So the subject of this talk is not directly related to random matrices, I will spend some time defining the setup of it. So the model I'm going to talk about is called the random walk in random environment. Then we spend a little bit of time defining it. So this is a stochastic process which corresponds to the movement of a particle on the hypercubic lattice. Well, it could be some other graph, but I will talk about it, I will focus on this graph on the hypercubic lattice. And the particle has jumps, for the sake of this talk, the jumps will be always two nearest neighbors, so the possible jumps will always be some unit vector in the lattice. So this is the L1 norm. And now the jumps of the process of this random walk are itself random. So this means that the jump probabilities will be random, and they will correspond to choosing at each side some set of numbers. So these numbers are the probabilities to jump in a given direction indexed by the jumps, so they have to be positive, or they will not be probability, they have to sum up to one. And this is what will define what I'm calling an environment in the title. So actually I will define what I will call the environmental space. This is just means setting at each side, or prescribing at each side some jump probability, which I am defining as this Cartesian product here. And what will we call the environment is just choosing, as I said, these probabilities. So at each site I have to make a choice. So this will be some element of the environmental space. And so the omegas are a set of numbers which are in this class of numbers which sum up to one and which are positive. So you can, okay, these are definitions which in a sense it's much easier to understand them making a picture here. So what I just said in all these definitions is that if I have a site, X, let's say on the lattice, and I want to know what's the probability that the random walk will jump from site X to X plus E, this will be given by the environment that I choose, and will be given by the number omega X E, that's what I'm saying. So actually the random walk itself will be defined by its transition probabilities. So for example, I will fix the starting, this is going to be the starting point, and the environment, and the random walk will have a certain probability of jumping from a site Y, let's say at a time n to a site Y plus E, which is just given by the environment. So this is for all points on the lattice. As you can see, the process I'm looking at in this talk will be discrete time, it's not crucial, and the process is starting always from site X in this case. So well, this is a Markovian process if you fix the environment at least, but what I want to do now is I want to put some randomness on this environment. So that, of course, is just given by, will be modeled by putting some probability measure defined, probability measure in the environment. So I will call it P, the double bar here means that I will distinguish the probability measure on the environment, which is going to be written by double bar, I say P with double bar, sorry, from the law of the random walk itself. So there are usually two ways of looking at the random walk. So one is, the first one is what's usually called the quenched, let's say quenched point of view, which means that the environment is frozen. So the law of this random walk I already introduced it is going to be called, I will call it P with a subscript X standing for the starting point and omega for the environment. So, but there's another point of view which is the, it's called usually the averaged, sometimes annealed, which means that somehow I want to study the movement of the random walk, but averaging what's going on with respect to the environment. So I'm just essentially taking an average here over the environment of the law. Both, the behavior of the random walk under both measures is not necessarily the same actually. And actually the second, the average law has the disadvantage that the random walk is not Markovian under it. Okay, so there is a behavior which is going to be important to distinguish throughout this talk and maybe I will make some assumptions right now. So the first one is that I will assume that the choice of the jump probabilities is IID under the law. And the second one is that I will suppose that all of these probabilities are larger than some constant. I cannot have any jump probability which is equal to zero. So I need to introduce a behavior which will be relevant in my discussion which is what's called the ballistic behavior of the random walk. So I'm going to introduce a direction. The idea is that I have some direction here. And there is a possible behavior of the random walk which is called transient or transience in direction L which means that the random walk somehow wanders a way to infinity in direction L. This point here is just the inner product or the projection of the position there on the walking direction L. So in the picture it would mean that you're on the walk and you move backwards somehow but at some point it will end up going to infinity, it could oscillate but its projection in this direction will go to infinity. The other behavior which will be relevant is what I already said is ballistic in direction L. So this means that the transient behavior occurs with a given speed so this is a much stronger statement than transience or transience, directional transience let's say. So well there's an open question which I'm not going to discuss in this talk I'll just say it which is that it's believed that or many people believe that in dimensions larger than one any walk which is transient in a given direction is necessarily ballistic. So that's still open. I mean I wouldn't bet on it actually but okay so I'm more interested in scaling limits of the random walk in this talk now so I need to go on to large deviations for people to understand or to prove let's say a large deviation principle for the random walk and random environment. In dimension one it was proven in the 90s but at least for any dimension it was only in 2004 that it was proven by Varadan and well he proved two things first he proved that if the law of the well I said that I was going to assume that the law is IED but it is the only exception in this point that I'm going to say something for a law which is not IED necessarily. So if P is stationary and ergodic then so Varadan proved that a quench the large deviation principle is satisfied so I don't have time I'm not going to define what a large deviation principle is I think everybody knows should know or knows but let me just say that in in particular one can one one can show that in this is not completely trivial the corollary that if I look at the probability at the quench probability that the random walk in is at a position xn which so which asymptotically converges to a point when you normalize it because the range of the random walk at time n is a l1 ball of l1 norm of radius n so if you look at if if if you look at this quantity this converges actually to the rate function at x so this is the rate this is the what's called the quenched rate function it's not random and he also was able to prove a an average large large deviation principle so analogous in an analogy with the limit I just wrote one can show that his average large deviation principle implies now that if you look at the same expression but under the average law this converges to minus another number which depends on x another function let's say which is the average rate function so both of these rate functions are in not too much is known about them very little let me say some of the things so this is a very basic property turns out that these functions are finite this is very simple to if and only if in you look at points in an RD which is I said already which have a l1 norm smaller than equal to one the same is true for the quenched so this this set D will be important in the rest of the talk in can imagine it as a diamond since okay or a square which is rotated and now in general these rate functions are not equal the annealed and the quenched rate functions are not equal in general but a for example for example in dimension one they are they are in dimension two it's known that they are different but there's a result of the 2010 of you must he was a student of Iran who proved that in there is a if if the random walk is ballistic so this is what I already discussed at the beginning well he assumes something not acceptable necessity but then in there is an open set where the rate functions coincide I'm not saying I'm not giving you all the information of the set let me give a little bit more in this picture the set the set D is the open set which is here inside is in is a set which contains an important point which is the velocity when the run the walk is ballistic the velocity is not zero and this open set you can imagine it with something like that this is the only thing which is known but he has assumed this very somehow I would say it's an annoying thing which is ballisticity I don't have time to explain why but it's just somehow ballisticity it gives a structure of the law to run the walk the annealed law which is where you can you can see some IED structure of the run the walk and the other there's another result of 2008 this is not Polish but in who was also a student of rather and I'm not even going to state that I'm just going to say that he gives a variational formula for the quench rate function and it is very non non explicit but say somehow an abstract variational formula which I would say has not been very useful conceptually it is interesting but it has not been so usual to to make computations in a sense of the of these rate functions so somehow the big problem is how if somebody could compute or give some formula for these nice formula for these race functions okay so the so recently with with other people so the Rodrigo Bassais and Chiranjeev Mukherjee and and we were able to in a sense get rid of this assumption but there is a cost which is very natural and somehow gives the right I would say the setting for the quality of the rate functions which is low disorder and I forgot to say the result of the mass is the valid only dimension larger than equal to four otherwise it's not true so here's all you will also you also need dimension larger than equal to four and a low disorder so I will define it right now so there is a you can define for given a measure a law of the environment p you can define it's disorder measured as in the maximal distance between the random probabilities at a given site with the their averages so this this q is the average of omega so this a number this epsilon is measuring how far the the maximum the the maximal distance that you can have between the randomness omega and the q so the theorem says that there is a for for any compact set which is in the boundary of this diamond so I'm looking at the boundary not only now there is an epsilon in an epsilon which depends on on the compact set such that in if the the disorder of the environment is smaller than this epsilon the rate functions agree on the compact set the the the first important point of the theorem is that a it gives the equality of the rate function without the assumption of elasticity if for the moment the the result is only in the boundary in the interior with it is work in progress and as a corollary the you you can you have the an explicit rate an explicit sorry formula for the for the both rate functions so this is a because the anneal rate function at the boundary is can be computed immediately so you get a formula for a formula for the quenched one and I used to say well that the the there is really a phase transition here it is possible to show and I'm not going to say the the whole the the full all the details of the theorem that in general if epsilon is large enough these rate functions are not going to be