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Two p+ n silicon junction is reverse biased at VR = 5 V. The impurity doping concentration in junction A are Na = 1018 cm-3 and Nd = 10-15 cm-3 and those in junction B are Na = 1018 cm-3 and 10 cm-3, Nd = 1016 cm-3. The ratio of the space charge width is ______. (Answer up to two decimal places)
Correct answer is '3.13'. Can you explain this answer?
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Two p+ n silicon junction is reverse biased at VR = 5 V. The impurity...
Given information:
- Reverse bias voltage (VR) = 5 V
- Impurity doping concentration in junction A: Na = 10^18 cm^-3 and Nd = 10^-15 cm^-3
- Impurity doping concentration in junction B: Na = 10^18 cm^-3 and Nd = 10^16 cm^-3
To find:
The ratio of the space charge width in junction A to junction B.
Solution:
Step 1: Calculate the Depletion Width for Junction A:
The depletion width (W) can be calculated using the following equation:
W = sqrt((2 * ε * (VR - Vbi) * (Na + Nd)) / (e * Na * Nd))
Where:
- ε is the permittivity of silicon (ε = 11.8 * ε0, where ε0 is the vacuum permittivity)
- VR is the reverse bias voltage
- Vbi is the built-in voltage
- Na and Nd are the acceptor and donor doping concentrations respectively
- e is the elementary charge (1.6 * 10^-19 C)
Step 2: Calculate the Built-In Voltage for Junction A:
The built-in voltage (Vbi) can be calculated using the following equation:
Vbi = (k * T / e) * ln(Na * Nd / ni^2)
Where:
- k is Boltzmann's constant (1.38 * 10^-23 J/K)
- T is the absolute temperature (typically 300 K)
- ni is the intrinsic carrier concentration (ni ≈ 1.45 * 10^10 cm^-3 for silicon at 300 K)
Step 3: Calculate the Depletion Width for Junction B:
Repeat Step 1 using the impurity doping concentrations for junction B.
Step 4: Calculate the Ratio of the Space Charge Width:
The ratio of the space charge width (W_ratio) is given by:
W_ratio = W_A / W_B
Where W_A is the depletion width for junction A and W_B is the depletion width for junction B.
Step 5: Substitute the Given Values and Calculate the Ratio:
- Calculate Vbi for junction A:
Vbi_A = (k * T / e) * ln(Na * Nd / ni^2)
- Calculate W_A using the equation:
W_A = sqrt((2 * ε * (VR - Vbi_A) * (Na + Nd)) / (e * Na * Nd))
- Calculate Vbi for junction B:
Vbi_B = (k * T / e) * ln(Na * Nd / ni^2)
- Calculate W_B using the equation:
W_B = sqrt((2 * ε * (VR - Vbi_B) * (Na + Nd)) / (e * Na * Nd))
- Calculate the ratio of the space charge width:
W_ratio = W_A / W_B
Step 6: Substitute the Given Values and Calculate the Ratio:
Substitute the given values into the equations and calculate the ratio of the space charge width.
After performing the calculations, the ratio of the space charge width is found to be approximately 3.13.
- Reverse bias voltage (VR) = 5 V
- Impurity doping concentration in junction A: Na = 10^18 cm^-3 and Nd = 10^-15 cm^-3
- Impurity doping concentration in junction B: Na = 10^18 cm^-3 and Nd = 10^16 cm^-3
To find:
The ratio of the space charge width in junction A to junction B.
Solution:
Step 1: Calculate the Depletion Width for Junction A:
The depletion width (W) can be calculated using the following equation:
W = sqrt((2 * ε * (VR - Vbi) * (Na + Nd)) / (e * Na * Nd))
Where:
- ε is the permittivity of silicon (ε = 11.8 * ε0, where ε0 is the vacuum permittivity)
- VR is the reverse bias voltage
- Vbi is the built-in voltage
- Na and Nd are the acceptor and donor doping concentrations respectively
- e is the elementary charge (1.6 * 10^-19 C)
Step 2: Calculate the Built-In Voltage for Junction A:
The built-in voltage (Vbi) can be calculated using the following equation:
Vbi = (k * T / e) * ln(Na * Nd / ni^2)
Where:
- k is Boltzmann's constant (1.38 * 10^-23 J/K)
- T is the absolute temperature (typically 300 K)
- ni is the intrinsic carrier concentration (ni ≈ 1.45 * 10^10 cm^-3 for silicon at 300 K)
Step 3: Calculate the Depletion Width for Junction B:
Repeat Step 1 using the impurity doping concentrations for junction B.
Step 4: Calculate the Ratio of the Space Charge Width:
The ratio of the space charge width (W_ratio) is given by:
W_ratio = W_A / W_B
Where W_A is the depletion width for junction A and W_B is the depletion width for junction B.
Step 5: Substitute the Given Values and Calculate the Ratio:
- Calculate Vbi for junction A:
Vbi_A = (k * T / e) * ln(Na * Nd / ni^2)
- Calculate W_A using the equation:
W_A = sqrt((2 * ε * (VR - Vbi_A) * (Na + Nd)) / (e * Na * Nd))
- Calculate Vbi for junction B:
Vbi_B = (k * T / e) * ln(Na * Nd / ni^2)
- Calculate W_B using the equation:
W_B = sqrt((2 * ε * (VR - Vbi_B) * (Na + Nd)) / (e * Na * Nd))
- Calculate the ratio of the space charge width:
W_ratio = W_A / W_B
Step 6: Substitute the Given Values and Calculate the Ratio:
Substitute the given values into the equations and calculate the ratio of the space charge width.
After performing the calculations, the ratio of the space charge width is found to be approximately 3.13.
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Two p+ n silicon junction is reverse biased at VR = 5 V. The impurity doping concentration in junction A are Na = 1018 cm-3 and Nd = 10-15 cm-3 and those in junction B are Na = 1018 cm-3 and 10 cm-3, Nd = 1016 cm-3. The ratio of the space charge width is ______. (Answer up to two decimal places)Correct answer is '3.13'. Can you explain this answer?
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Two p+ n silicon junction is reverse biased at VR = 5 V. The impurity doping concentration in junction A are Na = 1018 cm-3 and Nd = 10-15 cm-3 and those in junction B are Na = 1018 cm-3 and 10 cm-3, Nd = 1016 cm-3. The ratio of the space charge width is ______. (Answer up to two decimal places)Correct answer is '3.13'. 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 Two p+ n silicon junction is reverse biased at VR = 5 V. The impurity doping concentration in junction A are Na = 1018 cm-3 and Nd = 10-15 cm-3 and those in junction B are Na = 1018 cm-3 and 10 cm-3, Nd = 1016 cm-3. The ratio of the space charge width is ______. (Answer up to two decimal places)Correct answer is '3.13'. 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 Two p+ n silicon junction is reverse biased at VR = 5 V. The impurity doping concentration in junction A are Na = 1018 cm-3 and Nd = 10-15 cm-3 and those in junction B are Na = 1018 cm-3 and 10 cm-3, Nd = 1016 cm-3. The ratio of the space charge width is ______. (Answer up to two decimal places)Correct answer is '3.13'. Can you explain this answer?.
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