 You can follow along with this presentation using printed slides from the nano hub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. Let's begin. This is lecture 25 for short key diode and short key diode is like a PN junction but historically and also for many practical reasons it is something and also for pedagogical reasons this is a very powerful device and I want to introduce you of how it works. Now I will start by talking about the importance of metal semiconductor junctions. Now metal semiconductor the PN junction we talked about so far what was it semiconductor semiconductor junction right it could be same semiconductor homo junctions or different let's say germanium on one side silicon in another so that it could be but here definitely it is different metal and semiconductor and I want to explain how it differs we will again go through the same process we will first draw the band diagram any new device you have first thing draw band diagram so we will do that we will also explain why the things that I have told you so far about doesn't really apply when you want to calculate current in a heterostructure but what I am going to tell you today that is going to complete the picture in terms of how to calculate current if you know these three processes then any other device that comes in your lifetime about various carriers various devices you should be able to handle and then I will conclude now metal semiconductor diode is actually very simple diode in one of the oldest diodes that people know about one side is this green semiconductor it could be n or p-dope but in this case let's say this is n-dope because nd is the donor and the red are the metal contacts now many times historically people didn't put a metal down but rather if you take any wire any wire and then touch it on a semiconductor then that becomes a metal semiconductor diode so in the old days the first transistor the transatlantic transmission and all those those are all metal semiconductor diode the first bipolar junction transistor I'm going to say in a second that was really metal semiconductor diode not the thing that you read in your book we what we read in our book is very different from what was invented in 1947 and most people don't understand that so I'm going to explain to you that now most of the time again we'll be doing a one-dimensional cut and look at the one-dimensional flow just for simplicity but as you realize that when you have a wire touching a two-dimensional or three-dimensional semiconductor a chunk of material the current will come through that point and spread out so that's a little bit more complicated but for the time being we'll ignore that complexity the symbols again the symbols are very similar to the diode but you will recognize that there is this extra wings sitting on the corner that tells you about that this is a metal semiconductor diode they're talking about most of the output characteristics will be very similar to the p-n junction but the physics is very different okay it has a lot of applications metal semiconductor diode for example if you in many as I said in radio transmission in other cases in older days people used to use this metal semiconductor junctions to demodulate extract back the carrier signal from the signal from the carrier but there are many others for example this is a historical replica of the detectors of signals you can see the wire right you see that wire hanging from that sort of the cantilever and the semiconductor is like a paste the yellow paste late sulphide is a yellow paste and that's in the early days that's what people used to do as a diode that's the diode actually now there are of course this first original bipolar transistor that turns out to be a metal semiconductor metal transistor it is not the classical transistor that you study and I will explain and you will do a homework on this that how the original transistor worked so that's already in your in your homework then people talk about carbon nanotube transistors that you may have heard about if you see that there is looks like a bridge between this a single line bridge between the two sides the two sides the two islands are contacts these are metal contacts and the long wire is a semiconductor carbon nanotube so that's something that's very important people work about it all the time metal semiconductor junction and then of course the STM tip anytime you have a metallic STM scanning tunneling microscope where you put it down and drag it over a surface to see the surface topology or the surface electronic property it is essentially a metal semiconductor junction because the tip is often a metal and the base is semiconductor so of course the physics is very similar so this is actually what you are going to learn has a wide range wide range of applications so the yellow one yellow paste as I said this is called gelena or late sulphide type paste and for many materials for many years the people have used it as a semiconductor material so let's start with equilibrium band diagram now equilibrium band diagram we are what I want you to notice the reason I am putting up this topic map over and over again is when you study for this course or you want to really capture the what you have learned in this course you should go back and see that everything I'm doing under the equilibrium column it is always the same I'm always following the four rules and I will never break from the four rules and I will always get the right answer and similarly for DC he will see that I will be doing a same set of things over and over again AC large signal same anytime I see large signal you'll see I'm doing charge control so I want you to follow the pattern because these are of course devices people already know when you go for work they will not be asking you to do a MOSFET analysis they will ask you to do something else and again if you know the pattern you'll always be okay so the first thing is that anytime I want to do an equilibrium I'll have to