Hello, everyone. Now, we're starting the Lecture 5. Lecture 5, we're starting of with semiconductor device, which is one of the most important one in a PN Junction. So to start a PN Junction, we first look at the concept of the PN Junction Outline. This is the PN Junction. If you look at the PN Junction, there is a p-type semiconductor and n-type semiconductor. To draw this with a socket diagram showing in here. This means that current can't flow from the p-type to the n-type, and current blocking by n-type to p-type. So current only flow in this P to N directions. Here's a p-type semiconductor, and here's the n-type semiconductor. A lot of holes are existing in p-type region and minority electron is located in p-type region. Same thing for the n-type, a lot of electrons are existing in n-type region, minority-p existing in n-type region. If they make a junction between the p to n, then a lot of holes in p-type region will diffuse to the n-type region. Same thing for the a lot of electron are diffused from the n to p region. There is a depleted region, space charge region is formed. Inside it, there is only fixed charge. For p-type, negative boron, in n-type, positive arson is formed. Therefore, if there is feel of positive charge and negative charge, then there forms electric field in this positive to the negative region. If there is charge in depletion region, then there is electric field formed by the Poisson's equation, which we will learn in a little later. Also, if there is electric field, then integration of electric field is electrostatic potential, voltage is formed. Voltage is formed to the p to n region. Electrostatic potential, which is the voltage, means that holes are stable in p region. To move this hole to n region, you require this much energy. Core built-in potential or built-in voltage. This is the PN junction. But if you look at the MOSFET device, there is the metal oxide, and semiconductor, and more structure here. There is a n plus a source region and n plus drain region. This is a p-type semiconductor. There will be two PN junction, PN junction, PN junction. So if you draw band diagram between the n, p, n then PN junction, band diagram showing in here. So p-type, n-type, p-type, n-type. Most part also consisted without two PN junction. Let's do this one more time. So p-type semiconductor and n-type semiconductor. In p-type semiconductor is not positive charge. P-type semiconductor is the neutral, means that equal number of freely moving hole and equal number of the fixed to minus boron charge. However, they both are neutron. Same thing for the n-type semiconductor. Then equal number freely moving electron and equal number of the boron, same plus charge. If you're making a contact, then measure that hole p-time region, go to the n-time region by diffusion. Same thing for the electron. In n-time region go to the p-time region by the diffusion and there is space charges form of the fixed charge of the arson and fixed charge of the boron in p-time region. Because of the fixed charge, positive to negative, then electric field is formed, to positive to negative. Then the holes diffuse to the n-time region, they go back to original p-time region by electric field. Then how much electric field is formed or how much holes moving from p to n or n to p? In equilibrium condition where the no voltage apply, current should be zero, therefore, equal number of the hole moving from the P to N, and equal number of the hole move back to the N to P, so that the current is zero. Same thing for the electron. This is the band diagram. P-type semiconductor, formed energy located close to the E_V, and then this is the n-type semiconductor, formed energy close to the conduction band. If they're making a contact under equilibrium where the no voltage apply, then the formed energy always constant. N-type band diagram should be go lower. Then you connecting the E_C and E_V, this is the band diagram of the PN junction. Here, p-type semiconductor, here is the n-type semiconductor, [inaudible] energy constant and the connection of the E_V, E_I, and E_C. What is the meaning of the band diagram? Energy band diagram. Energy band diagram is exactly the same for the concept of the electrostatic potential. In previous slide, I draw the electrostatic potential of the PN junction showing in here, and then holes in p-type is stable in the lowest energy here, and they require this built-in potential to move hole from the N to P. band diagram is exactly the same, except that band diagram is based on the electron instead of the hole positive charge, means that electrostatic potential is based on the positive charge. Positive charge to move from here to here requires built-in potential. But band diagram is designed for the semiconductor device. Semiconductor device moving charge is not the positive charge, but the freely moving electron. So electrostatic potential should be expressed by the band diagram, which is based on the negative electron charge. These band diagram means that electrons are stable in n-time region and require this qV_0 of the energy to move electron to the p-time region, which is the built-in potential. As I said, band diagram energy is exactly the same thing to the electrostatic potential, except the fact that band diagram is based on the negative electron charge, which is the negative end q. That's why the potential is opposite. In electrostatic potential, energy is higher, but the band diagram energy