 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. So let's get started. Today we'll be talking about bipolar junction transistor and how it is designed. This is lecture 2 on that topic. And in the first lecture regarding bipolar junction transistor and how it is designed, we talked about the problems of classical transistor. Now I don't want to overemphasize it because of course before people got into this problem, there were 25 good years, meaning that around from early 1950s to late or early 1980s, transistors were designed exactly the way I described in the first class. Essentially all silicon transistors and three regions emitter, basin collector, essentially all silicon. Early days there were a few germanium transistors but subsequently silicon was more robust and more easily processed. So that took over. So this had a good run. This classical transistor had a good run but starting from early 1980s, this classical transistor ran out of steam and subsequently many new innovations, a series of innovations came along. One had to do with poly silicon emitter, replacing the emitter instead of having a silicon which is single crystalline, having multiple crystals of poly silicon and poly crystals of silicon. And this was an accidental discovery and it turned out and people could have easily thrown it out without realizing that they had a great thing and that in fact at IBM changed the ballgame all together again in early 1980s. But the short base transistor is that people kept improving on poly silicon transistors for many years but by mid 1990s then transistor has sort of hit its limit in terms of bipolar junction transistors of how thin a base you could possibly have. And I want to explain those testings for DC characteristics and talk a little bit about large signal and small signal high frequency response before I conclude. So we are talking about DC bipolar junction transistor characteristics. So I have already explained this particular expression for current gain. You remember beta is the common emitter current gain that if you put a certain amount of base current how many times does it get amplified before it gets to the collector, right? Base current is small and typically these numbers are on the order of 100 to 1000. You do not want it very high because then the transistor will become unstable in the circuit context. And we saw that this gain is related to some intrinsic transistor parameters like the doping of the emitter and the base, the width of the base and the emitter regions and of course the diffusion coefficient means how fast the electrons go. You see that is on the numerator D sub n. So if the electron goes fast that means I have a lot of collector current and if the somehow the holes which are back emitted base current if that goes slow then that is a good thing for me. And overall I will have a great beta DC based on these parameters. Now I have explained the tradeoffs in last class and I hope you remember the tradesoffs are such that first if you assume that if you have a given emitter doping start there then so that you want to increase as high as possible without band gap narrowing. What is band gap narrowing that if you dope it very high then the band gap that you originally had that begins to narrow because the dopant state itself then can carry current by so the electrons can hop from one dopant to the next and instead of really taking the highway it can essentially tunnel through the one state to the next one. And so therefore the effective gap you have is little less and you can see the gap enters this picture through n i squared because that has the gap in the exponent. As a result small changes in the gap even point one e v can have dramatic effect on gain because of the band gap narrowing that is something you do not want to have. The second is that once you have decided on a emitter doping then you make the base doping as low as possible but you shouldn't do it too low. The reason is then there will be current crowding it will be very resistive and therefore currents will only go through the corners of the emitter and it will not flow uniformly. Do you remember that's why we had interdigitated structures last time and similarly early effect early effect is that with little change in the emitter voltage the base width gets modulated quite a bit and therefore the gain is no longer constant. So we want to avoid that and finally let's say you have decided on a base doping based on these considerations and then you talk about collector doping. Again you want to have as small as possible but if you have it too small what will happen that this is an n p n transistor that means the depletion region in the collector is p type right it's n d plus and if you flow a lot of electrons in then it can completely suppress the charge and the base may be pushed out of the junction. As a result that's called the karke effect and that you want to avoid because that limits high frequency response as we'll show you today. And finally the base width as thin as possible because do you see WB if you make could make WB small then your gain goes up so that's something you want to make it small. Now given these considerations as I mentioned that there are not too many things you can do right you have you can play with the three doppings but sort of constrained by various various issues so then people looked began to look around and see if there there's something they could do to improve performance. So the one was graded base transport I will come to that where the WB you do something about the WB instead of making it a neutral region you keep make an electric field in it so that the electrons can zip through the base a little faster because you have electric field so that's graded base transport