 You can follow along with this presentation using printed slides from the Nanohub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. Alright, so let's get started. This is lecture 28 on bipolar junction transistor and its design. There will be three lectures on this topic and this is the first one of that series. I could have easily titled this lecture as being this engineers dilemma between rock and hard place and that's what you're going to see that how difficult it is to find right combination of doping so that you can have a great bipolar junction transistor. Now I'll begin by discussing and reminding you one more time that what determines the current gain of a transistor, the variables that determine it and will focus on DC current gain as sort of the parameter to hang various concepts on. So we will use that parameter and then we'll take look at it piece by piece to see how to optimize the performance of a junction transistor. Next we'll talk about this issue about base doping. We'll see that we need in classical homo junction transistor base doping to be as small as possible but when you try to make it too small then you run into trouble. Then we'll see that well the other option is to go for collector. We'll try to reduce it but then we'll get running to problem again and then we'll come back for the emitter and try to raise the doping and then there'll be trouble there again. So you'll see the designer bipolar transistor has this very small space in which you can actually make a transistor and finally make some concluding remarks. Now the considerations today will be mostly on homo junction bipolar transistor. These are silicon, silicon, silicon or germanium, germanium, germanium for emitter base and collector region. Because of these problems which we'll cover in this 20-odd slides modern transistors are no longer homo junction transistor. Hetero junction transistor will allow us to get out of this terrible bind of this doping issues and that is what most modern bipolar transistors are all about. So we'll get there in next lectures. Let me remind you that the ever small model in a very simple and elegant form sort of captures the essence of bipolar operation. We saw that the model essentially looks like this. It is a emitter base and collector. So it is drawn here in a common base configuration because base is common between input and output. You can identify this being as an NPN transistor because if you look at the diode on the left hand side between emitter and base it is looking towards the N side. That is how P to N is how the diode should look like. Base is P, emitter is N and the diodes essentially face away from each other. You can see because it is Np and then Pm between emitter base and base collector and essentially you can also see that there are two other current sources again essentially facing each other and opposite to the direction of the diode. So very easy to remember and one thing you will notice that the current source for emitter base depends on the current I sub R reverse current between base and collector. Do you see that? And similarly this is sort of goes crosswise. So alpha f I af depends on I af on the other junction and this is reflecting the fact that the current you push through emitter to base when you forward bias it the same current flows to the collector side. So that cross coupling is being reflected by this current sources and we looked into this derived formula for the emitter current and that had this diode component which is shown here in red and also the current source component shown here in blue. Look that look at this I f is by definition I f naught multiplied by the voltage dependent factor. The whole thing is called I f and correspondingly I r is the whole voltage dependent thing multiplied by I r naught. That is by definition and correspondingly we had an expression for the collector current. You should check the sign whether the sign look about right corresponding to the direction of currents that we have we have shown over there. And the difference between the collector and the emitter current as you can see from the diagram ever small diagram would be the base current right because the sum of the three current must be 0. So if you know these two currents and if there is no current loss electron cannot be electron cannot be destroyed it can recombine that is fine if it recombines that so one component will go up another component will go down but so long electrons are not destroyed they can recombine not destroyed then the sum of the current must be equal to 0 because whatever goes in in some form electron form or whole form it must come out. And this is the definition and that is the base current. Now in here I have multiplied with I a e area of the emitter and in the bottom area of the collector a sub c because the emitter area and the collector area need not be the same. Remember the double diffuse junction you had this three layers on the top emitter was tiny on the top right coming from the top and the collector was big. So the two areas need not always be the same but for the simple analysis will assume that they are about the same. Now one plot that is very important and that essentially captures the essence of what a good bipolar junction transistor is is given by this following ratio but before I get there first of all notice that the collector current depends on base emitter voltage right because I am forward biasing it that's how the electrons are getting over the barrier and also on the base collector bias because that is on the other side the reverse bias is pulling the correct electrons out into the collector. Now as soon as the VBC is beyond a certain few kT remember that's a negative voltage. So the drum that goes with the exponential of Q VBC that term will go to zero because it's e to the power minus a large number so that will go to zero and this corresponding term will be small