Electronic – Source voltage in discharging capacitor equation

With the simplifying assumptions that:

• the solar cell is an ideal power source, which produce a constant power at any combination of voltage and current
• there are no other losses

then the calculation is quite simple. The energy, voltage, and capacitance of a capacitor are related by:

$$ E = \frac{1}{2}CV^2 $$

Using this, you can calculate the energy already stored in the capacitor at the initial voltage, and the final energy required at your target voltage. The difference is the total work required, and the time required to do that much work is the work divided by the power of your charger.

$$ E_0 = \frac{1}{2}C{V_0}^2 \\ E_{end} = \frac{1}{2}C{V_{end}}^2 \\ \Delta E = \frac{1}{2}C{V_{end}}^2 - \frac{1}{2}C{V_0}^2 \\ t = \frac{\frac{1}{2}C{V_{end}}^2 - \frac{1}{2}C{V_0}^2}{P} \\ t = \frac{\frac{1}{2}C ({V_{end}}^2 - {V_0}^2)}{P} $$

Given

• \$P=5mW\$
• \$C=1F\$
• \$V_0=1V\$
• \$V_{end}=5V\$

then:

$$ t = \frac{\frac{1}{2}1F ((5V)^2 - (1V)^2)}{5mW} $$

$$ t = \frac{\frac{1}{2}1F (25 - 1)V^2}{5mW} $$

\$F=J/V^2\$ and \$W=J/s\$ by definition of the farad and watt so:

$$ \require{cancel} \begin{align} t &= \frac{\frac{1}{2}1\cancel{J}/\cancel{V^2} (25 - 1) \cancel{V^2}}{0.005\cancel{J}/s} \\ &= \frac{\frac{1}{2}1 (24)}{0.005/s} \\ &= \frac{12s}{0.005} \\ &= 2400s \end{align} $$

As a first approximation, model the solar cell with a Thevenin equivalent circuit as determined by the open circuit voltage, \$V_{OC}\$, and short circuit current, \$I_{SC}\$, of the cell at a given illumination.

Then, for that fixed illumination, the time constant will be:

$$\tau = \dfrac{V_{OC}}{I_{SC}}C $$

The time required to charge to some fraction \$\chi\$ of the final voltage is:

$$t = -\tau \ln(1 - \chi)$$