 you can follow along with this presentation using printed slides from the nano hub visit www.nano hub.org and download the PDF file containing the slides for this presentation print them out and turn each page when you hear the following sound enjoy the show and we'll continue to discuss the notion of energy bands but hopefully today we'll be talking about real crystals I shouldn't use the word real because what we had been discussing in the past few days actually those concepts also apply to one dimensional and two dimensional crystals as well which are of course as real but for three dimensions which is most widely used our discussion today part relates to that three-dimensional crystals so we'll talk about EK diagram and constant energy surfaces two concepts that we have come to learn in last class in lecture six talk about how to characterize EK diagram that is how to measure them you see we are learning all sorts of complicated theory but in practice often you'll have to go to the lab and make this measurement and these two notions bandgap and effective mass at the end of the day all this complicated calculation that we are doing Schrodinger equations and EK diagram in various directions and all others these two are the only concept that we'll have to carry forward in order to understand transistors MOSFERS and other things so out of all this that will be the final distilled concepts that we'll have to carry through remember the original problem once again I want to come back to this often so that you don't forget in the math the cloud this this dense math of different things the main purpose and the main purpose being the number of electrons that we calculate free electrons that we calculate simply by multiplying the number of atoms with number of electrons per atom that do not really support the conductivity measurements and as a result we are trying to see where the electrons sit and why they do not all contribute equally now you already know the answer a little bit right because remember the wall bands that are full although they may have a lot of electrons sitting there they carry zero current so you can see immediately the huge fraction of the electrons although they are there they are really not participating in a conduction so we shouldn't really think about them so you can begin to see the outline of why the conductivity of different material depending on their band structure could be quite different because the fraction of electrons available for conduction now let's start with 3d again repeat the process exactly that way we did it for 1 and 2d this is a let's say a real-space cubic lattice and only real-life material that has this cubic lattice is you remember polonium right so that has only that one of course silicon and gallium arsenide don't have this structure but any case you can get started here if this is the real space how should I construct the reciprocal space well a is the lattice spacing all three sides and so we should make it 2 pi over a right 4 pi over a 6 pi over a in the positive direction and of course in the negative direction minus 2 pi over a minus 4 pi over a x y and z in 3 directions and once we do that we'll get a corresponding lattice but now in the wave vector space for the three-dimensional crystal as well now of course you realize this is going in all three directions of course but cubic lattice so they are equally spaced in this particular case had it been a rectangular type lattice parallel piped in that case this would be slightly different again how do you make the brilliance on you will have to do the Wigner site algorithm but on the reciprocal lattice you know that and that gives you essentially one this a section which goes from 0 to plus pi over a and 0 to minus pi over a in each side right and in three-dimension so it goes in all three dimensions so I have a little cube set centered on one of the points sort of bisecting by planes which are perpendicular to the connectors to the neighboring points so that's my brilliance one for cubic lattice now one thing I want to mention here that although I have drawn this entire cube and we know that this is where all the solution would reside right you remember that all the equations all the energy levels that we talk about are going to stay within this region within this k vectors now one point is that you remember when we classified the crystal we classified them in seven types this monoclinic cube they remember there's seven types now remember at that point I say that the cube has the maximum degree of symmetry you can rotate it 90 degrees and still you get the same thing you can reflect it around the mirrored plane get the same thing you can invert it take a point in R and just bring it to minus R then you get the same thing in fact there are 48 operations you can do on a cube that gives you back the same thing so that says actually you just need 148th of the Brillouin zone to describe your whole solution because of the degree of symmetry just like in 1v you could just go with 0 to pi over a that is all the information you need to keep right because just in that one-dimensional plane you just have mirror reflection but here you have 48 reflections or 48 symmetries so therefore in this case you just need 148 so even this you do not need just 148 of a little wedge is all that information that you need okay now in real life of course we said that the carbon diamond lattice in silicon or germanium or gallium arsenide then in that case this becomes a face-centered lattice with if I combine the corner point and the one-fourth down the diagonal atom right do you remember those two if I form a basis then it becomes a face-centered face-centered lattice and you realize by doing the homework that if I take the left-hand corner point the blue left-hand corner point then connecting it to let's say six three and two those are the lattice vectors for the primitive cell the homework one that you just did and so you can see the reciprocal lattice in this case which will not going to calculate but you could if you just use that algorithm it will not be simply 2 pi over a because a is not the primitive cell remember the a is sort of a unit cell but it's