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A pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW cm-2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10-19 C. The average optical generation rate (in cm-3s-1) is
- a)0.84 x 1019
- b)5.57 x 1019
- c)1.04 x 1019
- d)83.60 x 1019
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A pn junction solar cell of area 1.0 cm2, illuminated uniformly with 1...
Nm.
To find the current output of the solar cell, we can use the formula:
Power output = Efficiency * Power input
The power input can be calculated by multiplying the incident power density with the area:
Power input = Incident power density * Area
Substituting the given values, we have:
Power input = 100 mW cm-2 * 1.0 cm2 = 100 mW
Now, substituting the efficiency, we can calculate the power output:
Power output = 0.15 * 100 mW = 15 mW
Next, we can calculate the short circuit current (Isc) using the fill factor (FF) and the open circuit voltage (Voc):
Isc = Power output / Voc
Substituting the values:
Isc = 15 mW / 0.7 V ≈ 21.43 mA
Therefore, the current output of the solar cell is approximately 21.43 mA.
To find the current output of the solar cell, we can use the formula:
Power output = Efficiency * Power input
The power input can be calculated by multiplying the incident power density with the area:
Power input = Incident power density * Area
Substituting the given values, we have:
Power input = 100 mW cm-2 * 1.0 cm2 = 100 mW
Now, substituting the efficiency, we can calculate the power output:
Power output = 0.15 * 100 mW = 15 mW
Next, we can calculate the short circuit current (Isc) using the fill factor (FF) and the open circuit voltage (Voc):
Isc = Power output / Voc
Substituting the values:
Isc = 15 mW / 0.7 V ≈ 21.43 mA
Therefore, the current output of the solar cell is approximately 21.43 mA.
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Community Answer
A pn junction solar cell of area 1.0 cm2, illuminated uniformly with 1...
Concept:
The fill factor of a solar cell is defined as:
Efficiency is the ratio of the output power to the input power, i.e.,
Also, the optical generation rate (GL) will be:
Calculation:
Given η = 0.25, FF = 0.8, Voc = 0.7 and thickness = 200 μm.
The optical generation rate (using equation (2)) will be:
Efficiency is the ratio of the output power to the input power, i.e.,
Also, the optical generation rate (GL) will be:
Calculation:
Given η = 0.25, FF = 0.8, Voc = 0.7 and thickness = 200 μm.
The optical generation rate (using equation (2)) will be:
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A pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW cm-2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10-19 C. The average optical generation rate (in cm-3s-1) isa)0.84 x 1019b)5.57 x 1019c)1.04 x 1019d)83.60 x 1019Correct answer is option 'A'. Can you explain this answer?
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A pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW cm-2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10-19 C. The average optical generation rate (in cm-3s-1) isa)0.84 x 1019b)5.57 x 1019c)1.04 x 1019d)83.60 x 1019Correct answer is option 'A'. 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 pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW cm-2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10-19 C. The average optical generation rate (in cm-3s-1) isa)0.84 x 1019b)5.57 x 1019c)1.04 x 1019d)83.60 x 1019Correct answer is option 'A'. 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 pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW cm-2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10-19 C. The average optical generation rate (in cm-3s-1) isa)0.84 x 1019b)5.57 x 1019c)1.04 x 1019d)83.60 x 1019Correct answer is option 'A'. Can you explain this answer?.
A pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW cm-2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10-19 C. The average optical generation rate (in cm-3s-1) isa)0.84 x 1019b)5.57 x 1019c)1.04 x 1019d)83.60 x 1019Correct answer is option 'A'. 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 pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW cm-2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10-19 C. The average optical generation rate (in cm-3s-1) isa)0.84 x 1019b)5.57 x 1019c)1.04 x 1019d)83.60 x 1019Correct answer is option 'A'. 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 pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW cm-2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10-19 C. The average optical generation rate (in cm-3s-1) isa)0.84 x 1019b)5.57 x 1019c)1.04 x 1019d)83.60 x 1019Correct answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice A pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW cm-2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10-19 C. The average optical generation rate (in cm-3s-1) isa)0.84 x 1019b)5.57 x 1019c)1.04 x 1019d)83.60 x 1019Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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