 Welcome to the first lecture of this course, electronic materials, devices and fabrication. In this course, we will look at semiconductor materials, we will study their properties, mostly the electronic properties. We will then use these semiconductor materials to form devices. So, these devices could be made from two semiconductors, they could be made from a semiconductor and a metal or even a semiconductor, an insulator and a metal. We will also look at devices for some optical applications like say LEDs and solar cells. Finally, we will look at the current micro fabrication industry, where we will form integrated circuits made of these devices from semiconductors. Towards the end of the course, we will also look at some alternate fabrication techniques, where we use a bottom up approach to fabricate devices, rather than the top down approach that is currently followed. When we think of electronic materials, the first property that comes to mind is resistance. Resistance or resistivity, if you will, can be used in order to separate materials into three categories. The first one is conductors, then you have semiconductors and then finally, you have insulators. Since, we will be mostly dealing with metals as conductors, I will also write them as just metals. When we think of resistance, the unit of resistance is ohm and the symbol is omega. Resistance is related to both the voltage and the current by what is known as Ohm's law. Ohm's law states the voltage is nothing but the current times the resistance. Another way of writing it is resistance is voltage over current. The unit of voltage is just Ohm's. The unit of current is ampere. Now, resistance depends upon the dimensions of the sample. So, if you have a thinner material or if you have a wire with a smaller diameter, then the resistance is higher. So, we need to think of a parameter that does not depend upon the dimensions of the sample and this parameter is called resistivity. Resistivity is typically given as rho and it is related to the resistance by where R is the resistance, A is the cross sectional area and L is the length of the sample. The unit of resistivity is Ohm meter. You can also have Ohm centimeter. 1 over the resistivity is called conductivity. The unit of conductivity is nothing but Ohm inverse and meter inverse. So, we have introduced two concepts, right. One is the resistivity, the other is the conductivity. We also said we have three types of materials, conductors, semiconductors and insulators and the difference between them is because of the difference is the resistivity values. So, let us look at some typical values for resistivity for all these three types. Let us start with metals or conductors. In typical examples of metals, copper, gold, platinum, silver, what are some of their typical resistivities? So, rho unit is Ohm meter. In the case of copper, it is 15.7 times 10 to the minus 9. For gold, it is 22.8. 10 to the minus 9 is common for all four. Platinum is slightly higher, it is 98. Silver is actually a better conductor than copper, but silver is also expensive. So, when we look at metals, we have conductivity or we have resistivity of around 10 to the minus 9 Ohm meter. Let us next look at semiconductors. Some examples of semiconductors, germanium, silicon, zinc oxide, gallium arsenide, some of the typical values of resistivity. Silicon is around 0.1 to 10 to the 3. Zinc oxide is 10 to the minus 2 to the minus 4. Slower, gallium arsenide minus 6 to minus 2. In all of these cases, I have given a range of values. The actual resistivity or the conductivity depends upon the impurity level and also the crystal in state, whether you have a single crystal or a poly crystal. So these are some of the values for resistivity in the case of semiconductors. What about insulators? Examples of some typical insulators, you could say wood, water. We say it is deionized water. So there are no dissolved salts, glass and quartz. These are some typical values. So 10 to the 5 in the case of glass, the resistivity is around 10 to the 10, 10 to the 14 depending upon if you have some doping agents in glass or not or some impurities. In case of quartz, it is even higher. So what we essentially see is a wide range of resistivity depending upon whether you are a metal or you have a semiconductor or you have an insulator. So the question is how to understand this difference in these values. In order to do that, we need to look at the band gap of the material and how the band gap evolves. Our focus will be mainly on metals and semiconductors, but whatever concepts we develop, we can equally apply them to insulators. So let us go ahead and look at how band gap evolves starting with a metal. When you have an individual atom, the energy levels are usually sharp and distinct. The question really is what happens when all these atoms come together to form a solid? So let us start with the simplest atom that we know. That is the hydrogen atom. Hydrogen atom has one electron and it has one proton in the nucleus. The one electron is in the vocation and the atomic configuration for hydrogen is 1s1. What happens when we have two hydrogen atoms come together? When we have two hydrogen atoms, you have two 1s atomic