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A silicon pn junction with doping profile of Na =1016 cm-3 and Nd =10 15 cm-3 has a cross sectional area of 10-2 cm2. The length of the p - region is 2 mm and length of the n - region is 1 mm. The approximately series resistance of the diode is
- a)62Ω
- b)43Ω
- c)72Ω
- d)81Ω
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A silicon pn junction with doping profile of Na=1016cm-3and Nd =10 15 ...
To calculate the series resistance of the diode, we need to consider the resistance contributed by both the p-region and the n-region.
1. Resistance of the p-region:
The resistance of the p-region can be calculated using the formula:
Rp = (1/qμp) * (Lp/Ap) * (Na^2/2)
Where:
- Rp is the resistance of the p-region
- q is the charge of an electron (1.6 x 10^-19 C)
- μp is the mobility of the holes in the p-region (assuming a value of 500 cm^2/Vs for silicon at room temperature)
- Lp is the length of the p-region (2 mm = 0.2 cm)
- Ap is the cross-sectional area of the p-region (10^-2 cm^2)
- Na is the acceptor doping concentration (10^16 cm^-3)
Substituting the given values into the formula, we get:
Rp = (1/(1.6 x 10^-19 C * 500 cm^2/Vs)) * (0.2 cm/(10^-2 cm^2)) * ((10^16 cm^-3)^2/2)
Simplifying the expression, we get:
Rp = 62.5 Ω
2. Resistance of the n-region:
Similarly, the resistance of the n-region can be calculated using the same formula:
Rn = (1/qμn) * (Ln/An) * (Nd^2/2)
Where:
- Rn is the resistance of the n-region
- μn is the mobility of the electrons in the n-region (assuming a value of 1350 cm^2/Vs for silicon at room temperature)
- Ln is the length of the n-region (1 mm = 0.1 cm)
- An is the cross-sectional area of the n-region (10^-2 cm^2)
- Nd is the donor doping concentration (10^15 cm^-3)
Substituting the given values into the formula, we get:
Rn = (1/(1.6 x 10^-19 C * 1350 cm^2/Vs)) * (0.1 cm/(10^-2 cm^2)) * ((10^15 cm^-3)^2/2)
Simplifying the expression, we get:
Rn = 9.26 Ω
3. Total series resistance:
The total series resistance of the diode is the sum of the resistances of the p-region and the n-region:
Rtotal = Rp + Rn
Substituting the calculated values, we get:
Rtotal = 62.5 Ω + 9.26 Ω
Rtotal = 71.76 Ω
Since the question asks for an approximate value, we can round off the answer to the nearest whole number, which is 72 Ω. Therefore, the correct answer is option C.
1. Resistance of the p-region:
The resistance of the p-region can be calculated using the formula:
Rp = (1/qμp) * (Lp/Ap) * (Na^2/2)
Where:
- Rp is the resistance of the p-region
- q is the charge of an electron (1.6 x 10^-19 C)
- μp is the mobility of the holes in the p-region (assuming a value of 500 cm^2/Vs for silicon at room temperature)
- Lp is the length of the p-region (2 mm = 0.2 cm)
- Ap is the cross-sectional area of the p-region (10^-2 cm^2)
- Na is the acceptor doping concentration (10^16 cm^-3)
Substituting the given values into the formula, we get:
Rp = (1/(1.6 x 10^-19 C * 500 cm^2/Vs)) * (0.2 cm/(10^-2 cm^2)) * ((10^16 cm^-3)^2/2)
Simplifying the expression, we get:
Rp = 62.5 Ω
2. Resistance of the n-region:
Similarly, the resistance of the n-region can be calculated using the same formula:
Rn = (1/qμn) * (Ln/An) * (Nd^2/2)
Where:
- Rn is the resistance of the n-region
- μn is the mobility of the electrons in the n-region (assuming a value of 1350 cm^2/Vs for silicon at room temperature)
- Ln is the length of the n-region (1 mm = 0.1 cm)
- An is the cross-sectional area of the n-region (10^-2 cm^2)
- Nd is the donor doping concentration (10^15 cm^-3)
Substituting the given values into the formula, we get:
Rn = (1/(1.6 x 10^-19 C * 1350 cm^2/Vs)) * (0.1 cm/(10^-2 cm^2)) * ((10^15 cm^-3)^2/2)
Simplifying the expression, we get:
Rn = 9.26 Ω
3. Total series resistance:
The total series resistance of the diode is the sum of the resistances of the p-region and the n-region:
Rtotal = Rp + Rn
Substituting the calculated values, we get:
Rtotal = 62.5 Ω + 9.26 Ω
Rtotal = 71.76 Ω
Since the question asks for an approximate value, we can round off the answer to the nearest whole number, which is 72 Ω. Therefore, the correct answer is option C.
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A silicon pn junction with doping profile of Na=1016cm-3and Nd =10 15 cm-3 has a cross sectional area of 10-2 cm2. The length of the p - region is 2 mm and length of the n - region is 1 mm. The approximately series resistance of the diode isa)62Ωb)43Ωc)72Ωd)81ΩCorrect answer is option 'C'. Can you explain this answer?
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