 Okay, here we start. Last time we were looking for diffusion of any impurity in other kind of papers and we suggested that in practice the diffusion of impurities is a two step process and this is particularly true for solid state diffusion. Even in implant, actually there is a step, first step which we say implant itself is a pre-deposition step and there is a driving ladder there as well. So basically you somehow put impurities first and then push them inside, okay. That is the method of any impurity incorporation. We did look into the feature that in pre-deposition we performs normally in the case of solid state diffusion processes, a constant source diffusion at temperature T1 for time T1 and for time T1. So we say D1 is the diffusion coefficient at temperature T1 for an impurity which is going to be incorporated in a bulk of another type. NO1 is the solid solubility or active solid solubility at temperature T1 and then we say Nxt1 is equal to NO1 error function, complement X upon 2 root D1 T1 and after a time T1, the net amount of impurities per unit area which you question was 2 NO1 under root of D1 T1 by pi. This we did, I am just trying to recollect what we did. You know we have a gap of so many days so suddenly people also are not aware what we were doing. So maybe just for the recapitulations, we also said that after pre-deposition step the source is stopped, then source is stopped and then the wafers are reintroduced in the furnace at temperature T2 for time T2 and this process is called driving, impurities are driven inside the silicon wafer. At time T2, at temperature T2 the diffusion coefficient of impurities in bulk is D2 and Dt product is now D2 T2 and if since the impurities are already fixed by us during pre-depositions they only get redistributed during the driving cycle and the new profile is a Gaussian in nature and if you have solved that we far from the diffusion equation, the solution turns out to be Axt1 T2 2 NO1 by pi under root of D1 T1 D2 D2 and exponential minus X square upon 4, this is what we did last time and we did say at X is equal to 0, if I make X is equal to 0 this concentration is essentially 2 NO1 by pi D1 T1 by D2 T2 under root which is called the surface concentration for the profile. At the maximum concentration occurs at X is equal to 0 and that is called NS, N surface. To calculate junction depth we last time did but we may do again but so this is the two step diffusion process. However in this all doing this we assume that the sheet charge approximation is valid and we did say that D1 T1 is much smaller than D2 T2 and therefore that thin sheet of charge is possible. However this assumption is only an assumption and can be taken care. This was the last time so now we start today something more. The assumption in two step diffusion was that D1 T1 is much smaller than D2 T2. However even if this condition is not met we can still evaluate the profile and the method is called using Smith's integral. Those who are very keen to know some old books on diffusion theory available in the literature you can Google them but I will just give the final answer. We define two terms alpha is under root D1 T1 by D2 T2 and beta is X square upon 4 times D1 T1 plus D2 T2. Then just take a case D1 T1 is very close to D2 T2 that is not much larger but closer to D2 T2 then nxt1 T2 is 2n o1 by pi root of beta to the power infinite e to the power minus y square error function alpha y dy this is called Smith's integral. Simplification as I said this is a theory of diffusion and this has nothing to do with devices or semiconductors. This you can read from Smith charts or Smith integrals otherwise I will suggest you some book on diffusion. Since I am rarely going to use this but however I may use it I am not saying I will never use this and so I will actually have this simplified version of this is 2n o1 pi 0 to alpha e to the power minus beta 1 plus alpha square upon alpha square D alpha this is essentially known as Smith integral. So the condition is D1 T1 is very close to D2 T2. The first cases we discussed was when D1 T1 was much smaller than D2 T2. So we say pre-departition step is for shorter temperature or shorter time for a smaller temperature, drawings are for larger temperatures and larger times. Now this method as I say will lead to a solution which once you write down I will just say you. So I have a integral which if you have noted down I will show you what is the way Smith integral is given. Since I have used alpha the table from Smith himself has a alpha instead they have variable called u the only difference between mine and theirs. This is a Smith integral. These are the values of alpha and these are the value of beta okay and then these are the values of Smith integral. So the whole integral value can be obtained from knowing alpha and beta. Please remember alpha is u in this table because Smith used the term u we have been using alpha so that is the term. There is no difference between their terms and my term. The reference is already given. This is from a heat equation semi-infinite solid with short table of important integral. This appeared way back in 1953 in Australian Journal of Physics by R.C.T. Smith whose integral is name is given to his name. So these depending on alpha values and beta values please remember alpha is up to maximum goes up to 3 whereas beta can vary up to 5 okay. So this is the same table is repeated for higher values of beta 1.3 to 5 okay. So depending on alpha and beta this is just to show you where from this values can be obtained. So I can calculate alpha, I can calculate beta if I know d1, I know t1, I know d2, I know t2. Since I know almost this all from two step diffusion process so I know alpha, beta and the value of integral can always be obtained. Is that clear? So in case d1, t1 is close to d2, t2 please do not use the earlier expression and there is another example we will say you. Now surface concentration at x is equal to 0 is 2 NO1 pi by tan inverse alpha and if you say alpha is small, alpha is under root d1, t1 by d2, t2. So what does that mean? d1, t1 is smaller than d2, t2 means alpha is small okay. tan inverse alpha is alpha when alpha is small. So if I substitute them here NS is 2 NO1 by pi under root d1, t1 by d2, t2. So you return back to a case when d1, t1 is smaller than d2, t2. In case it is equal to d2, t2 you should use Smith's integral and then use surface concentration as this value okay. I have plotted normalized NXT function with surface concentration of this for various values of normalized distance with dt effective and we find alpha is 0 is Gaussian, alpha infinite complementary function and in between is the value. So this is the graph which essentially I have plotted roughly but this may not be accurate. That means the upper bound and lower bound is Gaussian profile and complementary profiles okay. Why I said this because in complementary function we have said d1, t1 is smaller. In the other case we said d1, t1 can be as much as d2, t2. So upper bounds alpha 0 and alpha infinite or large value. So complementary error to Gaussian is actually the NXT profile will appear okay. However as I say 99 cases this may not occur, d1, t1 will be always used less than d2, t2 but do not take it my word because I may intentionally set a problem in which d1, t1 will be comparable. So do not just go by it and see whether it is equal and then go for Smith's integral okay. So this is what I say in case there is a situation in which d1, t1 is larger than d2, t1 equal to d2, t2. Smith's solutions can be used alternately if d1, t1 is smaller than d2, t2 then normal predepositions will give me a