You can follow along with this presentation using printed slides from the Nanohub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. Let's get started. So this is lecture 26 on short key diode. This is the second part of the lecture where we'll be talking a little bit more in detail about the short key diode. In the last class we talked about why short key diode is historically important. Remember this was the first diode that people used much before they knew anything about semiconductor or how semiconductor and metal junctions work. This is before quantum mechanics before spatial relativity. This is I think in 1890s or so. That's first when they started working on this. But still even today in scanning tunneling microscope and for wide variety of other situations this device is still very useful. And for many practical devices even inadvertently, this is important as I'll explain a little bit later. Now today we'll talk about in a little bit more detail about some aspect of short key barrier diode and let me get going. So the first thing I'll be talking about is a thermionic emission current. This current you'll use this formulation anytime there is a discontinuity in the conduction and valence band. If there is a discontinuity then you want to actually calculate current always using this thermionic emission formula where you take a flux from the right, a flux from the left and then use the ballistic transport essentially to connect the fluxes across the junction. This you always have to do anytime there is a heterogeneous junction. Now I'll do a little bit more detail derivation. The last class we did a simple one and that was good enough. We will talk a little bit more today. We'll talk about a little bit about AC small signal response. Now this is very important because it is fundamentally different from PN junction. This is a majority carrier device. You do not have any diffusion capacitance in this structure. Anytime you have minority carrier injection then minority carrier response you have a large diffusion current and therefore diffusion capacitance is a very important component. Nothing like that here. So we'll see that these diodes are actually very fast. Talk about a few more things and then especially surface states and then I'll conclude. So when you want to calculate current across this heterogeneous junction here the structure is such that the metal side has been grounded. The metal is on the left hand side and you see I have just shown one line because that's the Fermi level inside the conduction band. I have not shown the age of the conduction band which is much below and of course even in metal there is always a band gap but that's further down. So that's why we have not shown those things in this picture. You also know that there is a slight band bending on the metal side at the junction that I have not shown as well is because the metal concentration, carrier concentration is so high 10 to the power 20 per centimeter cube easily. So therefore the amount of depletion and the band bending in its minuscule has not been shown in this particular picture. Now in the last class we looked at the two fluxes and calculate the current easily. Let's look at this a little bit more detail so that you understand how these calculations proceed. If there is a this device has been forward biased because the metal has been metal side semiconductor side is n and the n side has been connected to the negative terminal of the battery that's why the Fermi level has been pushed up and the net barrier has gone down from VBI to VBI minus A VA is applied bias and you know these two fluxes and these red and the blue fluxes are of course going in both directions but as far as I am interested in the calculating the flux at the interface I just take half of the fluxes that are going in the opposite direction. So generally you should really calculate use a formula like this. It is exactly the same as a simple formula that I showed you before but a little bit more rigorous. Let's go through it slowly. I have shown here three integrals three integrals over kx, ky and kz. You can see volume divided by 4 pi cube. Where have you seen that thing before? That is the shell right. Remember density of state in that calculation we had a shell for k and k plus delta k and we divided into small cubes. Do you remember that this must have included spin because it is 4 pi cube not 8 pi cube because I have already multiplied by a factor of 2 for the spin and this therefore is indeed if I didn't consider anything else and if I wanted to look at the density of state at a given point kx, ky, kz then that would be my density of state and you have done this homework also right. Do you remember that instead of looking at a shell between e plus delta e there was also a homework where they asked k plus delta k, ky kx plus delta kx plus ky plus delta ky kz plus delta kz. You had that homework and this was in preparation for this calculation. Now that is the density of state at a given value of k then this is the occupation. This is the Fermi occupation. Now of course this is only applies if it is degenerate or sorry non degenerate because then you can make this Boltzmann approximation of the Fermi Dirac distribution beta is 1 over kT. Now the reason I write it out this way rather than simply calculating the density of state multiplying by the Fermi factor is because there is something special about the velocity for this electron and you can see that. Notice that if an electron is going from let's say right to left in the x direction then it has a chance that it will go over the barrier and will contribute to the current. But if I had an electron which is going along the y direction perpendicular to the junction well it can keep going forever as long as it likes but since it will never cross the junction so therefore it will not contribute to the current right. And similarly in the kz direction in the z direction the electron could go forever but because it is never crossing the junction it is not going to contribute to the flux. As a result I cannot simply just