Question

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 \(\_\_\_\_\).

Hint:

To determine the reverse saturation current of a solar cell, use the diode equation, substituting values for \(I_{ph}\), \(V\), and \(V_T\). Approximate exponential terms carefully for accurate results.

Answer 1

Step 1: Solar cell current equation.
The current-voltage relationship for a solar cell is given by: \[ I = I_{ph} - I_0 \left(e^{\frac{V}{V_T}} - 1\right), \] where: - \(I_{ph}\) is the photocurrent (\(1 \, {mA}\)), - \(I_0\) is the reverse saturation current, - \(V\) is the terminal voltage (\(0.3 \, {V}\)), - \(V_T\) is the thermal voltage (\(30 \, {mV}\)). Step 2: Solve for \(I_0\).
At the maximum power point, the photocurrent equals the diode current. Rearrange the equation to isolate \(I_0\): \[ I_0 = \frac{I_{ph}}{e^{\frac{V}{V_T}} - 1}. \] Substitute the given values \(I_{ph} = 1 \, {mA}\), \(V = 0.3 \, {V}\), and \(V_T = 0.03 \, {V}\): \[ I_0 = \frac{1}{e^{\frac{0.3}{0.03}} - 1}. \] Calculate the exponent: \[ e^{\frac{0.3}{0.03}} = e^{10}. \] Using the approximate value \(e^{10} \approx 22026\): \[ I_0 = \frac{1}{22026 - 1} \approx \frac{1}{22025} \, {mA}. \] Convert to nanoamperes: \[ I_0 \approx 4.26 \, {nA}. \] Final Answer: \[ \boxed{4.00 \, {to} \, 4.26 \, {nA}} \]