solve that equation right that's the Poisson equation and this graphical solution is called this band diagram that's the graphical solution of the Poisson equation I'll also have to solve these equation but what I'm going to show you that I have to do something special about solving this equation because this may not really directly apply to metal semiconductor junction I'll explain to you in a little bit why okay what is the rules what are the rules for drawing a band diagram first draw equals a Fermi level I know so a Fermi level flat across this region because the property of equilibrium no matter what happens is always that the Fermi level is flat from one side to the other side of the junction because had there been any gradient there would have been a current right you remember that the gradient of quasi Fermi level tells you how much current there is no current in equilibrium no net current so therefore it's flat the next thing the next thing is to draw the conduction and a valence band of the respective side now this is a metal in a metal Fermi level is above the conduction band in a given conduction one of the bands inside one of the bands and so therefore you see the band gap is right there the blue but the Fermi level it's like a degenerate semiconductor as if and so the Fermi level is inside one of the bands why is it you remember because it has generally odd number of electrons so therefore in metal generally the Fermi level is half full in one of those bands okay so that's fine now what about the semiconductor oh by the way so there's a phi sub m is a metal work function from the Fermi level how long how much energy does it take for electron to go out and get free and so that is the amount of energy phi sub m so you can draw that immediately because you know what the Fermi level is and then you can just draw the dotted line that gives you the vacuum level now let's come to the other side for the semiconductor side this is a n-type semiconductor I assume right I assume the n-type semiconductor so I should draw that line first because this is close to the conduction band and with the using the band gap of that material whatever that material is then I draw the ev so that gives me the band gap now the next step is that I have the chi I'm sorry so this this day from ec to the vacuum level that is chi that symbol I'll put it down in a second but that is chi so you first have to do the chi to get the dotted line and the blue and the red region that you see with the vacuum level that is essentially taking the right hand dotted lines and the left hand side dotted lines for the vacuum level and connected up continuously it doesn't matter how how you draw it just draw it anything but it must be continuous so you do that and the next rule is that once you have that thing from the blue region you copy it down and you copy that down that's one side and copy this down and copy that down because equal to the band gap remember and the blue side band gap and the red side band gap need not be the same right need not be the same and so then you connect it up that's your band diagram that's the device that's really a metal semiconductor junction diode you can see the electrons from the ec side will go over the barrier and get to the blue side and get relaxed down to phi sub m for the energy so we should talk about the transport in a second but that's the answer that's the band diagram now in general in the band diagram whatever the band diagram is and I'm going to show you that in general I you have seen that the blue region I have drawn the drop in the blue region I've dropped it small right here and red I have bend it up a little bit more now in principle there's no reason why I should do that a priori but what I am asking you to believe or see and I'm going to prove you this in a second that in a junction you know that charge on the left hand side must be equal to the charge on the right hand side so if the left hand side in this relationship we have seen before right na which is the acceptor xp which is the charge on the left hand side for a p-n junction and dn and xn is a charge on the right hand side so the metal is degenerate right huge amount of charge so therefore quote unquote na is very large if it is very large then you can see xp has to be minuscule right it has to be minuscule because the total amount of charge on left and right must be the same if it is minuscule in fact it turns out to be so minuscule that in most of the cases you'll see the blue in the books the blue band bending is not even shown in the books because it's so minuscule like maybe two to five angstrom at most on one side on the other side hundreds of angstroms right 10 to the power 17 and one side 10 to the power few times 10 to the power 20 on the other side on the metal side so in the you know textbook when you look at it you will see that they have just drawn a straight line but actually this is what they have done they haven't just simply told you so that's why you make that negligible and don't draw the band diagram this way that's the guy okay what is my first thing do i do after i finish drawing the band diagram qvbi built-in potential that's the first thing i do and the built-in potential remember only thing i care about left most part of the material device and right most you may have 273 devices or 273 junctions but only thing that you care about for vbi junction of the material on the left most side number one and the material on the right most side 273 so that's the only thing you care about in between it can do whatever it wants vbi doesn't care so this is how it works so for example how will you do the vbi you know that right because the formula level is flat all i have to do is to make sure that on left hand side and right hand side they reach to the same level so for example this you agree delta is how far the ec is different from ef chi of course you can see that's the vacuum level work function and then qvbi is the extra built-in