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Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016 atoms/cm3 at 300 K temperature. Determine the approximate resistivity of the slab.
(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010 /cm3; Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).
- a)0.48 Ω - cm
- b)0.35 Ω - cm
- c)0.16 Ω - cm
- d)1.25 Ω - cm
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Electric Field of 1 V/m is applied to a Boron doped Silicon semiconduc...
Concept:
For a semiconductor slab doped with excess carriers and placed in an electric field, the current equation is given by:-
J = σ E
Where J = Current Density
σ = Conductivity of the semiconductor defined as:
σ = qnμn + qpμp
n = excess electrons contributing to the current conduction.
p = excess holes contributing to the current conduction.
Also, the Resistivity is defined as the reciprocal of conductivity.
Calculation:
The given Si semiconductor is doped with Boron impurity.
So, Na = 1016 /cm3 (p-type)
q = 1.6 × 10-19 C
μn = 1300 cm2/V-s and μp = 500 cm2/V-s
The excess electrons is given by; 
Since, Na ≫ ni, we can assume that the conduction current exists only because of acceptor impunities, i.e. because of Na.
So, σ = q Na μp
ρ = Resistivity of the given Slab
Putting on the respective values, we get:
Putting on the respective values, we get:
Most Upvoted Answer
Electric Field of 1 V/m is applied to a Boron doped Silicon semiconduc...
Concept:
For a semiconductor slab doped with excess carriers and placed in an electric field, the current equation is given by:-
J = σ E
Where J = Current Density
σ = Conductivity of the semiconductor defined as:
σ = qnμn + qpμp
n = excess electrons contributing to the current conduction.
p = excess holes contributing to the current conduction.
Also, the Resistivity is defined as the reciprocal of conductivity.
Calculation:
The given Si semiconductor is doped with Boron impurity.
So, Na = 1016 /cm3 (p-type)
q = 1.6 × 10-19 C
μn = 1300 cm2/V-s and μp = 500 cm2/V-s
The excess electrons is given by;
Since, Na ≫ ni, we can assume that the conduction current exists only because of acceptor impunities, i.e. because of Na.
So, σ = q Na μp
ρ = Resistivity of the given Slab
Putting on the respective values, we get:
Putting on the respective values, we get:
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Community Answer
Electric Field of 1 V/m is applied to a Boron doped Silicon semiconduc...
X 10^10 cm^-3)
First, we need to calculate the electron and hole concentration in the Boron doped Silicon slab. Since the slab is doped with Boron, it will act as a p-type semiconductor.
Given:
Doping density = 10^16 atoms/cm^3
Intrinsic carrier concentration of Silicon at 300 K = 1.5 x 10^10 cm^-3
Since the slab is p-type, the majority carrier concentration (holes) is equal to the doping density:
p = 10^16 cm^-3
The minority carrier concentration (electrons) can be calculated using the law of mass action:
n_i^2 = p*n
n = n_i^2 / p
n = (1.5 x 10^10)^2 / 10^16
n ≈ 2.25 x 10^4 cm^-3
Now, we can calculate the resistivity of the Boron doped Silicon slab using the formula:
ρ = 1 / (q * μ * (p + n))
where,
q = 1.6 x 10^-19 C (charge of an electron)
μ = 1300 cm^2/Vs (mobility of electrons in Silicon at 300 K)
p = 10^16 cm^-3 (majority carrier concentration - holes)
n = 2.25 x 10^4 cm^-3 (minority carrier concentration - electrons)
ρ = 1 / (1.6 x 10^-19 * 1300 * (10^16 + 2.25 x 10^4))
ρ ≈ 6.1 x 10^-3 Ω-cm
Therefore, the approximate resistivity of the Boron doped Silicon slab is 6.1 x 10^-3 Ω-cm.
First, we need to calculate the electron and hole concentration in the Boron doped Silicon slab. Since the slab is doped with Boron, it will act as a p-type semiconductor.
Given:
Doping density = 10^16 atoms/cm^3
Intrinsic carrier concentration of Silicon at 300 K = 1.5 x 10^10 cm^-3
Since the slab is p-type, the majority carrier concentration (holes) is equal to the doping density:
p = 10^16 cm^-3
The minority carrier concentration (electrons) can be calculated using the law of mass action:
n_i^2 = p*n
n = n_i^2 / p
n = (1.5 x 10^10)^2 / 10^16
n ≈ 2.25 x 10^4 cm^-3
Now, we can calculate the resistivity of the Boron doped Silicon slab using the formula:
ρ = 1 / (q * μ * (p + n))
where,
q = 1.6 x 10^-19 C (charge of an electron)
μ = 1300 cm^2/Vs (mobility of electrons in Silicon at 300 K)
p = 10^16 cm^-3 (majority carrier concentration - holes)
n = 2.25 x 10^4 cm^-3 (minority carrier concentration - electrons)
ρ = 1 / (1.6 x 10^-19 * 1300 * (10^16 + 2.25 x 10^4))
ρ ≈ 6.1 x 10^-3 Ω-cm
Therefore, the approximate resistivity of the Boron doped Silicon slab is 6.1 x 10^-3 Ω-cm.
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Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer?.
Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer?.
Solutions for Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having a doping density of 1016atoms/cm3at 300 K temperature. Determine the approximate resistivity of the slab.(Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010/cm3;Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).a)0.48 Ω - cmb)0.35 Ω - cmc)0.16 Ω - cmd)1.25 Ω - cmCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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