 So, this session we will have the calculations or tutorials as solar cell it should be the tutorial, but we have just seen the p-n junction is solar cell when you put a light voltage gets generated and therefore, p-n junction behaves as a solar cell. And there are various parameters of solar cell that we have seen yesterday the short circuit current open circuit voltage fill factor. So, let us look this look start looking at the problems we look at the problems for calculating the drift current, calculating the diffusion current. Calculating I 0 will be difficult, but we will try it absorption coefficient, absorption length then open circuit voltage short circuit current and eventually efficiency. So, those are the thing we will do. So, one thing that I have told in the morning is that in any semiconductor under equilibrium that one relationship that is always true is n 0 times p 0 equal to n i square. There is one thing always if I say if I say that doping of boron is stands for 15 atom per centimeter which means that many holes are created. So, holes number of holes created is also stands for 15. If I say doping of phosphorus is stands for 17 atoms per centimeter cube which means the number of electrons is also stands for 17 fine. So, note down a problem and do it a semiconductor a semiconductor is doped with 10 to the power 14 boron atom 10 to the power 14 boron atoms 14 atoms per centimeter cube. Let us say let us say silicon is doped with boron atoms of 10 to the power 14 atoms per centimeter cube find out the majority and minority carrier concentration. So, what are the majority carriers? Who holes are majority carriers? So, p 0. So, what is the majority stands for 14 number per centimeter cube and what are the minority? So, minority are electron and their concentration is stands for 6 numbers per centimeter cube right simple. If the mobility now the second part of this mobility of electron in silicon you know how much it is we have seen in the last presentation about 1350 right. Let us say 1400 it should be stands for 8 sorry or no stands for 6 that is right. So, N i is for silicon is stands for 10. So, N i square is stands for 20 no that is not N i. N i is intrinsic carrier concentration. Intrinsic carrier concentration for silicon is stands for 10 numbers per centimeter cube. Silicon atom density is 5 times stands for 20 to atoms per centimeter that is silicon atom density N i is intrinsic carrier concentration. So, which means intrinsic electron and intrinsic whole concentration stands for 10 per centimeter cube. So, N i is intrinsic carrier concentration which is for silicon it is stands for 10 numbers per centimeter cube. So, that will give you p 0 stands for 14 that is the majority carrier concentration that will give you N 0 stands for 6. So, you may have this kind of scenario right in general happens that your p side is stands for 16 your N side is stands for 19. So, in this case your electrons are majority that case holes are majority. Once you know the majority carrier at the side you can find out the minority carrier. So, once you know the majority carrier at a given side you can find out the minority carrier N i square by. So, take a p n junction x problem here we will come back to the mobility thing again. Let us do this again. So, right now do not worry about this part right now do not worry about this number just look at this number. So, you have p n junction now go next. Now, find out the other part holes are majority and this minority. So, just look at this part. So, then the majority and the minority are electrons then you go to the other side you know the number changes you know the electrons which are minority here now becomes majority here and holes which are minority here becomes majority there. So, you know this number you know this number you find out this 2 numbers simple how much is. So, how is the hole constant in 10 for 16. So, then N electron constant will be 10 for 4 electron concentration is 10 for 19. So, hole concentration here is 10. So, let us say some other problem quickly we will work on many problems. So, your piece of semiconductor which is p type doped with the boron. Boron doping is let us say 10 for 16 atoms per centimeter cube and then you need a mobility of electron mu N is 1350 centimeter square per volt second and mobility of holes is let us say 400 centimeter square per volt second find out the conductivity of the semiconductor. You also need a cross section area which let us say 1 centimeter square what you need to do you need to find out conductivity of semiconductor. So, let me introduce MN Rao and Amrata Joshi they are my PHE student they are here to help you out for your problems. So, they will be around if you need any help you can take their help, but N N N then mu N. Now, what is the conductive plus Q P mu P, Q is charge on electron 1.6 in for just conducted with electron, but there are holes also it is you have to sum it up. So, actually I can solve it now here. So, and just cross conductivity, conductivity equation comes out to be Q N plus Q P mu P. So, what we have is p type