Question 14.5: Calculate the steady-state photocurrent density in a reverse...

We may calculate various parameters as follows:

\begin{aligned}L_{n} &=\sqrt{D_{n} \tau_{n 0}}=\sqrt{(25)\left(5 \times 10^{-7}\right)}=35.4 \mu \mathrm{m} \\L_{p} &=\sqrt{D_{p} \tau_{p 0}}=\sqrt{(10)\left(10^{-7}\right)}=10.0 \mu \mathrm{m} \\V_{b i} &=V_{t} \ln \left(\frac{N_{a} N_{d}}{n_{i}^{2}}\right)=(0.0259) \ln \left[\frac{\left(10^{16}\right)\left(10^{16}\right)}{\left(1.5 \times 10^{10}\right)^{2}}\right]=0.695 \mathrm{~V} \\W &=\left\{\frac{2 \epsilon_{s}}{e}\left(\frac{N_{a}+N_{d}}{N_{a} N_{d}}\right)\left(V_{b i}+V_{R}\right)\right\}^{1 / 2} \\&=\left\{\frac{2(11.7)\left(8.85 \times 10^{-14}\right)}{1.6 \times 10^{-19}} \cdot \frac{\left(2 \times 10^{16}\right)}{\left(10^{16}\right)\left(10^{16}\right)} \cdot(0.695+5)\right\}^{1 / 2}=1.21 \mu \mathrm{m}\end{aligned}

Finally, the steady-state photocurrent density is

\begin{aligned}J_{L} &=e\left(W+L_{n}+L_{p}\right) G_{L} \\&=\left(1.6 \times 10^{-19}\right)(1.21+35.4+10.0) \times 10^{-4}\left(10^{21}\right)=0.75 \mathrm{~A} / \mathrm{cm}^{2}\end{aligned}

Comment

Again, keep in mind that this photocurrent is in the reverse-biased direction through the diode and is many orders of magnitude larger than the reverse-biased saturation current density in the pn junction diode.