 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 this is lecture 30 on heterojunction bipolar transistor I'll have two lectures and this is the first one now this is some of the materials taken from Professor Mark Lundstrom's previous lecture on this topic he has worked a lot on this particular problem for many years so it's sort of from both of us together we'll start with an introduction that what is this heterojunction bipolar transistor you probably have some idea already and then again new device start with equilibrium solution draw the band diagram first don't worry about how it works band diagram number one we'll talk today about types of heterojunction for equilibrium condition the electronic transport problem that how electron flows what the gain is and those types of things we'll talk about in the next class but today you will see that there are much more variation in heterojunction bipolar transistors and then you will have in a classical silicon transistor and then we'll conclude now there are not too many great references you can this first reference heterojunction fund after study fundamentals this is posted on the web and also Harvard Cromer who won Nobel Prize for this he also has a great paper but I had primarily asked you to follow the lecture and not go around looking for too much new material because that is probably not necessary now as I as you remember the topic we are discussing is the dilemma of a engineer who is trying to build a great bipolar transistor and till 1980s people used to do it by all silicon silicon or for emitter base and collector all silicon transistors and you could manipulate the doping the emitter doping higher than the base doping the collector doping a little lower you could manipulate the doping to get great gain and subsequently people tried with poly silicon transistors right poly silicon suppresses the base current and therefore you have good gain but at the end as I have mentioned to you in the last class that at the end of course the poly silicon can only limit the current to b sub s the last term and on the denominator which is the sort of thus recombination velocity or effective recombination velocity to the poly silicon which suppresses the gain but it can only suppress the current base current only by a certain factor and the problem is that as you keep making your base smaller and smaller and smaller eventually you have that v thermal velocity limiting your transport or gain so at that point you really do not have any other options but to look the only and the sole remaining factor which is this intrinsic carrier concentration ni squared in the base and ni squared in the emitter this immediately gives you an idea that how to make a better transistor so for example if you could make somehow somehow if you could make the emitter band gap larger then you realize that on the denominator the blue ni squared at that will have a n c and n v you know these are effective density of state that's what quantum mechanics hides and all those things but the main point is the main factor is e to the power band gap multiplied by beta which is kt 1 over kt now the band gap for beta a band gap for the emitter is e sub g comma e and correspondingly you will write out the one for the base right you'll write out one for the base but you can see that effective density of state well more or less they're about the same 10 to the power 22 or so let's say now so the only remaining factor and the big factor at that is the difference in the band gap and you can immediately see if if the emitter band gap is larger then the blue factor will be larger than the rate factor in terms of the band gap and you will have an exponential increase in the game because of the increase in the band gap now this is an idealized case we'll see that reaching this limit is not really that easy so do you remember that essentially these devices are all vertical we take a one-dimensional cut rotate it 90 degrees and generally plot it out this way and then the emitter region I have indicated in red that the band gap of region 1 which is emitter is larger than band gap of region 2 which is the base now in this particular case we have the band gap for the base and the collector are the same you see e g2 and e g2 what does it mean it means the emitter is of different material but the base and the collector are made of the same material however there is something called a double hetero junction bipolar transistor where the emitter base and collector band gaps all three are different the problem with the previous version or the band of the single hetero junction bipolar transistor is that for the collector many times if you keep the band gap small then when you apply a large bias then the collector generally breaks it through impact ionization so there is a loss of control of charge so what people generally do is on the collector side they again uses larger band gap material such that even when you apply a large bias it doesn't break so easily break meaning through impact ionization the transistor doesn't lose control at easily but this again brings in additional fabrication difficulties as well as conceptual problems that we will see now most of the transistor of course doesn't look like this none of them the actual transistors with look something like this you will see it looks like a mesa you start with a semi insulating substrate semi insulating substrate means that this is made of semiconductor but you have so many defects that you intentionally introduce that the Fermi level is pinned in the mid-gap so it's almost like a insulating material so it's called a semi insulating material why do you need it because semi insulating material have very low conductance as a result because this transistors will be used in extreme high frequency do you remember in the last class we said that if you have too many junction capacitances that reduces the f t of the transistor do you remember that the junction capacitances must be suppressed and by using a semi insulating substrate to begin with you reduce the capacitance so therefore it will be go to a very high frequency you can see the n plus collector n plus sub collector which is the red and n collector and p plus base and then n these are essentially you h quarter you start with a larger device you sequentially h to define this various regions now there are many applications of course first of all in optical fiber communication this is when I was in Lucent I worked on 40 gigabit system so many