First, find the photocurrent, which is equal to the short circuit current:
"I_{op} = q\\cdot A\\cdot g_{op}\\cdot (L_p+L_n+W) , where A=2\\cdot 2 cm^2"
"I_{op} = 1.6\\cdot10^{-19}\\cdot 4\\cdot10^{-4}\\cdot 10^{24}\\cdot 10^{-6}\\cdot (2+2+0.5)=0.288mA"
Then, by the known short circuit current, we find the open circuit voltage:
"V_{oc} = \\frac{kT}{q}\\cdot ln(\\frac{I_{op}}{I_{th}} + 1)"
"V_{oc} = 0.025\\cdot ln(\\frac{0.288\\cdot10^{-3}}{10^{-9}} + 1) = 0.487 V"
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