# Solar Cell Problem

Relevant Equations:

• $V_{oc}=\frac{k_t}{q} \ln{\frac{\tau_p*G_L*N_d}{n_i^2}}$
• $I_{sc}=A*q*L_pG_L$
• $L_p=\sqrt{D_p\tau_p}$
• $FF=\frac{P_{max}}{I_{sc}*V_{oc}}$

Attempt at Solution:

I got $V_{oc}=.298 V$ and $I_{sc}=3.04 mA$. However, I am confused as to how to find the fill factor. I know it is the maximum voltage and current the load can take, but how do I determine this with the information given? I thought it might have something to do with the diode current equation $I=I_0(e^{qV/kT}-1)$ Any help is much appreciated!

• I added Mathjax to your equations to make it look nice. It was a little hard to see your equations but this should fix it up. Try to learn some Mathjax as it's a pretty neat feature on SE :) here's a good guide on how to do it – KingDuken Oct 24 '17 at 14:53

The power delivered by a solar cell operating at open circuit and closed circuit condition are zero. The maximum power happens at some other point.

Power is given by $$P=V\times I=V\left[I_L-I_0\left\{\exp\left(\frac{V}{V_T}\right)-1\right\}\right],$$ because, current is sum of diode current and the current because of optical generation ($I_L$).

At the maximum power point, $dP/dV$ will be 0.

$$\left.\frac{dP}{dV}\right|_{V=V_m}=0$$

This will result in an equation with only one unknown $V_m$. Once solved, then $I_m$ can be calculated using $V_m$ obtained by solving.

PS: You will get a closed form expression for $V_m$, but not an analytical expression in terms of known values. May have to use numerical solvers to find the values.

• This is what I discovered myself. However, how do I find the power equation based on the data? Do I use the ideal current equation? Because the exponential does not have a max or min. – user4826575 Oct 25 '17 at 13:30
• @user4826575 you have to use diode equation of current + optically generated current. Please see the edited answer. – nidhin Oct 25 '17 at 17:55