 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 we'll get started on lecture 12 this is on equilibrium concentration of electrons and holes in a dope semiconductor now remember from the very beginning of the course I had been emphasizing the fact that I need two things to calculate current one is electron number and the other is velocity at which electrons move and for the electron numbers we first took a wrong turn we just looked at the crystal number of atoms in a crystal and just try to multiply with the number of electrons a particular material has and we saw that that really doesn't work in terms of conductivity of the material so we went through this quantum mechanics and statistical mechanics you know for me direct statistics density of state and finally finally we are in a place where we can consistently explain how many carriers I have at any given temperature in given material and that's really fantastic until 1940s perhaps even 1950s despite all the development in physics what you are going to learn today and what you probably learned yesterday in the last class people couldn't do so these are actually in some sense very advanced and sophisticated calculations so we'll see but actually now in retrospect appears quite simple we'll talk about carrier concentration but the point I want to emphasize today is the temperature dependence of carrier concentration because we'll see that although at room temperature it's very difficult to get carriers on the conduction band and get good conductivity if you raise the temperature you can see that the electrons will jump from the valence band to the conduction band more easily right it has more energy and that will dramatically change the conductivity and other carrier concentration for that in this particular case so we'll look at that we'll also talk about multiple doping and co-doping and I'll explain what they mean and the heavy doping effects these three things and finally conclude now this is something I have discussed in the previous lecture that in any semiconductor let's say silicon for example if I dope it with a certain amount of boron which is a p-doped material acceptor and let's say a certain number of donors like phosphorus then I will have a change of carrier concentration from the intrinsic value intrinsic value is very small right 10 to the power 10 in silicon out of 10 to the power 22 atoms humongous free spaces but let's say if you dope it ND and NA then we can calculate the change in the carrier concentration we said that if the material is doped spatially homogeneously that's means on the average you know uniformly drop the whole thing then not only it's globally charged neutral you know P and N it is locally charged neutral also that if I go in a micron by micron micron square by micron square micron cube by micron cube I'll see that locally it is charged neutral as well and we wrote those four expressions now only thing you will see that's slightly different from the previous lecture is that instead of writing the simplified expression for the holes I have wrote I have written the whole thing you see NV the farming integral of half and everything correspondingly you have the I'm sorry this should be an NC this is a typo I corrected NC and and then the rest of the rest of the term now this is the full expression if you solve this and the only unknown here is the farming level F you see everything else is known if that is solved then actually I'm done with my problem I could go home but the thing is you can see that this looks a very complicated thing first of all I have found me Dirac integral capital F half in one side and then I have all this for me Dirac like functions on the right hand side looks complicated and I mean just visually looks complicated one quick thing I want to point out why on the one side there is for me Dirac integrals and why on this other side just for me Dirac function because in one case you see I had continuum states so I integrated them over them right because I wanted to see how they are all occupied for the donors and acceptors I assumed a single level so I didn't do any integral so therefore you can see in two sides I have slightly different functions okay that's no rocket science and then of course if the family level is significantly below the conduction or valence band or away from the conduction valence band about 100 million electron volts let's say then I can make this approximation but how would I know you know I haven't solved for EF how do I know where is it right the best way to do it is to start from the simplified expression in the bottom solve for the EF the Fermi function EF because that's that's relatively easy and then you check the argument that is it below 100 million electron volts from the conduction and 100 million electron volts above the valence band if it is then I am done if not then I go back to the top expression because that's more accurate right and that will always apply so that's more accurate now my goal is to calculate EF and then that will allow me to calculate everything NP the electron hold density donor acceptor I'm done actually I can I can do a lot of things so the purpose of lecture today's lecture at least for maybe 10-12 slides is just to show you how to calculate it in principle you could read it yourself is so simple but I will help you to guide you through it's not recent the next few steps are not conceptually difficult at all it's just a few algebra algebraic steps so let me help you along on that so the first point is does this expression which we just said reflect what we did previously without making this argument for example one thing we said that if a material is undoped then I should not have any donors or acceptors right undoped so I don't have them and so that's gone so in that case you see in that case only the first and the second term remains n minus p equals 0 I can equate them and I can then solve them for the Fermi level EF and once I solve them this is exactly the expression we did before remember this is the intrinsic semiconductor of course the more complicated expression will reduce to this very simple formula when I remove my acceptors and donors no problem here and once you have EF you could easily calculate the value for n which is now the intrinsic carrier concentration 10 to the power 10 centimeter cube and correspondingly also the value of holes now when you have doping of course things are a little