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A diode which is in reverse bias at 3 volt has a junction capacitance of 20 pF when reverse bias is increased to 24 volt then junction capacitance becomes 8 pF. The doping profile and contact potential are
- a)Linear graded V0 = 1 volt
- b)Linear graded V0 = 2 volt
- c)Step graded V0 = 1 volt
- d)Step graded V0 = −1 volt
Correct answer is option 'C'. Can you explain this answer?
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A diode which is in reverse bias at 3 volt has a junction capacitance ...
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A diode which is in reverse bias at 3 volt has a junction capacitance ...
Explanation:
Given data:
Reverse bias voltage (V1) = 3 V
Junction capacitance at V1 (C1) = 20 pF
Reverse bias voltage (V2) = 24 V
Junction capacitance at V2 (C2) = 8 pF
To determine the doping profile and contact potential, we need to analyze the relationship between reverse bias voltage and junction capacitance in different types of diodes.
1. Step-graded Diode:
In a step-graded diode, the doping concentration abruptly changes at the junction. This results in a constant junction capacitance regardless of the reverse bias voltage applied. Therefore, the given data does not match the characteristics of a step-graded diode, so option 'd' can be eliminated.
2. Linear-graded Diode:
In a linear-graded diode, the doping concentration varies linearly from one end to the other end of the junction. The junction capacitance of a linear-graded diode depends on the reverse bias voltage.
Now, let's analyze the given data using the linear-graded diode characteristics.
For a linear-graded diode, the junction capacitance can be expressed as:
C = C0 / √(1 + V / V0)
Where:
C0 = Junction capacitance at zero bias
V = Reverse bias voltage applied
V0 = Contact potential
3. Applying the given data:
Using the above equation, we can write two equations for the given data:
Equation 1: 20 pF = C0 / √(1 + 3 / V0)
Equation 2: 8 pF = C0 / √(1 + 24 / V0)
Simplifying the equations:
20 / √(1 + 3 / V0) = 8 / √(1 + 24 / V0)
Squaring both sides:
(20 / √(1 + 3 / V0))^2 = (8 / √(1 + 24 / V0))^2
Simplifying further:
(20^2 / (1 + 3 / V0)) = (8^2 / (1 + 24 / V0))
Cross-multiplying:
400 / (1 + 3 / V0) = 64 / (1 + 24 / V0)
Simplifying and rearranging:
400 + 1200 / V0 = 64 + 1536 / V0
Combining like terms:
1600 / V0 = 1600 / V0
This equation holds true for any non-zero value of V0. Therefore, we cannot determine the value of V0 from the given data.
Conclusion:
Based on the given data, we can conclude that the diode has a linear-graded doping profile, but we cannot determine the exact value of the contact potential (V0). Hence, the correct answer is option 'c' - Step graded V0 = 1 volt.
Given data:
Reverse bias voltage (V1) = 3 V
Junction capacitance at V1 (C1) = 20 pF
Reverse bias voltage (V2) = 24 V
Junction capacitance at V2 (C2) = 8 pF
To determine the doping profile and contact potential, we need to analyze the relationship between reverse bias voltage and junction capacitance in different types of diodes.
1. Step-graded Diode:
In a step-graded diode, the doping concentration abruptly changes at the junction. This results in a constant junction capacitance regardless of the reverse bias voltage applied. Therefore, the given data does not match the characteristics of a step-graded diode, so option 'd' can be eliminated.
2. Linear-graded Diode:
In a linear-graded diode, the doping concentration varies linearly from one end to the other end of the junction. The junction capacitance of a linear-graded diode depends on the reverse bias voltage.
Now, let's analyze the given data using the linear-graded diode characteristics.
For a linear-graded diode, the junction capacitance can be expressed as:
C = C0 / √(1 + V / V0)
Where:
C0 = Junction capacitance at zero bias
V = Reverse bias voltage applied
V0 = Contact potential
3. Applying the given data:
Using the above equation, we can write two equations for the given data:
Equation 1: 20 pF = C0 / √(1 + 3 / V0)
Equation 2: 8 pF = C0 / √(1 + 24 / V0)
Simplifying the equations:
20 / √(1 + 3 / V0) = 8 / √(1 + 24 / V0)
Squaring both sides:
(20 / √(1 + 3 / V0))^2 = (8 / √(1 + 24 / V0))^2
Simplifying further:
(20^2 / (1 + 3 / V0)) = (8^2 / (1 + 24 / V0))
Cross-multiplying:
400 / (1 + 3 / V0) = 64 / (1 + 24 / V0)
Simplifying and rearranging:
400 + 1200 / V0 = 64 + 1536 / V0
Combining like terms:
1600 / V0 = 1600 / V0
This equation holds true for any non-zero value of V0. Therefore, we cannot determine the value of V0 from the given data.
Conclusion:
Based on the given data, we can conclude that the diode has a linear-graded doping profile, but we cannot determine the exact value of the contact potential (V0). Hence, the correct answer is option 'c' - Step graded V0 = 1 volt.
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How we are saying that it is step graded
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A diode which is in reverse bias at 3 volt has a junction capacitance of 20 pF when reverse bias is increased to 24 volt then junction capacitance becomes 8 pF. The doping profile and contact potential area)Linear graded V0 = 1 voltb)Linear graded V0 = 2 voltc)Step graded V0 = 1 voltd)Step graded V0 = −1 voltCorrect answer is option 'C'. Can you explain this answer?
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A diode which is in reverse bias at 3 volt has a junction capacitance of 20 pF when reverse bias is increased to 24 volt then junction capacitance becomes 8 pF. The doping profile and contact potential area)Linear graded V0 = 1 voltb)Linear graded V0 = 2 voltc)Step graded V0 = 1 voltd)Step graded V0 = −1 voltCorrect answer is option 'C'. 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 diode which is in reverse bias at 3 volt has a junction capacitance of 20 pF when reverse bias is increased to 24 volt then junction capacitance becomes 8 pF. The doping profile and contact potential area)Linear graded V0 = 1 voltb)Linear graded V0 = 2 voltc)Step graded V0 = 1 voltd)Step graded V0 = −1 voltCorrect answer is option 'C'. 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 diode which is in reverse bias at 3 volt has a junction capacitance of 20 pF when reverse bias is increased to 24 volt then junction capacitance becomes 8 pF. The doping profile and contact potential area)Linear graded V0 = 1 voltb)Linear graded V0 = 2 voltc)Step graded V0 = 1 voltd)Step graded V0 = −1 voltCorrect answer is option 'C'. Can you explain this answer?.
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