potential that's what you have on the left hand side and equal to five seven on the right hand side that's it so in this case these values five seven chi these are given from tables we have tables they tell you that if it is copper this much five seven if it is aluminum then this much five seven so this value you just read off from a book and chi of course silicon germanium each has different amount of chi that's fine delta how will you know delta doping right the doping tells you how close it is close to the conduction band so you can see qvbi is known qvbi is that way by the way this chi sub m uh phi sub m minus chi you see that's really effectively equal to the barrier that's the from the green region to all the way to the top of the conduction band so that's the barrier they will call it phi sub b that's the barrier height so barrier height is different for each pair of pair of materials right and then you can see this delta delta is given by n d over n c uh given by by that by that factor now this is only true if it is a non degenerate semiconductor right if it is degenerately doped on the right hand side then I'll have to write a more complicated equation for the value of delta but that you knew so so don't worry about this one I know qvbi the built-in voltage now for this and this is a general trick you can do it for any devices that you want no no new things so now I want to do a few more things uh let's see in general what you want to do is again you have a depletion charge the blue is the depleted charge on the on the n side why this depletion because there's band bending and the charges have been pushed away and the red one that I have shown you is the charge on the on the metal side and you can see it's huge and therefore its width is very small again you can do this in the book there are this integration you start from uh from the left hand side gradually come and gradually integrate through the red region which increases the electric field and then gradually go through the blue region which takes away from the electric field and essentially you'll have it on the right hand side extreme right hand side again the electric field is zero again the same thing p-n junction there's nothing nothing funny here but you can see because the red region is so tiny and skinny therefore the electric field when you integrate it it almost looks like a jump jump in here because it in a small distance it gives a lot of integrated area under the curve therefore it jumps up and gradually on the right hand side and the blue side also it will gradually take away the uh electric field the red region generated eventually it will make it equal to zero so you can calculate the maximum electric field can you not because if you knew the donor density and somehow if you knew xn the how much it has got depleted then you divide by the epsilon then you are done but of course you don't know xn so that I'll have to calculate somehow by the way so after you are done here then what you will do is you will integrate this uh electric field to get the potential right and you will now realize why there is the potential starts from almost close to zero because look at that curve if you integrate that skinny triangle on the left hand side of zero that has almost no area it has a large of ordinate but no area so therefore the blue curve on the bottom you do not get any contribution to the potential drop and then of course on the right hand side the potential quickly picks up and this potential is vbi on the right hand side right that's what vbi is and so you can calculate emax and xn divided by two area under the triangle over there that's equal to phi so max or equal to vbi and then once you have this you flip it over right and if you flip it over then that's the band diagram that you will see that is often drawn uh in your textbook okay so let me go through this a simple calculation to show you how it works exactly the same spn junction diode I really don't have to do much look at this so at e equals zero plus you have n d and xn and e is zero minus you have n metallic whatever that number is and xp is whatever the depletion on the metallic side is the thing is that I don't have to know that I'll show you in a second so in general you can write because the charge has to be the same you can write n d xn equal to np insert p xp and similarly for q vbi it's the area under the triangle you write it the same way right e at zero minus xn divided by two and zero plus xp divided by two and you will have this square term on xn squared and xp squared now you know q vq vbi right you know q vbi nm well you don't know in a pn junction I knew what n a and n d was so I will not be able to solve this problem in general but fortunately it turns out that nm is so large that xp is negligible as a result I'll be able to drop this term and then from q vbi calculate xn and not use the top equation at all you see that's the trick that's the extra trick that that you have to do and so from here from here what I'm going to say that that nm is very large if nm is very large then do you see how the standard pn junction equation becomes independent of nm do you see that if nm is very large compared to nd then nd I drop and I have nm on the top nm on the bottom that takes a that can get cancelled and I have nothing on the right hand side for nm that makes sense right because it is so large that in that case the tiny region doesn't really make much difference and you can do the xp but you can see this time it will be zero close to zero do you see that again because n sub m is much much greater than nd so I'm going to drop that but now n sub m is in the denominator and when you have a very large value for that it's like a one-sided junction so that goes away so mathematically it's telling me exactly what I have felt intuitively that this should be this should be the answer okay so now I have told you about this band diagram equilibrium now the dc current flowing through this junction