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For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7
- a)0.68
- b)0.72
Correct answer is between '0.68,0.72'. Can you explain this answer?
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For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 ×...
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For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 ×...
And ND = 0, the depletion width can be calculated using the following formula:
W = sqrt((2 * ε * Vbi) / ((1 / NA) + (1 / ND)))
Where:
- W is the depletion width
- ε is the permittivity of Si (11.8 * 10^-12 F/cm)
- Vbi is the built-in potential, which can be calculated using the following equation:
Vbi = (kT / q) * ln(NA * ND / ni^2)
Where:
- k is the Boltzmann constant (1.38 * 10^-23 J/K)
- T is the temperature in Kelvin
- q is the electron charge (1.6 * 10^-19 C)
- ni is the intrinsic carrier concentration of Si at the given temperature, which can be calculated using the following equation:
ni^2 = Nc * Nv * exp(-Eg / (kT))
Where:
- Nc is the effective density of states in the conduction band (2.8 * 10^19 cm^-3 at 300K)
- Nv is the effective density of states in the valence band (1.04 * 10^19 cm^-3 at 300K)
- Eg is the energy gap of Si (1.12 eV at room temperature)
Assuming a temperature of 300K, we can calculate:
- ni^2 = 1.5 * 10^10 cm^-6
- Vbi = (0.026 / 1.6) * ln(2 * 10^19 * 1.5 * 10^10 / (1.5 * 10^10)^2) = 0.821 V
Plugging these values into the depletion width formula, we get:
W = sqrt((2 * 11.8 * 10^-12 * 0.821) / ((1 / 2 * 10^19) + (1 / 0))) = 0.634 μm
Therefore, the depletion width for a Si one sided abrupt junction with NA = 2 × 10^19/cm^3 and ND = 0 is 0.634 μm.
W = sqrt((2 * ε * Vbi) / ((1 / NA) + (1 / ND)))
Where:
- W is the depletion width
- ε is the permittivity of Si (11.8 * 10^-12 F/cm)
- Vbi is the built-in potential, which can be calculated using the following equation:
Vbi = (kT / q) * ln(NA * ND / ni^2)
Where:
- k is the Boltzmann constant (1.38 * 10^-23 J/K)
- T is the temperature in Kelvin
- q is the electron charge (1.6 * 10^-19 C)
- ni is the intrinsic carrier concentration of Si at the given temperature, which can be calculated using the following equation:
ni^2 = Nc * Nv * exp(-Eg / (kT))
Where:
- Nc is the effective density of states in the conduction band (2.8 * 10^19 cm^-3 at 300K)
- Nv is the effective density of states in the valence band (1.04 * 10^19 cm^-3 at 300K)
- Eg is the energy gap of Si (1.12 eV at room temperature)
Assuming a temperature of 300K, we can calculate:
- ni^2 = 1.5 * 10^10 cm^-6
- Vbi = (0.026 / 1.6) * ln(2 * 10^19 * 1.5 * 10^10 / (1.5 * 10^10)^2) = 0.821 V
Plugging these values into the depletion width formula, we get:
W = sqrt((2 * 11.8 * 10^-12 * 0.821) / ((1 / 2 * 10^19) + (1 / 0))) = 0.634 μm
Therefore, the depletion width for a Si one sided abrupt junction with NA = 2 × 10^19/cm^3 and ND = 0 is 0.634 μm.
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For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7a)0.68b)0.72Correct answer is between '0.68,0.72'. Can you explain this answer?
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For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7a)0.68b)0.72Correct answer is between '0.68,0.72'. 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 For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7a)0.68b)0.72Correct answer is between '0.68,0.72'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7a)0.68b)0.72Correct answer is between '0.68,0.72'. Can you explain this answer?.
For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7a)0.68b)0.72Correct answer is between '0.68,0.72'. 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 For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7a)0.68b)0.72Correct answer is between '0.68,0.72'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7a)0.68b)0.72Correct answer is between '0.68,0.72'. Can you explain this answer?.
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