 you can follow along with this presentation using printed slides from the nano hub visit www.nano hub.org and download the PDF file containing the slides for this presentation print them out and turn each page when you hear the following sound enjoy the show okay we'll get started and let others drift diffuse introduce this is lecture 18 on continuity equation and today we will finish essentially the development of the transport equation and for the first time we'll solve a few concrete problems now in some way you should realize that until 1950 early 1950s this was the state of art of understanding in fact for last 50 years most people have essentially just solved these equations for variety of context and only recently about 10 years or so ago that things have 10 or 15 years the things have began to change in some respect here that we'll discuss perhaps later on so we'll talk about continuity equation then we'll solve an example problem and then we will conclude now as I have mentioned several times before that there are these five equations in fact three equations if you count properly that is in the first equation which is the continuity equation for electrons and I'll explain why what do I mean by continuity then you can see that there is an expression for the current and the second equation simply says how the current responds to electric field and density gradient so you put the second in the first one in fact those two are one equation and similarly there are another electrons another continuity equation for the holes and finally the Poisson equation now we will talk about the second first and the third equation today now before I move on there and talking about and discussing this let me point out something important if you looked at the fifth equation this is called a Poisson equation now Poisson equation relates the charge to the electric field or here the displacement is D related to the electric field with dielectric constant multiplied by the electric field now when devices become very small molecular devices or so not transistors big transistors that we talk about this equation the Poisson equation might get modified in fact it does get modified because in any of these the electron-electron interaction very close electron-electron repulsion is not accounted for here as if one electron is moving and the rest of the electron is sort of in a background or the combination of protons and electrons is removing a response to that so this equation on more sophisticated physics and smaller devices might get changed but for most devices is pretty good similarly the second and the fourth equation which combines drift and diffusion when you have very small device in which you do not have any scattering for electrons going from one contact to the other then in that case these two equations will also be not be correct these equations are valid when you have a lot of scattering as you're going from one contact to the another so therefore in most many recent results if you see in the literature you will see that there are modifications for the expressions for the current Jn and JP when the scattering is not as strong and diffusion Einstein relation may not be valid within the device but the first and the third equations these are conservation equations valid always regardless of what detail physics you put in for Jn and JP is valid always and let me explain why because this is simply conservation law so long electrons are neither created nor destroyed in a global sense then there of course these relationships will hold so let's consider a semiconductor of an arbitrary shape shown here on the right a top view and the ash color region is let's say top view of a semiconductor and I have put two contacts there square shape contacts and if you apply a voltage then the current will flow from one contact to the other now assume that I have divided this region into many miniscule squares or in three dimension you can say many miniscule cubes of size a now this size is not of course atomic size a I'm talking about more like let's say 15 to 30 angstrom a relatively big region almost macroscopic in terms of properties effective mass is valid mobility is valid those concepts are valid but smaller compared to the device dimension in that case if you put some boxes in the green box the the green and blue and yellow you will see the electrons sort of streaming in to the box let's say the blue box and gradually streaming out of the box so say there is certain number of electrons which will get in and certain number of electrons which will get out so let's blow that region up and look at this three dimensional picture sort of looks nice and in which the electron is coming in and getting out so I will consider that box to be of size delta x the length of the box blue box in the middle let's call that delta x and let's call the effective area a equals one you know a can be any number any this is the width and the depth but we'll call it one in some units now if I counted the number of electrons that are coming in and out of this then you can say that let's say the certain number of electrons coming in in jx now in 3d it will be a component of x y and z I'm thinking one dimension for this simplicity here and there's certain number of flux certain number of electrons per second that's flowing in the box and certain number of electrons flowing out of the box now how many how would you do the book count a bookkeeping well you can write something like this on the left hand side what I have of this equation what I have written that how the number of electrons within the box change with time do you see let's say n is the number density right number per centimeter cube and if I multiplied by the a which is the