is a little complicated not complicated conceptually but it's a little different I cannot use directly this drift diffusion equation that I have derived in the first five week of class anytime I have a discontinuity in the conduction band a hetero junction right or in the valence band then I have to use a slightly different theory which is very simple called a thermionic emission theory so anytime you see a discontinuity in the conduction band on any material any device then you have to use this particular way of viewing things that I will explain so I'm talking about now the dc and the short key barrier now what I just said is that this way will not work now this theory only works generally when you have a continuous conduction band or a continuous valence band in that case that works when you do not have that then we will try to do things in a slightly different way but actually simpler than p-n junction you will see it takes just a few few minutes to explain how it works but you have to pay attention this is the band diagram in equilibrium I shouldn't have to draw that again right so you know that I just drew it a few seconds ago qvbi you realize and for this material I have just grounded the metal metal side so you can see that's the ground on the left hand square little square that's the ground you always need to ground a terminal a terminal of a device if you don't ground it then you cannot draw a band diagram everything will be confusing so I could ground right hand side or left hand side I decided to ground the left hand side anything is fine you can do the other way now do you realize that this is a bias device and you say forward bias device see the rules whether I have followed the rules I've grounded of course the left hand side one of the corners and on the right hand side I have applied a voltage now a priority I don't know whether it's a forward or reverse bias but at least given that this voltage is negative let's say in that case what I have done is that I have split the Fermi level right so the Fermi level is now the top one is fn and the difference between ec and fn that has remained the same because carrier concentration in the bulk region cannot change so I have that I also have the f sub p quote unquote f sub p because it's really f sub metal that also continuous but equal to where it was before and do you see where I have stopped I have stopped on the other side of the junction and I have not continued any further how to continue any further that's your homework problem there's a homework problem you will see that that's how I show that where I stopped beyond that point if you wanted to go on then how will you draw the quasi Fermi level that's one of your homework problem but other than that I have followed the rules I have told you before about pn junction so I haven't violated anything now the reason what what voltage this I said it's negative because only with negative voltage the electron ec will move up electrons don't like negative voltage sort of that's how to remember and it always moves up and as a result there'll be an electron current flow through that structure from the right side to the left side because the barrier is now a little bit lower right qvbi was the original barrier now it's a little bit lower so more current will go from this side to the other side the current from the green side to order from the metal side to the semiconductor that almost remains the same because you see there's wall it has to scale or the barrier it has to go is the same on both sides from from going from the metal to semiconductor but from semiconductor to metal you see the wall is a little bit lower effectively a little bit lower and you realize that this is a reverse bias because I have applied a positive bias on the n side and as a result I have pushed down the quasi fermi level for the electrons while keeping the the metal for quasi fermi level almost the same now you'll not see this type of thing in the textbook and you see in that because they essentially don't draw it for many different reasons but you should have to follow the same rules so that you don't get confused okay you see on the other side of the junction I do the same thing same rules I follow the same rules if you don't follow the same rules then of course you cannot if you want to do the recombination generation events and other things until you draw them properly you cannot really compute properly okay so that's the reverse bias so this I will not have to tell you at all right when you apply a bias then this qvbi whatever was the qvbi even apply a bias with a va that gets reduced and the forward bias in the reverse bias case when you apply a positive thing then the depletion region becomes much wider right that's the reverse bias side xp was zero to begin with so zero I don't really want to modify with a va that remains zero so there is very little band bending on the metal sign okay so now I want to calculate a few currents but first thing I want you to realize that whatever you saw in the pn junction semiconductor if you look at a metal semiconductor junction you will almost see the same things same thing you will see that there is this region 1 2 3 remember this 1 2 diffusion control and bipolar and then there is series resistance drop you will see the same thing in some of the diodes if it has a lot of defects in the junction then you will see this number 6 and 7 the isaki type tunneling and also this trap assisted recombination at low voltage that has half the slope right do you remember half the slope because the electrons was coming in the junction and recombining so that regions are exactly the same exactly the same physics is different but the features essentially exactly the same and in the reverse bias is very similar you'll again see a region 4 and 5 4 region is as a depletion is getting larger and larger with reverse bias then there is this generation current that goes with a square root of v right square root of v is how