equal and okay now the proof I I I don't I want to I mean that's not the only thing I want to say in this talk so I don't have too much time I just want to say about the proof that a in one a observation that is important about the proof is that in there is a martingale in which plays a role here which is what I'm writing over here if you're able to get a good estimate on on a norm of on the L2 norm of this martingale in it is it is possible to show that in that actually the martingale this this this thing is converging to zero and this will give you the equality of the Laplace transforms of the the random walk and then you use a you you can use a very well known theorem of Gartner and Alice to get the the equality of the rate functions okay so now let me and go on to what I wanted to say about what happens in dimension two and so in dimension two you you cannot expect a result like this or you have equality of the rate functions even if the disorder is is is low but but actually and it is possible to find out this I want to talk about a make a connection with here with a what's called the kpz universality class okay this this this is a a somehow a picture of a universal behavior describing the movement of interfaces two dimensional interfaces which separate a stable phase by from an unstable one and so it turns out that the random walk is in this class also I want to I'm not going to convince you because I'm not I'm not going to have time to to give you the proof but at least give you the statement and so we'll have I will have to be a little bit more precise about the disorder this this rises I mean this this connection arises when one looks at the random walk at low the sort again but we will look at environments which are of this kind so the sort of appears in a particular way there's a parameter epsilon and and these size are iad again so you have a set of the laws which are which are joined they have a joint distribution given by the size and they are parametrized by epsilon but for the effect of the scaling I will do I will put a square root of epsilon here it's just a convention not it's not a and so okay I will make a scaling and yes I didn't say this the these size are also bounded by one otherwise okay so we'll make a a scaling which is not so simple to define so first I will define I will I will I will choose some some point on the on this space where the rate functions are finite and please I think I still have 80 minutes or or nine maybe nine and so let's make a picture here I have the I think I'm not on the blind spot I hope so I will choose some some point here as I said and now the second ingredient I need is a I will I will have to define some curves which have a well this this equation is is I don't know if it somehow this is an equation which interpolates between the equation of a circle and the equation of a square so for for small in if if you rotate the coordinates this corresponds as I said to the equation of a square for the constant a large constants so for for small values of the constant this is very close to a circle but as you increase the value of the constant you begin to to look more more like like a square so here even it's like this and and in the limit it will really be the square so these these are these these curves in a constant so now I will choose maybe maybe for the sake of the explanation I need to choose this point somewhere else like that let's say w I will choose a a the unit tangent of the point I'm I'm pointing at the which is a tangent in w passing through the curve so this is unit tangent so what I what I'm going to do now is it's a scaling in where time the time of the random walk is going to scale like like well diffusively in a sense like epsilon square but space will have two way two coordinates of scaling one is the omega the omega the w one sorry the w one is going to scale like time so in a sense this is w is going to be played the roll of time and the orthogonal one will scale like like its square root and well of course noise as I defined the model is scaling like the square root so now I think yeah let me see the theorem so this is a joint work with with well Jeremy Quastel so the the theorem says the following it says that in well dimension two there is a there is a well there are some functions which I'm going to call the first one alpha sub epsilon alpha sub epsilon is something that will depend on the on the point which I chose here on on a time I have to fix time I have to fix space this is this x is one dimensional and there's another scaling normal or you say normalizing function which also that will depend on this on the three quantities and also a couple of constants these ones depend only on on the direction I choose so what's going on is that if you look at the transition probability this it's the same transition probability I was looking at before the quench the one sorry the probability that you run the walk at time in is let's sorry I should write here at time t over epsilon square I'm not saying of course you have to take the integer part otherwise this is not defined I'm looking at the street time and so you as I said there you rescales one directional space like time but the other so this is a vector remember the other direction like the square root and the result is that this converges to and you have to center by this landa this converges in distribution to a function which is it is a stochastic function which is the solution of the stochastic heat equation and so where as I said let me write it this is the solution of the stochastic heat equation and there's the the gamma here u and and then you have here well maybe I should write it like that is space time white noise so why I said that this gives a connection with the kbz equation because the the the kbz equation is the help called transformer of the solution of the stochastic heat equations in a sense this is giving the relationship I want to I want to point out that the result is it's what's relevant about actually a theorem I didn't say is that the it is a result about essentially a model which is a polymer which is not directed and most of the well I think all of the results giving the scaling to this stochastic heat equation were under the assumption that we have some directed model and maybe I should just say very briefly that there is a I didn't I didn't speak about this today but there is is there a worker problem is the worker problems with my friend the Jose Ramirez it's not Alejandro Ramirez where we we can apparently there for a certain choice of the law of the round the walk and round the environment you you you can find determinant formulas it has to do with the talk that the first class I'm not going to say more about that okay thank you


Three points at home. For the first time in four months football at home but you got on the score sheet when it mattered most. What does it mean to you? Yeah I mean this win for us all actually is a very important win. Giving us a chance of still making finals football. So yeah we're very proud.  What did it mean? Yeah I mean we always love playing at Bankwest even if we have a smaller crowd than normal. Yeah it's always a great feeling in this stadium. And for finals I mean you mentioned how big it is. We now know that we can defeat a team that is right up there in the mix at the top of the table. We're pushing for top six. I mean how much further can this group go? Yeah as I said previously we just focused on the next game. So as many points as we can get not looking into the finals but we're just concentrating as many points as we can get. But yeah the goal was good.  It's a massive event. Congratulations. Cheers.


Welcome to Learning English, a daily 30-minute program from the Voice of America. I'm Ashley Thompson. And I'm Dan Novak. This program is designed for English learners. So we speak a little slower, and we use words and phrases, especially written for people learning English. On today's program, Dan Friedel reports on much of the world's record heat. Gina Bennett has a story on Guinness's investigation into its world's oldest dog record. Gregory Stockle reports on business leaders that fear their companies will not survive the next 10 years. Mario Ritter Jr. has a story on new research that shows why some people are more at risk of multiple sclerosis. Later, Faith Perlow brings us the everyday grammar lesson on mental health, but first... Many parts of the world have experienced record high temperatures in recent days, but much of the United States is dealing with extremely cold weather, thanks to what is known as a polar vortex. A map showing world temperatures over the last several days had a dark red color in the Arctic, Asia, the Middle East, South America, and parts of Africa. The dark red means the temperatures are 7 degrees Celsius warmer than the average from just over 20 years ago. The United States is one of the only places on the map that is dark blue, meaning very cold. In the state of North Dakota, wind chills reached minus 56 degrees Celsius last week in the state of Missouri. A professional American football game took place as temperatures dropped to minus 20 degrees Celsius. On the other side of the world, the world's top tennis players are playing in unusually warm weather in Melbourne, Australia. The Australian Open is taking place as the city records temperatures nearly 7 degrees warmer than average. The African nation of Oman is in the northern hemisphere where it is winter. It recently set a record for the warmest January night at 26.4 Celsius. Parts of the southern hemisphere are experiencing warmer than usual weather too. Argentina set a record for warmest January night at 27.3 degrees Celsius. Climate scientists say these extreme temperatures are due to warming in the Arctic. They say warming temperatures in the Arctic affect the weather all around the world. The pattern called a polar vortex is not unusual in the Arctic. However, experts say as temperatures in the North warm, the polar vortex widens and affects Canada and the United States. Jennifer Francis is with the Woodwell Research Center in Massachusetts. She is a leading researcher in the theory of Arctic amplification, which links the periods of cold weather to climate change. She said, when the Arctic is off the charts warm like now, we are more likely to see frigid cold invade places like Texas that are ill-equipped to deal with it, she said. Francis added, rapid Arctic warming is one of the clearest symptoms of human-caused climate change. Judah Cohen works for Atmospheric Environmental Research, a commercial weather company near Boston, Massachusetts. He compared the polar vortex to an ice skater doing a spin. When a skater spins very quickly, the vortex winds stay close to the skater, but if the skater slows down, the winds lose their tight circle. All the cold air then gets released away from the center of the polar vortex, Cohen said. When that happens in January, the target of the cold air is the United States. Cohen said the current polar vortex is stronger and longer lasting than usual. Marshall Shepard is a meteorology professor at the University of Georgia. He said that right now there is a complex interaction within our climate system. Winter will always bring us cold weather, but like the warm season, it may be changing in ways that we understand and in ways that we are still learning about, Shepard said. The weather expert said the extreme cold in the U.S. will change to unusually warm weather in several days, but another polar vortex is expected to hit at the end of January. I'm Dan Friedel. Guinness World Records said on Tuesday that it has suspended and was re-examining its world's oldest dog, record. The publishing company had given the record to a Portuguese dog named Bobi, which died in October of last year. The move follows criticism from veterinarians who raised doubts about the dog's age. Bobi was a