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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?
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The built-in potential of an abrupt p-n junction is 0.75 V. If its ju...
So, answer is 2.5.
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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).
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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?.
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