GATE 2018 EC Q.44: solution and answer

GATE 2018 · ECQ.44 · 2 marks · NAT

A solar cell of area $1.0 cm^{2}$ , operating at $1.0$ sun intensity, has a short circuit current of $20 mA$ , and an open circuit voltage of $0.65 V$ . Assuming room temperature operation and thermal equivalent voltage of $26 mV$ , the open circuit voltage (in volts, correct to two decimal places) at $0.2$ sun intensity is ________.

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Answer and solution

ANSWER

The open-circuit voltage at 0.2 sun intensity is Voc2 ≈ 0.61 V.

Official answer: 0.59 to 0.63Source: Official answer key, GATE 2018 EC (IIT Guwahati)

Governing relation. For a solar cell, the open-circuit voltage satisfies $V_{oc} = V_{T} ln (\frac{I _{sc}}{I _{0}} + 1)$ , where $I_{0}$ is the reverse saturation (dark) current of the cell's diode, and $V_{T} = 26 mV$ is the given thermal equivalent voltage.
Scaling of $I_{sc}$ with intensity. Short-circuit current is proportional to the photogenerated current, hence proportional to illumination intensity. At $1.0$ sun, $I_{sc, 1} = 20 mA$ ; at $0.2$ sun, $I_{sc, 2} = 0.2 \times 20 mA = 4 mA$ .
Justifying the large-argument approximation. Using the 1-sun data: $0.65 = 0.026 ln (\frac{20 \times 1 0 ^{- 3}}{I _{0}} + 1)$ so $\frac{I _{sc, 1}}{I _{0}} + 1 = e^{0.65/0.026} = e^{25},$ an enormously large number. Hence $I_{sc} / I_{0} ≫ 1$ at both intensities (at $0.2$ sun this ratio is only $5 \times$ smaller, still enormous), so the " $+ 1$ " can be dropped: $V_{oc} \approx V_{T} ln (I_{sc} / I_{0})$ .
Difference between the two conditions. Since $I_{0}$ does not depend on illumination, it cancels when the two conditions are subtracted: $Δ V_{oc} = V_{oc, 2} - V_{oc, 1} = V_{T} [ln (\frac{I _{sc, 2}}{I _{0}}) - ln (\frac{I _{sc, 1}}{I _{0}})] = V_{T} ln (\frac{I _{sc, 2}}{I _{sc, 1}}) .$
Numerical evaluation. $Δ V_{oc} = 0.026 \times ln (0.2) = 0.026 \times (- 1.6094) = - 0.0418 V .$
Final value. $V_{oc, 2} = 0.65 - 0.0418 = 0.6082 V \approx 0.61 V .$

KEY IDEA

A solar cell's open-circuit voltage is $V_{oc} = V_{T} ln (\frac{I _{sc}}{I _{0}} + 1)$ . Since $I_{sc} / I_{0} ≫ 1$ (verified from the 1-sun data, where $I_{sc} / I_{0} = e^{0.65/0.026} - 1$ , an enormous number), this reduces to $V_{oc} \approx V_{T} ln (I_{sc} / I_{0})$ , so a change in illumination (which changes $I_{sc}$ proportionally, with $I_{0}$ unchanged) shifts $V_{oc}$ by $Δ V_{oc} = V_{T} ln (\frac{I _{sc, 2}}{I _{sc, 1}})$ .

WATCH OUT FOR

Assuming $V_{oc}$ scales linearly or logarithmically with intensity directly (e.g., $V_{oc, 2} = 0.2 \times 0.65$ ), instead of recognizing that only $I_{sc}$ is proportional to intensity, and that $V_{oc}$ changes by $V_{T} ln (intensity ratio)$ through the diode equation. Also, forgetting to first justify $I_{sc} / I_{0} ≫ 1$ before dropping the "+1" — without that check, the cancellation of $I_{0}$ between the two conditions is unjustified.

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