 Let us start with a brief review of last class. Last class we looked at a PN junction under bias. So we applied both a forward and a reverse bias and looked at the IV characteristics. We found that a PN junction is essentially a rectifier so that if you were to plot I versus V, this is V. The first quadrant is the forward bias, this one is the reverse. We found that in the case of the forward bias, the current goes exponentially as the voltage. Typical value of current is around milliamps. While if you have a reverse bias, we have a really small current that is of the order of nano amps or micro amps and that is really a constant. We also wrote down the expression for the current in the case of a PN junction. So the expression is J is equal to some constant J S naught which is your reverse saturation current times exponential EV over KT minus 1. So this is in the case of forward bias. J S naught is the reverse saturation current. This is equal to Ni square over E, E DH over LH and D plus DE over LE and A. So DH and DE are the diffusion coefficients for the minority carriers within the P and the N region and LH and LE are the diffusion lengths. So today we are going to start with an example in order to calculate some of these values of the current in forward bias and also the reverse saturation current. But before we do that, I want to see how this reverse saturation current will change if you change the material. So let me write this expression for J S naught one more time. J S naught is Ni square over E DH over LH and D plus DE over LE and A. Now Ni square which is the intrinsic carrier concentration is a material property. We know that Ni square is nothing but NC, NV exponential minus EG over KT. We can substitute for Ni square in this expression so that J, which is nothing but J S naught exponential EV over KT minus 1. So I am going to substitute the J S naught here and instead of Ni square, I am going to replace it with NC, NV exponential minus EG over KT. Just for the sake of argument, we will find out that it is true. EV over KT is usually much greater than 1 so that I can ignore this term the minus 1 here. I am also going to introduce another term VG which is nothing but the band gap divided by E. So if you do that, J becomes equal to E DH over LH and D plus E DE over LE NA, NC, NV exponential E times V minus VG over KT. So V here is the external potential that you apply during forward bias. VG is nothing but the band gap divided by E. So if you were to plot the current versus voltage for different semiconductors, we find that if you want a given current, the voltage will be higher if the band gap is higher. So let me plot I versus V for three materials. So we look at germanium, silicon and gallium arsenide. So in terms of band gap EG, germanium has a smaller band gap than silicon which is smaller than gallium arsenide. Some typical values, we know germanium is around 0.7 electron volts. This is in EV, silicon is 1.1, gallium arsenide is 1.43. So my current will be in milliamps. It is the voltage 0. So we find that the curve for germanium comes first, then you have silicon and finally you have gallium arsenide. So germanium, silicon, gallium arsenide. So that for a given value of current, so let us say I want the current to be 0.1 milliamps. The voltage is lowest for germanium, it is higher for silicon and it is even more higher for gallium arsenide and this is because you have VG which is the band gap in that expression. So let us now go ahead and look at an example of a p-n junction in silicon and calculate some values for the reverse saturation current and also the current through the p-n junction in forward bias. So we are going to start with the p-n junction that we have looked at before. We have Na is 10 to the 17 and we have Nd is 10 to the 16th. The material is silicon so the intrinsic carrier concentration Ni is 10 to the 10. So we can calculate the contact potential in this p-n junction, contact potential V0 and we did this last class is nothing but 0.775 volts. So we want to know what the current is. So we want to know the value of J when we apply a forward bias and let me take the value of the voltage to be 0.5 volts. So let us write down the p side and then N side. So the first thing we want to know is how many excess carriers are injected because of the forward bias. So from last class if you remember the current in the case of a p-n junction is due to the minority carriers so that you have electrons that are injected into the p side. You have holes that are injected into the N side that causes the current. If you write down the values ppo is nothing but Na is 10 to the 17th. The concentration of electrons will be just Ni square over Na so that is 10 to the 3. Same way we can write for the N side. The formula for the excess carriers so that p-n of 0 is nothing but ppo exponential minus e V0 minus V over kT. So V0 here is the contact potential. V is the forward bias potential that is 0.5. So pno is the excess carriers. So if you substitute the numbers and evaluate pno at 0 turns out to be 2.4 times 10 to the 12 per centimeter square per centimeter cube. So this number is much greater than the equilibrium concentration of holes on the N side which is here. So this is much greater. Similarly we can calculate the excess electrons on the p side. This is equal to 2.4. So these are the excess electrons and the excess holes that are injected due to the forward bias. Now these are still minority carriers. So ultimately they will diffuse through the material. They will recombine with the majority carriers and get eliminated. So in order to calculate the diffusion length we need to know the diffusion coefficient. So let me first write down the mobility. So mu e that is the mobility of the electron mu h which is the mobility of the hole. So mobility in the case of a semiconductor usually goes down with increasing doping concentration. So there are standard tables from which we can get these values of mu e and mu h as a function of the doping concentration. So once we