How to check realistic specs of small solar panel?
I have a small solar panel rated for 6V/DC, 150mA. I know that those specs are under ideal circumstances, and I would like to do some measurements in different weather conditions to see what I can expect.
To measure voltage I can of course use a multimeter. For current, I think I know what to do, but I just wanted to check. Can I simply make a circuit like below to draw a variable current, and adjust the potentiometer to find the maximum current that can be drawn before the voltage drops significantly? Or is there a better or simpler way?
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For a "better way", research Maximum Power Point Tracking (MPPT.)
The solar cell will produce some number of Volts and Amps. But the amount of power (V*A, Watts) a solar cell produces is not simply "when one is at maximum."
For instance, the most current will happen when the load resistance is lowest. But the Volts will also be really low, so little power will actually be generated. Conversely, very little load will give a higher voltage, but lots of volts and little load also equals little power. A MPPT draws the optimal current at any given illumination so that the most power is delivered from the cell.
Your proposed circuit will work, but "before the voltage drops significantly" 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/m2. 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.
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All good designs and design testing is based on information contained in the component data sheets. A solar panel is no different so, you need to establish the operating conditions for measuring voltage and current based on the information provided by the supplier of the product.
It's quite likely that the voltage can be simply measured with a DVM but, measuring current may be a little more complex and that is why we use data sheets for understanding the testing conditions. However, for current measuring, it reasonably likely that a short circuit test is what produces the 150 mA you state. Again, using a multimeter to measure current is the route to go.
You can do other tests too using a dummy load and two DVMs; one measuring load voltage and the other measuring load current. This last method can be used to find the maximum power for a given level of light hitting the solar panel.
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