*A Si solar cell with dark saturation current Ith of 5 nA is illuminated such that the short-circuit current is 200 mA. Plot ideal I-V curve for the cell. Assume that the cell has a series resistance of 1
. Replot the I-V curve for this case and compare it.*

We can base on the voltage within open circuit over the device defined by |STREETMAN| p.400 and transform it to short circuit which is practically the same.

(1) |

Therefore we can substitute with which is 200 mA.

(2) |

Basing on these photovoltaic laws and using Matlab, I get the following I-V curves (Appendix A contains the calculation algorithm):

Calculating the IV curve with an ESR of 1 Ohm at first glance seemed to be fairly simple, however there are two ways of interpreting the problem. If we look at the pv-cell as a black box capable of delivering 200mA short circuited, and following the PV equation, we could easily just calculate the voltage drop over the series resistor and replot the curve
.

However it seems like this is not the most accurate way of calculating it, due to the fact that the voltage drop over will effectively cause a voltage drop over the pv-cell itself, thus affecting the transfer function in a non-linear way following the IV-law in Equation 1. The current from the very same equation can be represented as:

(3) |

Where is the model's series resistance and is the load (shunt resistance). Figure 1 shows the I-V characteristic under 1 series resistance (blue curve).

By looking at the two curves and comparing them, if we integrate the area of the two curves, we can see that the internal ESR of the PV cells is crucial for the theoretical maximum power delivered (areas differ significantly). Therefore it is important to design the PV-cell with as low contact/junction series resistance as possible. In reality ESR also differs with temperature as well, thus the curves on Figure 1 would also scale down (areawise/the optimim load rectangle IV) with the increase of temperature.