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Your proposed circuit will work, but <i>"before the voltage drops significantly"</i> is rather wishy-washy.  I would at least use a second meter to measure the current.

What you really want to know is the power, which is the voltage times the current.  With two meters you can sortof eyeball it and find a pot setting close to maximum power.  Then you multiply the voltage and current readings and declare that the panel power for whatever conditions you are measuring.

If I had to do this, I'd probably have a microcontroller measure both the voltage and current, then compute the power in real time.  It would then be a lot easier to adjust the pot for maximum power.

However, as Andy already said, the first step is to read the datasheet carefully.  Most of what you want to know is probably in there.  It should be clear enough whether the quoted 150 mA is when the panel is also providing the quoted 6 V, or is instead when the panel is shorted.  The datasheet really should tell you the maximum power output with full sun.

Another quick sanity check is using the panel size and a rough efficiency guess.  Sunlight at sea level on a reasonably clear day is about 1000 W per square meter.  Figure roughly 15% efficiency, so you actually get 150 W/m<sup>2</sup>.  Your 6 V and 150 mA figures imply 900 mW, which implies about 0.006 square meters, or 60 square centimeters, or 9.3 square inches.  If your solar panel is about 1.5 inches square, or about 8 cm square, then 6 V at 150 mA is plausible.