cross-sectional area and delta x which is the length of that box that's the volume so that's how many electrons I have within that little box now how does it relate to the current well if that depends on whether the number is growing or coming down depends on how many is coming in and how many is getting out just like in this room let's say students are constantly flowing in and some of them are leaving so the number growth within the room will depend on the flux of electrons coming in and the flux of electrons getting out now I have divided it with minus q in the denominator because these are electrons and I want the number and so in the jn j sub n there is already a minus q remember that because it's electron current density okay so this is almost done if I didn't have any other process in fact this would be the equation that will tell me how many electrons are coming in and out now remember this j sub n are the fluxes that are coming from the contact sort of right these are drift and diffusion current but of course there are another current that we can also think of that if you shine light as that's a third contact from which electrons hoser can come from right so in that case assume that the top yellow one arrow downward it is as if light shining on that sample so electron nut not coming from the contact but coming from outside and that one and similarly I can have a recombination recombination of these electrons recombination to what? recombination to the holes so in that case the electron is not getting out of the device for the top one the generation photon is coming outside and in but we'll talk about that in a second so g7 and r7 represents electron generation and electron recombination through trapezius state tunnel trapezius state recombination or direct recombination that type of thing and so all you have to do is again multiply with the volume and the gn because gn what is that that's the number per centimeter squared per second per centimeter cube per second so that's the generation rate and correspondingly the recombination rate you can see is the same everywhere so you can just divide it out and once you divide it out you see 1d place there will be a delta x left what do you see that in the first term on the right hand equation everywhere else the volume got uniformly multiplied so everything has lost all this a and delta x in this process now this derivative gn at x and gn x plus delta x divided by delta x what is that that's the definition for derivative right that's the normal definition for derivative and so in general if you thought think about three three dimension you will call that a divergence divergence is how many particles going outward so if you think about it there's a minus cube down there in the second part or first term on the right-hand side if you put it in you will see the first term actually represents the number of net outflow of electron that's divergence and so that's what I have written on the right-hand side okay now you see in this way equation it is saying that the electron number within a particular box can stay flat you know in the somehow all the fluxes that are coming in the balance each other it can stay flat electron number can go up so how come electron is not being created or destroyed well the reason electron is not creating and destroyed is in the global material and I'll explain that in the next slide but within this one equation electron number can go up it can go down as a function of time it can change as a function of distance and within the volume no problem that can happen but still in the next slide I'll show that the net number the total global number that doesn't change so you will have one equation for electron you can write that that's easy and similarly you will have another equation for the holes now this whole g sub p and you realize g sub n for the electron this is the same flux actually because the same photon came in and generated electron hole pairs and so GN and g sub p these are generally the same there are special cases where it can be different but in this case we'll assume it's the same and similarly the Rn and R sub p this is rate the rate at which electrons disappear the holes must disappear at the same rate because it's recombining with the holes so therefore these two processes are actually coupled and correspondingly you will have a equation for the holes only thing that is different in the second equation is do you see this there is a negative sign in the first equation and why there is a negative sign because holes carry a positive q so when you write the equation on the denominator you will write a plus q and once you do the everything else because of that plus q in that term you will have a correspondingly negative of the divergence that's it and so therefore these two equations will be globally linked now the continuity equation in general as I said I want to reemphasize this that this generation process g sub n is always actually equal to g sub p because anytime you generate an electron pump it up by photons let's say you leave behind one hole so the number of electrons that is increasing is the same as number of holes that is increasing right and similarly the recombination process will they recombine with each other and therefore they get lost at the same way now so now one thing you should look that if I were to actually take a difference of this odd or sum them up dndt and dpdt you should convince yourself that net rate that net rate should be independent of the current that is flowing in because no net number of electrons can change holes have been created that means electrons have been pumped up but that does not change the total number of electrons when something got recombined an electron was in the