kind of dog called a Hafero do Alentejo. He spent his life in a village in Portugal. Guinness World Records said the animal lived for 32 years and 165 days. It broke a record held since 1939 by an Australian cattle dog that died at 29 years and 5 months. Bobi was declared the world's oldest dog in February. Hafero do Alentejo are traditionally used to watch sheep. They have a normal life expectancy of 12 to 14 years. Guinness World Records, or GWR, said the review into Bobi's record was ongoing. The company added that it would re-examine evidence, seek new evidence, and reach out to experts and those linked to the application. While our review is ongoing, we have decided to temporarily pause applications on both the record titles for oldest dog living and oldest dog ever until all of our findings are in place and have been communicated, the spokesperson said. Bobi's owner, Lionel Costa, said in a statement on Tuesday that after his dog's death, an elite within the veterinary world, tried to give people the idea that Bobi's life story was not true. Costa said some veterinarians were unhappy because he said Bobi's long life came from the dog eating human food, which he said was often suggested by those in the field. Everything would be different if we had said he, Bobi, ate pet food, Costa said, adding that all requirements requested by the GWR were met. Costa said GWR has not reached out to him. No action has yet been taken about any record holders, the publisher said. It added that any action would be decided by the review's outcome. Before his death, Bobi still loved walks but had become calmer, Costa told Reuters. His hair or fur was thinning, his eyesight had worsened, and he needed to rest more than in earlier years. I'm Gina Bennett. A growing number of business leaders say they fear their companies will not survive the next 10 years because of pressure from climate change and new technology. That information comes from a new opinion by the consulting company, PWC. PWC questioned over 4,700 chief executive officers, or CEOs, from 105 countries and territories. It released the survey results on January 15th. The survey's release came as business leaders, political leaders, and activists gathered for the World Economic Forum's yearly meeting in Davos, Switzerland. 38% of the CEOs said they were optimistic about the strength of the worldwide economy. That is up from 18% last year, when the world was dealing with high inflation, rising interest rates, and more. Even with more positive opinions on the economy, the CEOs say they are increasingly worried about their company's ability to deal with major future changes. The survey found 45% of the CEOs had concerns that their businesses would not survive the next 10 years. The business leaders say they are trying to make changes, but they must deal with difficult government rules, a lack of skills among workers, and more. Artificial intelligence, the survey found, is seen as a way to make business operations faster. But AI is also seen as a weakness. Nearly 75% of the CEOs said AI will significantly change the way their company creates, delivers, and captures value in the next three years, PWC said. More than half the CEOs said AI will make their products or services better. But 69% noted that their workers needed training to gain skills to use the developing technology. The CEOs also said they were concerned about how AI would increase online security risks and misinformation. Organizers of the Davos meeting warned last week that the threat created by AI-powered misinformation is the world's greatest short-term threat. The PWC survey shows that climate change is both an opportunity and a risk. Nearly 33% of the CEOs said climate change is expected to change how they do things over the next three years. More than 75% of the executives said they have begun or completed changes to increase energy efficiency. But only 45% noted that they have made progress on taking climate risks into consideration in financial planning. I'm Gregory Stockle. Genetic researchers at the University of Cambridge in Britain say ancient DNA shows why northern Europeans have a higher risk of getting a nerve disease than other Europeans. The disease is called multiple sclerosis or MS. The researchers said the risk lies in genes from horse-riding cattle herders who entered northern Europe about 5,000 years ago. The findings come from a huge project to compare modern DNA with ancient genetic material taken from human remains including teeth and bones. A study released last year said it identified the earliest evidence of horse-riding in people called the Yumnaya. The scientists say they lived 4,500 to 5,000 years ago during the Bronze Age period of human history. The Yumnaya moved from the grasslands of what is now Ukraine and Russia into northwestern Europe. However, the researchers say those people carry gene versions that today are known to increase a person's risk of multiple sclerosis. The researchers added that they believe the same genes protected those herders from infections from their cattle and sheep. The research was published in Nature, a scientific publication. William Berry is a genetic researcher at Cambridge. He helped write the study. He said everyone involved was surprised. These variants were giving these people an advantage of some kind, he said. The finding was made possible by a gene bank with thousands of examples of early humans in Europe and western Asia. That project is led by Esca Willerslev of Cambridge and the University of Copenhagen in Denmark. While MS can strike any population, it is most common among white descendants of northern Europeans. Scientists have been unable to explain why. The cause of the disease is not known. However, one theory is that infections could cause it in people who have certain genetic qualities. Scientists say they have found 230 genetic variants that might increase the risk of MS. The researchers studied DNA from about 1,600 ancient Eurasians. They used the information to develop a map of population movements in northern Europe. They said farmers from the Middle East began pushing out hunter-gatherers about 5,000 years ago. Then the Yamnaya moved in. They traveled with horses and wagons and herded cattle and sheep. The research team compared the ancient DNA to the genetic information of 400,000 modern-day people stored in UK Biobank in Britain. They wanted to see if MS linked genetic variations persisted in the north. That is the part of Europe where the Yamnaya moved rather than southern Europe. In what is now Denmark, the Yamnaya replaced ancient farmers, making them the closest ancestors of modern Danes, Willerslave said. Rates of MS are especially high in the northern part of Europe known as Scandinavia. The findings raise additional questions and suggest a need for more research. One of the writers of the study, Astrid Iverson, of Oxford University, questioned why a gene variant that seems to have strengthened immunity later plays a part in causing what is believed to be an autoimmune disease. Differences in how modern humans are exposed to animal germs might push the immune system out of balance, she said. I'm Mario Ritter Jr. In recent years, more and more people have been talking about mental health. The pandemic made many people feel isolated or alone. The isolation caused some people to struggle more with their mental health. Different countries and cultures have different ideas about mental health. In the United States, conversations about mental health and mental illness are changing. Many well-known people have come forward and shared their mental health journeys through books and social media. As ideas about mental health change, so does the language we use to talk about the issue. In today's everyday grammar, we will talk about the language of mental health. Let's start with the question, what is mental health? Mental health contains the adjective mental and the noun health. Together, the words become a compound noun, meaning the conditions around one's mental and emotional well-being.  Now let's look at mental illness. Mental illness is a compound noun with the same adjective mental, but with the noun illness. If someone has a mental illness, they have a medical condition that affects their mind, personality or emotions. The illness harms their mental functioning and can affect their thinking, feelings and behaviors. Examples of mental illnesses include depression and anxiety. The word disorder may also be used with some mental illnesses. For example, you may know someone who has an anxiety disorder. When we talk about someone's mental health, it is important to use the right words. We want to be respectful of the person and understanding of their situation. We can use language that is centered on the person rather than their mental health issues. One way to do this is to say a person is living with their condition. The structure for this is subject plus verb live plus preposition with plus noun form of illness. Kurt lives with anxiety. Another acceptable and simple way to describe Kurt would be to say Kurt has anxiety. The structure for that is subject plus verb have plus noun form of illness. We can use this instead of using an adjective form of their mental health condition. As in, Kurt is an anxious person. It is also important to avoid certain words in some settings. For example, words like psycho and crazy may seem informal and acceptable to use, but they can cause harm. The word psycho when used informally means an unstable and aggressive person. For example, her ex-boyfriend is such a psycho. Psycho is related to psychosis, a mental disorder in which a person has a disconnection from reality. Experts say using such words outside of a mental health discussion can lead to stigma. Stigma is a set of negative beliefs that a culture has about something. Stigma can prevent people from seeking help with their own mental illnesses. In today's everyday grammar, we talk about the best words and language to use when talking about mental health. We learned how to use person-centered language to talk about someone with a mental illness. And we learned that some words should be avoided to help end the stigma around mental illness. Time, Faith Perlow. You just heard Faith Perlow present this week's everyday grammar. Welcome back to the show, Faith. It's great to be here with you, Dan. This week, you wrote about the language we use when we talk about mental health. I really wanted to share this information with our listeners, because it is an important subject to talk about. And in many areas of the world, it can be difficult to talk about subjects such as mental health. And until recently in the US, subjects like mental health and mental illness were not really talked about publicly. I think our conversations around mental health had to change. Slowly over the years, with TV shows like Crazy Ex-Girlfriend dealing with mental health and substance abuse, and even celebrities like Selena Gomez coming out with their own struggles with mental illness has helped people to not only talk about mental health more openly, but more importantly, these things have allowed people to seek help and support. And according to the National Alliance on Mental Health, one in five Americans will experience a mental illness and one in 25 people live with a serious mental illness. That's right, Dan. So this is why it is so important to avoid negative language, such as generalizations and stereotypes surrounding mental health. These words can lead to stigma and prevent people from seeking help and support. Well, Faith, thanks for sharing a little bit more about mental health and language with us today. It was great to share these things with you, Dan, and with our listeners. Thanks. And that's our program for today. Join us again tomorrow to keep learning English through stories from around the world. I'm Ashley Thompson. And I'm Dan Novak.


