 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 so let's get started today today we'll be talking about bulk recombination and this means that if you have a bulk of semiconductor on which let's say you have part of the equilibrium by some means let's say by shining light then how the equilibrium is restored now bulk could be bulk of many things gallium arsenide silicon and all other that type of things but as you know silicon is sort of the dominant material indirect bandgap material that derivation is a little complicated and that's why I want you to guide you through that gallium arsenide and other direct bandgap material very easy so I'll keep it to for the end when you are a little tired maybe towards then so we'll begin by derivation of Shockley Reed Hall formula now three names are associated Shockley and Reed these are the two people very famous they wrote a paper I guess in 1954 deriving the formula and hall in fact one year earlier I think he derived the formula independently and so there is often called SRH formula short for Shockley Reed Hall formula and I'll show you how to use Shockley Reed Hall formula for various special cases now remember Shockley Reed Hall formula only applies to indirect bandgap material trap assisted recombination this type of thing for silicon and many other indirect bandgap material it's not a general formula that applies everywhere and then we'll talk about OJ recombination and direct recombination remember we said in direct bandgap material this is very important OJ is when two electrons bump against each other one electron goes down the other electrons goes up very high in energy so we'll talk about that and then we conclude now let's think about it slowly we are talking about an electron red electron that dotted line is a trap I have a generally I used to draw it short here I have drawn it long because I'll have to draw a few more things but remember that's a localized state at one point right it's a defect or a trap now what we see here the electron is coming down to the trap being captured by the trap waiting there till a hole shows up and then dropping into the hole at the end of the process an electron is lost from the conduction band a hole is lost from the valence band and nothing has stayed in the trap the trap it has returned to its original position so this is going down we'll call that process 1 and 3 just to give it a name because we'll be using this later on we'll just call this process 1 and 3 now there is always an inverse process that where an electron from the valence band now jumps up to the to the trap and from that electron eventually goes up to the conduction band so at the end of the day the net result is one extra hole generated in the valence band and one extra electron generated in the conduction band and nothing left behind in the trap so trapped temporarily hold it held it the electron but at the end of the day it is not really holding anything and we'll call that process 2 and 4 and many times when I'm working out this math it will be helpful if you consider the dotted line in analogy of the USA remember I had these boxes for USA Mexico and India that type of thing so you can think about conduction band being let's say India the valence band being Mexico and the trap level being let's say USA so then you can clearly see the analogy I'm talking about in that context okay now this view that I just told you about is fine but for mathematical purposes a slightly different point of view is often useful so you see what happened in the left hand side picture the left and the right are exactly the same picture but treated in a slightly different way in the left picture as I told you is a recombination process one electron from the conduction band was lost through the trap eventually one holes from the valence band was lost so at the end of the day one electron gone one hole gone that's the status of that look at the right hand side where I have drawn the process three process one is exactly the same but I have drawn the process three in a slightly different way what I have said here that instead of the electron coming down from the trap to the hole in the valence band instead what we'll think that as if a hole has jumped up hole has jumped up to the trap and they have met at the trap the electron and hole they have met at the trap now you see what the consequence of this at the end of the day one electron is gone right it came down to the trap one hole is also gone because it jumped up from the valence band to the trap and there they recombined so at the end of the day the trap is also gone there's no electron and hole left in the trap also exactly the same process right in terms of as far as electron number and hole number is also exactly the same process now what about the other process two and four so let's think about that two and four remember what happened one electron jumped up from the valence band to the trap and then eventually from the trap to the electron how do they get this energy by the way they may string up a bunch of phonons and then get a help of a photon to eventually go up there remember going up is very difficult coming down is very easy it's like getting good rates getting a D is probably easier than getting an A now the equivalent process on the right hand side is something like this I have put together you can see a red and a white on the right hand side so I have one electron and one hole but they are together at the end of the day one electron goes up from there it started from there both of them and the hole jumps down look at what happened at the end of the day isn't it exactly the same as it is from the left hand side for two and four why is it exactly the same because at the end of the day I have one extra electron in the conduction band one extra hole in the valence band right and nothing left behind in the in that trap level so these are exactly the same process so instead of the physical picture physically what is happening will work