conduction band that came down to the valence band but if you look at the material as a whole of course electron and hole remains the same no problem individual component can change easily electron and hole number but not the global number see it's a very important thing to remember an analogy an analogy since we are at Purdue we must have a Wabash river around and let's say this is some sort of lake for around that place where so the first term will be the rate of increase in the water level in the lake that is that little cubic box which is like a lake here that must be equal to the inflow and outflow the deep blue regions these are electrons coming in and electrons getting out or water particles getting in and water particles getting out rain is coming sort of this is generation rate and then the evaporation getting out is sort of the recombination rate so the reason I wanted to say that this is global this is sort of a very general law this you see here there's nothing about electrons or holes anytime you have a bunch of particles you can write a global continuity equation that really doesn't matter whether you know about quantum mechanics statistical mechanics doesn't matter the physics comes in one level below when you want to write equations for the inflow when you want to write for equations for the outflow and then like the evaporation the physics of evaporation will determine the rate right the pressure difference and everything the physics will come in in filling up individual terms but this equation as a whole this could have been written 2000 years ago and it's still be fine you don't need any quantum mechanics or any sophisticated thing before that right okay so that that's the equation general term remember that you should be always be able to write these equations if I put three contacts in here you should be able to write this equation regardless of how many contacts I put regardless of whether you know the physics of the problem or not okay so all the equations are in place we are in good shape we know drift and diffusion equations we know the continuity equation we just derived now if you have to solve it there will be a problem do you see why there might be a problem first of all how many unknowns do I have here do you think I have three unknowns do you see electron number whole number and the electric field or potential how so isn't there a recombination rate present there r7 or rp well yes but remember in shock read all recombination I will replace that term as np minus ni squared divided by the big denominator so that term inside it actually only has n and p so I can forget about that generation rate most of the time it's a it's a certain flux certain number that will be given let's say cosmic ray is hitting is your device it will generate a certain electron whole pairs so that will be a given number and you can see that mu n and d sub n these are constants right these are constants so I have three partial differential equation and that's not an easy problem to solve and most of the time you'll need a big computer to solve it but remember most of these problems were solved 50 years ago before computers were widespread use of computer was widespread so how did they do that well they used a lot of approximation now one thing before I go on I want to mention this that many times people equate approximation with a negative connotation oh he has approximated the problem I want to flip the thing around I say the people who can do approximation correctly may know physics better than most people because they can identify what 10% of the problem or issues govern 90% of the problem right so being able to identify that 10% requires understanding more physics not less so what I'm going to do as I show you how to make approximations and this has been done by great people you know many of them things are copied by from the days of Shockley and others so these are actually very great people know many Nobel prizes they have made this approximation so you can understand that why these have deep physical inside but you have to understand also what do did they do and in that process develop your own intuition it will take a little bit but you will be able to do that if you follow the logic so there'll be two methods of solution one is numerical these days there are various sophisticated simulators you go to nano help and you will be doing that you can solve the problem and that's new medical solution I will show you a little bit just like I showed you how to do Schrodinger equation right discretization and other I will show you how to do that but today we'll be talking about the analytical method of how to solve that easily so generally eventually the analytical solutions the solution of the Poisson equation will be given by a graphical solution and the graphical solution of the Poisson equation is called a band diagram and we'll see how to draw it now one important point is that if you don't learn how to do band diagram in this course you will never learn it and actually you have wasted your time in this course the single most important thing that comes I can say for this course and also for the exams is that you must learn band diagram how to draw band diagram and I'll show you step by step how to do that and then of course we'll be solving it very in various approximations this four equations in diffusion approximation minority carrier transport and bipolar transport on wide variety of approximations will be solving that complicated equations but the good thing about this approximations are that they are so simple you can then eventually solve it in three lines but of course you shouldn't approximate yourself out of the problem because then you'll not get anything so let me show you