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I have a 3V, 70mA solar panel rated at max 210mW. If I design a RC series circuit with it, can I increase the power outlook to about 2W? If so, how can I determine the time it takes to charge? How long will it take to discharge to get a 3V, 500mA output?
Below is an example of how I intend to set it up.
simulate this circuit – Schematic created using CircuitLab
You can't get power out of nowhere, no matter what you do. So no way you can increase power. Period. Charging time of the capacitor is 5T = 5RC. It comes from exponential equation, and after 5RC you have 99% charge, usually considered full charge (or discharge, it's symmetric). I guess you aren't exactly familiar with electronics, you should read a few things about how capacitors and resistors work, as well as Ohm's Law and power. You can also find some really friendly vids on the youtube. It's all information I can give you regarding your questions. Check out those vids or articles, it will cost you half an hour time, but you will already be able to answer your own questions, it's pretty basic stuff you're asking.
If you want 2W permanent output then of course it is impossible (it would be free energy.)
If, as I understand from your comments, you want to charge your capacitor over a "long" time, and then discharge it at higher power during a short time, then yes, it is possible.
The theoretical limit is that you cannot create energy (energy=power x time,) just store it. The best you might achieve is to get out all the energy you produced without losses.
If you produce Pi=0.21 mW and want Po=2W output, the upper limit is discharging during a fraction f=Pi/Po=0.21/2=0.105 = 10.5% of the time. In practice, you will have losses, and with simple circuits the solar panel will not always produce the maximum power. I would guess that the best you can hope for in practice might be something like charging 20 times longer than discharging.
Now, regarding your circuit:
For exact calculation of the charging-discharging of the capacitor, we would need:
You'll have to get more than 3V out of your panels and more than 3V on the cap/battery to get some seconds of 3V 500mA out of it. (Increasing DC voltage is very inefficient so go for regulating it down). This is because a capacitor voltage drops as it discharges energy.
$$Energy=Capacitance×Voltage^2/2$$
In other words, you need the capacitor to have 3V worth of its energy, plus the energy you need spend, plus any energy lost due to inefficiency (even the best switching regulators are not 100% efficient- in fact efficiency is usually a function of how far off your source is from the desired output since higher source equates to more switching).
3V and 500ma is 1.5W. You said 2W (which is good since it should cover some efficiency shortcomings) so let's go with that for 10 sec. So 20J. You showed a 1F cap which is pretty beefy, so let's run an example with that too.
$$BaseEnergy=1F×(3V)^2/2=4.5J$$ So we need 24.5J to be able to do a 10s burst.
$$ChargedEnergy=24.5J=1F×Vcharged^2/2$$
$$Vcharged=7V$$
3V 500 mA for seconds puts you in the battery range unless you get a much higher voltage panel. Let's suppose you wanted to up-convert your 3V panel and found one with 30% efficiency (I don't recommend it to getting a higher voltage panel). Let's also suppose your cap was at BaseEnergy and not fully discharged so you just need the 20J.
$$20J=0.3×3V×0.07A×Time$$
$$Time=317.5s=5.3min$$
And that's just to generate the energy. To actually charge the cap might be a bit more (depends on how low you can make the resistance). It gets even worse when we take into account a diode (you'll need one to prevent the capacitor from discharging through your panel as you've drawn) and those tend to come at a voltage drop cost, cutting into your prescious 3V and putting it in an even worse range for the DC up-converter.
