Hello, everyone. So far we learned PN Junction. Our objective of this lecture is learning the MOSFET devices. Too learn MOSFET, after the p-n junction, we need to study two more thing. First, metal semiconductor junction, then we're going to learn MOS capacitor, then finally, we're learning MOSFET device. In lecture A, we are going to learn metal semiconductor junction. Metal-semiconductor contact, there are two types of contact. One is the ohmic contact, the other is Schottky contact. Ohmic contact is normally formed on metal to n-plus or p-plus highly doped region in semiconductor devices. In this ohmic contact, contact between metal-semiconductor are minimum resistance, therefore, linear IV characteristic in both direction, current can easily flow better to semiconductor to metal. These are the MOSFET devices. Here they oxide this semiconductor MOS capacitors, and then here source and drain region. This source/drain regions metal to semiconductor contact here are ohmic contact, so that current can freely move in here. Schottky diode or Schottky contact is current flow easily in one direction but cannot flow the other direction like a p-n diode. Schottky contact is used in a fast-switching rectifying contact. This is the example of a Schottky contact, this is the MOSFET device, metal-semiconductor field effect transistor. You have a source and drain region with the ohmic contact, current is freely moving to source to drain to semiconductor, though in the gate region, instead of they have a gate oxide, there is no gate oxide but direct contact of the gate material and semiconductor material. However, there formation is the Schottky contact instead of ohmic contact. So current can be throwing in one direction, current cannot flowing on to the other directions. We're going to learn this in chapter 6. First, metal semiconductor, Schottky contact. To form metal to n-type silicon Schottky contact, we have to choose metalwork function is higher than semiconductor work function. Here, what function is defined by the energy required to remove electron in Fermi energy to vacuum level? In semiconductor, it's defined that energy required to remove an electron from the Fermi energy to the vacuum level. So this is the metalwork function and this is the semiconductor work function. Another terminology is widely used in semiconductor electron affinity. Electron affinity is energy from the EC to vacuum. In here, if you making a metal semiconductor contact, their forms built-in potential, which is Pi M minus Pi S. We will discuss this in next try. First, depending on the metal materials, work function of the metal is determined. For example, of the silver, metalwork function is 4.26, aluminum, 4.28, gold is 5.1, platinum, 5.65, titanium, tungsten, etc. Also, depending on the semiconductor material, chi can be defined, for silicon, 4.01, germanium, 4.13, gallium arsenide is 4.17. Now, if you making contact between the metal to n-type semiconductor for the Schottky contact, what's going to happen is that, first, electron in n-type semiconductor energy level higher than metal Fermi energy. So bend diagram is electrostatic potential based on the electrons. So electrons are more stable in lower energy. Therefore, electron in n-type semiconductor will diffuse to the metal side. In n-type semiconductor region, because electrons are diffuse, they form the fixed positive charge of the r sin. This is the depletion region. In contrast, in opposite to the semiconductor metal side, there is electron accumulation. There is charge of positive and negative, then there are internal electric field. Also internal electric field, then the electrostatic potential is changing or bend diagram is bending. How're they bending? Add under the equilibrium fermi energy always to be constant. So the n-type semiconductor bend diagram along with the EF go down to meet Fermi energy is constant, but the Schottky barrier at the contact doesn't change. So this much is going exactly here, and they are here. So in here we call this barrier is Schottky barrier. Then this barrier or bend lowering is called built-in potential. So built-in potential q bi is equal to the q metalwork function minus semiconductor work function. Why? Let's look at that. These area bending is go down. How much they go down? Until fermi energy becomes the equal. So this much they go down. What is the "this much"? They go down this much, which is the built-in potential. Therefore, built-in potential is metalwork function minus semiconductor work function. Schottky barrier, in contrast, is original contact point over here. Therefore, Schottky contact is metalwork function minus chi. Therefore, Schottky barrier is defined like this. As I said in previously, Schottky barrier is exactly same as a p-n junction, so that the current can be flowing in one direction, current is not flowing the other direction. Let's overview the p-n junction. P-n junction is a p-type semiconductor, n-type semiconductor, then electron move to the other side, then there is a fixed positive space charge of the r sin. Same thing for the boron, negative fixed charge in p-type region. If there is charge then by the [inaudible] equation, there will be electric field. They are integration of the charge, therefore, they are one first-order equation and they are negative because here is a positive, negative, electric field is negative. The integration of electric field becomes the electrostatic potential. Because electric field is first-order, the electrostatic potential is second on the equation like this. Our project of the electrostatic potential is bend diagram, which is based on the electron. In here, built-in potential can be expressed like this W, can be expressed like this. [inaudible] n-type Schottky contact is similar. In n-type semiconductor, electron move to the metal side and they become a depletion [inaudible] positive charge, and electron are accumulated in metal. Then this you can think is a highly doped p plus n, p-n junction. When there is charge, then there is electric field. An electric field integration becomes the electrostatic potential. As you can see, as I said, like a force relationship, if you have a very highly doped semiconductor, then less doped region has a much longer depletion region. In other words, very highly doped semiconductor is extremely little depletion width. So therefore p plus region or metal region, there is almost no depletion width because are similar to the highly doped. Therefore, in here, X_p is almost zero. Therefore, built-in potential becomes X_p equals 0 therefore becomes like this. X_p equals 0, but what about the X_n. X_n is in W, you have a very highly doped acceptor then this term becomes the zero. Therefore, X_n becomes like this. So what happen if Schottky barrier under forward bias and reverse bias, it is dissimilar to the PN junction. These are under equilibrium, there's our Schottky barrier, this is built-in potential. Same amount of the electron will go over this barrier following by the Boltzmann distribution and same number of electron go to the other side. In forward bias, which means that you are applying positive voltage to the metal, negative voltage to the n-type. So PN junction forward bias positive voltage is to the p, negative voltage is to the n. So here's the n-type semiconductor. So therefore, you should apply negative voltage to the n for forward bias and positive to the metal. Then if you're applying the negative voltage, then the energy diagram will go up because this is based on electron and then deducing the barrier from the built-in potential. So barriers of a forward bias is deduced from the built-in potential through built-in potential minus forward bias. So huge electrons carriers that has the energy of the deduced barrier which follows a Boltzmann distribution will go to the metal side. Current is flowing in one direction, but Schottky barrier height doesn't change. Very small amount will go, therefore, electron moving from n-type to the metal, current flowing to the metal to n-type semiconductor. For the reverse bias, if you apply reverse bias, then you should apply positive voltage to the n-type semiconductor. Then this the reverse bias, then original built-in potential will be increased from the built-in potential plus the reverse bias, and almost non electron can go over this reverse potential but very limited carrier. Exactly same carrier on the equilibrium forward bias can go over the Schottky barrier, so no current is flowing. As you can see this Schottky barrier and the same condition similar to the PN junction, if you're applying the either forward bias or reverse bias, voltage drop occurs at the junction region of the semiconductor, because the metal side is very conductive and semiconductor junction region is relativity resistive compared to the metal. So those voltage drop occurs at the relatively resistive area which is the junction area.
