 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. So today we'll be talking about heterojunction bipolar transistor. The current transport through such transistors and this is lecture 31. In the last lecture we talked about drawing band diagrams and there are three different types of hetero structures that we talked about. You remember type one, right? That where the one smaller band gap just nestles in the larger band gap. Type two where it's staggered and the type three is completely misaligned and you saw that depending on the situation it could be depletion on both sides for type one. Type two, one side accumulation, another side depletion and the third was both sides depletion. So there is lots of interesting thing that goes on in HBT is that you do not really normally see in homojunction transistors. Now let's talk today about HBT and the DC characteristics. So I will begin by talking about abrupt junction HBT where one material ends and abruptly another material begins. Emitter is one material, aluminum gallium arsenide, base is gallium arsenide. All of a sudden at the junction the material composition changes and that is in contrast with graded junction HBT's where the material composition let's say aluminum gallium arsenide 70%, 30% aluminum and gallium 70%. Now in abrupt hetero junction that should have remained constant to the base emitter junction. However if you keep changing the composition gradually lowering the aluminum composition so that it changes over to the base region just smoothly that will be the graded HBT, graded junction HBT and that's really a very good thing, more difficult to do but a much better transistor. I will talk a little bit about that when you have run out of steam even with HBT then a graded base HBT where the base composition is changed that helps a lot and not a lot, somewhat. We will talk a little bit about the disadvantages of having double hetero junction HBT before concluding. This you would notice is an abrupt junction HBT because the emitter, the larger band gap emitter continues seamlessly to or continuously to the base emitter junction and then you can see a rapid reduction in the or immediate reduction of the band gap in the base. Gallium arsenide for example aluminum gallium arsenide in the emitter. Now in this case you would remember that this is type 1 transistor because both the delta EC and delta EV both are positive in some definition in this particular way. Now what would be the current? What would be the current through such a structure? So let's go slow but remember first of all that this has an abrupt discontinuity in the conduction band. What type of current transport should we use here? We must use thermionic emission in some form at the junction away from the junction diffusion maybe. Where have we used this before? When you discuss short key barrier again do you remember that there was a discontinuous conduction band and therefore we have to match the two fluxes from the two sides. Now let's see whether we can understand why the electron current shown here in red with the arrow electron current from the emitter to the base has this particular form. For a second assume things are in equilibrium therefore in the expression for Jn which is the last term the last term in VBE that's 0 but even the rest of the term doesn't look right. Ni squared over nB that's the minority carrier in the base that's fine. V is the velocity but what is that exponential factor e to the power delta EC over KT doing there? This comes from the following consideration. You see do you remember that when you talk about metal semiconductor junction thermionic emission at that point that at equilibrium the fluxes from the two sides must balance that must be equal to each other. So instead of computing the rate current which is the electron current from emitter to base in equilibrium in principle I could have also calculated the corresponding current from the base to emitter that two current must be the same. In this case if you look at the magenta then you realize that why the current has that particular form because look at that ni squared divided by nB minority carrier V is the velocity with which the magenta electrons are moving and of course the exponential is suppressing factor because unless electron have that amount of energy well it's not going to go over the barrier. So only the fraction that has that energy and above that is the fraction that goes over. Therefore we now know the current that flows in equilibrium from base to emitter. Therefore by detailed balance we know the equilibrium current from emitter to base and of course if you want to look at with bias how much extra current that there will be you multiply with QVBE divided by kT net current you take out a minus 1 put the whole thing in a bracket then that's the net current that's flowing in. Okay now generally that is small so I haven't really written that minus 1 factor for J sub n. What about the whole current? Now in the whole current remarkably enough here you see that I do not have any exponential of delta EV over kT. Why not? Because you see once it has the holes have crossed over in the emitter side well there's no barrier for backflow unlike the electron one right there's no barrier for backflow so correspondingly when you write a diffusion profile on the emitter side this will simply be a triangular profile the holes will diffuse from the end of the emit base emitter junction to the emitter contact as a result we have exactly the same expression that will have in a homo junction bipolar transistor. So you have this and this is you remember primarily is the base current and the 1 Jn is generally more or less the collector current so you can calculate the gain from here. What is the gain? Well gain is Jn divided by JP you multiply this thing throughout so you see that there is this ni squared ni b squared and ni e squared that's the two factors in red and blue that's what's coming in but do you realize that because in this abrupt hetero junction HBT there is this delta EC over kT that's sitting there. Now that's very bad you can see that that will eat up in my gain and if delta EC even let's say 0.2 EV a small discontinuity what is kT? 25 mille electron volts let's say so even 0.2 gives you e to the power minus 8 