 You can follow along with this presentation using printed slides from the nanohub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. Okay, so today we'll be discussing the electrostatics of PN junction diode. This is lecture 20. Now this is the first five weeks of the class was all about developing a framework through which we will be able to analyze devices, any devices, semi-classical devices. And we have gone through this notion of quantum mechanics. We have encapsulated them in effective masses and band gaps. We have identified that there's this two, only two bands. There are many, but only two bands sort of are relevant, conduction and valence band. And we have included that information in density of state and eventually in effective density of state. That simply says that we don't really have to think about all the states, but rather just one will compress it all the states into one state near the conduction band age and another in valence band age. And that's about it. And then we talked about how electrons are distributed within those states, right? Fermi Dirac statistics. And from that we did this transport equations. And I have showed you two examples, analytical and numerical, how to solve them. Now, from now onward, so that was first five-week, that's the foundation. You will take many other courses if you are at Purdue. Purdue is very good in theory, of course in experiment also. But what we'll do from today onward is switch to devices. And I will show you how the five equations that we have developed, how we consistently solve them for wide variety of cases. And that will give you exercise about how to use them. Now, what you will see in many of these cases, that it will look like, how did they know to make this approximation? You see, this is really, many of these things that you're seeing, was done between 1950s to 1970s by very significant people. And there is no one path by which these approximations were taken. There's no a priori logic. You solve more of the problems. You understand more of the physics. You solve more problems numerically and gradually you develop intuition. So if you don't understand immediately that why did they take these approximations, you are not alone. Everybody in the field who starts this in the beginning don't see the pattern. It's like sculpting. When a sculptor takes away, chips away certain things, we don't see the final products in the beginning. But gradually, only in retrospect, everything seems logical. It seems like that's the only path you could have gone. So today, the PN junction will get started, but it will see how many of those features. So have patience, but then at the end, try to see the whole picture through. That's very important. We'll talk about PN junctions and then talk about solving the Poisson equation for PN junction. And I call that drawing the band diagrams. As I said, it's very important that you understand this very clearly. Now, you cannot do it like an undergraduate student does. Because actually, this has a few steps that you need to understand clearly and follow the steps always. You'll never get wrong. But you cannot just memorize and draw these diagrams. So I'll show you how to do this. We'll talk about how to solve the problem in equilibrium and then how things change as a function of applied bias. So now we are moving on to the other book. The other book is really undergraduate book. So there'll be a lot more material that I will provide. Focus on the class notes. You'll be fine. Now, what is a diode good for? Well, diode is good for many, many things. In present day technology, that looks like that there is a series of things that diode is good for. A solar cell is a diode, semiconductor diode. A Avalanche photodiode, which is used to detect very dim light coming in. And Avalanche will show how Avalanche can be used to detect a few photons coming in and starting an Avalanche. And by looking at the Avalanche, you know only a few photons have come in. So many of the things that you see in the underground, deep underground detectors, they want to detect cosmic rays of various types, mean minuscule amount, those are actually these type of diode, biostain, Avalanche region. So we'll see that how it works. Gallium arsenide lasers, in fact, all lasers are semiconductor diode, essentially. And gallium nitride lasers, again, gallium nitride lasers or gallium nitride lighting, all the incandescent lights, they are supposed to be. The people say that takes about 20% of the energy in US. And those are all supposed to be replaced by the solid-state lighting, again, a diode. That's a diode, nothing more. And then organic LEDs, this is from Sony, I guess, that your TV one day could be essentially a very thin piece of plastic and all the pictures coming in, again, just a diode. So if you understand how a diode works, you see a bunch of a significant fraction of the semiconductor industry, you have a very good handle on. So let's get started. A PN junction diode reflects a semiconductor which has been doped by two different dopants, spatially. So for example, here in light blue, I show that the original semiconductor, let's say it is, donor doped, donor doped means it has a lot of electrons. And then you may have a region where extra acceptor atoms. What are the acceptor atoms? This could be boron, for example, that has been diffused in. And as a result, there is one region where NA, which is the acceptor atoms, is much larger than the background ND. Not the ND has disappeared, but NA has superseded the value. And you can see the dark blue and the light blue, the region around it, the perimeter around it, that will be called a junction, because the majority dopants change from one type to other. And the top red and the bottom red, these are contacts. And for now, we will not worry about the, these are generally metal contacts or various types of highly doped silicon contacts. But for now, we will not worry about this. We will come back to this point a little bit later. It's actually a very important problem. We'll come back maybe five or six lectures down about the junction between red and blue. These are called short key barriers. We'll come back to that. Now, you can see that this device is actually a two-dimensional device, that if you start the electric field from the top red to the bottom red, you can see the electric field will spread out, funnel out from the two regions. But in the beginning, we'll start by thinking that as if the blue is very big, very big region. So essentially it's a one-dimensional device. We will bring back the two-dimensional aspect of it a little bit later. Because you can see on the corners of that region, in fact, the electric field could be very high and that's a big concern. And we will come back to that later on. But for the time being, one dimension. Now, the symbols, well, people would many times draw a symbol like this for PN junction. But more importantly, most of the times this you may have seen, that people draw a diode as a triangle terminated by a solid vertical line. And essentially this is a symbol for a diode. And in here, essentially what we'll show that the current can flow easily in one direction, more easily in one direction compared to other. Now, where does the symbol come from? I mean, if you look at the other device, this doesn't look like that representation of the original device. It actually comes from a historical reason because in late, let's say 1890s to 1920s, that time the diode didn't look like this. The first diode that was used for transatlantic communication, that diode is many times looks like a point contact diode where a metal pin or a metal wire is pressed against a