EC > GATE 2014 SET-2 > PN Junction
When a silicon diode having a doping concentration of NA = 9 × 1016 cm-3 on p-side and ND = 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
Correct : c
Given:
NA = 9 × 1016 cm-3
ND = 1 × 1016 cm-3
Total depletion width W = 3 μm = 3 × 10-4 cm
ε = 1.04 × 10-12 F/cm
Depletion width on p-side (xp):
By charge neutrality: NA × xp = ND × xn
xp ÷ xn = ND ÷ NA = 1 × 1016 ÷ 9 × 1016 = 1 ÷ 9
xp + xn = 3 μm
xp = (1 ÷ 10) × 3 = 0.3 μm
xn = (9 ÷ 10) × 3 = 2.7 μm = 2.7 × 10-4 cm
Maximum Electric Field (Emax):
Emax = (q × ND × xn) ÷ ε
Emax = (1.6 × 10-19 × 1 × 1016 × 2.7 × 10-4) ÷ 1.04 × 10-12
Emax = (4.32 × 10-7) ÷ (1.04 × 10-12)
Emax = 4.15 × 105 V/cm
Depletion width on p-side = 0.3 μm and Maximum Electric Field = 4.15 × 105 V/cm (Option B)
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