and e to the power minus 8 a factor of hundred gone right there. So this is a dangerous thing yes you get a lot from ni squared ratio by increasing the gain but you give back a little from delta EC the any advantage from delta EV you still have so if you put it back you remember that ni squared the ratio of those two ni squared gives you delta EG over kT that the change in the band gap now if you take back the delta EC what remains just the delta EV and so at the end in abrupt hetero junction you do not get the full advantage of the band gap difference only a fraction related to the valence band that should be clear right this is a very important point that why it's simpler to process you grow one material in molecular beam epitaxy then change over or emulsivity then you change over and then grow another material very simple process but the gain is not as much so what do you do well let's take take out that notch that's really causing me trouble now remember that we are expecting a full gain delta EG over beta multiplied by beta by the way there is two beta sitting so don't be confused beta on the left is the current gain beta on the exponent is one over kT so yes just remember that and but the point is that for abrupt hetero junction I got a gain which has delta EV over kT not the full advantage so I need to do something a little better see whether we can do that a little better now now what happens for abrupt hetero junction bipolar transistor you have a corresponding you draw first a band diagram then you look at the potential that's fine but the chi of x you see chi of x up to the junction which is the work function if the same material the chi remains continuous and constant up to the end of the junction and then after that the base also has a constant chi and this is the abrupt junction in terms of composition now if you wanted to do graded in a state then this is what you do you would grade the composition so you start still with 1.8 let's say aluminum gallium arsenide 30 percent aluminum then you will gradually reduce the aluminum mole fraction right let's say we do it for 20 percent 10 percent and by the time you are at the base junction this is really a gallium arsenide altogether so you can see therefore the band gap will change continuously and the composition is changing continuously what is the price you pay the price you pay is during growth when you're growing that material so you have just finished growing your base mbe by mbe let's say then in order to make it graded you'll have to make sure that the composition changes slowly as the film is growing you're putting more and more aluminum in and that requires very precise control because the source from which the aluminum is coming you'll have to continuously tune it so that this composition varies continuously so that's a more complicated and more expensive process but if you could do that in this case then you get a tremendous advantage because you see this time the backflow the magenta one that I had before I didn't draw it here but you realize that if the magenta current wanted to go from the base to emitter then there's no delta ec sitting there holding it back this is exactly like a bipolar transistor homo junction bipolar transistor electrons are injected in the base you have a triangular profile electrons diffuse out if a electron wants to go in the other side go ahead this goes just fine just like a homo junction bipolar transistor therefore the electron current I have j sub n doesn't have that delta ec sitting on the top just like a homo junction bipolar transistor that's what I have well base current no problem right base current is exactly the same as before because I do not have any delta ec sitting there now do you realize that these statements is not true if I had type 2 type 2 had a staggered band gap and therefore the suppression I show no suppression here for example for the base current but in a type 2 there might have been a suppression for the base current but no suppression might be for the emitter current so you have to do it case by case and look at look to see whether back injection is possible without barrier or not and then right corresponding expressions again well no delta ec and so I have the full advantage I have the full advantage of this band gap hetero junction discontinuity which is delta eg over kt and this is a transistor that will have huge gain of course more expensive right because the material composition was more difficult to do okay so that's essentially it that is what chromar was able to show long time back he started early but he stayed on and chromar is one of those people almost like Shockley ingenious the ideas he has he has the same blocks to play with you know but the things he makes out of those same blocks is really remarkable time after time his ideas are actually beautiful and as you stay in this field you will of course is a very important figure you will see many example of his work is just in the style of how people do things that makes work extraordinary it's not what you are given but what you do with what you are given looks like that people can make very beautiful things out of it okay so why do I do HBT I have explained that already what HBT gives you at the end of the day and I just want to summarize is that do you remember that the base doping this is a NPN transistor the base doping is NA and the emitted doping ND and if I didn't have this exponential increase in the gain because of the delta eg then you realize in the homo junction I needed to make the emitted doping larger than the base doping that is what I needed however I can now invert it I can have instead of ND being 100 times more than NA I could make the other way around I could make NA I could make NA much larger than ND yes I take a heat on that previously I had a factor of 100 now I have factor of 1 over 100 from that yellow factor factor on the yellow but delta eg you see did you see that previously it is 1.8 in one side and how much was it let's say 1.2 on the other side for that particular combination 0.6 sitting on the exponential over a KT will give you a huge huge gain as a result even if you invert the base doping you still end up with a huge gain and as a result that gives you a very good transistor physically why that's happening because you are actually suppressing the base current by that band gap quite a bit and emitter current you are not suppressing as much as a result you get this gain now what does heavy base doping give you heavy with the base doping prevents punch through you remember when you tried to pull the collector down there was depletion in the base region which was eating away from the effective base width and as a result there was the early effect the output current was changing as a function of VCE now all those will go away if you make the base very thick base base doping very high moreover you can make it thin because now the depletion will not punch through so you can make it thin and when you make it thin what happens transistors become faster right because W square divided by 2D that is how fast the electrons must go so therefore time takes for the electrons to go through the base so that's again a very good thing and you can also reduce the emitter juncture emitter doping and that reduces the capacitance do you remember where the capacitance comes in there was this q kt over q ic multiplied by the junction capacitance when as it's talking about 1 over fta calculation so anytime you reduce capacitance there are fewer capacitance to charge as a result once again this is a great thing all over okay but this is not all you can have you can have a little bit more and sometimes people do it but it's not all common that is that graded base hbt and this is how it works so do you remember that if you make a polysilicon emitter and if you make a very thin base at the end you are sort of pegged to this value of gain this particular expression of gain and you know that all about it now is there any way to sort of manipulate that velocity at which electrons are going how do electrons go in the base base of a transistor it comes in minority carriers it diffuses around right it diffuses around and then it leaves because it diffuses around doesn't go straight therefore you have this w square over 2d it's not w over a velocity it is w square divided by velocity because it's going a little bit forward coming back going forward by coming back and eventually leaving after a long time diffusion limited transport if you could somehow make it drift limited or drift meaning with the electric field if you could have put an electric field in it that would be great now how do you put an electric field this is how in the base region itself you can make the band gap variable in the base itself instead of having continuous one material gallium arsenide throughout the base you can also change the composition in the base region itself and therefore your band gap will be changed now this band diagram is for a undone material because the formula level is sort of in the middle and you have these steps as it's going through making from a larger band gap little smaller little smaller and on the other smallest so EF is right in the middle but what happens if you dope it what would happen if you dope it with let's say accepted doping because you have an accepted doping the formula level must be parallel to the valence band right like this the same material on the left but this time I have doped it doped it with a let's say accepted doping in that case the separation between formula level and EV regardless of the composition must remain the same right because is that delta between the blue and a black line on the bottom that delta is determined by the doping this is that the same you see and so the as a result since the doping is constant the separation is constant now in order to accommodate the variable band gap then therefore the conduction band must slope and it must slope strongly now this diagram is not 100 percent right correctly drawn what I'll ask you when you go home is instead of thinking about this is a continuous break it up into three band gap regions three materials with three different band gap try to draw it what you will see that on the valence on the valence band instead of being in three continuous line you will have little notches appearing as it goes back and forth it will undulating notches and correspondingly there will be notches also in the base side on the on the conduction band side see whether you can draw the band diagram and argue why the notches in the valence band does not matter but you should do that when you're home okay if you now do this composition now look at this what happened to the base base now has a tremendous amount of electric field so as a result if you now put the electron in well it will not be going to diffuse anymore with that amount of electric field it is going to drift because the electric field will simply pull it on the other side and it will go fast and as a result you will have a higher performance by polarity so the jn that's the classical expression this is the jp a classical expression for the jp but and this is you know this is the gain but look at that with ni squared i have put a bar on it do you see that it's not very clear on the numerator for jn i have nib squared bar because i do not have a constant ni squared anymore do i i have a continuously varying ni squared so i have a ni squared bar and therefore when i look at the gain i have a gain which is continuously changing now is this ni squared bigger or smaller than before this is bigger because the band gap is coming down with the band gap coming down your ni squared goes up right so corresponding to what i had before this ni squared is larger as a result i have larger gain this is simply saying that the time i required for the electrons to go that is faster so for the same time i can pump more electrons compared to the number of holes and that is why this base transit time w square divided by mu e that is much smaller than the diffusion time and that gives me enhanced gain right because anytime you have a certain amount of charge that goes fast goes through the base very fast that gives you enhanced current and that gives you enhanced gain so graded base therefore is a very important very important consideration although it is difficult to do you know you have a 200 angstrom base or a 300 angstrom base region to grade it properly within that region is not that easy and so only desperate people are probably less desperate people sometimes they do it