integrate them all out and call this is the density of state multiplied by this velocity. You see the reason is of course not all those electrons going in x, y and z direction are going to cross the junction and they will not contribute to the current. So I write it resolved in x, y and z and look at the velocity I have multiplied with vx because that is the only part of the flux that is the only part of the flux that is going to contribute to the current. Now you also see that I have put the look at the limits in particular other limits are infinity to minus infinity whether electrons go in the plus direction or minus direction at whatever velocity that is fine they are not contributing to the current directly. But look at the velocity that on that for the velocity along the x direction I have a limit which is minus infinity is fine that if you have a lot of energy lot of energy of course you can comfortably go over the junction nobody is stopping you but by the way this cannot be infinity really remember all bands are finite. So I am talking about bottom of the band to whatever is the top of the band that is what it is now in for semiconductor bands are about EV or so so that is about infinity. But there are many like organic semiconductors that people are using for displays these days which has a very tiny very tiny band distribution or band width in that case putting the right limit not minus infinity would be the right thing to do okay. But why about what about V min well V min is you can see that V min is essentially is talking about that unless it has a minimum amount of velocity or minimum amount of energy then an electron going there will be bounced back by the barrier and as a result it will not go cannot contribute to the current and as a result it should not be included in the flux so that is where the limits come from. Now this looks like a horrible integral actually it is not that bad especially if you know the answer ahead of time. Similarly for the blue side you can see that there will be a barrier which will reflect the carriers and that is independent of bias because no matter what you do to the right hand side because there is almost no band bending in the metal side so that barrier where the blue is turning back that barrier is almost independent of voltage we told about discuss this in the last class. So now let us look at the Fermi factor E minus EF E is some energy above EC and you can see EF and that is the occupation right Fermi function Fermi function or the Boltzmann function at a given value of energy that only depends on E minus EF so you can split it into two pieces E minus EF you could say okay I have two pieces one is E minus EC right from the age of the conduction band to the energy of the electron and EC minus EF just these two pieces now E minus EC what is that that is the kinetic energy do you remember that we did this kinetic energy and potential energy business before so E minus EC is the kinetic energy so I should be able to write it as half MV square now M star of course because this is effective mass right because it is taking into account all the electrons or all the atoms that are around it so M naught but other than that half MV square and I have kept the EC minus EF yes as is now EC minus EF that value only depends on the doping so long you have a doping of a semiconductor somebody told you what the doping is EC minus EF is a number you can take it out of the integral you do not really care about it so this is what it is so you can see I have this EC minus EF I have pulled it out of the integral to the left hand side do you see that half MV square is sitting on the exponential I should have written an M star here in general now then of course I have written the DKX KX is H bar K is MV because H bar K is the momentum MV is also the momentum so I can write it as M star VX divided by H bar that is it and I can pull it out I can divide this because you can see each integral although there is this three pieces of integral each you can put it in their separate boxes and try to integrate them out you can see the Y and the Z pieces look very similar they are exactly the same something sitting on the top of exponential DV and you integrate between minus and minus infinity and plus infinity and but the X integral is slightly different you see it carries with it extra VX the limits are funny from minus infinity to minus V mean so that we have to be a little bit careful about the remaining two pieces are actually very simple okay so what should the values be well turns out that if you look up in your integral book then you will see that the first one is actually just a square root of pi that's the first one and the second one with some factors with some factors and the second one after you because you know after you have e to the power x square then DX so within some factors is square root of pi and the last factor because once you insert the expression for V mean do you see why that should be the value of V mean because V mean is related to the barrier height half MV square is that V mean must be at least equal to VBI minus VA so therefore I have that expression I will put it in as the top limit and once I have done that if I do this integral I will pull back out this the limit will come here and the exponential and then that will be the final answer so once I pull everything down you know all those extra factors then down this is the final result the final result has a bunch of constant goes as T squared EC minus EF VBI and you can see I can put all those prefactors up front as A0 and QVBI QVA divided by KT very similar to the PN junction flux you see exponentially increases with applied bias very similar but also very different if you remember metal semiconductor if you remember the PN junction and the expression for it do you remember that up front for the minority carriers there was this ni squared divided by Na ni squared divided by Nd that was a pre-factor that pre-factor says that PN junction current is extremely temperature sensitive why is that because ni squared has exponential of EG over