on this picture because then you can see the things will get in and out of the trap level and that will be easier to keep the book keeping up okay now let's think about it the first is we want to look at the number of electrons so we're focusing on let's say India the conduction band on the top and we want to see how many people come out and how many people go in now the one the state that conduction band communicates with it changes electrons and holes with let's say is the trap it doesn't know as far as compared for the previous picture starting for the previous picture it doesn't even know holes I bear valence band exists right because valence band is only talks to through the trap so therefore only thing I need to draw when I'm thinking about how many electrons getting in and out of the conduction band I just need to think about the trap and that's it so how many electrons in the conduction band and how is it changing with time so I have to consider process 1 and 2 right and that's why I have written the subscript 1 comma 2 so these are the two processes I have to think about now think about the first process first process so remember in the last class I told you about these electrons are swimming around this black electron swimming around and then there is this trap sitting fixed in the real space and once in a while they capture if they are empty then they can capture an electron and then become filled right in the last class we told you about that and this is exactly that process look at that N N is number of electrons if you have more electrons of course it will get captured more P sub T P sub T is number of traps that are empty of course it has to be empty otherwise how is it going to catch in electron and C sub and that is the capture coefficient proportional to the velocity because if it is running around quite a bit then the probability of getting captured is more right so proportional to the thermal velocity you saw that in the last class why is there a negative sign the negative sign is because of process 1 the number of electron is reduced because anytime they are captured as far as electrons are concerned it has lost a partner so the number is reduced what about the process 2 well process 2 is you can see that first I'll explain what is E sub N is but NT is the number of electrons that can jump from the trap to the conduction band remember it was the process 2 just a little bit before that is proportional to the number of traps that has electron if it has no electrons how is it going to give up anything so it's proportional to N sub T and 1 minus FC means that when it's wants to jump to a state that state must be empty because because poly exclusion principle remember if it is not empty then it cannot jump so it's 1 minus FC and an E N is the rate at which it goes up it's a emission coefficient now we don't know we haven't given any physical explanation of E sub N we'll do it in a second we told you about C sub N in the last class why plus sign well electrons are coming back so I have it as far as conduction business concerned I have more electrons now plus sign okay now remember okay so and correspondingly I will not explain this a great deal but you can convince yourself that between valence band and the trap you can write the same thing 3 and 4 I'm focusing on the number of holes this time again C P P and N sub T P is number of holes N sub T is whenever they are occupied because holes are captured by occupied traps remember that's when it has large cross section in the last class and correspondingly it has the emission probability so I hope that you will try to understand this one that's it I don't need any more equation I will manipulate this equation a lot but this is the basic equation and I'm actually done so now think about this word look at the title detailed balance in equilibrium of course it's redundant because detailed balance can only happen in equilibrium out of equilibrium it cannot happen right what is detailed balance every country every pair of countries in that case India and Mexico let's say everybody has individually the net rate is zero that's detailed balance steady state doesn't have detailed balance remember so every state pair wise they exactly cancel each other so since these things in equilibrium I'm assuming since it's in equilibrium I will turn out the on the light a little bit later but since it is now in equilibrium sitting on your desk without any external stimulation then I can write this DN DTA because of detailed balance what should the result be there should not be any net change in the number right number of electrons should exactly remain the same independent of time so I should set the left hand side to zero I can only do it only do it in equilibrium cannot do it any any other time now do you notice that I have done something which is a little fishy here I have taken out one minus FC do you see the second term had the one minus FC I can only do it I cannot do it in general I can only do it if the semiconductor is non degenerate that Fermi level is way down you have few electrons in the conduction band remember ni is 10 to the power 10 so it's most of the places are empty anyway f is very close to zero so I have dropped it in general if I give you a new problem where the level is degenerate Fermi level is up in the conduction band don't take this make this extra step right now once I have this you see here I know if I if I somehow could find out this n sub t and p sub t all those quantities and I'll show you how to then a beautiful relationship between these two and detail balance says that e sub n the emission coefficient is actually related to the capture coefficient you see it has to be if the total thing has to be balanced then the rate at which you capture must have some relationship with late at you which emit and this detail balance will allow you to connect the two coefficients in fact detail balance is a way in which in any rate equation this is semiconductor physics