GATE Exam > GATE Questions > The built-in potential of an abrupt p-n junc...
Start Learning for Free
The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.
- a)3.4
- b)2.5
- c)4.7
- d)3.5
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The built-in potential of an abrupt p-n junction is 0.75 V. If its ju...
So, answer is 2.5.
Most Upvoted Answer
The built-in potential of an abrupt p-n junction is 0.75 V. If its ju...
Given data:
Built-in potential (Vbi) = 0.75 V
Junction capacitance at VR = 1.25 V (CJ1) = 5 pF
VR2 = 7.25 V
We know that the junction capacitance (CJ) is given by the formula:
CJ = (sqrt(q*ε*NA*ND)/(Vbi))*(1/VR - 1/Vbi)
where,
q = charge of an electron = 1.6 x 10^-19 C
ε = permittivity of free space = 8.85 x 10^-12 F/m
NA and ND = acceptor and donor doping concentrations
VR = reverse bias voltage
To find CJ at VR2, we need to find the value of NA and ND at VR2.
Calculation:
Given that Vbi = 0.75 V, we can find the doping concentration ratio (NR) using the equation:
NR = exp(q*Vbi/kt)
where, k = Boltzmann constant = 1.38 x 10^-23 J/K
T = temperature in Kelvin
Assuming room temperature (T = 300 K), we get:
NR = exp(0.75*1.6 x 10^-19/1.38 x 10^-23*300) = 4.4 x 10^15
Since this is an abrupt p-n junction, we can assume that NA = ND = NR/2 = 2.2 x 10^15
Using this value of NA and ND, we can find CJ at VR2 using the formula:
CJ2 = (sqrt(q*ε*NA*ND)/(Vbi))*(1/VR2 - 1/Vbi)
= (sqrt(1.6 x 10^-19*8.85 x 10^-12*2.2 x 10^15*2.2 x 10^15)/(0.75))*(1/7.25 - 1/0.75)
= 2.5 pF (approx)
Therefore, the answer is option B (2.5 pF).
Built-in potential (Vbi) = 0.75 V
Junction capacitance at VR = 1.25 V (CJ1) = 5 pF
VR2 = 7.25 V
We know that the junction capacitance (CJ) is given by the formula:
CJ = (sqrt(q*ε*NA*ND)/(Vbi))*(1/VR - 1/Vbi)
where,
q = charge of an electron = 1.6 x 10^-19 C
ε = permittivity of free space = 8.85 x 10^-12 F/m
NA and ND = acceptor and donor doping concentrations
VR = reverse bias voltage
To find CJ at VR2, we need to find the value of NA and ND at VR2.
Calculation:
Given that Vbi = 0.75 V, we can find the doping concentration ratio (NR) using the equation:
NR = exp(q*Vbi/kt)
where, k = Boltzmann constant = 1.38 x 10^-23 J/K
T = temperature in Kelvin
Assuming room temperature (T = 300 K), we get:
NR = exp(0.75*1.6 x 10^-19/1.38 x 10^-23*300) = 4.4 x 10^15
Since this is an abrupt p-n junction, we can assume that NA = ND = NR/2 = 2.2 x 10^15
Using this value of NA and ND, we can find CJ at VR2 using the formula:
CJ2 = (sqrt(q*ε*NA*ND)/(Vbi))*(1/VR2 - 1/Vbi)
= (sqrt(1.6 x 10^-19*8.85 x 10^-12*2.2 x 10^15*2.2 x 10^15)/(0.75))*(1/7.25 - 1/0.75)
= 2.5 pF (approx)
Therefore, the answer is option B (2.5 pF).
|
Explore Courses for GATE exam
|
|
Similar GATE Doubts
The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer?
Question Description
The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer?.
The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer?.
Solutions for The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.
Here you can find the meaning of The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer?, a detailed solution for The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.a)3.4b)2.5c)4.7d)3.5Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice GATE tests.
|
|
Explore Courses for GATE 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
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup