Time of charge of capacitor of complex circuit

As it is not a piecewise continuous system, you'll need to calculate, in two steps:

First: you'll need to calculate the time of charging the capacitor until it reaches

$$ (Vb-Vc)/R = Imax $$

with constant current of Imax. if the current is constant that capacitance does not change this is a simple straight ramp curve upto the point where the current is no longer limited by the constant current.

Second: you can use the equations for power supply without limited current power supply.

Which are the exponential curves from that point onward to the asymptote.

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:

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