you can take anything you can take geophysics you can take astrophysics anywhere this notion of detail balance is so powerful it will always connect two processes or couple of processes in equilibrium right to each other and thereby reduce the coefficient by a factor of two is a very powerful technique so one thing you see that emission is related to the capture coefficient multiplied by a bunch of constants and we'll think about what those bunch of constants are then I put a zero there why because this is in equilibrium so that's why all these numbers have carry a zero around okay and the whole thing n0 p sub t0 and the numerator together I will call give it a name called n1 I'll just give it a name it's no physical significance nothing I'll just call it so that I don't have to write too much and in general you see as soon as en and c7 emission and capture are related in the previous equation instead of having to carry around e7 I'll insert n1 multiplied by cn you see I haven't done anything so long I don't know n1 I haven't just a substitution haven't really done any made any progress yet but we will okay so 3 and 4 is your homework you should be able to see the same thing and again notice on the top line for the whole rate I have eliminated f sub v because that was the number of holes in the valence band I haven't written that because this is non degenerate had it been degenerate I must carry that around right and again you see exactly the same thing but this time whole emission and whole capture these two coefficients are related to each other and the valence band only talks to the trap level so they are four on the right hand side I have just kept them to I haven't even written up the electron number here right now let's think about it I just defined n1 I didn't know what it is it looks like a bunch of constants and I also will get correspondingly a values of p sub 1 now look at the magic here that if I multiply ends n1 and p1 I don't know what they are I don't know but if I multiply you will notice amazingly that everything cancels only thing that remains is n0 and p0 do you see that how it cancels the trap trap occupation how it cancels and n0 and p0 why am I writing it ni-squared because remember this is in equilibrium this is in equilibrium so the number of electron concentration and the number of hole concentration regardless of whether there is trap or not if there is trap good if there's no trap that's even fine it's always equal to ni-squared that's the mass action right remember that's why I said it's very important it really doesn't depend on whether it's doped undoped whether accept a donor trap doesn't matter this is always true and therefore since n0 and p0 in equilibrium is equal to ni-squared somehow if I can calculate n1 I'm done because I can calculate p1 by using this relationship remember right this is why the quantum mechanics hides ni-squared because it is effective density of state nc and nv and the band gap expand the exponential do you remember that so this is why the quantum we are carrying along the quantum mechanics just it's in a disguise so you don't see it but everything is coming with us okay so now I want to calculate n1 because once I calculate n1 I'll be in good shape so first of all in order to calculate and so on n1 I want to know how many traps are occupied now if I want to know how many traps are occupied did I write it correctly I should have multiplied f0 okay let me double check on this one I think I must have written it correctly but let me double check on this one but mt0 let's say I multiplied by f00 which is the number of when it's empty so this is fine when it is occupied means when it's not empty and f00 means when it's nothing is there right no electron is there so this is fine 1 minus f0 means when it is occupied and when it's occupied that's n sub t so I multiply remember the derivation from before for the donor level remember there's no difference between a donor and a trap a trap is a donor in the mid-gap why is it why they have different name I'll tell you a little bit later in the next class but I should be used able to use the same formula for donor had it been below the conduction band I should be able to use the same formula for the acceptors so it's the same same formula so I put some put that in g sub d do you remember what is that degeneracy factor right for conduction band for silicon it is 2 and for valence band for silicon 4 right so we remember that quantity here we know beta is 1 over kt and e sub t is a trap level we know that copper where is it or gold where that level is in silicon we know that so we can put it in now let's try to calculate this thing now do you agree with this statement that p multiplied p sub t is when the trap is empty when the trap is empty means it doesn't have any electron that means n sub t is a total number the fraction which has lost both its electron is multiplied by f 0 sub 0 0 which is the blue f sub 0 0 so that's the numerator and in the denominator I also have the expression for n sub t okay and these are all in equilibrium remember that's why I am carrying around the 0 on there in the denominator in the subscript okay now you already know the expression for f 0 0 so let me not worry about it you just insert the whole thing n and once you do you see you get an expression for n sub 1 what is that do you see you whether you know everything on this one n i do you know that you know that 10 to the power 10 right that you know g sub d you know degeneracy factor what about e t well you look at the table whatever the e t is do you know e i e sub i the intrinsic level do you know that you know that because remember we said in equilibrium n is equal to p and from that we calculated the value for e f and e f is e i for the intrinsic material do you remember that it's close to the mid gap I have e i this whole thing I know and therefore I know n 1 where is silicon hiding is in here silicon hiding is in n i because if you change it to germanium