orbitals that come together. These interact and give you two molecular orbitals. So your two hydrogen atoms have two atomic orbitals which give you two molecular orbitals. For sake of brevity, I will call atomic orbitals as AO and molecular orbitals as MO. Let us denote these atomic orbitals by the symbol psi of 1s. 1s refers to the fact that you have the electron in the 1s shell. In quantum mechanics, psi represents the wave function of the electron. If you have two hydrogen atoms A and B, you will have two atomic orbitals. Let us call them psi 1sA and psi 1sB. Now in order to form molecular orbitals, these two wave functions psi 1sA and psi 1sB need to be put together and there are two ways of doing it. You could add both of them. This is called a bonding orbital and it is denoted by sigma. You can subtract them and this is called the anti-bonding orbital. In terms of energy, the bonding orbital sigma has a higher energy than the anti-bonding orbital and I am going to denote this as sigma star. So I will call this sigma star. We can also depict this pictorially. So let us take two hydrogen atoms. So at the center, you have the nucleus and outside you have the electron A and B. Each of them has an atomic orbital, so an atomic wave function. So this one is psi 1s of A, which is the atomic orbital for A and psi 1s for B. In both of these cases, I have chosen an exponential function for the atomic orbitals. There is a way of showing that it is actually an exponential function but we would not go into it. As far as we are concerned, we have an orbital that is exponentially decreasing away from the nucleus and we will keep it at that. So now we have two atomic orbitals. We can either add them or subtract them. If you add them, the resultant wave function looks something like this. So you have a minimum between the two atoms and then you have a maximum at the two nucleus. So this is your bonding orbital. We will call it sigma. You can also subtract them, in which case this is the anti-bonding orbital that is sigma star. If you look at it, sigma does not go to 0 at all. It goes to a minimum between the two atoms. On the other hand, sigma star goes to a 0 between the two atoms and this is called a node. So the sigma has zero nodes and the sigma star has one node. So we say that the sigma has a lower energy than sigma star. The bonding orbital is more stable than the anti-bonding. Another way to represent this compactly is to consider your two atomic orbitals A and B. Each has one electron. When these two merge, they form both a bonding and an anti-bonding orbital. So your bonding orbital has a lower energy, that is your sigma. Your anti-bonding orbital has a higher energy, that is sigma star. Each orbital can take two electrons of opposite spin. So both electrons go to the sigma. Thus, when you form a hydrogen molecule from two hydrogen atoms, the overall energy of the system is lower. So it is favorable to form H2 when two hydrogen atoms come together. You can also show this diagram by means of an energy versus bond length diagram. In that particular plot, you will have energy on the y-axis and we will have bond length on the x-axis. So energy is bond length. When the two hydrogen atoms are far apart, typically we say that the hydrogen atoms are at infinity, then what we have is simply the atomic orbitals. So I will show them here, which is the 1S orbital and it has one electron. Then you start to bring your hydrogen atoms together. So as you start to come together, they will see the effect of each other. Ultimately, the atomic orbitals will mix and give you your molecular orbitals sigma and sigma star. So for this, we will have some equilibrium distance. So let me call this equilibrium distance. So it is the equilibrium distance for the two hydrogen atoms in the molecule. So you now have the atoms come together. They form two molecular orbitals. One is sigma. The other is sigma star. So sigma is the lower energy, which is the bonding. Sigma star has a higher energy that is anti-bonding and both electrons go into sigma. So this is the picture as far as hydrogen is concerned. We can use the same diagram to explain why helium 2 will not form. In the case of helium, it is a noble gas. It has two electrons in the 1S shell. If two helium atoms come together, so you have two electrons here. You have two electrons here. Once again, both the 1X, you can think of mixing and forming a bonding and an anti-bonding. So this is your sigma. This is your sigma star. So now we have a total of four electrons. Two electrons will go into sigma, but two electrons also go into the anti-bonding that is sigma star. So the overall energy of helium 2 is higher than just two helium atoms. So that is why you will not have helium 2. This picture is good enough as long as we have two atoms. Now what if I had three hydrogen atoms? Let us look at that. So let us say we have three hydrogen atoms A, B and C. So your atomic orbitals are psi 1S A, psi 1S B, psi 1S C. When we have two atoms, we said they form two molecules. Now when we have three atoms, you are going to get three molecular orbitals. There are three ways of obtaining these molecular orbitals. The first one is to just add all three. This