Hello. I hope you've been learning lots from the lectures so far. We're getting closer to the actual structure we will need for a practical PV device. So, let's keep going. In the previous lectures, you saw how absorption of light in a semiconductor can create non-equilibrium populations of electrons and holes. Let us now consider this topic in more detail by working through an example question. The question is stated as follows. Consider a crystalline silicon wafer with a thickness of 200 microns. It is a p-doped wafer with a dopant concentration of five times 10 to the 16 per cubic centimeter. Both surfaces are perfectly passivated, but the bulk minority carrier lifetime is 50 microseconds. This wafer absorbs an irradiance of 50 milliwatts per centimeter squared, but which is composed of only red photons with a wavelength of 630 nanometers. Using this information, calculate the minority carrier density and the implied open-circuit voltage. If you wish at this point, you may stop the video and try to solve the problem on your own. If not, let's continue. First, let's deal with the excitation. We need to know how many photons per second are being absorbed in our wafer and are creating electron-hole pairs, and we need to get everything into the same units. First, we can use this top expression, which states that the product of a photon's wavelength, expressed in microns, and its energy, expressed in electron volts, is equal to 1.24. This means our red photons each have an energy of 1.97 electron volts. This is equivalent to 3.15 times 10 minus 19 joules per photon. Notice in this case that I used the energy per photon and not the gap of the semiconductor. However, once this more energetic photon is absorbed, the electron hole pair it creates will thermalize down to the valence and conduction band edges within a few picoseconds and that extra energy is lost. We can now turn this into a volumetric generation rate, which should have the units of events per second per cubic centimeter. We can do so using this equation. Within this equation, we recognize our radiance of 50 milliwatts per square centimeter. Now expressed as 0.05 Joules per second per centimeter squared. Our energy per photon and our wafer thickness. We notice that the units work out and crunching the numbers, we obtain our generation rate of 0.79 times ten to the 19 per second per cubic centimeter. That generation rate, which is the rate at which the electron hole pairs are created leads to excess carriers in the semiconductor. The carrier concentrations of electrons and holes in situation can be split into the equilibrium carrier concentration, N naught and P naught and the excess carrier concentrations, delta n and delta p. However, since our wafer is p-doped, the equilibrium holes can be ignored. Likewise, the p-type doping of the wafer will completely dominate any holes created by illumination. In this case, we can approximate that there are only excess minority electrons and equilibrium holes equal to the acceptor dopant concentration. We can later check that these approximations were true. In this wafer, there's no extraction. So, to obtain balance, the generation rate must be balanced out by recombination, the bane of photovoltaics. Furthermore, you saw that in the lectures, that the recombination rate will be determined by the excess minority carrier density, delta n and the minority carrier lifetime in the wafer, in this case, tau m. As it is delta n that we don't know, we can reorganize the formula and plug in the numbers. This gives us our excess minority carrier density, 3.95 times 10 to the 14 per cubic centimeter. This is indeed much less than the doping level and so you can check that it is also much larger than the equilibrium minority carrier density. So, we were correct in our initial assumptions. We now have the answers to the first part of the question. But what do these answers mean for the implied open-circuit voltage. To get that answer, let's walk through what is happening in our wafer. Initially, in the dark, we have a single fermi level, describing electron and the hole populations. However, upon irradiation the single Fermi level is no longer valid. Rather, quasi Fermi levels are present for electrons and holes. As the hole concentration doesn't change the hole quasi Fermi level, FP doesn't move. However, the electron concentration changes enormously. So, the electron quasi Fermi level, Fn approaches the conduction band edge. To know the exact position of his levels, we again use the carrier concentration equations. Notice that for the electron and hole concentrations, I've used the approximations we made before and used the excess carrier concentrations for electrons and the doping concentration for holes. Plugging in the numbers tells us that F_n is 0.294 electron volts from the conduction band edge and F_p is 0.153 electron volts from the valence band edge. The last thing we want to know is the magnitude of the Fermi level splitting. This indicates the maximum amount of work per electron that we can extract from this system. From the diagram, it is clear that this value will just be the band gap minus the two distances that we just calculated, giving a quasi-Fermi level splitting of 0.653 electron volts and therefore an implied open-circuit voltage, VOC of 0.653 volts. This value is the maximum open-circuit voltage we could expect to obtain from a device made out of this wafer and under this illumination condition. Through this work problem, I hope you've gained some insight into how to calculate the non-equilibrium carrier concentrations in a semiconductor under illumination. Thank you for your attention and see you again soon.