n i will be different g sub d this is where e these are various places the material constants are hiding if I know n 1 do I know p 1 well of course because of this and because of that I know p 1 as well I'm done you see this is math there's no deep physics in here just have to keep track of a few lines of algebra okay I'm done with this one n 1 I know p 1 I know I can go forward now now this time I look at USA which is sort of the trap level previously I looked at the conduction band and the valence band this time I'm looking at the yellow region that how the electrons are coming in the trap and getting out of the trap same for the holes right so how would you do that so let's look at this so I'm looking at how the trap concentration is changing trap occupation is changing as a function of time first of all see whether you agree with this I have written is a minus of the change in the electron concentration is this does it sound right it is right because anytime a electron is lost from the conduction band that loss is the trap levels gain because trap got one when the electron conduction band is unhappy but trap is happy it got an extra one they have a negative sign what about holes the holes the same way because I'm looking at occupied number of traps anytime the whole concentration went up right how did it go up because the electron jumped up to the conduction band that's why its whole concentration went up because it jumped up as a result I will have a plus sign for this one please try to convince yourself that this is a correct statement right if I give you two trap level in exam I should not that complicated perhaps but you should be able to do this detail balance right the equation at least of the trap level right instead of having one level now I already know the expressions for dndt I just derived it in the last few slides I know that and I also know that for the dp dt I'll just flip the sign and put it in here right and the en and cn e p and cp well they are all related so I'll put them all in here so I'm getting an expression here I essentially know everything do you see that look at this equation cn I will calculate it multiplying by sigma the capture cross section from the last class and the thermal velocity do you remember last class cn I know if I know cn I also know c sub p now n and n1 I know I just calculated and n sub t is what I'm trying to calculate and p sub t is one minus n sub t or nt minus so except from nt I know everything on this equation I should be able to solve this so now I'm talking about that's the general case if I solve that one I should be able to solve in every case transient steady state I should be able to do any problem that is a general equation always true single trap level non-degenerate always true that equation but if I insist that it's in steady state then what's going to happen in steady state what will happen that although the trap level talks to the conduction band and to the valence band right it talks to many different people individually they will not be balanced anymore why because this is no longer in equilibrium right they are no longer in equilibrium so they will not be individually balanced however globally they will be balanced that there will be no net change in the population in the trap as a function of time so I said that thing to zero so now I can solve for n sub t this I can easily solve for this you can just work out the algebra this is one line of algebra I will not go through that and let me just look at the net rate of recombination so remember this is in steady state that means all rates are the same the number rate now rate at which electrons are disappearing the same rate at which the holes are disappearing right net it has to be the same so if I just wanted to calculate the flux that is at which it is disappearing I will just say the recombination is minus dp dt and from the previous expression if I substitute the value from the previous slide substitute the value of n sub t I have the whole thing very complex looking equation this is what terrorizes solid state students but it's not that bad I will show you do I know everything here in order to calculate the trap number if somebody gave me n and p if somebody did then everything else is known n sub t n sub t the number of traps right I know the number of traps I know all the capture coefficients n i squared I know so I should be able to calculate this and that is I will be going to give you some example but before I proceed let me quickly ask you something you see in this one it says in equilibrium if I wanted to look at equilibrium in equilibrium the rate must be zero right and then the only way this whole thing can be zero is if np is equal to ni square so this is an important statement when we derive them a law of mass action we just multiply two things and it came ni squared came out and it didn't really say how violently electrons are moving around but you can see actually np equals ni squared actually comes from a detailed balance condition like this here the electrons are coming down from the trap they are going up through the trap they are all around and at the end in equilibrium np is ni squared see so that is somehow is getting reflected in this expression so equilibrium is not a retired person's life equilibrium is a very active person's life but all the processes are balanced it looks stationary and this instead of carrying it around we will call this whole quantity tau sub n this is minority carrier recombination time for electrons and tau sub p this whole thing in the bracket just to look it make it look nice that is the whole recombination time okay let me show you a few examples derivation done so let me show you a few examples how to use it let us consider a semiconductor in which you have shine light on it right on line on but not very strong light just a few extra electron hole pairs so let's say on the order of maybe 10 to the power 12 10 to the power 13 in silicon ni is 10 to the power 10 