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Problem: Solar Cell

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!

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  • \$\begingroup\$ 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 \$\endgroup\$ – KingDuken Oct 24 '17 at 14:53
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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.

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  • \$\begingroup\$ 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. \$\endgroup\$ – user4826575 Oct 25 '17 at 13:30
  • \$\begingroup\$ @user4826575 you have to use diode equation of current + optically generated current. Please see the edited answer. \$\endgroup\$ – nidhin Oct 25 '17 at 17:55

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