 You can follow along with this presentation using printed slides from the nano hub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. So today we'll talk about surface recombination and generation. This is lecture 15 of the series. Now in the last class we talked about bulk recombination that if you have a piece of silicon sitting on your table and you put a flash light on it then we saw that excess electron holes are generated and if you turn it off then they recombine and they have various channels of doing so. They can go through traps and in that case remember they need a phonon to help them match their wavelength because generally the bottom of the conduction and bottom of the valence band bottom of the conduction and top of the valence band they are not aligned. Phonon helps them or the trap helps them to provide the extra momentum. OJ recombination two bumping against each other one going down and the other going up in energy and gradually relaxing down. All those channels essentially conspire together contribute together so that the electron can come back to the equilibrium state. Now that was for the bulk happening inside the semiconductor. Now you know that most transistors are actually very very thin right very thin and surface is a very important property. We'll see MOSFET transistors are primarily surface where the electrons flow is slither below the surface. So surface is actually everything and how electron and holes recombine on the surface not in the bulk bulk was last class how they recombine the surface is an important property that we have to understand. So we'll talk about nature of interface defects in the beginning then we'll see how to adapt the formula that we have already done adapt to interface charges interfacial states and how the recombination now can proceed through the interface. And we'll give you some example that how to use this formula just like last class right the formula itself was very complex it has a lot of terms np minus ni squared a bunch of term in the numerator but under appropriate conditions they simplify and in that case we can just get a quick feel about how the recombination proceed based on a very simple formula. So that's something we'd like to do especially under depletion region again this depletion region is something that occurs in the MOSFET a lot and that's why we are in anticipation of a future problem we are solving it here here you may not see the context yet but eventually you will see every problem I'm solving every one of them will be used later so we should we are preparing ourselves for it and then I'll conclude. Now there is this is a the problem I'm talking about today was one of the fundamental reason why MOSFET semiconductors which is in your computers although it was a sort of invented or proposed beforehand patented beforehand 1930s but because of this reason alone this type of interface traps region or interface defect region it didn't really fly the first one came around about bipolar transistor that was the first one in 1947 that was what was invented and eventually came to the market ahead of time but even for the bipolar transistor there was a big problem and the big problem is the following and now I'm not going to go through the details of this I will show you this again at a later time the only thing I want to show you here that this is three different regions let's say an emitter the yellow region the green is the base region and the current the red arrow is showing that current is supposed to flow from somewhere one terminal to the other why the flow and under what conditions we'll talk about that later on but the problem was that you see there are lots of exposed surfaces these exposed surfaces on the corner electron can not only go through the bulk and recombine in the process but they can also creep along the surface and surface is exposed to the outside world do you remember in the early days in the first or second class we talked about not only about volume that what happens number of atoms per centimeter cube how to compute that but we also talked about in 100 surface how many atoms do you have in 111110 surface how many atoms do you have right you have done homework this is in anticipation of this problem because depending on the surface you have you will have different number of atoms and in the process there can be recombination and it was a big deal a big problem and that's why many people do not know that the original transistor that was invented in 1947 that never went to commercial production because of this problem and the first person to solve it really in a in a significant way or in a proper way is by actually covering that region up and I will show that again in a later time instead of keeping the side region exposed actually covered that up and then in the process make these transistors robust and manufacturable and this is the right hand side that is what made integrated circuit possible if had it been things so they're like on the left hand side that would have been a disaster no integrated circuit would have been possible bottom line surface recombination is a very very important important thing and you may not have heard about this person but this is one of the leading person this is one of the eight people who are sort of responsible for the initial stages of transistor development significant contributions and so bottom line is that surface recombination is important and we'll see it many many times during the course so again just in another point of view that you remember this particular picture that it was silicon on the bottom single crystalline silicon and then there was an amorphous oxide right do you remember in the middle region here I show 16 angstrom these days most of the transistors have about 12 angstrom of oxide this even thinner thinner than a dna dna double-stranded dna is 22 angstrom and this is half almost half the size of the dna and then poly silicon transistor now if you look at the silicon you remember that the silicon is tetrahedral structure right and we pushed it flat on the surface and that is how we have drawn it and the circles are each silicon atoms in the crystalline phase and the double lines are essentially the lines of the covalent bonding