right per centimeter cube very weak light i have that's why the word low level injection so the number is not very high how will i know how the electron and hole after i turn it off how will it disappear let's look at this the way it will disappear is by something like this so i have this electron and holes now i have written n as n naught plus delta n now delta n here does not mean it is small in general don't always assume that if i have a delta it always means small in our context it will not be so in general i can always write it n naught equals n equals n naught plus delta n and p equals p naught plus delta n why do i write delta n not just delta p because i shine light in it and the number of electrons that was generated must have left behind is same number of holes and so the number of electrons and number of holes in the spatial case is exactly the same so therefore i write delta n and delta p equal to each other that's what i have written now it can be simplified a little bit more and let's say whether you agree the first term on the numerator n naught multiplied by p naught what is that ni square and you can see that will take care of the last term on the numerator ni squared is gone and you can see if you cross multiply this then you will have a delta n multiplied by a constant and delta n square and you have the same bunch of things on the denominator now if it is small low level injection if delta n is small delta n squared is even smaller right so i'll get rid of that similarly if this is a p type material p type means it has a lot of acceptors then it has a few electrons right relatively few electrons and so which term can i get rid of i can get rid of n naught why because it's a p type material lots of holes few electrons and that's why i get rid of the n naught compared to p naught you see you'll have to make relative comparison and once i have done that i will simplify in every case for example i can get rid of n naught from here and delta n is small compared to p naught right and so i'll get rid of that the whole term and only thing that remains is p naught because p naught is much larger than delta p because of it's a low level injection so i'll get rid of this now this step you'll have to follow step by step when you go home follow step by step but at the end of the day look at that expression delta n divided by tau 7 so how will it decay as a function of time this is dn dt right that's the rate of change this proportional to delta n over tau n so it will go down exponentially as a function of time right that's that's how it will going to decay so the few electrons you have pumped up in the conduction band they see a sea of holes all around it because it's a p type and they bump in the holes all the time through the traps and recombinates keeps recombining so therefore you don't see the expression for anything related to the holes here you see there's delta n over tau n because holes are so numerous anytime it wants a hole it will get a hole so as a result i do not have anything in the final expression related to the holes you see that's not a rate limiting supply you have to understand the meaning of the equation so that you remember it exactly high level injection same procedure but now you have shown a very strong light let's say you have put a laser on a bunch of say a little piece of semiconductor semiconductor was intrinsic to begin with huge number of electron hole pairs so therefore what are you going to do you will expand it the same way but this time unfortunately you cannot leave out delta n square in fact delta n square is much larger than the first term around right because ni let's say 10 to the power 10 or so and i delta n if it is 10 to the power 18 then this whole second term will be significantly larger the number t tiny amount of electrons and holes you had in the beginning doesn't matter because the light has generated so many and so you will put it in and this time you'll see that instead of just tau sub n you sub have tau sub n plus tau sub p why is this why is does it physically is like this because now the electron and hole number are almost equal to each other right they are nobody is dominant so therefore it will depend on how quickly they can find each other it is not that anytime electron once there is always a hole or the vice versa it is here both are limiting you have to have find both at the same time in order for the recombination to occur so therefore you serve have tau sub n plus tau sub p high level injection right okay and the previous one was low level injection by the way this very quickly so in organic solar cells a topic i am working on recently this type of processes are very important organic solar cells are essentially intrinsic right but the sunlight comes in and it generates a huge number of electron hole pairs when they recombine this is the process with which they will recombine not the the low level injection process now let me ask you this is which one is higher in one case i have a delta n in the numerator and another case i have the you know a little bit bigger numerator a denominator compared to the other one but you see high case has a bigger denominator does it mean at a higher level of injection you have lower recombination that makes no sense so what's wrong exactly right so the delta n for high level injection although they are both delta n one delta n is 10 to the power 18 another delta n maybe 10 to the power 12 so although it looks like that the first case is actually looks like smaller actually first case is orders of magnitude higher because they will recombine very quickly right there's lots of electrons and holes so that's why the expression looks funny but it is correct now what happens if somehow and this will come later on when we talk about diode somehow you can take out all the free electron holes take them all out now if you take them all out then n is 0 p is 0 because i took them all out so if i take them all out then i have n i squared on the