so we'll drop it in the normal active region not always in the normal active part because VBC is reverse biased so we drop that term small current. So let's focus on the collector current I sub c as a function of VBE and let's focus on the red line where I show the IC is increasing with VBE now do you realize that if I had it in a log plot log linear plot then it should be a straight line you do see that right because if you take a log of IC over a a being the area which I am assuming the same for both emitter and collector then expansion whatever is the pre-factor that will become a constant and Q over kT that would be the slope and so that is my first curve you can see at high current level the IC current gradually flips over or gradually rolls over that's the ambipolar ambipolar type transport that is the high injection that those type of things so that is something we are not showing there so the current will gradually roll off now what about the base current well the base current from the previous slide you can also calculate the base current and correspondingly the base current will also depend on VBE because as soon as you forward bias a base emitter junction electron goes up easily and the hole also goes also goes to the emitter easily and so therefore they will have the same dependence on the base emitter voltage that's very good and that's what you see in the blue line is if you just look at this expression up to this point it is not obvious why the blue should be below the rate right in just an expression if you take a log I can understand why it should go linearly up but it still doesn't say that yeah whether the blue should be higher or lower than the rate but you realize one thing that if it is a good bipolar transistor and you are taking a signal from the air right from the antenna which is the base current let's say and pumping it in your loudspeaker or whatever speaker in that case if IC is not bigger than IB then you are not going to listen to any music so what I want and I haven't done that yet is to I have to make sure so that IB is less orders of magnitude less than IC and the ratio would be my current gain that how much amplification do I get as a result of this transistor action and this we will call beta the beta DC is because this is a DC current gain how good is my amplification how strong my music comes out when the radio station is really weak that is the beta DC the current gain and is called a common emitter I'm sorry this is be a common base current gain so this is no this is fine common emitter current gain because the emitter is common between input and output base current is the input and the collector current is the output okay so let's see what that beta DC is okay I can insert those complicated expressions but you realize that VBC on the numerator that term as soon as you have it in a reverse bias I just dropped it two seconds ago right in the previous slide the second term on the numerator and as a result ah now I have a very simple expression do you see why because on the numerator I have Q VBE divided by KT on the first term look at the denominator again I have Q VBE divided by KT minus 1 because both are being governed by the base emitter junction electron goes easy holes flow easy also as a result the exponential factor will drop out and of course the VBC as I just said reverse bias strong negative number so that number essentially drops out it simply collects the electrons that is coming from the emitter so I have this simple expression you can see there are some diffusion coefficients of electrons and holes floating around there is this WB and WE how long the emitter region and the base regions are and so on so forth so we will take this expression apart term by term and we'll see how to make a great transistor now if you wanted to do a common base configuration well that's not difficult either for example in that case you will define another quantity called alpha DC which is the DC transfer gain which is the ratio of IC over IE look at the top one beta is IC over IB right IB is a tiny amount of current but IC over IE should that number be larger than or smaller than one the current is coming from the emitter and flowing in the collector right so it could be at most one and most likely less than one if there is any recombination and other things less than one as a result you can and also you can relate it to the beta DC so you don't have to calculate them separately because of this because I know beta DC is IC over IB IB is IC minus IE right that's the difference and so if I just take a ratio divide throughout by IE then IC over IE on the numerator will give me alpha and correspondingly there will be an alpha on the denominator and you can see that how beta will be very large right if alpha is close to one then how large would beta be beta will be humongous because look at the numerator because one minus alpha and alpha being close to one I will get a huge gain and typical gain of a transistor these days would be on the order of a hundred to a thousand so that would be that range of amplification that we are talking about okay so the bottom line of this is no rocket science here is as soon as you know one quantity alpha you know the other quantity beta that's it so you don't have to calculate them separately so we'll focus on beta and then do the whole analysis if I asked you to do about alpha you'll come back and simply put it in this formula and you'll be done and generally this gain and generally this gain has this various pieces of the same expression I told you about now you see this is I have plotted beta dc and you can see this x axis is about 100 in the bottom I have it plotted against as a function of IC which is essentially VBE because as soon as you increase VBE IC goes up so that's your x axis and the y axis is beta you can see the gain is pretty flat in the middle region in the very large part it is