semiconductor. And that is the visual representation. So it's a historical remnant of that picture. So that's why diodes actually look this way. Okay, and we will do this again in the short key barrier chapter. We'll show that what that type of diode is. And that's really older type of diode are all metal semiconductor diode. Not semiconductor semiconductor diode, you see here both sides are semiconductor. That's the distinction. So as we go through the various topics in this next eight to 10 weeks, I want to make sure that you follow the trends because the goal is not to teach you about diode short key barriers and other things. That's good. But the point is that I want you to see a pattern that every time I solve a problem, I use the same approximations over and over again depending on the problem. And once you see the pattern, the goal is that once you walk out of the course, then they give you a new device. You essentially step through these various steps, various approximations, you'll always come out with reasonable answers. Not probably always correct, but always reasonable answers that will be a good first approximation. Then you solve it numerically and see how good you are. But at least you can get started. So today we'll start by talking about diode and talking about when a diode is in equilibrium, that I haven't applied any voltage across it, a diode sitting on your table on its own. Again, as I said, that we'll be solving these five equations over and over again. Now the first thing is in equilibrium, uncontacted diode, do I have any current? Of course not, right? Then of course then, therefore the four equations on the bottom, I can forget about it for the time being. Now that does not mean there is no electric field. It doesn't mean there's no diffusion. It simply says there's no current. That means in equilibrium, there is this detail balance. Equilibrium doesn't mean that nothing is happening. Yes, it is sitting on your, the diode is sitting on your desk doing nothing. But internally a violent storm is raising. Electrons jumping to the conduction band from conduction band, electrons coming to the valence band, electrons going back and forth between one side of the junction and the other side of the junction. Huge electric field inside the device. But outside, from outside, everything looks calm, no current coming out. And if there is no current coming out, then we will not worry about the continuity equations and the associative. So my goal is today, the lecture, is to show you how to solve this problem in equilibrium. Now you could say I have already solved this because didn't I solve it in one of those classes where I said P minus N plus ND minus NA equal to zero? And that from that I calculated the Fermi level and from that I calculated the N and P and ND. Didn't I do that? I did that all. What's the difference? Well, the main difference that I'll show you here is that in that calculation I assumed that it was a homogeneous sample, that the number of dopants at every point is homogeneously distributed. Therefore I could say that every point when it's globally zero, not only it's globally zero, but locally it was zero at every point also. That assumption doesn't work anymore because here it is not ND and NA are not the same everywhere. So I will have to take back what I said before and solve the problem, a new and in this particular case. And then I will show you that when I apply a bias, then how we solve for various cases the DC small signal and transient response. But let's set that aside and we'll come to it when time is. I want to remind you about what a N type material and a P type material is. Do you remember from chronic penny model after we solved the Schrodinger equation we got a bunch of bands. Although I had this continuous line for the four bands these are not continuous, right? But there are points which are very close by. What is that? 2 pi over LA, right? That was our L, 2 pi over L, the dimension of the sample very close and so therefore we draw it as a continuum. And we have also told you about this conduction and valence band. So I am taking that picture and essentially representing without the case side I'm representing it in the first set of diagrams which is you can see that there is this yellow region green conduction band and a blue valence band. And of course there are bands, red bands underneath which I have not drawn. Now if I say that this is a, oh by the way the straight line or straight black line on the top of this, this we'll call the vacuum level. Vacuum level means that if an electron can go up to that far then it will leave the semiconductor, right? Photoelectric effect do you remember? Shine light on it, the electron jumps up and if it has enough energy then it gets out of that material. So that black line is called a vacuum level because the electron will then be free from the semiconductor and can go outside. Now if it is a n-type or n-doped material means Nd then why should the Fermi level be? You think Fermi level be close to the conduction band. Is that right? And there will be more electrons and there will be fewer holes. I haven't drawn any holes in the blue band but you get the idea. Generally there will be few holes in there and the deep blue is the number of electrons. So initial part is full of electrons. Okay, no problem. Now what about the other region? If it is a p-doped, p-doped means NA, right? Lots of acceptors. Then I would have the Fermi level close to the valence band, okay? And I have shown you deep blue, lots of holes, very few electrons, I haven't drawn that. I should but I haven't drawn that. Now when these two materials are in isolation not talking to each other then this is where things are. Fermi level is flat, vacuum level is sitting wherever it has to and that must be the same vacuum level if both sides are the same. Both sides silicon, same vacuum level. If one side is silicon and other side germanium, a different material, of course the band will be different and the vacuum level will also be different, right? The amount of energy you need to eject electron from the germanium, of course is not the same as it is in silicon. So it might be different but for the time being, let me assume that they are the same. Now if I bring these two material close together then one thing you can say that one side has a lot of electron and the other side has a very few electron. So they will try to come to this side by diffusion and similarly one side has a lot of holes, the other side doesn't have enough holes. So it will try to go to the other side. But in principle, if you bring these two material together, if you tried it in your lab, just bringing n-type material and p-type material together, nothing will happen. Actually nothing will happen because these surface states that we talked about, there are actually in every material lots of surface states and when you bring the normal two material together, then the surface states will eat up all the extra electrons. There will be no electron transfer from one side to another. That is why you cannot simply bring an n-material and a p-material and sandwich them together and to get a p-n junction there. That's not that easy. Then it would have been done many years ago. Rather, you need to diffuse things out so the junction doesn't have too much defects and only then that's what we did, right? Diffuse atoms in and therefore the junction doesn't have too much defect only then interesting things will happen as we'll see. Assume therefore that this is a diffuse junction. The junction has been created and there's no surface defects. Then in that case, what will happen? In the picture above what I have shown is