Gaussian profile as we discussed. There is another problem which comes, please remember which I did not say so obviously but maybe I can go back in my sheet if I have written on this okay. I said after the predepositions step we go for, we remove the wafer from the furnace, we will see what are the kinds of furnace we use and impurity source is dropped and then wafer are removed and in fact what we do is what we say there is a small glass of phosphor silicate sitting on the wafer because phosphorous or not phospho silicate, impurity silicate. It may be arsenic silicate, it may be boron silicate or it can be phosphorous silicate, glass is sitting on the glass means O2, O3 of that okay. So if phospho silicate glass is sitting there then first we edge that glass okay, first we edge the glass. Why because if we keep that glass the impurities are within that glass. So if that glass is retained it is this constant source diffusion because then impurities will keep coming. So the first thing we do is for dry windage to remove that upper layer of glass, glass is oxide and therefore we know silicon dioxide Hs in Hf where silicon does not. So we just put it into Hf diff and remove the glass. So that any source of impurities then are removed from the surface. Then there is a constant because source is removed so it becomes then a fixed charge kind of impurities inside silicon at the surface and they are driven in. But I already said also that impurities can come out okay, I mean there is a isotropic situation at high heat temperature both side impurities will go okay, random. So we must stop impurities going out. So first thing we do is we start the dry wind process in oxygen ambient. We push the wafer in furnace at a given temperature T2 and start oxygen inside furnace. So it starts oxidizing the silicon surface and silicon dioxide is a mass for impurities to come out okay it does not allow. So this step is very important that we create another glass which is not dope glass but silicon dioxide layer and then as the time proceed impurities will go down as the time you want to do for okay. So this step that first you remove the glass is very crucial for oscillate glass or borosilicate glass because that will remove the impurity source from the surface okay which is not very clearly stated by many. You say okay you do pre-dependent dry wind and then it will be pseudo situation because some impurities will be keep coming from the glass itself okay during your dry wind cycle. Then that smith integral why I showed you this because then that smith integral situation may appear because you are also having a source and also driving it okay. So it may have a situation in which D1 T1 can be as close to D2 T2 and therefore this is a situation which you should normally be avoid we remove the glass immediately okay. Okay what we do is in a real process cycles implant a diffusion may be done may be fourth third or fourth step or fifth step even for the channel stripper it may be second step. Since there are too many diffusion after every diffusion cycle there is oxidation but there are number of such temperature cycles before the wafer is completed for a IC. It goes through a large number of temperature time cycles. As many steps you go through in process that many DT products are actually appearing. Every new step you create you have a DT for that process. So the first one which you did it is not only seeing the D1 T1 but also seeing D2 D3 T3 D4 as many DT products it sees it keeps on adding additional DT to the earlier one. This is called DT effective. So as many process steps you go through temperature cycle time cycle those many DT products should get added to the first one for the second one third four should be added for the third one fourth whatever number of cycles you have gone through. So every temperature time cycle adds to the first diffusion some kind of D2 T2 products should get modified which is called DT effective. So here is a some kind of example shown here is a I am making an NPN transistor. For example the we have a diagram concentration of N for example which is 10 to power 16 per cc and then we are going to do a base diffusion which I did at which has a pre-deposition driving cycle done at D1 T1 D2 T2. This p tau diffusion which is I am performing sorry yeah this is the diffusion I did. However nothing happens to base because base is a constant uniformly dope wafer so no concentration gradient so it does not change. However when I do a emitter diffusions which is this which may be only error function profile it sees another DT product this base diffusion has additional DT coming from error function timing okay that must get added to this second profile. So what will happen this profile will get modified by this profile. If I have not done it the junction which was say here actually is not here but since the profile has changed it has changed to slightly higher values now. So please remember since the profile actually is going deeper now the junction depth changes the base width is different collector widths are different and therefore it has to be understood that any time temperature cycle will push the next earlier cycle further ahead inside and that may change the junction depth okay. If it changes the junction there what it also will change the surface concentration for that area and therefore in your evaluation of device parameters you will have to know what exactly was the base widths what exactly was the junctions doping at the surfaces and that is a very crucial issue in getting a beta of a transistor particularly in NPN beta is decided by the base width larger the base width smaller is the beta. So we are trying to control this so what is it means for a given beta I should apply we know how many thermal cycles I am going to go through and for that value of final I must get my XJ or the base width correspondingly so that I get a beta of my choice. So this has to be a priorly decided how much initial I should do second time and it pushes it how much it will go and is that base width is sufficient for my beta which I am designing okay. So process design also is related to device parameters and therefore very crucial because you if you do not take care of second diffusion third diffusion your base widths will keep changing and larger the base width larger smaller is the beta and all whether it is digital or analog we prefer beta larger analog of course it is a gain function why in digital we need larger betas it is a delay time larger currents at the same this so it will give a smaller delay okay or fast speed so larger beta is a requirement for most cases of course in some current sources we may not need but otherwise we will prefer to have larger betas. So typically this may look like emitter base collector this is the junction this is my XBE and this is my XBC this is my base width and this is of course collector and this is my XC in case you need because and this has to be heavily doped because n plus emitters are doped so you can see why therefore error function is to be used firstly emitter should be shallow this is necessity for some reason we will show you later and secondly it is a concentration has to be n plus higher if I put it second garden cycle what will happen it will reduce is NS further so I want the largest NS available for the emitter so I will actually have a shallow diffusion with high solid solubility so it is actually performed at little higher temperatures so that solid solubility is higher there okay this is and for a very short durations same thing can be achieved in the case of implants and we will see then we look for implant process we