A silicon p-n diode has a doping of $N_D = $ cm^-3 and $N_A = $ cm^-3. What is the depletion width on the n-side $W_n$, and the depletion width on the p-side at 300 K?

For Si: n_i = 6.4 × 10⁹ cm^-3, ε_r = 11.9.

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<p>Within the depletion width approximation, the depletion widths are calculated on the page, <a href="http://lampz.tugraz.at/~hadley/psd/L6/abrupt.html">Abrupt pn junctions in the depletion approximation</a>. The formula for the total depetion width $W$ is,</p>

$$W= \sqrt{\frac{2\epsilon_r\epsilon_0(N_D+N_A)V_{bi}}{eN_DN_A}}.$$

<p>Here $\epsilon_0 = 8.854187817\times 10^{-12}$ F/m is the permeativity constant and $V_{bi}$ is the built-in voltage. The formula for the built-in voltage is,</p>

$$V_{bi} = \frac{k_BT}{e}\ln\left(\frac{N_DN_A}{n_i^2}\right),$$

<p>where $k_B=1.380658\times 10^{-23}$ J/K is Boltzmann's constant, $T$ is the absolute temperature and $n_i=\sqrt{N_c\left(\frac{T}{300}\right)^{3/2}N_v\left(\frac{T}{300}\right)^{3/2}}\exp\left(\frac{-E_g}{2k_BT}\right)$ is the intrisic carrier concentration. The intrinsic carrier concentration depends sensitively on the temperature. If we use the temperature dependence of the band gap given on the <a href="http://lampz.tugraz.at/~hadley/psd/L6/abrupt.html">Abrupt pn junctions in the depletion approximation</a> page, the intrinsic carrier concentration at 300 K is $n_i = 6.4\times 10^9$ cm<sup>-3</sup>.  From the doping concentrations and $n_i$ the built-in voltage can be calculated to be,v<script>document.write(Vbi);</script>

<p style="text-align:center">$V_{bi} = $ <script>document.write(Vbi);</script> V</p>