Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > In a p-type semiconductor, p = 1016/cm2, and ...
Start Learning for Free
In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” is
- a)-400 V/cm in -z direction
- b)-200 V/cm in +z direction
- c)-200 V/cm in -z direction
- d)-400 V/cm in +z direction
Correct answer is option 'D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a...
The force acting on a charge q placed in a magnetic field B and an electric field E is given by
F = qv x B
The velocity of the hole placed in an electric field is
Now, F - + q E ...(ii) (+q = charge on a hole)
Comparing equations (i) and (ii), the electric field due to the Hall effect will be -400 V/cm in +z direction.
F = qv x B
The velocity of the hole placed in an electric field is
Now, F - + q E ...(ii) (+q = charge on a hole)
Comparing equations (i) and (ii), the electric field due to the Hall effect will be -400 V/cm in +z direction.
Most Upvoted Answer
In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a...
The acceptor concentration is Na = 1018/cm3. The electron concentration is given by the equation:
np = ni2
where np is the product of the electron and hole concentrations, and ni is the intrinsic carrier concentration.
The intrinsic carrier concentration is given by the equation:
ni2 = (Nc * Nv) * exp(-Eg / (2 * k * T))
where Nc is the effective density of states in the conduction band, Nv is the effective density of states in the valence band, Eg is the energy gap between the conduction and valence bands, k is Boltzmann's constant, and T is the temperature in Kelvin.
Assuming typical values for silicon:
Nc = 2.8 * 10^19/cm3
Nv = 1.04 * 10^19/cm3
Eg = 1.12 eV
k = 8.6173 * 10^-5 eV/K
T = 300 K
Plugging in the values:
ni2 = (2.8 * 10^19/cm3 * 1.04 * 10^19/cm3) * exp(-1.12 eV / (2 * 8.6173 * 10^-5 eV/K * 300 K))
= 3.1 * 10^20/cm6 * exp(-1.12 / 0.0519)
= 3.1 * 10^20/cm6 * 0.000002
= 6.2 * 10^14/cm6
Since np = ni2, the electron concentration can be calculated as:
np = 6.2 * 10^14/cm6
= 6.2 * 10^14/cm6 / (1/cm3)
= 6.2 * 10^14/cm3
Therefore, in the given p-type semiconductor with p = 10^16/cm2 and Na = 10^18/cm3, the electron concentration is approximately 6.2 * 10^14/cm3.
np = ni2
where np is the product of the electron and hole concentrations, and ni is the intrinsic carrier concentration.
The intrinsic carrier concentration is given by the equation:
ni2 = (Nc * Nv) * exp(-Eg / (2 * k * T))
where Nc is the effective density of states in the conduction band, Nv is the effective density of states in the valence band, Eg is the energy gap between the conduction and valence bands, k is Boltzmann's constant, and T is the temperature in Kelvin.
Assuming typical values for silicon:
Nc = 2.8 * 10^19/cm3
Nv = 1.04 * 10^19/cm3
Eg = 1.12 eV
k = 8.6173 * 10^-5 eV/K
T = 300 K
Plugging in the values:
ni2 = (2.8 * 10^19/cm3 * 1.04 * 10^19/cm3) * exp(-1.12 eV / (2 * 8.6173 * 10^-5 eV/K * 300 K))
= 3.1 * 10^20/cm6 * exp(-1.12 / 0.0519)
= 3.1 * 10^20/cm6 * 0.000002
= 6.2 * 10^14/cm6
Since np = ni2, the electron concentration can be calculated as:
np = 6.2 * 10^14/cm6
= 6.2 * 10^14/cm6 / (1/cm3)
= 6.2 * 10^14/cm3
Therefore, in the given p-type semiconductor with p = 10^16/cm2 and Na = 10^18/cm3, the electron concentration is approximately 6.2 * 10^14/cm3.
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. Can you explain this answer?
Question Description
In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. 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 In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. 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 In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. Can you explain this answer?.
In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. 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 In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. 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 In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. Can you explain this answer?.
Solutions for In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” isa)-400 V/cm in -z directionb)-200 V/cm in +z directionc)-200 V/cm in -z directiond)-400 V/cm in +z directionCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup