# Test circuit for a solar panel used to power a supercapacitor

I was hoping to power a device with a supercapacitor that would provide power when there isn't light available, so essentially I just need the solar cell to charge the supercap as fast as it would discharge. How do I calculate this? When I use the charging formula for a capacitor, I get an undefined value (shown below). How do I calculate the time to charge the supercap?

$$\\ t=-R*0.47*ln(1-\frac { 5.5}{5.4})\$$

The supercapacitor I would be using is 5.5 V, 470 mF (link: https://a.co/d/2dwhxAC) and the panel has a Voc of 5.4 V and Isc of 36uA at 200 lux. The MPP voltage is 4V and the current is 28uA (link: https://www.solems.com/wp-content/uploads/FTI_photopile_10072048.pdf).

• How will you connect the solar panel to the supercapacitor? What is the solar panel's MPP voltage and short- circuit current? Commented Feb 3 at 19:02
• "When I use the charging formula for a capacitor, I get an undefined value." How so? Can you edit your question to show us your math? You mention the capacitor and give partial specifications for the solar cell. Please include a link to the data sheets for both cap and solar cell when you edit the question. Commented Feb 3 at 22:29
• @ocrdu I edited to add the missing information. Commented Feb 3 at 23:08
• @TimWescott I edited to add the missing information Commented Feb 3 at 23:08
• But regarding the datasheet of the solar panel the 36uA are inside the house, right? Outside it will be from 5mA to 26mA, what sounds more reasonable.
– Gos
Commented Feb 4 at 7:59

As Michal Podmanický pointed out, your capacitor is going to take thousands of seconds (hours) to charge, and they showed this from an energy perspective. That's fine, but it doesn't account for the gradually reducing charge current over time, assuming instead that the power being delivered to the capacitor is constant.

Your formula for the time it will take to charge to 5.4V via resistance $$\R\$$ is close, but not correct. The formula for the time it takes for a capacitor's voltage to go from some zero to some value $$\V\$$, on its way towards asymptote $$\V_A\$$, is:

\begin{aligned} t &= -RC \cdot ln\left(\frac{V_A - V}{V_A}\right) \\ \\ &= -RC \cdot ln\left(1-\frac{V}{V_A}\right) \\ \\ \end{aligned}

I think you swapped $$\V\$$ and $$\V_A\$$. Anyway, this gives us:

\begin{aligned} t &= -R \times 0.47 \times ln\left(1-\frac{5.4}{5.5}\right) \\ \\ &= R \times -0.47 \times -4.0 \\ \\ &= 1.9R \end{aligned}

You must choose R to pass whatever current the panel can provide at about 5V, which seems to be about 30μA. This is initial current, current that will flow when the capacitor has 0V across it, and the remaining 5V is across resistance R. By Ohm's law:

$$R = \frac{V}{I} = \frac{5V}{30\mu A} = 167k\Omega$$

Plugging that value for R into the time equation:

\begin{aligned} t &= 1.9R \\ \\ &= 1.9 \times 167k\Omega \\ \\ &= 317,000s \\ \\ &= 88 hr \end{aligned}

Sorry to disappoint. On top of that, due to the exponential nature of the charge/discharge curves (using a fixed resistance), charging is extremely slow near the end of charging. That means capacitor voltage will fall much more quickly in the early stages of discharge than it will rise when power returns.

It's not all bad news though. You don't have to use a fixed resistance, and assuming the load draws a similar current, once the capacitor is charged it will power that load for days.

Here's a design that will charge the capacitor with about 10μA constant current, right up until full charge (as opposed to ever-diminishing current as the capacitor charges):

simulate this circuit – Schematic created using CircuitLab

With a constant charge current, capacitor voltage rises linearly, instead of exponentially, and charges fully somewhat quicker, even though charge current is significantly less than 30μA. It also means that even when nearly fully charged, charge current doesn't diminish, and the capacitor charges at the same rate, right up until it's full. Here's a plot of capacitor voltage (orange) as input voltage (blue) from the solar panel varies:

Just be aware that whatever current the capacitor is taking (during charging), that current is not available for the load, so with 10μA going to the capacitor, out of the 30μA available, the load can draw a maximum of only 20μA.

• If I use an energy management chip (with an LDO regulator, buck and boost converters, etc) can I decrease the charge time? Commented Feb 4 at 15:43
• @zenizenitsu You have about 150μW of power. Whichever way you look at it that will take a minimum of 11 hours to charge that cap to 5V. You might approach that with an efficient DC-DC current source, but any linear solution (like mine here) will take at best 22 hours. With some tweaking my circuit can get most of the 30μA through the cap, which would take about 24 hours of daylight. The only way to charge it faster is with a different, stronger, current source. Commented Feb 4 at 17:36
• @zenizenitsu Have you considered using a smaller capacitor? It could charge in minutes, and discharge in the same time if the load draws the same current as you use to charge it. Commented Feb 4 at 17:36

Calculate energy of cap when is at 5.5V:

$$E_{5.5{\rm\,V}} = \frac{1}{2} \cdot C \cdot V^2 = 7.11{\rm\,J}$$

Calculate energy of cap when is at 5.4V:

$$E_{5.4{\rm\,V}} = \frac{1}{2} \cdot C \cdot V^2 = 6.85{\rm\,J}$$

Subtract it:

$$E_\Delta = E_{5.5{\rm\,V}}-E_{5.4{\rm\,V}} = 0.26{\rm\,J}$$

Calculate the time for charging of 0.26J from power of solar cell (btw, 1J=1Ws):

\begin{aligned} E_\Delta &= V \cdot I \cdot t \\ t &= \frac {E_\Delta} {V \cdot I} = \frac {0.26{\rm\,J}} {4{\rm\,V} \cdot 28{\rm\,\mu A}} = 2321{\rm\,s} \end{aligned}

That is a lot of time but even worst you have to maintain Mppt to charge the cap within this time.