but most of the time you'd probably try to find some other way to improve the game okay so that's about all about single hetero junction bipolar transistor abrupt or continuous abrupt because in emitter base side you had a discontinuity that gave you a lot of things a lot of good things now what about the collector side well on the collector side you can also have larger band gap just like on the emitter side you have a larger band gap on the collector side you could also have a larger band gap what type of transistor would it be you will start with aluminum gallium arsenide and emitter base is gallium arsenide well the collector is again aluminum gallium arsenide now what would it buy you do you remember from the last lecture that when we applied a large bias to the collector then there was impact ionization and the current essentially increased abruptly and that we did lost transistor action now do you also remember that the impact ionization depends on the band gap if you increase the band gap then it will be much later the voltage that you can apply in the output will be much larger than before as a result increasing the band gap is a good idea it's a smart and good idea now why do you want to do it here not in silicon base silicon transistor first of all in silicon you do not have this technology right of continuously varying the band gap you don't have that but more importantly many of this hbt's are used in high power application where the output voltage is could be quite large as a result if you don't increase the band gap then the thing might break very easily and that's something then you cannot get the power gain that you need as a result hbt's for high power applications often will have double hetero junction bipolar register but it has another advantage which I will come in a second first is this symmetric operation because you know that base emitter sorry emitter base collector if you make just once junction hetero junction then you cannot flip the emitter in a base you couldn't flip that anyway for silicon transistors also do you remember because emitter doping was large base doping was little bit small collector doping even smaller right now you cannot make a emitter a collector and collector and emitter right you cannot do that because then you will not have have any gain the collector which is now your new emitter will have a smaller doping compared to the base you would not have any gain so therefore a classical homo junction transistor is not symmetric that you cannot interchange the base and a collector but if you could do that that would really simplify circuit operation many times the interconnections amongst the transistors will be considerably simplified if you could make it symmetric it is possible through the double hetero junction bipolar transistor because now doping is sort of plays a secondary role you could dope the emitter and the collector with same doping because you know and dope the base very heavy and the gain will come from the band gap and as a result you will have symmetric structure symmetric operation can interchange base in the collector which is great now there is also the no charge storage in the base collector junction simply means on the collector side charge storage is significantly less because the band gap is larger ni squared is smaller now the point I wanted to and the final point about higher collector breakdown voltage I already explained right what is the energy do you need approximately in order to have a breakdown voltage three half cg approximately approximately and so therefore increasing the band gap helps you but the point I wanted to make is this reduced collector offset voltage this is a major of the asymmetry and let me explain what it means this is a very strange thing that you wouldn't really expect now if you think about the collector current as a function of vce the collector to emitter voltage now you know how the characteristics look like why does it look like this by the way do you remember this is sort of the reverse bias side reverse bias side of a base this is a pn junction diode you remember forward size is exponentially increasing on the reverse side the current begins to saturate and why does the current subsequently increase well that's because the emitter current is coming in and joining the collector current so therefore this things this thing spaces out now if you have this characteristics if you look at the left hand side it looks like all of them are going through 0 right on the very left hand side it looks like when you said vce equals 0 the current becomes approximately equal to 0 is that right well not really if you magnified this region you will find that even when you said vce equals 0 current is not actually equal to 0 so you do not have you have not put in an output voltage had it been a symmetric device you see the current from base to emitter and current from base to collector would have been the same because you do not have any output voltage it looks the same but because of the change in the doping base emitter and collector are not the same right in normal homo junction it's not the same as a result there will not be there will be an asymmetry in the current the current flowing from base to emitter will be different from the current flowing from base to collector and as a result you will have a net output current flowing in the structure and that is something you do not want so how do you suppress it this will happen in homo junction this will happen in single hetero junction bipolar transistor so this is how to calculate that current you know the j1 j1 is the base emitter current forward bias you have I have written vbe I should have written vbe which is vb minus ve let us say j2 well this time I am not neglecting it because in this case I am really applying a zero voltage in the output it is not like a large collector voltage where the back flow has been completely been suppressed right it is zero voltage so this time I cannot ignore the back flow the j2 now you have done this homo in the homo remember there are equivalent circuit a bar small model in that case for various bias configurations you have you accounted for all these currents actually j1 j2 and everything now throughout the course I was just talking about forward active