 you can follow along with this presentation using printed slides from the nano hub visit www.nanohub.org and download the PDF file containing the slides for this presentation print them out and turn each page when you hear the following sound enjoy the show let's get started okay so today we'll be talking about bipolar transistors and this is lecture 27 for the solid state device course now a famous picture for bipolar transistor that you have may have often seen in many many publications is the one shown on the left hand side you will see in the towards the bottom a triangular shaped wage region sort of held together by a spring like thing from the top and you can see two wires coming in those two wires will be the emitter and the collector of this transistor and I will explain what and then there's a flat slab of semiconductor which is again connected with a wire towards the right and that will be the base of the transistor now this is the first bipolar transistor that you hear about from in 1947 which was invented in 1947 this was a germanium transistor it was not a bipolar transistor in the ordinary sense it was a metal semiconductor metal junction point contact transistor more like a short key barrier transistor and in your homework problem you are really asked to solve and explore this particular problem now the way this particular transistor works is that this is ingenious the way it was done this wage shaped region the person Britain who was one of the inventor of the transistor he took a very thin foil gold foil and wrapped it around this the region next to the wage he wrapped it around took a blade you know this is how I take things out and cut just underneath on the bottom of the wage so that the metal film separates out by a little bit and then he pressed it down on the base which is the green and that is why the electrons will flow from the emitter to the collector through that base region so this is how the first transistor was made now this is the sort of the symbol of the transistor now this is a very famous picture and this is one of the most misleading picture ever that you might see the person on the right with this full set of hairs and on the right with the tie that person know all three have ties those days looks like but on the right is Britain who actually is the maker of things he actually can put together experiments very very quickly and he's a beautiful experimentalist the person and the desk is the Shockley which is the person in the front is sitting on is actually his desk retains desk the person on standing with the thinning here that person is Bardeen and Bardeen is one of those people he's a theorist in the last class we talked about interface surface states and why when you put a metal next to a semiconductor why it is difficult to have modulation of the Fermi level Fermi level pinning in the last class Bardeen was the one who explained it and he is one of those theorists who can look at a very perplexing set of experiments and can make sense out of it he's had this one of those rare set of people he of course went on he came back to Illinois and went ahead and he got another Nobel Prize one of the first people to get two Nobel Prizes in physics the second was one superconductor and these two are really the inventor of this transistor the person sitting in the in the front is Shockley Shockley is the boss and he is the visionary but for this project he had nothing to do but he essentially wanted all the credit and you can here you can see he's sitting on the microscope which he would have really sat over there so this is a the picture that in fact Brayton and Bardeen never really liked and afterwards they didn't even talk to each other for long period of time now one of the things that you may not know that in that time in 1945 46 47 Purdue was in the epicenter of the bipolar transistor transistor work in fact this is the work by Ralph Bray from Purdue in the physics department who essentially prompted the Bellas people to eventually publish the result they have invented their transistor effect a few months earlier but when Bray presented some results in a conference the Bell Labs team these these people who all worked in Bell Labs they realized that they cannot hold back on a sort of public announcement of this discovery anymore and so they rapidly went back and had a news conference and published the result so Ralph Bray was I have met him several times he used to when as a student here he stopped by a few times a wonderful graduate student at that time and did fantastic would research and so Purdue could easily have been the place where transistor was born but this was the right time for transistors because even in Europe this is in Nazi Germany the transistor research was going very fast because they had to this rudder in Second World War was becoming very important and all the rudder was ground-based and the Germany didn't realize that the Britain was working on an airborne version something that you can put in a plane a small rudder and that would be very effective they thought that you know only second-order pilars they depend on instrumentation the first order pilars can do dogfights and other things and they can bomb somewhere very effectively but by the time they realized in 1944 when they captured or a downed British plane they realized how far rudder researchers come in by that time they threw in everything they had in rudder research and for rudder research they needed a solid-state switch which was not that heavy and in order to do that they had a huge amount of resources but of course it was too late and the person who was responsible he eventually moved to France and within about two years by 1948 transistor was even demonstrated in France at that time so this was something whose time has come and you know in many cases people of course find the same thing all over in many as various context so it's a beautiful fantastic period of creative time for electronic devices now on more mundane things now the transistor that I just showed you is point contact transistor that is not the transistor which eventually won out the version that won out is shockly the person who is sitting in the front he was so unhappy that he didn't invent this particular transistor that he went ahead and for one month isolated himself and came up with the idea of a transistor and which is called a junction transistor which is what you study and he came up with this ideas and that is eventually what became accepted as the transistor in 1954 that's the first commercial production of transistor came along that is actually shockly is designed not the original design the original design actually didn't fly at all it was too noisy and I'll explain later on why the transistor this particular version you can see the n plus region is the emitter where the electron is coming in the P plus base the green region the base that is where the word base comes from because you know in the original configuration the semiconductor was sitting as a slab on the base so that is where the word base comes from and the collector which was on the previous configuration on the other side of the wage in this particular case configuration it is this n plus region which is sort of going from the bottom this is called a double diffused junction transistor is because the white region is the n plus collector in which you first diffuse by high temperature the P plus base and the second diffusion is the n plus emitter region so that's why it's called a double diffused bipolar junction transistor and we will be talking about just this vertical section one dimensional but transistor bipolar transistor by definition is a two dimensional two dimensional device so you take that one rotate 90 degrees and this is what the configuration will be talking about that you can see the n plus emitter the green base and the collector and then again n plus sub collector on the far right hand side one thing I want to mention very quickly that in this simplified configuration the base contact on the green side the black base contact is actually very close to the to the original semiconductor the P plus region is strictly speaking this one dimensional version cannot work I'll again explain later on down the road that the base contact has to be several minority carrier diffusion length away from the junction and you see on the top side on top picture the green base base connection is really a little bit up away from the main junction region where the current is flowing that is essential but for the time being we will stay with this particular simplified picture this is how a modern bipolar transistor works and you can easily see once you understand what to look for how the various pieces work you can see this wing shaped emitter region and many times this will be polycrystalline silicon I'll explain again why you can see the green base region you can see and many times these days these are silicon germanium one of the questions I might have for you during the exam is why this is silicon germanium what advantage does it give you and of course I will explain and then there's the dielectric trench to isolate one transistor from another because you know you'll have put millions of transistors together and if you don't isolate them current will flow through the substrate from one contact to another rather than through the electrical contact themselves and these are some of the pictures you should take a look and you can see very easily that there are this this region this region on the bottom is very similar to the emitter region the three top columns coming down those are electrical contacts for base emitter and the collector region so I would encourage you to actually look at the picture on the bottom and then compare it with the wonder on the top so that you can easily isolate individual pieces I couldn't mark on this on this particular picture in the bottom because that will be too difficult but this idea is essentially the same transistor shown in a slightly different way so the main point I want to make you will see several difference one is the emitter is now polysilicon the base silicon germanium so why did these things happen why is it that we couldn't stay with Shockley's original version and move on with it why is it that we needed all these versions of very variations in order to make modern transistor work so we'll explain that very quickly then for symbols and conventions these emitter many times as I said this is polysilicon emitter briefly noted as the abbreviated as poly emitter means polysilicon emitter the dope the base will be low-doped and I'll explain why and the collected doping optimization of it is a very important consideration and I will have three lectures from tomorrow from from the next class explaining the physics and the mechanics of optimization of bipolar transistor now you I get the basic idea that if you have three currents emitter collector and the base current when they all flow out this is Karshav's law that the sum of the current in DC in DC will sum to zero of course if you have AC then the divergence of current is equal to dndt or dpdt is a charge built up inside in DC the sum is equal to zero and correspondingly the sum of all three voltages e to b emitter to base base to collector and collector to emitter you know circular voltages where they would sum up to zero again these are elementary rules that you must you must have learned in your undergraduate circuit classes symbols well symbols for NPN transistor is essentially three terminals you can see very easily the base emitter and collector only thing to note is on the on the emitter side the arrow is pointing outward and the reason I meter the emitter arrow is pointing that that way outward is because as I'll show in a few minutes that the current flow in an NPN transistor is outward through the emitter and therefore you have that arrow go a pointing up or outward for PNP transistor the emitter current flow is inward and therefore you can see that arrow which is coming in from the emitter side is pointing inside and as a result that would be the symbol for PNP transistor so this you may have seen many times in your undergraduate years maybe so again anytime we have a new device band diagram is the first thing we do and band diagram is a solution of the Poisson equation it's a graphical solution of the Poisson equation so let's get started so we'll talk about equilibrium bipolar junction transistor hetero junction bipolar transistor we'll come back to that later on that will be like four lectures down so this is the solution of the Poisson equation that we are looking into and this is the idea again I ask you to follow the rules don't try to do it too fast because if you do then you will make a mistake not in a particular one but when a general I give you a new example or a new problem you'll make a mistake so let's stay with the basic idea the first rule is that you draw a flat quasi Fermi level and that is the dashed line going through through the whole device in the middle in the middle and you can see the dashed black line the second is that I have here an NPN transistor and so I put the EC on the very left hand side equal to the from the Fermi level depending on the doping I separate it by a certain amount and on the very left side the EV is then determined from the conduction band by the banger and I also say write out the with difference of Ki one the work function the dotted line over there that's the vacuum level on the left hand side I follow the same rule for the right hand side also for the collector now you may see that the blue lines on the left and right are exactly the same in this particular picture that need not be in general depending on the doping this levels will be at different places but this for particular for the convenience I have drawn it this way the base region you see that the band has been shifted differently because it's a p region and therefore the Fermi level is closer to the balance band so that's the first line the balance band line is the first one you draw and then you draw the conduction band line through the band gap because you know the band gap and the chi two well from the conduction band you put the chi two that gives you the vacuum level and once you have the three vacuum levels you make them continuous and that's how the vacuum level lines have been drawn and subsequently you copy directly the blues from the emitter and the collector bring it down to the balance band that's where they are and then you copy it down and also for the conduction band and you know this right this is this is not difficult now the particular transistor I'm talking about is a homo junction bipolar transistor silicon silicon silicon or germanium germanium germanium all three regions same material remember the double diffused so it was the same material in which you diffuse twice so that's a homo junction double homo junction device and therefore when you copy the red the red doesn't have any discontinuity because essentially the same band gap it flows through and then similarly on the conduction band you don't see any delta EC or delta EV you see a continuous flowing line from emitter base to collector now again this is particular thing drawn this particular way is not difficult but if you don't follow the rules then you will make a mistake where you will not identify delta EC or delta EV correctly so please follow the rule even when you know the answer okay so electrostatics in equilibrium that simply means that now I can determine the various depletion with you know in equilibrium because these are actually back-to-back p-n junctions so I between emitter and base I have a n and p junction and from base to collector I have another n and p junction so essentially two back-to-back p-n junctions now I know all the physics about p-n junction like spend a lot of time so answer I can just open up the relevant lecture note and copy and you know that this particular one xn on the emitter side the n region the how far it has been depleted is given by that particular expression nb is the doping in the base ne is a doping in the emitter that's the n type doping and nb is a p type doping vbi is a built-in voltage now I want to mention something here little bit down the road but let me first finish up you know the other side of the junction you know how much it is going to be p because x sub p is because this is in the p region you know base is the p region and then therefore again you have ne and nb you know all those how this plays out in a junction now you focus on now you focus on the collector side again collector is n type and you get the depletion corresponding depletion in the base because of the collector side and that would go with the NC and n sub b now you can see that therefore the base region is sort of invaded from both sides and one side emitter is depleting a little bit because it wants some charges and on the other side the collector is depleting a little bit so the real base which is sort of the intrinsic region inside is actually getting smaller and you don't want it very small and I will explain why one thing I want to mention that I when I define vbi before and this vbi is not exactly the same when I told you about vbi before then I said that you know you have 25 region you take region one and you take region 25 and you essentially look at the chi one and chi two from the both region and the band gap and get the vbi that way right that is one way of doing vbi and that vbi in this particular case would be zero right in this particular case because the same band gap same chi one chi two vbi is zero if I just took emitter and the collector being the end points however for this calculation vbi is taken pairwise that it is from between n and p you get a vbi because you when you are calculating the depletion region for the emitter base in that case even the collector is so far out on the right hand side that you say that my vbi for this pair of junctions this pair of materials is not affected by something on the right hand side so this vbi is because between n and p that pair of regions right so that you have to be careful that this is different from what I told you before now let's talk about current in the bipolar junction transistor will go slow and you will see how it works out but you realize that again I'll have to start with DC solution of the band diagram and let's see whether you remember how this is done first anytime you want to draw a non-equilibrium band diagram you always have to ground one terminal first thing is grounding and you can see the black arrow which is sort of in the middle here I have grounded the base meaning grounded meaning holding the potential to a particular value let's say in this particular case 0 so I must keep one of the terminals at 0 potential before I draw do anything and correspondingly that Fermi level for that material or that terminal will not change from its equilibrium value that's what