we'll talk about it in the next lecture not this one. The second one the topic that I will talk about today is how polysilicon transistor effectively suppresses d sub p effectively reduces it and as a result your gain goes up. As I said that that was an accidental discovery in 1978-79 at that time. The third one is manipulating this NIB and NIE and that will be done with hetero junction bipolar transistor and you know couple of years ago there was a Nobel Prize for this work just for that term manipulating it and therefore all the cell phones and very high frequency circuits you have many times use a version of a transistor which is called hetero junction bipolar transistor and those chromar got Nobel Prize for it and we will discuss it in the next two classes and then of course the classical Shockley transistors and let's me start by talking about the polysilicon emitter and the innovation that went went with it. Now all modern transistors have polysilicon emitter the wing shaped blue region that you see in the top is actually a polysilicon so polysilicon means that instead of growing it at high temperature where the material is single crystalline do you remember all the atoms in their places the Bravais lattice do you remember those things long time ago so those Bravais lattices and all those things so we are talking about the green region that you see just below the wing and a little blue region over there so those are single crystalline but the wing region itself is a polysilicon material or polycrystalline material do you see how the base contact is coming through through the green and you can also see how the collector contact contacts the bottom substrate and therefore the current flows vertically so we will again turn it over flip it 90 degrees and look at these characteristics now the heritage region that I show here is the polysilicon polysilicon emitter so now emitter has two pieces one piece is this darker blue n plus that's single crystalline and another piece is this heritage region that is polycrystalline material and we will see how it works so this is how it works but polysilicon transistor does is the following once you have you remember is we are what we are trying to do is increase gain but we are in trying to increase gain by keeping the collector current the same but by suppressing the base current so what I am going to show you is how to suppress the base current because you know gain is either you can increase the numerator or you can decrease the denominator either way it's fine so long you are just working for looking for gain and so what we'll show you here is how to suppress the base current now you know how the base current comes around right so you have the holes coming in from the base contact which is shown here in the middle in the p region green p region and okay so once there is an animation which will come a little bit later I assume before I get there so the holes will go from the p to n plus region and flow out through the emitter contact now when it does so if I look at the hole profile then I see at 1n I have p1 this is the exponentially increasing part because the barrier has been reduced but the really interesting issue is you see the p2 is not zero if I had a metal contact then I know that the p2 which is the other end the concentration should have been zero what polysilicon does is that it limits the velocity at which electrons can go so it's like a surface recombination velocity and it is not infinite unlike metal as a result this because it has finite vs the p sub 2 which is on the other end at the polysilicon junction is not zero so in that case you can immediately see the current will be a little lower right because the slope of this p1 minus p2 for a given width that has actually been actually smaller and as a result I will have less current that's the physics there's nothing more to it I will do a few lines of algebra but that's the physics that I have been able to reduce the slope so first of all if you have polysilicon emitter I have i sub p because I'm talking about holes I have a e in the subscript because I'm talking about holes going into the emitter and poly because I am in the presence of poly so that's that's those three things are do you see minus q d p sub p p1 minus p2 divided by we what is that p1 minus p2 divided by we that is dp dx right do you remember minus q dp dp dx so p1 minus p2 divided by w that's my current now that current must be equal to whatever rate at which polysilicon take can take it out and polysilicon takes it out as minus q v sub s which is the surface recombination velocity here it's not really recombining at the surface it is very different from the previous case that we saw so right surface shock lee reed hall do you remember in that case we had a surface recombination velocity it is not that it's just limits the velocity at which electrons can escape to the polysilicon multiplied by p2 because that's the concentration I have can I solve for p2 now I think I can that's the ratio p2 divided by p1 and that's related to the diffusion coefficient and the surface recombination velocity do you agree with this statement does it look right do you see if the surface recombination velocity was very large like metal then in that expression vs should have been infinity right that's a relatively speaking if that's the case what would we have been the value for p2 that would have been 0 do you see that and that would have been 0 that says that indeed if it's a metal then the p2 must be 0 okay now we can now easily calculate because I have p2 I can put it in that expression minus q v sub s multiplied by p2 can put it in and can calculate the current do I know p1 by the way what is p1 this is ni squared divided by ne exponential of that