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A sample of germanium is doped to the extent of 1014 donor atoms/cm3 and 7 x 1013 acceptor atoms/cm3. The resistivity of pure germanium is 60 Ω cm. If the applied electric field is 2V/cm, then the total conduction current density is given by
- a)52.24 mA/cm2
- b)24 mA/cm2
- c)92 mA/cm2
- d)12.12 mA/cm2
Correct answer is option 'A'. Can you explain this answer?
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A sample of germanium is doped to the extent of 1014donor atoms/cm3and...
The resistivity of pure germanium is 60 ohm-cm.
To find the resistivity of the doped germanium, we need to consider the effect of the donor and acceptor atoms on the conductivity.
The conductivity of a material is given by σ = nqμ, where σ is the conductivity, n is the carrier concentration, q is the charge of the carrier, and μ is the carrier mobility.
For donor atoms, the carrier concentration is given by n = ND, where ND is the donor concentration.
For acceptor atoms, the carrier concentration is given by p = NA, where NA is the acceptor concentration.
In a semiconductor material, the total carrier concentration is given by n = p = ni^2 / ND, where ni is the intrinsic carrier concentration.
The resistivity is given by ρ = 1/σ.
Given:
ND = 10^14 donor atoms/cm^3
NA = 7 x 10^13 acceptor atoms/cm^3
ρ = 60 ohm-cm
First, we need to calculate the intrinsic carrier concentration ni.
Using the equation ni = sqrt(NC * NV * exp(-Eg / (2kT))), where NC and NV are the effective densities of states in the conduction and valence bands, Eg is the energy gap, k is Boltzmann's constant, and T is the temperature.
For germanium at room temperature (T = 300 K):
NC = 2.8 x 10^19 cm^-3
NV = 1.04 x 10^19 cm^-3
Eg = 0.67 eV
Plugging in these values, we can calculate ni:
ni = sqrt(NC * NV * exp(-Eg / (2kT)))
= sqrt((2.8 x 10^19) * (1.04 x 10^19) * exp(-0.67 / (2 * 8.617333262145 x 10^-5 * 300)))
= 2.48 x 10^13 cm^-3
Next, we can calculate the carrier concentration n = p = ni^2 / ND:
n = p = (2.48 x 10^13)^2 / (10^14)
= 6.17 x 10^11 cm^-3
The total carrier concentration is n + p = 2n = 2 * 6.17 x 10^11 = 1.23 x 10^12 cm^-3.
Now we can calculate the mobility of the carriers.
The mobility for electrons is typically higher than the mobility for holes in germanium. Let's assume the electron mobility is 3900 cm^2/Vs and the hole mobility is 1900 cm^2/Vs.
The conductivity σ = nqμ = (1.23 x 10^12) (1.6 x 10^-19 C) (3900 cm^2/Vs)
+ (1.23 x 10^12) (1.6 x 10^-19 C) (1900 cm^2/Vs)
= 7.56 x 10^-2 + 3.69 x 10^-2
= 1.23 x 10^-1 (ohm-cm)^-1
Finally, the resistivity ρ =
To find the resistivity of the doped germanium, we need to consider the effect of the donor and acceptor atoms on the conductivity.
The conductivity of a material is given by σ = nqμ, where σ is the conductivity, n is the carrier concentration, q is the charge of the carrier, and μ is the carrier mobility.
For donor atoms, the carrier concentration is given by n = ND, where ND is the donor concentration.
For acceptor atoms, the carrier concentration is given by p = NA, where NA is the acceptor concentration.
In a semiconductor material, the total carrier concentration is given by n = p = ni^2 / ND, where ni is the intrinsic carrier concentration.
The resistivity is given by ρ = 1/σ.
Given:
ND = 10^14 donor atoms/cm^3
NA = 7 x 10^13 acceptor atoms/cm^3
ρ = 60 ohm-cm
First, we need to calculate the intrinsic carrier concentration ni.
Using the equation ni = sqrt(NC * NV * exp(-Eg / (2kT))), where NC and NV are the effective densities of states in the conduction and valence bands, Eg is the energy gap, k is Boltzmann's constant, and T is the temperature.