between silicon right so we have four neighbors tetrahedral structure and you can see every atom on the planarized version of it still has four four surfaces now this of course continues up and down had it been a bulk this would have continued up and down but just think about the row just on the surface just on the surface then what would happen this is what is going to happen now anytime you pull the silicon up the silicon that was sort of terminating providing an electron participating in the double bonding it will take its electron out now if it takes it electrons out then the remaining a silicon on the surface now are in trouble because they cannot no longer have the four electrons from the neighbors that was necessary to complete this bond and so then each one of them is sort of deficient of one extra electron right do you see that and as a result what will happen by the way these are called dangling bonds now itself it's not dangling it is fine it's on but it is anticipating waiting for another electron to come along now you can you can see that if this was the case if we didn't do anything to this problem then this is what's going to happen as if you think about two electrons the fate of two electrons then one electron could just go no problem so it can go from one part to another travel that's fine but what would happen sometimes once in a while what would happen that this electron would come in and instead of going where it is supposed to go it will find a local bond nearby remember it's supposed to be a double bond so it's very energetically favorable it will just go and stop there and if it goes and stops there and then comes back it will be a very horrible noisy transistor because sometimes it goes sometimes it doesn't moreover if the charge is sitting there sometimes it says sometimes it doesn't i'll show you later that something called a threshold the electrostatic potential of the silicon underneath would be fluctuating wildly that's no good and that's a big problem in transistors these days so in fact a significant fraction of my research is based on what happens reliability physics what happens when bonds are broken in a random way and this is one of those problems a very significant problem actually and this was problem was again even eventually was solved in fairchild in 1965 to 1967 that is when this problem was solved before that no MOSFET was possible MOSFET came along that time because after solving this surface passivation problem okay so the how does the surface states look like so think about so i have what i have done in that picture on the top with circles all around i've rotated the previous picture from the previous slide 90 degrees so you can see single bonds facing towards the right now think about the dark blue line i have taken a cut and i'm looking at the potential electronic potential going down and in the bottom what you see is this electronic potential up and going up and down do you remember chronic penny model the chronic penny model was series of this right going up and down up and down but the only difference here is that it now it has reached a surface now when it has reached a surface you can see on the bottom the blue dark blue one that potential has gone up because it doesn't want electrons to go out it is like having a table now electron or electronic wave function should not be able to jump out of the table otherwise the table would dissolve right so close to the surface then there is a potential barrier and as a result what i'm showing here in the green is the electronic wave function and you can see it's more or less periodic but towards the edge it's not periodic anymore so when you solve Schrodinger equation for this problem chronic penny equation for this problem not only you will get normal bands conduction valence band and all those but you will in addition get bands in which there'll be extra levels now these are called surface states now this is very different from this trap levels right trap levels were foreign atoms like gold copper coming in silicon so this was physically a different material that gave a different trap level here no foreign material it is just that i have terminated the barrier and therefore i have introduced a level which is solution of the original Schrodinger equation but with a new boundary condition you see and as a result as a result what will happen is that we'll have an ek diagram and this ek diagram in addition will have a level shown here in a blue level in between right and this is one of the surface levels surface states now that is the ek press after solving the Schrodinger equation you will have an ek diagram and correspondingly in the real space close to the surface the dark blue do you see in the middle between conduction and valence band that's the surface states these are not foreign atoms any time you cut a bulk material you are going to introduce one of those states now what is going to happen is through those levels through those levels electron from the conduction band will be able to jump to the valence band through the help of this trap levels or surface state now this is not a physical state in the sense of a trap level but of course it function as if it was one so we have one at a given energy and where is it and where is it that depends on where the surface is terminated that determines it now assume that instead of cutting on the top electron is cutting coming somewhere near the bottom again 90 degrees same way they're rotated now I have shown here something which is a made surprising a little bit on the left side on the left side what you see is the electrons where sort of silicon atoms are coming down then it sort of got shifted and then it came down coming down again what is that all about this is because anytime you have a surface you cannot have absolutely plain surface not possible anytime you cut it or anytime you break it what happens it goes flat a little bit and then a extra layer of atoms are gone sometimes an extra layer two layers of atom are still there so on the atomic level there is modulation of the height going up and down a little bit on the order of a few angstrom and this is present anytime you try to terminate a surface right so on the bottom side what