Nd is donor and the squares plus squares are donor atoms fixed in space because donor atoms cannot go around and that's a fixed in space and it has given away an electron and that electron next sitting next to it is in a blue circle with a negative sign. That's the electron. So every donor has given away one electron and therefore here you can see every green square is matched with a blue circle on the left hand side on the donor side. And similarly for the acceptor side you can see that the green squares are negative. These are acceptor atoms sitting fixed in a real space and it has captured one electron. Where did that electron come from? From the valence band and therefore it has left behind one hole in the valence band and those holes are plus with a plus sign with a red circle. Now the circles can move around. Electrons and holes can move around but the green things those cannot move around. Those are atoms. So those will be called space charge because they are fixed in space and therefore those will not be able to move around. Now let's see if I wanted to look at the charge and the doping in this junction. Then how would it look? In the, what I have plotted here is the green. The green line is essentially sorry, the blue line is ND. You see I have a fixed number fixed up to the junction and then going to the other side of the junction this has a different N line. And why it is I'll explain in a second. And similarly if you look at the number of holes from the other side, you can see the holes are more or less fixed under this condition, more or less fixed going from the right, right up to the junction. Then it drops down on the other side and becomes a small value. Now how small? This you already know, don't you? In the extrinsic part, you know there's freeze out and intrinsic but we are thinking about the extrinsic part where all the atoms are ionized or all the donors are ionized. So P is equal to NA. You agree, right? That number of holes is equal to the number of acceptors and what will be the minority carriers in that part? In the acceptor dope part, it will be Ni squared divided by NA. When does this relationship apply? In equilibrium, right? Only in equilibrium. If it's off equilibrium, we shouldn't use this at all because then it doesn't apply in equilibrium. And similarly on the other side you can see the red on the donor side will be Ni squared divided by ND. Then you are done. And the product of these two are the same. So therefore you can see that that's how I have drawn the diagram. That the product of these two values are approximately the same, equal to Ni squared, let's say. Now when you have this junction and there is huge number of holes, red holes sitting on the red circles holes sitting on the right hand side then they will not stay like this. What they will try to do is go to the red circles, we'll try to go to the left side because there are a few holes. So this goes. Now I didn't draw it. I should have just put it on the other side but you see the other side has so many charges that a few holes crossing over will get diluted in this big region. So I have not drawn that. And similarly if you wait a little bit more few of the electrons from the other side will try to get to this side, right? This is diffusion happening. Now why isn't anybody preventing the electrons to come to this side? Is because in the beginning there's no electric field. Everybody whole thing is balanced. The charge of the whole thing was charged neutral. So when a electron wants to come to the right hand side nobody is saying that don't come. Just diffusing out. Now this will continue a little bit. You see this red circles disappearing and the blues are electrons disappearing. Disappearing meaning is sort of coming to the right hand side. I haven't drawn that. But you see then it will stop because when the next series of electrons further down wants to come you can see on the donor side there are these exposed negative acceptor electrons acceptor atoms which are sitting there which says don't come anymore because now you have a bunch of negative charge sort of guarding the junction. And as a result electrons will not be able to come anymore. And the same fate for the holes. You see hole wants to go to the left hand side. There are lots of holes on the right, few on the left. It wants to go but there are all these positive squares sitting on the left hand side of the junction which essentially repel the holes. So therefore a field has built up which will prevent the diffusion from occurring anymore and there will be a nice balance at the end. So if you look at the end concentration what will happen? This is how it will look like. Do you agree with this? The number of electrons which are the circle it sort of stays flat up to a point then it drops down. Drops down because there are fewer holes on the other side and similarly the red holes essentially stay up to a certain point and then it drops down away from the junction. A little bit away from the junction the carrier concentration has dropped down. Now what is the, now this is not really how it happens. What happens is that it will not drop down all of a sudden it has a fixed value and then it goes to zero. That's not going to happen. Actually what happens that it drops down very gently over and similarly for the holes it drops down gradually from the one side to another as the potential has built up. But these approximations where I have sort of approximated with a sharp drop this is called a depletion approximation. Remember this is a log plot. So although it looks like that it is continuously dropping down the first point may be at 10 to the power 18. The next point may be 10 to the power 17 third point 10 to the power 16 by a few tens of angstrom carrier concentration has dropped by 100 factor of 100 1000. And we'll see from one side of the junction to the other side carrier concentration may drop by 12 orders of magnitude. So therefore this step like picture is completely appropriate and this is called a depletion approximation because you assume that as soon as the depletion region starts the carrier simply vanishes. Let's see think about the charges a little bit more that what is the net charge? I have talked about the individual pieces of charges. What is the net charge? Now the green I have shown is the donor atoms, right? The square green squares and you can see the donor's atoms stay flat up to the junction. And then it goes to zero on the other side. Why on the other side I don't have any donor, right? So it goes to zero. Now what about electrons? Electrons from the left hand side. Do you agree with this? That electron was flat up to a point and then it dropped because it has diffused away to the other side and it has stayed on a low value on this. What is the net charge in this? The net charge you can see will be QND approximately, approximately. Because in one case up to that depletion region beginning of the depletion region N is perfectly balanced by ND, right? Is that right? And therefore you have zero net charge. So therefore I haven't drawn anything up to that point. And then when you subtract then the subtraction from the green to the blue in the depletion region is essentially equal to ND because most of the electrons have gone away from that region. Now this region is called a space charge region. Why? Because the charges are fixed in space and they cannot move around and the space charge will essentially dictate the electrostatics. Now what about on the right hand side? The holes have gone to the right hand side so there should be a little bit of charge on the right hand side also, right? If you subtract the green from the blue you have