not the primitive cell so in this case the from you see going from the corner point to three this is be related to a over 2 something like this and so therefore the corresponding reciprocal lattice you do if you do it will not look like the cubic one you see it will look like a little bit more complicated now the first thing you notice that the age of this is not pi over a to minus pi over a because the atoms are actually closer together it's not pi a apart in general as a result the first thing you will notice is it is as if it is pi over a over 2 pi over a over 2 because the a over 2 is actually close to the spacing point and so the age of the brilliant zone in this case is 2 pi over a not pi over a as we have seen before and then this looks a little complicated that's not really a major point but I want to point out a few things here and that you will get clarified a little bit later also the center point the center of the brilliant zone in this case has a name gamma point so we'll say that and if you go in the kx k y direction you will hit a square rhombus surface on one of the one of the planes remember they are just doing Wigner site algorithm based on the inverse lattice so therefore you see all these planes and the point where it hits it is called a x point you can see here x point and how many x point should I have I should have six because I have six faces and so therefore I am going to the six sides I have six x points now if I go along the diagonal if I go along the diagonal I will hit a hexagonal face how many diagonals do I have going from the center well I have eight is that right because I have eight corners and if I start from the center and trying to join them eight car eight lines and therefore I have eight points so I have is that right one two yes so I have eight and the distance from the gamma point the center point is point eight seven two pi over a does it look complicated actually not really this is square root of three over two right point eight seven square root of three over two why did the square root of three come because you have in order to go along the diagonal body diagonal you have one x side one y side and one z side that gives you square root of three but you are going from the center to the end not the end twin so therefore you have square root of three over two that's point itself okay and then there are this all sorts of other points going from gamma to x halfway in between you call it delta and all sorts of other points you will do a complicated homework next time and when you do that you will learn all this very clearly about how to use them actually now you are learning how to define them but how to use them you will see this is an important point that people often miss that for this lattice it is not pi over a to the end but 2 pi over okay now I haven't really solved the problem I'm saying solution reside in this brilliant zone so that means that every point I'll have to see whether there is a solution or not that's coming next now before I get there let me just give you an analogy so that you can understand this complicated diagram you know many times we see pictures like this in children's books like this is from my daughter's daughter's book she is in fifth grade or fourth grade and this is a picture that I took she's learning about art and other things and so what is plotted here is a four-dimensional information in some way density at every point as a function of x y and z in fact the solution we are going to look into energy as a function of kx ky kz is very similar to that four-dimensional information so if I could actually understand how to represent this information in a simpler way then that I will apply also to the ek diagram how would I do that well one simple thing to say that this is a very difficult thing to draw and also it's difficult to read off even so one thing I could do I could take a theta theta one and phi one and I could take a cut along that line and I could say density as a function of r as a function of the radial coordinate changes in a particular way along that theta one and phi one now if I represented this theta one by a set of angles right and phi ones and also by a set of angles so take this cut along a whole series of lines in that case I could plot let's say 50 of them and by then two-dimensional projections I could essentially capture the information that I have in the four-dimensional plane right you see this now you see the problem is much sigma or less complicated much less complicated because remember I just said 148th of the whole thing because of the symmetry is all you care about so in fact you wouldn't need a 100 cuts only a small piece if you could represent it by these two-dimensional cuts that's all you care about okay now let's see this and the next slide how it how it works out in real crystals so again just very quickly so this space just like XYZ as the three coordinates over there kx ky and kz are the three coordinates and we just like here we are looking for density on the other one we'll be looking for energy that energy at every point if there is a band gap there will be no energy in that kx ky kz okay so now just like I said because it's a complicated thing it's difficult to draw that picture four-dimensional picture over there we would rather take a cut along a line and plot it so our first plot I am showing you is going from gamma to x point along that line and along that line you see on the right hand side for germanium I have shown you also a corresponding solution of the Schrodinger equation from gamma to x point now you'll notice that below I have written 100 why is it I say 100 here it's of course 100 direction but didn't we define all these things in a real space remember the crystals it was all the planes of the crystals and other things we represented it here of course this is not real space these are reciprocal space but of course vector doesn't matter here 100 is not the same 100 as the other one the other one was 100 perpendicular to real space planes certain x certain y and certain z here it is certain kx certain ky certain kz so these two don't confuse these are not the same thing this is same direction is the same