can rearrange things and this is the result that I will get I'll get W that's the inverse of 1 over the thermal velocity and 1 over diffusion coefficient times critical length alright that's just saying that T is lambda over lambda not plus the channel length now because under high bias it's just the critical bottleneck region that's appropriate so lambda over lambda plus script L doing some algebra this is what we get so the way to think about that is we think about this little bottleneck it's sort of like the base of a bipolar transistor current is injected from the source into this bottleneck region it's got to get across the bottleneck region if there's some scattering it's going to do some kind of random walk once it gets across that bottleneck region it just gets swept out into the drain we can always write the current as width times average velocity times charge charge is given by MOS electrostatics average velocity is just given by those terms in the previous slide it's the slower of two velocities D over L physically represents the average velocity that carriers are diffusing at now there's trouble if L gets too short diffusion is random thermal motion you can never diffuse faster than the thermal velocity because you're undergoing a random walk as carriers are just thermally diffusing so if L gets too short the maximum velocity that you can go across this region yet is this ballistic injection velocity and this expression just makes sure that in the two limits you approach the right limits actually it works in between too there was a question about this this expression in lambda over lambda plus L is something that you can derive from a simple model Professor Dada does it in his green book I think it's discussed in my online nano transistor tutorial too it really works very very well in practice not only at the diffusive limit and at the ballistic limit but it works in between very nicely so you can smoothly go all the way now this is something it's not new in 1970s I think about 1972 I first remember reading a paper people were making bipolar transistors and you could make bipolar transistors with very very thin bases and a bipolar transistor you always think about the current being limited by diffusion across the thin base people started to worry about this problem in bipolar transistors that they didn't want to get unphysically high velocities in the base and unrealistically high f sub t's and they started describing it this way it's the same kind of thing so Ed Johnson was right the MOSFET really is a bipolar transistor in disguise it's really very useful to think about this bottleneck region at the beginning as the base of a transistor of a bipolar transistor and to think about the rest of the channel where the electric field is high like the collector of a bipolar transistor it's a really useful way to think about it ok, so just to wrap up what have we learned so the important point is that the MOSFET the way you understand how you make a transistor is you control current flow by manipulating energy barriers and you design the device such that at the top of that barrier the potential is strongly controlled by the gate influenced by the drain that's what we mean by a well tempered MOSFET people are working very very hard to do that that's why you start hearing more and more discussion of double gate devices nanowire devices it's all an attempt to control the electrostatics and control that potential by the gate and not by the drain voltage there is this short bottleneck region that's really the important region for determining the on current if electrons can get across that region it doesn't matter if they scatter or not the transmission we can estimate simply in terms of the mean free path and the length of that bottleneck region the length of that bottleneck region is a little more difficult to get I mean there we really need a simulation you know that's controlled in detail by what the shape of the potential barrier is the scattering, the self consistent electrostatics but we can get the gist for what's going on and then if we have strong gate control the increases in drain voltage just make that collector a better and better collector but don't really influence what goes on here there are second order effects things called DIBL those of you that know about MOSFETs know about now what have we left out? we've left out quantum mechanics these are some simulations in the model about MOSFET where we're showing the energy resolved current so where it's so the vertical axis is energy the horizontal axis is position so here's the source here's the barrier in the channel here's the drain going down here so red means this is a ballistic device so it tells us where the current is flowing in energy so the device is supposed to be off and the small leakage current that's flowing over the top of the barrier okay that's a 13 nanometer channel length so these are some simulations done by my colleague Matthew Luizier so in the next simulation he made the channel length 10 nanometers you can see that the current still flows over the top of the barrier so we've been neglecting quantum mechanical tunneling or anything like that so our model is good at least till 10 nanometers if you go to 7 nanometers the model you start to begin to push the model you can start to see some current going underneath the barrier actually it's not too bad you can still get by but boy you start going lower and it really gets severe if you go to 4 nanometers the electrons don't even know that there's a barrier there now it's really difficult to make a transistor because a transistor is all about manipulating energy current by manipulating energy barriers if the electrons don't know there's an energy barrier there it's really hard to make a good transistor so ultimately this may set a scaling limit now I say may because in practice the most severe challenge for people is usually the electrostatic challenge arranging the 2 and 3D electrostatics such that there's a region that's strongly controlled by the gate and not by the drain in practice that's usually more severe it's still not clear that you're going to be able to have an electrostatically well tempered device at 4 nanometer channel lengths if you could then you would have this quantum mechanical problem ok so just to leave you with the thought these transistor characteristics they're primarily a function of the electrostatics of the problem it's important for us to realize as we make the channel shorter and shorter that the current doesn't scale as W over L it approaches a limit and the on current is not controlled by any high field saturation velocity but by a ballistic injection of velocity and it's really interesting the velocity does saturate in a ballistic MOSFET but it saturates where the electric field is 0 not where the electric field is high and that has a strong influence on the shape of the IV characteristics ok if you want to know more about this story I can point you to a short course from an earlier summer school and so I hope that gives you a flavor I think for the next couple of days Professor Dada is going to be talking about a set of fundamental issues in a lot more detail I hope this gives you an idea of how you apply some of those ideas it's really 10 or 20 years ago these were things that physicists worried about at low temperature now there are issues that really every working engineer needs an understanding about as you're scaling devices to their limit all of these issues are becoming more and more important for us


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