The figure shows the I-V characteristics of a solar cell illuminated uniformly with solar light of power $100 \text{ mW/cm}^2$. The solar cell has an area of $3 \text{ cm}^2$ and a fill factor of $0.7$. The maximum efficiency (in %) of the device is ___________
The efficiency of a solar cell is given by:
Efficiency \( \eta = \frac{P_{max}}{P_{in}} \times 100\% \)
Where \( P_{max} \) is the maximum power output and \( P_{in} \) is the input power.
Step 1: Calculate Input Power \( P_{in} \)
The input power per area is given as \( 100 \text{ mW/cm}^2 \) and the area of the solar cell is \( 3 \text{ cm}^2 \).
\( P_{in} = 100 \text{ mW/cm}^2 \times 3 \text{ cm}^2 = 300 \text{ mW} \)
Step 2: Calculate Maximum Power Output \( P_{max} \)
The maximum power is given by:
\( P_{max} = V_{mp} \times I_{mp} \)
Where \( V_{mp} \) and \( I_{mp} \) are the voltage and current at the maximum power point.
We use the open-circuit voltage \( V_{OC} = 0.5 \text{ V} \) and short-circuit current \( I_{SC} = 180 \text{ mA} \).
The fill factor \( FF \) relates to the maximum power as:
\( FF = \frac{V_{mp} \times I_{mp}}{V_{OC} \times I_{SC}} \)
\( V_{mp} \times I_{mp} = FF \times V_{OC} \times I_{SC} \)
\( P_{max} = 0.7 \times 0.5 \text{ V} \times 180 \text{ mA} \)
\( P_{max} = 0.7 \times 0.5 \times 180 \times 10^{-3} \text{ W} \)
\( P_{max} = 0.063 \text{ W} \) or \( 63 \text{ mW} \)
Step 3: Calculate Efficiency
\( \eta = \frac{63 \text{ mW}}{300 \text{ mW}} \times 100\% = 21\% \)
Validation: The calculated efficiency of \( 21\% \) is within the range of 20.5 to 21.5.
Which among the following Electronic component works with the principle of Light to Voltage conversion?
Solar cells are made of
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