rolling off why is it rolling off because the collector current because in the high injection part that's rolling off but the base current is much smaller so that doesn't roll up that doesn't get into the high injection part so that's why it rolls off on the higher side in the lower side because of the base recombination the recombination you should go back and take a look in the previous slide because of the base recombination the base current is very high compared to the collector current as a result there is a roll off also in the lower side so this is our main object of analysis and from next four classes I'm just going to bring this formula over and over again bring this formula up over and over again and everything that you have to understand about bipolar how it operates all is in here there are in fact at least two noble prizes maybe even three in this formula so let's start with a given emitter you know I have to start with somewhere so let me assume given emitter thickness a certain dimension on the order of let's say 0.2 micron something on the emitter for a given emitter land what should I do in order to get a large gain I must make the base as short as possible you can see be a space in the numerator why is it what why does it make sense because if I make the base smaller then you can see the gradient of the electron concentration will be higher right because from one side you are injecting from the emitter on the other side you are collecting it so the gradient will be higher current will be higher that's why you want short base in 50s it was even few millimeter remember that the triangular waist shape the first 1947 transistor I showed they took a blade and cut this gold foil that was going between source and the collector emitter and the collector fraction of a of a millimeter or so that was what it was in 1950s today if you open up many of these very high performance circuits then this will be about 200 angstrom and that is how it has been for at least for last 10 years even when I was a student about 15 years ago when I worked on this it was about that that's right it doesn't change too much since then now let me assume for the time being that the material for the emitter and the base are the same so therefore the nib which is intrinsic carrier concentration in the base region and intrinsic carrier concentration in the emitter region what does it depend on by the way depends on the band gap do you remember ni squared is n capital N sub c N sub v e to the power e g over kt so therefore we'll assume that both the emitter side and the base side is the same material same band gap so I'll drop the ni squared term now if I assume for a given value of any let's start there for a given value of emitter doping in order for to get high gain I must have emitter doping higher than the base doping you see this is a central piece if I don't have the emitter doping higher than the base doping because this is an exponential factor 10 to the power 18 10 to the power 16 I so I have a large multiplier here on the doping well it's very difficult to make base very thin so it's much difficult to control that one emitter doping looks like a easy thing to do I'll just put some dope and n and diffuse it that will be much nicer so that is how in 1950s if you're an engineer at the time that is how you think and we'll start by looking into making base doping as small as possible in order to pump up the beta dc so you want high gain reduce the base doping that would be your prescription unfortunately so this is what you do you take the emitter doping whatever it is the maximum value you can get at the furnaces at that time and you will keep the base doping the green one lower than that maybe two orders of magnitude then you get a factor of hundred in gain the collector one I have shown it here lower than the base doping but really that expression isn't telling me anything about collector doping yet I will get to that in a little bit later because from that expression the NC could be anywhere we'll get to that in a second but this would be your recipe ne larger than nb in order to amplify or boost a signal from the air to your mu to your speaker so let's see what's wrong in making the base doping small that's a major problem here this is what's wrong the analysis we do have done so far is all one dimensional we have just assumed that that is vertical cross section a one dimensional card that's not really how a transistor operates you'll be there'll be many more surprises when you actually do an actual transistor by the way do you realize that this double diffuse junction configuration automatically ensures that the emitter will have a higher doping than the base because if you had the n a in the base the green region if the emitter doping was lower then that then you will not have any emitter to begin with because there is n a already you put a few n a nd and if nd is less than n a then of course that region remains a p type region so as soon as you have a n type region that means the doping is automatically higher than the green region otherwise nd is not larger than n a and you cannot have a transistor so this configuration automatically ensures that now think about it i have applied a bias vbe between base and emitter right emitter is in the middle base is where the rate circle is where the electrons are flowing in now this electron comes in and essentially recombines or come goes out through the collector or through through the emitter right that's the holes the holes coming in from the base base contact they come in and then through the reduced barrier go out through the emitter contact now when they do you can realize that whatever was the potential difference between the base and emitter contact at the contact point as the holes are coming down of course there'll be a voltage drop and if you keep the base doping low the green region doping low that means it has a very