GATE 2024 EC Q.65: solution and answer

GATE 2024 · ECQ.65 · 2 marks · NAT

The photocurrent of a PN junction diode solar cell is 1 mA. The voltage corresponding to its maximum power point is 0.3 V. If the thermal voltage is 30 mV, the reverse saturation current of the diode (in nA, rounded off to two decimal places) is ____.

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Answer and solution

ANSWER

The reverse saturation current is I₀ ≈ 4.13 nA.

Official answer: 4.00 to 4.26Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

The solar cell obeys the illuminated diode equation $I = I_{L} - I_{0} (e^{V / V_{T}} - 1)$ , with $I_{L} = 1 mA$ (photocurrent) and $V_{T} = 30 mV$ (thermal voltage).
Power delivered is $P = V I$ . At the maximum power point, $\frac{d P}{d V} = I + V \frac{d I}{d V} = 0$ .
Differentiating the diode equation, $\frac{d I}{d V} = - \frac{I _{0}}{V _{T}} e^{V / V_{T}}$ . Substituting into the MPP condition: $I_{L} - I_{0} (e^{V / V_{T}} - 1) - \frac{V I _{0}}{V _{T}} e^{V / V_{T}} = 0$ .
Rearranging: $I_{L} = I_{0} [e^{V / V_{T}} (1 + \frac{V}{V _{T}}) - 1]$ .
At the maximum power point, $\frac{V}{V _{T}} = \frac{0.3}{0.03} = 10$ , so $e^{V / V_{T}} = e^{10} \approx 2.20265 \times 1 0^{4}$ .
Then $I_{L} = I_{0} [e^{10} (1 + 10) - 1] = I_{0} (11 e^{10} - 1) \approx I_{0} \times 2.42290 \times 1 0^{5}$ .
Solving for $I_{0}$ : $I_{0} = \frac{I _{L}}{11 e ^{10} - 1} = \frac{1 \times 1 0 ^{- 3} A}{2.42290 \times 1 0 ^{5}} \approx 4.127 \times 1 0^{- 9} A$ .
Rounded to two decimal places, $I_{0} \approx 4.13 nA$ .

KEY IDEA

At the maximum power point of a solar cell's I–V curve, $d P / d V = 0$ , which gives the condition $I + V \frac{d I}{d V} = 0$ ; substituting the diode equation $I = I_{L} - I_{0} (e^{V / V_{T}} - 1)$ and solving for $I_{0}$ links the given photocurrent, MPP voltage and thermal voltage directly to the reverse saturation current.

WATCH OUT FOR

Forgetting the product-rule term $V d I / d V$ and instead setting $I = 0$ (an open-circuit-style condition) at the "maximum power point," which drops the $(1 + \frac{V}{V _{T}})$ factor and gives a badly wrong (much larger) value of $I_{0}$ .

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GATE 2024 EC Q.65: solution and answer

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