bit more complicated so let's see how it works let me assume a case where it is n type when somebody says that I have a n type semiconductor what they mean that they have only donors that gives away electron so that the material has a lot of electron right so that's why it's called n type material and in that case we will assume that there are not too many acceptors too many n a is approximately equal to 0 that I have not put any boron in I have only phosphorescent and correspondingly you could do the whole thing for p type doping okay so you say if I don't have any acceptor I should get rid of that term right should get rid of that term and you remember that the term I have shown here in red circle I've circled this this is actually n n written in a slightly different way more expanded expanded function okay now I want to let me assume the problem has been solved you see you put it in a mat left function or you somehow solve it I'll also solve it in a few minutes but let's say I have solved for EF first from this expression only unknown and I have inserted the value of EF back n to calculate the value of electron concentration let's say I have already done that if I did that this is what I'll be saying what I have shown you here in the y-axis is how many electrons I have as a function of how many donors I put in let's say I put in 10 to the power 18 number of donors phosphorus n in silicon and I have 10 to the power 14 number of electrons so on this plot it will be 10 to the power minus 4 so there'll be a point in 10 to the power minus 4 in general if I take a temperature as in the x-axis I will see a plot something like this that at very low temperature the number of electrons will be minuscule even when I have doped it you see this is not intrinsic but even when I have doped it still it's very small somewhere in between you see in that intermediate temperature range and that changes with every semiconductor I have the number of electrons equal to the number of dopants as if approximately it's number these two numbers are the same and finally at some point the number of electrons becomes larger than the number of dopants now how can that happen I put some dopants in how can I have the electrons which is larger than this we'll explain in a second and there are you know in principle if you solve the equation in the previous page in a computer you'd have gotten the whole car in a single shot you just put the temperature in and get a Fermi level out put a temperature in get the Fermi level out and you could have plotted the whole thing in a matlab function in a single call the whole thing but if I wanted to do it analytically then it's better that I you know put it in different chunks so that I can solve them easily so one region will be calling it freeze out I mean in Indiana looks like this is the appropriate region of operation the other part we'll call extrinsic the yellow part you see will name it extrinsic where the electron concentration is approximately equal to the donor concentration now remember this is the part I want my device to operate on right most of your computer if you take it to Alaska that will be in the freeze-out region let's say or if you took it for a solar mission then it will be on the right-hand side let's say which would be the intrinsic part now the most of the time I really want to when I design a computer or design and I see I want to make sure that my devices really work in the extrinsic region the yellow region not in the green or in the blue part and what happens in the extrinsic region that although you had a lot of dopants in the beginning but if the temperature is high enough let's say 600 degree Kelvin or 700 degree Kelvin it's a very high and this can routinely happen when you have by a computer sometimes if things are not properly grounded if you touch it then the ESD pulse just like when you have touch a door knob the amount of voltage that can flow in in the transistor can be huge in fact in that case temperature rise can easily be several hundred degrees see of course it will destroy the computer but if the temperature rise is significant in that case the number of electrons from the valence band will jump to the conduction but then it let me show you some pictures and in that case that will take over so let me explain so in the freeze-out region all the dopants are there but it at least need about 100 milli electron volts to donate right it's bound level remember the hydrogen levels if you don't even want to give it hydra 100 millivolts then nobody is giving any electron and therefore the there'll be no dopants and no dopant will do you nothing right you haven't given any temperature so let's say 10 degree Kelvin so you don't have any electron now a little bit later when you increase the temperature a little bit more then you'll see the donors will easily give out electrons to the conduction band because it needs only 100 milli electron volts but there'll also be a few from the valence band valence band will be few why because it's such a big jump one EV or so in order to jump up so therefore in that region it will be approximately equal to whatever number of donors you have now if you went farther increase the temperature even more well donors have given whatever they could remember in the valence band what is the effective density of state on the order of 10 to the power 20 lots of states they are lots of electrons so when you increase the temperature if it could give all the way jump up it could actually give huge number much higher than what donors can give right so at some point you can see it can easily overwhelm the number of electrons from the donors and that's why the ni of the material can exceed the donor density and that's when it takes over and it's no longer a dope semiconductor per se it's like intrinsic semiconductor because as far as at that temperature electron density is concerned it doesn't know about donors at that point you see okay so these three regions let's try to make some calculation how many how to get that curve analytically so what I did here in the first one you know this that n is by definition effective density of state about 10 to the power 20 per centimeter cube multiplied by that beta is one over kt by the way anytime you see beta in this one easy minus e f depends on where the family level is so you can rewrite it you see NC e to the power beta e f on the right hand side and you can flip the capital