semiconductor. So, our minority concentration N comes out to be 10 raise to 4 per centimeter as majority concentration P is 10 raise to 16 per centimeter cube. So, actually we can say this conductivity will be approximately equal to Q P mu P value and we can actually calculate that. Because N is very small as compared to P the first term is going to be negligibly 0. And you can always in fact, whenever you are talking about semiconductor with electrons and holes, this always happens one of the term is going to be small as compared to other term. So, you can directly neglect the first term and therefore, conductivity is just equal to Q P mu P. We can substitute then it comes out to be 1.6 into 10 raise to minus 19 P value P value is 10 raise to 16 and mu P value is 400. So, we can accordingly get the value. So, it will be 6.4 into this is 18 9 and 6 4 ohm centimeter 6.4 into 10 raise to minus 1 per ohm centimeter fine. So, you can find out the conductivity whether it is p type or N type or whatever is doping very simple you need to know the mobility. You do not need area by the way area is not required. So, if a parameter is given you do not have to necessarily use it. Now, let us extend this let us let us do the second part of this problem. Yeah, assume for the same conductivity of the material let an electric field of 1 ohm centimeter is applied across the p type material. Can you guess what will be the current density? So, an electric field of 1 ohm centimeter is applied across this region p region. So, what will be the drift current when electric field is applied drift current will result. So, what should be the drift current when 1 ohm centimeter of electric field is applied across p region which is having the given doping. So, what we have as a current density will be sigma into electric field. So, sigma value we have already found out to be 0.64 and 1 volt per centimeter is the is our conductivity. So, we will get it to be 0.64 per centimeter square this is the current density value. So, if we have area as it is given in the previous sum we can actually find out the current value as well. So, if area is 1 centimeter square then your current is 0.64 ampere. Now, this 1 volt per centimeter is a small current a small electric field typically current electric field can be of much higher and particularly within the junction the electric field goes to as high as I think as for 5, as for 6 volt per centimeter very high electric field. So, actually 1 centimeter square cross section area can carry lot of current this current is very small because electric field is very small here. So, 1 volt per centimeter is a small electric field fine. Now, so drift is ok you can find out the drift electron and hold drift etcetera. Now, let us go to the problem with the diffusion. So, the problem with the diffusion is there on your screen you can read it instantaneous carrier concentration in a n type semiconductor at 2 points separated by 100 micrometer each from each other or 1 point 10s for 17 per centimeter cube and 10s for 15 per centimeter cube. If the diffusion coefficient of electron in the semiconductor is 30 centimeter square per second find out the diffusion current density in the semiconductor. If you any doubt Amrita and Rao is here you can ask them any time you know how to approach this problem you need to find out diffusion current for diffusion current of electrons right electron gradient is given and you know the diffusion current is proportional to the concentration gradient and diffusion coefficient and charge right. So, you can actually use the expression that we have discussed in the morning before lunch session and find out it is very easy again. If you need any help regarding anything you can ask the 2 TAs here they will help you out. Is that clear to everybody? Yeah, this is what we need to calculate the profile instead of profile what they have mentioned is 2 point the concentration 2 different points which are separated by 100 micrometer. Now, they have brought down everything to the centimeter this is per centimeter cube so this 100 micrometer stands for minus 2 centimeter and this term is negligible as compared to this term you know 100 times smaller. So, therefore, make it nearly 0 so therefore, overall it is stands for 17 by this. So, the gradient is stands for 19 per centimeter times 4. Now, what they have done is they have used very large number but if it is a minority carrier concentration same answer you will get same answer you can get if the numbers are smaller and the distance is smaller. So, if the carrier concentration gradient 100 microns is very large distance but if that has to happen in within let us say 0.1 micron 100 nanometer this number will go even higher. So, the gradient can be higher but then you can put this here and you can get the diffusion current due to the electron for a given concentration gradient. What I can say I can also describe my N x as a function right I can say ok my N x is x square minus 2 x plus 10 from function and then find out the diffusion current do it quickly 2 minutes of. So, this is my concentration