And I know plenty of multi-millionaires who drive old cars, they mix in. You know, you won't know them from anybody, but they're happy with their own skin. They're happy with their own success. They don't want to boast about it. But they know they've welcome to a life, a stress-free life, because they've studied the job properly and they've done it right and succeeded. We don't need to tell anybody about success there, because everybody today on the internet can look happy when they take a photo of that phone. Anybody can smile for 10 seconds, take a picture. OK, the world's all right. And when the phones put down, they cry. And it's so easy now to be a plastic millionaire, isn't it? There's all this taking by Rolex. Watch for £80 a month. You can't get the best of a car for £400 a month. You can look the part. Oh, are you? What are you when you look at it in colour today? A fraud.


So, we continue our discussion on SOI MOSFET that is silicon on insulator MOSFET. We have already discussed how to realize the SOI wafers. We have also discussed what are the merits of SOI MOSFETs in as a discrete device or as a CMOS circuit. So, all the advantages we have discussed last time. Now, we will see how this device functions differently or how is it different from the classical MOSFET. So, this is the cross section of the SOI MOSFET. The top portion is actually the conventional structure except that you have the gate only in this portion junction only here. So, source channel drain and the gate oxide then the gate metal or doped silicon. Now, in the case of SOI MOSFET you have got the buried oxide or back oxide. We can call it as such back oxide and the substrate on which you have this oxide is grown the device is fabricated. I have shown this thing here the substrate it will be usually thick to hold the device there. So, you can see that I just to see the terminology is you can use it as a dual gate double gate front gate with the front gate oxide thickness T oxide f instead of T oxide you have the T oxide f back gate and the back oxide with the substrate acting at the gate region and back oxide is or T ox B. Here I have shown it differently we will do the analysis for different thickness of oxide, but ultimately you want to have double gate MOSFET where it is symmetrical that sort of thing will final decision will be having on that if possible to in this lecture or subsequently. Now, symmetrical double gate there is a DG double gate is DG MOSFETs G 1 G 2 are identical materials and T ox f is equal to T ox B. Now, there are another class of devices you call this is a double gate MOSFET it can be symmetrical or non-symmetrical the way I have put it it is not symmetrical. Now, undoped to this layer channel region can be doped or even you will see that we can even have it undoped. So, undoped ultra-thin body MOSFETs that is U T B MOSFETs will have T silicon that is the thickness of the silicon layer T silicon which is 1 to 10 nanometer that is the ultimate. Now, what we will see will be how to go about looking analyzing this how to understand the performance of this. You can have basically two types of SOI MOSFETs one is partially depleted SOI MOSFET other one is fully depleted SOI MOSFET. Lot of material on SOI MOSFET is available in a single book authored by Collinge silicon on insulator technology materials to VLSI. Cluer publication 1991 first edition, 1997 second edition lot of basic materials are there. What I discuss will cover some portion from there some portion from the papers which are which appeared. So, now let us see what is partially depleted and fully depleted do not be overwhelmed by this particular diagram. It is this portion see this is the same double gate what I have shown top gate front gate voltage V G F we can apply back gate is substrate V G B we can apply your front gate oxide. Then between the front gate oxide and back gate oxide you have got the silicon the whole thing is silicon. You can see the N plus layer is there source going from between the front gate oxide and back gate oxide similarly drain is there. Now what we are talking consider now is the partially depleted MOSFET means if you take the bulk MOSFET you have the only the front gate front gate oxide and the entire substrate the substrate is depleted. How much is the depletion layer width depends upon the doping level. So, at threshold the depletion layer width is maximum. Now we are considering a when will it be partially depleted or fully depleted. In the case of bulk MOSFET it is always partially depleted only one channel. For example, if I cover this portion below this line that is actually the conventional MOSFET where when I applied sufficient voltage this channel will invert there is there is inversion below the inversion layer there is a depletion layer. And the depletion layer width depends upon the surface potential. When the front channel let us forget about back channel when the front channel is inverted the surface potential is twice Y F and that depends upon the doping concentration. So, phi F is K T into logarithm of N A by N I. So, higher the doping more will be phi F. To give an example when the doping is 10 to power 16 phi F is 0.35 volts and that will give maximum depletion layer corresponding to that is given by this relationship. We have already discussed entities that I have calculated and put it is about 0.3 microns. Why I brought this number is supposing the thickness of this soil layer is 1 micron. When I apply gate voltage to the front gate when the front channel is inverted depletion layer width will be 0.3 microns. So, if this is 1 micron thickness entire layer below that will not be depleted. That means channel this whole thing is channel T silicon which is shifted down here. T silicon is that thickness T silicon is only partially depleted when it is inverted. So, below this layer there is no depletion layer, but now I use this back gate also to deplete from the back side and invert the back channel. So, what you have shown is this portion is 60 maximum depleted due to front gate. This portion is 60 maximum depleted by the application of back gate. That means at a T maximum you will have an inversion layer here and inversion layer on the top, but now you can see there is no difference between the bulk MOSFET and this. You have got one MOSFET on the top, you have got another MOSFET on the back. It is like putting two MOSFETs in parallel. There is no coupling between this region and this region. They are isolated. That is partially depleted. So, in this example if the T silicon is total from this layer is 1 micron, the top depletion layer will be 0.3, bottom depletion layer will be 0.3 and you will have 0.4 micron thickness undepleted. So, this is a partially depleted case. So, basic device equations for a partially depleted that is PD, SOM MOSFETs are same as that of bulk devices. There is no difference. So, we do not have to discuss any more. So, this is not used very often except that you can still use it for radiation hardware devices. It may not be, you can have two channels and the device is isolated from substrate fully. That sort of thing is used, but this is not a very popular device today for ICs. Two inverted channels can exist in parallel, but remember these numbers, 10 to the power of 16, 5 to the power of 0.35, depleted depletion layer with maximum 0.3 microns. So, T silicon is 1 micron which is greater than twice the XT maximum. So, it is a double gate operation. So, in that case when the double gate operation is there, if T silicon is more than twice the depletion layer width, it is partially depleted. If it is single gate that is there is no back gate, then T silicon should be greater than XT maximum, then that is partially depleted. The idea is clear. Fully depleted device, it has removed all these, no material is there, but there is a depleted layer to show that I put this diagram like this here. So, you can see here front gate and back gate. So, if I take same 1 micron thickness, of the SOI layer and if I use instead of 10 to the power of 15, 10 to the power of 14 doping per centimeter clove, P type doping, then phi f corresponding to that is 0.458. I am sorry that is twice phi f. See let us remember phi f here was 0.35 and doping is lower, phi f will be smaller. So, what I have here is twice phi f. You have to correct it. It is not phi f, twice phi f. Phi f is this divided by 2. So, twice phi f is 0.458. So, corresponding to twice phi f that depleted layer width is 2.465 because go back to that one. Twice phi f is lower, but n a is lower by 100 times. Twice phi f in 10 to the power of 16 cases is 0.7 volts. 10 to the power of 14 cases is 0.45 volts. Not much different, but this doping is reduced by factor of 100. So, depletion layer width will be more by factor of about 10. So, you can see that the extreme maximum here is 2.46 micrometer. If there is a thick as silicon layer, silicon depletion layer will be 2.46 as compared to 0.3 micron is 10 to the power of 16 cases. So, now what we say is if the thickness of the channel is 1 micron, the entire layer will be depleted. It will go beyond that. Even if you have this thickness of channel is 2.46 micrometer, when it is inverted, the whole thing will be depleted. So, the thickness is less than 2.46 micron. It is definitely inverted. Both front and back channels can be inverted. What happens in between? We will see some of the analysis. So, that means the case is different from that of bulk MOSFET. You have got the entire layer depleted. You have got the front channel inverted, back channel inverted. In fact, you will see that if you change the front channel potential, it will change both front channels into the potential and back channel potential. In other words, you will see that there is coupling between the front gate and the back gate, because there is no charge here. Oxide depleted layer oxide. So, whole thing is coupled. If I change the whole thing between the two, the field between the two will change. So, that is the meaning of coupling. We will understand more about that later on. Now, let us go back to the 10 to power 15 case. Can I make a fully depleted SOI MOSFET in 10 to power 15 doping? HT maximum is 0.3 micron. Suppose, here I make this thickness of the SOI layer 0.1 micron. It is a fully depleted case. So, if I go to thinner and thinner layer, I can use slightly higher doping. Now, this is the idea of fully depleted and partially depleted. This operation of this is different from the bulk MOSFET, whereas partially depleted is same as the bulk MOSFET. So, we need not go into discussion of that. We will get down to discussion on the fully depleted SOI MOSFET. Let us take a look at that. This is a diagram, which tells you the energy band diagram. All of you are familiar about energy band diagram. This is the bulk MOSFET. Bulk MOSFET put upside that way, gate, oxide, substrate. So, all that is put like that. So, gate is there, oxide, shaded region is oxide. This is the depleted region. You can see the depleted region, band gap, energy band diagram bends upwards because the field is in that direction. So, you have got below the depletion layer, there is no band bending. There is no band bending. There is no voltage drop. The entire voltage drop is across this depletion layer. So, this is the band diagram of the bulk MOSFET. And if I draw the intrinsic level, that will come from here like that. Intrinsic level will be below the permeable that will be inverted. Now, let us take the partially depleted SOI MOSFET. So, this is the MOSFET now. I put it like that. Front gate, front gate oxide, silicon, back gate oxide, back gate. So, you have got partially depleted. This is oxide. This is the depletion layer. So, please remember wherever depletion layer which is there or wherever depletion layer is present, the band bending takes place plus to minus. Now, this is the region which is not depleted. To go back, we are coming from here to here. Depleted, no depletion. Depleted from backside. Depleted, no depletion. Depleted because of applied voltage plus here. So, field is in that direction here. Here, the field is in this direction. The band diagram is like this. Now, let us say supposing this portion is removed. That means, I have made the thickness equal to sum of this and this. This is removed. So, you put them together. No depletion. I am sorry, no undepleted portion. Fully depleted. So, this goes up like this. This goes up like this. And the electric field here will be 0.  