NC and EC on the other side okay now you also know this right do you know this ND plus is ND one plus two you remember to the degeneracy factor why did that come from because I had one bound level talking to one conduction band that's why I had two if we talk to two valence band I had four right that's that too and e sub D is the donor level remember hundred milli electrons volts below the level so that's what I have but you see I also have a in the magenta I have shown an e f but from the top expression I could replace that I what all I did you see from the top expression the magenta power e f I have replaced it in the bottom one now look at this expression carefully what is unknown in here that the EC is known right conduction band is the e sub D is a donor level known NC effective density of state in silicon I know that so in fact if you look at it except for n that's actually a constant I mean I know ahead of time for the material so I will write it everything that is known already in n sub xi now of course this is a material dependent constant for silicon it will be one n sub xi for germanium which is a different n sub xi and the effective quantum mechanics and everything is hiding through n sub C effective density of state has effective mass right effective mass knows about quantum mechanics so all those are hiding in this little n sub xi okay so from here I can calculate look at this that's the full expression I don't write na minus because it is a n type material right I don't have any accepted now do you agree with this statement I have written the previous one in the expression below and it has a several trick that you should remember the first one is this you see the first term on the left hand side that goes with the valence band and the number of holes that expression p multiplied by n on the left hand side equal to ni squared anytime you have a device doped or undoped but in equilibrium you see I haven't applied any voltage here right doped or undoped in equilibrium that one is always correct p multiplied by n and so you can see in order to get the value of p I can simply write ni squared divided by n no problem right second one well n is n no problem and the third one I just derived now this is a equation the only unknown here is n and if I somehow can calculate n I can go home so far there's no approximation but that's not really true right because the law of mass action how did I derive that I assume the material is non degenerate do you remember I multiplied this p expression and vn expression and I got ni squared only when the material is non degenerate so that is an approximation already built into it when I came here but other than that I haven't done anything so far so let's look at the freeze out region and see why it goes like this let me assume first of all it's a donut dope material so it has a lot of electron actually and not as many holes you'd assume that and we'll check that in a little bit so we'll drop the first term just to get started assuming the number of p the number of holes is small and we'll check that once you have done that drop the first term ni squared divided by n on the top equation then you will get something simple and you can see that will lead to a quadratic equation for n ni n squared plus n xi multiplied by n and the whole thing and once you solve for it you'll get a expression like this which goes with n xi and number of donor density now don't don't worry about these things right this is so simple don't even have to think about it now after you have calculated n however from this expression you should put it back in the original expression with ni squared divided by n to check whether the p you know the number of holes is indeed small or not it will be but whether it's not correct or not that assumption you should check okay so this is I'm done and in the freeze out region if I just plot it out plot this expression out I'll be able to get the first part of the car that is something you can easily put it in your computer and see how it works by the way where is the temperature where is the temperature in this expression it is in n sub xi why because n sub xi has effective density of state sitting in there right you remember an effective density of state is density of state which is temperature independent multiplied by the width of the region which was occupied and that changes at every temperature so the place where the temperature is hiding in here and why that gives you that particular curve is because n sub xi is temperature dependent okay so we could plot it out that's not a problem now what about the transition region from the freeze out to the extrinsic well in that case we can take this expression which we just looked into and sort of look at its behavior at higher temperature that where does this curve go at the end of the freeze out region what I have done from the step on the top to the step on the bottom is I have assumed the number of donors is actually smaller than n sub xi how do you know that n sub xi you can easily calculate and because at every temperature you can calculate it and nd is some value that you put in by hand right the number of dopants you put in so you expand that term and you can see the factor half which was under the square root that becomes on a tailor on a on a binomial expansion becomes a half upfront now do you see that what this value is going to be one will cancel you see that you see that the other piece will cancel also this n sub xi will cancel and this becomes equal to n sub d you see this and only at that point only at that point the n will become equal to nd now there is no insight here you see I just did a math few lines and no insight but the point you sort of can see as you are raising the temperature right n sub xi will get larger and as it gets larger then you will be able to make this expansion and then that will equal to n sub d what happens at lower temperature you see at lower temperature was going to happen n sub xi will be a smaller number if it is a smaller number you will not be able to make that expansion and when you cannot make that expansion you will see the carrier concentration going to 0 right so this is something you should check it out okay how will you calculate the whole density anytime you know the electron density always use this mass action rule n multiplied by p is a n square in equilibrium remember because we will see many examples where it is not true