APhO 2001, Експеримент — Задача 3. Question (3)

• b. $I\left(1+\frac{R_s}{R_{sh}}\right) = I_{ph} - \frac{V}{R_{sh}} - I_d$
• c. Official solution: $I = I_{ph} - I_d = I_{ph} - I_0(e^{\beta V} - 1)$; $V = 0$ gives $I_{ph} = I_{sc}$; $I = 0$ gives $V_{oc} = \frac{1}{\beta}\ln(I_{sc}/I_0 + 1)$.

(3) (Question (3) : 3.0 points)

Assume that the solar cell can be modeled as a device consisting of an ideal current source (light-generated current source), an ideal diode, a shunt resistance $R_{sh}$, and a series resistance $R_s$,

a. Draw a correct equivalent circuit diagram for the solar cell under light illumination.

⊙Correct circuit diagram (1.5 points)

b. Derive I-V relationship for the equivalent circuit. Express the result in terms of $R_{sh}$, $R_s$, light generated current $I_{ph}$ ,and current passing through the diode $I_d$.

Applying Kirchhoff’s loop rule, one obtain

⊙Correct equation (0.5 points)

c. Assuming that $R_{sh}$ = infinity and $R_s$ = 0 and can be neglected, find the I-V relationship and prove that it can be written in the form as given below:

where $V_{oc}$ is the open-circuit voltage, $I_{sc}$ is the short-circuit current, and $I_0$, $\beta$ are constants.

proof: with $R_s = 0$ , $R_{sh}$ = infinity, the equivalent circuit becomes

Hence $I = I_{ph} - I_d = I_{ph} - I_0(e^{\beta V} - 1)$

for close circuit $V = 0$ $\quad \therefore I_{ph} = I_{sc}$

for open circuit $I = 0$ $\quad \therefore I_{sc} - I_0(e^{\beta V_{oc}} - 1) = 0$

⊙Correct proof of $V_{oc}$ vs $I_{sc}$ relation (1 point)

Partial credit may be given for appropriate intermediate results.