Hello. I hope you've been learning from the lectures so far. In the previous lectures you were presented the current voltage characteristic for a solar cell. Let us consider this topic in more detail now by working through an example question. The question is stated as follows. For the 100 centimeter square cell presenting the current voltage curve below, as measured under AM 1.5 illumination, determine the following photovoltaic parameters. Open circuit voltage, short circuit current, fill factor, and power conversion efficiency. Along with this question, we are presented a current voltage curve or I-V curve for a solar cell. 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. To correctly extract the photovoltaic parameters from this curve, we first need to be aware of a few critical points concerning such curves. The first important point concerning I-V curves is that two of the important reference points occur at the values of zero voltage and zero current. So on the lines indicated here and here. Always look carefully as these points may or may not be located on the edges of the graph. These zero crossings will give the position of the open circuit voltage when the current is zero, and the short circuit current when the voltage is zero. The second point of which to be aware is that there are four equivalent ways to present such an I-V curve as shown here. These may look quite different, but they all represent the same data. The only difference between these representations is the convention used for the positive directions of voltage and current in the equivalent circuit. Current flows in, current flows out and voltage is defined by a positive and negative measurement point. However, all four present the same information. So look closely when presented with such curves. Look for the zero crossings and think about what is represented by the curve at which you are looking. The two most common representations of solar cell I-V curves are shown here. And an important point to keep in mind is that in only one quadrant of these graphs is the device supplying power to a load. That is to say, correctly functioning as a solar cell. If forced to operate in one of the other two quadrants of operation, the solar cell will be absorbing power from some other source. Going back to our original curve, we can start to extract the photovoltaic parameters. At the point of zero current, we can extract the value of open circuit voltage, which in this case is 0.62 volts. At the point of zero voltage, we can obtain the value of short circuit current. In this case, 3.5 amps. Knowing the size of our solar cells, we can also express this as an aerial current density. For our 100 centimeter squared cell, we calculate 35 milliamps per centimeter squared. For the next step, we need to know the point on the curve at which the device provides the maximum power. To do so, we plot this information another way, as the power versus voltage where power equals voltage times current. It's informative to note that this curve crosses zero at two important points, one where the voltage is zero and another time at open circuit voltage when the current is zero. In between these two points is the maximum power point located in our case at a voltage of 0.4 volts, giving the voltage at maximum power point Vmpp. We can now use this point along with the original current voltage curve to locate the current and maximum power point Impp which has the value of 3.1 amps. To summarize, we have now extracted the open circuit voltage, the short circuit current, the voltage at maximum power point and the current and maximum power point. We can now use these four values to obtain the fill factor using the equation shown here. We obtain a value of 57 percent. It should be noted that the best solar cells provide fill factors upwards of 80 percent. We can also use the voltage and current and maximum power point to obtain the maximum power that can be extracted from the cell. In this case, 1.24 watts. You can go back and check that this is also the maximum power we observed on our power versus voltage curve. To obtain the power conversion efficiency, we also need to know the instant power on the cell. Knowing that this cell is being measured under AM 1.5 which is equivalent to 100 milliwatts per centimeter squared or one kilowatt per meter squared, our 100 centimeter squared cell receives 10 watts of power. This means that the power conversion efficiency of this cell is 12.4 percent. This is quite a low value compared to the most standard technology crystalline silicon, which can be easily around 20 percent at the cell level in production. Through this work problem, I hope you've gained some insight into how to accurately interpret a current voltage characteristic and extract the critical photovoltaic parameters from it. Thank you for your attention.