I'm trying to show on the left hand figure what I'm trying to show is a consequence of the height modulation and what will happen as a result what will happen as a result that the potential looking to the going from the left to the right will not be exactly the same and as a result you will have a corresponding for the red you will have a defect state which will not be exactly in the same place as the blue one it will be slightly different because of course the potential is slightly different so when you solve the Schrodinger equation the state is in a slightly different place now as a result what will happen that there will be a corresponding electronic state but so see here the red is slightly different position in the conduction band you see that and as a result now again you will have another electron recombination path but there will be it will be through a different state so what is happening here is that the surface is sort of non-smooth non-smooth had it been an absolute smooth surface only one defect or only one surface state because of this surface oscillation there will be a series of them essentially populating the entire band gap that's why surface states are so dangerous you know copper or gold gives you one level sits in the middle it helps a certain recombination no problem in the surface there is a series of them for throughout the whole thing and therefore lots of ways for the electrons to come down to the balance band and recombine recombination we don't want for the timing there are devices where we wanted but for these we do not want and therefore a series an entire series of levels it will be full remember each state is coming from the consequence of this thickness modulation or height modulation now one thing again I have on the left hand side I have rotated the picture again back to 90 degrees just for one thing I want to make sure that you understand that recombination can only involve a single level you cannot jump look at the version I have written on the right called wrong right do you see right uh red uh shown here in red wrong in which what is wrong I am trying to show an electron as if it is jumping first to the red level then to the blue level and then to the balance band instead of making one intermediate stop it is making two intermediate stops that cannot happen why not the reason is you remember although I have drawn this picture in this particular way for my convenience where all the trap levels are looks like they are in the same position they are really not look at the picture on the left hand side because the blue state came from a region which was vertical blue cut on the left side on the right other hand the red surface state came from a region which is corresponds to the vertical a red line on the right hand side these two states are far apart they are not from the same place do you see that and therefore an electron cannot start going down to the conduction band or sorry through the balance band and then all in a sudden jump 20 atoms further to go to the another level and then go down that's not possible because this is like one high rise you start going down and then you have to jump to a different high rise building and then go to the ground floor impossible so actually double levels double stops triple stops impossible always a single stop as we are going from the conduction band to the balance band is this clear this is a very important point because if you don't understand this you will not understand the derivation i'm going to do in a second but say very important point that physically you should make sure that you understand why this doesn't happen so let's try to now use because this one trap level or one surface state in between conduction and balance band as far as math is concerned you know what's the difference what whatever we did for the bulk we shall just use it for the for this problem but in the bulk we had one level here we have a bunch of level we'll sum them up and then we'll go home so let's start let's start remind reminding ourselves what the formula was for single level bulk trap last class do you remember this horrible looking formula np minus ni squared in equilibrium n naught p naught is equal to ni squared no generation no recombination in equilibrium means huge amount of generation and recombination no net generation or non-racial recombination you can see the trap level sitting there nt number of traps you have more the number of traps you have is the denominator the way it is more the recombination will be now i will have to do a sum a little bit later so let me do this a trick here so what i have done is i have flipped the nt up that's the only thing i have done nothing nothing else is done i have flipped the nt up now for a single level now i have nt is the number of traps per centimeter cube right 10 to the power 16 per centimeter cube but remember that i have interface traps throughout that energy region so instead of writing it at nt as a single number which is integrated over energy why not i write it as effective density of traps per unit energy d sub t of e you know certain number per centimeter cube per e v right multiplied by e v delta e the small region i'm thinking about so dimensionally these two are exactly the same but i'm talking about a certain energy e that's it i haven't done anything here right okay now if i know that's the recombination for one surface state at energy e what do i have to do in order to sum them all up just do an integral because i'll just integrate over all the all those all those energy levels of the single surface states why do not i have double integrals where electron comes down once at one level makes another stop and then goes down is because what i just explained in the last slide no two you're not allowed two stops going down so it's a simple single integral over energy of that original formula and you are done okay so let's think about a case in which it's a example case about how to evaluate that formula and let's try to do that this is a donor doped donor dopes means primarily lots of electrons right lots of electron donor gives a lot of electron in so that we have a little bit excitation let's say photo generation i have and as a result i have delta n s zero