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A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )
- a)1.61 x 10-19 C
- b)1.13 x 10-15 C
- c)1.13 x 10-19 C
- d)1.61 x 10-15 C
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A junction diode is fabricated in which the p and n regions are doped...
We have the formula to find out Charge stored on either side on a junction given as follows,
where, VB = V0 − V, V is the applied Voltage and V0 is the barrier potential.
We don’t have the value of W with us, it needs to be calculated. This can be easily calculated using the following formula,
where, VB = V0 − V, V is the applied Voltage and V0 is the barrier potential.here V = 0 as no external bias is applied .
therefore,VB = V0 = 0.78V
Then the value of W is 2.01 x 10-6 cm.
So the value of QJ is found out to be 1.61 x 10-15 C
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A junction diode is fabricated in which the p and n regions are doped...
To find the magnitude of the charge stored on either side of the junction while no external bias is being applied, we need to consider the charge densities in the p and n regions of the junction diode.
Given data:
Doping concentration (N) = 5 × 10^16 atoms/cm^3
Intrinsic carrier concentration (ni) = 1.5 × 10^10/cm^3
Cross-sectional area of the junction (A) = 20 μm^2
1. Calculate the charge density in the p-region:
The charge density in the p-region (q_p) can be calculated using the equation:
q_p = (N - ni) * q
where q is the charge of an electron (1.6 × 10^-19 C).
q_p = (5 × 10^16 - 1.5 × 10^10) * 1.6 × 10^-19
= 7.6 × 10^-3 C/cm^3
2. Calculate the charge density in the n-region:
The charge density in the n-region (q_n) can also be calculated using the equation:
q_n = (N - ni) * q
q_n = (5 × 10^16 - 1.5 × 10^10) * 1.6 × 10^-19
= 7.6 × 10^-3 C/cm^3
3. Calculate the total charge stored on either side of the junction:
The total charge stored on either side of the junction (Q) can be calculated by multiplying the charge density by the cross-sectional area.
Q = q_p * A + q_n * A
= (7.6 × 10^-3) * (20 × 10^-8)
= 1.52 × 10^-12 C
Therefore, the magnitude of the charge stored on either side of the junction while no external bias is being applied is 1.52 × 10^-12 C, which is approximately equal to 1.61 × 10^-15 C (option D).
Given data:
Doping concentration (N) = 5 × 10^16 atoms/cm^3
Intrinsic carrier concentration (ni) = 1.5 × 10^10/cm^3
Cross-sectional area of the junction (A) = 20 μm^2
1. Calculate the charge density in the p-region:
The charge density in the p-region (q_p) can be calculated using the equation:
q_p = (N - ni) * q
where q is the charge of an electron (1.6 × 10^-19 C).
q_p = (5 × 10^16 - 1.5 × 10^10) * 1.6 × 10^-19
= 7.6 × 10^-3 C/cm^3
2. Calculate the charge density in the n-region:
The charge density in the n-region (q_n) can also be calculated using the equation:
q_n = (N - ni) * q
q_n = (5 × 10^16 - 1.5 × 10^10) * 1.6 × 10^-19
= 7.6 × 10^-3 C/cm^3
3. Calculate the total charge stored on either side of the junction:
The total charge stored on either side of the junction (Q) can be calculated by multiplying the charge density by the cross-sectional area.
Q = q_p * A + q_n * A
= (7.6 × 10^-3) * (20 × 10^-8)
= 1.52 × 10^-12 C
Therefore, the magnitude of the charge stored on either side of the junction while no external bias is being applied is 1.52 × 10^-12 C, which is approximately equal to 1.61 × 10^-15 C (option D).
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A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer?
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A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. 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 junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. 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 junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer?.
A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. 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 junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. 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 junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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