 In the first few classes of this course, we have looked at electronic materials. We started with intrinsic semiconductors, so these are pure semiconductors but we found out that things could be a lot more interesting if we added a small amount of dopants to these semiconductors. So the next thing we looked at were extrinsic semiconductors where we could have selectively either more number of electrons or more number of holes. In the next few classes, we are going to focus on electronic devices. In order to form a device, we need to put materials together and when we put materials together we will form junctions. So in this first lecture on the series of electronic devices, we are going to start by looking at junctions. We will first start with looking at a junction between two metals. So we will form a metal-metal junction and we will use this to understand the concepts of band alignment. From there, we will form a junction between a metal and a semiconductor. There are two types possible here and you will consider them. One is your short key junction, the other is called omic and from the metal semiconductor we will go to your semiconductor junction and the most famous of this is your p-n junction. In today's class, we will focus on the metal-metal and the metal semiconductor junction. So the most important rule when a junction is formed is that the Fermi levels must line up at equilibrium. So whenever you have a junction, the Fermi levels must line up and this is at equilibrium. Equilibrium here means there is no external potential applied to the system. So consider the case of two metals. For this particular example, I will choose platinum and molybdenum. So let me just draw platinum. The metals are characterized by having a continuous balance and a conduction band so that there are filled and empty states and EF that is the Fermi energy represents the gap between the filled and the empty state. So this is platinum, these are all filled states, these are all empty states, EF separates the filled and the empty. The distance from the Fermi level to the vacuum level to the topmost is the vacuum level is the work function. I am going to call it phi of platinum and platinum has a work function of 5.36 electron volts. So I am going to form a junction between platinum and then molybdenum. So once again this is EF, this is phi of moly. The work function of molybdenum is 4.2 electron volts. So here we see that platinum has a higher work function than moly. So we are going to form a metal-metal junction by putting platinum and molybdenum together. We will worry later how these junctions are actually formed. Right now we will just say that we formed the junction. So if you think about it molybdenum has a whole bunch of electrons close to the Fermi level and there are a whole bunch of empty states in platinum. So when a junction is formed electrons from molybdenum with the lower work function can move into platinum. So when this happens there will be a net positive charge on molybdenum and because it is losing electrons there will be a net negative charge on platinum because it is gaining electrons which means an electric field will be set up when a junction is formed between platinum and moly. So let me form the junction and I said that whenever you form a junction the Fermi levels must line up. So I will put the Fermi levels at the same energy level. In this picture you can think of the molybdenum levels going down so that the EFs line up or you can think of the platinum going up so that the EFs line up. Either case is okay. So this is platinum and then this is moly. Let me just erase that extra portion here. So this is the work function of platinum. This is the work function of moly. So in this diagram we said that when we form the junction between platinum and molybdenum electrons move from moly to platinum. So when electrons move a net positive charge is created on the molybdenum side and when these electrons move to platinum a net negative charge is created. So because you have this positive charge and negative charge there is an electric field E. The field goes from positive to negative and there is a contact potential. Potential goes from negative to positive. This contact potential V0 is equal to the difference between the work functions. So EV0 let me write it here is nothing but phi of platinum is phi of moly. If you put in the numbers platinum we said was 5.36 moly is 4.2, 1.16 electron volts implies V0 is 1.16 volts. So we can depict the contact potential here which is the difference between the work functions. So this represents EV0. This contact potential here opposes further movement of electrons so that we have a junction that is in equilibrium. So when we have two metals that come together with different work functions and they form a junction once again we do not look into the mechanics of how the junction is formed. We just say that we have a junction that is ideal which means there are no defects. In such a case electrons will go from the metal with the lower work function to the metal with the higher work function. This in turn leads to a contact potential at the surface and this contact potential opposes further motion of electrons. The value of the contact potential depends upon the difference between the work functions of the two metals. So where do we use these metal-metal junctions? One particular example is in the case of thermocouples. To understand that we need to look at something called the C back effect. So consider a piece of metal, one end of it is hot, the other end of it is cold. So your cold could be the room temperature, the hot end could be something that is placed in a furnace at elevated temperature. So if you have two ends, one that is hot and one that is cold, electrons on the hot side have a higher thermal energy because yesterday we saw that thermal energy is equal to 3 over 2 kT which means higher the temperature, higher the thermal energy or higher the velocity. So these electrons will tend to drift towards the cold end so