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I have researched about capacitor charging time, with its charge being at 63% after 1 time constant, being t = RC. I am trying to repeat this by myself, using a 16V 1000µF capacitor, a 1KΩ resistor, and a constant power supply (~7.3 volts, ~1.9 milliamps) from a solar panel, which, as I understand capacitor charging (I may be totally wrong about this, if so please correct me) one time constant should be equal to t = RC = (1KΩ)(1000µF)= 1 second.

Theoretically, after 1 second, the capacitor should be at 63% of its capacity. For it to be at 99% of its capacity, 5 time constants must pass by. I tried this, and I waited for 5 seconds, disconnected the capacitor from the solar panel and resistor, and using a multimeter measured the capacitor's voltage, which didn't even reach 1.5 volts at any of the many times I tried.

I don't understand it, is it that the way I am trying to find the charging time is wrong, as a consequence of informing myself totally wrongly, or that the capacitor that I am using is defective?

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    \$\begingroup\$ What's 1.9 milliamps ? If you limit the current, you cannot expect it to behave in line with (this specific) theory. Your initial current when charging with 7.3V should be 7.3mA. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 28, 2018 at 16:48
  • \$\begingroup\$ That is the current that I measured using a multimeter, in the 2m section, and the output current that it measured was 1.9 milliamps. At the back of the solar panel there is a label that specifies that the output current should be 150mA when the voltage is 9V (supposedly the maximum), but in this case (I am using my lamp for this) I have 7.3 volts in average and an output current, as measured by my multimeter 1.9mA, or 1.9 milliamps. What should be the equation to calculate charging time, with current being considered? \$\endgroup\$
    – PlanetGazer8360
    Commented Feb 28, 2018 at 16:55
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    \$\begingroup\$ Average? You said the supply is constant. If it is not, your theory does not apply. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 28, 2018 at 16:56
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    \$\begingroup\$ And I didn't get the "I am using my lamp for this" part. Draw your circuit, please. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 28, 2018 at 17:00
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    \$\begingroup\$ Well, this setup is very far from ideal experiment conditions. Too many uncontrollable variables. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 28, 2018 at 17:05

1 Answer 1

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Your understanding of the capacitor is correct, your understanding of the power supply is wrong.

At initial connection the cap will try to draw 7.3mA through that 1K resistor assuming the voltage is a solid 7.3V.

Unfortunately, the solar panel can only supply 1.9mA.

On it's own, at that level the capacitor will take a lot longer to charge. (>28S)

However, what also happens is the voltage at the panel will fall since it is being over-loaded. That further reduces the charge current extending the time even longer.

In other words, basically, the solar panel is presenting a large terminal resistance at the currents you are demanding, which affects the time constant.

How you are illuminating the solar panel also matters. Depending on the type of lamp, there may be a significant AC component to the light. That can cause the solar panel to be "dark" for a fraction of the cycle, at which time, it acts as a diode shorting out your circuit.

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  • \$\begingroup\$ I also tried for up to one minute, and the voltage measured did not surpass 2 volts \$\endgroup\$
    – PlanetGazer8360
    Commented Feb 28, 2018 at 16:59
  • \$\begingroup\$ May it be that the capacitor is defective? \$\endgroup\$
    – PlanetGazer8360
    Commented Feb 28, 2018 at 16:59
  • \$\begingroup\$ @PlanetGazer8360 probably not, solar panels have some odd characteristics. Try adding a diode before the resistor. \$\endgroup\$
    – Trevor_G
    Commented Feb 28, 2018 at 17:01
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    \$\begingroup\$ I think you are just doing something wrong. Either your "charging" circuit is wrong or you are not measuring it properly. What is this battery giving 5.31V? \$\endgroup\$
    – Eugene Sh.
    Commented Feb 28, 2018 at 17:29
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    \$\begingroup\$ So you are using a dead battery. Great. Come on, if you want to conduct an experiment, you should use a proper equipment. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 28, 2018 at 17:34

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