Electric field is 0. From here out it is 0. Merge them together. Up to this point, the electric field is in that direction. Belong to the front gate. And from this side onwards, from this back, right side to the left hand side, electric field is towards the left side. So, energy band diagram bends up. Where they meet? Where the two depletion are meet? Here, field is in that direction. Here, the field is in that direction. There, the 0 field is there. And where is the 0 field is there? There is a potential minimum. Comes down. Minimum increases that way. More about this we will see as we go further on this discussion. So, this is the energy band diagram of power. Fully depleted SOF MOSFET. This will be inverted if this potential is twice SOF. This will be inverted actually if that potential is twice SOF. Notice also, you are applying that voltage with respect to this source region. Here, after the analysis becomes slightly different, slightly different marginally, but you will have to be careful in looking at the entire situation. So, this is again the diagram of the SOF MOSFET. Now, these have shown this back oxide, T oxide B thicker compared to a T oxide F, not symmetrical. Symmetrical case, they will be equal. And this is the T silicon, T S, we call it in the notation. And this is the T oxide F source range. We are now talking of fully depleted MOSFET, no more partially depleted. So, you can have in this case several type of operations. One, you want to have MOSFET operation, there must be at least one channel inverted, so that electrons in the N channel device electrons are created and you can transport it here like this. So, one case is front channel inverted and back channel can be accumulated. You can have the entire layer depleted, but still you can have, how do you accumulate the back channel? Apply negative voltage to the gate. So, if you take a MOSCAP p-tap substrate, apply plus voltage to the gate, you will deplete. It keep on increasing plus voltage, it will be inverting. Suppose, you apply negative voltage to the gate with respect to the substrate or the source, plus charge will accumulate, it is accumulation. So, I can have accumulation layer here on the back channel by applying a large minus voltage to the gate, back gate. But at the same time, I can invert this front gate by applying plus voltage to the front gate. You can see the flexibility that you get. That is one operation. This is only more for academic interest understanding purpose. In some cases, they use it in some special applications. Otherwise, it is very rarely used, but you may come across such situations when you make use of it for an integrated circuit sometimes. Second case is front channel is inverted. Apply a plus voltage large enough, so that you can invert. How much is the voltage that you must apply to the front gate to invert depends upon what is the condition on the back channel. That also we will see. See the bulk mass set, your threshold voltage is decided once the doping is fixed and the oxide thickness is fixed. Now, here another parameter, you can control the threshold voltage by using a back gate. You can do that also in the bulk mass set. Back gate bias can change, but that will only increase the threshold voltage, reverse bias if you put. If you cannot apply plus voltage to go to the back gate, because then there will be follow bias across the junction. So, you can only increase the threshold voltage. You can control the threshold voltage some extent marginally. Here, there is an oxide. You can apply plus minus without hurting this junction. So, if you apply minus here, inversion. If I apply plus here, depletion. So, front channel can be inverted with back channel depleted. Second case, third case is I apply sufficient plus voltage to the front gate, sufficient plus voltage to the back gate, so that both channels are inverted. So, front channel inverted, back channel inverted. Now, you can see when you have both channels inverted, you can draw a line at the center. You can say half of the top portion of the silicon is and the depletion layer there belongs to front gate. It is not shared between the two, both are inverted and half of the layer below this middle line here, that is belonging to the back gate. So, whatever depletion layer charge is there, it belonging to front gate there or if T silicon, T silicon, total charge per centimeter square is Q N A into T S. So, Q N A T S by 2 will belong to the top gate, other charge will belong to per centimeter square will belong to the back gate. There is another class of devices, which is more soft dedicated version or another version of this third version, that is volume inversion. Ultrathin body, U T B is ultrathin body, double gate S Y MOSFET. So, today we talk of ultimately ultrathin body and in this analysis, unlike in the bulk make MOSFET, you will talk of T ox C oxide. C ox is capacitance per centimeter square for gate oxide. Here, you will have talk of front gate oxides C ox F, which is epsilon oxide into epsilon 0 of course, divided by T ox F, that is per centimeter square. And you can also talk of C T silicon, what is T silicon? If the entire silicon layer is depleted, the capacitance corresponding to that layer, silicon layer, that is C silicon is epsilon silicon into epsilon 0 divided by T silicon. Instead of oxide, you have got that silicon permittivity, silicon thickness. You can talk of that, that is when it is fully depleted, when it is fully depleted, you can look into it as a capacitor. When you go to sub threshold slope, you will see that that capacitance complete temperature, that is why I just brought in here. C ox B. Now, when you go into the analysis of this structure, what we will do with B? We will assume that this P plus, P layer is P plus. That is the stability, so that whatever voltage I apply is available here. To simplify, it will modify it only marginally with a small drop in the substrate, if the doping is present there. So, when you go to double it, it is in fact, the situation will be, that will be held up, this will also be held up. So, we consider that case to avoid more complication. In other words, you can say this is a metal here, this is a metal there, there is no drop here in the silicon. Now, let us take a look at how to, what we are interested in is, what are the operational modes and what are threshold voltages. Once you know the threshold voltage, you can use the MOSFET equation, same equation you can use to describe the MOSFET current, transconductance. So, the entire analysis is hovering around the threshold voltage. You can do the same equation, you see ox A W L P G minus V threshold whole square or W Q into V into V into V into V into V, that is the ultimate equation, velocity into charge into W and Q depends on what is the threshold voltages and what is the gate voltages. So, now, just before we go into that, you can see that this is bit more complicated structure than the other one. Notice, when you apply gate voltage in the bulk MOSFET, what happens and how it will be? The bulk MOSFET, I am going to be slow here because slightly different from the concept there. Gate, black, oxide, depletion layer width, when a plus voltage and of course, inversion is there. Now, here how did you recall, how did we find out the threshold voltage? Is that at threshold voltage V G equal to V T N and V G gets shared between the oxide and the silicon and in silicon, it is shared between the, it is part of a depletion layer. So, at threshold voltage V silicon is twice fair, the drop. So, depletion layer corresponding to that is Q depletion by, I am sorry, oxide drop. To make it very clear, gate voltage is shared between the oxide and silicon and at threshold voltage, what is the silicon drop? Phi phi f, that is very clear. And once, phi phi f is the drop in silicon, you know how much is the depletion layer width from this formula. So, it is phi phi f, so depletion layer is this much. Once you know what is the depletion layer width, you can find out what is the, what is the voltage drop across the oxides, that is this one. Charge in the depletion layer width is Q N A into X T maximum divided by that, divided by C oxide is the voltage. So, what we did was, V silicon is twice fair, corresponding to that X T maximum is given by this formula, corresponding to that, you know Q D is that much per centimeter square. Q D by C oxide is the voltage drop X of oxide. Take a capacitor, charge on the other side of the capacitor is Q D minus Q D. Q D by C oxide is the voltage of X of oxide. This is how we did the calculations. So, how did you arrive at that? I am doing that, because you may have multiple layers beyond this point. If you have multiple layers beyond that point, you may not be able to easily compute this. Here, you know once is drop across silicon is twice fair, you know depletion layer width, you know charge. Charge by C oxide is the voltage of a cross oxide. Now, look at the other way. What is the charge related to, how is the charge related to the field and the surface? If you apply Gauss law to this surface, top of the silicon surface, you apply Gauss law. It is part of the infinite surface. So, you apply Gauss law to that portion per centimeter square, part of that per centimeter square, charge is Q N A X T maximum. Charge divided by epsilon s is field Gauss law. The field lines permeating on the surface is charge beyond the surface divided by epsilon s. Epsilon r epsilon 0, that I have put in epsilon s. So, you took that once centimeter square of the sphere, then you say the electric field is that much. So, what we are telling is, the Q D by epsilon s is electric field in silicon. Now, from the continuity of flux, if you take, I do not know whether you have studied electromagnetic field, you must have studied or basic electrostatics. If I have studied, here look at this. I have oxide here, I have silicon here. Electric field here and here will be different. What is continuous is a flux, del dot d into d of s is constant or the flux is