one example so consider that I have a crystal and this blue and as the crystal is growing I'm incorporating different amount of dopants maybe or the temperature of the crystal is changing as I'm growing it so the bottom line is that because of the growth process I have the this regions have different properties so the red region is terminated by the metal contact and similarly the green region on the right third region is also terminated by metal contact you must have two contacts so that the currents can flow in the middle yellow region light is shining right so this is a problem and this is an unpassivated surface that what that simply is giving you a clue that you should recall consider surface recombination that might become important right so every statement here is trying to give you a clue of how to make the approximation so first thing it says this problem I'm solving a problem let's say it's an exam problem that it is accepted dope so you immediately in your head you realize that this is essentially all acceptors and P the number of holes in the extrinsic region is equal to NA right that's the first thing you immediately realize by as soon as you hear the word accepted dope now it said the light has been turned on in the middle section and it reaches a steady state the right region is full of mid-gap traps what clue I'm trying to give you here full of mid-gap traps means there'll be a lot of shocky reed hall recombination right because mid-gap traps are responsible for it so I have to consider in region 3 in the green part recombination process right why did that happen because when I was growing let's say for some reason if I had some unintentional impurities incorporated in the green region then I may have a lot of traps there now I say the interface strap is on the end on the right side and the left region which is red or magenta that region is strap-free so I should not have any recombination there and these are contacted by metal contacts okay now you may not realize if you had gone to the talk by Mark Pinto that this is in some simple form an example of a solar cell in a very simple form solar cell has a little bit more complexity but then you get the idea that sunlight striking a section of the solar cell and the currents are being collected from the outside so it's a baby solar cell let's say because it doesn't have all the junctions and other things but other than that let's say we can get started here how do I solve a problem like this okay the first thing is I'll recall I'll ask you to recall how we solve the Schrodinger equation because this is exactly essentially the same procedure because at the end of the day same mathematical equations you know so once you have a second-order differential equation whether it's a Schrodinger equation or whether it's a diffusion equation once it is down on your paper piece of paper who cares it's the same method of solution and same technique so everything that you have learned there it will you will be able to use it here for example remember that when you had analytical solutions we had the second-order equation and in each section we solve them individually right we solved a sin kx plus b cosine kx or the exponential form we had these individual solutions in various places what was the next thing we did we had the boundary condition on the two sides these are the metal contacts and this problem these are metal contacts on two sides and that's sort of the plus infinity minus infinity equivalent to the boundary condition then there was all this continuity relationship right you remember that allowed us to stitch up the solution from one side to another we'll do the same here the three regions one two and three you know in the last last one we'll stitch up the solution in individual interfaces and that will give me the global solution now there is no equivalent of four and five but that's all right we should be able to get up to three and then that should give us the carrier concentration profile so very quickly I just want to remind you that this is how we did it in various regions we stitch up the solution applied the boundary condition and had the whole thing in place so let me now show you how the drift diffusion equation are solved exactly in the analogous way we'll see so we will take one piece at a time not all three pieces one piece at a time solve it and then eventually bring them all back together and just stitch it up or stitch the solution up using boundary condition let's talk about region the middle section region one little section has light in it right light shining on it the arrows so I have that first equation dndt now I want to make some approximations what approximation can I make think about it and listen to me carefully because these approximations has a logic which you need to reproduce when you solve problems as well because I will give you a slightly different problem to solve of course this material is accepted dope and I have not talked about any variation in doping density that means the number of dopants is uniform throughout if it is uniform throughout that means the P the whole concentration that has been is there that's also uniform throughout if in equilibrium before light shining if it is uniform throughout what diffusion gradient do I have I do not have any diffusion gradient right it's uniform if I do not have any uniform gradient so you can see the diffusion term must drop the gradient of n10 must drop what about electric field does it have any electric field well I haven't applied any voltage here and the carrier concentration is uniform so if the electric field comes in from outside because the carrier in the total the net charge here and I'll show you in the last slide our last equation is also 