GATE 2020 EC Q.34: solution and answer

GATE 2020 · ECQ.34 · 2 marks · MCQ

A $p n$ junction solar cell of area $1.0 cm^{2}$ , 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 $= 200 μ m$ . The charge of an electron is $1.6 \times 1 0^{- 19} C$ . The average optical generation rate (in $cm^{- 3} s^{- 1}$ ) is

(A)
$0.84 \times 1 0^{19}$
(B)
$5.57 \times 1 0^{19}$
(C)
$1.04 \times 1 0^{19}$
(D)
$83.60 \times 1 0^{19}$

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Answer and solution

ANSWER

The average optical generation rate is $G \approx 0.84 \times 1 0^{19} cm^{- 3} s^{- 1}$ , option (A).

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

Since fill factor $F F = \frac{V _{oc} I _{sc}}{P _{in} \cdot ( area )}$ 's numerator equals the maximum output power, and efficiency $η = \frac{P _{ma x}}{P _{in} \cdot A}$ , combine both definitions of $P_{ma x}$ : $P_{ma x} = η P_{in} A = F F \cdot V_{oc} I_{sc}$ .
With $A = 1.0 cm^{2}$ and $P_{in} = 100 mW cm^{- 2}$ : $P_{ma x} = 0.15 \times 100 mW cm^{- 2} \times 1.0 cm^{2} = 15 mW$ .
Solve for the short-circuit current: $I_{sc} = \frac{P _{ma x}}{F F \cdot V _{oc}} = \frac{15 mW}{0.8 \times 0.7 V} = \frac{15}{0.56} mA \approx 26.79 mA$ .
Since $A = 1 cm^{2}$ , the short-circuit current density equals the current numerically: $J_{sc} \approx 26.79 \times 1 0^{- 3} A cm^{- 2}$ .
Assuming every photogenerated carrier across the cell thickness is collected, the average optical generation rate $G$ (per unit volume, per unit time) relates to $J_{sc}$ by $J_{sc} = q G t$ , so $G = \frac{J _{sc}}{q t}$ .
Convert the thickness to centimetres: $t = 200 μ m = 200 \times 1 0^{- 4} cm = 0.02 cm$ .
Substitute: $G = \frac{26.79 \times 1 0 ^{- 3} A cm ^{- 2}}{1.6 \times 1 0 ^{- 19} C \times 0.02 cm} = \frac{26.79 \times 1 0 ^{- 3}}{3.2 \times 1 0 ^{- 21}} cm^{- 3} s^{- 1}$ .
$G \approx 8.37 \times 1 0^{18} cm^{- 3} s^{- 1} = 0.84 \times 1 0^{19} cm^{- 3} s^{- 1}$ , which is option (A).

KEY IDEA

Link the solar cell's efficiency/fill-factor rating to its short-circuit current density ( $P_{ma x} = η P_{in} A = F F \cdot V_{oc} I_{sc}$ ), then use $J_{sc} = q Gt$ (full carrier collection over the cell thickness) to back out the average generation rate.

WATCH OUT FOR

Forgetting to convert the thickness from micrometres to centimetres (or forgetting it at all), and mixing area/current-density units when area is exactly $1 cm^{2}$ — either error shifts the answer by orders of magnitude, landing on option (D) or a value between the given choices.

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GATE 2020 EC Q.34: solution and answer

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