epsilon s into epsilon. So, here epsilon s is, epsilon s is the field in silicon is E s and epsilon s is the permutivity. So, epsilon s into E of s should be equal to epsilon oxide into E oxide. That is the continuity of flux. This is a D, D is epsilon into E. D in silicon is D in oxide. Sorry, D in silicon is D in oxide. So, now you can see, I have drawn the electric field diagram here. E oxide is higher than in the silicon. If I know E oxide, I know voltage of across oxide. See, I am just trying to get this relationship slightly differently, because you may not have that replacement derivative precisely, but you can find out. If you can find out the free peak field on the surface, you can find out the voltage of across oxide, because if I know the peak field E of s in silicon, I know the oxide drop, oxide field from this relationship continuity of flux. Now, if I know the oxide drop field, I know what is the voltage of across oxide is. It is exactly a derivative actually. I am just recalling that. So, E oxide into T oxide will be the oxide drop. E oxide is from this relationship epsilon s E of s divided by E oxide is epsilon oxide is that. This is a flux in silicon divided by permittivity in the oxide gives me the flux in the oxide, electric flux, electric field. Flux is product of the two. If I divide by this, I get electric field in oxide. That into T oxide gives me a voltage drop. So, what is this quantity E of s into epsilon s here? E of s is Q d by epsilon s charge beyond that surface. Whatever charge is there, divide epsilon s is the peak field. So, E of s into epsilon s was the charge here. That is the charge. So, same relationship you got differently. Why I derived that is, if you want to find the voltage pressure voltage, all that you have to do is, what is the voltage at a pressure across oxide when the surface potential is twice y f. So, potential near the surface is twice y f. You can find the voltage drop across oxide by knowing the field near the surface of silicon. That is because epsilon s into E of s actually is the charge beyond that point. Whatever is the number of layers divided by C oxide will give the oxide. I think it is, when you go through this, once more you will actually see. So, by doing this, I have shown that what you write here is the same as this. It is writing Q d. I wrote it as, I showed that it is equal to, Q d is equal to whatever charge beyond that point. It does not be depletion layer. Epsilon s, that is, permittivity of the material layer, silicon into the silicon field at the surface. That is what we get now here. Threshold voltage is twice y f plus this. So, if you go to the same aspect, all that you have to do is, if I want to find out threshold voltage of this type of structure, all that I have to do is, at threshold voltage of front gate, I know that the potential here is twice y f. So, I say potential twice y f. That is the condition for immersion. Why is twice y f? And if I can find out the field on the surface multiplied by permittivity divided by C oxide, I get the oxide drop. That is all what we are doing. It is as simple as that. We are trying to find out where assuming twice y f is the potential there. So, gate voltage is twice y f with respect to ground. So, potential here with respect to ground is twice y f. Beyond that, only the gate to oxide drop is there. That is just epsilon s into E of s divided by C oxide f. If I forget to say, I have assumed that it is at oxide. Now, let us take a look at it a bit more. It is a bit slow on this. So, this is the case of bulk MOSFET on the left hand side. Take a look at those figures now. This diagram I have drawn already in the previous page. This is oxide. There is a gate. This is silicon. This is no-tone. So, putting all like this, I put it like that. Gate, oxide, silicon, bulk. So, in silicon, because it is informally doped, electric field will always be linear. We have seen it number of times. So, this is the extreme maximum. At inversion, we are considering threshold voltage. So, we are considering a situation where there is inversion. That means, we are considering a situation where the surface potential, that is potential of the surface of the silicon is twice phi f. I think in the previous series of lectures, you would have called that twice phi b. If it is called, sometime people call twice phi b, I call it twice phi f. Both are same. So, you have the potential as phi f s. Electric field is linear and integral of that will give a parabolic term. So, voltage drop in silicon will be decaying like this parabolicly. That is going back to basics. And the potential is continuous. In the silicon, in the oxide, there is voltage drop. Electric field is constant, but there are no charges. If the electric field is constant, the potential is integral of that, linear. So, you can see that voltage rises in a parabolic manner in silicon, bulk MOSFET, then it is linear. Electric field rises linearly and it is constant. There is no charge. In between, there is no charge. Here, it is linear because as you go from here to here, the charge keeps on increasing because depletion layer. The charge from here to here is, if you start 0 from here, I am sorry, you charge 0 from here, this end. As you go, the charge is going on increasing and voltage is going down to 0. So, let me, without going into that, linear becomes parabolic and constant becomes linear there. And you see here, the oxide is higher than the silicon here because of the difference in the permittivity. Lower permittivity is the factor of 3. So, this oxide is 3 times this quantity. Now, let me now go to a better example. A bit more involved with situation, SOI MOSFET with the back channel accumulated. Front gate oxide, back gate oxide. These colored portions are oxide. This is the T-silicon. What you are telling is, we are talking of fully depleted case. That means, by the time it has inverted, the entire silicon layer is depleted. But if I start with the back gate bias VGB negative, see this diagram, this full line here shows the potential. When I apply a voltage across this VGB, back gate, just go back to this. I will show you. See here, when you apply VGB, you may be wondering with respect to what you are applying there. You are applying with respect to this one point or some point somewhere in the silicon, somewhere outside. There is a neutral point. With respect to that, you are applying. In the case of MOSFET, also you are applying a voltage with respect to the gate to the source. Source is connected to the substrate. Here also, the similar thing situation you can have. So, when I apply voltage here, negative with respect to that, this will be accumulated. Back gate is made negative with respect to ground. That means, this channel will be having plus charges. P becomes more P plus. It is accumulated. Now, the potential here corresponding to that will be how much? In an accumulation layer, see in a depleted layer, because of a depleted layer and band bending, you get charges sufficiently. You get a region over which charges spread. And the potential can be 0.1 volts, 0.2 volts, 0.3 volts or even 0.5 volts depletion layer drop. For example, when you take 10 to power 16 doping, you have got twice 5 is 0.7 volts. That means, voltage drop is 0.7 volts. Now, when you have accumulation layer there, you can accumulate the charges with very small voltage surface potential. So, the potential of this region with respect to the substrate will be very small, plus here, minus here, plus there. In other words, you will have a charge sheet with very little drop in silicon. So, the assumption here again would be that it is a charge sheet with no drop across that charge sheet. Like what you do in the case of inversion, you take the inversion layer, you do not transfer a drop across that, because charge is large. So, here you can say that plus charges are there, which is actually decided how much you have voltage applied and how much is the back gate thickness. So, if you take the back gate here, you will have a field in the direction, plus here, minus here. But the voltage here will be 0, because if you have a large plus charge there, that can come with very small voltage across, voltage in silicon, because after all the charge is exponentially related to whatever background charge is there. Whole concentration is 10 to the power 16 into e to the power of 5 by B T. So, let us say 5 is, even if it is 2 millivolts, e to the power of 5 by B T will raise this particular carrier concentration from 10 to the power of 16 to double that value very easily. 25 millivolts if you take, e to the power of 25 millivolts by 25 millivolts, 2.27. So, if doping it 10 to the power of 16, 10 to the power of 16 into e to the power of 1, 2.27 times 10 to the power of 15. So, carrier concentration is very large there. So, at that voltage that surface potential there is just 2 millivolts, 25 millivolts. So, what we are telling is by applying 90 voltage, the entire voltage will drop across this particular region and the potential of the surface will be raised with respect to the source by very small quantity, 25 millivolts, 30 millivolts of that order. So, we are neglecting that and seeing the potential is 0. We can take into account, you will see that even the analysis will still hold good. So, I am neglecting that voltage surface potential, taking it as 0. That is what I put here, psi S B is 0. So, when you do the entire analysis, you will see that if you want to consider back channel accumulated, you can put psi S B equal to 0. Front channel inverted, you will put psi S F, psi S F is equal to twice by M. This is understanding that with which we are doing the analysis. See more simplification, but holds good quite a bit. So, if I draw the electric field here, I fix the voltage as V G B, there will be electric field like this. Then we charge here plus charge here, minus charge here. This plus charge can be very large because doping is not there and this will be negative charge. Now, when you started with that bias fixed, we are not varying that voltage, we are not varying the field. When I apply voltage to the front gate, what happens in the initially? I am applying it with respect to substrate, that is with respect to source. So, applied voltage exactly in this case, I am sorry, exactly like this, it will start appearing, you know, if it is not depleted fully and the voltage is very low, some voltage appears across the oxide, some voltage across the silicon. So, you will have initially, there will be depletion layer starting from 0 here. Entire layer is not depleted initially, there is 0 potential here. Depletion layer will start widening and at one point, the depletion layer will come and merge here. That is the dotted line. The dotted line is the situation when the depletion layer from front side has, from the front side has widened and ultimately merged with this point. Where was the negative charge come from at the time? From the depleted portion. When I apply plus voltage to the gate, there is no charge here. Negative charge come from this substrate as the depletion layer widens because of more and more plus charges at the front gate, more and more negative charges from the substrate, more and more widening of the depletion