mode where the output is strongly reverse bias the input is forward bias so I neglected this j2 before cannot do it anymore so that would be your j2 and you can again calculate the current which is between vbc and that is the current flow on the side and the current j3 will correspondingly have a structure now let me assume that this is asymmetric that is you have j3 either to the emitter or to the collector but you have a current because the structure is asymmetric right so in that case one current will be larger than other now what you want the collector current flowing out what will be the collector current flowing out it will be j1 minus j2 from the electron side right because two are opposing fluxes and j3 which is the base current flowing out through the collector so these are the three currents should I add or subtract now that depends on for the electron current do you see there is a strange sign here this minus do you see why it should be minus because the electron current j1 minus j2 the current is going to the right from left to right and the actual sorry the electron is moving from left to right but the current is flowing in the opposite direction so therefore when I want to collect it with j3 j3 what is j3 that's the whole current and when whole current moves to left to right it really the I am sorry when the whole flux moves to from left to right in that case the whole current as well moves from left to right so do you see why therefore I must have this j1 j2 and j3 in this particular configuration if you just look at the fluxes I should have added a plus sign at j3 but you get the idea right okay now in order to get zero output current what voltage at which the current becomes zero because we know that at zero we see current is not zero so I have to put some voltage so that the current becomes zero what voltage is that so let's do that now if you do that you know set those three currents and sum them to zero you will get expression like this it sort of makes sense right you can see all the nib squared and nyc squared those are appearing you see that there is no ni e squared because I am just summing up the current on the collector side so I don't have those those factors now how can I make you can work it out when you have a few extra moments but how can I make what's my goal my goal is to make the offset voltage as close to zero as possible because then I can invert the emitter to collector right I can invert it I have symmetric operation how do you have that what do I need I need the gamma R to be very large if I could make the gamma R very large then you see I will have just a log of 1 log of 1 is zero and therefore I will have the offset voltage equal to zero which is what I want now how do I make the gamma R which is the reverse gain how do I make it large well you can see just like the emitter to base by making the emitter junction larger a emitter band gap larger I was able to increase the gain correspondingly by making the collector junction larger for the reverse flow I can increase the gain but this gain is gamma R so therefore by increasing the emitter collector band gap I can increase gamma R and make the whole thing disappear this offset voltage disappear do you see why where it's coming in do you see where it's coming in because you see n i b squared and now you have n i c squared right and from here if n i c has a larger band gap then what would happen this n i b and n i c squared that will again give you e to the power delta e g over k t right now between base and a collector and as a result this will become a huge factor gamma R will become a huge factor the offset voltage will disappear and the transistor will become symmetric that will simplify your interconnections as you are connecting one transistor okay so I will just end with a few comments about modern design in the next few minutes so at the end this is the expression you have now do you see how hbt helps you which term does hbt help first of all you can pump more current do you agree I see can be larger compared to compared to compared to before for homo junction bipolar transistor why because remember I no longer need to keep the emitter current I'm sorry the collector doping small because previously I needed collected doping small why because my base doping was small my collected doping had to be smaller in order to control the rd voltage right so that the depletion goes on the other side the price I paid for that was the kark effect that when electron came in it flooded the junction and the junction was lost I do not have that problem anymore I can put a whatever doping I want in the collector side if I do that then my kark effect can be pushed down at a higher current it's not so easy with so much doping to overrun the junction as a result my current IC could be larger if my current IC is larger therefore the whole turn will become smaller and therefore my ft will go up right what about the capacitances by decreasing the emitter doping you essentially increase the depletion is that right increase the depletion because depletion is inversely proportional to the doping if you increase the depletion then what happens to the capacitance capacitance goes down because epsilon not a over d is the separation of the plates so that goes down if that goes down this term will go down but if you increase the collected doping too much of course you are going to pay a penalty on the other term so don't do too much don't be too greedy but between these you can see that you can now drop this term quite a bit higher performance what about wb because now you can increase the base doping quite a bit so therefore you can make your wb much smaller without worrying about punch through without having worrying about that these two junctions will short as a result your wb can be much smaller and that gives you higher performance you could also do the grading right grading on the base and that would have even increased that base transistor even faster so all around this is a generally a better transistor more expensive of course but a better transistor and as a result you will go much higher frequency car defect will occur at a much later stage so you should be able to clearly articulate if somebody asks you what is you know how does hbt improve performance just