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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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