top and i you can see that i have taken out n and p from the denominator and this is a constant i know the important point i want to show you here is in the previous cases i always had a plus sign r was delta n divided by tau n in this case i have a minus sign what does it mean it means the previous case there are lots of electron hole pair they were recombining and they were going away the number was going away this time this is being generated because there's no electron and hole it wants to go back to the 10 to the power 10 right so it will keep generating until it can reach the equilibrium and therefore i have a minus sign here it is inverse of recombination generation so that is what i have here now this again i'll show you in devices everything i'm doing every step will be used in the subsequent lectures so it's important to know how it works now the two remaining processes i'll talk about is direct and og recombination direct recombination i'll not even try to may i'll just make an analogy and let it let it go with that you see in the recombination generally what we had in the denominator we had sorry numerator we all had n p minus n i squared that says in equilibrium that thing must vanish in the denominator for the shock lead hall we had things proportional to the trap concentration right n t capture cross section and all those when you have direct recombination and electron coming in recombining with holes i don't have any traps or anything so the whole thing that i had in the denominator i'll just replace it with something called a b now of course i can derive it properly this depends on the band gap and other things but that's another class maybe 659 they will will derive it properly but for now let's assume that it's a simple formula or life is simple and if you want to handle low level injection in gallium arsenide direct band gap material then you will just do this it's a p dope material let's say you'll expand it and you will drop terms you will say delta n is equal to delta p the second order square term you will drop and eventually you will get an expression for proportional to delta n the excess generation you have what is tau direct by the way tau direct is a rate so this first term b multiplied by p naught that is the rate or time constant with which the extra electrons and holes will disappear this time you see it has a number of holes because if you have more number of holes of course that allows you to recombine faster similarly you can say what what happens if i completely remove electrons and holes from this from a given region okay this is this is no problem n and p is zero so there is a net generation upwards and given by b and i squared why do you find b well this is listed in the tables in every material somebody if you just look up a handbook they will tell you what the value of b is for a given material oj recombination again by analogy look look whether you can understand does it look right let's start with the red one the red first term i write it as n squared p first focus on that n squared p does it look right because two electrons have to bump in against each other n squared and p well it has to find a hole in order to get in right so therefore it's multiplied by p now in principle i should have written a one minus fc for the second electron to go up but that that level everything is empty so therefore only thing i have is n squared p right now why do i write n i squared n on the right hand side on the second term the reason is this term must vanish in equilibrium does it vanish in equilibrium do you see because in equilibrium n will be replaced by n not n not by multiplied by p not what is that n i squared and you can see that this term will vanish in equilibrium so this is necessarily it's not a derivation but it looks right do you see on the other hand on the second term we're multiplying c sub p that there is a p squared multiplied by n this says two holes bouncing each other so it's so much energy one electron going in the conduction band and another going deep down in the valence band that's the second term and that's where it comes from these constants are measured people and in fact i spend many years i've not placed few years in my life working on semiconductor lasers and i really have to go through the intricacies of the physics here but for you one slide is more than enough hopefully low level injection you should work it out again put these conditions in and once you put this condition in you will see that if it is a p-doped material you will see that p squared will become n a squared because p is equal to n a the number of holes equals to the acceptors so you can put it in and you will see that the recombination in this case the og recombination if the doping is very high n a is very high in that case og recombination would be the dominant recombination mechanism in many semiconductors right because you see n a square why did the n a squared come from because that's the p squared over there you need two holes and when you have high values because it doesn't have to consume momentum directly remember that was the problem same wavelength that was the problem so when you have lots of electrons you can bypass that conservation problem and therefore have very rapid recombination through the og process so let me i think let me show you just one slide and then i will end because i think the i will pick it up in the next in the next class so now consider we have all this complicated derivation so let's take a step back the semiconductor doesn't know that you are grouping things into shock lid hall direct og when an electron goes up yeah when you shine light it will go down in any path it can find direct available good if it doesn't have a direct path it will take the shock lid hall path it will take any way it can go it will go down so when you shine light and let the light be off turn it off then it's going to go down from experiment i'm saying and in experiment if you look at the output