that, that's true but again the argument that it is very small. Why? Because on this side it is ni squared divided by na. If na is 10 to the power 18 the minority carriers is about 100 in silicon. And so in comparison we will not draw the blue part. We'll say that okay, this is too small for us to worry about it. Let's worry about the main charges. So we will not draw anything on this side. Now what about the other side? I have just told you about the donor. Acceptor must do the same, right? And so, oh by the way, so this region we'll call it xn, the depletion region in the n side. And there is this bulk region, bulk region is charge neutral. Although there's lots of electrons and lots of holes but individually they are charge neutral. And this you can easily see what will happen to the other side. In the acceptor side similarly you have a certain value and similarly you have the whole concentration, whole concentration depleting away from the junction. And this time around however, because of the sign the net charge will be in the other direction. You see that because na is, na is negative, right? Na is negative and I have depleted a bunch of holes close to the junction. So when I subtract look at the net charge the rate should go in the opposite direction. So that's it. After the charge exchange, then this is the net charge I see in the junction and that will govern everything about this diode, this little charge is going to govern this. One thing you realize that the area under the blue and area under the red, they have to be the same, right? Because of the charge balance. They have to be the same because if it were not the same, you see there'll be a net electric field coming out of this region. If you allow Gauss's law and take slices, if these two pieces were not the same, it will become the same. Until it becomes the same it will not stop. So this is a very important statement that even when you have a device non-uniform at the end the whole sample as a whole is charged neutral, right? And that's the statement. The area under the blue and area under the red are exactly the same. And that's QNA. That's again an approximation. These are all depletion approximation happens because you assume a sharp drop. You will do a homework where you don't assume it but that's the difficult problem that you get to solve. And this is called a depleted region, right? Depleted not of all charges. Depleted just of mobile charges. And this is also called a space charge region because these fixed charges are fixed in space and that's what you have. Now I'm going to show you a bunch of things. You just pay attention. This is very easy, but you have to pay attention. So let's say this is the charge I have that I have gotten from the PN junction. You can see at various places. I'm sorry, this LPL fixed it. It should be XP on the bottom and XN, but I'll fix that. If you have the charge, do you know how to calculate the electric field? You know that. Very under this curve, up to a given X is the electric field. This is the electric field. Because you have a rectangle for the charge in the bottom, as you start integrating from the left hand side, gradually come in, every new slice adds a constant amount. So an integration of a constant function becomes a straight line in the electric field plot. And because there is net charge on the other side, negative charge on the other side, once you begin to cross the junction and begin to integrate, this will subtract because there's a plus charge for the blue and negative charge for the red and this will gradually subtract. And do you realize that eventually when you are end of the junction, sorry, this point should be here. I'm sorry. So when you are end of the junction, then it should be equal to zero. Now, this statement where the blue and the red are the same in the junction, that only happens for a very special case. I'll explain in a second. If both sides are silicon, but if one side is silicon and another side is aluminum, you cannot make that blue electric field and the red electric field continuous at the junction. I'll explain in a second. But the electric field, the maximum electric field, how will you calculate that? Well, you will calculate it as the area under the curve, area under the blue curve. How much is that? You will have NA, sorry, ND multiplied by XN, right? That's the area under the curve, distance multiplied by height and you have to divide by this epsilon so that you get the electric field, right? That's the charge and you will get the electric field and similarly for the red, you will also do the area under the curve for the red and you can see how much is the area under the curve on the red, XN and, sorry, NA and XP, right? These two, and you divide by the epsilon in order to get the electric field. So that's the peak electric field that you can easily calculate. Now in the book and in all the other places, you are supposed to do a lot of integration. If you have to do this integration every time, you will not solve any problem. It is better just look at the geometry of the problem and you should be able to draw this in five seconds rather than going through this individual integration. When you need the integration, you do the integration but for this type of simple problem, you should be able to do it in single shot. I have the electric field. How do I get the potential? Area under the curve. But area under the curve for this, again, up to a given X so I can look at the electric field that way, right? Area under the curve. It doesn't go to any zero anymore because area under this curve, blue and red curve for the electric field, of course, that will be a net positive quantity and how is potential related to the electric field? With a derivative and a minus sign, is that right? And so I have drawn the potential below zero, right? These three pieces are nothing, right? You know this, you have done this many times. Now, if you have the potential that way, I haven't solved the problem, I'm getting there. We'll call this arbitrarily VBI and it will be called a built-in potential. I don't know how to calculate it, I will find out but for the time being, we'll call it VBI, BI for built-in potential because the field exists in the absence of any battery are connected outside. So it's a field inside, built in, that's why it's called a built-in potential. Now, once you know this potential, can you draw the band diagram? Of course, because EC will go with V with a negative sign. So I have drawn the EC, you can see and then correspondingly the EV with a flip of the sign and you can see now how this device is not like the ones that we have done before. In the bulk region on the far to the left, it looks like a typical N device. In the bulk region far to the right, it looks like a typical P device, donor dope acceptor dope device, it's when something is happening. But the issue here is that this line which is like an arrow, this is the Fermi level. In equilibrium, Fermi level must be flat. That's the statement. And then off equilibrium we'll see how it comes about. Now let's talk about how to draw a band diagram. How can we get to the last result without really having to go through all this? How can I do that directly? So I have a, so I want to first show you a general way of doing it. Then we'll get to specific methods. So you remember that I have this P and N region. There's this depletion region. And the first thing I have to draw, first thing when you have a blank sheet of paper, first step one is to draw a horizontal straight line. And that's the Fermi level. Don't have to draw anything else. First is straight line through your page. That is Fermi level because you know in