The p-n Junction — Question 5

Question 5

Applied bias changes both depletion width and peak field

For a fixed abrupt junction, $W\propto\sqrt{V_j}$ and $E_{\max}\propto\sqrt{V_j}$. At zero bias $V_{bi}=.800\,\mathrm V$ and $W_0=.400\,\mu\mathrm m$. Use $V_j=V_{bi}-V_a$, with positive $V_a$ for forward bias.

1. Find W under .600 V forward bias and under 2.40 V reverse bias.

2. Find the peak field magnitudes in both cases using $E_{\max}=2V_j/W$.

Question 5 - Solution

Applied bias changes both depletion width and peak field

(a) Method and calculation

$$\boxed{W_F=W_0\sqrt{(.800-.600)/.800}=.200\,\mu\mathrm m}.$$$$\boxed{W_R=W_0\sqrt{(.800+2.40)/.800}=.800\,\mu\mathrm m}.$$

(b) Interpretation and further calculation

$$\boxed{E_{\max,F}=2(.200)/(.200\times10^{-6})=2.00\times10^6\,\mathrm{V/m}}.$$$$\boxed{E_{\max,R}=2(3.20)/(.800\times10^{-6})=8.00\times10^6\,\mathrm{V/m}}.$$

Physical check

Reverse bias widens the depleted region while increasing its field. The formulas exclude breakdown and strong forward injection that invalidates the depletion approximation.