0 and as a result no electric field in this material also do you see why because number of holes that is positive right and the number of acceptors is negative because they have given they have given those electro or the holes so therefore the combination of them has 0 charge this is charge neutral since it is charge neutral therefore I do not have any electric field great because then my Jn is actually 0 and because it is 0 my big first term goes away on the on the right hand side of the of the top equation now the n after light has start shining on it I can divide it into two pieces n0 plus delta n what is n0 before light coming in that's n0 after light has come in I have that extra delta n where is coming from from the valence band the valence band electrons have jump up the total amount being delta n now this will assume in the beginning that this is a small injection light is weak and so the recombination term I can write it as minus delta n divided by tau n do you remember where this comes from this was the shock to read hall recombination in the case of when light small injection right minority carrier injection that's where the first term comes in and the G will remain whatever it is that's the light coming in okay now you have seen that I have crossed out the value of n0 well n0 is a constant because remember this is before light came in it is uniform and of course before light came in nothing was changing with time it is only this little delta and that's changing with time so the n0 is a uniform time independent so as soon as you take the first derivative with respect to time that one goes as well right now you can see that why this equation is very easy to solve okay so that's the first equation for the electrons what about holes holes the same way same way no current uniform semiconductor no electric field and therefore you essentially drop that current also and in this case you do the exactly the same thing for the holes as well now the electric field one I could see from some of your faces that you didn't really understand that why I said the electric field to be zero now let me convince you that I didn't pull any trick electric field always has to come from the Poisson equation right Poisson equation has P minus n plus Nd plus and Nm minus I can understand that before I have light turned on that everything is balanced charged neutral remember piece of semiconductor sitting by itself every point the same one what about after turning the light on steel zero the reason is the delta n the extra electrons that you have created is exactly balanced by the extra holes that you have created as a result even after shining the light on in non equilibrium even the gradient of the electric field is zero now gradient is zero that doesn't mean the electric field itself is zero but remember I do not have any contacts here so my input electric field is zero and the gradient is zero therefore throughout the electric field is zero do you realize this and that's why I was able to drop the term okay now this is a high school equation in a good way hoping that you'll remember so you can solve this right this one equation I can solve and you can plug this equation back this is by integrating factors and other things that you can check but the one way it is to do it is by simply plugging the solution back in so I'm solving for the middle region and if you do that you can convince yourself that the solution is indeed indeed right now I have delta n x sub x comma t I shouldn't really have to put an x there right it's uniform so I really didn't need that for the time being but I just wanted to make sure that that I put the boundary conditions in here so what are the boundary conditions at time t equals zero what is delta n no light delta n is zero minus zero right delta n is zero because I haven't started generating any carriers as if delta n is zero at t equals zero then a plus b must be equal to zero itself and a is equal to minus b you can see what will happen at t equals infinity after I have shined the light for a long period of time what will happen you can see that it will reach a constant and what is the value of the constant is simply a that is the value of that of that constant okay okay and do you agree with this statement this at t equals infinity then things have gotten steady state right so that left hand side of the top equation what is d delta n dt at t equals infinity nothing is changing so therefore it must be zero right no change it must be zero so do you see that delta n at t equals infinity will be equal to g multiplied by tau n from that fast equation do you see that that's how I have inserted in in this last equation so I know the value of a do I know the value of b now of course I know the value of b and so that's my solution I know everything here tau of n is the that is where the number of traps and the velocity the capture cross section do you remember all of those things hide in tau of n so if you have a lots of traps it will recombine faster so that's why the toss of n sits in the g is the light that's coming out from outside and so that's the number somebody will give you the flux that is coming in and from here actually you can solve the whole problem so we are already partway there but let's all think now think about the other two pieces so this is the function of time you realize that it will come in and it will saturate what is this value g sub g multiplied by tau 7 that's the value when things have saturated now let's think about the first region you see what is going to happen that the in the yellow part light will come in generate electron hole pairs and they are going to get out through the red red side and through the green side they are trying to get out of the device because there's too much carriers so by diffusion it will try to get out of the device so let's think about that again accepted dropped so we know that let's see whether we can solve it again the full equation now this time unfortunately will not be able to take out jn anymore first of all can I take out this recombination and generation generation I can take out right no light how come I take out the recombination because I said trap free no traps no mid gap levels therefore I can take out the recombination okay that's fine dndt why did I take it out because I said steady state so therefore that's giving you a clue that I want to know in general well how the current flows so that takes out electric field is again zero I'll explain in a second but this time gradient is not zero why because from one side from the yellow carriers are coming in delta n is not zero but on the left hand side in the metal contact the concentration is zero and so therefore there's a concentration gradient and as a result we'll have when you insert the expression for the current into the first equation do you see that you will have a second order equation for diffusion you see that why what happened to the q the q cancel right you see the q cancel and therefore I only have dn and second derivative of n now I'll to solve that equation in this region one let's assume that that region is a whatever a is may be 200 micron right do you agree with this statement that the solution of this equation is a simply linear solution that's right right because second derivative of that will must go to zero so therefore this must be a equation remember I have said it at x prime and x prime is going in this direction going towards left so I have to just be careful when I put the values for x prime in now at x equals a the delta n is zero now this statement I have not explained I will do that in a second but let's say it's zero now why it is zero is something that you need to understand very clearly why in a metal contact anytime there's a metal contact delta n is zero why is it the reason is once the electrons get out of the semiconductor and gets into the metal metal has very high mobility electrons go very easily right when you have a copper wire put a voltage in it doesn't a lot of current flow so electrons have very high mobility and very high velocity within the metal now if it has a very high velocity in order to support a certain amount of current how much delta n would you need it will need a very small amount of delta n right because the velocity is very big as a result what will happen that the delta n because the velocity is very large delta n is approximately equal to zero as a result I have any time size semiconductor is contacted by a semiconductor metal at that point of contact I will assume the boundary condition to be equal to zero it's an approximation but a very good one you realize this okay so if I have that equal to zero I can put it in and relate C with D and I just put x prime equals a so that's one and at x equals zero delta n is will be equal to C right because I put x prime equals zeros and I get this and as a result if you put those two things in two boundary conditions in you get a relationship of delta n varying linearly with distance something like this you realize this of course that's the solution C plus DX prime was the solution and after I have gotten the value at x prime equals zero I will have delta n and it will go linearly to zero now this is something as soon as you see a trap free minority carrier in your head you should immediately be able to draw this picture even before solving all this equation you should solve it many times to get a practice but trap free without recombination well you immediately have a straight line going to the metal contact now be careful if I don't say there's a metal contact there you should not be immediately be using this boundary condition being able to equal to zero that you shouldn't use but it should be always be linear anytime it's trap free okay alright the last piece the last place is full of traps right this region so I cannot do what I was doing before exactly I do not have any generation steady state so I take out delta n delta t I get take it get out generation why no light in that region no light so therefore I take out the generation now this time I cannot take out the recombination because this has a lot of mid gap trap levels copper gold these are mid gap trap levels in silicon so somehow they have gotten incorporated so I cannot take it out the electric field is zero and correspondingly only thing that is remaining is a diffusion term so I put it in put the second equation in in the first one and you can see that that will give me these two these two terms the minority carrier recombination delta n divided by tau n right that's that you remember now remember the various approximations we did that applies in various cases had it been high injection what would have been this value delta n divided by tau sub n plus tau sub p remember I was asking that question that how I how come in high injection the recombination this looks smaller so under appropriate conditions you will have to replace it with appropriate expressions for recombination right if it's direct recombination you'll put the direct recombination term in be multiplied by delta n so whatever it is as I tell you from the problem you should be able to pick it up and put it there now the n naught originally before any injection was was there this was a uniformly dope material as a result n naught was also a constant and therefore independent of position as a result I should be able to take out that n naught because it doesn't have any derivative and that's my equation and this equation