Calculate the open-circuit voltage of a silicon pn junction solar cell.
Consider a silicon pn junction at T=300 \mathrm{~K} with the following parameters:
\begin{array}{ll}N_{a}=5 \times 10^{18} \mathrm{~cm}^{-3} & N_{d}=10^{16} \mathrm{~cm}^{-3} \\D_{n}=25 \mathrm{~cm}^{2} / \mathrm{s} & D_{p}=10 \mathrm{~cm}^{2} / \mathrm{s} \\\tau_{n 0}=5 \times 10^{-7} \mathrm{~s} & \tau_{p 0}=10^{-7} \mathrm{~s}\end{array}
Let the photocurrent density be J_{L}=I_{L} / A=15 \mathrm{~mA} / \mathrm{cm}^{2}.
We have that
J_{S}=\frac{I_{S}}{A}=\left(\frac{e D_{n} n_{p 0}}{L_{n}}+\frac{e D_{p} p_{n 0}}{L_{p}}\right)=e n_{i}^{2}\left(\frac{D_{n}}{L_{n} N_{a}}+\frac{D_{p}}{L_{p} N_{d}}\right)
We may calculate
L_{n}=\sqrt{D_{n} \tau_{n 0}}=\sqrt{(25)\left(5 \times 10^{-7}\right)}=35.4 \mu \mathrm{m}
and
L_{p}=\sqrt{D_{p} \tau_{p 0}}=\sqrt{(10)\left(10^{-7}\right)}=10.0 \mu \mathrm{m}
Then
\begin{aligned}J_{S} &=\left(1.6 \times 10^{-19}\right)\left(1.5 \times 10^{10}\right)^{2} \times\left[\frac{25}{\left(35.4 \times 10^{-4}\right)\left(5 \times 10^{18}\right)}+\frac{10}{\left(10 \times 10^{-4}\right)\left(10^{16}\right)}\right] \\&=3.6 \times 10^{-11} \mathrm{~A} / \mathrm{cm}^{2}\end{aligned}
Then from Equation (14.10), we can find
V_{o c}=V_{t} \ln \left(1+\frac{I_{L}}{I_{S}}\right)=V_{t} \ln \left(1+\frac{J_{L}}{J_{S}}\right)=(0.0259) \ln \left(1+\frac{15 \times 10^{-3}}{3.6 \times 10^{-11}}\right)=0.514 \mathrm{~V}
Comment
We may determine the built-in potential barrier of this junction to be V_{b i}=0.8556 \mathrm{~V}. Taking the ratio of the open-circuit voltage to the built-in potential barrier, we find that V_{o c} / V_{b i}=0.60. The open-circuit voltage will always be less than the built-in potential barrier.