layer. At a particular point, you can call it as the punch through voltage, where the depletion layer is completely merged with this point. That is the depletion distribution there. Similar to this, but this field will be much smaller than that because this bit was large, 1 micron, let us see and that is not depleted. This is, if it is 10 to power 16, this let us say 0.2 micron. It has depleted fully, but if it is point, if it is 10 to power 16 doping and if this width is 0.1 micron, surface potential is not by 5. When this has depleted here, just go back to that graph and show you. Here, here itself we can see. If it is 10 to power 16, surface potential will become twice higher if the depletion layer width has become 0.3 microns. Now, if I cut it from here and keep it only 0.1 micron, it whole thing would be lowered like this. So, that is you have got the edge of the depletion layer here, potential is much lower than that. All that you have to do is pull this side down, so that field is lower. So, when the depletion layer is not 0.3 micron, the voltage of a cross silicon is not twice higher, it is less than that. So, that is the situation here. It is not twice higher and oxide drop is actually corresponding to this peak field. You have got the oxide drop, which is lower than that. You have to follow very carefully, because this is bit slightly different from what we usually discuss. Now, at this point, whatever plus charge was available was coming from this negative charge here, depletion layer. Beyond that, if I increase the voltage, plus charge I put here, the minus charge will come from, where can it come from? It can come from the other end. So, what happens is, the field lines from here rise parallelly. I wish I had another diagram here. See, at this point, electric field is something maximum here. Here, it is 0. Now, if I increase the electric field at this point, that is 100 lines increase from here to here. That is it. All the 100 lines do not find the negative charge here. It will have to go to the other end of the gate, where there is negative supply is there. So, the extra 100 lines, all of them will cross this. Over simplification the electric field has increased from here to here by some delta E. The delta E is 100 lines. All the 100 lines, delta E will cross this. That means, 100 lines which are extra crossing here will cross this also. That means, the electric field will increase from here to here by delta E. So, what we are telling is, this distribution is not going to change, where there are no charges to terminate. See, electric field distribution, if I see electric field will go up here, if there are charges here, but there are no charges here. So, whatever extra charge is coming here will cross here. So, the entire field line will go up here, it will like this parallel. That is the meaning of drawing this line mean. There are no charges in between. All the charges will come from here. So, what will be the voltage distribution? Voltage here is not changing. Why? Is your held voltage constant by the back gate? That is remaining same thing. But this field is remaining same thing as before. It is not changing. But what has happened is, the plus charges here have reduced because the negative charges, let us say there are 1000 lines here originally, out of which now here it is only 1000 charges, if there are field lines are constant. Now, when these extra 100 field lines come here, this has not changed. That is because the field lines which are coming, 100 lines coming here, remove those plus charges because they are coming from here now. So, you have got that extra 100 lines coming from here. There is no need of that plus 1000 charges. There will be only 900 charges. I am giving the numerical number because 1000 charges here, 900 charges here, 100 charges coming extra from there. Charge you tells is maintained. So, this field is not changing. This voltage drop is not changing, but field lines have moved up here. So, the distribution now will be this potential is 0 here. Why? There are plus charges here with very small potential here. That is this till the accumulation here is maintained. So, long as there are plus charges here, accumulated charges are there, where are the plus charges coming from? They are accumulated. So, long as there are plus charges are present, the potential is 0. If I go on increasing that, if the plus charges disappear, then it will start depleting, but there are plus charges at 0. So, from here to here is what is the potential here? We are considering situation where it is inverted twice y of. So, now we know the potential in the silicon, which is less than this 0.3 microns. Now, it is 0.1 micron. The potential drop across silicon is more is twice y of, but the length over which the potential drops is smaller. So, evidently we will see that the field here will be higher. So, that aspect you can see from here, corresponding this field here will be higher, field here will be higher. So, this is a serve MOSFET in the back channel accumulated. We will discuss about the voltage calculation in the next presentation, but this is a crucial thing to understand. We can before you go to the next lecture, we can go through this once more and you will see appreciate that lecture, next lecture after seeing this only, again going through this. So, is the accumulation condition, but here the field will be higher, surface potential is twice y of. Now, let us see the other condition. I will finish these two before we wind up for today. Back channel grounded, see in this case, back channel accumulated by applying voltage. This is slightly simpler. I should take this first. In a complicated thing, finish first, go to simpler things afterwards. Here, I have grounded this, connected back channel to ground. No voltage applied. So, as I start applying from voltage from the back gate, Dipliesion layer keeps on moving. Forget about the solid line, observe the dotted line. That is the situation when the Dipliesion layer has completely depleted this channel. Initially, Dipliesion layer will be narrow. It will keep on widening and come up to this point. Just I will go to one slide. It will be clearer to you here, for example. I have grounded this point, let us say. I keep on increasing this voltage. Then, you will have the Dipliesion layer, which will be narrow initially. Higher voltage is widened. This is the electric field. Widen the electric field like that. That is the situation. Here, I have plotted the dotted line where the Dipliesion layer has just reached this point. No rolls from the back gate. You have grounded that. So, you can calculate what is the applied voltage is. At this point, what is the electric field here? Q N A into T silicon is the charge divided by epsilon silicon is the electric field here. From that, you can find out this one. That is, in this case, you can straight away see Q N A T of S is the charge. You have C oxide is the voltage of across oxide. Now, once it is depleted just here, beyond that, I increase this gate voltage. So, remember this is grounded. So, applied voltage now shares between this oxide, this silicon and this. Once it is just depleted fully, if this were silicon, what would happen? The Dipliesion layer would have moved further down. But now, again the same number. I have increased the voltage so much that the surface potential with respect to ground becomes twice twice. This is 0. There is a potential drop here. There is a potential drop here. There is a potential drop. Just when it is punched through, there is drop across oxide and silicon. When I increase to beyond that point, at inversion, there will be drop here, drop here, drop here. And across the front oxide, it is linear drop because there is no charge. Across this is parabolic because charge is constant across this linear drop. But this is 0. So, there is a rise of potential and the potential is psi S B and this psi S F. So, if you look into the electric field then, psi S B minus 0, that is the eox B. And front oxide is gate voltage that is applied is threshold voltage minus twice that is the oxide. How much voltage I have to find? How much is the electric field here? I can find out or voltage drop I can find out if I find out the surface field. So, if you want to calculate the threshold voltage of this device, all that I have to find out is how much is psi S F? How much is the electric field here when this potential becomes psi S F? That calculation we will see next time. So, what will you say about the field here compared to the previous case? Previous case, the drop across the silicon was twice psi F because this is 0, this twice psi F. In this case, drop across silicon, same silicon thickness, 0.2 micron is twice psi F minus psi S B. So, the field here is smaller compared to this. Now, let us once again go back to this. Compare the situation here. Twice psi F dropped across 0.3 microns. That was the replacement area width. Now, this thickness is smaller, but the voltage drop across that is much smaller. Twice psi F minus psi S B. This can be as close as to that. So, the field here will be smaller. See, why I am pointing out this field is, we had said that in the bulk MOSFET, the mobility is affected by doping. The mobility is affected by the field in the vertical direction. So, vertical direction in this case is like this, from the gate down to the channel. So, you can see that in this case because of the difference in the voltage drop across the vertical direction is less, the field in the vertical direction will be less than the case of bulk MOSFET. So, we can see this. Already, there is a pointer telling you that if I use a server MOSFET, the vertical electric field can be reduced. We have to go back later on C, where we can reduce the doping concentration. Also, we have to see. So, this is the situation. Now, let us take the other case, where back channel also is inverted. When the back channel is inverted, what is the potential there? Means, I applied sufficient voltage at the back gate. So, that is twice way up. I applied sufficient voltage at the front gate, that is twice way up. So, electric field lines will be terminating here to 0. Part of this belonging to top channel, part of this belonging to the back gate. So, electric field lines will come in this direction, left to right and on the right hand side, electric field lines will be coming from right to left. So, it is actually extension of this. I have already biased with the gate bias for the differential layer with this here, let us say. So, you have the, just when it is whole thing is depleted, you will have this situation. I will think I will discuss this in the next lecture. What we have seen here is, if both channels are inverted, both potential will be twice way up. So, what is the drop across the channel? That is, difference in the potential is 0. So, drop is low here in this case. Field is low. So, what we are trying to point out here is, in the case of a soy MOSPET, you can have a situation where the both channels are inverted and the vertical electric field will be smaller. If the electric field is small here, electric field is small in oxide also. You can have that type of situation. Actual threshold table will discuss in the next lecture.


