Your current .4W cells should provide 80 milliamps in full sun, or 160 milliamps for the two in parallel.
Yes, but they are not supposed to meet those ratings under load, as far as I understand. I measured their open circuit voltage at around 6 V and their short circuit current was around 82 mA.
The highest charging speed I measured so far was 80 mA, but that was with both cells connected and no window in between the cells and the sun.
The charging speed in the location I want to use the solar light in is much slower than that (~40 mA peak charging speed).
Voltage is Q/C and Q is current times time. Your 3 hour charge should then provide a charge of 3 x 60 x 60 x 0.16 or 1728 Coulombs. Dividing that by 400 says that starting from zero volts you get 4.3 volts.
Can I really calculate it like that even though I have a series resistance of ~ 50 Ohm in between the solar cell and the capacitor?
It is worth checking to see if you have other loss factors, and to see if you can reduce the ones which are actually there.
I will try to optimize the position of the solar cells some more, but I think there is no way around a bigger cell. I think I will try some amorphous thin film cells instead of the polycrystalline ones I am using right now.
The questions I ask myself right now are:
* Should I go for low voltage solar cells to keep the voltage difference between the solar cell output and the target voltage as small as possible?
* Or should I better shop for solar cells with the lowest price per watt and use a buck converter (like the Pololu one) to make use of the extra voltage I get from the solar cells?
I will have to shop around for some more parts to play with anyways. Both you and BrianHG have suggested interesting things I would love to try out.
Edit: I would have calculated the charging speed like that (assuming that we are not current limited by the solar cell).
RC (time constant) = 50 ohm * 400 farad = 20000
Vs (supply voltage) = 2.7 volt
I takes t = 3382 s (~1 hour) to charge the capacitor from 0 V to 0.42 V:
Vs * (1 - e^(-t / RC)) = 0.42 V
It takes t = 15402 s (~4.3 hours) to charge the capacitor from 0 V to 1.45 V:
Vs * (1 - e^(-t / RC)) = 1.45 V
That means it takes ~3.3 hours to charge the capacitor from 0.42 V to 1.45 V.
try replacing the diodes with a good npn transistor
I don't have any of those. I will try to get my hands on some more parts next week.