voltage minus 1 right that's the p1 the junction so I already know p1 v service is a material property and the diffusion coefficient and the emitter width these are all all I know so therefore I could easily calculate this current now if I did not have polysilicon then if you go back what you what will you have done is p1 minus 0 instead of p1 minus p2 I would have p1 minus 0 divided by we again the same thing so the difference between when you have poly and when you have just pure silicon is that this instead of having just the diffusion coefficient divided by we that velocity I have a more complicated combination over here now if I take a ratio of these two currents that if I have polysilicon emitter versus if I don't in that case what would be the ratio I simply take the divide this one from the top to the one in the middle and I get this ratio now first of all does it make sense assume first vs is infinity that is I am really my poly is so good that it's taking all the electrons out if vs is infinity what is this expression this expression is 1 so therefore you haven't improved anything on the other hand assume that vs is very small if vs is very small that is poly is blocking the all the current in that case the second term on the numerator or denominator would be 0 compared to the first one and you can see vs is very small then the poly current the base current will be miniscule so in the process I have suppressed the base current that's what we are trying to do now for as far as electron current is concerned going over from the emitter to the base to the collector well no problem because I haven't really changed anything there's no poly over there in the junction and so in that case I will simply do slide the same expression n1 minus 0 divided by wb that is dn dx and then q multiplied by dn so therefore I will have that current and correspondingly the n1 value you already know that n1 value so therefore I am ready to compute the gain in the presence of poly silicon transistor now one thing I will ask you to think about is that why is does the poly only suppress the whole current which is going in that direction but not doesn't do anything to the electron current because electron current is also coming through the poly and why it didn't do anything that's something you should think about so that was the supposed to be the animation that which is supposed to say that the electron actually flows from the base side to the emitter side through that contact and that's the expression we try to derive okay now let's calculate the gain so the whole current is essentially all that is there is to it to the base current why is that because the amount of base current flowing to the collector is miniscule do you agree because base is collector is actually reverse bias so it has a huge barrier on the collector side but a very little barrier to the to the emitter side and as a result the base current is essentially entirely is due to the flow of holes and this is something I have already shown you in the last slide now let's think about it so I want to calculate the gain of it in the presence of poly silicon it is I see the collector current divided by IB in the presence of poly the base current in the presence of poly I could divide and multiply with the base current in silicon you see in this last term I have IB SI that's in silicon so called divide and multiply it's the same thing but fortunately I know the expression for the first one IC divided by IB at silicon that's the standard expression right in the presence in the absence of poly silicon that's the extended expression I have but because of this say poly silicon and silicon suppression of poly silicon I have this extra factors on the very end which has this red diffusion coefficient and small this blue surface recombination velocity so that ratio I'm just plugging things back in from over the top from the from the table or from the box on the top now do you see that in this case you can have a humongous gain because if BS is small then let's say close to zero then on the term on the right the numerator the blue on the numerator will go to zero but look at the denominator there is a v sub s is sitting there and therefore this could actually be significant it could be a huge gain in this particular respect and physically that makes sense if your vs is small that means there is no whole current flowing to the base so by even a tiny amount of base current you can have a huge amount of collector current and that's why this amplification is so large and that's the point I wanted to make now again just to tell you a quick story that this came about accidentally because the person who was supposed to grow this crystal and this is a sort of a story it's a rumor story but presumably it's true the person the technician who was supposed to grow the poly silicon or the silicon for the emitter accidentally that reactor malfunction so the temperature was not as high as it was supposed to be the temperature was little lower and therefore instead of having the crystal silicon crystal grown in the emitter they accidentally had poly silicon grown so they were actually planning to throw it away right because you know there's a malfunction in the reactor so better throw it away but it turned out they just tested it and amazingly it turned out that that has a gain the one that sort of malfunction with the malfunction reactor had a gain which is significantly better than anything they were making as a result this poly silicon transistor essentially subsequently become the standard in the industry I have posted a paper on the website about the history and you can take a look and this is a very nice story about about this whole thing okay so do you understand physically forget