# Time to charge a capacitor with a power limiting supply and ideally no resistance

So I have a capacitor being charged by a solar cell which has a varying (but readable) power output. This is being fed directly into a 10 farad cap with a solar panel voltage of about $2.8V$. For the sake of this question let's assume the output of the panel is $5mW$. Also, at the time of calculation, the voltage on the cap is known.

I would like to calculate the amount of time it would take to reach a voltage on the cap from some starting voltage on the cap (as a function).

Using a bit of algebra I've figured

$$t = CR \cdot \ln(1-V_C/V_S)$$

where $V_S$ is the open circuit voltage of the solar cell and $V_C$ is the voltage on the cap. And the resistance is

$$R = P_S/({V_C}^2)$$

where $P_S$ is the power output of the solar panel.

but the current and voltage vary based on the power output and charge and therefore the apparent resistance also changes (well I would assume). I'm a bit rusty on my calculus and am having a hard time coming up with an equation to express this relation.

• Erm or would it simply be t = C * Ps/(Vc^2) * ln(1-Vc/Vs) dvc/dt? – Lindenk Aug 12 '13 at 19:28
• What's the short circuit current? – Chef Flambe Aug 12 '13 at 19:37
• It was about 2.1mA – Lindenk Aug 12 '13 at 19:48
• Why are you calculating the resistance, if you are interested in the ideal case of no resistance? – Phil Frost Aug 12 '13 at 20:03
• Also, do you want to model your solar cell as an ideal constant power source? This probably isn't accurate unless your charger is fairly sophisticated. – Phil Frost Aug 12 '13 at 20:05

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}

• Ah, I forgot physics was a thing. Thanks, I was thinking of this all wrong. – Lindenk Aug 12 '13 at 20:32
• @Lindenk as Sheldon Cooper would say, everything is physics. – Phil Frost Aug 12 '13 at 20:41

The constant power assumption is not realistic, because the panel will simply not put out 5mW regardless of load.

An ideal constant power source hooked up to a capacitor will charge a capacitor without limit to an ever higher voltage, because it keeps putting out energy and the capacitor has to keep storing it, thereby becoming more and more charged.

Even the most crude model of the situation has to take into account the limiting process: that the voltage on the capacitor will not exceed the applied one, and so the charging has to slow down as the capacitor reaches the target voltage, and therefore the power flowing to the capacitor has to diminish.

We're better off constructing a schematic of the situation based on the equivalent circuit model. If you substitute values into this, it can be subject to calculations or simulation:

simulate this circuit – Schematic created using CircuitLab

Of course, you have to somehow work out the values for the components of this theoretical model which match your given solar panel.

According to this model, things are somewhat complicated. The solar cell can generate electricity when it's unloaded. When the load C1 is open, current still flows and dissipates energy in RSH and D1. There is also some internal series resistance RS that wastes energy and creates a load-dependent voltage drop.

The presence of D1 can be the basis of a simplified view, since D1 can be regarded as regulating an approximately constant voltage. Suppose that RSH is subject to an approximately constant voltage thanks to D1. This means that a constant current flows through it RSH. Then when we subtract that current from IL, the remaining current is distributed between the RS circuit and D1. D1 dumps whatever the load does not draw, without significantly changing its voltage.

And so, if RSH has an approximately constant voltage, it means that RS and C1 form a simple RC circuit, where we can apply the rule of thumb that the capacitor is around 99% charged after five RC constants (or else use the exponential charging formula to work out the time to reach a specific voltage).

Here is where we can also apply a constant power approximation in a way that is justified, though we don't need it. If the voltage on RSH is roughly invariant thanks to the diode, and since constant current is flowing into this node from the current source, it means that the power being delivered by the internal current source is roughly constant. (Constant current times constant voltage.) Some of this power ends up stored as energy, integrated over time into C1, and the rest of it is wasted in dissipation through RS, RSH and D1. Once C1 stops charging, all of the power is thereafter wasted in just RSH and D1.

• I disagree with "and so the charging has to slow down as the capacitor reaches the target voltage, and therefore the power flowing to the capacitor has to diminish". There is no reason the charging has to slow down. This is not an RC circuit. In fact, as the capacitor charges, then per each unit of voltage added to the capacitor, the amount of energy added increases. An analogy would be a stack of books; initially adding a book does adds little energy because it need not be lifted high; later, each book adds more energy, proportional to the square of the distance lifted. – Phil Frost Aug 12 '13 at 22:09
• Also, that this is an equivalent circuit does not mean a solar cell is actually like this. There isn't actually a current source in parallel with a diode, even though it can be modelled as such, by Norton's theorem. – Phil Frost Aug 12 '13 at 22:13