far the depletion changes and so you see region 4 region 5 you'll have impact ionization if you apply large enough bias then electrons will come in they will multiply within the junction in huge amount of our current will flow so you will have 5 also same things basically the same physics so I'll not repeat many of them because you can you can already understand it yourself now let me now calculate the current for one and that I will use then to explain the rest of the things so the current for one then let's see whether I can calculate the current this way j sub t is the net current that's flowing across this junction I'm not using the drift diffusion theory right previously I used the drift diffusion theory minority carrier transport and all those there's no minority carrier here on the right hand side full of electrons on the left hand side full of electrons also I do not have any minority carrier so short key barrier is a majority carrier device right and so I'll have to calculate it slightly differently and you can see that there's a discontinuity in the conduction back okay let's let's look at this very simple so the blue part is the current flowing from metal to semiconductor at a applied voltage of v a and the red one is from semiconductor to metal now I want to calculate it a voltage applied voltage v a so I have applied put a volt v sub a in both cases a for applied now the next one is interesting because next one doesn't happen in general the next line says that the metal to semiconductor the current that's flowing the blue one that one whether you apply a bias or not doesn't matter do you agree with that statement because you see what has happened that although the vbi is moving up and down in response to v a the charge the depletion region in the other side in the metal side is so tiny that even with bias it's not going to change much it's going to stay zero so the barrier that you see is starting from the metal side metal fermi level all the way to the top of the junction that is independent of bias going from the metal to the semiconductor side so therefore this doesn't depend on voltage so instead of writing j m to s metal to semiconductor at v a I could just say that thing doesn't change with respect to voltage and therefore I can set it to voltage equals zero same value that's what I'm saying okay now at zero bias that's what I want to calculate so I want to get an expression for what was the current from metal to semiconductor at bias zero well what was the current at zero the current must have been zero net current must have been zero because of the detailed balance right any process that is going from left to right at zero must be balanced by a separate one going from right to left as a result I can say the net current when the voltage is zero is also zero and as a result the net current from metal to semiconductor at equilibrium must have been equal to each other right going from semiconductor to metal you see that at the voltage zero that makes sense because at equilibrium there is no net flux if there were net flux the fermi level couldn't have been flat right fermi level being flat means no current and no current means the component must balance out each other so that's what we have and so I'm almost done so this is detailed balance and so what I have here now you can see the metals to semiconductor at bias zero I could replace it with semiconductor to metal at bias zero and so all I have to do is to calculate the current from semiconductor to metal add various biases and I don't really care about the metal because I have already thrown that away you see in both cases the metal is the most important thing but in some way the depletion region I don't care about how much it depletes because it's tiny and similarly in here by using detail balance I have sort of forgotten about the blue region and only the red I care about so let's calculate it it's not difficult so the metal to semiconductor current is q as that's the charge nm over 2 now nm is the number of electrons which I don't know because it's not like doping or accepted that I know about so metal I do not know divided by 2 why because half of them going in the from light to left and the other half is going from from the blue half of them is going in the positive direction half in the negative direction so I only take the half that is going to cross the junction and so I have that too and of course not all those blue electrons will be able to go only those that have energy above the barrier phi sub b is a barrier only those that have above the barrier only those will be able to go so I have multiplied with exponential of q phi sub b divided by kt that's the boseman distribution right so the which is barrier is very high only then I can do it if it is very low then of course I have to do this for me did I can other other other issues that I forget and then the thermal velocity the velocity on the average the electrons move so I put that in that's from one side what about the red side now from the red side I have ns divided by 2 n is the number of electrons on the semiconductor side that I know right n is equal to if it is nd doped n is equal to nd right so ns is equal to nd whatever is the doping in the extrinsic region if it is in this freeze out or intrinsic region then I will calculate it accordingly do you see that the barrier this time for this red electrons is q vbi minus va and if the electrons have energy above that only those will go if it has below that it can go but it will be reflected turned back by the mountain well turned back by the barrier and so they cannot go and they will not contribute to the current so I have q vbi minus va divided by kt and thermal velocity once again the same now the two thermal velocity should be the same or not they cannot be the same right because remember velocity is half mv squared is equal to kt and v half m for metal is a effective mass which is different from the effective mass of the semiconductor