will see that nothing much is changing normally same profiles Gaussian profiles should be obtained in the case of implant there is no error function profile it is a naturally Gaussian system and therefore the only difference will be Gaussian systems but otherwise profiling is the identical to what it is there we will see how much depth impurity should go or why not they should be here or there how to control them all these issues we will look into it okay. So I just give an example so please remember the subsequent time temperature cycle must get added to the driving times of this because they it will keep adding to that earlier value in the case of I already showed you to calculate the junction depth in a two step process we say this is my surface concentration X is equal to 0 this is my surface concentration 2a no 1 by pi root d1 d1 by now look at it I made it dt effective how many temperature time cycle it has gone through it is dt effective same as d2 t2 should be replaced by 4 dt effective when this is my NS and what is the junction definition that at junction the the concentration of the wafer at X is equal to 0 should be same X is equal to XJ must be same as the bulk concentration or base concentration. So NB is NS exponential minus XJ square 4 dt effective is that clear at junction where is the junction occurring if this is your diffusion profile let us say I am making PN so this is my XJ below is NB and there is a profile here so this concentration at X is equal to XJ is equal to the base concentration there where the net impurity concentration is 0 and junction occurs and therefore I substitute NXT this is NB equal to NS this function NB is known NS is known how NS is known d1 t1 is known dt effective is known NO1 is known pi is known therefore NS is known dt effective is known so what we can calculate XJ that is the junction depth evaluation can be done I repeat this is the surface concentration NS is 2 NO1 by pi what is surface concentration at X is equal to 0 whatever is the concentration is the surface okay at the surface so this is surface concentration I know everything of it pre-reversion temperature therefore d1 pre-reversion time so t1 any thermal cycle how driving it sees dt effective d2 t2 d3 t3 d4 whatever cycles it has might have gone through so this is my surface concentration then the profile is written as NS exponential X this at X is equal to XJ for example in the case of base collector as shown there the concentration at that point is the base concentration where the net impurity concentration becomes 0 when this is equal to this concentration is 0 this minus this is 0 that is what it essentially means junction if that happens NB is NS exponential minus this and now from there what do you think XJ will be from here quickly some LN term will appear LN term exponential is that noted down of course these these are not specific to any book any book gives this so there is nothing great about this or you can yourself solve it okay so not very difficult maths going on okay so XJ square is 4 dt effective LN NS pi in B please remember NS is much larger than in B so this will be a positive number and dt effective is also positive number so XJ will be positive number it will not be minus kind of things anyway so let us in the case of base collector junction we said 2 dt effective LN NS by NB is the junction depth for base collector okay at base collector junction this much base width actually you have to subtract XC from here so that you get the base width okay so I can always evaluate junction depth is that clear to you I can always evaluate the junction depth if I know temperature time cycle through which the wafer has gone through during pre-deposition driving and any other temperature time cycle with no source okay if there is a source in between then that will be a Smith's case okay you will have to recalculate most of it then okay okay now when I go I do this how do I monitor in the lab what has happened okay firstly I am interested to know for a given time temperature cycle the profile itself but given a profile is not very easy okay the way we do for example normally I will just show you if my this is my profile let us say it is a Gaussian for some reason 2 step or something so what I do is I first so let us take this step this is only junction depth I have shown you the substrate I am not showing so first I monitor something on this surface at X this is my X is equal to 0 okay then so surface concentration is here the edge this surface by some amount which is known to me delta X I edge that top layer by amount delta X then on this I do now new measurement okay I keep doing this by step and reach junction is that clear every time I get a surface I have top surface I removed it new surface I removed it new surface and I will keep getting some measurement value at every surface okay and the measurement quantity which I will monitor is the sheet resistance okay what I am going to monitor at every edge area is the sheet resistance so I please remember so essentially what I am doing is discretizing it as if and I am creating new surfaces and keep monitoring see if I know my profile at this point this point this point this point this point this point then I can plot the profile is that clear since I will find out this concentration for this surface method profile at that point so at every time set I will get NX okay every edge steps so I keep plotting NX1, NX2, NX3, NX4, NX5 and once I get I get the profile this is called step profile I mean this is called the edge profiles you can always get the NXT in real life there are other techniques of profiling etching is done in what we called as secondary ion spectroscopy sims as it is called and the sims also basically may be some other day all spectroscopy in one day I will tell you what basically they do same thing in the different Raman spectra this spectra is basically same process different sources different outputs okay so is that clear to get a profile you will have to start from the surface and start etching and so only thing is what is the guarantee should have that your etching should have uniform etch rate for a given time you will etch okay so you should be assured that that delta X is same for all the steps you have gone through okay the time step has to be such that it exactly gives delta X every time otherwise you know this if it is shape and you do not know how much then you have problems so in normal sense this also is an issue in real lab people do not want to use H and every new and now and then so if you dip it in the same solution next time already this pH value of the solution has changed okay so the etch rate of the next for the same time is not same so there are catches catch words and how do you do it some other day when technology starts okay so is that clear how to profile so to get it our ultimate time is to get a profile let us say okay and in many cases I do not need even a profile what I need is sorry NS and let us say this is my NB value itself this junction I am only interested in the surface concentration that profile whatever I got where is the XG I am satisfied with most derivations I need to do for my real life situations but in case you need to profile there is a technique of course there is a optical technique the Raman specter many techniques in which profiling can be done this is the easiest one can do by H method the only catch word here is even there etching is done by some kind of sputters which also does not have constant rates so whichever technique you do this guarantee that H rates are constants are not very good so some sheath is always there in the actual