you see here in red and the red will only go to the other side up to the end of the junction right on the other side so in the emitter side it goes to the end of the other junction and simply on the collector side goes to the end of the junction the red line I should have drawn it a little bit more you can see the junction and the collector is a little bit more but that that's that's the first thing the second thing is let's assume that I have put a negative voltage on the emitter this is NPN transistor and if I have a negative voltage which way the band would go goes up right any time I have electron I have a negative voltage it goes up because it doesn't like it and so the voltage when I put a negative voltage on the N side or negative voltage on any side when time I put a negative voltage the Fermi level will go up by V EB in this case again the Fermi level quasi Fermi level the FN the blue dotted line stays a flat to the other side of the junction in the base side in this case the current flow I'm assuming is small therefore I do not show explicitly any gradient in the blue or in the red remember the high injection regime in which we had this quasi Fermi level drooping and therefore the junction potential was a little less than the applied voltage so I'm assuming low applied current a low voltage here now if I apply a positive voltage on the collector then the Fermi level would go down this time and it will go down by the amount VCB V sub CB base to collector to base and you can see again the blue region is going down and going to the other side of the junction so this is the basic configuration of the emitter base collector junction in a normal operating mode again very easy if you apply a bias what will be the junction there V BI must be replaced by V EB V BI minus V EB because the voltage is a little bit less now right because you have applied a forward bias do you realize why it's forward bias because of course in general when any time you connect end to end end material to the end terminal of the device and P material to the P terminal of this is forward bias so that's why it's I am saying is forward bias that you know that correspondingly the base width and then for the collector side you have the same thing but except do realize that you have VCB because this time the relevant bias is actually the base collector bias I'm talking about the right hand side and if you have correspondingly for the base you can get the corresponding depletion region for that one as well assume for a second in a state of having it been a homo junction had it been a hetero junction what would what would the what things would have changed one is you wouldn't have one the kappa s or k sub s right you'd have case one case two and that is something you have done in your homework and also I've shown in the hetero junction diode before so that's one thing that that we have changed because two materials two dielectric constants you would have V BI if you have let's say silicon germanium and a third material so this pair the first pair would have one V BI and the second pair would have a different V BI so the V BI in this case would also change but other than that this is the expression nothing there's nothing fancy about it so just when you're reading this remember that how you do a hetero junction transistor in the same way but I will show you in subsequent lectures also how does transistor work in this normal operating condition because when you forward bias a diode you realize the base emitter barrier has been reduced compared to the equilibrium value in equilibrium drift and diffusion balances each other but when you reduce the bias in that case the reverse electric field cannot push you back with as much field and so the diffusion wins over and you can see the green electrons sort of flowing over and similarly the you can see a few red holes essentially moving to the emitter side also when the barrier is reduced but the interesting thing is that the electrons the green electrons which has crossed the emitter base junction those will actually now they cannot really come back and recombine with the base if the base region is very short they will simply get out to the collector so by pushing a certain number of red holes in and this I will explain later you can control the junction bias thereby you can control the flow of the green electrons and therefore the green electrons then come come to the output side and run your amplifier so the red electrons are your antenna signals that are coming in with tiny amount of current and then the green electrons are flowing in into your speaker where they give you a amplified signal so this is how a bipolar transistor would work okay and in for convenience I'll just write this down because many times you know instead of writing N sub B at E that number of donors in the emitter if you know it's a NPN transistor why carry around the D you know that the emitter is donor doped so in that case you simply write N sub E so in the many cases this simplified convention is a little bit easier than other one is the diffusion coefficient the diffusion coefficient only thing you care about is a minority carrier diffusion in a particular region right what is the minority carrier diffusion in a in a base region these are electrons for NPN transistor as a result you can see D sub B in the middle column in the middle element D sub B has been written as Dm so instead of writing Dm D sub B you write because electrons are actually the minority carriers in the NPN transistor so you substitute a few symbols so that you don't have to carry around so many indices okay so let's see whether we can calculate some current now you actually know all this so if you just stay with this you'll see that you know all these answers because you know how a diode works so let's assume that I do not have any recombination and I want to collect calculate the current going from the emitter to the collector now I could calculate current anywhere I could calculate current at x equals 0 in the far left hand side in the collector in the emitter region but there electron is the majority carrier I have lots of electron a tiny amount of field I don't like it that like it like it to calculate there similarly I could have calculated on the very right on the collector side same problem so anytime you