is lower in n-time region because electrons are more stable here. Let's look at the movement. Holes are moving from the p to n-region by the diffusion, then hole current should be same thing, same direction. Movement hole is at this direction, current move way is the same direction. There's electric field is formed, then the holes went to the n-region, they go back to the original p-region by electric field. Therefore, hole movement is n to p. Current is the same direction. For electron case, electron diffusion occurs from the n to p region. Current is opposite direction, to the right direction. Then because of the electric field, electron drift is from directly to the hole direction, but current direction is opposite. Let's explain this one more time. We are repeating this in a different perspective. Here's the p-type semiconductor, here's the n-type semiconductor. A lot of the holes are located here, a lot of electrons are located here. Because there is a concentration gradient, a lot of electron located here, they move to the less concentrated area of the n-region by diffusion. Same thing for the electron, a lot of electron in n-type region, less electron in p-type region, they will diffuse. This electrons are diffused to the p-type region. Then in depletion region, because electron moves, there is fixed ions charge in junction area, which is the positive charge. Same thing for the p-region, fixed boron charge of negative charge formed at the junction region. Because of the fixed positive charge and negative charge, there's electric field is formed. Because of this electric field, the holes that moved from the p to n by diffusion, they go back to the origin p-region by the hole drift. How much diffusion of the hole and how much diffusion of a drift of the hole is formed? Which means, they're under equilibrium where the low voltage apply, those should there be zero so that the current should be zero under equilibrium conditions. Same thing for the electron. Because there is charge formed, then under the equation of the electric field is the derivative of energy band diagram, dE_i per dx. Then electric field is formed and then Fermi energy goes down to meet the Fermi energy equal. As I said previously, band diagram energy is exact the same concept of electrostatic potential. This means that there's electric field is formed, then band diagram should be changing by distance. That's the concept of the electric field, which is the electrostatic potential difference in dx or band diagram energy by dx. Therefore, potential should be changing and then the changing that the E_c, n-type region go down until when? Until Fermi energy is equal, which means our equilibrium condition. So this is the band diagram and then Fermi energy constant in space charge, there is a fixed charge in n-type asking plus, p-type region boron minus. Then electric field form the n to p direction. Then the hole that diffused to the n-type region, go back to the original hole of p-type region by electric field. Those numbers should be equal so that the current is zero. Here's an important question. What is the slope in depletion region or curvature? This curvature means there is electrostatic potential is changing by distance between those p-n region. So potential changing by the distance means electric field. So in depletion region, there is electric field, determined by dark curvature of the n-type region. If you look at p-type neutral region and n-type neutral design, they are equal number of the hole and equal number of the positive charge and negative charge, they are neutral, there is no charge, then there is no electric field and band is in that banding. That's the concept of the p-n junction. Now, we are going to solve the equation. As I said, p-n junction has an electrostatic potential, requires this much built-in potential is formed so that the holes in p-type region move to the n-type region. What is the built-in potential of the p-n junction? So electrostatic potential can be expressed by the band diagram, energy band diagram of the E_c, E_i, E_v and Fermi energy, which is the based on the negative electron. So negative electron is stable in n-region. This much built-in potential, qV_0 zero is required so they'll require so that electron move from n to p-region. To calculate the built-in potential qV_0, qV_0 is the, how much n-type region go down so that the Fermi energy becomes a constant. This doping effect of qV_0 is addition of the Fermi energy, my n-type region minus the E_i plus Fermi energy minus the E_i in p-type region. So addition of the Fermi energy minus the E_i plus Fermi energy hole minus the E_i. These two addition, built-in potential is lower. Then you already know that n-type region, Fermi energy minus E_i is kT ln n-type doping concentration per n_i, which is the 1.510_10. Then p-type region E_i minus the Fermi energy p is kT ln p-type majority hole in p-type semiconductor, which is expressed P_p per n_i, which is the intrinsic carrier concentration, 1.510_10. So if you are adding these two, then qV_0 becomes the kT ln n_n per n_i plus ln P_p per n_i. Then final conclusion is, kT ln n_n times P_p per n_i square. So this equation you need to memorize, qV_0, built-in potential is kT ln n_nP_p per n_i square. N_i square per n_n becomes the n_p. So n_n per n_i square becomes the p_n. So it becomes like this. Then P_p per n_i square becomes minority carrier n_p. So this equation can be expressed in a two way too.