equilibrium, Fermi level must be flat. So that's number one. Just follow these four rules. You'll never be wrong for any device that will come in your life. The second one is that get to the bulk region, far away from the junction. And you know in the bulk region, if it's N equals ND, then where the Fermi level has to be. You know that, right? All of you know that. And so therefore I have drawn the blue conduction band because it's donor. So I'll first draw the conduction band. If it is our acceptor, then I'll first draw the valence band. So I've drawn the conduction band in the donor first. The next what I have to do, if I know what the material is, of course, let's say silicon, I know the band gap. So I will draw the valence band line. Okay, that's one side taken care of in the bulk region. What about the other side? I already know EF, I've drawn it on the piece of paper. And the first thing I should draw in the P region, which one did it come? Okay, oh, and that's the chi one is that solid black line. That's the vacuum level. So I also have to draw the vacuum level first. Okay, now correspondingly on the P side, I'll have to do this. First I have to draw the valence band. Do you realize why? Because it's P region. I can immediately get where the valence band is. Now I also know the band gap for this material. And if I know the band gap, then I can draw the conduction band because that's separated by that amount. Next thing is that for that material, whatever that material is, I can also draw the vacuum level because vacuum level is a property of a material. You open your textbook or any reference book, they will tell you for germanium, what's the vacuum level? For silicon, what's the vacuum level? So you just read it off from a textbook or from a reference book. Okay, now the next step is making the vacuum level continuous. So you see on the top side, I have this red-carved line and the blue line. It doesn't have to be exact. I just make the vacuum level from one side, go to the other side without any discontinuity in it. It could be completely wrong, but just draw a line which will go from one side to the other. Okay, now the next step is this. Let's focus on the left hand side of the conduction band, left for the donor side. Whatever vacuum level line you have drawn, you just copy and paste like this. You see, the whole region is chi-1. Do you see this? The whole region, you have parallelly shifted it down, keeping the chi-1 the same throughout. And similarly, you copy it down all the way down to the valence band. How do I know where should I end? Because it's continuous with EG, the band gap I have. I should come and stop at the junction. Same thing on the right-hand side. Copy the blue, copy the red, keep the chi-2 distance the same and bring it down here. Why have I done it for the valence band first? Because it's a P-type material. So I have to do the valence band first. And then, keeping the band gap, I have, I will bring down this. You're actually done. This is the solution of the Poisson equation in such a complex material. One side could be silicon, other side could be germanium. It could be any material that you can think of anyway. This is the rule. If you follow this rule, you will never, for any material, you will go wrong. But you have to follow the rule. If you try to memorize this, for a combination of material, inevitably, you cannot make it work. Now, these material, this will be called a type 2 material, but that's a different thing. For many systems, there are this type of strange junctions that form. Now, let me then talk about how to calculate the built-in potential. I have the solution already, right? I have drawn a graph and I have solved the Poisson equation. Now, let's see whether you agree with this statement. Maybe I try one more time. That in equilibrium, if you say that you want to calculate QVBI, but you realize in equilibrium, Fermi level is flat throughout. And if I have delta 1 plus chi 1 plus QVBI, that height must be equal to chi 2 plus EG2 minus delta 2. Do you see that? It is because, sorry. So this comes in because it completes the circuit and essentially it has the same value, so you can make this statement. This statement is always true. In equilibrium, this statement is always true. That on one side you go to the vacuum level, from another side you go to the vacuum level. EG2 brings you down too much. And so with the delta 2, you get back where you have to be. Do you know delta 2 and delta 1? Of course, because as soon as you know the doping, you know how far away the EF is with respect to EC, so therefore you know that. You know chi 1 and chi 2, right? Your reference book told you so. So you know that. The only thing you don't know, well EG2 you also know. The only thing you don't know is QVBI. So you're calculating QVBI. Now I want to mention one point. Here I have done one junction. If you had 25 junctions, this rule will always hold true. You will have to just take the leftmost junction, leftmost point and the rightmost point. In between anything can happen. That's fine. But this is giving you, remember just in Schrodinger equation, I have minus infinity and plus infinity for boundary conditions. This is the same way. QVBI for any complicated material is the same. So long you look at the first region and the last region. This statement will always hold true, right? Okay. So this one will have that. And you can calculate what QVBI is. Delta two, you know the accept adhesion. Nv is the effective density of state for the holes, for valence band and you calculate these values. You know all this. And from this, you know all these quantities. Therefore, you know QVBI. Well, once you know QVBI, you have actually solved the problem. Another thing about interfaces. Remember, displacement is continuous. If you have the same material on both sides, then the electric field as well is continuous. But if you have two different material with two different dielectric constant, then electric field is not continuous. It is only the displacement is continuous. So be careful when you draw the diagram, whether it's continuous or not, that will depend on whether both sides is silicon or not. If it's different material, then you make sure that there is a discontinuity. Make sure of that. When you know, this is the basic elementary statement that Kappa one and Kappa two essentially are dielectric constant, relative dielectric constant of the two sides and you take care of that. Okay, if the both sides are the same, then what's going to happen? Then you follow the rules one more time. Draw the first Fermi level. Draw the conduction and valence band for silicon on one side. Draw the chi one is this vacuum level and correspondingly draw the valence band on the other side and draw the chi one. Now both sides have chi one, do you see? Both side because same material, homo junction. And then again, you will make it continuous and you will copy. Now do you see in this time, the junction will be continuous. Why? Because both side is chi one. And so when they will copy them, they will come together without having a discontinuity in. So homo junction doesn't have this. Now if you're an undergraduate student, you will just learn about this part. You know, in three or five and others, just a continuous line. But this is underneath is hiding all these rules. They don't teach you there, but you must know now as a graduate student. Now in this case, of course, you can put the original formula in, original formula is valid, but now chi one is equal to chi two, right? So that will get away. Chi one is equal to chi two. And then you see that nv and c e to the power e g over kt, that expression in the