why there is an s because we are talking about surface quantities so in this edition of zero i have a s sitting there and i have shine light on it shown light on it there have been this extra delta n and extra delta p generated i have turned off the light and i'm seeing how the electrons are going down from the conduction to the valence map do you remember the top formula that that's something you should remember right i have split the total concentration out into the equilibrium concentration and extra concentration and delta n and delta p of course in general can be the same now essentially you have we have done this in the last class minority carrier injection remember in the bulk case only thing we have done n naught and n naught multiplied by p naught what is it always equal to always equal to ni squared so you can see why i have gotten rid of that ni squared and also it's a minority ionization or minority injection and as a result the delta n multiplied by delta p that turn is gone right so i have just you see i have a delta p sitting on the top and the trap levels that effective density of traps at a given energy those are it now the numerator is something to think about so let me take one second to explain it first of all you see this time i'm not really immediately ignoring delta n and delta p right so first thing i am ignoring is that this is a donor dock material is that right donor dot means how many holes do you have let me give you an example you'll understand how to think about this what is ni 10 to the power 10 per centimeter cube ni squared 10 to the power 20 per centimeter cube keep this number in your head 10 to the power 20 per centimeter square for silicon now let's say i have doped something with 10 to the power 18 right if i doped something with 10 to the power 18 how many minority carriers i have np is ni squared n is 10 to the power 18 ni squared is 10 to the power 20 i have 100 i have 100 of the other type of carrier is that right so therefore think about it 10 to the power 18 10 to the power 2 which one is bigger so that is how i drop it that p s not term that is why i drop it because it's 100 compared to 10 to the power 18 okay now one the other thing i have done in this expression is you will see i have pulled out the n s 0 in that expression i have there are six terms so i have dropped that ps ps not term but i have kept that ni ni s and the pis terms this was the n1 and p1 do you remember what is the multiplication of n1 and p1 always also equal to ni squared do you remember from the last class so i have actually just kept them but other than that everything has been pulled out ns0 has also been pulled out there is a reason why we are doing it in this particular way and i'll show you in a second because we'll have to evaluate an integral i'm preparing the integrants in a way so that my life will become simpler now you can see that ns0 on the numerator and the denominator in the second line that will cancel so i have an expression that looks something like this now for next three or four slides i'll say i'll focus on the denominator because if i can do something with the denominator during the integral i'm in a good shape so we'll see so this is the denominator looks horrible there's three terms well one is not too bad i have ni s divided by ns0 so that's what i have won and i also have p sub is divided by n sub s0 what is n sub s0 equal to the donor doping this is a donor doped material so that's why i'm carrying around those terms you had it been accepted doped material i would have to just do the other thing okay now do you see the substitution i have made ns0 in the extrinsic region remember that extrinsic region freeze out extrinsic intrinsic region so i have replaced in the extrinsic region ns0 with n sub d so that's a constant only thing i have to worry about now everything i know now somehow i have to take care of nis and p sub is you know this is just math i'm helping you around when you start reading through your textbook and going through the note you will see this in five minutes so don't worry about it if you are lost a little bit just look at the general concept the tricks i make now ns0 i can always write it as for the first second term in that last line i can write it in terms of ni because this is something you have done before right ef minus ei and beta is what is beta 1 over kt 1 over kt and similarly the n1 n sub 1s that one you can correspondingly write an expression for this now do you see that why this expressions could be correct because if you multiply n sub 1s with p sub 1s again you see that it will become ni square the magenta and the blue if you multiply do you see that it will become ni square so essentially i'm writing down the expression just like i did it in the previous previous class and when i cancel these terms i get an expression which looks something like this complicated but this is the real form do you see e minus ef e is the trap level now right surface one of the trap levels ef Fermi level Fermi level i can find out why will i find the Fermi level from do you remember that we have solved n minus p plus nd plus minus na remember that the whole thing said to equal to zero that's why we found the Fermi level from so let's say we find the Fermi level Fermi level is known e is whatever trap i'm talking about so it is 1 plus ex plus some constant which is the ratio of the capture cross sections and e to the power minus x do you see the minus sign in this one e minus ef and ef minus e so there's a minus sign well i'm in good shape now i'll be able to integrate this so let's try to do that in the top i'm showing one level this is conduction and valence band in the blue curve in the real space electrons coming down from the conduction to the valence man in the bottom curve which is green i'm plotting the denominator you know the formula that i just derived 1 plus ex and plus e minus x that one i'm drawing in the blue x is energy and y is the magnitude of the denominator how does it look i'm just looking at the function so you have seen this so let's try to evaluate that denominator at energy level e equals ei now when energy is e equals ei at that the from the previous expression you can see that the second term will