profile to the measured profile okay okay is that point clear how to monitor profiles every surface step you find RS value so what is this RS we are already done earlier but let me repeat again we are defined a sheet resistance for uniformly dope semiconductor we say R is equal to RS times L by W where RS is defined by rho by T and rho is defined by for entire material it is Q mu and N, N is in D dope this I know is donors which is N since if I know my aspect ratio length to width for semiconductor bar then I know knowing RS I can find the resistance but in real life I am not interested to find even right now resistance what I am interested is to know only RS so what is the problem with rho by T assumption is this N is constant mu is constant so rho is constant when this will occur during crystal group we make it uniformly dope materials okay there N is constant and therefore mu is also constant so there are rho is fixed whatever doping I have done that much rho is known but when I diffuse this N is not constant N is a profile of donors or acceptors NX this NXT whatever you are doing is essentially like this the profile of carrier concentration is same as doping concentration so this profile since it is a profile it is not a constant value this is varying function and if it is varying function I must get some average value so I say RS is then equal to 1 upon 0 to XJ sigma X DX what is sigma X DX means sigma is the inverse of rho so Q mu N X DX from 0 to XJ I multiplied by XJ both sides so XJ upon this just not done so this is the RS XJ is some kind of this function which I will write down again why I wrote NX minus NB because at up to XJ the profile has some concentration minus NB is the resistance available for you okay sigma X again is Q mu NX minus NB substitute here okay I am not sure whether Plummer has done the such a detail this but final answers are available so if I say 1 upon RS XJ is Q mean N and P I wrote it because it can be P type dopender N type so either of them can be NX NB may be P type or N type depending on what doping you start and what this NX minus NB and this is 1 upon XJ can you think quickly this is integral of something is averaging 1 upon X 0 to X FX DX is the average of value so this is average value so Q mu N average is essentially what we are getting is 1 upon RS is that point clear 1 upon X integral 0 to X FX DX is the average value of FX okay very standard method so this is the average conductivity Q mu NX 1 upon is given by upon RS XJ so what we do is essentially there is a NB variation in real life there will be different base concentrations substrate may have a different concentration you may have a NX any profile for that matter right now we assume mu is constant but in real life even that needs to be modified so what we plot is surface concentration versus RS XJ at different NB values is that point clear I plot surface concentration for two state diffusion we know so surface concentration versus RS XJ at different value of this if I can get actually I am resubstituting from here this okay so what we are essentially saying that if I monitor RS and if I monitor XJ what will I monitor I will monitor RS for the surface concentration and I will monitor XJ by some other technique so I will know this is that clear I already started with base concentration or background concentration I have decided so for a given background concentration for given RS XJ depending on what kind of diffusion you perform you could have done p type error function n type error function complementary and p type Gaussian or n type 4 possibilities so for all of 4 I can do these graphs these integrals could be put into this graphs itself okay I will show you the example and once I know RS XJ value I know NB value I can calculate the surface concentration of the profile is that point clear to you I monitor RS I monitor XJ so I and normally this is explained as ohm into microns RS XJ product is defined as ohm into microns so do not make it centimeters there this is plots are ohm into microns against per CC of the surface so I monitor RS which is ohm per square I monitor XJ I get a product of RS XJ for with this 4 possible profiles I will show I can get my surface concentration what surface concentration in Gaussian 2 step diffusion 2 NO 1 by pi under root D1 T1 by D2 T2 is your NS is that clear so if I monitor this and I know D1 D1 D2 T2 NO 1 so I can get RS NS from there essentially now I am saying if I do not know one of the times and if I monitor this I will know that time then is that point clear if sir I do not know D2 T2 okay since I do not know T2 but I now found NS by this measurement so NS is 2 NO 1 by pi D1 D1 D2 under root of this so D2 is known D1 is known T1 is known but T2 is not known but since NS was known I calculate the time for driving is that correct so this is the inverse calculations can be performed for a known required value of NS I can actually decide in design how much driving I should perform is that clear to you this is the inverse process I am talking this is straightforward but in real life I will be asked to find what time I should drive in so that I get this much NS value okay for given XJ okay I want for given XJ that NS value so I must so how much driving time I should have okay so I say okay I know D1 D1 I know the temperature of driving but I do not know the time for which I should do so I monitor as if on the device this can be calculated also and plot on NS this and get value of NS which will give me T2 time so this is how the design of process starts given something we go back and figure out how much temperature time cycle I should adjust so that I get the value of my choice technology is here technology is not analyzing this technology is something has to be unknown and said this is what I want if I want what do I do okay you could say if there is a possibility in some case problem may appear in this book time for pre-deposition is also not so some more data must be provided so that I should be able to get T1 value or D1 is known D1 temperature values so all possible combinations are done by some measurements and then we can figure out what should have been done then or what should be done so that these values are attainable I will solve the problem and solve this okay for example in the case of constant source diffusion nx is NS error complement error 2 root dT please remember in the case of error from complementary function what is surface concentration the solid solubility limit NS n0 okay which you get from the graph which I will show you okay I will come back to problem then I will assure you so since I know n0 from a graph itself or from the tables I know this value for the Gaussian it is this there the surface concentration is this so either of the data missing could be calculated if I measure RSXJ and saw see it on NS versus this graph so one of the missing data could be obtained is that clear all to say you actually did a diffusion you monitored even T1 everything and monitor RSXJ and for that you find our NS if that value of NS and this value of this value does not match that means there is additional machine if you have done somewhere time temperature cycle exceeded or reduced depends on which is lower higher values so there is some calculations which you earlier did is has a mistake which resulted in something different from what you actually monitored okay so there is a you can also verify your result from so what is computer simulation cat tool does it for this process simulation given a data final process data I can find the process cycles or for a given process cycle I they can give me what values I will monitor