about the math yeah this is high school algebra but the the poly silicon how it suppresses the base current with by slowing down the holes and as a result increases the gain and subsequently after 1980s this is this became the standard in the industry but of course like all good things it comes to an end and it came to an end sort of I would say in mid 90s or early 2000 when so they this game went on for a while so they got a lot of gain and then gradually they started shrinking the base over a period of time to get more and more gain but at some point you cannot play that game anymore for this reason for the same reason and in fact it flip side of the same reason so this is the gain I have for the poly silicon do you see in the end I have one over v service that's the that gives me the huge gain in for for the poly silicon transistor now let's let me tell you about then the base transport and how in fact that time over time got poorer not better but poorer and that limits the game in the following you see if you make the base very small so it started with let's say several hundreds of microns in the beginning gradually you're making the base smaller and smaller why do you make it smaller because remember the base width is in the denominator because shorter the base width more the gradient faster is the current and therefore more gain now if you keep making it smaller and smaller and smaller the diffusion velocity which is d over wb will keep increasing right because the gradient is more it will go far away faster but there is a point beyond which electrons cannot go any faster even if it doesn't have any scattering there's a maximum velocity electron can go and that's called the saturation velocity you cannot electrons even if it didn't scatter ballistically went from one side to the other it will not go any faster as a result when you make the base very very very small in that case the maximum velocity it can get out of the base is this v thermal thermal velocity now I'll explain in a second why I didn't talk about it before but let me first get you an expression for it so now instead of the current being simply n1 divided by wb and the other side being equal to 0 because I have denied velocity on the on the collector side I will have this v thermal multiplied by n2 because that is the velocity at which electrons can escape and it cannot go any faster again I will do the same same game see n2 over n1 I have the same ratio just like before five minutes before and correspondingly I will have this if it is ballistic if the transport is ballistic then I will have the same ratio just like in the polysilicon I had it I had a finite velocity in the polysilicon I have a finite velocity in the ballistic transport case so I have the same ratio exactly the same thing but this time it's bad I'll explain first of all you see okay let me get to the next slide so I'll play the play the same ratio I'm just copying it from the previous slide but let me write this out and see whether you agree so let's go slow so first is that if I have both polysilicon and the base is so short that electrons are zipping through it ballistic no scattering no diffusion coefficient right diffusion coefficient requires scattering no diffusion no scattering no diffusion coefficient electrons zipping through then do you agree that I could split it up into three pieces so I could compare how the ballistic collector current compares I see ballistic divide by I see silicon when it's not ballistic so I can think about that then I can see how the collector silicon normal standard the green part how it relates to the base current you know the classical one and finally I can see how the base current in the standard classical transistor is related to that one in the poly you see if you cross multiply you will see that exactly what I have written I see comma ballistic and I be poly it's exactly the same thing I just expanded it out no rocket science here so I will just copy those expressions now if I copy this expression do the do you see the symmetry in the third term and in the first term right look in one case I have v sub s in the denominator in another case I have v thermal on the numerator right because it's a finite velocity and the corresponding diffusion coefficient come into play now what will happen let's see what will happen so let's say my v sub s is very small okay so if the my v sub is very small then in the third term on the right in the the equation preceding equation v sub s the blue one is negligible compared to the red one so I will drop that term and correspondingly bring it in but look at the first term if my v thermal is much smaller much smaller than the diffusion one this is in the black the first term in that case what will happen correspondingly exactly the same way I have to drop v thermal and the magenta v thermal on the numerator that is the only one that's going to survive as a result you can see now the gain is gives is given by the ratio of the thermal velocity divided by the surface recombination velocity now thermal velocity is a material constant if it is silicon you have a thermal velocity therefore you can see there's no more place to go if you decrease your base anymore you see you cannot get any more gain now does this sort of recons does it go with what I had been saying before before what happened that the diffusion coefficient for the electron d over wb when wb was large in that case that was a small quantity compared to v thermal in the first term in the magenta one and so generally in the long term nor long base transistors d over wb you drop that if you drop that one then v thermal on the numerator v thermal on the denominator so that will give you one and that is the standard transistor however in this case as