profile now and as a function of x is given as x square minus 2 x plus 10 and find out the diffusion current. So, again you find a gradient ok. So, you find a gradient 2 x minus 2 but then you have to find the diffusion current. So, then J N diffusion is again Q D into 2 x minus 2. So, that is your diffusion current simple. So, basically you need to know the gradient from some way or other you can have the diffusion profile or you can have the carrier concentration at two different points or you can have some expression which is giving you the diffusion profile. Once you do that you can find out the diffusion current ok. So, now when there is a electric field as well as there is a concentration gradient you will have 4 component of current right. In the first case there is only gradient there is no concentration difference no gradient. So, therefore, only drift current in the second case this example there is only diffusion current because there is only there is no electric field electric field is 0. But there may be example when both diffusion when there is a electric field as well as the concentration gradient the current component will be due to both concentration gradient of both diffusion and drift. Fine it is clear the diffusion problems you can find out again the diffusion for the electrons and holes. So, we can actually create more complicated situations in which there is a concentration gradient there is electric field and this there are the minority carrier concentration which is not very small minority carrier concentration can also be large and things like that. So, depending on the situation you take appropriate approximations and then solve the problem fine depending the situation take approximate take appropriate approximation and then you solve the problem. You need to do lot of practice by the way before get used to it fine what is next? So, this problem is about absorption a beam of photons of 500 nanometer wavelength is falling on a piece of indirect band gap semiconductor the absorption coefficient of this wavelength of photons is 10 power 4 per centimeter. So, what is the maximum distance the photon would be able to travel in the semiconductor before getting absorbed? Yeah it is very simple. So, the probability of interaction of a photon with a semiconductor depends on its wavelength or energy, but it does not depend on intensity. Mean now I have given you only one photon is incident I mean photon of particular wavelength is incident means you can also calculate the energy for the particular wavelength e is h c by lambda. So, if photon of this energy is incident. So, it goes through it goes through some distance before it gets absorbed in the solar cell or in a material. So, that distance you can estimate by it is inversely proportional to 1 by absorption coefficient. So, assume this as a semiconductor if pi is the flux that is incident on the semiconductor. So, by flux I mean the number of photons per unit area per second. So, these may number of photons per unit area per second or incident let at the surface the photon density the flux is pi naught. So, as it goes through the semiconductor it will be absorbed and it gets reduced. So, it reduces in the form of pi naught e power minus alpha x as we go as we go through the distance x increases and the pi naught value decreases the pi value decreases mean all these photons are observed in the before the travel a distance of L. So, when x equal to L when x equal to 1 by alpha which is nothing, but L which I have shown. So, then pi naught will be pi will be pi naught e power minus 1 or pi naught by e times the flux is reduced by e times. That means, pi naught minus 1 by pi naught minus the rest of the flux pi naught minus pi naught by e is observed in this distance L or x equal to 1 by alpha. Am I clear or basically one more point I want to make clear absorption depends on absorption coefficient. So, I mean to find to what depth a photon travels we need to consider the absorption coefficient and it depends it is property of the material. So, the another important mechanism that happens in solar cells is recombination. So, the question is like this the minority carrier concentration in an n type material is about 10 power 3 centimeter cube due to the light incident on the material the minority carrier concentration has increased to 10 power 16. If the excess minority carrier they travel a distance of 2 micrometer in 2 microseconds before recombining calculate the recombination rate of minority carriers capture coefficient absorption coefficient it is not that complex as light falls electron hole pairs will be generated. So, initially the minority carrier concentration is 10 raise to 3 per centimeter cube when light has when light is falling the concentration has increased to 10 raise to 16 per centimeter cube. So, this excess concentration will not remain like that it will get recombined and what decides this recombination. So, it basically depends on excess carrier concentration the more the excess concentration the more the recombination. So, I will show you the equation. So, the