denominator. What is that equal to? Ni square. And so therefore, so long you know, n a and n d, you are done. You can calculate the built-in voltage. Okay, now we'll solve it. Solve this equation properly and see how it goes. Let's try to see how this equation would proceed, how the solution would proceed. In the first case, I have just drawn the previous picture that we sketched out for the charge, for the electric field and for the potential. I have just flipped it around, drawn the charge in the top and have gone down onward. Now the electric field first talk about, let's talk about the electric field. The electric field in a junction, do you agree that the electric field at zero minus and electric field at zero plus is given by charge divided by the dielectric constant, n d and xn, there's a total charge. And similarly, na and xp, there's a total charge on the red side, right? These two must be equal. Now if both sides are silicon and only if they are the same material, then dielectric constants are both sides are the same. And as a result, I don't have any dielectric constants. So in this equation, how many unknowns do I have? I have unknowns of xn and xp, n d and na I know, okay? So I need another equation. Step down one more time. QVBI, how much is QVBI? Well QVBI is the area under the curve under the electric field. And what is the area under the curve? The blue strangle, it will be half multiplied by the height multiplied by the base. That's the first term on the QVBI on the right hand side. You see this half for the triangle xn and E0 minus. And similarly, E0 plus xp divided by two, because that's the red triangle. And when you sum it up E0 minus and E0 E0 plus, you have already calculated on the top, you put it in. Am I done? Actually I am done because you already know QVBI. I calculated it two seconds ago. Nd and na you know. How many unknowns, how many equations? Two unknowns, xn and xp, two equations, I'm done. Now you can solve for it. This is even I can solve, xn and xp is done. And you can see this QVBI where all the chi and other things are hiding, the band gap and other things are hiding. This is where the quantum mechanics is na and nd. So all the quantum mechanics of this whole problem is hiding in the built-in voltage. That's why it is. And you can get an xn and xp. As soon as you know xn and xp, you can calculate the peak electric field, right? Because it is nd multiplied by xn. You can calculate the peak electric field. You can do all other things that is necessary, right? From here. This formula is sort of easy to remember because you remember the denominator, is na plus nd is the same for both cases. For xp, the numerator is nd. So for p, it's sort of the other one is there and then correspondingly na goes in the denominator. So many times it's easy to remember this for quick calculation. But you don't have to. You can easily calculate this. Now, would you solve this problem of xn and xp if the two sides are not the same? Will you do it for the homework? I mean homework means it's not real homework. Means you should practice it to see you understand. If you really need to calculate the electric field at a given point and that is where I will end today, then you should in principle, you can write out the electric field at various point. That's the charge. Charge is equal to ddx and you can integrate between these two points to get the electric field at any point. That's what is done in the textbook. And you can get the electric field at any point. And similarly, you can integrate on the other side between any point zero and x and calculate the corresponding electric field. You can do the whole integral and find out the exact shape of the potential. But the essence of the problem has already been solved because now you know how the potential will proceed. So the next topic we want to discuss is the band diagram with applied bias. And we have so far solved the equilibrium problem and equilibrium we realize that there's no current flow. So the only problem you had to solve was just solving the Poisson equation. And that gave us the band diagram essentially tells us how the Poisson equation is solved. And as a result, how the potential varies in space. Now the question is that a material without bias or without a battery connected to it is not a very interesting device. And the question is what happens when I apply a battery next to the terminals of PN junction diode. Now when you do that, the situation is no longer in equilibrium or device is no longer in equilibrium. So from now to probably next several slides maybe even a couple of lectures. Today we'll get started but next couple of lectures will be on DC characteristics of diode. That once you apply a bias, how current flows, what is the magnitude? What is the physics? It's although it's a very simple device, you'll find that it's actually a very complex and yet very beautiful physics hides in this very simple device. So we'll see. Now one thing people immediately notice once they apply a bias across a PN junction that obviously current flows in this system and the current flows in one direction, the current flows very easily. And I have plotted the logarithmic of the current I through the diode as a function of the applied bias VA and the current flows very easily because there's an exponential rise in the current. On the other side and this we'll be calling forward bias. In the next slide I'll explain what do I mean by forward bias? But in forward bias current flows very easily, current increases exponentially. But in contrast to a simple register, when you take a register and apply a battery across it, you find that the current flows equally well or equal as easily in the positive direction or if you flip the battery in the negative direction it flows as simply or as easily. Unfortunately or fortunately for this matter, for a diode, for a diode, okay. So in the forward bias let me just suggest that although it's a logarithmic increase in the current you will see that it is not a simple logarithmic curve. Had it been a simple logarithmic curve then log I would have increased with applied voltage as a straight line, as a simple straight line. But you can see that there are curves, twists and turns that I have leveled as one, two, three, and six and seven at these various regions. And these various regions are sort of very interesting. It contains within them a lot of physics that as a graduate student we need to know. Now on the reverse side what is interesting that if you switch the terminals of the battery same diode connected on the other side then we see that there is still a current flow but first of all the current flow is typically much, much smaller. And then it doesn't really look like the forward bias current because forward bias current is shown here in green region and you see that that has it characteristics which is not reflected, reflected in the reverse bias the ash colored region because you see the first current increases it remains for most flat for quite a bit and then all of a sudden without any apparent cause looks like that the current simply becomes very, very large all of a sudden. And again, this part will be called reverse bias I'll explain it in a second but the main point I wanted to make in this slide that the various regions associated with this current flow asymmetric current flow is what we are after we want to know why it happens and what is the physics behind it and there's a long list for the regions, various regions and physics that governs that transport in that region we'll go through them one at a time so you don't worry about it we'll understand them all. Now had you been an undergraduate student only thing