[MUSIC] We have introduced the properties of the transport of charge carriers in a semiconductor. We'll now move on the study of the essential device for the understanding of the properties of a solar set, the p-n junction, what is called a diode. First, we will look at your p-n junction at thermodynamic equilibrium. Then, we'll see the p-n junction out of thermodynamic equilibrium to reach the photovoltaic effect. Firstly some words on the electrostatic potential in an inhomogeneous semiconductor. We will consider a p-n junction that is to cell system in which part of the semiconductor is n doped, another part p doped. There is an inhomogenuity of concentration in the physical system. Since there are in this semiconductor inhomogenous carriers of n on p, which vary depending on the position, deficient currents will appear. In the semiconductor, electrons on holes diffusion before reaching the thermodynamic equilibrium without external electric field. For instance, the electron current is zero, so there will be equality between the drift on diffusion currents. These two currents are opposed on vanish for electrons on holes. Since there is balance between the two contributions, one sees that the consequence of the concentration inhomogeneity is the appearance of an electric field E in the semiconductor. The electrical neutrality cannot be retained throughout the semiconductor due to the appearance of this electric field even at thermal equilibrium. The relationship between the electrostatic potential and the charge density are all the dependant of x is given by the Poisson's equation. Which therefore relates the second derivative of the potential of the charged density. So charge density in a semiconductor depends on the mobile carriers, electrons in the condition bond n of x holes in the valence bond, p of x. But also fixed ions which are linked to doping atoms donors or acceptors. Therefore, the contribution ND plus on an A minus correspond to fixed ions that do not contribute to the electrical transport towards external circuit. In the semiconductor, solving the Schroedinger's equation gives bond structure with EV and EC limits. The presence of an additional electrostatic potential, energy level will be shifted by a quantity E 1, dependant of x, and the chemical potential, or the Fermi level is constant spatially across the system since the chemical potential is the characterisation of the thermodynamic equilibrium. Their energies are plotted as referred to the vacuum. Recall that the vacuum level is the required energy to remove the electrons from the semiconductor solid close to the surface, and to leave it in the vacuum with zero velocity. We will now turn to the study of the p-n junction, which consists in contacting the p doped semiconductor to n doped semiconductor as shown in the figure. So we will consider an n semiconductor which is on the right side. The Fermi level is close to the level of the donor as we have seen previously. We'll assume that the n doped semiconductor only donors are introduced without any acceptors, and the same in the p doped semiconductor. Therefore the Fermi levels are different in both sides. Since in the end side, the Fermi level lies close to the donor level and the close to the acceptor level in the p side. We will now connect the p doped semiconductor to the n doped ones. In the n doped semiconductor, the majority carriers are electrons, and in the p region majority carriers are holes which reveals inhomogeneities of concentration in the material. Consider, for example, the electrons. They will move from the right where the chemical potential is highest to the left side, p side, until equality of the chemical potential across the system. The holes are moving in the opposite direction as compared to electrons. So the electrons will move from right to left. That is to say, to us the p area in which the holes are the majority carriers. Therefore, the electrons will recombine with holes. And with thermodynamic equilibrium, the chemical potential will be constant throughout the system. In addition, we have created what is called a space charge region. In these, electrons leave inside to reach the p zone, where they recombine. So we have less positive ions relative to the donor effect. It refers to the first phase atoms close to the junction on the right that have lost electrons leading to positive ions. Likewise, negative ions appear on the left relative to the acceptor atom. There is therefore a creation of a space charged region with fixed charges that are positive or negative ions. This diffusion process finally stops due to the formation of a dipole, positive and negative charges on either side of the junction. This dipole formation will be correlated with the creation of a permanent electric field, which will therefore prevent the diffusion of electrons for example from right to left. The direction of the electric field is such that the preference of diffusion of electrons from right to left and holes from left to right. So a permanent electric field has been created on the chemical potential. It's constant throughout the system, so total current being zero. So there is the electric field as the interface. So there will be few movable charges in this space charge region since they are drifted by the electric field. In terms of potential thermodynamic equilibrium, the chemical potential in constant is a physical system throughout the semiconductor. The top of the valence bonds on the bottom of your conduction bond were offset by an amount of E phi dependent of x. The electrical field being the derivative of the potential. So a potential barrier appears which prevent the flow of carriers. The space charge zone is a depletion region with a few free charges. This is the resistive zone. Another way of seeing the chemical potential is this area of space charges is far from the bond edges. Indeed, as the probabilities vary as exponential minus Ec minus mu. So there are very few carriers since the Fermi level as the interface zone is far from the bond edges. I have presented a quantitative approach of the p-n junction at equilibrium. I invite to refer to appendix three for firstly, a quantitative assessment of the potential barrier. And you will see that this potential barrier is still slightly below the bond gap, about 0.6 eV for silicone. This appendix will also evaluate the width of the depression layer, and it is seen that in the case of silicone, it is a very thin layer of the order of a micron thick. Thank you very much. [MUSIC]