is the lowest energy. You can also do one subtraction. This is the lowest energy. This has the highest energy and then you have one more in between them. So in this particular case, the contribution from the second hydrogen atom is neglected because it is times 0. So all you have is psi 1S A and minus psi 1S C. So let me just write that down. So psi 2 is nothing but psi 1S A minus psi 1S C. You can depict this pictorially again by taking three hydrogen atoms. So in the first case, you are adding all three orbitals together. So these are A, B and C. This is psi 1. So you are adding all three. This one, the function does not go to 0 anywhere. So you have zero nodes. In the case of psi 2, we said that the function goes to 0 at atom B. That is why there was only contribution from A and C. This is psi 2. So it has one node. And in the case of psi 3, we have two nodes. So this is the lowest energy with zero nodes. Then you have one node which has a higher energy. Finally, psi 3 that has the highest energy. For this case also, we can draw an energy versus bond length diagram. If we do the energy versus bond length for three hydrogen atoms, what we have are three orbitals, psi 1 with the lowest, psi 3 with the highest and then psi 2 in the middle. Now we have three electrons. Two electrons will go to psi 1 and one electron will go to psi 2. So this is what will happen if you have three hydrogen atoms. Now if you have four hydrogen atoms, you will have four molecular orbitals. You will have four lines here. If you have five, you will have five lines here. So thus, as the number of atoms increases, you will have more lines on your energy versus bond length diagram. So ultimately, we can say, if you have n atoms, it could be n hydrogen atoms, then you will have n molecular orbitals. Each molecular orbital can take two electrons of opposite spin. So 2 times n energy states and the number 2 comes because you can have two electrons of opposite spin. So this is the picture you have developed using hydrogen. We also showed how helium will not form. What if we increase the complexity a bit further and then look at lithium? When we have lithium, lithium has three electrons. So the electronic configuration is 1S2, 2S1. The 1S shell in lithium is an inner shell. So this is an inner shell. The inner shell is usually not involved in bonding. So we would ignore the effect of the inner shell. So what you are interested in is the 2S1 electron. So this again has one electron in the 2S shell. So this is similar to a hydrogen atom. So we can use the same argument for lithium as in the case of hydrogen. So if you have n lithium atoms taking only the 2S electron into account, you will have n atomic orbitals which will give you 2n energy states. Again why 2? Because each energy state or each molecular orbital can have two electrons. Once again, we can draw an energy versus bond length diagram for lithium. So we have energy on the y-axis, bond length on the x-axis. So energy, bond length. Then you have the equilibrium spacing here. This is the equilibrium spacing of the lithium atoms when they come together to form a solid. So you have the 1S2 which is an inner shell. It has two electrons. As I said, the inner shell does not take part in bonding. So it remains a straight line. It does not form a band. Now you have the 2S1. If you have n lithium atoms, the 2S1 of all of those lithium atoms come together to form n molecular orbitals. So once again, we can draw this picture where we will have n lines here. In this particular case, I have drawn 1, 2, 3, 6 lines. So in this particular case, n is 6. The spacing between these lines depend upon n and it also depends on the equilibrium distance. So what is a typical value of n? n is usually a very large number. For example, if we take lithium, lithium has an atomic weight of 7 grams per mole. So if you have 7 grams of lithium, you have 1 mole and 1 mole has 6.023 times 10 to the 23 atoms. This is nothing but the Avogadro number. So your typical value of n, if you had 7 grams of material is approximately 6 times 10 to the 23. So if you have 6 times 10 to the 23 lithium atoms coming together to form a solid, you have 6 times 10 to the 23 molecular orbitals. Each of them can take 2 electrons. So you have 10 times or 12 times 10 to the 23 energy states and this is a really large number. So what this means is the spacing between the individual energy states is really small. If you go back to the diagram, the diagram where we drew here and had n equal to 6 and drew 6 lines, if you have n equal to 23, 10 to the 23, you are essentially drawing 10 to the 23 of these lines. So what I am trying to say is for really large ends, these become continuous and what we have is an energy band. So we can redraw the energy band diagram for lithium taking into fact that you no longer have individual molecular orbitals, but you have an energy band. So in this case, your energy is on the y-axis. We reference an energy level called 0, which is the vacuum level. So let me call this the vacuum level and this has the value 0. So everything can be referenced to the vacuum level. So you have an energy band that is half full. So in this case, these are all the full states and