of this very high performance systems are actually made of HBT's now of course there are a to D converters and mostly these days these are military applications for very high frequency very high frequency equipment and these days people are talking about even a Terahertz transistors but all based on HBT so if you go and take a job if any of the defense contractors this is what they will have you work start working on so they're having a Terahertz power gain cutoff frequency FT remember oh no this is power gain so power gain what was the factor there was one was FT and another was you remember F beta I don't remember I just I'm losing it for a second but I will come back to that so on that one it's a power gain not the normal gain of the current itself so there was a second definition and so that is the second cutoff frequency that one has to define so you can see how a actual transistor looks like let's focus on the left looks like the one of those Mexican temples right you see subsequent measures defined one top one top of each other and you can also see you can see emitter base and a collector and you can see that the emitter and the collector and base the contacts are coming in like a flying serpent right it's like a and the reason is of course reducing capacitance is everything because in by removing the material from underneath you are removing the line capacitance because unless you remove the line capacitance at these high frequencies before the signal comes to your device most of them will bypass and will return back to the source and will not reach and will not be amplified by this transistor and you remember I mentioned it few times that chroma on Nobel Prize for defining the concepts of by of HBT of course the original concept is still due to shockly shockly did everything he was of course a genius of the highest order and but the actual potential of how to use it and wide variety of ingenious techniques are all due to chroma and the chroma is essentially worked out all the details and so he is primarily credited with this technology so after all these interesting stories get getting back to business of equilibrium solution so the first thing is that we'll have to grow material one top of each other right for example we might have an aluminum gallium arsenide emitter and the base could be gallium arsenide or we could have indium phosphide as the emitter indium gallium arsenide could be the base so these are different materials now you cannot simply take a material and put another material on top of it because as soon as you try to put another material this there will be defects forming unless their lattice constants are about the same for example the aluminum gallium arsenide emitter gallium arsenide base look at this you see aluminum gallium arsenide that has a certain band gap band gap let's say about 2.12 something like that and then then you have gallium arsenide over there both having on the x-axis a lattice constant which is about something between 5.65 or so since they both have the same lattice constant when you grow aluminum gallium arsenide on top of gallium arsenide they essentially nicely match up there there's atoms nicely match up and so there's no strain and no defect generation so therefore these you can use for making an HBT now many times you would also like to use let's say indium phosphide do you see in the middle indium phosphide is right in the middle with the lattice constant of about about maybe 5.9 a little less and band gap of 1.3 or so let's say now in that case if you want to have another material right underneath for the base the right one would be a combination of let's say indium arsenide and gallium arsenide do you see the bow shape line and the rate point sort of sitting in between so you'll have to take a combination of right combination of indium arsenide and gallium arsenide mix them and then have a material whose lattice constant is just about right for indium phosphide and then on top of it grow indium phosphide this is why we studied remember all the lattices we studied the lattice constant in the beginning of the semester this is why we studied this because without this many of the technological features would be impossible right so lattice and the lattice type and structure is very important to define a particular technology now these days there are wide variety of combinations for for various materials for transistors for solid-state lighting and all sorts of things so this would be an indium phosphide indium gallium arsenide transistor so let's solve the Poisson equation by this graphical method and that's the band diagram and rules are exactly the same I cannot remind you too many times that you draw the draw the Fermi level one side is n the larger bandgap side let's say on the emitter is n so you put the larger bangers the bandgap side as n and let's say for this particular case aluminum gallium arsenide this is 1.8 eb let's say you will put the vacuum level with chi one similarly you draw than the other side you can see that this other side is this other side is p type and also has smaller bandgap remember this is going to be the base and base must have smaller gangap compared to the blue emitter so I'm putting this red as a smaller bandgap p type chi 2 of course you have a certain work function depending on the material whatever material you have gallium arsenide you will have certain chi 2 that you will read off from the book and then it will make this whole thing continuous copy them down and that completes your band diagram that's it now you see I have written a word called type one this is a type one heterostructure what it means that the bandgap of the small one fits right inside the bandgap of the large ones so therefore you can see there is a drop from coming from emitter to the base from the n type to p type the electron going down and similarly coming from the n type to p type the holes also there is a positive delta ev there's a change in delta ec and delta ev so the bandgap of one is fitting right inside the other one this is called type one hetero junction because soon soon we will see if each cases where the one bandgap and the other bandgap will stagger such that in one case the delta ec is positive but the delta ev is actually negative so those will be other types that you can see there are possibilities here that homo junction transistor silicon transistor of course wouldn't have so okay so you know how to draw band diagram then and these would be called abrupt hetero