that there will be a net positive charge on the hot side and a net negative charge on the cold side which means there will be an electric field through your metal and there will be a potential. Let me just call it V. This potential is dependent on the temperature difference between the hot and the cold end and a coefficient that is called the C back coefficient. So delta V, let me write this as delta V here which is the potential difference between hot and cold is nothing but S times delta T. The integration is from T 1 which is the temperature of the cold end to T 2 which is the temperature of the hot end and S is called the C back coefficient. Now C back coefficient is essentially a material parameter so if you change the material the value of the coefficient will be different. So how do we use this in order to understand thermocouples? The case of a thermocouple a junction is formed between two dissimilar metals. Give you an example, consider a junction between aluminum and nickel. So this material is aluminum and this is nickel. So you essentially form two junctions at the two ends. So one end of this junction is placed in the hot side where the temperature we want to measure. The other end is cold so typically this could be room temperature. So let us say the temperature of the hot end is T h, temperature in the cold end is T c. Now whenever we have a junction between two dissimilar materials we said that there will be a potential which depends upon the work function. So there will be two potentials here, one set up at the hot end and the other at the cold end but there will also be a potential within the material because of the C back effect and this potential will be different for aluminum and nickel because the C back coefficients are different which means there will be a net potential in the system that we can measure. This potential is usually very small, it is of the order of microvolts and this net potential in the thermocouple so let me call it delta v depends upon the difference in the C back coefficients. So if you have two materials A and B, in this case it is nickel and aluminum, this S A minus S B delta T and the integration goes from the cold end to the hot end. So depending upon the temperature and the difference in the C back coefficients you will have a net potential in the thermocouple. So usually there are tabulated values for this potential for a given pair of materials. This will again depend upon the operating range of these materials. So if you want to find out the temperature of an unknown furnace or an unknown sample by measuring the potential and then using standard tables we can calculate the temperature. So this is an example where metal metal junctions are formed and whenever you have these junctions there will always be a contact potential. Next let us move to a metal semiconductor junction. So we will consider the case of a junction formed between a metal and a semiconductor. So this is useful in the case of when we form devices on a wafer these devices have to be connected to an external circuit and this is usually done by forming metal contacts with these devices. So there we will have metal semiconductor junctions. So in the case of a metal and a semiconductor let phi m be the work function of the metal and phi semiconductor phi semi be the work function of the semiconductor. So there are two options possible one you have phi m you have phi m there is greater than phi semi this type of a junction is called a short key junction. On the other hand you have a situation psi m is less than the work function of the semiconductor this is called an ohmic junction. So we will start by looking at the short key junction first. So a short key junction is 1 psi m greater than phi semi. So let me draw the band diagrams for the metal and the semiconductor separately then we will put them together and draw the band diagram in equilibrium. So we will follow the same procedure that we used when we form two junctions between metals. So I have the metal on my left. So this is my metal it has a work function of psi m this is the Fermi energy the top of the metal is your vacuum level. So let me form a junction with a semiconductor for this example I will take an n type semiconductor but we could use any other material as well. We actually write that below so I can draw the diagram. So in the case of a semiconductor we know we have a valence band we have a conduction band there is a band gap so let me just erase this. So this is the band gap EG we will denote the top of the valence band as EV the bottom as the conduction band as EC. So this is an n type semiconductor so the Fermi level will be closer to the conduction band so let me call it EFN and the distance from the Fermi level to vacuum is your work function so this is the work function of the semiconductor. So now we are going to form a junction between this metal which has a work function which is greater than the work function of the semiconductor. So if you look at this look at this picture you can see that this is an n type semiconductor so there are whole bunch of electrons in the conduction band of the semiconductor there are whole bunch of empty states in the metal so when a junction is formed electrons can move from the semiconductor to the metal and when this happens there is a net positive charge on the semiconductor and there is a net negative charge in the metal. So once again we will have an electric field and then we will have a contact potential and this contact potential will oppose any further motion of electrons. So when we draw the band diagram between a metal and the semiconductor we have to show the bands bending in the case of semiconductor to explain this contact potential. The rule that is to be followed is that band bending goes up in the direction of the electric field so we will see in a minute what this means. So when we form the junction the first thing we are going to do is to line up the Fermi levels and then we are going to bend the bands in the semiconductor