once again I should be able to solve do you know the solution of this equation that's how it looks almost like a Schrodinger equation right does it look almost like a solution of the Schrodinger equation no surprise because look the second derivative the first term of the second derivative and like the potential term on the next one they're exactly the same equation of course they should have exactly the same solution so I have that else of n is called a recombination length this is the square root of dn and tau n so that's that's what it is if you put it in and solve it I have else of n is I have made it short and written in that form again terminated by metal contact delta n is 0 and so that relates f with e and at x equals 0 remember this x and the previous x prime I'm not the same thing right x prime was going the other direction so with x at the metal contact it again becomes 0 that allows me to relate e and f and that's it that's the solution in here if you just you know there is a bunch of exponential floating around you put this values of e and f n in the first equation and this is the solution at the end you will get right what is the unknown here unknown only thing unknown is delta n but other than that do I know else of n of course I do because I just said it is square root of d dn why did I get d from Einstein's relationship because I knew mobility I measured mobility do you remember all those four probe measurements I got mobility through Einstein's relationship I got d and from d I was able to calculate and tau sub n number of traps the capture cross section the thermal velocity from those all those things I know and therefore I can calculate l sub n and this whole thing is known except this delta n and delta n I will find by stitching the solutions together in various places okay now do you see it looks like a some sort of bridge look in the middle but look at this is a carrier concentration in the middle uniform light shining so what is the value of delta n in the middle section g multiplied by tau sub n right a certain rate it is being created and it's recombining at the same rate and so steady state is g tau sub n and on the left hand side trap free going to the metal contact it is getting linearly to zero so that's one side the other side is full of traps and therefore it is disappearing faster and therefore before getting to the contact by the way where are those electrons going they are recombining with the holes and the holes are being pumped through the contact so in fact in the green region holes will flow in to support the recombination of the electrons as every electron is moving in holes will flow in so that with one extra one and therefore the current will flow because in the circuit whatever electron gets out from on side that must be balanced by the hole current from the other side okay now do you see how to calculate delta n delta n is already known at x equals 0 and x equals 0 prime that is the interface between the magenta and the yellow and the yellow and the green right so I put those things in and that will be equal to g is tau sub n I know everything right how do I know a and b well I agree with the material how thick they are I was I agree with the materials or I so I know a and b so in fact after this once I know the light the g the flux of light then I can calculate delta n and as soon as I know delta n then I can calculate also the current because if I know delta n I can look at the derivative the derivative of the car gradient of that region will give me the current right so therefore I can calculate the current as well in most cases what is going to happen is a reverse is going to happen let's say you have a little detector you have just built and the cobase satellite or some satellite is going out in outer space you have designed your little semiconductor detector you put it in the spacecraft it goes out and then there is this cosmic showers coming down right and in the cosmic shower coming down and you are you have a on the outside of the device you have a little emitter the emitter is telling you how much current is flowing from that you back calculate what cosmic ray radiation is coming through you see so what will happen j n is what you're going to measure through the emitter and what you are going to infer from it is the rate g at which the flux is coming in right and then you can reorient your satellite depending on if it's a too much cosmic radiation then you can take various control action around it solar cell panels you remember this big wings on the satellite they have on Hubble telescope and others same physics but over there of course you want to catch more sunlight as possible again that is the physics of all this in any simple form but you see we didn't solve any complicated differential partial differential equation still we know so much about about these problems okay so in the last two slides you have seen that I have neglected the electron concentration or is the electric field and dropped out the drift term and I have only retained the diffusion term now this is only appropriate for minority carriers for the majority carriers I cannot really get out the drift term and therefore the electric field in that case I cannot assume it to be zero so what's happening you remember that in equilibrium if it is a homogeneously dope sample as was the case for the example I just told you about in that case the electric field will be zero because n naught and p naught and n d plus and n a minus they will conspire together to make the electric field locally at every point equal to zero so that was not a problem now remember that this was an accepted dope region right that's how we set up the problem in the accepted dope region I can say that