KT and the band gap because NC capital N C capital N V e to the power EG over KT right and the band gap itself is extremely temperature sensitive as a result the pre-factor is very temperature sensitive on the other hand here I do not have the band gap explicitly sitting there of course band gap is hiding somewhere and we'll talk about that in a second but in general since there is no ni squared term so therefore although the expression looks almost the same it is much less temperature sensitive now what about now you should try to work it out because you know I have just shown you a square root of pi over there but I haven't accounted for all the constants that we'll need do this when you go home and you will try to see whether you can pick up the all the constants now that's one thing that difference between PN junction and a short key barrier there is of course you can always calculate the current remember that at zero bias the current from semiconductor to metal must be exactly balanced by the current from metal to semiconductor so you set a VA equal to zero so the current from metal to semiconductor will simply be A0 and so the net current is A0 e to the power q VA beta minus 1 beta is 1 over kT now just wait for think about a second about so quantum mechanics is hiding where is the information about doping and and where is the information about the fact that I have a pair of metal let's say aluminum with germanium where is this information in this expression okay so one is you can see m star m star is sitting there so quantum mechanics is there and that will contain the information about whether it's silicon or germanium whatever the semiconductor is now where is and you can see the doping where is that EF minus EC that is where the doping of that semiconductor is hiding that's fine now what about the band gap how do I know that this is a particular metal or particular metal semiconductor configuration that has a certain band gap why do I get that information that is hiding in VBI if you remember the about VBI you remember that had on one side chi sub m from the Fermi level to the vacuum level on the other side we had the chi plus VBI plus the doping so that actually contains the information about where what pair of material it is and as well as what type of barrier and other information you have you know deriving an expression these are not complicated things but you should be able to come back and say that if I change my material let's say somebody saying I want to do now I don't have aluminum deposition I want to do copper deposition today for some reason you should immediately be able to say what is it in this expression that you are modulating of course you cannot modulate VBA that's your triple A battery from outside but you can modulate the constant up front that's your design of the devices so what I explained so far is this IV characteristics in the diffuse or in the thermionic part of the device right and you can see the slope of that region is q over kt as soon as the applied voltage is a little bit high because then the minus 1 will be negligible and then the you can do in a log linear plot and that will have a slope of q over kt very similar to the diffusion in the p-n junction do you remember that okay now you can also do other things many times these devices might have a lot of traps between metal and semiconductor side and so in that case what will you do you will just take the previous expression exactly the same expression for p-n junction and realize that dndt will be your p-n junction just like your p-n junction the recombination and you will integrate between 0 to wn because that's the depletion width just like we did before and you will do the recombination in the forward bias junction now this is something you have also done at homework that when you have a forward bias junction what the depletion width is and correspondingly what is the integrated Shockley-Riedhol recombination so you have done this before and that will give you a recombination generation dominated region that has half the slope of the thermionic current region right why is it because it's the electron and hole concentration both has to balance at the point of maximum recombination therefore this extra 2q over kt that's where that comes from what about on the reverse side reverse side again you know that this is exactly the same this time this is reverse bias so instead of recombination in this depletion region you will have generation so the generation is what is that equal to minus ni divided by tau or 2 tau right so so essentially we will have just have to integrate that and once you integrate that only thing you realize that this is increasing the width of the depletion region is increasing with square root of v and as a result my current will also keep increasing as a square root of v what about that last piece impact ionization or what was the other thing zener tunneling well both can happen here what's wrong if you have higher bias on this positive let's say you put a high reverse bias on the semiconductor side the electron can easily tunnel in you can see the triangular barrier don't you see that on the left hand side I have huge amount of electrons full of electrons on the metal side and it can easily tunnel through that triangular barrier to the semiconductor and therefore I can have a huge amount of current just like a p-n junction similarly I can have impact ionization that is a carrier may be emitted or generated in the depletion region and then it may start rolling down the barrier and in the process if it has high enough energy that will then it will generate its its pair and this process will continue and then you will have an avalanche multiplication exactly exactly the same physics that we have discussed for p-n junction so I wouldn't go into that you will see exactly the same thing so just remember this is the one sided junction that is the important thing because there is no depletion on the other side so apart from that everything is exactly the same now let's talk a little bit about ac small signal now in the