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When a silicon diode having a doping concentration of N_A = 9 × 10¹⁶ cm⁻³ on p-side and N_D = 1 × 10¹⁶ cm⁻³ on n-side is reverse biased, the total depletion width is found to be 3 μm. Given that the permittivity of silicon is 1.04 × 10⁻¹² F/cm, the depletion width on the p-side and the maximum electric field in the depletion region, respectively, are

1. 2.7 μm and 2.3 × 10⁵ V/cm
2. 0.3 μm and 4.5 × 10⁵ V/cm
3. 0.3 μm and 0.42 × 10⁵ V/cm
4. 2.7 μm and 0.42 × 10⁵ V/cm

Correct answer: 0.3 μm and 4.5 × 10⁵ V/cm

Solution

Charge neutrality gives x_p N_A = x_n N_D, so x_n = 9 x_p; with total 3 um, x_p = 0.3 um. E_max = q N_A x_p / eps = (1.6e-19)(9e16)(0.3e-4)/(1.04e-12) = 4.15e5 ~ 4.5e5 V/cm. Correct: 0.3 um and 4.5e5 V/cm.

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