and this band bending will go up in the direction of the electric field. So let me put this junction together so the first thing I am going to do is line up the Fermi levels so this is EF on the metal side and EFN on the semiconductor side so we can say that the semiconductor comes down so that the Fermi levels line up this is your metal. Far away from the junction the semiconductor behaves like an n type so you have a conduction band and you have a valence band let me just erase this line and at the junction or near the junction we will find that the bands bend so this whole region is your n type semiconductor. So let us go back to this picture we said that when the junction forms between the metal and the n type semiconductor electrons move from the semiconductor to the metal so there is a net positive charge on the semiconductor side and there is a net negative charge on the metal side which means there is an electric field which goes from the semiconductor to the metal or there is a contact potential. Now in the case of a metal the electron density is of the order of 10 to the 22 in the case of a semiconductor the electron density is usually lower so if we talk about a typical n type semiconductor your electron densities can be around 10 to the 16 to 10 to the 18 per centimeter cube. This means when the electrons move from the semiconductor to the metal they not only move from the surface but they also penetrate a distance within the bulk of the semiconductor. So there is a region within the semiconductor where electrons are lost as they move into the metal and this region is called the depletion region. So let me redraw this diagram and mark the various regions here so we have formed the junction between the metal and the n type semiconductor the junction is at equilibrium. So let me just redraw that figure so this is the metal side and at equilibrium your Fermi levels line up far away from the junction it still behaves like a typical n type semiconductor but closer to the junction because of electron motion you have band bending. So this portion is the metal this side is your n type there is a net positive charge on the semiconductor side and there is a net negative charge. Now because of the difference in electron densities electrons from the semiconductor not only move from the surface but they also move from a certain region within the bulk. So this distance from where electrons from the semiconductor move to the metal is called your depletion region. Another name for the depletion region is called the space charge layer. So we now have an electric field and a contact potential between the metal and the semiconductor. Just like we had in the case of two metals this contact potential depends upon the difference between the work functions. So ev0 is nothing but phi m-phi it is an n type semiconductor so I will just write it as n slash semi if you mark those two values this is phi m and this is phi n. So this contact potential represents the barrier for the electron to move from the semiconductor to the metal. So at equilibrium you have a barrier that is set up that prevents any further motion. There is also a barrier for the electron to go from the metal to the semiconductor. This barrier is called phi b which is called the short key barrier. The short key barrier is given by the work function of the metal minus the electron affinity of the semiconductor. So if the top if the bottom of the conduction band is EC electron affinity is from the bottom of the conduction band to the vacuum level. Another way of writing this is that this is equal to phi m-phi n which is the contact potential plus the energy difference between the bottom of the conduction band and the location of the Fermi level. So if you have a metal and an n type semiconductor junction we have the work function of the metal greater than the work function of the semiconductor and when this junction forms under equilibrium you have a contact potential that prevents further motion of electrons from the n type semiconductor to the metal. So let me just draw this schematically here. So this is the metal side, this is the n type semiconductor side, I will draw this with a dotted line to show the depletion region. So the metal has a net negative charge and there is a net positive charge on the semiconductor. So this is the case when you have this system in equilibrium. Now things become more interesting when we try to bias. So when we try to apply an external potential to this metal semiconductor junction. Now there are two ways of biasing the junction. First is called forward bias and the next is called reverse bias. When you bias a junction the system is no longer in equilibrium. You have an external potential that is applied which means you have electrons or holes that are being injected into the system. So in bias the Fermi levels will no longer line up but will be shifted depending upon the applied potential. So let us look at forward bias first and then we will consider reverse bias. So let us look at a short key junction in forward bias. So once again let me draw the metal and the semiconductor. So in a forward bias the metal is connected to the positive side and the semiconductor is connected to the negative side of an external potential. So V is the external potential, this is positive, this is negative. If we look at the external potential we find that that potential is opposite to the contact potential that is being set up that is set up within the junction. So V0 is the contact potential, V is the external potential and V is opposite V0. So the effect of the external potential is to reduce the total contact potential at the interface and you can show this in the band diagram by shifting the Fermi level of the semiconductor up. So if I want to show the band diagram of a metal semiconductor junction under forward bias this is your metal side EF. So if you have a junction in equilibrium you would show that the Fermi levels line up but now we have a junction where you have a forward bias. Which