I am considering a solar cell powered project but I have some questions.
What happens during low light levels. Does the rated volatage decrease or only the current, or both. If I want to implement a battery backup what kind of circuit will switch between the two. Are solar cells shortcircuit proof? Thanks
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Both the voltage and the available current go down. The voltage roughly proportional to the log of intensity (so small decrease) and the current in proportion to intensity, if you use power at the optimum rate.
Usually, the circuit just runs from the battery, and the solar cells connect to the battery through an anti reverse diode, or regulator.
Sure. They just get as warm as if they were laying in the sun with nothing connected. This allows them to be regulated with a shunt dummy load that soaks up all the excess current once the battery reaches full charge. For a reliable operation over many months, the cells have to provide a full day's charge in just a few hours, to allow for a run of rainy weather.
-- John Popelish
John, Is an anti-reverse diode a special type of diode. Can I use a IN4148? Any advantage of using a schottky?
It is nothing special. It must stand the battery voltage in reverse (when the cells are in the dark) and carry the solar cell current in the forward direction. The schottky type wastes about half as much voltage compared to a junction diode with the same current rating, but will also leak more current when the solar cells are in the dark. A
1N4148 is fine for up to about .1 amp maximum cell current.You may not need an anti reverse diode at all, if your cell array leaks a small enough current when it is in the dark, and still connected to the battery. Put an ammeter in series and measure the leakage or put a small resistance in series, measure the voltage drop and calculate the current with ohm's law. For example, if you put a
1k resistor in series and had 0.1 volt drop across it when the cells were dark, that would mean that the cells leaked about .1/1000=.0001 ampere (100 microamps).If that current times the number of hours of darkness is a small fraction (say, .01) of the forward current times the number of hours of daylight, then the leakage will not use a significant fraction of the charge energy. Adding the anti reverse diode would lower the dark leakage, but would also lower the daylight charging current, so it might be a net loss to have it. Measure the two cases and find out.
-- John Popelish
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