but of course in a different space okay so going from gamma to x I see that I have plotted a bunch of solution now this is not all just like in one dimension I had four bands remember four meaning actually I had quite a few I just plotted four similarly here what they do they have just plotted the ones that are important for electrical engineers there are bands below this there are bands above this and the bands that have been plotted are close to this 3s and 3p levels this one is 2s all those levels those have not been plotted 3s and 3p are the plots that have been shown here you can see a few features for example in the one-dimensional case everything was going first if it's going concave up the next one was concave down and this sort of alternated but you can see here on the bottom side the two bands both are concave down right so in three dimension when you have a complicated material it's no longer necessary that that alternate route should follow similarly you can see that there are other bands and we'll learn about them a little bit later shown here on the this is above 0 so 1e v or 2e v and most of the time we will be only be concerned with the bands between let's say minus 1 to plus 1 or plus 2e v nothing more than that we don't really need anything more for the time being but the point is this is a one section of the solution along gamma x how to use them will come back now let's say I want to draw it in another direction so let's say I want to draw it along that direction gamma to L now that's another direction so I again should have another panel of solutions right in today I'll have another panel and you can see gamma to L shown here in the bottom do you also see one one one written there shouldn't it be one one one because it is going along the body diagonals of course it should be one one one of course the gamma to x and gamma to L has been drawn approximately the same but of course not the same remember 0.87 was the actual direction in along the body diagonal so the point is that these are just representative things when you actually draw calculation or do calculation you will have to remember that all sides are not the same gamma to x and gamma to L they might be different and you can see also the band look very different the solution along the diagonal is actually very different than solution around gamma to the x point we'll use them a little bit later also these characteristic features of how bands look these are essentially all generic always remember the plots that you see is only part of the solution space that are irrelevant for electrical engineers there are other for opto there are people who are going to opto electronics and other fields they might be interested in completely different bands so in that in their book you will see the same thing but focused on a different region let's compare so you can see germanium silicon in the middle gallium arsenide on the top more or less what you see on the bottom side below 0 this this EV V stands for the valence band where most of the electron states are occupied a few holes exist in that case you see all three materials because these have the diamond lattice or zinc blend lattice right they are very similar so their band structure also look very similar other material will have other bands you can see in the valence band side there are two or three bands so here for example there are two or three bands and for all of them more or less the same information and we'll also see for the conduction band side it is more complicated but the only thing I want to emphasize in here is the notion of a band gap do you see that it is the okay so definition of a band gap is the minimum point maximum point in the valence band and the minimum point on the conduction band they may or may not occur at the same wave vector point for example if you look on the gallium arsenide side you will see that let's say the maximum of the valence band is looking like it is at 0 0 EV if you look on the left hand side on the other hand the minimum value for the conduction band also occurs at the gamma point the K equals 0 point and you can see that it occurs since it occurs at the same point therefore this is a direct band gap material why it's called direct I'll come back in a second it is a direct band gap material only light can directly excite this electrons in a gallium arsenide but if you look at silicon or germanium what you will see that the maximum point for the valence band occurs at K equals 0 gamma point you can see that but what does it occur for silicon why does it occur for the minimum occurs for the conduction band it occurs along the X point remember the six faces so it occurs close to that that point and since it is not occurring in the middle panel since it is not occurring at the same K point it is an indirect band gap material similarly germanium would be an indirect band gap material but the lowest point in the in the conduction band is along the L point because you can see from that picture why this written EG and again this germanium will be an indirect band gap material okay so these are some of the features there is something called a split-off valence band there are two valence bands one top of each other you can see and then one is a slightly such as if it has split off from the group and so there is something called a split-off valence band very important the lasers semiconductor lasers that you work with when you talk in telephone most of them actually have to account for that valence band otherwise you cannot design lasers so very important obviously for indium phosphide systems and many other systems for optoelectronic properties now let me move on to the other points now I want to say a few more things about about this gallium arsenide one just as a matter of definition you need to understand the definition clearly because then we'll be just using them often saying oh something is happening in the gamma point there are in the x-point some you have to then remember all these definitions so that there's no confusion definition of the language you can see that there are this gamma point but there is a subscript like gamma 6 and then there are x6 l6 so then this is