means the Fermi level of the semiconductor will be shifted up. So this will be EF of the semiconductor if I show the bands, let me just redraw this to show the bands. This is the conduction band, this is the valence band. So this shifting up is proportional to the applied voltage. So the contact potential of this junction now is nothing but E times V0-V. So this contact potential we saw earlier was the barrier for the electrons to move from the semiconductor to the metal. By applying a forward bias we have reduced this barrier. Which means it is easier for the electrons to go from the semiconductor to the metal so we can have a current. And the magnitude of this current will depend upon the magnitude of V. So higher the value of V, lower the contact potential and higher the current. We can write an expression for the current. J represents the current in forward bias, it is nothing but J0 exponential EV over KT-1. V here is the applied potential. So the current depends exponentially on the applied potential and higher the current and higher the potential, higher the current. So this is the case of a metal semiconductor junction, a short key junction in forward bias. So what happens if we apply a reverse bias? In the case of a reverse bias the metal side is connected to a negative and the semiconductor is connected to a positive. So this is the metal, this is the semiconductor, so this is reverse bias. In this particular case the external potential V is in the same direction as the contact potential which means the overall barrier goes up. So in forward bias we showed that the Fermi level of the semiconductor will go up. In the case of a reverse bias the Fermi level of the semiconductor will go down. So if you draw the band diagram for this, this is your metal, this is EF, this is the Fermi level of the semiconductor it has gone down which means the overall barrier has now gone up. So the barrier is EV0 plus V. In the case of reverse bias there is a small current that goes from the metal to the semiconductor and this current which is your reverse current depends upon the short key barrier. Here we can draw an IV diagram for a short key junction and IV diagram is one where we plot current on the y-axis and voltage on the x-axis. So the first quadrant is your forward bias and the last quadrant is your reverse bias. In the case of a forward bias we found that you have a current that exponentially increases as the applied voltage. So this shows the exponential increase. In the case of a reverse bias there is a reverse current which is small and a constant and depends upon the short key barrier. So this we can show like this. So this is the IV characteristics of a short key junction. So in this case you have a short key junction that will conduct in the forward bias because you see a current and will act as an insulator when you have a reverse bias. So a short key junction will act as a rectifier. So let us next consider the omic junction where the work function of the metal is smaller than the work function of the semiconductor. So phi m is smaller than phi of the semi and you will use the same n-type semiconductor as an example. So this is the metal side, this is phi m, this is the semiconductor side. So the valence band, EV is the top of the valence band, EC is the bottom of the conduction band and it is n-type. So EFN is close to the conduction band edge. So in this case it is the reverse of the short key junction. So we have a large number of electrons in the metal side. These electrons can move to the semiconductor. So an electric field will again be set up and the field will be in the direction opposite to that of a short key junction. But because you have electrons going from the metal to the semiconductor, you will have an accumulation region. Remember in the case of short key junction, we had a depletion region where electrons moved away from the semiconductor. Now we have electrons moving in so that you have an accumulation region. We can draw the omic junction in equilibrium. In equilibrium the Fermi levels will line up. So now the bands will bend the other way because the electric field is in the opposite direction. We just redraw this portion to show the band bending. So this is your valence band with EV, this is your conduction band with EC. So now you have bands that bend down because the electric field goes in the other way. So in the case of an omic junction, this is the metal, this is the n-type semiconductor and you have an accumulation region. So this is a region where excess electrons go from the metal to the semiconductor. We can look at the behavior of an omic junction in the case of a forward or a reverse bias and this behavior is entirely different from that of a short key. Now because you have an accumulation region, the behavior of this device under a bias basically acts as a resistor and the resistance is given by the n-type semiconductor. So an omic junction will behave as a resistor under forward or a reverse bias. So the junction will conduct whether you have a forward or a reverse bias and the conductivity is determined by the resistance of the n-type semiconductor. This is opposite to a short key junction which we saw earlier behaves like a rectifier. So that it will only conduct in forward bias but will not conduct during reverse bias or you can say that it has a very small reverse current in reverse bias as a rectifier. So that it only conducts in forward bias. So this difference between these two junctions depends upon the difference between the work functions of the metal and the semiconductor. All the examples you have worked out today were with an n-type semiconductor but we could draw similar band diagrams if we had a p-type material. So we can have similar omic and short key junctions. So today we have seen two types of junction, one between two metals and the other between a metal and a semiconductor. In next class we are going to form junctions between semiconductors and the first one we will look at is the p-n junction.