you had to worry about was number one, the diffusion limited transport the rest from two to seven well that's not something you would have worried about had you been an undergraduate class but as a graduate student you really need to know all of them because this will be the basis this diode will be the basis of a MOSFET will be a basis of a bipolar junction transistor all the devices that comes in you will see that this is a series of PN junctions in various configurations so as soon as we understand one PN junction the physics of it what happens under various configurations the other devices you will see will just cut and paste various concepts in various places and we'll all be done now the definition of a forward and a reverse bias how do I know which side is sort of for current easy to flow and which side it is not let me first define it and we'll understand the physics a little bit later the simplest way to remember what forward bias is is to remember that you should connect the P side of the junction P doped side what is P doped? accepted doped side of the junction to the P terminal of the battery P to P and similarly the N side of the junction should be connected to the N terminal of the battery and you can see the large for the battery I have shown with this a small vertical line and a large vertical line you can see that and the large vertical line is essentially the P terminal of the battery so this is forward biased and in this case current flows very easily and what is reverse bias? Well reverse bias is the opposite that is that if I take the N side of the semiconductor and connect it to the P terminal of the battery then this would be called a reverse bias diode current doesn't flow very easily in the reverse bias diode and this is asymmetric property is a central thing you see without this we couldn't take an air wave let's say you are hearing NPR in the morning national public radio the reason let's say you want to use this type of non-linear device is you want to demodulate the signal and the way you demodulate the signal that's that would might I might explain it down the road but the point is without this asymmetry you might not be able to demodulate the signal so it's very important actually for practical applications so again anytime we have a device we again go time and again to the same set of equations and in the same set of equations we solve it so first is we would like to first solve for the band diagram solve the Poisson equation but this time the difference from wave previously is that I have a battery connected to it equilibrium didn't have a battery now I have a battery so I'd like to draw the band diagram in the presence of the battery I'll have a few more rules the four rules before you know flat was a Fermi level and all other the conduction and valence band in the right places those will remain the same you should always draw them but after you have applied a bias it will change things a little and I want to show that however we will not be talking about current flow in this class in this class means in this lecture we'll take it up in the next lecture now this is the rule you can see here that this is the band diagram in equilibrium how do you know this is a band diagram in equilibrium is because the quasi Fermi or the Fermi level E sub F is flat throughout and you know that when the quasi Fermi level or Fermi level is flat throughout then there is no current flow in the system now I didn't go through all the intermediate steps that I had to go through but this is a homo junction that both sides of the diode are silicon right and in that case you you saw that the conduction and valence band they essentially line up each other with each other in general of course that's not the case right if you had a silicon in one side and germanium in another then in that case of course there will be this notches in the conduction band in the valence band so here I am taking a simple picture but behind behind this of course there are all those rules by which to construct the band diagram so you should always remember that don't draw this blindly but let's say I am thinking about a homo junction that both sides have the same material in equilibrium now if I apply a bias for this device that is connect a battery to the two terminal how should I modify the band diagram so that the Poisson equation is solved in the presence of the battery that's what I want to know and the rules are pretty simple actually by the way before I go in and solve that problem let me first ask you this question what is the maximum value of QVBI approximately that one can have you know QVBI we calculated right with the electron going up all the way to the vacuum level from one side and the electron going up all the way from the vacuum level on the other side we did the math but physically what is the maximum value you can have you see typically the maximum value you will have is when EC is getting closer to EF because you remember that in a degenerate semiconductor it's very difficult to push EF above EC so approximately I should say the EC minus EF getting close to zero is sort of on one side how close the Fermi level can be to the conduction band similarly on the other side the Fermi level can be approximately be equal to the valence band you know a little bit more a little bit less so what is the VBI then do you realize that it is essentially equal to the band gap so when you have a new semiconductor and you just want to quickly solve for a problem and you don't have the effective density of state this that and everything you can quickly get a QVBI simply by equating it equal to the band gap okay now next question is VBI is built in potential in equilibrium now I want to apply a bias the first rule for applying a bias is that you have to take any one of the material in the device that you are interested in and ground it without grounding you do not have a reference so what you see on the right hand side with a blue rectangle is my ground terminal so I am saying that I am going to ground the P side of the device now I don't have to I could easily equally well grounded the N side any side is fine but then you have to be consistent so I ground it on the P side and keep the Fermi level for the P side in the same place where it was before in the top diagram so the same place it stays in the same place now I have applied a bias so therefore applied a let's say a negative bias on the negative terminal of the battery to the N side forward bias do you remember negative terminal to the N side and so if I have applied a bias V minus V then the electron energy will go up by that amount it will go up by that amount because the electron essentially will not like to stay in the negative part so it will go up if I applied a positive bias then it would go down by the same amount now the next thing is to realize this following statements first focus on the right hand side of the valence band and the Fermi level you know the difference between the Fermi level and the valence band must stay the same why? because it's a bulk region the number of holes that you have must be equal to the number of donors or number of acceptors and the only way they can be the same is if the gap between the valence band and the Fermi level remains exactly why it was before similarly on the left hand side in the negative side although I have pushed up the Fermi level that is the red dotted line but the gap between the red line and the black conduction band must remain the same because N the number of electrons must be equal to Nd or the donors even in the presence of bias as a result what's going to happen the gap between them must remain the same except that