basic equation recombination rate is excess carrier concentration by life time. Life time is the time before which the excess carriers get recombined. So, these 10 10 raise to 16 electrons per centimeter cube they get recombined after 2 microseconds. So, they travel till 2 microseconds and then recombine. So, the recombination rate can be calculated using this. So, you might be knowing that recombination is one of the most degradation mechanism that happens in a solar cell. We do not want it to happen, but it happens but somehow we try to avoid it so that we can improve the efficiency. So, these excess carriers that are generated because of light incident they have to be collected as soon as possible before they get recombined. But in the means of collecting they have to travel through the semiconductor and they cannot travel more than 2 microseconds. So, they get recombined before they get recombined we need to collect them. So, we can improve the recombination performance if you have more life time. And the another term which can be used for this recombination is diffusion length. So, diffusion length is the distance the distance the carrier can travel before it get recombined. So, both this life time and diffusion length they are the measures of the performance of the device. So, the more the life time the more the diffusion length the better the better will be the performance of the cell or efficiency of the solar cell. This problem is about short circuit current density or the current density that will be produced when light is incident on a solar cell. So, when light is incident you are given that the generation is 10 raise to 20 electron hole space per sorry it is not odd terms. So, carrier generation is 10 raise to 20 electron hole space per centimeter cube per second. So, because of these generation will be having these electron hole space are generated and they need to be collected before they get recombined. So, if we can collect them without getting recombined we will get all the current, but as shown by this equation. So, diffusion length mean l n and l p play important role in the performance of the device. The carriers all the carriers that are within one diffusion length will only contribute to light generated current. So, you have equation q g l n plus l p plus it is one of the important parameters for solar cell and the another important parameter is open circuit voltage. So, the expression for that is so you have already got some light generated current density assume j naught which is called reverse saturation current as 10 raise to minus 13 ampere per centimeter square and you can find open circuit voltage for that particular light generated current. So, please go on finding please find open circuit voltage for this generation and this reverse saturation current and k T by q you can assume as 2 6. So, on the units of k T by q are it is it is volts k T is electron volts divided by charge is volts, ln j l by j naught plus 1. Another important parameter for solar cell is fill factor. So, it indicates the amount of series resistance and shunt resistance that are happening that are occurring in the device. So, once we know open circuit voltage short circuit current density and the fill factor we can proceed finding efficiency of the solar cell, open circuit voltage is given short circuit current is also given and fill factor is given as 75 percent can find efficiency from this equation mean as for the earlier problem. 10 is the first value light generated value which we have calculated. Short circuit current density is same as j l light generated current. So, we have already calculated VOC this as this is some measurement generally according to efficiency output by input is our formula. So, input here is the light incident. So, it is the power in the incident light or power radiation. So, it is as it is mentioned it is standard testing conditions. So, we can take it 1000 watt per meter square. Similarly, we have units of incident radiation as 100 milli watt per centimeter square. So, we need to find efficiency we need to use short circuit current density. So, that we can have consistency in units, VOC is in volts and JAC if we consider that in milli ampere per centimeter square and fill factor is not having any units and the incident radiation is having units of milli watt per centimeter square. So, to maintain consistency we need to use short circuit current density instead of short circuit current. So, assume an area of 1 centimeter square. So, this is so as we have milli in both numerator and denominator we can neglect it or if we take we should take in both numerator and denominator. 1000 is watt per meter square when it comes to milli watt per centimeter square. So, it comes to be 100. Yeah, maintain the consistency of units. If you take meter square you should also take short circuit current density in meter square per meter square then you can get this value. So, this is the final so you should tell approximately 12 percent. So, we are about to close this session then we shall leave. Thank you. Thank you very much.