just to say that the six subscript essentially saying that there are six such bands as you'll see but the more important thing is that they all belong to the same band you can see l6 gamma 6 and x6 they will all belong to the same point and similarly you can see gamma 8 and correspondingly each is sort of just to indicate that they belong to the same band they'll have the same index and in this particular case you can another thing is important to see for gallium arsenide is that the l-point and the x-point for this one is actually very close to the gamma point within about point 2 or point 3 ev now for many 3 5 semiconductors that is true that the gamma point the difference between gamma point and l-point and that of the x-point is actually very similar and that's very important in terms of electron transport properties in about lecture 13 or lecture 14 I'll have to come back and see the implication of this we'll explain the implication of this at that point for the carrier transport okay so that's one way of looking at it right ek diagram where you do the panels the other way to look at it is constant energy surfaces you remember in 2d that's another way of looking at it so for example at a given energy or for example at a constant density these are like contour plots so at a constant density you can say how the kx ky and kz how they essentially change across the constant energy plane and in fact if you wanted to represent the original plot original density plot you do a series of constant and density plots and that will give you an idea about how the original structure in 4 dimension how it was changing so the corresponding one is shown here now this picture and the ek picture are exactly the same we'll have to will do a homework in which we will go between this one and the previous one same picture actually because you are representing this e as a function of kx ky and kz in 2 different ways so of course they are the same so let's focus on the germanium one germanium do you remember i said that the minimum occurs at the l point right in the previous picture and you can see along the 1 1 1 direction and you can see the constant the constant energy surface at low energies essentially they occur along the 1 1 1 direction for silicon why did they occur they occur along the x point and so you see that they have the constant energy surfaces along expand and why did it occur in gallium arsenide on the gamma point these are all conduction bands so therefore we have one band there occurring in the and you can represent it mathematically by those energy versus wave vector relationship and from those you could calculate the effective mass right because effective mass is the second derivative here i show you i and j in fact it will represent what should it represent i could be x y and z so if i wanted to know that how the electron go in the x direction right what should i do i will make i equals x j equals x so it will be a second derivative of energy with respect to kx i can take a derivative from there and i can calculate what the various effective masses are that if it has to go in different directions at what mass should it go very simple right you can see now in case if it is more symmetric like in gallium arsenide you know it's like a little sphere so you can see the effective mass in all three directions should be the same right and there should not be any off diagonal term which is dkx dky if you see if you try to take a second derivative with respect to one and then two so if you take a first derivative with respect to one you see the two and three will go out right then if you try to take another derivative with respect to two even the one will go out so no off diagonal terms do you see this and so you can see the one d effective masses you will have along x direction y direction and z direction they are the same equal to two a no off diagonal component what does it mean of not having an off diagonal component what does it mean it means that if you apply an electric field in x direction electron will not try to go in the y direction if you have off diagonal component generally that's true right if you apply an electric field in x direction electron should go in the x direction but sometimes if you have off diagonal component what will happen you are trying to put in force in x direction but the electrons wave space is such our energy space is such that x direction is blocked so it sort of goes through along a channel in the y direction and so in those cases the transport will be more complicated not in this case not for this material okay now in constant energy surface one thing to notice that let's say this is for which material is it this is for germanium right germanium has along eight of the body diagonals and but one thing you have to be worried about is that you shouldn't just take eight you have to take half of eight four because this constant energy surface is only half of it is inside the brilliance on remember anything inside the brilliance zone other real solutions we are interested in anything outside we don't care about so therefore in this case half of this space you see so this is the ek diagram and this is the corresponding constant energy surface i have l point is the minimum point so i have drawn eight l ellipsoids on the constant energy surface however you can see the minimum occurs exactly at the l point so therefore the minimum of the energy also occurs exactly on the surface so the surface cuts it into two and therefore you have only four effective ellipsoid inside the brilliance zone you see that okay so let's try out another thing where the result is not exactly the same okay now this one for the valence band i was expecting a silicon one but that's fine for the valence band you should check out for silicon silicon is slightly different but for the valence band the valence band can be more complicated you can see that the structure is more complicated and in that case even you can represent the e and a k you cannot do it as simply as it was before and you have a slightly more complicated relationship in fact this can be analytically derived in your advanced courses you will learn how to analytically derive