okay and days if I monitor in real device and if I get then that means my steps are correct is that here both ways I did a process on computer okay this is called the variety of one is center center are there is old program called supreme then there is a dios so there are many process simulator similar what process simulators need all the maths available to you which you are doing by hand I should be told to computer here we take some assumptions there you need not because any numerical solution for them for a system is not very difficult you need to know which process someone has already written those programs okay so you just have to tell okay I did this what is RSNXJ I should get you got it you went through that process in real life and verified whether it occurred or it did not and to a great surprise it may not occur that is the best part of all of it so this is called the first time around you do something and you do not match then you start guessing which step is overstepping it so you retailer it on the program till two loops one loopy loops you actually get the same values then you know what is the new value you get for which you should do the process the reason is why this doesn't match because the models which I assumed is I assumed silicon doesn't know it so it did behave otherwise so I had to figure out what how did it behave so I keep more changing my models to suit the result okay that's what all modeling people do okay due regards okay so how do I monitor RS there is a standard technique which is called 4 point probe okay on the dope silicon wafer please remember there is the RS is only monitored on silicon dope area and not on the oxide so first thing you have to do is H out the oxide wherever it is wherever you want to measure the resistivity let's say these are four probes okay they are normally phosphor this material which I forgot so these are very sharp tip these are of course steel but phosphor bronze sorry these are phosphor bronze tips below the advantage of phosphor runs they do not have large tensile strength so they do not actually penetrate but they do give very good contacts so if I we take a case of three points separated by distance R1 and R2 and let's say I pass through the current I in this this is the nodal point and let's say this plus minus from which current is flowing and I want to know potential at point P is separated from V plus and V minus by distance R1 and R2 this is a case of solution of Poisson's equation which some if you are I think you are doing professor Vasis course n times Poisson's equation will appear n times continuity equation will appear so you will know more about it in case there is an issue there compared to what I wrote do call me normally I also teach devices many years so hopefully I am right okay so psypes I RS of course as I say I derived it and I didn't want to derived for you just to take values RS is the sheet resistance so I can say the potential at psi is essentially given potential name so psi at P is I RS upon 2 pi ln R2 by plus integral constant because d psi by dx is rho by x a rho by epsilon is that clear this is the Poisson's statement so we solve this in this now for this I know from two point source Vx V plus V minus I can find potential at any point P which is why I didn't take straight line because even if at any angle it doesn't matter for it so I just made it any angle this distance and this distance should be known this theory has been utilized in for is that why did I show this this Poisson's equation solving is actually utilized in solving this problem in a 4.4 point probe case separated by distance s each probe is separated by distance s so this is my plus and let's say the way it is in last two probes I pass a constant current source okay this small R which is introduced there is essentially to limit the current please remember you need to limit a current infinite current should not flow so it is a constant current source and R is limited there okay so I pass a current in the outside probes okay so to say current is entering like this okay then please remember I want to calculate potential at 1 and potential in between two probes I applied a voltmeter between the center probes I applied a voltmeter so I want to know voltage difference between these two okay is that here the outermost I put it a constant current source I and in center two of them please remember from here this probe is how much distance to s from this it is only s inverse is equal to 2 from here it is 2s here it is s same way here it is 2s and here it is s this is symmetry is what is I am looking for that's why I use uniform space probes okay so I calculate 7 potential at this this I call 1 first point which is from I plus it is s distance so Simon is irs upon 2 pi ln s upon 2s is that clear s upon 2s plus some integral constant a if I calculate potential here with reference to this ln R2 by R1 I am taking a ratio difference I am calculating so psi 2 is irs ln for this it is 2s s so it is 2s by s plus a the difference of potential between 1 and 2 is psi 1 minus psi 2 which is the voltage voltmeter is going to monitor okay if I subset this I get irs by pi ln 2 a removes because that will get subtracted from here irs is pi upon ln 2 into V by I is that clear so what what did I do I took a 4 probe which are fixed on this jig you just push it down put your current source connect between the last two probes and monitor voltage between the center probes okay that is your VI I is known from a constant current source which is what I am fixing so I know I I know measured voltage V ln 2 is numerical number pi is the numerical number so I can monitor my RS is that correct 4 point probe is the easiest technique of monitoring sheet resistivity or sheet resistance RS okay now this is that clear I diffused I can calculate by Poisson's equation potential at any point separated by R1 R2 from two points current I is flowing through them for this probe potential at this point is irs 2 pi s upon the other where from your measuring this R1 R2 this is R2 this is R1 okay so s upon 2s plus a for the other side if I want to find potential here so I should s 2s upon s okay by using the same method so I get psi 2 is irs upon 2 pi ln 2s by s ss cancels so it is ln by 2 and ln minus ln by 2 but minus minus will become plus so that 2 pi will become pi so irs by pi ln 2 is the voltage okay or RS from here is pi upon ln 2 into V by I since pi is known ln 2 is known okay how much is ln 2 okay log 2 is 0.3010 2.303 multiply whatever it comes so and 3.1415798 is the pi value if you want or 555 by 111 mini numbers okay there is a good mathematical research going on even on pi value think of it okay how you define pi okay okay so is that clear how to monitor RS so one of the parameter of that RS XJ was monitoring RS so which I did okay is that okay everyone understood okay of course there are methods which I didn't say 3.probe 2.probe 6.probe but some other time when you are in lab you actually do many mischiefs okay in the class I do not want to show you many other mischiefs which we could do and still get correct answers okay okay now the next value which I want to evaluate is the junction depth okay this is called screwing techniques screw this has a radius R okay so this is my junction XJ this is my substrate so let us say this is my wafer and I put this screw and actually grew through it okay and the tip of that R is known to me okay tip of this screw R is known to me so what will it will do it will etch the doped area like this is the area so this area will actually go which area this whole area will actually go because I will screw remove the