have a current flow in a complicated region always check out where there's minority carriers because minority carriers of the region where it's easy to calculate current but since it's a DC current current flow is continuous once you calculate in one position same at every position so this is sort of you choose this base region because it is simplest simplest to calculate not because this is the only place you could calculate it okay I do not have any recombination so therefore I have a straight line for the diffusion current right remember no generation recombination steady-state second derivative of n equals 0 and ax plus b is the solution okay now on the one side of this triangle is I have from the forward bias junction p-n junction over there do you see that I have an exponentially increased minority carrier concentration you can see q vb e beta is 1 over kt q vb e is how much it has been forward bias by and ni squared divided by nb that's in the equilibrium how many minority carriers I have that's something you already know now don't be misled by the the right hand side the right hand side I didn't mean it to be 0 it did not be 0 it could be any value and that value is given by simply the p-n junction on the right hand side right the p-n junction the corresponding voltage is vbc right base to collector whatever that voltage is that's the p-n junction voltage so you have that do you see again you have ni square divided by nb because that's the base region minority carrier one thing I'm carrying a subscript in the b in ni squared then I have a subscript b because for homo junction of course band gap everywhere is the same ni square everywhere is the same but if it hadn't been a hetero junction then ni square in different places could have been different so I'm carrying that one will not be useful now but will be useful later now vbc this is a large negative bias do you agree this is a large negative bias and so essentially what's going to be the delta n will be some negative quantity be less than 0 delta n is less than 0 n is 0 that's what it's going to be okay so I have my solution actually delta nx is a x plus b but I could write it as x plus b well straight line I can write it also in the second form c equals 1 minus x divided by wb plus d is multiplied by x divided by wb I can write that right because you can collect the term in x and then that will become your a and you can collect the term again the constant terms and that becomes your b so it's the same form but the advantage of this form is the following because do you agree that that is really c because when x is 0 when x is 0 then the term d will go away right the term associated with d will go away and therefore c is delta nx 0 so that gives you the value c what about this red one do you see what the red one is going to be at x equals w this term associated with c will go away and therefore only term associated with d will remain so that's d just by inspection you can write it and so therefore that's your answer all I did was to put the value of c put the value of d and I'm done this is the you can see that if you increase the vbe to a certain value that will cause a forward junction current to flow and vc bn corresponding with all the voltage dependence will come out of it okay good so that's actually done I'm actually done I can calculate various things based on this so for example I wanted if I wanted to calculate current okay no problem calculate the current so I wanted to calculate the collector current and j sub n collector current the electron component of the collector current because this I'm calculating just electron how would I do that I'll just take the derivative of dndx and evaluate it at wb right now of course this is a triangle and the derivative is the same so therefore in this case it really doesn't matter I could have taken the derivative at any point and the value would have been the same but if I had a recombination in that case I'd have to be the amount of current flowing in through the emitter would not be the same as is leaving in the collector sign so then I have to be careful but this definition is obviously correct because I want to do it for the collector point and I have taken derivative at wb okay and then if you take a derivative you see that the x derivative first derivative so the wb will come out in the front and with a common with a negative sign and the second term there'll be also a wb coming out with a positive sign fine so now you can see that the current has a negative sign does it make sense electron flowing from left to right right and so therefore current is flowing from right to left therefore I have a negative sign therefore remember in that npn transistor the arrow was flowing out of the collector I'm sorry out of the emitter right the arrow and you can see why because of that little minus sign current is actually flowing out through the emitter junction now by the way that's not the only current flowing in the device because when the base was forward bias the holes also want to flow out and when the holes want to flow out it's the bottom side the blue the blue triangle that you see over there is the holes flowing out as a minority carrier through the emitter and again you will have the same same thing you can do a x plus b on the metal side on the far left contact between emitter and the metal contact what is the boundary condition is a 0 and you also have the excess minority carrier and that will go as e to the power qvb divided by kt minus 1 and you have done this in diode so I let me not go through in detail take the derivative you have the current again this current has a negative sign why holes going this way now going from right to left holes are positive and therefore any direction holes go so does the current and therefore the emitter current is some of these two currents the collect the electron current and the hole current both flowing out through the emitter emitter junction now in reality if you plot out the previous expression if you plot out the previous expression for the collector current as a function of vce or vcb then you will get a set of curves like this these are called normal active region and you can see how it sort of works that anytime your base current goes up that means your vbe has been forward biased and that means the first part of