junction transistor because one material abruptly stops emitter stops and then another material begins the composition of the material doesn't change as a function of position what is the composition you can see on the top that 30% of the gallium atoms in gallium arsenide has been replaced by aluminum so that's what it says aluminum 0.3 gallium 0.7 and arsenic that's for the emitter region and that's 1.8 now again the first thing you do with these things is VBI you calculate VBI right I'm doing the same thing over and over again there should be no mystery how do you do VBI no rocket science again delta 1 plus chi 1 plus q VBI is equal to EG the bandgap on the second side right on the red plus chi 2 and you have sort of gone over so you have to take out minus delta 2 you know this minus delta 2 business and you know this delta 1 and delta 2 how do you calculate that you know the doping and so therefore you can easily calculate the how much the delta 1 and delta 2 are going to be so this is very easy actually you put in the values for delta 1 and delta 2 and remember the two sides do not have the same effective density of state right two different material so how you don't have the different you you'd have different effective density of state you can see Nv2 and NC1 and so on so forth EG2 and that's second care of and chi 1 and chi 2 of course you read it from your table or from a book from that material this is tabulated how do they get chi 1 and chi 2 by the way these are all this photoelectric experimentry remember which allows electrons to be pumped up into the free region so that it can be collected so from the first part of the course okay so you know QVBI if you know QVBI then the rest of the things are easy do I remind you one more time that if you have two different material then you have this kappa 1 and kappa 2 and electric field is not continuous right electric field is not continuous what is continuous displacement D the displacement is continuous displacement is continuous only by the way and specially in a conjunction bipolar transistor if you do not have any interfacial charge if you have charge between base and the emitter let's say the defective layer you have have not been able to grow it very well a defective layer then D1 minus D2 right will be equal to the charge that you have in the interface if you do not have any charge then these are the relationships that's what you're assuming okay these we have done so many times we should be experts now you again realize that NB XN and NA XP which is the depletion on the two sides are inversely related to the doping on the two sides so we know that right so you can we always like this and you see this first relationship is essentially charged conservation relationship it really doesn't care about whether you have kappa 1 one side kappa 2 on the other side it really doesn't care but on the second one it does so when you have VBI then in order to have VBI you have to have the area under those triangles for the electric field right for the for the blue line and the red line and correspondingly you have this electric field at 0 minus multiplied by XN and electric field at 0 plus multiplied by XP divided by 2 is a triangle and so you can correspondingly calculate insert the values of the electric field you can get the expression for VBI two unknowns two equations well you have done it many times you know how to do it XN is only the same expression as before except you see this two relative dielectric constants sitting there does it look about right had it been the same material had both sides been silicon then the two kappa should have been the same right on the emitter and base do you see if the kappa had been the same on the numerator you would have been kappa square on the denominator you would have pulled out a kappa get rid of one of the kappa old result for silicon that is how it should be and then correspondingly you can get a value for XP and therefore now you can correspondingly find the depression reason one thing you realize because I have this gain through band gap difference I'm no longer constrained with having the base doping lower than the emitter doping because now emitter doping could be lower than the base doping but so therefore in general the gain could have been less than one but because I have this band gap difference that is giving me so much that I can spend it for inverting the doping sequence right and that is the whole point about a conjunction by puller transistor I do not have any problem anymore with early effect because my doping base doping is large so no problem with that and correspondingly my carc effect is also suppressed because now I can raise the collector doping so all those problems are gone because I can take advantage of the gain get a lot of gain by the band gap difference you see that is the central point why we do this okay so few more types fights of hetero junctions and then we will be done what is the difference between the picture that I just draw a few minutes ago and this one the same aluminum gallium arsenide gallium arsenide system except it looks very different why does it look so different because it is actually P type for the emitter and N type for the base do you see that you see that the green is again the flat Fermi level and on the base side do you see that it is on the on the N side and gallium arsenide side the conduction band is very close to the Fermi level you see that and correspondingly balance band is very close to the Fermi level on the N side by the way one comment about the symbol anytime you see a capital P that means that's a larger band gap region because otherwise you know many times just from the material given unless you look up in a book you didn't know what the band gap is right so that indicates to you that the larger band gap is whatever has this larger capital symbol that has the larger band gap now it's again you can see the delta EC and delta EV both are positive of course this is the original combination of materials and therefore it's still of course type 1 junction now let's talk about type 2 type 2 junction this is another lots of places there are various applications so again I'm doing an n type on the emitter same rules but this time you see the band gap does something like this now that's a type 2 because you can see the red one is not fitting inside it's not fitting inside the blue one but rather delta EV is negative the going the other way delta EC is positive just like the previous type one