I put a 0.5v 1A monolithic silicon solar cell on the front end of the Joule Thief circuit. Added a schottky diode and 2000uf cap on the output. I wanted to see how long it would take the Joule Thief to charge the cap bank. There are two 1000uf/50V caps in parallel to get the total 2000ufs. In full sun it took 3 mins 34 sec. to charge to 21.8 V.
Took voltage measurements every 15 seconds. Here is the data:
Here is the Joule Thief circuit that shows the connections for this experiment (to see a full size image click image then click image on following page):
Solar Joule Thief Circuit
Cheers!
From my own experiences with Joule Thief at low voltages, I would say that the half volt solar cell is barely turning on the silicon transistor. If there were two solar cells in series for 1V, the caps would charge in a matter of a few seconds. We can infer from the capacitance and time, how much the current is, and it is a very low current, less than a milliamp. The output and charging current would benefit greatly from changing the transistor to a germanium.
Also, as the output voltage rises, the voltage from the feedback winding to the base also rises. The transistor is rated for a negative 5 or 6V on the base, and it is important that this negative voltage is not allowed to get higher than -5V. A 1N4148 diode with its cathode or banded end to the base will limit the voltage.
This circuit actually starts oscillating at ~460mV and runs down to about ~380mV.
This is a study to see if the Joule Thief can charge a capacitor from a single solar cell. The stored energy then can be used to boot strap another circuit which is currently being tested, more on this in the future. The problem with using multiple solar cells in series is that if you shade one cell the output of the string is severely degraded.