look at this expression think about it term by term this has everything now modern transistors are of course horrendous because and these are actually many of them are commercial transistors you have to have look at this I just wanted to give you a feel because what you just saw is not really the whole story let me start from the top near the context so it's the indium phosphide indium gallium arsenide transistors many of the lasers when you talk and telephone are based on indium gallium indium phosphide indium gallium arsenide uh tribe transistors why why is this indium phosphate because the band gap they have in indium these transistors many of that band gap is the same that you need for the laser and that laser when it pumps light out that light must match the optical fiber so therefore because you know light emission is related to the band gap and so in order for the light emission to do match the transmission window of the silica which is this optical fibers it must be done in indium phosphide and therefore when you want to integrate your amplifier and the laser into the same module then in that case this indium phosphide technology is very useful so in telecommunication integrated circuits you use all the time why is the top one indium gallium arsenide what is doing there on the top 400 angstrom on the top it is because if you use that higher band gap with the metal contact on the emitter no electron will come in it's such a big band gap difference so therefore although everything I said is fine your emitter contact will kill you as a result you want to start it easy smaller band gap high doping so that the electrons can gradually flow in that's why you use a smaller band gap highly dope 3 times 10 to the 19 400 angstrom what is silicon doing there by the way silicon is a end open silicon is a end open indium phosphide you remember that table group 4 is silicon if it replaces indium which is 3 then it will become a donor do you remember that so silicon is a donor and that's why this is like a end emitter of the side you can see the indium phosphide you can see the grading indium gallium arsenide you can see in the emitter region and then you have the setback region that 3 times 10 to the 16 is to increase the depletion region you increase the depletion region why because you want to reduce the junction capacitance and also you don't want the silicon to flow in in the base region it's so thin you know you put silicon in raise the temperature the silicon the end opens might flow in the base so in order to prevent that you have the setback layer is it a single hetero structure or a double hetero structure do you think it looks to me indium phosphide indium gallium arsenide and then indium phosphide so it's a double hetero hetero structure BJT's right and you can see it from here also that you have a larger band gap collector larger band gap emitter and you have this base region and you see indium gallium arsenide on the left hand side ease the flow of electrons from the contact and then there are many other transistors I just wanted to show you some of this so that you know how complicated some of the designs are you will not find it when you go to work and this is already 10 years old when you go to work you will not see what I taught you you will see things much more complicated but if you understand how to draw band diagram and this you know this flow of current diffusion and a drift and thermionic emission all this are nothing looks more complicated it's nothing more than that so you can see that correspondingly you again have a double hetero structure on this indium gallium arsenide phosphide that one and on the collector side you have a larger band gap but you also see that they have graded the base collector junction so that there is no notch sitting there piling up electrons in the in that region so this has advantage of large band gap but also no charge build up in the base region and corresponding with the other one so let me summarize so I try to explain to you why despite all the difficulties HBT's have taken over and that's because it allows heavy base doping and allows you to invert the doping and it allows moderate emitter doping emitter doping reduces capacitance a good thing now wide band gap collector has many advantages I already explained the one I explained is essentially was a symmetric operation by reducing the offset voltage to zero so in beyond large breakdown voltage that's a very important consideration that I gives you symmetric operation simplifies and band gap engineering this is a general word that you will hear many times where you change the band gap to get some extra performance advantage that's what they call band gap engineering and band gap engineering has many uses but this one is a particularly technologically relevant one there are many other every time you go to upstairs or maybe in a bar most of the professors are trying to do one form of band gap engineering or other trying to manipulate different material when they manipulate different material they are actually manipulating essentially the band gap of the device and by changing the band gap you have many advantages but this HBT is one of the most commercially important band gap engineer devices we talked about heterojunction launching ramps which is this notch single hetero junction so in fact you have a little extra energy to launch electrons from sometimes that helps we talked about compositionally graded base what is that when you have a grading in the base region it pushes the electrons quickly through the base the increases the gain and if you eliminate the band spikes then you get there full advantage of delta EG otherwise in a abrupt junction what do you have just the delta EV whatever was that that part and so that's it and as I said the most important research topic these days if you go in any military research lab you will see the commercial ones except at TRW and few other places except that most of the work is now in the military and they are really looking at terahertz karta frequencies and now that's not easy silicon is not going to get there but there is a very good chance that bipolar will will not bipolar the HBT scan can do it it's not still a done deal but perhaps it will do it okay so let me stop here thank you