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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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Given information:
Doping concentration of donor atoms = 1014 cm-3
Doping concentration of acceptor atoms = 7 x 1013 cm-3
Resistivity of pure germanium = 60 cm
Applied electric field = 2 V/cm
Calculation:
1. Calculation of conductivity:
The conductivity of the doped germanium can be calculated as follows:
σ = q (pμp + nμn)
where,
q = charge of an electron = 1.6 x 10-19 C
p = concentration of holes (acceptor doping) = 7 x 1013 cm-3
μp = mobility of holes = 0.05 cm2/Vs
n = concentration of electrons (donor doping) = 1014 cm-3
μn = mobility of electrons = 0.39 cm2/Vs
Substituting the values, we get:
σ = 1.6 x 10-19 (7 x 1013 x 0.05 + 1014 x 0.39)
σ = 0.57 (S/cm)
2. Calculation of resistivity:
The resistivity of doped germanium can be calculated as follows:
ρ = 1/σ
ρ = 1/0.57
ρ = 1.75 Ω-cm
3. Calculation of current density:
The current density can be calculated using Ohm's law as follows:
J = σE
where,
E = electric field = 2 V/cm
Substituting the values, we get:
J = 0.57 x 2
J = 1.14 (A/cm2)
The total current density can be calculated as follows:
Jtotal = Jp + Jn
where,
Jp = current density due to holes (acceptor doping)
Jn = current density due to electrons (donor doping)
Jp can be calculated as follows:
Jp = qpμpE
Jp = 1.6 x 10-19 x 7 x 1013 x 0.05 x 2
Jp = 0.028 (A/cm2)
Jn can be calculated as follows:
Jn = qnμnE
Jn = 1.6 x 10-19 x 1014 x 0.39 x 2
Jn = 0.82 (A/cm2)
Substituting the values, we get:
Jtotal = 0.82 - 0.028
Jtotal = 0.792 (A/cm2)
The current density in terms of milliamperes per square centimeter (mA/cm2) can be calculated as follows:
Jtotal (mA/cm2) = Jtotal (A/cm2) x 1000
Jtotal (mA/cm2) = 0.792 x 1000
Jtotal (mA/cm2) = 792 mA/cm2
Therefore, the correct option is (A) 52.24 mA/cm2.
Doping concentration of donor atoms = 1014 cm-3
Doping concentration of acceptor atoms = 7 x 1013 cm-3
Resistivity of pure germanium = 60 cm
Applied electric field = 2 V/cm
Calculation:
1. Calculation of conductivity:
The conductivity of the doped germanium can be calculated as follows:
σ = q (pμp + nμn)
where,
q = charge of an electron = 1.6 x 10-19 C
p = concentration of holes (acceptor doping) = 7 x 1013 cm-3
μp = mobility of holes = 0.05 cm2/Vs
n = concentration of electrons (donor doping) = 1014 cm-3
μn = mobility of electrons = 0.39 cm2/Vs
Substituting the values, we get:
σ = 1.6 x 10-19 (7 x 1013 x 0.05 + 1014 x 0.39)
σ = 0.57 (S/cm)
2. Calculation of resistivity:
The resistivity of doped germanium can be calculated as follows:
ρ = 1/σ
ρ = 1/0.57
ρ = 1.75 Ω-cm
3. Calculation of current density:
The current density can be calculated using Ohm's law as follows:
J = σE
where,
E = electric field = 2 V/cm
Substituting the values, we get:
J = 0.57 x 2
J = 1.14 (A/cm2)
The total current density can be calculated as follows:
Jtotal = Jp + Jn
where,
Jp = current density due to holes (acceptor doping)
Jn = current density due to electrons (donor doping)
Jp can be calculated as follows:
Jp = qpμpE
Jp = 1.6 x 10-19 x 7 x 1013 x 0.05 x 2
Jp = 0.028 (A/cm2)
Jn can be calculated as follows:
Jn = qnμnE
Jn = 1.6 x 10-19 x 1014 x 0.39 x 2
Jn = 0.82 (A/cm2)
Substituting the values, we get:
Jtotal = 0.82 - 0.028
Jtotal = 0.792 (A/cm2)
The current density in terms of milliamperes per square centimeter (mA/cm2) can be calculated as follows:
Jtotal (mA/cm2) = Jtotal (A/cm2) x 1000
Jtotal (mA/cm2) = 0.792 x 1000
Jtotal (mA/cm2) = 792 mA/cm2
Therefore, the correct option is (A) 52.24 mA/cm2.
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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 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 1014 donor atoms/cm3 and 7 x 1013 acceptor 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 Electronics and Communication Engineering (ECE) 2024 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 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 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 Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 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 1014 donor atoms/cm3 and 7 x 1013 acceptor 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 Electronics and Communication Engineering (ECE) 2024 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 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 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 Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 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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