them but do you see here that there is a possibility that not only you will have diagonal terms right second derivative with respect to x y or z but you can also have off diagonal terms here right second derivative of e with respect to kx and ky you see you have off diagonal terms because in this case once you apply an electric field once it tries to get out here it may not have any space to go it will have may have to come out in a different direction and therefore there are off diagonal terms in the valence band but not necessarily in the conduction band and that was the corresponding to that point this valence band looks so complicated in in that point now let me talk a little bit about all theories are good you know theory people like them to calculate complicated things but actually unless you measured them experimentally none of these things are useful because theory people often make a lot of mistakes so let's start with two things that you really need to know about one is a band gap remember if my calculation is correct all the gap between energies they should come out right and if my calculations are correct the curvature which gives is given by the effective mass those should come out right these two things i want to discuss now first about band gap consider that that's flat line in the middle is the Fermi level remember we haven't explained it but we'll discuss it later and focus on the blue point and let's say light has coming in if light is coming in that blue point the photons are coming in photoelectric experiment remember and what this will do that light once it comes in the photon will be absorbed and that energy if it matches if it matches and the electron will jump to the conduction band right that blue point and when it's jump when it jumps light has been absorbed so if i'm looking on from the other side all in a sudden i will see no light coming out so i will see an absorption of light if the photon energy is right so i'll see an absorption in light and that will tell me that there must be two bands now here i say two and three but in general pair of bands are are here if a photon of a slightly different energy comes well i cannot the blue point cannot do anything anymore because if it tries to jump they doesn't find a receiving state but the red point which i just showed that might have a right match and in that case that electron will jump but one thing i want to point out here immediately that between state one and two there cannot be any absorption because both states are full you see electron when it jumps it has has to have a empty receiving space so only thing this experiment is going to tell you are the between states that are full partially full at least and partially empty the ones that are important for transport if the states are completely full let's say bottom lower level states those will not participate in this absorption process so we are getting the right band gap that we are interested in so you see the red point and you can continue doing this experiment and if you continue doing the experiment you will see that there is a corresponding energy spectra that is absorption spectra and do you see how this might come about the bottom point you see because that's the lowest energy gap the green point corresponds to from the edge on the one side jumping up all the way to the top not the green point over there and that's the green point here and then there'll be a gap because electrons from the two cannot go to four directly there's much a gap in between so there's a corresponding gap here and then the starting with four then other band will continue now one signature i will explain all this later on of a direct band gap material is that the absorption spectra must go with square root of e right square e being the photon energy on the other hand if it a indirect band gap material what are indirect band gap silicon germanium or indirect band gap material right so in that case you will see a slightly different feature that it has this complicated shape that goes the square of the gap i haven't explained it i will explain that later but for the time being if you see in your experiment that this is happening then you know that this is con direct band or indirect band gap material another thing to see is that when we talk about solve the Schrodinger equation there is no notion of a temperature right there is no temperature because all the atoms were sitting still in their place a a part and everything but if you allow the electrons to essentially vibrate at higher temperature the band gaps that people calculate so at 300 degree c 600 degree c then they are not just a part they are vibrating a little and the band gap that people calculate for example look at silicon at very small temperature zero degree it was about let's say 1.2 ev as you raise the temperature electrons are vibrating and the bands are spreading out a little bit the problem that we have not solved and in advanced courses they do by 600 degree c you can see the band gap is getting close to 1 so as a function of temperature which we have not accounted for so far the band gap changes and it's an important change to account for for actual devices now finally i have three more slides and we'll talk about how to measure effective mass now in order to measure effective mass we have to do a second experiment first one was absorption experiment the second experiment and the experiment that we have to do is sort of the following what are we trying to get at we are trying to get the curvature of the band gap of a particular band near the bottom because that is how fast the electrons move when you apply an electric field so that's what you're trying to get and on the other hand which is equivalently we are trying to get the curvature for these ellipsoids because that gives us the effective mass along a particular direction so this is how what the next measurement is very elegant actually you see what they do they have a microwave source at 24 gigahertz around 24 gigahertz they fix it so i have a device a microwave is coming to that device with 24 gigahertz then they put a magnetic field with a knob let's say i mean i'm just simplifying it now do