small signal case so everything I discussed is like dc you know all the currents look the same but the thermionic emission current is a majority carrier that's why not very temperature sensitive relatively speaking compared to p-n junction all other parts this region 1 2 3 4 5 6 those are all exactly the same as we have done in p-n junction now the ac response is very interesting because very different actually this is a again a metal semiconductor diode n-type contacted with the metal the dashed region and you see that you have a triple a battery that's the va and let's say you have the radio your favorite radio signal that is in a microvolt or so and that is a small va and as the bias is modulating with that small va sometimes it's a little bit more forward bias sometimes a little bit less so it's oscillating right the native bias is oscillating and so therefore there will be a current that will oscillate in response and again in principle you should have all this conductance at that bias at that bias governed by the dc bias the junction capacitance well that I might have and but the thing is that I will not have a diffusion capacitance here and that's what I want you to understand why I don't have a diffusion capacitance and therefore this device is very fast actually so how will you calculate the conductance well the rule is the same because you know the equivalently you have the same expression beta is kt and m depends on whether you are in the thermionic emission part you know that's then it's one or you are in the trap recombination generation dominated part the name is two and in that case so you will provide put the relevant expression for m in the region you are operating right whatever you use your dc bias that point will all automatically tell you what is your m at that point of operation right so you have a value of n you take a derivative first you take a log bring the i not on the other side and take a log and then then take a derivative with respect to the current and as a result you can get an expression for the forward bias conductance and forward bias conductance you can see that has a series resistance rs what is this series resistance coming from this is for the electrons as they are coming from the contact all the way to the junction all the resistance in between that is the series resistance because remember that's not what we have accounted for so far only thing we have accounted for so far is what happens in the junction but if i if i have a one meter long wire copper wire of course even before coming to the junction that one meter long wire will have certain resistance same for the for the semiconductor side so i will have that and you can correspondingly calculate what the forward bias conductance would be okay what about the junction capacitance well do you remember the junction capacitance is always associated with majority carriers that as soon as you start sort of changing the voltage across the two ends of the junction in the majority carrier side the pulse immediately comes to the junction side given by this what is this time called dielectric relaxation time because when you have a majority carrier the way the response propagates is not a electron doesn't go all the way through this electron tells its neighbor that i have been pushed help me next that one says to the next and that propagation is extremely fast that's why when you turn on the light in the room then the you don't really have to wait before the electron himself goes in the bulb and says let's turn on but rather each pushes the other and eventually this propagates very very fast on the order of less than a picosecond so unless your computer is operating at a terahertz then you really don't have to worry about this propagation time because that's very fast and the junction again the bias when it's a little bit more your junction is a little bit less when it's a little bit less your junction is a little bit more and so this on one side there is metal electrons coming from the one side the other side the majority carriers are coming and this bounces back and forth at the junction majority carrier phenomena and therefore you calculate the majority carrier capacitance a epsilon a over w but w is just the depletion with whatever the depletion with and you know i have we have calculated the w that if you apply a forward bias w goes down right and then if you have a reverse bias the w will be larger but that's the point you generally get the idea you can easily calculate this now the main point is that there's no diffusion capacitance in short key diode this is because of this following thing in short key diode there's no minority carrier if you don't have a minority carrier nothing is diffusing because as soon as it comes in in the other side where it could be a minority carrier it immediately joins the majority carrier you know you sort of you come to a new country and instead of keeping your identity for a while you just you sort of blend in with the majority and in that case you can enjoy all the facilities or the transport advantages that you have with the majority it's the same thing here and the reason it happens is because as soon as the electron comes in when it's this much high in energy it has no band gap or anything right in this region phonons will immediately scatter it down as if it has recombined with the majority carriers immediately as a result it is not sitting up there sort of slowly responding and diffusing as a result you do not have any diffusion charge and if you don't have any diffusion charge as soon as the charge comes in here then it responds with the dielectric relaxation because it's a majority carrier and I do not really have to wait for the electrons to gradually go by back and forth and respond with the minority carrier lifetime so this is about short key diode and this is I want to emphasize very different from the p-n junction diode where you can see the red triangle on the left hand side is the minority carrier unless you have a lot of traps sitting there then it stays on the minority carrier side for a