that jn sub c that's a certain value and correspondingly on the right hand side at a given vbc then you will have a particular value so this is essentially the plot that you see with the series of red lines are for various vcbbc and for various vb if you plot it out in a matlab you see that this is exactly what you will have I'll explain why that one has a slope because if you plot it out this particular curve you'll see that that will not have that particular slope it will go to a certain value and then it will become flat why it has a slope is something interesting and that has to do with the depletion region in the base you know the wb in this expression I'm assuming it to be constant right I'm assuming a certain but you see as soon as I reverse bias the base collector the wb will gradually shrink because the depletion region will get bigger as a result there will be an extra modulation and that is early effect that I'll explain later this is the log of jc with vb e so let's say for a particular value of vbc the right hand side the second term on the top is a constant now if you keep changing vb e the first base emitter voltage then the first term will change right first term will change and it will change exponentially now when it changes exponentially this is a curve you have seen before have you seen this curve before why it sort of rolls over you have seen this before because this is nothing other than the simple diode this region 1 2 3 4 remember I explained the physics of why it rolls over why it has half the slope the ambipolar part exactly the same physics because with keeping the vbc fixed I am just collecting whatever the my diode is providing the first diode whatever is providing I'm just collecting it as a result I will have the exactly the same physics and in all this in all this regions now many times for circuit analysis people would summarize this results in something called an a bus bowl model very useful actually to remember this so a diode is represented by a transistor is replaced represented by this simple model see whether you can understand this first of all do you see that there are three contacts emitter e on the right I have a collector C and on the bottom I have base B so first three three contacts that anything now if I didn't know anything then at least I could say that emitter to base is a p n junction so that is my diode number one on the left hand side with the triangle looking to the left the first p n the second p n junction well from base to collector I have another p n junction so that's my triangle looking to the right that's the diode looking to the right the only thing I have to worry about that what is this current source is doing alpha or IR which sort of looks to the other side I'll to the to the first is to the second diode and alpha f if what is this all about that is something I need to understand but other than that you can see it's a simple diode now one way to remember this diagram is the following NPN diode right NPN transistor so N to p do you see one of the diode should be moving why it should be looking to the from the emitter base diode why it should be looking to the emitter side right because p is in the middle n is on the left as soon as you know how one of the diode would behave then you immediately know the second diode would look the other way they don't like each other and then as soon as you can see then therefore the current sources is also looking the other way compared to the diode and so this is how you remember how the four pieces as soon as you know is NPN or PNP as soon as you have one of the diode fixed the rest of the things have a specific configuration among them okay so that's my collector current now I see I have multiplied with a area a because I'm talking about total current not current density so I have multiplied the area okay that's fine other than that you can see the whole thing sort of I recognize from the calculation I just did except the one on the right the far third term on the right going with VBC you see that what is that the first two terms are actually the electron current influenced by the emitter base bias and the base collector bias that's the first two terms I have already derived it but of course I should have also considered the minority carriers minority carrier flow in the collector because I'm looking at the collector current and I should have drawn a triangle on the collector side also and computed the current and that is exactly what has been done and that the final term is the whole diffusion current in collector that I didn't show you but that must be there now you see the first bunch of constants this DN divided by WB NIB squared NB all those things I can call it a constant alpha F IF not so that I can only focus on the voltage part of it and this big term on the second side I will call it IR 0 and the voltage term remains as is so that will be my whole the whole thing now that's a constant but that's a constant at a given temperature right do you realize why look at that that NIB depends on only temperature that depends on temperature because NC NVE to the power EG over KT remember and EG depends on band gap depends on temperature so therefore these constants are only at a given temperature these are good approximations I can encapsulate them in a constant and I can put in there okay so that explains you see that you can see the current source the second term is a diode current IR 0 exponential of QV VBC beta minus 1 that's the diode current and the first term alpha FIF not that's the current source where is it coming from this is coming from the other side so the collector is being influenced by the emitter and emitter is being influenced by the collector so that is where the other term comes in the same thing for emitter do you realize that once again the emitter current again I will have two components one component is from the electron current another from the hole current and you can see the first two terms one of them is the hole current which one is it by the way B sub P that is the minority carrier hole current flowing out through the emitter region the first term on the bottom and again the two terms you can put them in a constant I will call that IF not I'm sorry alpha IR not and then correspondingly the second term the smaller term I can call it at IF not and