silicon from there okay however what is open to me there I will see this distance I will 3 this distance and I will also see this distance this is what I will see from the top is that if I grow it like this this portion this portion and a flat portion this is what I am going to see okay so let us calculate some XJ in terms of the available since R is the radius of curvature of the screw down okay so R is constant everywhere okay let us say point where it at the surface to the center point is a distance A it is identical both sides okay it is a universe uniform screw okay let us say from the junction this point this point will be visible to you because the etching will do but this this portion will this point to this point is B same as this is B we call this is R but we call this flat portion where you see actually the distance is called C2 and from the top surface to this origin I call distance C1 okay this is my maths nothing great assuming curve are large enough comparatively so that can be treated linear in most cases I could say R square is A square plus C1 square hypotenuse square is this distance square plus this distance square so C1 is under root R square minus A1 square by same logic if you look at this B square of course I should have done this okay I should have drawn another line here so R square is equal to B square plus C2 square what I am doing is this which I did not draw so this is again R this is B and this is C2 so C2 square plus B square is again R square okay so C2 is under root R square minus B square so C2 minus what is junction depth C2 minus C1 is the junction depth so C2 minus C1 is J R square minus B square to the power half minus R square A square to the power half now the way we choose we have made a choice of the screw is this distance is A are much smaller A and B are much smaller than the radius of curvature of the screw which you are used okay which is large enough so by you can use binomial expansions if A and B are smaller than R we rewrite this term by expansion so R1 minus B square by 2R square minus 1 minus A square 2R square so by subtracting correctly I get XJ is A square minus B square by 2R is given to me from the screw I used R is known to me so what I have to monitor to get XJ A and B is that clear so if I monitor A and B I have the value of junction depth okay now how do I monitor A and B this portion is known to me is that correct this portion I will see assuming this is linear roughly okay because this curvature is large enough so it will this can be treated as a straight line okay so what I do is I take this H screwed where silicon has gone out I dip it into some copper sulphate plus SCL solution okay if I dip this into SCL plus copper sulphate SCL removes much of the oxygen around and copper sulphate actually plates the silicon this is called electro less plating okay this is called electro less plating so this portions where silicon is available to you and also this lower portion you will get coating there now from the microscope I can actually see two annular rings in fact on the top one has a B dia 2B dia other is 2A dia so I can actually monitor 2A minus 2B or 2N2B and therefore I know A and B and since I know A and B I will be able to find out XG this is called junction lab this is called screw techniques there are other methods which are more advanced method like Sims method optical methods this book our plumber's book has given many other techniques much more advanced technique but I just thought to you that this is how I did in 1973 or 72 12 even now okay so another few things we should be allowed to monitor when you start a wafer this was a question I thought you will ask how do you know NB is P type or N type let us say generally wafers are given with all kinds of data so when you take out of a box you know which wafer what orientation what doping is given but let us say X man has taken yesterday some wafers and put it on a rack and this man was asked to do the process in the night now he never told him which wafers you took so how does he know that which P type or N type you started okay so he has to I figure out what is the kind of wafer so this is a hot probe method in which I can find out whether the wafer is N type or P type example given is an N type the way it is done is you have a silicon wafer and you have two probes which is connected by a voltmeter centers voltmeter that means the arm is in the center it can go on the right or it can go on the left okay one of the probe shown here I can actually has a many a time this probe was used by the what is the probe I can use of course tip I will have to change what could be heater here the soldering iron itself is efficient it itself heats okay so you can fix a tip there so this can become your hot probe the other terminal of that is connected to a cold area through a voltmeter now since it is a N type yes since it is an N type hopefully it has larger number of electrons than holes that is why it will be called N type okay so if I heat the electrons below this hot area or hot volume there actually will energize the electrons okay because you are heating the way this area of course this is small tip so it does not expand to the whole wafer or something a small area gets heated and the electrons below that area gets excited since they get excited actually so they start moving okay as they start moving equivalently positive charge they start putting after all whenever electron leaves the lattice the ionized donors come there so if electrons start moving donor starts appearing ionized donor positive charge okay as this become positive and this essentially is cold so most of this they actually dissipate their energy and they are collected here since electrons are negative so this probe becomes negative this probe become positive so this is positive so the displacement of the on the left so you know if the wafer is N type the deflection would have been there now the question is very interesting if it is a P type we still argue that holes move away acceptors are released and the meter will show opposite polarity so I know it is P type okay electrons are known particles okay so I am 100% sure if I heat they will move okay holes are not particles holes are absence of electrons okay they are in a valence band these are quantum electrons they only move in valence band so when I heat why should holes move these are bound electrons to the valence band so how do they move and if they do not know why should pole meter show the other direction that means they move this is done alive to move at best what will do they will jump in a valence band from one strike to the other side but there is no hole motion in the valence band quantum jumps so how a hot probe still show you a P type material which by theory is not valid okay the another experiment which I had to do before I this I many times want to know the for a standard dope papers uniform dope papers I want to know the mobility of the carriers and I also want to know the carrier concentrations okay well of course is a sheet resistance measurement and you can always get R is equal to RS L by W but that is understanding that it is a uniform square rectangular this in real life there is nothing rectangular though we do solve problems for a good Cartesian coordinates okay okay so we know well by resistance so here is a hole sample okay it is not exactly like this but almost like this I have I apply a potential across this which pushes a current I X and we say V X by I X is an area of course this is my W this is my thickness this is my length so R is equal to R L by A which are V X by if I find the potential here then it is this