# Solar Cell

A solar cell has a saturation current $I_s =1 \mu A$. When it is illuminated it produces a short-circuit current $I_{SC} = -50 mA$. Being given the working temperature $T =290 K$ and Boltzmann constant, compute the following:

a) Internal resistance of the solar cell when the illumination tends to zero.

b) No-load voltage generated by the solar cell, at the illumination from the problem.

c) The maximum voltage on a load resistance $R_m = 5 \Omega$, which corresponds to the maximum power supplied by the solar cell to the external circuit.

The analytical expression for the current-voltage characteristic is:

$I =I_s\left ( e^{(e U)/(k T)}-1 \right )-I_L$ (1)

where $I_L$ is the current when the solar cell is illuminated.

From above we obtain:

$exp\left(\frac{e U}{k T}\right)=\frac{I+I_L+I_s}{I_s}$

By differentiating with respect to I we obtain

$\frac{e}{k T}\frac{d U}{d I}exp\left(\frac{e U}{k T}\right) = \frac{1}{I_s}$

The internal resistance of the cell is thus:

$R =\frac{d U}{d I} = \frac{k T}{e}*\frac{1}{I_s}*exp\left ( -\frac{e U}{k T} \right )=\frac{k T}{e I_s}*\frac{I_s}{I+I_L+I_s}$ (2)

With no light one has $I=I_L =0$ so that

$R = (k T)/(e I_s) =2.5*10^4 \Omega$

The no-load voltage $U0$ is found by imposing the condition $I =0$ on the above equation (1):

$I_L =I_s\left ( e^{(e U_0)/(k T)}-1 \right )$

from which we get

$U_0 =\frac{k T}{e}*ln\left ( 1+\frac{I_L}{I_s} \right ) =270 mV$

For maximum power to the outside circuit external load resistance is equal to the internal resistance of the solar cell. By taking $U =U_m$ into equation (2) one obtains

$R_m =\frac{k T}{e I_s}*exp\left ( -\frac{e U_m}{k T} \right ) =R_0*exp\left ( -\frac{e U_m}{k T} \right )$

Therefore

$U_m =\frac{k T}{e}*ln\frac{R_0}{R_m} =213 mV$