Electrical – Charging a capacitor with low current

I have had several students who have used super cap chargers. They provide a lot of nice features like variable current limit on the charging, the particular one that I linked to will also make it so you could use 2 caps with lower voltage on each one and it will make sure you don't over voltage one.

The only down side is the chips for these are VERY small and can be difficult to solder if you aren't very experienced.

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:

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