you make the base it's smaller and smaller and smaller at some point you get no benefit but making it any smaller or any thinner and that is what happened in mid nineties in fact and that is the point beyond width they have not scaled the base width anymore about 100 to 200 angstrom in the highest performance transistors today now I will go through the frequency response these are all about DC characteristics but simply because you have a very significant gain of a transistor does not mean that you have a great transistor and that is what I want to explain in the next few slides so this will be a small signal now this will require some following through so please pay attention and then we will see whether you can understand it you may have to go back home and read it one more time so what I want to know is actually I have a great DC gain that it amplifies the DC current well but do I have a great AC gain that is if I have a small signal coming from a radio station what is the highest frequency that I can amplify if you try to do that what you will see that when your frequency is low that's the rate curve and you know x axis is f y axis is log of gain when the frequency is low the dc one but that's great you have a lot of gain so it could be 100 or 1000 but as you try to amplify more and more higher and higher signal then what's going to happen that then at some point the gain is going to drop like drop like a stone towards the end because it's a log plot c and even for one order of change in the frequency there may be a factor of 10 decrease in the gain so you started with 100 now if your frequency gain is always on the southern 10 and very very soon one then you have don't have a transistor anymore you cannot catch the corresponding radio signal or signal you're trying to catch let's say wireless signal and amplify it anymore so let's say what is the cell phone signals these days you know it's about 2.4 gigahertz right so 2.4 gigahertz or so that's the cell phone transmission band so you're you have to design a transistor let's say which at least amplifies comfortably up to that frequency so your f sub beta at the point up to which things are flat must be 2.4 in this case if it are for some other application then it may be lower or higher so the question is what determines f sub beta or what determines f sub t when the gain becomes one that's what I want to know and the thing I'm going to derive in next 20 minutes is this complicated looking expression that it depends on many things you see on the left hand side I have 1 over ft so I want to calculate ft so I have 1 over ft and you see on the right hand side I have a bunch of things I have the base width so I want to make this thing as small as possible then 1 over ft will be large right or sorry ft will be large and so if I make w be small then that enhances my frequency performance there are a bunch of capacitances do you see that there are these are all junction capacitances those are also very important on the second hand side on the right hand side do you see I see somewhere second term on the denominator if I could have a lot of I see if I can forward bias it which are very high I see then the second term will gradually drop out and as a result the whole sum will become smaller and as a result I will have a larger ft right so for larger ft I should have as high a current as possible as low a capacitance as possible as thin a base as possible right so these are the things I want now how does this expression come about so let's see how it comes about now if you did your homework and or at least the one that I suggested in the class then you might have remember might remember that if you take the the ever small model which I drew it in the common base configuration and convert it to a common emitter configuration do you remember few class then you should have these particular circuit where you had two diodes one going from the base to the collector collector is on the other side and another the c pi and c mu are essentially the capacitances various capacity junction capacitances so this is something you should have had if you don't know how to do this perhaps you will go back and look at that one one more time but for the time being assume that this is what it is let's proceed with this one now this is a dc equivalent circuit or dc or this is a generic equivalent circuit I'm only interested in small signal equivalent circuit so therefore I can simplify this significantly more in the following way first of all I drop that diode why do we do that because of course is a reverse bias reverse bias junction tiny amount of current is flowing right base to collector isn't it reverse bias tiny amount of current is flowing that diode isn't important I drop it but of course I shouldn't drop the capacitances the capacitances are still important okay now after I have dropped it first of all look very quickly c pi in both cases c mu dc on the right hand side it's about the same only thing the first diode I have replaced it with r sub pi that's one thing I have done and the last current source I have replaced it with dm multiplied by vpe how did I do that in the following way you know that the forward current is if not an exponential of that we know that so first of all let's see I have a small signal so I'm changing the base current a little bit right with little bit on top of the dc bias so I must know how is my base current changing with respect to vb as my input signal is changing how is my base current changing so if I take a derivative then do you see that ib is essentially 1 minus alpha f multiplied by if because I don't have the diode so all the dc current or any current that is flowing in is flowing in through the diode itself right therefore that's my base