Hi everyone, in this video, I'm going to be going over question 8 from the Compound Data 2 workshop. So this is quite a fun question because we're actually implementing all of these things that we're learning like passing an array into a function, designing a function and things like that. So this is quite a fun question and to do this, we have multiple steps we need to do, so we're going to plan a lot before we implement our function. It's always good to plan what you want to do before just going straight into it and coding. That can be quite dangerous and you might get some errors. So let's get started. So we're writing a function called n zero sub arrays that when passed a two-dimensional integer array returns the number of sub arrays that have only zeros in them. So there are two different ways you can go about this question. I might start with the easier way and then have a look at creating helper methods. So we know that's going to be our name. We know what the structure of a function looks like. We've got a return type, we've got a name, primitives like this, so I'll just have that just to get my structure there. I know my name and what's my return type? I want to return the number of sub arrays that have only zeros in them. So that will be an integer value because we're returning a counter for the number of items and then if we have a look at our parameters, we're being passed in a two-dimensional integer array. I'll call my data. Okay, great. So now we'll need to do some planning. So what we need to do is have a look at each individual sub array, work out if that sub array has only zeros and if it does increment a counter value that we can set up. So let's write some comments to create a structure and create some planning. So create a counter to hold the number of arrays with all zeros and at the end I want to return that counter. So I'll add that at the bottom. So what do I need to do? I need to loop through each index of the main array and then I need to loop through, let's indent this a bit, loop through each index of each sub array and this is where we're going to be doing a lot of our work here. So for each sub array we need to check if each value is zero. So as soon as we come across a value that's not zero, then we can stop looping and we don't need to check if we'll say in this one here, if we've checked this first number here, it's not a zero. So there's no point checking the remaining items in this sub array because we're only incrementing our counter if every item in that sub array is zero. So we need a condition that says stop looping once we've found a non-zero number. So perhaps we can create a boolean that will have true or false as to whether we've found any non-zero numbers. So that would be up here because it will be for each sub array. So create boolean to hold whether we've found a non-zero number and that could even be part of our loop condition. We can have two parts to our loop condition and then we need to check if the value is non-zero and if it is non-zero change our boolean. So once we've exited our inner loop here, what we need to do is have an if condition to check our boolean and our boolean is going to represent if we found a one. So, sorry, if we found a non-zero, so if that value is true or perhaps if that value is false that means that we've found only zeros and will increment our counter. Okay? So that looks like a plan. Let's start implementing these bits one by one. So first I want to create a counter to hold the number of arrays with all zeros. So that count will be an int. I'll call it total and it could be zero to start with and then I want to return that counter at the end. So I'll go return total. I want to loop through each index of the main array and we know how to do that. It's that same loop we've done many many times. Great and we need to close that there. Here I want to create a boolean to hold whether we found a non-zero number. So boolean non-zero is equal to false because at this point we haven't checked any numbers, so we haven't found any non-zero numbers. Okay, then we want to loop through each index of our sub array. So, I forgot something, data i dot length. And we've forgotten something here, this loop. We also said that we want this loop to stop looping as soon as we find a non-zero value. So we need to add another part to our condition here. So the index must be within that range within zero. It's initial value and data i dot length and we want non-zero. Perhaps let's let's change this. Let's say all zeros and it will be true initially. And so when we find a value that is non-zero then we change our boolean to false. And if it's true we increment our counter. So let's end all zero. Here I can say all zero is equal to true but all zero evaluates to a boolean already, so we don't need that equals equals true. Well we know that it is a boolean. So I'll just say all zero. So these two conditions must be true for our loop to loop and then we need a condition to check whether this value is non-zero. So I'll say if data i m if that value is not equal to zero then I want all zero to be false. So now if that was the case if the item was not equal to zero then this would be false and this loop would stop running because all zero will be equal to false. And so this whole boolean we know that whenever we have an and in our boolean expression that the whole thing if one of them is false then the whole thing will evaluate to false. So this will terminate our loop and now we need a condition here which we said if condition to check our boolean and increment our counter. So if all zero if it's true then that means that they're all zero then we increment our counter. And again as I said before all zero is true. So we don't need to say equals equals true because if all zero is false we've got false is equal to true which is false. So we don't need that equal sign there. We can just say all zero. And Yeah, this looks okay. We want to be careful. Oh, no, that's okay because each time we loop through the main array we're entering this outer loop and each time we enter the outer loop this all zero is being reset to true. So if you initialized declared and initialize this boolean variable up here you'd have some issues because you're not resetting it. So it's important to have it here. But if you needed access to it outside the scope of the loops then maybe you would need it up here and you can just reset it here because we know that any variables declared inside a certain scope only exist in that scope. So if I declared it here it would only exist within these brackets and perhaps in any in a different circumstance where I needed to return the boolean I'd need to have access to that outside of this scope. And so having it out here is I guess a little bit safer, but considering the function we are trying to design then that's not necessary. Great, so let's double check that this is working with this test value that they've provided us. So let's create this is integer array R equals this. So let's print out, well actually let's assign that to a variable and zero is equal to the output of. So that should give us three. So I'll go print line let's have a look awesome what if none of them had all zeros that should give us zero. Great, and what if they were all zeros and of different lengths? We want to create lots of different test cases just to ensure that our program is working correctly. Should be zero, should be five because there are five sub arrays here. Great, and because we were careful with how we did our loops and we made them dynamic by having data dot lengths rather than hard coding it to five because there are five sub arrays and hard coding this to two because in this particular case each sub array length was only two we've made this scalable so that it doesn't matter what dimensions our two dimensional array has this function will always be successful which is good we want to create functions that can be scalable it's good practice. So that's the first way of doing it another way of doing it would be creating a helper method a sub method because we can see that this function here there's a lot of things that it's trying to do so we can split that up into two different functions so that we have two different responsibilities. So perhaps what we want is a helper method that will return true or false as to whether each item in here is non-zero so that would take all of this out into a sub method. So it will return a boolean we'll say all zeros and its input will be just a regular array because we're only checking a sub array so what we can do is we can say four zero i is less than a dot length i plus plus and as soon as we come across a non-zero value if it's not equal to zero we can return false straight away and if we get to the end of this loop then we can return true so this is looping through each index of that sub array and if that item is not equal to zero returns false straight away otherwise returns true so now what I can do is I can remove all of this here and I can just make a function call so I can say boolean all zero can be equal to all zeros and I'll pass in data i which is that sub array there and if all zero is true total plus plus otherwise we don't so we could even make this a little simpler rather than having a boolean value called all zero we can just put this function call in as our condition so I'm making this even simpler now and that is a little simpler because it has two different responsibilities maybe my function name isn't that great but it's easier to debug now so this is still working should be five let's go back to our initial one just to make sure that everything's working so this should be three great and we could definitely have a look at some more test cases but I'm quite confident that this should work so yeah lots of different ways to implement this method either way that you do it is fine um yeah I think I prefer using helper methods just because it's easier to wrap your head around what's happening and separate the responsibilities and put your code into modules I find that's helpful