long time as you bounce your potential up and down then the the string the blue wavy line that says that how it diffuses each individual electrons needs to go and that takes a lot of time so therefore we say we say actually a slower device now you could make the p-n junction diode almost as fast as the short key barrier if you do this many times people will intentionally put in put in defects on the minority carrier side if you do that then what's going to happen that as soon as the minority carriers goes there then the defects will allow it to join to provide a sort of a by way or a shortcut to join the majority carriers and if it can join the majority carrier first then the amount of stored diffusion charge on the minority carrier side is very small and as a result this could respond as fast and that's what many times people do in terms of in terms of if they wanted to make something very fast a p-n junction diode very fast this is something people have done historically people have done i do not know whether they still do it they might still do it for some specialty devices i do not know exact examples where they are still used so let me talk about a few things about this metal semiconductor junctions metal semiconductor junctions are everywhere if you have a metal let's say even when you talk about a p-n junction even when you talk about a p-n junction or any junction whatsoever you always have to connect at that end of the day with metal right and if you are connecting with metal right there's a wiring if you want to wire things up then you will always have a always have a barrier now the question is that if you have that barrier then how is current going to flow you don't want that anytime you put a metal down that there is a barrier because then current may not flow unless you apply a bias so what do people do if you do not want the metal semiconductor junction to bother you because if that is not the main purpose of your device let's say you are doing a normal p-n junction diode normal and then you have to put metal on both sides now then another two short key barrier diodes short key barriers are sitting on the top and the bottom you don't want that you just want the p-n junction in that case only records you have is to dope the n side very heavily that's why do you see that in the purple region I have shown n plus n plus means very heavy doping and the what it does is that when you have very heavy doping then the electrons don't have to go over the barrier they can just tunnel through the barrier and as a result the resistance may be minuscule in that case because they can just go through the very thin triangular barrier region now that's for one type of carrier do you see is although it has problem with those electrons if you wanted holes to come in and out you do not have a barrier for the holes to come in and out do you see on the bottom side that if the hole wanted to come in and out there's no barrier so the whole transport is always omic omic means linear dependence with voltage no exponential dependence with voltage so on the one side for the minority carriers in this case no barriers so they can easily come in and out now for the major majority carriers there is a barrier but you can get rid of that effect by doping it very heavily and that's what the tunneling is in an ultra thin barrier electrons will tunnel and so although it has a barrier but it will behave as if it had nothing and that's called an omic contact so people will often ask you that are you sure that you have an omic contact they're asking that whether you have treated the contacts properly so the short key barriers is not disrupting your flow of electrons there's another thing that is also sort of you'll see in various textbooks that people talk about although I do not know how important this is and that is something to call lowering of the short key barrier and this is what it means when you have an electron in the top let's say a red electron trying to get out to the metal side from the semiconductor I have so far assumed that they can this is like a free electron going from one side to another but you remember that I'm going to a metal do you remember that if I have let's say from undergraduate years forget about this particular situation that if I have a metal then if I have a charge sitting on top of the metal then the electric field must terminate vertically on that surface because if I had even in a tangential component current will flow in the semiconductor in the metal and current cannot flow there cannot be any potential difference so it must terminate vertically that means there's always sort of an effectively an image charge on the other side anytime I have a charge on top of a metal so that is what I have shown here that anytime the electron is moving the red electron is moving towards the metal it induces the surface charges on the metal right because it must terminate perpendicularly as a result I could in a state of replace the metal with a corresponding equivalent image charge so if I have an electron on the red what will be the sign of the yellow one that must be positive right image charge is always the other polarity and so I have a positive and a negative charge they must attract each other as a result effectively speaking as a result the red electron will get an extra pool to pull it out of this of the semiconductor as a result what's going to happen that look at the barrier when you combine this extra electrostatic pool then as if the barrier has gone down effectively because the electron can pass out from this here a little bit easier and so that's why that barrier is a little bit lower than what you'd have otherwise would have thought about if the other side was a semiconductor then you wouldn't have this effect as a result this extra lowering of the barrier is sometimes very important remember this is exponential in the barrier height the current exponential in barrier height so although it's tiny it may be like the several hundred millivolts but that on an exponential can change your current quite a bit so therefore many