type one transistor okay that's fine now the homo junction hetero junction transistors also allows you to do isotype isotype meaning N and N you know previously in silicon if you had N and N that would be very interesting right it's a bunch it's a sort of uniformly dope material if you had a little bit of doping change maybe this is slight variation not too much but for hetero junction transistor even the isotype meaning both sides N could be interesting you can see here both side is N type and you could also correspondingly have both sides as p type and correspondingly draw band diagram now I'm not stepping through the steps to create this band diagram right but when you are in your home you should check to see whether you can draw these band diagrams yourself now one thing you realize on this band diagram that although this both sides are N type but in one side you have a depletion region right because charges have been pushed back in one side so you have a depletion charge however on the other side you have accumulation on this little notch you will have electrons and this therefore you cannot use the depletion approximation we used to calculate the calculate this XN and XP remember those are supposed to be purely space charge that the mobile charges have been pushed away only the donors have been exposed and the acceptors have been exposed and in that case we could calculate those charges here however on one side yes charges have been pushed back and so therefore on the larger bandgap side yes you can have this depletion charge but look at the other side in fact charges have come in because this has gotten closer to the Fermi level the rate Fermi level so therefore I cannot use my standard standard depletion approximation QVBI and all those formulas now how do you calculate it well that will be another course where you will learn how to do this but for the time being you can either you just understand it and if you do numerical simulation you know through nano hub they will do all these things automatically so you don't have to worry about it now by the way one thing people often say that if you have two metals this is a exam problem I often give that if you have two metals do you have a barrier and most people say well I have one metal another metal lots of electrons no barrier not really because of course metal similarly have different work functions and when you bring two metals in you will have correspondingly a corresponding discontinuity that's how you make thermoelectric coolers or because you have many times two metals that comes in you know your picnic coolers sometimes you take to picnic in order to keep it cool many of the junctions thermocouples for example two metals coming in you heat one junction cool another junction by passing current through it that happens only because when you put two metals in they do not have the same same band line up so when the electron goes over from one metal to another it can lose energy or it can gain energy through that notch so it's a very practical thing that you see and use all the time now there is a type 3 also and type 3 is really bizarre let's let's look at it so this time that you can have a band gap like this range so let's say I have this gallium antimunide and indium arsenide this two I have so I'm drawing the same band diagram and you'll have to sort of put it in same continuity of the valence band sorry vacuum level you can do that and when you do you see that this again is let's say p type on the n p type on the emitter side and n type on the base side let's say do you see now the band gap through two sides band gaps even are not inside each other at all it has been completely shifted out and you can correspondingly see that in this region very strange things are happening there's no depletion region per se on both sides carriers have accumulated on one side what has accumulated on the left side left side of the junction a bunch of holes a huge amount of holes have accumulated on the right hand side what has accumulated electrons because look at the Fermi level the green Fermi level has gotten inside the back that means that this is now has become degenerate so yes you always have that balance of charge on the you have p charge on one side the holes and n charge on the other side the electrons you have the balance of charge you have the junction but these charges are not space charges these are mobile charges and so therefore this behaves in a very different way than corresponding to a normal junction a norm quote unquote normal junctions that we that we know about so let me conclude on this part so we are talk we talked about hetero junction transistor and it offers the solution to the gain problem associated with polysilicon transistor with extremely short base because you see we got stuck we couldn't get any gain and the base doping if I needed to the base doping to be lower than the emitter doping we were in trouble so therefore we got out of the trouble by manipulating ni square and that gave me a lot of gain that I could trade off against other thing and steel could have a fantastic game now equilibrium solution of HBT of course this is just like normal BJTs bipolar junction transistor but of course I have much many more options type one type two type three depending on how the bands been what materials you have now if you do not follow the rules right there's no hope that you will get this band diagrams right because this is complicated enough that you can never memorize these things if I give you a combination of material from let's say pick something up from one of the papers in the literature if you don't follow the rules of these things consistently which takes a few more minutes then just memorizing it then you will be in trouble so let's follow the rules now as I said three different types of hetero junctions the first one is type one where the larger band gap material completely nestles the smaller band gap material the other one is staggered where the valence band is out of alignment but the conduction band is still inside but it could go the other way or so valence bands could be inside the conduction might could go outside that's all type 2 and type 3 well they don't even talk to each other they are essentially completely misaligned now for type 2 and type 3 we wouldn't use the classical depletion approximation in order to get the band band liner now we haven't talked at all about current transport the gain and all other things that we will do in the next class