if I substitute in this I get row WT by L V X by I X if semiconductor is N type rows Q means N 0 plus Q mu P P 0 but P 0 into N 0 is an I square so it is an I square by N 0 but if it is N type N 0's are larger than P 0 so we neglect that so you get Q mean N 0 okay the directions I already said this is X this is Y and this the top portion is Z okay vertical is that this is X this is Y okay is that okay this is Hall's sample okay Hall's sample is slightly different from this it has to be made actually which one it can be either side the carriers will go on left side or right side okay so if I do this analysis which I will not right now do long way we know that if I apply a magnetic field orthogonal to the motion of electrons going from V plus to V minus then flowing a current I X and therefore there is an electric field of V X by length which is the E X field electric field which gives a force of E Q E is the force so Q V X by L is the electric force but when I apply a magnetic field it also experience a Lorentz force which is actually Q V cross B in three dimensions okay it is a curl of V and B but in case our case it is V Y B Z and the net force if the electrons do not move because if the current is 0 instead of state Q V Y B Z is Q V X must be equal to 0 okay so E X from here is minus V by Z but V Y I just can calculate I X upon the last term I calculate it is this so I now know electric field is proportional to magnetic field through term of this so what will happen it will create essentially what if you have written down what I am going to do is the electrons were moving in X direction but as soon as you apply the Lorentz force is orthogonal so it will try to push the electrons on the one side whichever is the electron that contact will become minus the opposite contact will become plus okay if the opposite the electrons may move outside and this minus plus will change so that is not important is that okay so what Hall is trying to do is to relate this B Z term with E X and I know how much magnetic intensity of the magnetic field I am going so many Gauss okay there is a standard this Hall probe available to you so you do not really do anything fix B fix I also everything is fixed you just put your sample okay but how do we get it is the numbers is that okay so the Hall voltage that is across if you see the I am sorry since the electrons can be either because this they wherever electrons go sorry this should be minus plus so you have a potential difference between because the space charge this is again Poisson's equation you can find the V Y V Y is integral ex dy and you calculate all of it so you get this term T V Y upon B Z I X is called Hall coefficient T is known V Y you monitor B Z and I X you are forced this is a constant current so so this R H is called Hall coefficient okay so if I substitute back here I get concentration N or P is 1 upon Q R H depending on V Y is plus or V Y is minus okay R H is positive for Hall's negative for electrons since I know my N or P whether the machine this I use this resistivity term Q Mu N 0 for N type where Mu N therefore is corresponding to this is R H upon rho or R H upon sigma since sigma can be monitored by 4 probe uniform is very easy as to monitor that RS L by W so I know sigma I have just monitored RH so I know the mobility of electrons or holes so I can get concentrations and I can also get the mobility of carriers so Hall effect is a very powerful tool there is a version of Hall this is called Van der Pohe samples those who are in technology come to me I will show you what is Van der Pohe okay so these are some measurements I have left many measurements there are N number of measurements we can do right now let us say if it is 1 0 1 1 1 plane there is a method called X R D X ray diffractions one can find the distance between planes okay so I know X R D so I know the planes which I am getting because 2 D sin theta is known to me if I want to know the concentration I have many ESCA method Sims method may I can do many optical spectroscopy techniques I can do optical techniques okay it measures epsilon it measures the dielectric strengths okay so I can I have many measurements possible both optical electrical non-electrical which FTIR for example if I want to know whether it is S I O bond is correct so if I drew FTIR FTIR stands for Fourier transform infrared spectroscopy so if I do that I know each bond has some energy associated so absorption is maximum there so if I pass through and I see a peak down I know yes I have a bond has come so oxide is there okay so there are many techniques in VLSI technology you need to actually ascertain then you want to see the material how it looks texture as it called TMS you want to monitor areas then you can do SCM scanning electron microscope so if you really go to a good VLSI lab technology lab there has to have so much analytical instrumentation apart from standard CV IV which all of us are great about IV and CV that is what we think is end of it but you are needed to do many many measurements before you arrive at something tangible okay so please remember we all whenever we make a process we have wonder power samples hall probes hall areas we are measurement for sheet resistance areas we have many test areas one chip is dedicated sometime five four corners one center just to monitor the process okay this is that test chip so one of the most important thing in designing at the process is to design that test area how do I know what I did is okay individually and that is most important in the case of test so some of the test which we do is shown here these are least complete but I have given you how many possibilities and if you are really going to someday TSMC or something you will never know the other technique because you will be in a group you do only simulation and you so I am in technology in TSMC or any foundry you are not allowed to move out of some 10 10 by 25 area your cocoon no one knows what is next to okay any foundry global foundry is coming I think many some of you after you are in first year second year student may appear so most foundries do not allow too much leeway only labs like us you can move but here also now there is so many committees you cannot enter here you cannot enter so you probably do not know what is going on in the next room okay unless your friend tells okay which he may not know it is he will hide it okay I have an example to solve quickly before we quit for the day I actually did a solution for this user of graphs okay so I have a p type doping to be done in n type and I started with n type dopant n type wafer which has 4 into 10 to power 16 per cc as background concentration I did a pre deposition T1 at temperature T1 of 950 degree centigrade these values are typically for a 5 micron process they will be to scale down from many ways when I go to a nanometer scales why I use this 5 nano 5 microns because this gives number some big enough to see something happening okay and I did pre deposition for 10 minutes okay now you do not write down right now first I will find out so I have figured out I want to find out D1 at 950 degree centigrade so as I said the graph is in Kelvin's so 1000 upon 1223 is 950 plus 273 is 0.817 or 8 to roughly so here is a graph which is a taken from fairs paper trace temperature it is visible good this is ratio 1000 by 1223 as for example I am looking this is the diffusion coefficient and this you can see this line is the line for phosphorus and boron any other impurity will have to go for other this what is very implanted what is we are diffusing boron in the entire wafer so for this curve roughly 0.2815 or whatever it