become ni divided by nd ni is 10 to the power 10 nd is 10 to the power 18 so that number is gone and the second term similarly is ni over nd that term is also gone at the intrinsic level only so the magnitude of the denominator at the intrinsic level is one essentially one so i have one you can see i have shown here in the red graph on the right hand side at the point where is ei that the magnitude is one the denominator is one now what happens in the denominator at e equals ef now if e equals ef then from the previous slide you should put the expression e equals ef and the value for x which is a difference between e and ef that will become zero so x is zero so if you put x equals zero then you will see that at that point the value of the function will be two right and similarly you can get another point ef prime sort of on the other side where the function becomes two as well you know you're plotting it if you plot it in a computer one second it will give you a plot right in a calculator if you do the function plot so it will do the function plot and if you do that you will see the function looks like this very denominator is very small in the middle but as you are going close to the surface or to the conduction and valence band the denominator x actually blows up right now you see I can now therefore approximate this in the following way because it is blowing up on the two sides at ef and ef prime instead of carrying around that denominator and the whole integral why not I approximate it with two functions that is one function which is sort of infinity at the two ends and one in the middle now remember these are the time when before a lot of computing power is available this paper was written when this was written almost in 1953-54 so therefore all these tricks and other things these are historical things yeah you may not really worry too much about it but there is a very important point I want to make which is something you need to know and this one explains it but let me first finish it so essentially I can replace the denominator with one in this range and with infinity outside now do you see now why any if a level a trap level is not in the middle then it cannot participate in the recombination process you see that because think about a donor level which is very close to the conduction band in the donor level that one the denominator is infinity so surface recombination or any recombination will actually go to zero denominator is infinity right so therefore a donor cannot be a recombination center had it been a recombination center all electrons will immediately disappear through the donor levels it's a very important point that you need to understand that why even when you have 10 to the power 18 number of donors or acceptors why is it that none of them participate in the recombination process at all that was a question last time right somebody asked this question I was trying to explain it in the board okay so this is a approximation for the denominator now I am all set because in a state of doing all those integrals from ev valence band to conduction band and looking at the complicated integral that's my integral why is it because the denominator is one but instead of going from ev to ec conduction to valence band do you see my limits now ef and ef prime because that is the only part where the denominator is one outside infinity well if you put it infinity in the other one that gives you zero contribution gone and as a result this one you know if d sub e d of e if I assume that is more or less a constant throughout the gap as it is shown here in a in a picture like this on the left hand side is taken from the book the left hand side is a valence band and the right hand side is a conduction band this is plotting from various experiments various references and experiments the number of trap levels here shown how many do you see 10 to the power 10 per centimeter squared per ev that's the typical unit right effective effective defect levels and it goes maybe go below and 10 to the value 11 and 10 to the power 9 yeah let's let's assume for the time being that is flat and if I assume it's flat if you want to put it in in the computer and integrate properly well that's no rocket science either but you can assume it's a constant and then you are done then you see that all the traps between ef and ef prime multiplied by whatever is the average defect density d sub it what is it interface defects at the interface or surface levels at the interface and that's what you have and the excess carrier concentration delta p is there why isn't there any delta m or any m because this is donor doped lots of electrons just a few holes here so therefore when you only see delta p you don't see it's not restricted by the number of availability of number of electrons so you don't see any electrons here right and that whole term in the beginning is called a surface recombination velocity by definition the three terms c the captured coefficient c sub ps dit and ef minus ef prime that together you see that's a material constant given a doping given a certain doping that's a constant dit is just a material parameter capture cross section is a material parameter ef depends on the doping but more or less this s sub g is called a surface recombination velocity it has a dimension of a velocity and that is a number that you will see mentioned in many of the handbooks and it's a very well known number people will say oh my i have done a new processing but my surface recombination velocity is horrible they made a measurement so let's say they will shine light and they'll see how the current is decaying and from that they will extract back the rate r on the left hand side and from that they know what this delta p is so they will extract back a surface recombination velocity if it is horrible that means they have to do something to the surface because the surface still not good enough for a good electronic devices so bottom line is somehow or other suppress s sub g by improved processing right that's something one has to do now just to give you one example in terms of surface recombination recombination in the depletion region what is a depletion region depletion means it has been depleted of any carriers means