GATE 2016 EC – Question 46
Consider a silicon p-n junction with a uniform acceptor doping concentration of $10^{17}$ cm$^{-3}$ on the p-side and a uniform donor doping concentration of $10^{16}$ cm$^{-3}$ on the n-side. No external voltage is applied to the diode. Given: $kT/q = 26$ mV, $n_i = 1.5 \times 10^{10}$ cm$^{-3}$, $\epsilon_{Si} = 12\epsilon_0$, $\epsilon_0 = 8.85 \times 10^{-14}$ F/m, and $q = 1.6 \times 10^{-19}$ C. The charge per unit junction area (nC cm$^{-2}$) in the depletion region on the p-side is ________
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Correct answer: -5 to -4.6
Explanation
The built-in voltage is $V_{bi} = 0.026\ln\frac{10^{17} \times 10^{16}}{(1.5 \times 10^{10})^2} = 0.757$ V. Using the printed value $\epsilon_0 = 8.85 \times 10^{-14}$ F/m, which is $8.85 \times 10^{-16}$ F/cm, we get $\epsilon_{Si} = 1.062 \times 10^{-14}$ F/cm. The total depletion width is $W = \sqrt{\frac{2\epsilon V_{bi}}{q}\left(\frac{1}{N_A} + \frac{1}{N_D}\right)} = 3.3 \times 10^{-6}$ cm. The part on the p-side is $x_p = W\frac{N_D}{N_A + N_D} = 3.0 \times 10^{-7}$ cm. The charge per unit area is $-qN_A x_p = -1.6 \times 10^{-19} \times 10^{17} \times 3.0 \times 10^{-7} = -4.8 \times 10^{-9}$ C/cm$^2$, which is $-4.8$ nC/cm$^2$.