people discuss this as an important effect now this is a topic I have not worked personally I most of the other things I have worked in my career sometime or other so what I'm saying is sort of a little bookish in the sense that that's what they discuss in the book I do not know for sure that how important some of these things are there's a reference in the Z's book and page 143 that you can look up finally it's something about called Fermi level peening and the Fermi level peening is a very important recent phenomena and let me explain to you that why anytime you bring a metal to a semiconductor and push them together you may not always get conduction at all conduction you have to do something very special about it the reason is that semiconductors are a very special thing when you terminate a semiconductor right make it finite do you not have a lot of interface states do you remember the surface recombination velocity and all those states sitting in the mid-gap and the recombination that we talked about of course those are all there I mean simply because I haven't drawn it doesn't mean they do not exist and so what happens that near the age of any semiconductor there's a lot of surface states as if you have this NT NT is the number of traps right like you have a lot of traps like as if you have a lot of doping in the end of this region because if you solve for the position of the Fermi level in the presence of the surface states then you will see the Fermi level is very close to the middle because we have so many traps as a result you see that the red region I have shown with sort of a Gaussian car that is somehow to represent the presence of the surface states and when you have that amount of surface states it pins the Fermi level to the middle of the band gap regardless of what you did internally it may be a heavily end up region really it doesn't matter because EF could be close to EC near the middle but near the edge the doping essentially they have been taken over by the surface states as a result what happens often that if you are not careful and don't treat this surface space states properly then even if you so let's say in one side the metal on the right hand on the right hand side metal has a given work function on the left hand side let's say metal has a different work function but it wouldn't matter the reason is that even when you bring them in in contact what's going to happen that the interface region will exchange carriers with the metal and will not tell anything to the semiconductor because that has enough electrons as the result was going to happen that the barrier the net barrier will not be modulated by the choice of your metal which is a horrible thing right because you wanted to design a transistor that changes with the metal as you change the metal but if it doesn't respond because it's like a gatekeeper you know this surface states is like gatekeepers and the gatekeepers are all absorbing and talking to the metal on the other side not letting the bulk of the semiconductor know as a result the depletion doesn't change and as a result the modulation doesn't occur so this is a very bad thing historically Bardeen was the first one who explained it in 1940s as why many of the devices do not work but even today even this year if you go to conferences you'll hear this term Fermi level pinning because many modern transistors they are trying to put hi-k dielectric on the device and then on top of it metal and they are worried about Fermi level pinning because most companies cannot solve it properly and that's why many companies cannot introduce hi-k products with metal gate in the in the marketplace the same problem but there are solutions of various sort but this is something important to know about metal semiconductor junctions so let me conclude so as I said short key barrier has wide range of applications and this is very practical device but more importantly metal semiconductor junctions are everywhere you always have to connect your transistor whatever carbon nanotube spintronic device whatever you have you always have to connect them with metal and so metal semiconductor junction whether you like it or not it's always there and you have to handle it properly now one distinguishing feature from p-n junction is a majority carrier device no diffusion capacitance and therefore and also no strong temperature sensitivity which is a good thing and you realize that because I had this hetero junction not because it's metal semiconductor but because I had a hetero discontinuity in the conduction band I had to use thermionic emission anytime next time you see a discontinuity regardless whether it's a metal semiconductor or two semiconductors you know germanium and silicon a discontinuity in between near the near the junction thermionic emission current that's how you should calculate it now one final point I want to make about this I have talked about semiconductor semiconductor I've talked about metal to metal I'm sorry metal to semiconductor the question is can you draw a grand diagram if you had metal to metal copper to aluminum if you put it will there be a barrier you think actually there will be a big barrier if it didn't work then your thermocouples wouldn't have worked how why should be this why this band gap would be coming from because one side you have phi of m1 and another side you have phi of m2 you you'll use the same rules I have told you about flat Fermi level and then vacuum level on the two sides make them continuous pull them down the vbi that you will see even in a metal metal case is always there that's why when you pump a current through metal metal junction you have this thermocouple wire energy can be gained or lost you know there's various types of thermocouples people people can buy so that part I will not discuss in detail because for semiconductor physics that's not important but as a concept that's very important that you understand semiconductor semiconductor semiconductor metal and metal metal how they behave together okay thanks


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