GATE 2021 EC – Question 52
A silicon P-N junction is shown in the figure. The doping in the P region is $5\times10^{16}\ \text{cm}^{-3}$ and doping in the N region is $10\times10^{16}\ \text{cm}^{-3}$. The parameters given are
Built-in voltage ($\Phi_{bi}$) $=0.8$ V
Electron charge ($q$) $=1.6\times10^{-19}$ C
Vacuum permittivity ($\varepsilon_0$) $=8.85\times10^{-12}$ F/m
Relative permittivity of silicon ($\varepsilon_{si}$) $=12$
The magnitude of reverse bias voltage that would completely deplete one of the two regions (P or N) prior to the other (rounded off to one decimal place) is ________ V.
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Correct answer: 8.16 to 8.24
Explanation
The depletion widths satisfy $x_nN_d=x_pN_a$, with $x_n=\frac{N_a}{N_a+N_d}W=\frac W3$ and $x_p=\frac{2W}{3}$. The N region (0.2 µm) is depleted first, at $x_n=0.2\ \mu$m, i.e. $W=0.6\ \mu$m (then $x_p=0.4\ \mu$m$<1.2\ \mu$m). Using $W^2=\frac{2\varepsilon_{si}\varepsilon_0(\Phi_{bi}+V_R)}{q}\left(\frac1{N_a}+\frac1{N_d}\right)$ we get $\Phi_{bi}+V_R=\frac{qW^2}{2\varepsilon}\cdot\frac{N_aN_d}{N_a+N_d}=\frac{1.6\times10^{-19}\times(0.6\times10^{-6})^2}{2\times12\times8.85\times10^{-12}}\times3.33\times10^{22}\approx9.04$ V. So $V_R\approx9.04-0.8=8.2$ V.