you know that anytime you put a magnetic field on a semiconductor material the electron starts going around it cyclotron frequency right that's cyclotron frequency now once it starts going around it of course is going around it at a certain frequency now the point exactly at point when that frequency matches with the microwave frequency 24 gigahertz let's say at some point is going at 24 gigahertz per second right so you will see from the other side all in a sudden the microwave signal has been absorbed because it's like a pendulum when pendulum is swinging when i push my daughter in a swing then when she is swinging at the same rate that rate i am pushing then is exactly the amplification is maximum right absorption is maximum so from the other side you will not see any microwave coming out there'll be an absorption of microwave from that i know that my field magnetic field is right right value which is giving me this 24 gigahertz frequency right okay now let's see the next step what you will first see in this experiment that first there'll be an absorption but then a little bit later when will that absorption be think about gallium arsenide so now if you have the valence band now in the case of a valence band what's going to happen they have a slightly different mass right so they are going to go at a slightly different point you may have to use a slightly different magnetic field in order to push it at 24 gigahertz and so at that magnetic field there'll be another absorption and just by looking at these two absorption peaks you can figure out what is the conduction band effective mass and what is the valence band effective mass by using this formula the frequency what is that 24 gigahertz nu naught and i'm going to derive it the equation in a second it is proportional to b naught and inversely proportional to m star and you can see that if i measure a b conduction for the conduction at which there was an absorption peak the red one let's say i know nu naught 24 gigahertz i know the effective mass for the electrons and similarly i put b val which is a for the valence band and divide it out i get the valence band effective mass if it matches with the theory my theory is good if it doesn't match theories should have to go back and do their work again how does that equation come about which i use very simple actually don't get confused or by all this you see when something goes around you have a centripetal force that's the m star v square divided by r naught you know this formula right from college and when you have a magnetic field and the electron moving at a velocity v you have the Lorentz force on the right v cross bz why z because magnetic field is in a vertical z direction and the electrons are moving in x y plane you multiply you will get u v v z right from that can you calculate v on the left hand side v squared right hand side v you can calculate the value of v and if you know the velocity it's going around then what will be its frequency very easy see whether you understand this the time to go around a circuit is 2 pi r naught divided by v 2 pi r naught is the distance it has to cover in a single shot v is the velocity so this is the time it takes to complete one circuit and that is given by the first equation or in the blue blue a background then if that's the time what's the frequency if one circuit takes a certain time let's say one nanosecond then it's the inverse of that that tells me the frequency that's the new north and that immediately gives me the formula I am after that's it so essentially I can then calculate from this formula it is telling me that if I know new north microwave source 24 gigahertz and if I know b naught the point at which the whole thing is in resonance then I can calculate m star and that is what can you understand from here also that remember I had four bands partially filled partially empty why wouldn't I get any information about band one because band one is completely full electrons cannot go anywhere they want to move but they cannot move band four no electrons of course there nothing will move so the ones that will move are exactly the ones that we're interested in that are either partially full or partially empty because then the electrons can go around changes k value changes velocity you see so let me conclude with on this one that we have talked about e k diagram and constant energy surfaces now these are original things four dimensional complicated to draw but we draw these panels of information along a particular direction like x point the l point to represent the information we always look at a particular energy range right not the full thing depending on what we're interested in for germanium for silicon we are particularly interested in in bands close to 3s and 3p if you go back and see the top electron levels that is where the 3s and 3p levels would be and these have all same energy yes 3s 3p and they mix and because they mix this called hybridization sp3 hybridization therefore the band looks so complicated you know there's going some going up some going down it's no longer as simple as one dimensional uh chronic penny model now as I said the e k diagram and energy diagram contain equivalent information and you should learn how to use them in everyday work because for many new materials you cannot just take a m star from looking at a book many times you'll have to go to a theory friend and ask okay can you draw calculate the band structure for this particular material completely new material let's say you are your advisor and you are working on in that case from that material you'll have to extract this m star information and you will do an homework that will clarify this concept and finally experimentals measurement two right one is microwave absorption measurement for effective mass and the optical absorption experiment to get the band gap of the relevant one the balance band and the conduction band that takes part in power conduction process those are the things you have to measure because without those actually you don't know where theory is correct or not okay so we are really getting close to finally getting to the transport problem but let me end uh lecture seven here