so they are not necessarily the same but I am going to forget about the blue one anyway so I have I shouldn't have to worry too much about it now this one I have just split it up into two pieces q vbi and va over kt and this in equilibrium this in equilibrium must be equal to this do you realize why because when you said va equals zero when you said va equals zero this quantity is exactly the same as the blue current on the other side because then this two must be the same and as a result I can write it in this particular form this particular form is essentially j semiconductor to metal at zero voltage because the let me see whether I explained it properly in this case this is the current at non-zero bias right at zero bias at zero bias this va should be equal to zero and in that case then the at zero bias the metal and because of the detail balance metal semiconductor and semiconductor to metal those two currents must be the same so if you hide this va part you can see this is the metal semiconductor part and therefore when you have the applied voltage you bring back this exponential and that's the current so you take that thing out and do you realize now that from here I can calculate a current which will be a constant on the left hand side there will be a constant which I do not know a priory but it will be some sort of constant and I have pi sub m and then it goes exponentially well voltage so if you put a log log on the voltage versus current then also log linear so semi log then what slope do you expect at a few if va is a little bit higher than kt then it's going to be q over kt right and so that's the region one that we have discussed before so it has very same dependencies but look at this pre-factor the pre-factor is not like ni square divided by na and ni square divided by nd remember in the p-n junction I had that so the pre-factor has changed but the rest of the things have not changed at all the same exponential dependence I have it here now what I want you to do is to when you go home let me check whether you really or check for yourself whether you really understood how this calculation was done I did it for a metal semiconductor junction right and I got this result whatever the result that I just showed you before now if you do it for a p-n junction generally we did it with the minority carrier you know electron coming in and in a small signal things oscillating up and down and all those things we did you can get exactly the same result by doing thinking about as if you have two electrons two fluxes of electrons one coming from p to n side another coming from n to p side and then for the electrons that are coming from the blue electrons there is always this variable vbi minus va and then you subtract the flux you will exactly exactly get back to the p-n junction theory or p-n junction derivation that you did so this is slightly different way of doing the same thing that we have done before but this is in some way more general because this allows you to calculate current anytime you have a discontinuity the minority carrier transport and other things that you had cannot directly be applied when you have a strong discontinuity you have to do some more modification right so thermionic emission theory will be useful in this case by the way if you look at recent literature of how people handle carbon nanotubes or ballistic transport you know you hear about this the devices are getting so small electrons goes from one contact to the other without ever scattering do you hear about these things what is the channel length these days of transistors so they are working on nowadays maybe 22 nanometer the channel length maybe 12 nanometer 120 angstrom from source to drain and many times they will not scatter at all if you calculate the value of lambda mean free path do you remember mean free path if you calculate it you will see that many times it is becoming comparable or larger than the channel length separation how do you calculate current you don't have scattering if you don't have scattering how do you talk about mobility and diffusion coefficient right without scattering but even in that case if we apply a bias use the thermionic emission theory flux is going from right to left left to right subtract it you'll get a beautiful answer no problem so what you're learning here is even today is exactly the same of what people do in their research literature okay so to conclude I have talked about short key barrier diode it's a majority carrier device do you understand why because it is always the majority carriers are playing transport no minority carriers here very important distinction say historical importance but as you said carbon nanotubes and modern MOSFETs these are actually treated by the same theory so it's also relevant now history repeats itself in some very interesting ways there are similarities and differences in p-n junction similarities in terms of band diagrams and other things right exactly the same but in terms of calculating current we use thermionic emission theory so we should be we should be able to understand this difference and apply them in both ways so that you understand the full power of the of the method now the trap assisted current and avalanche breakdowns in a tunneling all are the same exactly the same I'll show you a little bit in the next class but exactly very similar to the junction diode you have a depletion region well you have shockly redollary combination and then you integrate over the junction you have the same result so what I'm saying that these two pieces short key barrier and p-n junction read this side by side and read this closely that why something happens here doesn't happen here you should be able to clean this up and once you have done this then for the rest of the devices everything MOSFET bipolar and all those things are a series of p-n junctions a series of metal semiconductor and p-n junction combinations if you understand this rest becomes very easy okay thanks