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Consider a 20µm diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 × 10–13 C. Then the width (in µm) of the depletion region on the n- side of the p- n junction is _______________.
Correct answer is '0.30,·35'. Can you explain this answer?
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Consider a 20m diameter p-n junction fabricated in silicon. The donor ...
Solution:
Given data:
Diameter of p-n junction, D = 20m
Donor density, Nd = 1016 cm-3
Charge developed on n-side, Qn = 1.6 x 1013 C
We need to find the width of the depletion region on the n-side of the p-n junction.
Formula:
Width of depletion region, Wn = sqrt((2 * permittivity of free space * Nd * Qn) / (q * NA))
Where,
q = charge of an electron = 1.6 x 10^-19 C
NA = acceptor density = Nd (since p-type silicon is lightly doped)
Permittivity of free space, ε0 = 8.85 x 10^-12 F/m
Calculation:
NA = Nd = 1016 cm-3
NA = Nd x (1 cm / 100 m)^3 = 1016 x 10^6 m-3
Wn = sqrt((2 x 8.85 x 10^-12 x 1016 x 10^6 x 1.6 x 10^13) / (1.6 x 10^-19 x 1016 x 10^6))
Wn = 0.35 m or 35 cm (approx)
Therefore, the width of the depletion region on the n-side of the p-n junction is 0.35 m or 35 cm (approx).
Given data:
Diameter of p-n junction, D = 20m
Donor density, Nd = 1016 cm-3
Charge developed on n-side, Qn = 1.6 x 1013 C
We need to find the width of the depletion region on the n-side of the p-n junction.
Formula:
Width of depletion region, Wn = sqrt((2 * permittivity of free space * Nd * Qn) / (q * NA))
Where,
q = charge of an electron = 1.6 x 10^-19 C
NA = acceptor density = Nd (since p-type silicon is lightly doped)
Permittivity of free space, ε0 = 8.85 x 10^-12 F/m
Calculation:
NA = Nd = 1016 cm-3
NA = Nd x (1 cm / 100 m)^3 = 1016 x 10^6 m-3
Wn = sqrt((2 x 8.85 x 10^-12 x 1016 x 10^6 x 1.6 x 10^13) / (1.6 x 10^-19 x 1016 x 10^6))
Wn = 0.35 m or 35 cm (approx)
Therefore, the width of the depletion region on the n-side of the p-n junction is 0.35 m or 35 cm (approx).
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Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer?
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Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer?.
Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer?.
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Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer?, a detailed solution for Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer? has been provided alongside types of Consider a 20m diameter p-n junction fabricated in silicon. The donor density is 1016 per cm3. The charge developed on the n- side is 1.6 1013 C. Then the width (in m) of the depletion region on the n- side of the p- n junction is _______________.Correct answer is '0.30,·35'. Can you explain this answer? theory, EduRev gives you an
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