that's it that's my that's my current flow and these are the definitions once I have that the only thing is that during circuit analysis these bunch of things yes complicated but you can calculate it once and then go home because you can see that B sub n ni square NB WB well lots of numbers but you punch in your calculator once get in our certain number whatever voltage you are applying for the transistor it doesn't care right so therefore you put them in a constant and just look at the voltage dependence of the transistor action so you can correspondingly write the this is a common base configuration for example in this case because the base is common between input and output everything that we have been talking about so far because you can put your radio signal between emitter and base that's your antenna and your high powered speaker on the between collector and base that's your amplification the whatever current is coming out that's your amplification and the corresponding a personal model will look like this this is exactly the one that I just explained between emitter collector and base I have put two extra things in here CBE and CBC after you put this extra capacitors in you can do a large signal analysis on this one also if you go from 0 to 1 volt 1 to 0 volt small signal large signal you can put it in here what capacitances are these by the way junction capacitance and the diffusion capacitance do you remember so those capacitances are here this is a diode of a minority carrier it has a minority carrier so both diffusion and junction capacitance are important had it been a metal semiconductor metal transistor remember the 1947 transistor was a metal semiconductor metal the two wires coming down were two metals and this base sitting in the in the bottom the green part that was a semiconductor in that case what capacitance I wouldn't have the diffusion capacitance right because there was no minority carrier in that transistor as a result this capacitance values will be different but the diodes will look about the same look about the same so that is what I wanted to mention that the capacitance values will not be the same apart from that the transistor model will look exactly the same from outside now many times I am interested you know the previous one did this common base configuration gives you power gain but doesn't give you any current gain means whatever current comes out of the emitter you saw those electrons coming in the same electrons flowing out through the collector so you didn't whatever current was coming in you get the same current out on the output side so in that case if you put a big register on the collector then you can have a power gain right because you can have a lot of current flowing through a large register but you do not have any current gain but in order to have current gain many times you would put them in the common emitter configuration do you see why it's called common emitter configuration because the emitter is common between input and output and signal would come through the base current and the collector current correspondingly will have the amplified signal present here how would you then do an ever small model on that one since it is getting late I'll ask you to do it by yourself you see this original one you see this last one in the previous slide I just have rotated it 90 degrees haven't done anything more and you can see now the base is so to the input side and there is the collector and correspondingly the emitter but if you work on this a little bit so there's nothing here right this is just a copy of the previous one but if you work it out a little bit here I want you to show that that particular circuit in the middle can be simplified to this particular circuit on the right and that has for example have one current source and that current source depends on IEF and IR you can see that there should be some dependence you can see there is a C sub pi and that will depend on CBC and CBE and the corresponding the C mu C mu and C pi those two are dependent on the capacitances that you have this is really five lines of algebra but you have to sit down and do it so this is a practice problem because generally professors like in their exam put some sort of variations of this and ask you to calculate certain quantities for the transistor so have a practice at this one and then we'll see if you have any questions okay let me conclude so the physics of bipolar junction transistor is most easily understood in terms of physics of individual junctions you know pairwise this is best understood now the equations are most effectively used if you put them in equivalent circuits and once you put them in equivalent circuit you can do anything on them you can do AC DC put a large signal in there and could do all sorts of analysis now what I have told you so far is something an undergraduate student should know because you know that's at least the beginning but what you haven't really understood yet is how more much more complex this transistor is if you just go outside make a three region terminal three terminal device you know it will not work as a transistor it has to have a particular combination of doping in a nb and n sub e and n sub c in for for this transistors to work and that is a little bit more tricky thing and that's what I'm going to do in less three lectures explain to you how to understand and fabricate and design a transistor because what you just learned is not really enough for a graduate student and finally there's a very beautiful book you know you you are spending you'll spend many years learning about this topic lots of homeworks and so you should have some fun this book by called crystal fire this has a terrific history of the invention of transistor the story I told you about Bray Ralph Bray and the tremendous fights Shockley had with Bardeen and Bretain and the implication of it even in the Nobel ceremony most of the time Shockley would stay aside he will not talk to Bretain and Bardeen they were very good friends both of them are very good friends and so this was a tremendous tremendous thing and Shockley was a very controversial person very controversial I don't want to get into that that will be a political incorrectness for the in the class but you should read it and this is a beautiful book okay all right thank you