current so I have 1 minus alpha f multiplied by if now do you see if you take a derivative of if with respect to vb look at the expression on the very left then what do you pull out a q over kt you see that over in the left you pull out a q over kt and so that's what you see if you pull out a q over kt so 1 over r pi is qib multiplied by kt no problem this is simple and ib you could write it as ic divided by beta dc other than that this exactly the same expression what about gm the transconductance well you could again calculate alpha f multiplied by if why do I just talk about alpha f multiplied by if I am talking about the current source what is the value of ir in the blue in the equivalent ir is 0 right where is reverse biased so the only part in the blue current source is alpha f ir so therefore in the transconductance I am just taking with that with respect to vb again you take the derivative if you take a derivative you pick up a 1 over kt q over kt and that's your final expression and so if you wanted to know how does the current source change when you change your base signal a little bit then you will just cross multiply and write it as gm multiplied by vb and that's what is done in the equivalent small signal equivalent circuit on the top this magenta on the top okay so I have that this circuit you know even second year student undergraduate could solve right hopefully hopefully if you haven't forgotten so let's see how to solve this one now the way you solve this one is what you call a short circuit current gain so you will short the emitter to the collector without any load and you want to know how much current would flow when I send in a little bit of base current how much current will flow from emitter to the collector to the output circuit that's the dotted line I could write it in one sentence I see over ib look at ic first ic has this two current component right when the current is coming in one is this current source going down gm vb that's one piece and another is flowing through the capacitances what is that capacitances impedance j omega c multiplied by the voltage difference which is the collector to base voltage difference do you see on the numerator I have vc b and multiplied by c mu so that's what I have on the numerator now on the denominator ib has these three components one component is vbe divided by r5 that's the first term on the numerator second term is I have a c pi and it will again be multiplied by vbe and that's my second term and do you see what the third term is c mu multiplied by vbc that's the third term that's it there's nothing nothing more there two nodes and a few elements are here how simple can it be so then I want to know this is my gain at any frequency you see as you change your omega this ratio will keep changing and as a result my gain will be changed now at what point does the gain go from 100 to roll off and then become one that's whatever input signal I put I don't get any amplification so that is the point when I will call that frequency f sub t and that must be equal to one right by definition so I want to know at what f sub t will my gain become one now do you see that the expression I have from the top I'm just taking the absolute value now given normal transistors the c mu term on the top the capacitance term on the top is generally small now there's no reason a priori reason that is people have historically have seen so you drop that term so you only keep gm and similarly 1 over r pi it's a forward bias diode current that's small so you drop that one also so you have that particular frequency so that's my answer I have 1 over omega t 1 over 2 pi ft is equal to c pi plus c mu divided by gm it looks nothing like the long expression that I am after remember that long expression I told you about a few seconds ago but it will become that long expression in a few minutes because the gm two minutes ago in the previous slide I showed you that is kt over q i c first thing so that you can see that that's the term in the rate okay the c pi and c mu these are junction and diffusion capacitances do you remember the diode equivalent circuit that there was a diffusion capacitance there was a junction capacitance do you remember so I have just split it up cj is the junction capacitance cg is the diffusion capacitance I have two junctions I have two capacitances done now the collector side I do not have any diffusion capacitance right why not because this is reverse bias heavily reverse bias I do not have many minority carriers so I will drop this term so I will drop that term I'm sorry I should have dropped the other term this term not this one and correspondingly so you have that first term that is already in the final expression I have that so I will just leave it alone this one somehow will become w square divided by 2d and other things so I will see how that becomes but before I do that let me convince you that the capacitance divided by the current that has can be written as the qb divided by ic do you agree with this statement because you can see that q ic divided by kt couldn't I write it as d ic over dbb remember that is how I pulled out q over kt a few seconds ago and if I flip dbe on the denominator in the numerator then c multiplied by v how what is that that's the charge so therefore this becomes the charge because I have flipped this voltage up in here and so this become the charge and d ic comes in here so instead of expressing calculating the diffusion capacitance and all I could just evaluate this quantity and that will give me the second term the rate term on the right but let's see how to calculate that that's easier now in order to do this this is where you need the charge control model so the charge control model if you go back you will see during