know mu we can calculate the diffusion coefficient thing but k t mu e over e. So if I substitute all the values this gives me the diffusion coefficient of the electrons. So 3.1 centimeter square per second can also get the diffusion coefficient for the holes. Now if you want to find the diffusion length we should also know how far or how long these minority carriers can travel before they recombine. So we need to know the carrier lifetimes. So let me take the values tau h is 417 nanoseconds tau e to be 5 nanoseconds. So these are again values that are known. Typically these will also depend upon the doping concentration. So tau h is the lifetime of the holes that are traveling through the n side of the p n junction. Tau e is for the electrons that are traveling through the p side. So we can then calculate the diffusion length. So we can substitute the values we have d h here we have tau h here. So this is 21.8 10 to the minus 3 I am sorry 10 to the minus 4 centimeters or 21.8 micrometers. L e is nothing but d e tau e which is 1.24 times 10 to the minus 4 centimeters 1.24 micrometers. So we have all the values that we need for calculating the reverse saturation current and the current during forward bias. So J s naught we can calculate. So you have calculated all these values the diffusion coefficients and the diffusion length. So we can substitute the numbers and evaluate J s naught. So J s naught turns out 11 ampere per centimeter square. So this is the reverse saturation current. So this is the current that will be flowing through if I have reverse bias the p n junction. So the current during forward bias is nothing but J s naught times exponential p v over k t minus 1. The voltage that we applied is 0.5 volts. So J comes out to be 3.03 times 10 to the minus 3 amperes per centimeter square. So this is the forward bias current. So earlier we said that a p n junction is a rectifier and we can actually see that it is because J is much greater than J s naught. In these calculations we have assumed that the length of the device is larger than the diffusion length. So let L h and L e be the lengths. So this refers to the physical lengths of the p and n side. So we have assumed that L h and L e are larger than the diffusion lengths. So this kind of a diode is called a long diode. So that the diffusion lengths are what goes into this equation. If the physical length is actually shorter if L h and L e are smaller than the diffusion lengths then it is called a short diode. The only difference in the calculation is that in the equation for the reverse saturation current instead of the diffusion lengths in the case of a short diode we will put the physical lengths of the p and the n side. So let us again look at the IV characteristics but let us look at the reverse bias side. We said that in the case of a p n junction in reverse bias the current is a constant does not depend upon voltage. So we said that the current is a constant which is equal to the reverse saturation current but it turns out at really large voltage values the diode breaks down so that you have a large reverse current flowing through the material. So this voltage where this happens is called your breakdown voltage this happens in reverse bias. So when this happens we said that the p n junction has broken down and there are two mechanisms for this. One is your avalanche breakdown. So this occurs for p n junction with low doping so you have low doping concentrations on the p and the n side so that you have a wide depletion region. So if I were to draw this p side that is my n side this is under reverse bias. So in this particular case an electron that is being accelerated by the field can essentially interact with a silicon atom and because it has sufficient high energy because of the large external potential that is applied that electron can ionize the silicon atom and produce more electrons. This in turn can interact with another silicon atom so that you have an avalanche of electrons that are produced. So this effect where the electron hits a silicon atom and ionizes it is called impact ionization and the effect of this is that you have a large current. So one mechanism of breakdown is called the avalanche breakdown occurs at low doping concentrations. Another mechanism of breakdown is called the Zener breakdown. So the Zener breakdown occurs at large doping values. In this particular case the depletion region is thin or that is narrow. So if you were to draw the energy diagram for this we just draw slightly. So this is my p side that is my n side these are the Fermi levels. So we are in reverse bias so that there is a large barrier but because your depletion width is small you can have electrons tunneling from the p to the n. So this electron tunneling is called the Zener effect and because of that you have a large current so you have a breakdown of the junction. This leads to so we have looked at a p n junction first in equilibrium and then in the case of a bias both forward and reverse you also looked at the two breakdown mechanisms that are possible in reverse bias. So far in the p n junctions you have considered you have considered the material to be the same. So you have the p and the n type and they are both the same material. So they could be silicon or germanium or gallium arsenide but the material is same. The next thing we are going to look at briefly is what happens if you have two different materials so that you have a hetero structure. So now you have a hetero junction so that you have a junction formed between p and n type of different materials. Once again when we have such a hetero junction we are going to assume that you have an ideal junction with no defects. So this imposes a restriction on the types of materials that we can choose. In order to have an ideal junction with no defects we must have lattice matching between the two materials or if you want to put it in another way the mismatch must be minimal. So depending upon the degree of mismatch we can control the thickness of the second layer on