p naught is equal to n a right in equilibrium in equilibrium I can say that now in equilibrium I can also say that the minority carriers which is the number of electrons is equal to ni square divided by n a in the extrinsic region right I'm assuming full ionization not in the freeze out or intrinsic region okay n naught is a very small number right for example if p naught is 10 to the power 18 and if it is in silicon what is n naught 100 because ni squared is 10 to the power 20 and so n naught is actually a tiny tiny number fine but it is flat uniformly dope and together they make the electric field zero but you remember for the problem that we are solving the carrier concentration in non-equilibrium in equilibrium non-equilibrium case was not zero and then what will happen to the electric field that's what I want to explain so you remember that there were in the middle section light was shining it generated electron hole pairs g multiplied by tau so when that was the extra carriers and it was sort of coming out in region 1 and region so region the two contact sides it was coming out and let me assume that it was coming out with some sort of in the trap free region with a linear profile there my carrier concentration is no longer n naught it's a little bit more than that delta n right now in principle then I cannot neglect the electric field because I have a delta n but it is not a photo generated region so I do not have a corresponding delta p here as a result how come my electric field be equal to zero that's the problem right that's that's the problem that's the part I didn't explain as clearly now what is going to happen is because any in any system the system wants to minimize the electric field because the energy goes as e squared the electric field square at every point so it always tries to minimize the total amount of charge on the right hand side so what's going to do as a result that it will pull in it will increase the electric field a little bit in that region a little bit very tiny amount and it will pull in holes from the contacts to exactly almost exactly compensate the electron concentration so it's not photo generated it is coming in in order because it wants to minimize that quantity on the right hand side so it will pull in a little bit of little bit of holes to sort of match the profile now do you see here that this region is almost charged neutral at every point almost but it's not density neutral at every point the density is no longer the same right at every point density is non-uniform but charge-wise is essentially more or less the same and as a result that delta p flowing in from the contact this is almost equal to zero now why couldn't it pull make delta n exactly equal to delta p well if delta n is exactly equal to delta p then you don't have any electric field and if you don't have any electric field then the holes cannot come in right it in order to make the holes flow in you will have to have an electric field and if you do a calculation this electric field on the will be on the order of a micro volt as a result for minority carriers I was able to get rid of the electric field term however for the majority carriers 10 to the power 18 sitting in the first term for the number of holes a tiny my electric field can still have a huge effect so drift is not negligible for majority carrier drift is only negligible for minority carriers that is how this thing works out now actually if you have done the homework you already know this because you have already seen this electron and hole exactly balancing each other in your plots but you may not have realized it but when I returned the homework then you should double check and see whether you agree with the statements that I just made in terms of understanding why you can neglect the electric field okay so let me conclude here what I began discussing today is the continuity equation is a basis for semiconductor device analysis because this was like a capstone for all the five equations that you needed to think about these are actually done and this is generally continuity equation is always true you should be able to take any device any problem and without consulting any textbook or anything be able to write the continuity equation be it for phonons photons electrons whatever it is you can write it the full numerical solutions are possible and I will cover it in the next class a discussion in the next class there are commercial simulators that you can buy with hundred thousand dollars that people and when I was in Bell Labs and also at a gear systems I have used this a lot these are like eighty thousand ninety thousand lines of code solving those five equations of a variety of situation but that doesn't give you insight so being able to solve a complicated problem in a computer well computer knows most than you do most of the cases and you cannot catch your mistakes let say you put a wrong value of thousand when you wanted to write ten to the power minus five in a state you accidentally wrote ten to the power minus fifteen well computer will give you some answer but if you don't understand the analytical way that what the result should be then the computer will give you a completely ridiculous result you will take you to your boss and we'll get fired and the final one is nothing from personal experience now and the final one is analytical solutions provide a great deal of insight and this is something requires years to develop but once you develop it this becomes powerful you almost without solving the equation the solution floats to you you can just look at the problem and you know what the solution is going to be and without solving any any problem in on a long derivation or simulation okay alright that's it thanks