dc conditions d qib divided by d ic essentially just qb over ic let's see whether you agree this we did in the class before did I qb q sub b is the area under the triangle is it ballistic do you think this is ballistic transistor not really right you see because it's a classical transistor on the other side for the green triangle has gone to zero had it been ballistic then it wouldn't have gone to zero it would have moved with a certain velocity but anyway so you have this area under the triangle that's your qb and the current ic well ni divided by wb so you see the expression I got is essentially wb square divided by 2 dm this is a very famous expression because this tells you how short when you make the base small how short the base transit time is there's one more piece and then we are done you see so once it comes the end of that the base collector junction or then it has to transverse this huge depletion region very high field and electrons will go with the saturation velocity through that through that junction how long does it take is it simply wbc divided by v sat not really do you remember the image chart we talked about a few few days ago right if you have image chart then it will go a slightly faster and what is going to happen that as soon as the red electron gets into the depletion region it will be imaged on the collector side and as a result the current will begin flowing even when the red electron has not arrived to the collector as a result the current would look like a triangle but of course at the end of the day you have only transferred the charge q so you have half multiplied by i i is the current and that's changing with time multiplied by tau the average time for the electrons to go that must be equal to q and as a result the time it takes for the electrons to collect that depletion region is wbc over 2 v sat it goes a little faster because the image charge on the collector essentially pulls it in sort of and so it goes a little faster then you dive otherwise expected okay so i'm done you can see the junction capacitance part right remember the junction capacitance part in the denominator part in the in the first part this is the dqb divided by dic two pieces how long it takes for the electrons to go through the base and how long it takes for the electron to go through this depletion region add them up that's the whole thing as a result what happens that when you begin to increase the collector current the i sub c let's say you're increasing it the second term begins to drop because you can charge the capacitance fast you have a lot of current flow so you can charge the capacitance first and so second term begins to go small and your 1 over ft becomes gradually smaller ft becomes larger so you can see the frequency is gradually increasing as you make the collector current more however what happens if you push too much collector current at some point it overwhelms the collector doping do you remember what is that effect called carc effect right where it overwhelms it and the base used to be wb this tiny region and when so many electrons sort of has flooded in it has destroyed the junction and as a result the base has been pushed out all the way to the sub collector now i have this humongous base instead of the tiny one right as a result my wb at some point when i make i see too big my wb will be so big that any advantage that i had here will be compensated over compensated by the first term as a result my frequency will begin to drop because i record now more time this whole thing has to go through this long pushed out base this is exactly what happened in mid 60s and carc people used to make this transistor measured them in their lab and always they will see a turnover nobody understood it the best mind at that time you know all this big industrial lab they had sony they had ibm they had ti scientists are looking all over that what could it be that i cannot push my frequency anymore and carc explained that effect and that's why the effect is named after him he explained it because he was the first one to suggest that how excessive collector current can overwhelm wb and as a result it will roll over so it's called i sub k that's the car so in general i told you about all those things tau sub b is w square divided by 2d and the collector transit time is wbc divided by 2 v side that's fine generally people have to add this extrinsic resistance the resist resistances that you have from outside that becomes an important part also so that's something one has to check and also sometimes you're interested in maximum power gain not only the frequency gain i'll not derive that but that can be calculated as shown in the bottom bottom expression so let me summarize so the basic idea is that the basic transistor the standard transistors of 1950s have actually gone through many modifications most important of them are this poly silicon transistor right that was a great innovation without it transistor scaling bipolar would have stopped at 1980s because things have stopped by them but it was 15 more good years because of 15 20 more good years because the poly silicon but then the short base essentially killed it because you cannot improve it any further you'd not have any more any more gain so these days people play with the junction capacitances they try to make it as small as possible by making good isolation and other things thereby increase the frequency now i showed you how to do the small signal i made a lot of uses of the diode results so if you didn't understand diode results properly this is the time to go back and review them that will make things clear because i assume that you know but hopefully it will become clear with time now the bottom line is at the end is that classical homo