high resistance right low doping means low conductivity high resistance and so when the hole flows in there'll be a potential drop and as a result in the middle of the region you will have less potential difference than the edges of the edges of the transistor as a result there'll be a non-uniform turn on for this transistor characteristics you can see what i have shown is effective vbe plotted as a function of the horizontal axis within the green region you can see as the potential whatever you apply through your battery you know a AAA battery or whatever that's between the black two contacts as the current is flowing in then there is gradually potential drop and so the vbe the difference between them is gradually getting smaller and remember it's vbe that allows electron to be injected in the base or in the emitter junction as a result i have to be very careful about this only when my base doping is low right then then i am i'm getting into this trouble so the beta you know now the problem with beta is that it is no longer you can calculate a ratio but it is no longer uniform because current at every point within this and x is the horizontal axis depends on a vbe well that depends on the position how far away is it from the contact because it's a low doped base as a result what's going to happen is this forget about the picture on the left this is from the from the book so that i am just putting it up here so that when you go back home and refer to it you can refer to that section of the book but the point is this as a result of this non-uniform potential drop there will be more current from the corners right because the vbe is larger there so that you can see the blue that current will be there will be much more current there because more forward bias do you see the rate in the bar middle the potential difference is much smaller now compared to what you put in the contact it has been less forward biased and as a result the current is less and this is called current crowding because current will sort of crowd near the corners and most of the you can have a huge emitter region but because of this base resistance issue low base doping issue your transistor will sort of look like there is only conducting at the edges of the emitter in the middle region no current and that's not very good of course this will burn the transistor if you want to pump this up an amount of current it will burn it and so therefore low base doping you can see error gets us into a lot of trouble what's the solution these days when you look at many of this wireless stations wireless stations base stations you know when you drive through 65 and others we see these big towers right on the big towers over there there are these high powered about 90 watts this amplifier that sends our signals all these cars are getting the radio from that antenna and bipolar transistors over there generally are very high powered and those high power transistors have emitter and base essentially interdigitated so that you do not have a one big chunk of emitter and one big chunk of collector it is all interdigitated so that the distance between emitter and base is always minimized and that way you do not have too much resistance between them so this is a solution all modern transistors would have because of this base doping issue so you cannot make it too low that's that's what I'm saying there's more problem if you make the base doping low so let's say I do that I deduce the base doping now when I reduce the base doping do you agree with this statement that I will have a depletion region base emitter and base to collector I will have a depletion region but do you see there's WB what was my original base you know that width of the green region that is not really the base because of this depletion because of this depletion in the base emitter junction we have done this formula before right the amount of this depletion region because of VBE for the bias whatever region I have this is not much of a problem it's a forward bias junction if anything is better compared to zero bias it's a little smaller depleted region smaller but what about the collector welcome collector you have applied a huge reverse bias and as a result it will be sort of eating up eating up in that region a huge fraction if you dope it low you can see do you see in the formula that the NB is in the numerator NB is the numerator it sort of makes sense right if you dope it low it has less charge it's much zipline more in order to provide the charge balance so the NB is on the denominator and if you dope it low it will deplete a huge amount and at the age of that depletion region the carrier concentration will go to zero right that's the reverse bias part as a result this effective base W is actually becoming small and at large enough bias it will be completely pumped through it will be completely depleted and in that case base has no control you can see the W on the third figure on the bottom has is has W equals zero that means it doesn't have any base it's no longer a transistor as a result low base doping is not a very good idea okay so so what are you going to do first of all that WB what we had as a WB in that original formula you know the whole formula I shouldn't be writing as WB I should be writing as WB prime whatever is left after the depletion has eaten away my original metallurgical base region so WB prime is WB minus the two depletion width from the two sides and I said the collector doping or the collector side essentially it's a way collector bias it's a way lot of the region from the base and that's not a good thing now it's not a good thing because of this following reason first of all punch through is lose transistor action altogether so that's one problem but even before that even before that there are problems so by the way do you see that this gain depends on the voltage how so because generally in the original expression there