is D is 10 to power minus 14 centimeter square per second is that okay any other value you will have to go up and monitor okay is that any other 0.75 or 0.75 so this is the value this will be around 1.5 into power minus 13 depends on the ratio yeah it will be this will be true so it will be slightly less okay log no log please remember each is also a log log of log okay so you will have to also think how far you are away 50 percent point is around 1.3 okay of that 60 percent is 0.4 or 0.5 so some guesswork told us okay so this is the first graph you will use to find out what is D1 so I got D1 is equal to 10 to power minus 14 temperature I have fixed but time I said 10 minutes which is 600 seconds make it seconds so D1 D1 is 6 into 10 to power minus 12 or under root D1 D1 is 2.45 into power minus 6 centimeters so this is pre-diversion data one more data I missed but maybe I will come back that you know then I did driving at after of course as I say I removed glass and then start driving in oxygen some nitrogen is also added okay and that I did at 1100 degrees centigrade for 60 minutes driving is too big actually I am doing too deep is junction right now this number was just chosen so that junction depth is big enough okay so again I will what is the temperature 1373 degree Kelvin ratio of 1000 to 1376 is 0.7 or something so we go back on the graph again and whatever here is 1100 degree centigrade also so correspondingly you find the D this is centigrade so you can use centigrade come to this graph read correctly from the correctly in the sense nearest to that you cannot correctly monitor it but roughly close to the value 1.4 and 1.5 is okay but 2 I am not agree because 2 is a larger number there okay is that okay so given a temperature I can find diffusion coefficient for any species particularly from graph boron or phosphorus so if I do this I get D2 4 into 10 to power minus 13 centimeters per second T2 is 3600 seconds 10 is how much 60 minutes so 3600 seconds so D2 T2 is product of these two which is 1.44 and for minus 9 centimeter square or root D2 T2 is 3.79 into 10 to power minus 5 centimeters so I know under root D1 T1 I know under root D2 D2 I also want to know NO1 I also want to know NO1 that is for pretty passion temperature 900 okay 950 this is linear scale please remember the lower scale is linear and this is log scale okay so 950 somewhere here which species which graphs I should use the this one or this one lower one active concentrations so for phosphorus at 950 roughly you read from here 950 this value okay no no we are we oh sorry sorry thank you so for boron maybe I might have taken then wrong number so just check it correctly so you think it is p n junction instead of p n junction okay tell you I in night I did mischief maybe but okay so you read this so no no I think I read correctly so I get 950 degree boron has sorry for boron roughly it is okay not very wrong okay actually I took boron data but wrote first so you started with ND so this 320 per 20 per cc is your surface concentration of this for the surface concentration in solidity same now I want to evaluate RS from RS XJ graph okay I have done Gaussian profile two step diffusion okay I have four graphs okay but these graphs are now all loaded on the site so you can choose this please download and get it prints this is your two step diffusion profile these are four graphs I may show you this is for n type complementary function I am plotting surface concentration versus RS XJ at different NV this is for p type error function this is n type Gaussian and this is p type Gaussian so whichever is the this you should correspondingly choose the graph so in other case it is p type Gaussian which we have to use so what we do is just I will come back to get RS first I should know NS from this graph if I know NS for given in B what can I calculate RS XJ is that correct for given NS how come how well I get NS 2 NO by root D1 D1 upon D2 T2 is surface concentration for Gaussian so I I know that value for this I calculate RS XJ is that clear so once I get RS XJ value from this graph then I can calculate XJ how I calculate XJ 2 root D2 T2 upon ln n surf upon NB to the power half this is the Gaussian profile at X is equal to XJ and it then be so I saw this and I get XJ of 4.35 microns is that clear previous slide okay sorry no no no so RS XJ I will monitor from that graph because in surface I know NB I know so from Gaussian profile for p type I know how much is RS XJ and what is the unit oh microns please remember is that okay from the graph what will monitor surface concentration known for given NB go down and figure out what is RS XJ okay that value you are noted down from the graph okay now question was always asked every year by many that sir if so much simulation is possible center is does so well no whether centers does correct or not who will know because program is only a program okay so this is called analytical starting point we know where we are okay so very far you are right okay these days you do not have to do everything is not correct you cannot do 100 percent on system you have to know what system is doing okay even that you do not know you just run programs so then also you do not know what models they are used okay is that okay so let me write two slides I will show is that okay so I know this profile so this is my surface concentration so I can get RS XJ is that okay all of you so I calculate XJ from 2 root D2 LN surface I know all values so I get 4.35 micron as my junction depth and I also have evaluated RS XJ from the graph for given N surface and NB is that clear so what is to be found out now RS XJ is known XJ is known so RS is known just note down I will show you the final answer so all of you will bring these sheets during exam in this water say till mid-sem if I finish oxidation so bring oxidation also if I do not then you bring only diffusion if I finish implant then you also bring implant profiles so all this data is already created from various papers I have copied these are available on our course website please download and print I am not going to supply you any of these sheets I have forgotten so be forgotten okay that is it okay is that okay so since RS XJ was found as 200 microns from the graph for this 400 power 16 and surface concentration of 1.2 minus 90 into 19 per c see RS is 2 upon 4 see I am doing it because I am first time doing it so I I will have to go through steps you can directly write the solutions I also write step because from there roughly I know I am right in calculations on that okay so if I I can directly one time substitute I never do that if you have seen I have to put steps the reason is obvious that at least mistake in one I figure out the rest is correct otherwise we do not know which place it went wrong okay some system you should create okay is that okay whole point 35 so RS XJ from the graph we figure out is 200 on micron divide 4.35 to 200 sheet resistance is declared as ohms per square okay 200 by 4 point which is 46 roughly 46 ohm per square is the sheet resistance okay inverse would have been what I monitored RS I monitored XJ then calculated NS and then saw something is missing in pre depression driving cycle figure it out what was that okay is that the method clear to you so these graphs help you to actually get these are integrals so it is difficult to solve every time integral so these are put into graphs directly okay plotted for you okay this finishes the diffusion techniques tomorrow we start with techniques this is only diffusion maths physics we did maybe do little physics level do it and we will look into how actual diffusion is performed