anytime you apply a voltage okay so we just saw this nd again once you can once you know the temperature is large enough then that factor e f minus ed is it a positive number or a negative number you think e f is if the whole quantity is negative you can see at large enough temperature the whole factor will go to 0 at the whole exponent will go to 0 and therefore the whole factor no it will go to minus infinity and the whole factor will go to 0 so you can you should check that step out also this should the whole term should go to 0 and again you have the same expression but this time because the whole factor goes to 0 you can throw out the denominator replace it with 1 you see this that once you do that again you have a quadratic equation and once you have the quadratic equation you can solve it see so once you know this nd value of nd and ni squared you can calculate once again the value of the electron concentration in the yellow region this part is sort of just map there's no physics in here but I'll just lead you through so what happens when nd is much much greater than ni when can that happen when the temperature is low enough so that the number of electrons from the valence band do not jump to the conduction band as easily as the number of donors I have if I have that assuming then do you see do you agree with this statement that n will become equal to nd ni squared you drop it and therefore they take the square root and that gives you nd you see that nd over 2 d plus nd over 2 so I will have nd if the it is much larger than the intrinsic concentration around here in this space now what happens if my temperature is very high very high in that case temperature is very high means lots of electrons are going from the valence band to the conduction band my ni is bigger than nd and in that case do you see why this nd is negligible compared to ni squared take the square root I'll have equal to ni and that's why in this intrinsic carrier concentration will exceed the dopants and in that case the material will become intrinsic once again even even in the presence of doping now the question the reason why this is a very important discussion is because many times let's say you have your you work for NASA and you are asked to design an electronics during the lift off period you are asked to design an electronics which will survive the high temperature during the process either you can insulate it of course but if it is somehow exposed to high temperature then let's say you designed a computer just fine with all the dopants to the carrier density everything working at room temperature but as soon as it goes to higher temperature you can see because there is so many electrons going to the conduction band the whole thing will become intrinsic again your transistor is not no longer a transistor because all the doping all the design you did those are all gone and this is just a chunk of useless silicon and of course your computer isn't going to work so high temperature electronics many times in the furnaces and other places when you want to design this silicon will not work one ev is simply not large enough so people here there are professors who spend their lifetime working on silicon carbide for example a little bit larger band gap material lots of people use you'll see if you look at professors interest they will say high temperature electronics and then the material they list next to their interest you will see always has a higher band gap so therefore the range of temperature they remain in extrinsic is much larger because of course electrons cannot go up that high as easily and that is how depending on the operation or depending on the use condition the temp the material that to be used will be different okay this is important point so if I give you an exam problem or a homework problem in which I specify a temperature and ask you to calculate this you shouldn't always put n equals nd immediately and start doing the calculation in the undergraduate years you could do that not now okay so once you know n that's what you did last few slides right solve for n once you know n you could calculate if because EF and n are one have one-to-one relationship and once I know EF I know whole density I know donor concentration I know everything so I'm done actually I can I can calculate everything and then if I multiply apply electric field electrons will begin to move from here now the points I wanted to make a few for your points so I'm done with simple undergraduate level discussion about how to dope how to calculate the number of electrons and I'm essentially done but you should be a little bit careful about a few more things because I made a series of assumption which may have escaped you the first is nobody says that you just have to have one type of donors in principle in a given material you could have two donors two different type of donors now if you have two different type of donors let's say phosphorus and something else some other material then you will have not one level you will have two levels right red and blue both will give electrons how would you solve for that problem just by generalizing this a little bit so what I have if they are you have equal amount of doping then I have ND but you see I have a ED1 in the denominator of the third term shown here in red that goes with the donor level of that trap right what about the blue one well this time I have another term which is the fourth term from the bottom in which I have E sub D2 because the electrons have to jump from conduction from the blue level to the conduction band so I will have to introduce another term here right so in general if I have multiple levels I will just add up multiple levels here and then I'll again you see only unknown here is what is unknown yeah everything else I know I will again use exactly the same procedure that we just discussed and calculate the electron number whole number and everything now in D two and these I have shown here on the third and fourth term I have assumed them to be equal need not be equal if they are not in D1 and D2 and you can see how close now this is if you have multiple levels I'm sorry I'm sorry I have to backtrack on this one so this is a multiple levels of the same donor that's why I have kept this to end is the same because if I have just phosphorus let's say two levels if I had first for it doesn't have two levels in silicon let's say in that case I will just do it because