The p-n Junction — Question 8

Question 8

Minority injection at the depletion edge grows exponentially

For an ideal forward-biased p–n junction, minority electrons on the p side satisfy $n_p(0)=n_{p0}e^{V_a/V_T}$, where $n_{p0}=n_i^2/N_A$. Take $n_i=10^{16}\,\mathrm{m^{-3}}$, $N_A=10^{23}\,\mathrm{m^{-3}}$, $V_T=.0250\,\mathrm V$, and $V_a=V_T\ln1000$.

1. Find equilibrium, biased, and excess minority densities at the edge.

2. For $D_n=.00200\,\mathrm{m^2/s}$, $L_n=100\,\mu\mathrm m$, and $e=1.60\times10^{-19}\,\mathrm C$, find the injected electron-current magnitude $eD_n\Delta n(0)/L_n$.

Question 8 - Solution

Minority injection at the depletion edge grows exponentially

(a) Method and calculation

$$\boxed{n_{p0}=10^9,\quad n_p(0)=10^{12}\,\mathrm{m^{-3}}}.$$$$\boxed{\Delta n(0)=n_{p0}(e^{V_a/V_T}-1)=9.99\times10^{11}\,\mathrm{m^{-3}}}.$$

(b) Interpretation and further calculation

$$|J_n|=eD_n\Delta n(0)/L_n.$$$$\boxed{|J_n|=3.197\times10^{-6}\,\mathrm{A/m^2}}.$$

Physical check

The excess minority density remains far below the majority density, consistent with low-level injection. The exponential boundary population leads to the ideal diode current law.