these are all the empty states. So how do we arrive at the fact that the energy band is half full? If you go back to the energy versus the bond length diagrams, energy versus bond length, you have lithium which is 2S1, you have a 1S inner shell, you have 2S1. So this is the energy band here and there are a total of 2N states, but each lithium atom will only contribute one electron. So you have only N electrons. So hence only half of this band is full. So we can erase the top half. So you have a situation where you have half of the band that is full, half of the band that is empty. This is the same thing I have shown here. You have half a band that is full and half a band that is empty. This energy state that separates the full and the empty regions, I have labeled it as EF. EF is called the Fermi energy. Now this is a very important term. The Fermi energy at 0 Kelvin separates the full and empty states. So Fermi energy separates the full and empty states. This is the most often used working definition for Fermi energy. When we come to semiconductors, we will try to modify this definition a little, but as far as you are dealing with metals, we will use this. So let us just put some numbers here. So in the case of lithium, this energy is 0. The Fermi energy is at minus 2.5EV, which means you need 2.5EV of energy in order to excite an electron from the Fermi level to the vacuum level. EB is the bottom of the valence band. In the case of lithium, EB has a value of minus 7.2EV. So the units of energy that I use here EV is called electron volts. EV is called electron volts is related to the energy in joules by the expression 1.6 times 10 to the minus 19 joules. So 1EV is 1.6 times 10 to the minus 19 joules. This is the picture as far as lithium is concerned. So in the case of lithium, you have a 2S shell that is half full. So you also have a band that is half full. What about magnesium or beryllium? So beryllium has an atomic number of 4. So it has an electronic configuration 1S2, 2S2. Magnesium is below beryllium in the periodic table, has an electronic configuration 1S2, 2S2, 2P6 and 3S2. Once again for both of these elements, these are the inner shell. So your outer shell is essentially an S2 configuration. This configuration is the same as in the case of helium. We saw that helium has a 1S2 configuration. So the question is how will beryllium or magnesium form metals or form solid because we know they do and helium does not. The reason is in the case of both beryllium and magnesium, you also have the P shells and the P shells are empty. So here you have a 2P0. In the case of magnesium, you have a 3P0. So what you have when you form a solid is a mixing between the empty 2P0 and the full 2S2. In the case of magnesium, you have a mixing between the empty 3P0 and the full 3S2 and that is the reason why you have an unfilled state and why these are metals. So this once again, we can show this pictorially using the energy versus the bond distance diagram. So let us use the example for beryllium. You have energy on the y-axis and then bond length on the x-axis and we have some equilibrium distance. The 1S shell is an inner shell that will not take part in bonding. You have a 2S2 shell that is full and you have a 2P0 that is empty. So when you draw the bond diagram, this is your 2S shell which you also have the 2P0 shell. So what you have is a mixing of the 2S and the 2P so that this whole thing which is a mixture of the 2S and the 2P forms a band. Then you can fill in the electrons in this band so that these are all the filled states and these are all the empty states and the energy separating the filled and the empty states is your Fermi energy. So that is the reason why beryllium or magnesium even though they have a full 2S2, electronic configuration are still considered metals because they have mixing between the S and the P so that now they have a band that is half full and half empty. This band which is caused by interaction of the electrons in the outermost shell is called the valence band. It is caused by the interaction between the outer shell electrons. There is also a conduction band. In the case of metals, the valence band and the conduction band are both the same. So in the case of metals, you have both valence and a conduction band and they are both the same. In the case of semiconductors, you will find that the valence and the conduction band are different and there is a gap between them. This is what we call the band gap. So in today's class, we have looked at how metals form. So how you can start with individual atoms and then these atoms come together to form molecular orbitals. The case of metals, you find that these molecular orbitals ultimately form a band and these bands are not completely full. There are always some empty states are available and this is why metals are such good conductors because these electrons which are there in these bands can acquire energy and go to these empty states and thus can move freely within the metal and this makes metals very good conductors. In next class, we will start to look at semiconductors and we will see how we have a band gap in semiconductors. We will also look at different classifications of semiconductor.