I have the same number of donors but if it is co-doping co-doping means two different type of dopants each having a single level then I will have ND1 and ND2 and in that case again the same expressions I will extract an EF and calculate everything from this point right and why do I have two over there is this degeneracy factor because they are talking to the conduction band if I wanted to talk to the valence band via co-doping then each of them will carry a factor of four so now the next point I want to talk about is heavy doping what is heavy doping and light doping I already said that the number of typical dopants the typical amount of doping is 10 to the power 17 maybe 10 to the power 18 per centimeter cube number of atoms 10 to the power 22 one atom in a million host so you see this think about it how the proportion this Lafayette has how maybe 250,000 people and you are only one foreigner that's let's say and then multiply it by a factor of 10 or so that is how dilute even a reasonably dope semiconductor is very very few they don't even know about that there are other foreigners or other dopants around in the whole area so far apart therefore I already mentioned that they each have a discrete level where only one can sit two cannot sit because of the coulomb interaction and everything you have you know that however assume that in modern especially in source drain of modern MOSFETs the doping level can be as high as 10 to the power 20 or even sometimes five times 10 to the power 20 very high and almost every 10th or every 50th atom one could be a donor now in that case you see the donors I have shown here with the green levels right now typically and the three ones with the multiple color those are host atom but now the donors are close enough that electrons can jump from donor to donor without first going through the host atom they are close enough and that's why you can see the green level if they were completely isolated they would have two green levels just you know one in the bottom one in the top two levels because they keep to themselves but now because they have a neighbor to whom they can go to therefore that level have split have you seen that each band now each green band has been split into two because it has a neighbor to which it can go to back and forth so it is sort of forming a band that is on you see compared to a band of the host it is sort of a band of the donor or the band of the dopants so what will happen is that in typical semiconductors if I have a ek diagram like that like the one shown in the right hand side you know this right the density of state as a funk or the ek diagram over there in addition what you will have and for the case in the bottom is you will have some additional states near the age of the conduction band shown here in green and age of the valence band shown here also in green and this is because the donors are talking among themselves and as a result the band has essentially narrowed a little bit see so this is something these are called band tail states because they stay slightly below but now there is a continuum because the donors can talk among themselves but fewer right donor atom is still one-tenth whether 100 so therefore fewer than the normal density of state but no longer completely negligible no longer just one level like a trap that has shown you or dopant as I was showing you before this EG1 and EA1 remember those levels those levels no longer one level it is a continuum now okay so what is the distinction the distinction is that there is a difference between band transport and hopping transport and this is how if you take a normal semiconductor which is lightly doped 10 to the power 18 let's say then the electron generally jumps up so first point is that because let me come back come to that point in a second because the levels the gap has essentially narrowed because of the green states now the band gap is no longer easy it's a little bit less when light comes in shines on it the jump will be a little bit less and so the first thing to check notice is that the law of mass action P multiplied by n is no longer the band gap of silicon 1.1 eb but rather it is the effective band gap another effective and it is EG star and the EG star is that the green top of the green and the bottom of the green for the conduction and valence band that reduced levels is the effective band gap and so you will see all your relationships will change a little bit yeah not completely change a little bit but also interesting is this that properties of how carriers move along so when you have a normal semiconductor electrons donate to the conduction band and electrons go through the conduction band it is like highways right we go on to 65 we take a local exit and get to the highway and that is how we go from city to city there's no local road per se right but if the donors are sort of dense enough and if they are a local road around then you see what it will do it will jump from atom to atom to atom without ever going to the conduction band I mean you can go to the Indianapolis by taking the local roads also right without going through the highway if there are enough cities so these two types of transport band transport and hopping transport are very important now if it were I was teaching it 15 years ago I wouldn't even have mentioned this because electronics was all about band transport silicon you know your Pentium or at that time whatever version of computers Intel had or other companies had no longer because this type of transport in modern solar cells right if you go and work for let's say any solar cell company today they make the solar cells are made of not crystalline materials they are mostly made of amorphous or micro crystalline material and most of the groups now in Purdue they their research is gradually focusing on this you know energy is a big thing these days and in those materials transport often is not via the band but rather through the localized states hopping from one side to another and the properties of transport how it walkers you will learn it in more advanced courses but at the very least you should know this because that's very useful and relevant for the devices that we are presently interested in in fact your flat panel displays the things that you go by this 46 inch 52 inch the pixel each one of the pixels is driven by transistors that are I'm most of the time made of amorphous silicon which doesn't have a