the first. So if you look at epitaxial growth the case of epitaxial growth the layer you are trying to grow has the same lattice constant as that of the substrate. So that if there is a lattice mismatch there is an inherent strain in the material. So the thickness of the layer you are trying to grow is inversely proportional to the lattice mismatch. So more the mismatch then the thinner the layer you can grow ultimately if your mismatch is large you are just going to have a large number of defects at the interface. So if A e is the lattice constant of the epitaxial layer A s is the lattice constant of the substrate then we define mismatch delta A e minus A s divided by A e. So this is the mismatch. So the thickness of the epitaxial layer that you can grow T c is proportional to A e over 2 delta. So this is an approximate expression. So this is equal to A e square over 2 A e minus A s. So let us take an example of silicon and germanium. So let us say I am trying to grow a germanium layer on silicon. The lattice constant for silicon A s is 5.43 for germanium. So that is my epitaxial layer is 5.66. In this particular case the mismatch delta if you try to put it in percentage is 4 percent. So it is nothing but A e minus A s divided by A e. So the critical layer that you can grow the thickness if you use the formula A e square this works out to around 7 nanometers. Which means you can grow a layer of germanium on silicon up to 7 nanometers there will be some inherent strain in the germanium but if you go beyond that you are going to form defects in the material. The case of germanium and silicon you actually have island formation in this kind of growth is called a Stransky-Krastanov growth. Instead of germanium silicon let us say I have gallium arsenide growing on aluminum gallium arsenide or the other way round. In this particular case the lattice constants are much closer. So this is 5.65 this is 5.66. So here the lattice mismatch delta is much smaller it is typically around 0.2 percent silicon germanium it was 4 percent. So it is easier to grow epitaxial layers here. So in the case of a hetero junction we want to choose two materials with different band gaps. We want different band gaps because we want to exploit this difference for some interesting electronic properties but at the same time we want a good lattice match. So let us now look at a band diagram in the case of a p-n hetero junction. So I am going to consider a p-n junction. So the first let me draw the n side and I am going to say that the n side has a higher band gap than the p side. This is my n side. This dotted line represents vacuum. This is the conduction band that is the valence band and this is the fermi level. So I will just call this put a subscript 1 so that this is material 1. So we can define an electron affinity. Electron affinity is nothing but the energy from the bottom of the conduction band to the vacuum level can also define my work function. This is going to form a junction with a p-type material which has a smaller band gap. This is the p side. So this is EC2, EV2, this is EF2 and once again you can write down values Sky2 and the work function. So now I am forming a p-n junction. So I have EG1 greater than EG2. I have the electron affinity going the other way and the work function also going the other way. We can also define an energy gap between the conduction bands of the two materials. So that is delta EC. You can also define an energy gap between the valence band delta EV. So we want to put together this p-n junction. So the first rule is at equilibrium the fermi levels must line up. So let me draw the p-n junction. So the fermi levels must line up. I have EF1 and EF2. So I have material 1 which is my n-type semiconductor. This is the material with the wider band gap and then I have my p-type material with the smaller band gap. So if you look at this p-n junction you see that electrons go from the n to the p-side. So that there is a net positive charge on the n-side. There is a net negative charge on the p-side. So the electric field goes from n to p. This is the same concept in a regular p-n junction and we know that bands bend up in the direction of the electric field. So the bands have to bend up on the n-side and the bands have to bend down on the p-side. So we also said that there is a difference between the energies of the conduction bands. So a delta EC and a difference in the energies of the valence band delta EV. So when the bands bend we must make sure that that difference is preserved. So that this gap is delta EV. Let me just extend this and this gap is delta EC. So now if I join these two we just we can get the p-n junction under equilibrium. So if you look at this, this is different from how a p-n junction would look. You have the same material. Specifically if you looked at the conduction band there is a region where there is an energy minimum. So that electrons in the p-side can essentially accumulate in a region near the junction. There is a region of energy minimum which means electrons can accumulate near that region. Also the energy barriers are different for the conduction band and the valence band. So the barriers are different for the electrons in holes. So this can again affect the conductivity of the heterojunction. So heterojunctions have some important properties when we look at optical properties because of the fact that you can have electron accumulation. If instead of a p-n junction where the n has a higher band gap, if you choose a p-n junction where p has the higher band gap, you will see accumulation of holes near the junction. So with this we are done with p-n junctions. So p-n junctions are an example where we have an interface between two semiconductors. So they could be the p and the n of the same material or they could be the p and the n of the different material. In next class we are going to look at devices where we have more than one junction. So an example of such a device is a transistor. So the next class we are going to start looking at transistors.