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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?
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 /cm(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:
Free Test
Community 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 /cm(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:
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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?
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