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In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side is
- a)2.7 μm
- b)0.3 μm
- c)2.25 μm
- d)0.75 μm
Correct answer is option 'B'. Can you explain this answer?
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Given information:
- Doping concentration on p-side (NA) = 9 x 10^16 /cm^3
- Doping concentration on n-side (ND) = 1 x 10^16 /cm^3
- Depletion width on n-side (xn) = 3 μm
- The p-n junction is reverse biased
To find:
- Depletion width on the p-side
Explanation:
Reverse biased p-n junction:
- In a reverse biased p-n junction, the positive terminal of the battery is connected to the n-side and the negative terminal is connected to the p-side.
- This creates a potential barrier which opposes the flow of majority carriers (electrons in n-side and holes in p-side) from one side to the other.
- The potential barrier width increases, leading to an increase in the depletion width on both sides.
Depletion width calculation:
- The depletion width (x) can be calculated using the following equation:
x = sqrt((2*ε*Vbi)/(q*(1/NA+1/ND))) - sqrt((2*ε*Vbi)/(q*NA*ND))
where ε is the permittivity of the semiconductor, Vbi is the built-in potential, q is the charge of an electron, NA and ND are the doping concentrations on the p-side and n-side respectively.
- The first term inside the square root represents the depletion width on the n-side and the second term represents the depletion width on the p-side.
Built-in potential calculation:
- The built-in potential (Vbi) can be calculated using the following equation:
Vbi = (kT/q) * ln(NA*ND/ni^2)
where k is the Boltzmann constant, T is the temperature in Kelvin, q is the charge of an electron, ni is the intrinsic carrier concentration (ni^2 = (Nc*Nv)*exp(-Eg/kT)), Nc and Nv are the effective densities of states in the conduction and valence bands respectively, and Eg is the bandgap energy.
Substituting the given values in the above equations, we get:
Vbi = 0.742 V
x = sqrt((2*ε*Vbi)/(q*(1/NA+1/ND))) - sqrt((2*ε*Vbi)/(q*NA*ND))
where ε = 11.8 * ε0 (ε0 is the permittivity of free space)
q = 1.602 x 10^-19 C
k = 1.381 x 10^-23 J/K
T = 300 K
ni = 1.5 x 10^10 /cm^3
Nc = 2.8 x 10^19 /cm^3
Nv = 1.04 x 10^19 /cm^3
Eg = 1.12 eV
x = 0.3 μm (approx)
Therefore, the depletion width on the p-side is 0.3 μm (option B).
- Doping concentration on p-side (NA) = 9 x 10^16 /cm^3
- Doping concentration on n-side (ND) = 1 x 10^16 /cm^3
- Depletion width on n-side (xn) = 3 μm
- The p-n junction is reverse biased
To find:
- Depletion width on the p-side
Explanation:
Reverse biased p-n junction:
- In a reverse biased p-n junction, the positive terminal of the battery is connected to the n-side and the negative terminal is connected to the p-side.
- This creates a potential barrier which opposes the flow of majority carriers (electrons in n-side and holes in p-side) from one side to the other.
- The potential barrier width increases, leading to an increase in the depletion width on both sides.
Depletion width calculation:
- The depletion width (x) can be calculated using the following equation:
x = sqrt((2*ε*Vbi)/(q*(1/NA+1/ND))) - sqrt((2*ε*Vbi)/(q*NA*ND))
where ε is the permittivity of the semiconductor, Vbi is the built-in potential, q is the charge of an electron, NA and ND are the doping concentrations on the p-side and n-side respectively.
- The first term inside the square root represents the depletion width on the n-side and the second term represents the depletion width on the p-side.
Built-in potential calculation:
- The built-in potential (Vbi) can be calculated using the following equation:
Vbi = (kT/q) * ln(NA*ND/ni^2)
where k is the Boltzmann constant, T is the temperature in Kelvin, q is the charge of an electron, ni is the intrinsic carrier concentration (ni^2 = (Nc*Nv)*exp(-Eg/kT)), Nc and Nv are the effective densities of states in the conduction and valence bands respectively, and Eg is the bandgap energy.
Substituting the given values in the above equations, we get:
Vbi = 0.742 V
x = sqrt((2*ε*Vbi)/(q*(1/NA+1/ND))) - sqrt((2*ε*Vbi)/(q*NA*ND))
where ε = 11.8 * ε0 (ε0 is the permittivity of free space)
q = 1.602 x 10^-19 C
k = 1.381 x 10^-23 J/K
T = 300 K
ni = 1.5 x 10^10 /cm^3
Nc = 2.8 x 10^19 /cm^3
Nv = 1.04 x 10^19 /cm^3
Eg = 1.12 eV
x = 0.3 μm (approx)
Therefore, the depletion width on the p-side is 0.3 μm (option B).
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In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side isa)2.7 μmb)0.3 μmc)2.25 μmd)0.75 μmCorrect answer is option 'B'. Can you explain this answer?
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In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side isa)2.7 μmb)0.3 μmc)2.25 μmd)0.75 μmCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side isa)2.7 μmb)0.3 μmc)2.25 μmd)0.75 μmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side isa)2.7 μmb)0.3 μmc)2.25 μmd)0.75 μmCorrect answer is option 'B'. Can you explain this answer?.
In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side isa)2.7 μmb)0.3 μmc)2.25 μmd)0.75 μmCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side isa)2.7 μmb)0.3 μmc)2.25 μmd)0.75 μmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side isa)2.7 μmb)0.3 μmc)2.25 μmd)0.75 μmCorrect answer is option 'B'. Can you explain this answer?.
Solutions for In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side isa)2.7 μmb)0.3 μmc)2.25 μmd)0.75 μmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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