light that is coming out you will see it is decaying as a function of time so x axis is time y axis is delta n and i am t equals zero in this one is when i have turned the light off just light turned the light off and i'm looking how the photons are coming out so from experiment what i will get is something called a tau effective that's how i'll see the whole thing decaying now from that how am i going to understand get this various coefficients c sub n the capital B for direct recombination how am i going to do that so in order to do that you'll have to understand how these recombinations actually depend on various things so as i said every all processes will work simultaneously and this all recombination rates will be together but i just told you that for each one of them remember this is low level injection so every one of them is proportional to delta n i have a shock lid hall time constant i have a direct time constant and i have a oj time constant this i just derived in the previous three slides and this each one of them has a certain dependency the first term the shock lid hall it depends on the number of traps the direct it doesn't care about number of traps because no trap is involved only thing it cares about is the number of dopant atoms right why because it gives you a certain number of holes or certain number of electrons oj yes it depends on the number of dopant atoms but as a square because two of them have to bump against each other so they have different dependency so if i do a set of experiments in which i'm changing the dopant level systematically then this decay will be different for each one of them and from that i should be able to pick out what these different coefficients are you see so this is what it is let's this is from your book and i want to explain how did they get this one so y axis is tau effective so from experiment somebody shine light on it on silicon dioxide and they saw how the whole thing is decaying and they just put one data point on each one of them x axis number of dopants so that's the x axis so on the left hand side is close to intrinsic on the right hand side heavy doping effect remember heavy doping effect we talked about so that's happening on the the right hand side let's see whether we can understand this the tau effective should be the sum of all these things now in the left hand side n sub d is close to zero right it's close to intrinsic so that means the second and the third term will drop out at the low low side it will it will drop out so from there if i know the number of traps i should be able to calculate c sub n that's that's good now if you go at a very high doping side then you see the first and second term will drop out why because these are actually smaller terms the third term will now be humongous because it's n d square now remember this is a log log plot so if you take a log log on the tau effective on on this on the right left and right hand side what should be the slope of that line minus two right log if you take a log and if you look at this curve you will see exactly minus two is exactly the value on this curve uh the slope and the high sign is exactly the value on this curve and experimentally people have seen exactly very similar that's theory that's the theory so you put various values of n d you can generate the curve but experimentally i have shown you from two different papers that indeed at high density this is 10 to the power 20 very high density right one atom in every hundred uh host atoms very high density and you can see how the things are falling off so from here if i give you a curve like this from here by looking at the slope and looking at the intercept you should be able to calculate back the oj recombination constant is that right and from our lower values you should be able to calculate the number of traps because actually from that you should be able to deduce all this okay all right so this is all about recombination that how equilibrium is restored we perturbed it with light and we said that the shock lid hall recombination this is one of the things you have to learn unfortunately in order to earn a six-figure income later on but hopefully this was not too painful and this is an important recombination mechanism that's a dominant thing that's how you design your anytime you carry a stick memory stick many are or DRAM memory you buy for your computer all of them actually if the people who design them understand how the this recombination mechanism work because you are allowed to lose only one electron in every 10 years because some of the memory that has so few electrons in it that you cannot lose any electron because if you lose one in a year or one in a few years that's also even too many so this recombination mechanism somehow or other you have to have extreme control over before you can have a viable product that you can sell it is looks a little complicated but if you follow through the steps and why don't you do this exercise where instead of having one trap level you have a pair of trap levels and see whether you can derive the detail balance for this you know it's no rocket science here very simple very simple thing now direct band to band and og recombination I use the very similar formula they have their independent derivation but generally for direct band gap material shock treated hall is not that important this direct band gap and og recombinations are the dominant are the dominant things especially for laser og is is the dominant dominant process and this is not just some theory for last 50 years hundreds and hundreds maybe thousands of engineers and physicists have done beautiful measurements you know those looks like few data points but who did it really had to work hard to set up the experiment shine laser on it get the electron hole pair out and see how they are decaying as a function of time a beautiful experiment that has been done hopefully some other course will have an opportunity to discuss them okay all right thank you