Supposing I have a small solar panel rated at 5V and a supercapacitor rated 5.5 V 1 F, would it be possible to charge the cap fully?
Or in general, can a cap be fully charged with a low current, constant voltage source?
\$\begingroup\$
\$\endgroup\$
5
-
3\$\begingroup\$ yes, of course. \$\endgroup\$– Marko BuršičCommented May 5, 2016 at 14:08
-
\$\begingroup\$ Fully charged means it has attained a terminal voltage of 5.5 volts, your solar panel only produces 5V. \$\endgroup\$– Andy akaCommented May 5, 2016 at 14:14
-
7\$\begingroup\$ Read the cap's datasheet, looking for specifications for leakage current. If the leakage current exceeds the charging current, then ... no. \$\endgroup\$– user16324Commented May 5, 2016 at 14:15
-
\$\begingroup\$ This is what I was looking for, leakage current! Thanks! (I'll accept it as an answer if you add it). \$\endgroup\$– Ivan VorasCommented May 5, 2016 at 14:23
-
1\$\begingroup\$ The cap may discharge through the solar cell when the sun goes away. Consider to include a Schottky diode. \$\endgroup\$– JimmyBCommented May 5, 2016 at 17:25
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
Take this datasheet as an example.
The leakage current used in their example is 2μA, though I cannot find any real tabular data. This is because for most of these types of caps it is in the micro ampere range. (Although from unknown sources, I would also not be surprised with dozens of μA or more)
That means your source needs to supply only 5μA to be sure to eventually charge it.
What is a much bigger problem/risk, with these kinds of capacitors you can also see in the datasheet: They do not like high currents!
If you take more than 1 or 2mA out of one the life will be very significantly reduced and if you charge it directly with a voltage source without a series resistor it will also not last long, due to the initial peak currents when charging.
The datasheet shows you how long it will take to charge for a 30 Ohm and a 100 Ohm resistor for 5.5V and 1F.
\$\begingroup\$
\$\endgroup\$
For all practical purposes, Yes. Bear in mind that the charger rate will be progressively slower as the capacitor approaches the max voltage, so theoretically it approaches, but never actually reaches that number. Generally the rule is 5 time constants, (RC), is the time to "mostly" reach Vmax. Take a look at this little tutorial...
http://www.electronics-tutorials.ws/rc/rc_1.html
The only thing that will alter this is the capacitor's internal resistance. Considered as a resistor in series with the internal resistance of your source, this creates a small voltage drop and acts as a limit to the ideal "5 time constant" rule.