junction transistors can only go so far and nowadays then hetero junction bipolar transistor has essentially replaced the technology and in the next two classes i will tell you how hetero junction bipolar transistor works okay now i have three slides as an aside on collector base breakdown voltages that i want to cover one thing i have not discussed is what happens when you apply a large bias in the collector now you can see the base collector junction is essentially a diode when you apply a large bias on a diode what happens eventually it breaks down and it breaks down by two methods right what are those zener tunneling electrons could tunnel from the base to the collector or impact ionization it could break both ways i want to explain to you why the breakdown of a bipolar junction transistor is a little bit more complicated slightly more complicated than a simple p-n junction because there is an extra amplification involved let me explain first let me explain what is the essence of current gain in a bipolar transistor in a very simple way you see this is a common emitter configuration right you put a base current certain amount of base current let's say and the base current flows out through the emitter that's what happens that's the blue blue arrows now the way to think about it is you know the expressions for the base current you have a bunch of constant but the main thing is you have a qvbi beta is 1 over kt qvbi minus 1 so when you push a base current then the transistor must accept that current because you are forcing it to only way it can accept the current that it will keep lowering the barrier in the base emitter side until it has lowered the barrier enough so that the whatever hole current you are putting in through that blue triangle up triangle that can flow out to the emitter so you are forcing your input is base current you're forcing a voltage in response now when you force a voltage in response what it does it forward biases the junction and the electrons are just sitting there it says well my barrier is lowered and so electrons decides to go on its own so now you can see for electrons the vbe becomes the input because the barrier has been forward bias so the current will flow and as a result you can see that red arrow going from the emitter to the collector region so therefore this is how the base current controls the collector current this is how the gain occurs right now if you understand this part now you will understand why the breakdown is fundamentally different in various base configuration various transistor configuration let's first talk about the case where my there is a common base configuration now in a common base configuration the emitter current is constant I fix the emitter in that case how will the breakdown occur I have used a large collector bias you can see a huge difference and a large electric field on the base collector junction again the base current will come in it will forward bias the junction current will flow over now it will once it starts getting into the collector region it has trimmed this amount of electric field as a result there will be impact ionization right there will be how much energy do you need approximately three halves the band gap you cannot have just the band gap a little bit more three halves the band gap let's say and as a result you will the green one is a second electron that has been generated sort of and the red one that you see are extra holes that have been generated now the holes like to float up right these are like bubbles so it floats up so it floats up now the thing is the collector current it doesn't want anymore because that's the common base configuration so in that case it will get out through the base base current is not controlled as a result this breakdown completely depends on the base emitter voltage the base collector voltage it doesn't care about what is happening to the emitter the larger voltage that you apply on the base collector side that's when the breakdown occurs this in contrast with the next thing the next thing is common emitter configuration in common emitter configuration the figure on the right shows the iv characteristics it breaks down much more easily do you remember it has a lot of gain right i c over ib beta lot of gain but you see the problem with this because this breaks down much more easily and this is why look at that again current coming in base current coming in flowing out through the emitter no problem electron flowing in trying to impact ionize but now this time base current is fixed because this is a common emitter configuration base current is fixed so this extra holes that are coming in the base doesn't want it anymore base has already decided how many current how much current it can go as a result it cannot have that as a result the base this extra excess current that has been generated now must get out through the emitter now the more emitter current it has to push what will happen to vbe now the vbe must be larger or that the barrier must be reduced so that it can accept that current right as soon as it reduces the barrier what will happen that correspondingly it will forward bias the junction more electrons will flood in and more electrons flooding in means more impact ionization very soon because the presence of that first barrier first pn direct the breakdown will occur very quickly and you can see on the right hand side that how much faster it's not a it's almost begin to break down at a very low voltage even by the 20 volts gradually currents are increasing quite a bit so this is a very important distinction between a just a single diode versus a bipolar transistor and that you something that you should remember okay and this is the basic point about breakdown voltage that's it thanks