whole conduction band on the N side will be pushed up a little bit now you can see that there's something a big change here because the barrier which is preventing an electron to go to the other side is no longer QVBI but it has been reduced by the applied voltage V and the current might be able to flow from one side to another easily so this rule this extra rule of drawing diagram is something you have to remember but there is another tricky piece which I want to mention next you see I have not drawn the red red dotted line and the blue dotted line all the way to the other side you may notice that the rule is and a rule that I'm going to explain to you later is that you have to stop the red line must start from the N side and then continue all the way to the other side of the junction and then stop similarly for the blue one you start from the P side you know whatever is the dominant side defining the Fermi level you continue all the way and on the other side of the junction then you stop beyond that point you don't continue how to calculate the Fermi levels beyond that point is something I will show you later but for the time being we'll stop at the end of the junction going to the other side and if you remember their homework problems if you go back in the recombination generation problem that I assigned you will see that this is exactly how they have drawn the diagram now one another thing you will see here that instead of writing EF I have written F sub P as the alternate symbol and this will be called a quasi-Fermi level so this is the Fermi level in the presence of a bias so the F sub P is either called a quasi-Fermi level or IMMREF Fermi spelled backward and similarly on the other side you also see an FN in a state of EF because EF is a property of the equilibrium it's no longer in equilibrium I have to reply the bias as a result I will have this modifications another rule is that when things are out of equilibrium then N at any position X is not simply Ni e to the power EF minus EI multiplied by beta beta is what is beta? 1 over kT so I'm writing beta as a short form and instead when once you have EF then once you have once the device is non-equilibrium then in that case EF must be replaced by F sub N because that's the new name and similarly for holes which is PX we'll write it in the blue you can see F sub P which is the replacement of EF for the holes and FN and FPR of course not equal to each other anymore in equilibrium they were they were both equal to EF no longer and a very interesting consequence of that is that when you multiply N with the P any point in the device then you realize that it is no longer simply equal to Ni square Ni square was the rule for equilibrium and what happened at that time the Fermi level that you had on both sides they cancelled now of course FN and FPR not equal to each other and so when you multiply this will not continue and therefore this difference will multiply the Ni square this is a rule that we will need later on now let's calculate some depletion width now these things will become very easy now because you know when you apply have a device you have applied a negative bias to that end side I want to know how XN and XP the depleted part which is depleted of all mobile charges right space charges are all there all there how that changes with in the presence of bias so first of all you again say the same rule X Nd multiplied by XN is equal to Na multiplied by XP do you remember that this is only for homo junction because if you had a hetro junction I would require that the displacement to be continuous so this is one rule and this is the second rule that for the potential the area under the triangle will have the same same issues as before only thing in the equilibrium case only thing that I have changed is the built-in potential now in the presence of bias is not QVBI but rather QVBI minus V because my barrier has gone down by a certain amount now do you realize that why current would flow in that case you understand that the current would simply flow because previously diffusion and drift were exactly balancing each other right lots of electron on one side very few little electron on the other side they want to go in one direction and the electric field built up because of the depleted charge they pushed it back it's like a having a river lots of water then then a bank of the river which is a potential barrier so although you have lots of water in the river and no water in the dry part of course by diffusion it would want to come but because the bank of the river essentially it's preventing the water and keeping it confined when you put a bias then it destroys this equilibrium and as a result diffusion now wins over because the electric field has gone down right same distance but QVBI minus V is a reduced barrier so therefore the diffusion current will dominate over and as a result current flow will stop okay but you have this two relationship and as I mentioned you can solve this problem Xn and Xp and that will give you this expressions that instead of QVBI or VBI you will have a VBI minus V now if you have a forward bias in this case will Xn and Xp be smaller or larger do you think if V is going up then VBI minus V will be smaller value and so therefore Xn and Xp will become smaller so depletion region becomes smaller in forward bias this same analysis also applies for the reverse bias in the reverse bias case the V will be minus V and so as a result the Xn and Xp will get get significantly larger in the reverse bias again I wanted to point out that it is only for the homo junction that you have the dielectric constant on the both terms the same if I give you a problem for heterojunctions I hope you will be able to solve that problem as well do it just for when you have a few extra moments very quickly to show how this works it's not very it's not to scale very much but it sort of gives you the idea so if you are in equilibrium let's say you have a certain amount of charge a certain amount of electric field by integrating the rectangle and a certain amount of potential integrating the triangle that you have the red and blue triangle if you apply a forward bias then essentially it shrinks the depletion region and as it shrinks the depletion region you have a smaller amount of charge do you realize the electric field will also be smaller the triangle has gone down because you are integrating a smaller area so the electric field goes down when you forward bias a junction and as a result of course the barrier also goes down and then the reverse bias side so it essentially gets widened out more depletion and when you have more depletion what happens the electric field also goes up that's very important because we will see in the reverse bias side remember the one part where the current was increasing tremendously with a high bias that is because that high electric field there will be impact ionization in the diode and as a result that's the physics of that part so we'll do all this a little bit later okay so to conclude this part this lecture we just learned how to draw a band diagram and this is one of the most important topic if you haven't learned how to draw a band diagram you actually have wasted your time in this course band diagram is something you must learn everything else is secondary you should I should have told you that before right and if you consistently follow the rules for drawing the diagram you will never go wrong but if you try to memorize then you're certainly will go wrong and this is something I want you to follow the rules even when you think you know the answer in that case follow the rules that will save you a lot of trouble down the road