crystalline property and in that case transport could often be through a hopping transport the organic also electronics hopping transport so lots of parts of electronics is actually being governed by this type of transport through the blue states sorry green states okay so few more things now so there is what was I was trying to get to in the previous slide that remember they have shown you this slide before in the beginning a lecture one I wanted to wrap it up on this side I emphasized how beautiful the silicon substrate is look at that every atom in this place and this is just one part and these are essentially how far apart on the order of a few angstrom apart and think about a 12-inch wafer 12-inch wafer this type of periodicity I mean this sort of is unbelievable that what people can do in terms of crystalline structure and that's why you have Pentium says this high speed in a silicon transistor but of course there are material which are amorphous oxide random material and then there are polycrystalline material now how what you have learned so far translates to these materials so first of all let's talk about polycrystalline material and let's think about the fate of an electron going like that so in the beginning let's say it was just to make an argument 1 0 0 then a little bit in the middle it goes through 1 1 1 let's say and then a little bit later it goes to 2 1 0 let's say in various directions going now you know that the band gap at different directions is different do you know that in crystalline material along the L direction and along the x direction the band gaps are not the same are they right so therefore on its route going from one side to another the electrons will see a variable band gap as it is going through no sharp band gap and therefore what will happen that there is no notion of one single band gap anymore in this particular case but rather you will have the electron will go through every time it goes through a grain boundary it will get scattered because the band gap has changed and as a result the transport will be a little bit more difficult but the notion of a band gap steel exists the density of state steel exists it is just that the band edge is a little bit more amorphous I mean not clearly defined anymore depending on which direction you are coming so it is sort of the band gap for different electrons going coming from different directions the band gap will essentially look isotropic on the average and there will be an increase in scattering but apart from that everything is just like silicon now what about the final point about amorphous material it's very important that we understand this in the top side I have shown amorphous regular structure now in the homework I asked you to look at the structure on the top where you view the silicon crystal in a various angles do you remember the nano hub exercise and so it is a periodic crystal and let's say tetrahedral structure you can take a look and the atoms are where they are supposed to be I have flattened in a A and D case I have flattened it out the four three-dimensional structure so you can see the regular periodic structure and we have solved the chronic penny model and we have defined the band gap and everything and we know how we have all these discussions about calculating the electron number of electrons and other things now what happens when you have an amorphous material now many people call amorphous material to be random material not really because if you can if you look at how the amorphous material shown here in section B in the very bottom on the left hand corner left hand column you will see that the distinction between a regular crystal and the amorphous is not very much in the amorphous case sometimes you have a ring of five sometimes you have a ring of three in the regular case you have always a ring of four but you never have a ring of eight or so it is a little bit amorphous a little bit away from the periodic crystal but not dramatically so that is why even in amorphous silicon right or amorphous silicon dioxide band gap is still a completely valid concept right so this statement I want to emphasize because most people leave this course with the assumption without understanding this final statement that the periodicity is sufficient right periodicity if you have a silicon germanium it is sufficient for band gap development of a band gap but it is not necessary because even in a amorphous silicon like this so long the structure is approximately periodic you will have a band gap if you didn't have a band gap you couldn't look out of the window you see the reason you couldn't look out of the window because the window is made of silicon dioxide right it's a it's a form of glass now it's a amorphous material of course because it was randomly processed and it's a amorphous material light still shines through that means that there must be a band gap otherwise how does the light come through now if there is a band gap in a amorphous material then surely it is band gap is not a property of silicon alone you see so this is a very important point that you should not miss simply because you used chronic penny model and other things we shouldn't assume that that is a requirement for band gap density of state semiconducting properties none of them are consequence of periodicity so let me conclude this so we started with the charge neutrality condition on a uniform semiconductor and from that we used we obtained a Fermi level that's that's simple equation we had we obtained the Fermi level and once I know Fermi level I know everything else NP number of ionized donors and acceptors I know now the point I wanted to make that every all the discussion I had was for a field-free region I was no had didn't have any electric field if I had that then this condition of charge neutrality then the condition of this mass law of mass action NP is equal to ni square all those have to be thrown out of the window not completely thrown out of the window have to be modified and then we'll have to use a something called a Poisson equation instead of the simple equation we had to calculate electron and hold density slightly more general and we'll see that and finally I wanted to emphasize the heavy doping effect and amorphous structure play a very important role in modern electronics that wouldn't be relevant probably 20 years ago but now fundamentally important when you graduate and look for a job okay thank you