is no voltage anywhere you know doping diffusion coefficient there's no doping but no voltage but now you can see that through the depletion region the width of the depletion region well that depends on the voltage right so therefore your gain will now start depending on the voltage and that's not very good and I'll explain in a second so ideally this is the rate characteristics you're supposed to have but experimental data is on the top and what you see and what I exaggerate that in practice the collector current even at large vbc is keeps on increasing and this is because of this depletion issue so let's see how it works out by the way do you see the problem here first of all physically forget about the map so let's say you're trying to listen to rock and roll or something big humongous sounding music I don't think you hear rock and roll anymore but let's assume that you hear it now if the music is very low then you have a certain gain and then if the music is very high amplitude is very high then all on a sudden you should have a proportional rising gain if you don't have a proportional rising gain anytime the input is small then you have a certain gain let's say hundred factor of hundred anytime the input is a little higher then if you don't have hundred then the dynamic range from the inside will be distorted on the outside because whatever the singer is doing on the inside there they have a range but if it's not faithfully reproduced on the outside if the gain depends on your input signal level that's not a very good music isn't come going to come out very well right as a result this increase this difference in ic increase in ic as a function of vbc because remember vbc is the my output output voltage that's what the my amplifier my speaker is and if that is changing with the input bias then that's not a very good thing so we don't want it so that's the bottom line now how bad it is how will you know how bad it is well this is how to do it as I just told you that the collector current will depend on wb and the wb wb prime and the wb prime depends on vbc because that's the remaining depletion with that's what you're talking about now let's see how bad it is in order to know how bad it is ideal is if my slope is zero or that would be my ideal right if it is not ideal then I want to know how bad it is I can do it by this way I can say how is the collector current changing as a function of vbc that's the derivative right and the derivative is the slope now the left hand side is easy to see do you see the right hand side of the equation ic divided by vbc plus va in the following way let's say I draw a line this blue line going all the way over there do you see that at a given point let's say I have ic and in the base of that triangle what is the base of the triangle va plus vbc that's the base of that of the triangle and ic over that now for a good transistor this should be very vbc would be small compared to va because if it were the red line what should have been the va that should have been infinity right because d ic that the slope should have been zero and the only way slope is zero is va is equal to equal to infinity that's the only way it could have been zero so compared to that I can always neglect vbc vbc on the order of maybe a volt or so so I can drop that and therefore in a good transistor in a reasonably good transistor I can say ic over va being the measure of the slope of the of this bad effect of low doping and this is called it's after gm early and this is called an early voltage the v sub a is called an early voltage so if some somebody says to you I have a large early voltage that means he has just designed a transistor which doesn't have this base doping problem that's what he's trying to trying to tell you now let's try to see how to evaluate this formula because I don't know va right I don't know va so how am I going to calculate it so this is how it works you see what I have done in the derivative I have divided by q nb wb on the first line and correspondingly do you see the blue q nb wb so top and bottom I have divided they haven't done anything now do you realize in the first term first derivative I can pull the nb out q nb out I have done that and the q nb wb what is that by the way q is charged nb is the doping right nb is the doping in the base wb is the width of the base so that is the amount of charge that is the amount of majority carrier charge I have in the base region so I'll call that q sub b all right now I know that q sub b right because I know nb and I know wb from just from the geometry of the transistor now do you see how dq dvc what is that that's the capacitance because capacitance by definition is a derivative of the charge with respect to the voltage so that's my that's my capacitance but why is the d ic over dwb is ic over wb and where did I get that from here now it will take a second yes just take a look so I have an expression for ic which is inversely proportional to wb if I take a derivative what's going to happen with respect to wb I will pick up a one over wb square now the whole expression divided by wb square the first wb one of the wb I can pull back in ic and the second wb well that's what what remains there so therefore the derivative you can convince yourself in a second is indeed that formula okay so I'm I'm getting there so I combine the two pieces I have just evaluated the derivative and if I equate them on the both sides I get an expression for v a which is the early voltage and do you see what early voltage depends on q nb wb right and divided by the capacitance of the collector base junction this is what it depends on now do you see the problem what do I want here v a to b as large as possible right that's good because then I have almost flat characteristics now if I drop nb that's not a good thing because if I drop nb vlb is small so therefore my gain will not remain constant and as a result