For germanium at room temperature (T = 300 K):
NC = 2.8 x 10^19 cm^-3
NV = 1.04 x 10^19 cm^-3
Eg = 0.67 eV
Plugging in these values, we can calculate ni:
ni = sqrt(NC * NV * exp(-Eg / (2kT)))
= sqrt((2.8 x 10^19) * (1.04 x 10^19) * exp(-0.67 / (2 * 8.617333262145 x 10^-5 * 300)))
= 2.48 x 10^13 cm^-3
Next, we can calculate the carrier concentration n = p = ni^2 / ND:
n = p = (2.48 x 10^13)^2 / (10^14)
= 6.17 x 10^11 cm^-3
The total carrier concentration is n + p = 2n = 2 * 6.17 x 10^11 = 1.23 x 10^12 cm^-3.
Now we can calculate the mobility of the carriers.
The mobility for electrons is typically higher than the mobility for holes in germanium. Let's assume the electron mobility is 3900 cm^2/Vs and the hole mobility is 1900 cm^2/Vs.
The conductivity σ = nqμ = (1.23 x 10^12) (1.6 x 10^-19 C) (3900 cm^2/Vs)
+ (1.23 x 10^12) (1.6 x 10^-19 C) (1900 cm^2/Vs)
= 7.56 x 10^-2 + 3.69 x 10^-2
= 1.23 x 10^-1 (ohm-cm)^-1
Finally, the resistivity ρ =
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A sample of germanium is doped to the extent of 1014donor atoms/cm3and 7 x 1013acceptor atoms/cm3. The resistivity of pure germanium is 60 Ωcm. If the appliedelectric field is 2V/cm, then the total conduction current density is given bya)52.24 mA/cm2b)24 mA/cm2c)92 mA/cm2d)12.12 mA/cm2Correct answer is option 'A'. Can you explain this answer?
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A sample of germanium is doped to the extent of 1014donor atoms/cm3and 7 x 1013acceptor atoms/cm3. The resistivity of pure germanium is 60 Ωcm. If the appliedelectric field is 2V/cm, then the total conduction current density is given bya)52.24 mA/cm2b)24 mA/cm2c)92 mA/cm2d)12.12 mA/cm2Correct answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A sample of germanium is doped to the extent of 1014donor atoms/cm3and 7 x 1013acceptor atoms/cm3. The resistivity of pure germanium is 60 Ωcm. If the appliedelectric field is 2V/cm, then the total conduction current density is given bya)52.24 mA/cm2b)24 mA/cm2c)92 mA/cm2d)12.12 mA/cm2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of germanium is doped to the extent of 1014donor atoms/cm3and 7 x 1013acceptor atoms/cm3. The resistivity of pure germanium is 60 Ωcm. If the appliedelectric field is 2V/cm, then the total conduction current density is given bya)52.24 mA/cm2b)24 mA/cm2c)92 mA/cm2d)12.12 mA/cm2Correct answer is option 'A'. Can you explain this answer?.
A sample of germanium is doped to the extent of 1014donor atoms/cm3and 7 x 1013acceptor atoms/cm3. The resistivity of pure germanium is 60 Ωcm. If the appliedelectric field is 2V/cm, then the total conduction current density is given bya)52.24 mA/cm2b)24 mA/cm2c)92 mA/cm2d)12.12 mA/cm2Correct answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A sample of germanium is doped to the extent of 1014donor atoms/cm3and 7 x 1013acceptor atoms/cm3. The resistivity of pure germanium is 60 Ωcm. If the appliedelectric field is 2V/cm, then the total conduction current density is given bya)52.24 mA/cm2b)24 mA/cm2c)92 mA/cm2d)12.12 mA/cm2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of germanium is doped to the extent of 1014donor atoms/cm3and 7 x 1013acceptor atoms/cm3. The resistivity of pure germanium is 60 Ωcm. If the appliedelectric field is 2V/cm, then the total conduction current density is given bya)52.24 mA/cm2b)24 mA/cm2c)92 mA/cm2d)12.12 mA/cm2Correct answer is option 'A'. Can you explain this answer?.
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