n is zero s is zero right so that's that's something it's important and how how things happen there well that's what i have done here why am i doing this again i have done all this complicated calculation and i just got a formula why am i doing this again is because remember in the very early part of the previous derivation i assumed it's a minority carrier injection right few carriers donor doped that's how i did the whole integration my if my original assumption is wrong then i'll have to redo the calculation but fortunately it will not take as long i'll start have to start from the original one but i should be able to drop n and p right because depletion region i have that the same thing and then i will be able to just rearrange the terms a little bit and you can see that now this is a function on the numerator i have e to the power x and in the denominator i have e to the power 2x now when you are in high school do you remember this type of integral e to the power x divided by e to the power 2x plus 1 then you can make a substitution and get the integral do you remember what will be the answer tan right tan theta it will take a minute i mean high school had been a long time ago right now you can do this integral in five seconds this will you can pull everything out why have i pulled the it out uniform i have assumed it uniform i have pulled it out n i is of course i can pull it out this little simple integral you can do in your sleep and then the answer is by the way just one second so in order to do this integral i need to take the limit to infinity plus minus infinity e c and e v to the plus minus infinity i can do that because most important traps are always in the middle region right in the middle region so conduction and valence band are many k t away many k t away because one e v even if it is like a 10 k t away so that one for the exponential it looks like infinity so therefore i substitute with plus minus infinity so that i can do the integral easily and that is my integral do you see what the surface recombination velocity would be it would be the geometric mean of this electron and hole capture number of interface levels you have beta is one over k t right pi over two that comes from the tan tan in between integrated between plus and minus infinity again very simple formula something that you can probably almost remember how to derive now do you see a sign problem here not a problem is the right sign what sign do i have it's a negative sign and why do i have a negative sign because when things are depleted it wants to go back to the equilibrium as a result instead of recombining it is actually generating so that it goes back to the equilibrium values therefore i have a negative sign there and compared to the previous case we had a positive sign okay now this is something i have already mentioned to you that and most students probably go out of probably finishing graduate school without realizing this is that why the donors and acceptors don't act like a recombination center is because the place where the donors and acceptors are the denominator at that level at that point e sub d donor level few kts down right from the conduction band at that place the denominator actually blows up becomes infinity and therefore no recombination generation through the trap level and that's why you should always remember why they don't participate the trap levels or interface levels on the other hand don't provide much electron as donors or acceptors those are very bad donors acceptors in the mid gap region but very good recombination centers so both has the donors and acceptors has the same problem okay so this is a quick summary of various types of recombination for example in depletion what is something i just derived the pi over two you remember the integration of tan inverse between plus and minus infinity minority carrier interface recombination derived in the beginning of the class and compare them with the bulk minority recombination do you see dimensionally they are the same do you see that because n sub t in the third one for bulk minority it is number per centimeter cube right per unit volume on the other hand di t multiplied by the energy di t is per unit energy and that one that multiplicate energy range gives you energy so together this again gives you per centimeter squared or per centimeter cube so dimensionally they are exactly the same the slight difference here and there but exactly the same okay so looks like that this is why we should end now let me quickly point out a few things so we'll go on move on and then attach contact to the transistor and then we'll see how the current flows this is all about recombination and generation what is very important that this is sort of a detour in terms of we want to know how the current flow this is a detour but these steps are very important because when a device is good you know somebody like a you have a silicon processor if you join intel today and they know how to passivate the surface how to reduce di t to a value of let's say five times ten to the nine very little value no problem so if some of the process is already done you really don't have to worry about it almost negligible but when you are a researcher or a new scientist working on a new material in that case you will ask a technician to go and measure you some data you will set up you know a surface remember this various orientations we talked about so you will ask them go and make this measurement and come back to me then you will have to deduce from that what was the surface recombination velocity under what condition and how many do i have and if you have too many you will have to count the number of states on the surface and see what type of processing you can do to suppress the level so when things are mature you don't need to know anything others have already done the dirty work for you but as a new scientist as a new engineer in that working on new materials then it becomes central that you understand this can and solve and debug this problem you see so that's why although it's a little complicated math please take a few moments so that you can understand them it's exactly clearly step by step and you can apply it to a new situation right okay thank you