I'll be in trouble so what is the only option do I have here if I want to keep the base doping low what is the only option we have do something about the capacitance if I can do something about the capacitance even with low nb I can steal I still have some hope that the cbc will work out and that is how the collector doping comes into the story now let's talk about the collector doping so what to do about the collector doping here I said that in order to then keep the base doping low I have to do something about the collector capacitance and keep the early voltage high and the capacitance by the way depends on this do you see this capacitance between base collector this is epsilon at a over d whatever was a depletion width and the denominator is whatever the depletion width is so that's that's what my capacitance should be this is by the way what capacitance is this this is junction capacitance majority carrier capacitance why did I not write a diffusion capacitance here because remember this is reverse biased reverse bias junction so I do not have a diffusion capacitance here I only have majority carrier junction capacitance so that's what I only think I wrote so we'll have to do in if I want large capacitance then I will have to reduce this depletion and in order to reduce the depletion I must have I must increase this depletion so in order to increase this depletion then I must reduce the collector doping that's my only only option that I have left but there's always a but and this but is that you cannot really do that if you cannot really drop the collector doping arbitrarily low and that's because of the car effect this requires a little thought please follow me because if you don't you will spend a lot of time reading the books and still may not understand it so it says something easy but just stay with me for for a few minutes I have only drawn the base and the collector part of it of the transistor and I have a do you see what type of transistor I have NPN and I have this collector you can see the depletion is positive right ND is positive and the green I have this depletion in the base so base size so I have that standard transistor okay now let's say I start pumping more and more current into into this transistor by the way so I know this formula right that NBXB and NCXC that depletion on both sides must be the same I also know VBI minus VBC must be equal to the square of these distances this you have already done so there is nothing in here but notice something that happens when you start pushing a lot of current by forward biasing the emitter based junction what happens this is what happens you see you see the original green and the ash colored region that's the equilibrium but as you start putting more and more current by forward biasing the base emitter junction your doping sort of effective doping and effective depletion width changes to the magenta color region why because of this when the electrons are coming lot of electrons are coming so it will go like this the next electron will come the next electron will come so the depletion region which originally didn't have any charge right it was completely space charge region fixed charges now there'll be electrons flowing and with a lot of electron the number could be almost comparable almost comparable to the background doping in that case I cannot neglect that N and P right the small N small p electron number I cannot neglect in the Poisson equation and in that case my new equation should be NB plus N you can see the green the height of that has gone up because it was NA negative electron coming in negative so totally NB plus N has gone up on the collector side it was a positive ND just positive electrons flowing through effective charge is a little less than it was before right so therefore you have NC minus N and on the other side you have NB plus N because of this change in the charge the depletion region will also not remain the same XB prime now and XC prime you can again do VB minus VBC because VBC is your battery that has not changed VBI is a material dependent thing that has not changed so now when you solve this two pair of equations what you will see is this and this is a few minutes takes a few minutes go home and do it the XC prime how far it is how big is the depletion region that has a complicated expression but let me just focus on one issue and then I will move on is this N N you can write it as you see the collector current is Q V sat N which is on the left hand side why is it because it's mostly drift current flowing through the base collector junction very high field do you remember what happens when you have a drift at a very high field from the first part of the course do you remember the current comes up and then it saturates what was the value about 10 kilo volts per centimeter right do you remember so here when you apply a volt or two volts on the collector and you have this tiny depletion region the field is really much higher see in that case your current will be simply Q V sat multiplied by N so you can relate the carrier concentration N to the collector current J C and you can put it in here that's exactly what I now comes the punch line oh sorry I have another slide on here um so first of all what's going to happen is at low bias this is going to be an electric field and depletion as you pump more current you can see that your base emitter junction is shrinking and the collector junction is increasing base junction is decreasing and as you pump more current the base emitter junction will disappear completely it will completely disappear because there is no charge it will be completely disappear and that's what is called base push out because base has been pushed out all the way to the collector now you have a humongous base lots of recombination your transistor is all gone this is all a consequence of low collector doping that's why you get into this rubble and in your book you will see that a series of curves in the beginning you have this base emitter and base collector fields you know this constant electric fields on the both sides and as you pump more and more current eventually you see on the base a base emitter side the junction has completely disappeared electron field has completely moved to the other side now at what current does it happen you can see from that formula what current it will happen at what current do you think xc prime will become infinity because you don't have any junction anymore and you can see that will happen when the numerator goes to zero and what when will the numerator go to zero well numerator will go to zero when the current is so much that it overwhelms the collector doping because whatever you had your this is the collector doping n a plus n d plus the electrons coming in it is completely overwhelming the total number of carriers and as a result you don't have any junction and this j sub k is in honor of kark who in 1965-66 first explained this effect because because of this people couldn't push the transistor gain to much higher frequency that I will come later but this is your problem you see in here if you drop and see too much what are not going to happen this effect will come at a very low voltage so you cannot really drop your collector volt doping too much so you see the problem base doping you try to reduce it you get into trouble then you say okay I'm going to reduce the collector doping now you get into trouble here so you go back and say well what can I do perhaps high doping in the emitter maybe because I cannot drop this maybe I can do something on the emitter sign unfortunately you would already know the answer you cannot do that first of all because when you have high doping do you remember band gap narrowing I talked about this band tail states coming in when you dope very much so that is one problem that will effectively make this ni squared much smaller than ni's ni b squared I have a type over there I'll fix it on the numerator so that's one problem but the more important problem is the more so here here it is so you can see that because at high doping essentially your gap will be narrowed your gain will essentially become very small very quickly if you dope it too high and the other is this tunneling issue and do you remember this you must remember this right if you try to dope too much the electrons is not going to go over the barrier and will be controlled by the base barrier it has a easier path it will simply tunnel through the emitter to the base and will essentially get out of the base it will not go to the collector at all right because it can tunnel it's much easier and as a result you will have a horrible transistor so you can see that yes emitter should be larger than the base base should be larger than the collector but you have a very narrow space to navigate and to keep your job when the manager says design me a high frequency transistor so let me summarize now the basic transistor thing that you learn in undergraduate is not too difficult that is something pretty easy but when you join a company who is designing a bipolar transistor based circuits very soon you realize that the teacher and our professor in the undergraduate class didn't really tell you enough of the complicated story optimal design is difficult now in general the rule of thumb is always have to have emitter gain emitter doping larger than the other two doping that's that's that's generally the case but the point is that you have this trade off what let me let me quickly summarize them if you reduce the base doping what happens base series resistance current routing and as a result non-uniform turn on so that's a big problem and I also have trouble with if a low base doping I have trouble with early voltage my gain is not uniform with respect to the input so then I go to the collector side that I need a large depletion with and reduce the collected doping so the depletion mostly proceeds in the collector side but then I have this problem that as soon as I try to pump up my emitter current so that I have good gain in that case the electrons goes through and it overwhelms this depletion region and very soon the depletion charge is equal to the electrons that is flowing in and so there is no charge to balance the base charge base gets pushed out and you have a base with you know you're supposed to have very thin base for good gain your effective base now is all the way up to the sub collector region which is a horrible transistor so that is the trouble and finally if you try to dope too much on the emitter side tunneling and band gap narrowing that kills you so it's a difficult story but hopefully there are many ways to get out of it otherwise you know modern technologies wouldn't have really relied on bipolar so much so we'll get to that in the next four lectures thank you listen one thing about this I know these are not easy things right there are no good books on this on unfortunate there are books where you'll get bits and pieces of it especially Z's book SMZ's book you can take a look the book your textbook that has also but what I really want you to do sit down with this slides maybe with a group of two or three or come to me sit down and this is to me very logical it will not be logical to you in the beginning but I'm sure if you sit down and flip slide by slide you will understand it if you go for looking for books hunting for books in the library well here I get a little bit there I get a little bit you are going to be in trouble because first of all symbols are different people make different approximations so things are not always consistent so it's very important that you just sit down with the notes don't worry too much about the book yet first understand the notes then go back and read the book you'll be fine these are not difficult things at all but do it systematically okay